The Flat Earth Society
Other Discussion Boards => Science & Alternative Science => Topic started by: jimster on May 23, 2021, 10:09:09 PM
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Do ICBMs exist? Who aims them? Do they know the distance to the target? What map do they use? FE or RE$, gonna hit in a very different place.
V-2s in WW2 certainly existed, works fine on FE, not precision guided and short range, so earth curve doesn't matter. Starting in 1946, Werner Von Braun at White Sands and the USSR started improving the V-2, gradually improving the range and guidance until precise over thousands of miles. Tested and pr'd, from Vandenburg to south Pacific test range. I see no reason why ICBMs would not work on FE, but the aiming equation would be very different.
I found a diagram of ICBM trajectory that I can't manage to link to, but if you extend the line from the launch site straight as in FE, the range error has got to be hundreds, maybe thousands of miles.
Do ICBMs exist?
Have they been tested for accuracy?
Does the equation use FE or RE math?
The details of this story on RE are well known. Can someone fill in the FE details? Programmers secretly know FE, but are under threat to keep it secret? ICBMs do not have accuracy? What's up?
Perhaps USSR, UK, USA, China, etc all know FE, all agreed to keep it secret, and all want to have ICBM credibility when no such thing exists?
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V-2s in WW2 certainly existed, works fine on FE, not precision guided and short range, so earth curve doesn't matter. Starting in 1946, Werner Von Braun at White Sands and the USSR started improving the V-2
Yes, lets all trust a Nazi war criminal who was protected from prosecution by the military to boast claims for its own military space weapons that everyone desperately wanted at the time for national security. Wonderful source there.
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Do ICBMs exist? Who aims them? Do they know the distance to the target? What map do they use? FE or RE$, gonna hit in a very different place.
V-2s in WW2 certainly existed, works fine on FE, not precision guided and short range, so earth curve doesn't matter. Starting in 1946, Werner Von Braun at White Sands and the USSR started improving the V-2, gradually improving the range and guidance until precise over thousands of miles. Tested and pr'd, from Vandenburg to south Pacific test range. I see no reason why ICBMs would not work on FE, but the aiming equation would be very different.
A ICBM wouldn't really work well on a FE with UA. The equations in use depend upon a decreasing gravitational attraction as the rocket gets more distant from the center of the earth. Under UA that wouldn't make a difference and the altitude VS propulsion strength would be different and the rocket's range would be shorter. Perhaps North Korea's rocket man thinks the earth is flat. That would be a good thing because his aim would be wrong. North Korea does have rockets because one of them were seen going over the ship I was on while we were transiting between Korea and Japan several years ago. Things were a bit tense a few weeks prior to that while we were in Honolulu loading cargo because of some threats from rocket man.
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On NASA Marshall Space Flight Center director Wernher von Braun -
A survivor's account from Wernher von Braun, the SS, and Concentration Camp Labor: Questions of Moral, Political, and Criminal Responsibility: (https://web.archive.org/web/20140327145803/http://epizodsspace.no-ip.org/bibl/inostr-yazyki/german-studies/2002/neufeld_25.pdf)
Like the good Nazi he was, he immediately started shouting that it was sabotage, when just at that point VON BRAUN arrived accompanied by his usual group of people. Without even listening to my explanations, he ordered the Meister to have me given 25 strokes in his presence by an SS [man] who was there. Then judging the strokes weren't sufficiently hard, he ordered that I be flogged more vigorously, and this order was then diligently carried out, which caused much hilarity in the group, and following this flogging, VON BRAUN made me translate that I deserved much more, that in fact I deserved to be hanged, which certainly would be
the fate of the "Mensch" (good-for nothing) I was.
A quote from an article called The Rocket Man’s Dark Side (http://content.time.com/time/printout/0,8816,220201,00.htm), published by TIME:
Indeed, some 20,000 died at Dora, from illness, beatings, hangings and intolerable working conditions. Workers, scantily clad, were forced to stand at attention in the biting cold during roll calls that went on for hours. Average survival time in the unventilated paint shop was one month. One prisoner told of being bitten on his legs by guard dogs. Presumably to test the effectiveness of a new medication, one of his legs was treated, the other allowed to fester and deteriorate.
For reasons best known to von Braun, who held the rank of colonel in the dreaded Nazi SS, the prisoners were ordered to turn their backs whenever he came into view. Those caught stealing glances at him were hung. One survivor recalled that von Braun, after inspecting a rocket component, charged, "That is clear sabotage." His unquestioned judgment resulted in eleven men being hanged on the spot. Says Gehrels, "von Braun was directly involved in hangings."
Hangings were commonplace, and Dora inmates remember von Braun arriving in the morning with an unidentified woman, having to step between bodies of dead prisoners and under others still hanging from a crane. These were not ordinary hangings, Gehrels says, "not hanging that breaks the neck of the prisoner, but they were slowly choked to death with a kind of baling wire around their neck."
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Something I agree with Tom Bishop!
Werner Von Braun was morally reprehensible. We could talk about ethics and history, but it happened, and all I was doing was saying what happened.
The OP question was when and what is the discrepancy between what the govt says we have in missiles and what we actually have. When did they find out FE and the conspiracy begin? Does the aiming process use FE, has it been tested, or is it all a lie that multiple countries engage in because they want us to think they can nuke anyone?
Many engineers participating in the program would have realized that gravity was not as their calculations expect and the earth is flat, and the range would be way different. They did test shots from Vandenberg to south Pacific. I worked in the dp center at the Air Force Flight Test Center, believe me, they know every little detail, telemetry, radar, inertial guidance and gps. Must be quite a few engineers in the space program that know.
https://en.wikipedia.org/wiki/Missile_Test_Project
They would trace the shot from start to end and study it in great detail. Using radar, gps, inertial guidance telemetry (all checked to verify that they match), A plot in 3 space, displayed in a graphics program.
The FExplanation possibilities are:
1. The entire ICBM program is phony,
2. The engineers are monumentally incompetent.
3. You take a job offer out of college to work on the ICBM program, and day 1 a NASA thug takes you in a room and threatens to kill your family if you reveal FE, you leave this meeting and act the same, carrying on as normal, never getting drunk and telling someone, never acting odd, going to work in FE and home to RE for years. You help your kids with their astronomy homework and never let on in any way, because you know a NASA thug will kill them.
4. The engineers have been manipulated by false data from the conspiracy. There is a web of people collecting and processing this data, and multiple teams doing that in other countries. The data has to correlate, there is a lot of it, the process is complicated, multiple copies of data processed different ways. The complexity and effort level to do this is unimaginable.
Anyone got an explanation that is plausible to a person who doesn't care about FE? Anything without giant insanely complicated yet super secret conspiracy? Maybe everything appears as RE says, but advanced math and science reveals that light and radio waves bend exactly as they must to make a flat earth look round? Any proofs, experiments, equations for those forces?
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It wasn't only Von Braun, Operation Paperclip involved many Nazis:
https://www.aish.com/ci/s/Operation-Paperclip-The-Truth-about-Bringing-Nazi-Scientists-to-America.html
On July 6, 1945, the US Joint Chiefs of Staff issued a top-secret memorandum that was so explosive it was never even shown to President Truman. Titled “Exploitation of German Specialists in Science and Technology in the United States”, it outlined a program for “procurement, utilization and control of specialists” – in other words, a plan to recruit Nazi weapons scientists and bring them to the United States. The Military Intelligence of the War Department, a unit known as G-2, was given control of the program.
It was a plan conducted in secret, that not even the President knew about.
Another Nazi, Arthur Rudolph:
When he was granted entry to the US under Operation Paperclip, Rudolph was described by American officials in Germany as an “ardent Nazi”. He’d personally overseen slave labor and been present when prisoners were executed. West German and American officials classified him as a war criminal. Yet these accusations were quietly erased from his official file, and Rudolph worked for NASA.
In the 1960s, Rudolph became a key engineer at the Marshall Space Flight Center in Huntsville, Alabama, managing teams of scientists working on the Saturn 5 rocket that in 1969 launched the Apollo rocket in the first manned flight to the moon.
So it's not only the director; in the above we see that we have a Nazi managing teams that worked on the Saturn V.
Clearly, when you are trying to get knowledge from captured enemy Nazis you should put them in direct management of your critical rocket programs rather than a consulting role.
Many were brought to the US:
In total, about 1,600 Nazi scientists were brought to the US under Operation Paperclip, as well as their families. They evaded justice and in many cases were able to erase all mention of their Nazi pasts in their official biographies. For most of these scientists, justice never caught up and many died in America, seemingly innocent workers whose neighbors and friends and coworkers never knew the gruesome secrets in their pasts.
Maybe America isn't such an upstanding country after all:
For years, Nazis “were indeed knowingly granted entry” to the United States the OSI concluded. “America, which prided itself on being a safe haven for the persecuted, became – in some small measure – a safe haven for persecutors as well.”
I guess we have to trust literal Nazis and the War Department on this one. Think what you will, but I have a hard time trusting the Nazis and the War Department on anything.
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When icbms are tested, they are tracked by multiple radar sites. Each site knows the bearing and distance to the rocket. The coverage overlaps. Matching up the data from the overlapping parts of the data will locate the stations in 3 space in relation to one another. In fact, two radar stations a few hundred miles apart tracking the same object forms a triangle with two known angles and two known side lengths, so the distance between the radars can be calculated.
Yet another way to know distances, and one that would be critical to missile testing.
Tom Bishop, did you decide the earth is flat and on every challenge make up an explanation, or did you carefully consider all the evidence and decide? Do you understand RET? I think RET explains day/night, seasons 24 hour days at the poles, tides, the apparent motion of stars and planets, gyrocompass, sextant, eclipse, etc.
Does it explain all those things, even if the earth is flat, does RE geometry "work"? Would a person on the surface of a 8000 mi diameter sphere spinning 1 rpm, 93 M mi away from a really bright light see the sun come over the horizon, take 12 hrs to cross the sky, and then appear to sink beneath the opposite horizon only to rise again 12 hrs later? If tilted slightly, would there be areas near the poles that at some times of the year got 24 hour sun and other times 24 hour night, the border of which would be the exact angle of the tilt?
Does RET "work"?
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I am of the belief that RE geometry does not truly work. The Wiki explores that. My best evidence is that the RE scientists admit that various parts of RE don't work or that the data suggests something else; and they are not misquoted. When you put it all together RE doesn't work. All the Wiki does is put it together and cite with minimal commentary, to say the things they can't.
FE has mysteries, contradicting models, and much to be explored, certainly, but there is a zero dollar research budget there and relies on research that can be found online, which is quite limited when you are trying to find some specific things.
ICBMs are more of an unknowable quantity that we just have to trust the Nazis and War Department on. If you want to trust them, fine with me. I will opt to explore the possibility that they are nefarious.
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I am of the belief that RE geometry does not truly work. The Wiki explores that. My best evidence is that the RE scientists admit that various parts of RE don't work or that the data suggests something else; and they are not misquoted. When you put it all together RE doesn't work. All the wiki does is putt it together and cite with minimal commentary, to say the things they can't.
Ship's navigators believe that RE geometry really does work. Celestial navigation has been proven to work, countless times. GPS is just an automated form of celestial navigation and has been proven to be accurate each and every day while at sea. Gyroscopes indicate that the earth is a sphere. This fact can be demonstrated in the Zetetic fashion every time it's tried. It doesn't matter how many books you read or how many 'scientists' you give reference to, those who make their living traveling internationally have to believe in a spherical earth. It isn't 'brain washing' it's the practical application of the theories taught. When theories are taught and you can't make them work when everything is on the line, then even the most adamant RE'er would get suspicious. In 20+ years of traveling around the globe this has never happened to me. What have I missed? Where does RE geometry go wrong? It's important that navigators know this kind of thing when we have to depend upon what we are taught to make a safe arrival at the next port.
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Celestial navigation is allegedly based on the theory that the Round Earth is the center of the universe and that everything is revolving around it.
But I'm not sure if even that model exists. We have tables of the Moon based on historical data, not a mathematical revolving model; at least one that doesn't use epicycles.
The great RE mathematician Isaac Newton used epicycles in his math of the Moon going around the Earth:
https://wiki.tfes.org/Astronomical_Prediction_Based_on_Patterns#Newton.27s_Epicycles
Newton's Epicycles
Historian of Science William Whewell (bio (https://en.wikipedia.org/wiki/William_Whewell)) informs us that Newton used epicycles for the Moon:
History of the Inductive Sciences (1846) (https://books.google.com/books?id=vdQyAQAAMAAJ&lpg=PA174&ots=aNPf5Bu6CP&dq=as%20we%20have%20already%20said%2C%20of%20the%20ancient%20epicycles%2C&pg=PA174#v=onepage&q&f=false)
“ 3.— The Epicyclical Hypothesis was found capable of accommodating itself to such new discoveries. These new inequalities could be represented by new combinations of eccentrics and epicycles: all the realand imaginary discoveries by astronomers, up to Copernicus, were actually embodied in these hypotheses; Copernicus, as we have said, did not reject such hypotheses; the lunar inequalities which Tycho etected might have boen similarly exhibited; and even Newton36 represents the motion of the moon’s apogee by means of an epicycle. As a mode of expressing the law of the irregularity, and of calculating its results in particular cases, the epicyclical theory was capable of continuing to render great service to astronomy, however extensive the progress of the science might be. It was, in fact, as we have already said, the modern process of representing the motion by means of a series of circular functions. ”
Take a look at this model of the Moon in this paper (http://sites.apam.columbia.edu/courses/ap1601y/Moon-Earth-Sin%20RMP.70.589.pdf).
See the illustration on page 600 and the caption in the image, and note that the basic model was "adopted ever since."
V. THE MANY MOTIONS OF THE MOON
A. The traditional model of the Moon
A plane through the center of the Earth is determined at an inclination g of about 5 degrees with respect to the ecliptic. The Moon moves around the Earth in that plane on an ellipse with fixed semi-major axis a and eccentricity « of about 1/18. The Greek model was quite similar, except that the ellipse was replaced by an eccentric circle.
The plane itself rotates once every 18 years in the backward direction, i.e., against the prevailing motion in the solar system, while keeping its inclination constant. The perigee of the Moon, its point of closest approach to the Earth, makes a complete turn in the forward direction in about nine years.
The following picture (see Fig. 1) emerges: first we fix the direction of the spring equinox or some fixed star near it as the universal reference Q in the ecliptic: counting always from west to east, we determine the angle h from Q to the ascending node, i.e., the line of intersection for the Moon’s orbit with the ecliptic where the Moon enters the upper side of the ecliptic; from there we move by an angle g in the Moon’s orbital plane until we meet the perigee of the Moon; and finally we get to the Moon by moving through the true anomaly f. All these three angles have a double time dependence: linear (increasing for f and g, while decreasing for h) plus various periodic terms that average to 0.
(https://i.imgur.com/HpCH7Pn.png)
~
D. The evection—Greek science versus Babylonian astrology
The Babylonians knew that the full moons could be as much as 10 hours early or 10 hours late; this is due to the eccentricity of the Moon’s orbit. But the Greeks wanted to know whether the Moon displays the same kind of speedups and delays in the half moons, either waxing or waning. The answer is found with the help of a simple instrument that measures the angle between the Moon and the Sun as seen from the Earth. The half moons can be as much as 15 hours early or late. With the Moon moving at an average speed of slightly more than 308 per hour (its own apparent diameter!), it may be as much as 5° ahead or behind in the new/full moons; but in the half moons, it may be as much as 7°308 ahead or behind its average motion. This new feature is known as evection.
Ptolemy found a mechanical analog for this peculiar complication, called the crank model. It describes the angular coupling between Sun and Moon correctly, but it has the absurd consequence of causing the distance of the Moon from the Earth to vary by almost a factor of 2. In the thirteenth century Hulagu Khan, a grandson of Genghis Khan, asked his vizier, the Persian all-round genius Nasir ed-din al Tusi, to build a magnificent observatory in Meragha, Persia, and write up what was known in astronomy at that time. Ptolemy’s explanation of the evection was revised in the process. In the fourteenth century Levi ben Gerson of Avignon in southern France seems to have been the first astronomer to measure the apparent diameter of the Moon (see Goldstein, 1972, 1997). Shortly thereafter Ibn al-Shatir of Damascus in Syria proposed a model for the Moon’s motion that coincides with the theory of Copernicus two centuries later. The crank model was replaced by two additional epicycles, yielding a more elaborate Fourier expansion in our modern terminology (see Swerdlow and Neugebauer, 1984).
With the improvements of the Persian, Jewish, and Arab astronomers, as well as Copernicus, the changes in the Moon’s apparent diameter are still too large with +/- 10%. As in Kepler’s second law, the Fourier expansion (12) has to include epicycles both in the backward and in the forward direction, in the ratio 3:1.
"The crank model was replaced by two additional epicycles, yielding a more elaborate Fourier expansion in our modern terminology."
"With the improvements of the Persian, Jewish, and Arab astronomers, as well as Copernicus, the changes in the Moon’s apparent diameter are still too large with +/- 10%. As in Kepler’s second law, the Fourier expansion (12) has to include epicycles both in the backward and in the forward direction, in the ratio 3:1."
That epicycle Moon model was used in antiquity before gravity, and was used by Newton with his gravity model, and appears to be still used, judging by the quotes on the Astronomical Prediction Based on Patterns (https://wiki.tfes.org/Astronomical_Prediction_Based_on_Patterns) page in our Wiki. So I'm not entirely sure that the revolving model you think exists actually ever did exist. It appears as if they couldn't really get it to work cleanly even before gravity theory.
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Celestial navigation is allegedly based on the theory that the Earth is the center of the universe and that everything is revolving around it.
I can see that you are really confused. At least you used the term 'allegedly'. The old Polynesian navigators did look up at the heavenly bodies and learned how to use them as 'trail signs' to navigate across the Southern oceans thousands of years ago. The pattern of ocean waves was also used as another very useful indicator. These days the art of celestial navigation has advanced a bit. Just get a copy of Dutton's Navigation and Piloting for a more up to date version of how celestial navigation works today. You will see that it's now assumed that the earth circles the sun and so on. All the bunk about 'historical data' , 'epicycles', and the 'three body problem' are just diversions and/or trolling and used when necessary when you can't effectively refute the real world experience of today's navigators. We can all understand why this must be done but it doesn't keep you from looking a bit shady & disingenuous. The bottom line is if you get on a ship with a current ephemeris and a good sextant and use your celestial navigational skills based upon spherical trigonometry you can keep pretty good track of where you are anywhere in the world. I don't need to look up 'references' for this statement, it's in my inventory of personal experiences. It is true that this is a obsolete art as my iPhone has a very accurate GPS receiver that matches up perfectly with the ones on any ship that I was working on. No, GPS does NOT depend upon any land based systems. My phone works perfectly in the middle of the Pacific, Atlantic, or Indian Oceans where there's no land within 1000 miles. It does take a couple of minutes for the GPS receiver to come up with a good position and that time is some shorter when you can get a signal from a land based tower. That's the only function of a land based station when using GPS is to shorten the time necessary to get a usable position.
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Celestial navigation is allegedly based on the theory that the Earth is the center of the universe and that everything is revolving around it.
I can see that you are really confused. At least you used the term 'allegedly'. The old Polynesian navigators did look up at the heavenly bodies and learned how to use them as 'trail signs' to navigate across the Southern oceans thousands of years ago. The pattern of ocean waves was also used as another very useful indicator. These days the art of celestial navigation has advanced a bit. Just get a copy of Dutton's Navigation and Piloting for a more up to date version of how celestial navigation works today. You will see that it's now assumed that the earth circles the sun and so on. All the bunk about 'historical data' , 'epicycles', and the 'three body problem' are just diversions and/or trolling and used when necessary when you can't effectively refute the real world experience of today's navigators. We can all understand why this must be done but it doesn't keep you from looking a bit shady & disingenuous. The bottom line is if you get on a ship with a current ephemeris and a good sextant and use your celestial navigational skills based upon spherical trigonometry you can keep pretty good track of where you are anywhere in the world. I don't need to look up 'references' for this statement, it's in my inventory of personal experiences. It is true that this is a obsolete art as my iPhone has a very accurate GPS receiver that matches up perfectly with the ones on any ship that I was working on. No, GPS does NOT depend upon any land based systems. My phone works perfectly in the middle of the Pacific, Atlantic, or Indian Oceans where there's no land within 1000 miles. It does take a couple of minutes for the GPS receiver to come up with a good position and that time is some shorter when you can get a signal from a land based tower. That's the only function of a land based station when using GPS is to shorten the time necessary to get a usable position.
Interestingly, there is now an increasing interest in automated celestial navigation, as the filtering and sensor technology gets smaller and more usable. Not quite as easy to use as GPS, and doesn't see through cloud, but can't be jammed and potentially very accurate.
https://www.popularmechanics.com/military/research/a36078957/celestial-navigation/ (https://www.popularmechanics.com/military/research/a36078957/celestial-navigation/)
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All more unfound claims from RE here.
All you need for any missile to be fired to any target is the right amount of fuel and a simple quadratic equation which works perfectly fine on any x/y coordinate surface.
In other words, FLAT.
End of story.
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All more unfound claims from RE here.
All you need for any missile to be fired to any target is the right amount of fuel and a simple quadratic equation which works perfectly fine on any x/y coordinate surface.
In other words, FLAT.
End of story.
Source? Or is this on your own experience.
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All more unfound claims from RE here.
All you need for any missile to be fired to any target is the right amount of fuel and a simple quadratic equation which works perfectly fine on any x/y coordinate surface.
In other words, FLAT.
End of story.
ICMBs could certainly work on a flat earth but you'd need to know the distance between places, which you don't.
And the trajectory would be significantly different. So all the people who do the calculations and write the software which make it all work are presumably in on it or lying or being fooled or something?
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All more unfound claims from RE here.
All you need for any missile to be fired to any target is the right amount of fuel and a simple quadratic equation which works perfectly fine on any x/y coordinate surface.
In other words, FLAT.
End of story.
So if I give you the lat/long of two places some distance apart, one being our ICBM launcher and the other our target, how would you go about calculating the distance between them, in order to feed the necessary range into your simple equation?
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@TomBishop
There's a really simple way to end all of this nonsense. Simply produce an accurate Flat Earth map. That's all you need to do and this all ends. It truly is just that simple.
We'll wait.
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@TomBishop
There's a really simple way to end all of this nonsense. Simply produce an accurate Flat Earth map. That's all you need to do and this all ends. It truly is just that simple.
We'll wait.
Something something, no research budget, something something... (at a guess).
But the point is you don't need a research budget. There are web sites where you can track ships or planes in real time. All you need is to be near a port or an airport to verify that the planes or ships arrive when those sites say they have. I can go to Heathrow if it helps and he can go to his local international airport and we can verify when places leave from there and arrive at Heathrow, or the other way around, see if it matches what these web sites say. And sure, we can't verify whether the plane is where the sites say it is in between times but if we verify that it is at the start and the end of the journey then it seems reasonable to be confident that these sites know where these things are at all times.
It's bizarre that Tom thinks that RE geometry working is a "belief". There are global airline and shipping industries which rely on it working. We all use GPS routinely these days and can verify from personal experience that it works. It doesn't get much more Zetetic and empirical than that. I just had a look at my phone and I am where it says I am. Are we supposed to believe that this doesn't work at sea? RonJ has provided his personal experiences of how it does work and he'd be in trouble if it didn't. I'm sure many of us have been on flights and used the "Time to Destination" feature where you get a map, it shows you exactly where you are, how fast you're going, how far it is to your destination and how long it'll take to get there. Is the suggestion that all this is faked and doesn't really work?
It's all too silly. You can trivially prove that no FE map is possible given known distances, so the only way of maintaining FE belief is to claim that those distances aren't really known. But that's a difficult claim to justify given the above.
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It's bizarre that Tom thinks that RE geometry working is a "belief". There are global airline and shipping industries which rely on it working. We all use GPS routinely these days and can verify from personal experience that it works. It doesn't get much more Zetetic and empirical than that.
To Zetetic or not to Zetetic. That is the question.
The thought kinda came to me after I posted the above. Time and time again, simple observations of everyday things point to RE yet rather than embracing what is witnessed thousands of time daily all over the world, FE 'Zetetics' choose to make absurd excuses and theories to try to explain them away while not being able to produce simple things like a FE map that fits the observed view.
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All more unfound claims from RE here.
All you need for any missile to be fired to any target is the right amount of fuel and a simple quadratic equation which works perfectly fine on any x/y coordinate surface.
In other words, FLAT.
End of story.
Source? Or is this on your own experience.
Why do I need a source for a patently true statement?
V2 rockets work.
Rocket attacks happen in the Middle East all the time.
What we have not seen as of yet is the use of a supposed intercontinental booster with a functioning warhead.
Regardless, with such a weapon, precise accuracy isn't necessary. The people using such things could ultimately care less for anyone else's life but their own. Collateral damage? Not a concern.
Like I wrote, all you need is the right amount of fuel and a simple quadratic equation.
End of story.
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All more unfound claims from RE here.
All you need for any missile to be fired to any target is the right amount of fuel and a simple quadratic equation which works perfectly fine on any x/y coordinate surface.
In other words, FLAT.
End of story.
ICMBs could certainly work on a flat earth but you'd need to know the distance between places, which you don't.
And the trajectory would be significantly different. So all the people who do the calculations and write the software which make it all work are presumably in on it or lying or being fooled or something?
You do not even know if ICBM's work on a round earth.
You've never seen one in action.
So stick your nonsensical thinking back under your hat.
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All more unfound claims from RE here.
All you need for any missile to be fired to any target is the right amount of fuel and a simple quadratic equation which works perfectly fine on any x/y coordinate surface.
In other words, FLAT.
End of story.
Source? Or is this on your own experience.
Why do I need a source for a patently true statement?
V2 rockets work.
Santa Clause came down the chimney last Christmas and put a nice present under the tree for me. The fact that a present was there is patently true. Why would I ever need a source to believe it was Santa who delivered it? Santa Clause delivers, and so does the Tooth Fairy.
Anyone see a problem with the logical progression of this? There is NO simple quadratic equation that will describe the targeting of an ICBM. If that were so then that same mathematician would also be able to write an equation to solve the 3 body problem!
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You do not even know if ICBM's work on a round earth.
You've never seen one in action.
Well I know that rockets are a thing. I've seen plenty of launches on TV and I've seen a shuttle launch in person.
So I'm pretty confident that the technology to create an ICBM exists.
From what I understand an ICBM has never been fired in anger, but with anything I haven't personally witnessed I use my model of reality to assess how credible it is. I've never been to Australia but I know people from there, I know people from here who have visited and I have seen plenty of photos and video of the place. I don't have to have seen it with my own eyes to believe it exists.
If ICBMs were said to teleport to their destination then I would be highly sceptical, I don't believe that technology exists. But I know that rockets exist so it seems perfectly credible that ICBMs are possible with current technology. Either multiple countries are lying about having them and no other countries are calling them out on it, or they're a thing. The latter seems far more likely. And as I said and you agree, they'd work on a FE. But you'd need a working map to aim them correctly and we both know why you don't have one.
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You do not even know if ICBM's work on a round earth.
You've never seen one in action.
So stick your nonsensical thinking back under your hat.
I'm going to plagiarize Pete here for a sec.
If you're going to resort to this tired and worn out tact, please do the intellectually honest thing and choose not to respond.
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All more unfound claims from RE here.
All you need for any missile to be fired to any target is the right amount of fuel and a simple quadratic equation which works perfectly fine on any x/y coordinate surface.
In other words, FLAT.
End of story.
ICMBs could certainly work on a flat earth but you'd need to know the distance between places, which you don't.
And the trajectory would be significantly different. So all the people who do the calculations and write the software which make it all work are presumably in on it or lying or being fooled or something?
You do not even know if ICBM's work on a round earth.
You've never seen one in action.
So stick your nonsensical thinking back under your hat.
Well let's expand my previous, and as-yet-unanswered question a little then. You're right, nobody has used proper ICBMs in anger yet, but plenty of shorter range ballistic missiles have been fired. Take the Scuds used by Iraq in the Iran-Iraq war and the later Gulf War(s). Typical range of shots was in the region of up to 500 miles or so, with a max altitude just shy of 100 miles. So if you were an Iraqi, aiming your Scud at Tehran (35.7N 51.3E) from somewhere near the border in northern Iraq at, say, 35.5N 45.7E, would you use the round earth great circle distance of 314 miles, or some other number from your 'simple' but as-yet-unrevealed FE map? From my previous post (https://forum.tfes.org/index.php?topic=13948.msg237441#msg237441 (https://forum.tfes.org/index.php?topic=13948.msg237441#msg237441)) the 'SteelyBob ratio', or the correction factor needed to convert from the RE map to the monopole FE map shown in the wiki, is 1.17 at 35 degrees north, meaning those two places would be an 368 miles apart on your monopole FE map.
So which one would you use? Or do you have some other distance in mind?
We can talk about corrections for earth rotation, reductions in g with altitude and another minor irritations later...
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All more unfound claims from RE here.
All you need for any missile to be fired to any target is the right amount of fuel and a simple quadratic equation which works perfectly fine on any x/y coordinate surface.
In other words, FLAT.
End of story.
Source? Or is this on your own experience.
Why do I need a source for a patently true statement?
V2 rockets work.
Santa Clause came down the chimney last Christmas and put a nice present under the tree for me. The fact that a present was there is patently true. Why would I ever need a source to believe it was Santa who delivered it? Santa Clause delivers, and so does the Tooth Fairy.
Anyone see a problem with the logical progression of this? There is NO simple quadratic equation that will describe the targeting of an ICBM. If that were so then that same mathematician would also be able to write an equation to solve the 3 body problem!
Simply and absolutely ridiculous.
All missile equations having to do with trajectory are quadratic in nature.
You have no business even commenting in the thread as it is quite clear the subject matter is well over your head.
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You do not even know if ICBM's work on a round earth.
You've never seen one in action.
Well I know that rockets are a thing. I've seen plenty of launches on TV and I've seen a shuttle launch in person.
So I'm pretty confident that the technology to create an ICBM exists.
From what I understand an ICBM has never been fired in anger, but with anything I haven't personally witnessed I use my model of reality to assess how credible it is. I've never been to Australia but I know people from there, I know people from here who have visited and I have seen plenty of photos and video of the place. I don't have to have seen it with my own eyes to believe it exists.
If ICBMs were said to teleport to their destination then I would be highly sceptical, I don't believe that technology exists. But I know that rockets exist so it seems perfectly credible that ICBMs are possible with current technology. Either multiple countries are lying about having them and no other countries are calling them out on it, or they're a thing. The latter seems far more likely. And as I said and you agree, they'd work on a FE. But you'd need a working map to aim them correctly and we both know why you don't have one.
We also both know the ICBM's have never been used, so like I wrote.
The entire thread is pointless.
Just more BS from RE supporters.
They would work on an FE because of the reasons I stated.
First, pinpoint accuracy is not necessary as anyone using such a thing could care less who dies as long as it is not them.
Second, a quadratic equation is exactly what is used in determining missile trajectory.
End of story.
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All more unfound claims from RE here.
All you need for any missile to be fired to any target is the right amount of fuel and a simple quadratic equation which works perfectly fine on any x/y coordinate surface.
In other words, FLAT.
End of story.
ICMBs could certainly work on a flat earth but you'd need to know the distance between places, which you don't.
And the trajectory would be significantly different. So all the people who do the calculations and write the software which make it all work are presumably in on it or lying or being fooled or something?
You do not even know if ICBM's work on a round earth.
You've never seen one in action.
So stick your nonsensical thinking back under your hat.
Well let's expand my previous, and as-yet-unanswered question a little then. You're right, nobody has used proper ICBMs in anger yet, but plenty of shorter range ballistic missiles have been fired. Take the Scuds used by Iraq in the Iran-Iraq war and the later Gulf War(s). Typical range of shots was in the region of up to 500 miles or so, with a max altitude just shy of 100 miles. So if you were an Iraqi, aiming your Scud at Tehran (35.7N 51.3E) from somewhere near the border in northern Iraq at, say, 35.5N 45.7E, would you use the round earth great circle distance of 314 miles, or some other number from your 'simple' but as-yet-unrevealed FE map? From my previous post (https://forum.tfes.org/index.php?topic=13948.msg237441#msg237441 (https://forum.tfes.org/index.php?topic=13948.msg237441#msg237441)) the 'SteelyBob ratio', or the correction factor needed to convert from the RE map to the monopole FE map shown in the wiki, is 1.17 at 35 degrees north, meaning those two places would be an 368 miles apart on your monopole FE map.
So which one would you use? Or do you have some other distance in mind?
We can talk about corrections for earth rotation, reductions in g with altitude and another minor irritations later...
There is no "round earth great circle distance."
That is a bunch of ridiculous crap inserted into the discussion.
If I was to have a bird's eye view at Yankee Stadium on any particular Sunday, I could see both the Yankee's pitcher and the opposition pitcher launching curve balls toward home plate in a "great circle," fashion over a distance of 60 feet 6 inches from the pitching rubber to home plate.
Flat and level surface.
Same thing with the missile use in the all of the Middle East wars. If I drove from Beirut to Tel Aviv or from Damascus to Tel Aviv, I could use that distance to fly the missile very accurately toward its intended target.
NO RE necessary.
Just stop with all these meaningless objections.
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So how far is it, then?
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@TomBishop
There's a really simple way to end all of this nonsense. Simply produce an accurate Flat Earth map. That's all you need to do and this all ends. It truly is just that simple.
We'll wait.
In turn, we will wait for you to produce an accurate RE map.
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So how far is it, then?
I do not know.
I have had no reason to drive from Damascus or Beirut to Tel Aviv.
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So how far is it, then?
I do not know.
I have had no reason to drive from Damascus or Beirut to Tel Aviv.
I've given you two lat/long coordinates for an example launch site and target. It should be a simple case of plugging them into your equation, right?
So, how far is it?
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@TomBishop
There's a really simple way to end all of this nonsense. Simply produce an accurate Flat Earth map. That's all you need to do and this all ends. It truly is just that simple.
We'll wait.
In turn, we will wait for you to produce an accurate RE map.
Take your pick.
https://www.ultimateglobes.com/
Of course accuracy will improve with the size of any map so this would be more accurate.
https://en.wikipedia.org/wiki/Eartha#:~:text=Eartha%20is%20the%20world's%20largest,41%20feet%20(12.5%20m).
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@TomBishop
There's a really simple way to end all of this nonsense. Simply produce an accurate Flat Earth map. That's all you need to do and this all ends. It truly is just that simple.
We'll wait.
In turn, we will wait for you to produce an accurate RE map.
I like any WGS-84 chart. We used them at sea every day. They were always accurate and always based upon a spherical earth. If you can find an inaccuracy in one of them then PLEASE tell us. We need to know to enhance the safety of the ship's & crews out there every day that depend upon them. Thank You!
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So how far is it, then?
I do not know.
I have had no reason to drive from Damascus or Beirut to Tel Aviv.
I've given you two lat/long coordinates for an example launch site and target. It should be a simple case of plugging them into your equation, right?
So, how far is it?
And as I stated, I have not driven that route and have no reason to drive it.
I am sure the people that were firing the rockets over there had plenty of opportunity to drive it plenty of times and I am sure they had distances that were accurate enough, given just how "deadly," those Scud missile attacks were. Might as well been the North Koreans firing the missiles, LOL!
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@TomBishop
There's a really simple way to end all of this nonsense. Simply produce an accurate Flat Earth map. That's all you need to do and this all ends. It truly is just that simple.
We'll wait.
In turn, we will wait for you to produce an accurate RE map.
I like any WGS-84 chart. We used them at sea every day. They were always accurate and always based upon a spherical earth. If you can find an inaccuracy in one of them then PLEASE tell us. We need to know to enhance the safety of the ship's & crews out there every day that depend upon them. Thank You!
You have no clue whether the WGS - 84 chart was based on a spherical earth or not.
ZERO clue.
"Somebody told me they were," is a more accurate statement.
And for the final freaking time, as has been demonstrated in previous threads, long distance navigation of any sort is broken down into easier to manage short trips, with frequent stops to check systems and position.
Nothing, and I mean, NOTHING, has fundamentally changed concerning which routes are taken over long distances, either by sea or air, for over 100 years. That is fact. And none of it, I mean of NONE OF IT, requires RE to work.
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And as I stated, I have not driven that route and have no reason to drive it.
I am sure the people that were firing the rockets over there had plenty of opportunity to drive it plenty of times and I am sure they had distances that were accurate enough, given just how "deadly," those Scud missile attacks were. Might as well been the North Koreans firing the missiles, LOL!
So those two points were a random place in the desert in northern Iraq and Tehran - a typical Scud firing from the Iran-Iraq war.
Why would you need to have driven a journey to know the distance when you have the 'x-y' coordinates you keep telling us are so simple?
Why can't you just calculate the distance? Should be pretty easy, right?
Unless...you can't?
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@TomBishop
There's a really simple way to end all of this nonsense. Simply produce an accurate Flat Earth map. That's all you need to do and this all ends. It truly is just that simple.
We'll wait.
In turn, we will wait for you to produce an accurate RE map.
Take your pick.
https://www.ultimateglobes.com/
Of course accuracy will improve with the size of any map so this would be more accurate.
https://en.wikipedia.org/wiki/Eartha#:~:text=Eartha%20is%20the%20world's%20largest,41%20feet%20(12.5%20m).
When you are able to tell the difference between a globe and map, let us know will you?
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Tom Bishop, do I correctly understand your position?
"Celestial navigation is allegedly based on the theory that the Round Earth is the center of the universe and that everything is revolving around it."
Where did you get that? Completely wrong. Please take a course in elementary astronomy. You don't have to believe it, but we have to talk about the astronomy taught in school, in textbooks, all over the net, not your straw man version based on quotes from pre 20th century obscure scientists. We are RE, we get to say what RE is today, and that model is consistent all over the place, schools, textbooks, internet, astronomers, gps engineers, celestial navigation.
Epicycles, from the wiki page:
Epicycles worked very well and were highly accurate, because, as Fourier analysis later showed, any smooth curve can be approximated to arbitrary accuracy with a sufficient number of epicycles. However, they fell out of favour with the discovery that planetary motions were largely elliptical from a heliocentric frame of reference, which led to the discovery that gravity obeying a simple inverse square law could better explain all planetary motions.
I searched looking for contemporary astronomy using epicycles, but after many pages, I could find only Ptolemy, Copernicus, Aristotle, all history telling the story of how Kepler came up with a better explanation. Zero contemporary references to epicycles. Explained here:
https://www.spitzinc.com/blog/epicycles-and-discovering-bad-theories/
As I asked earlier, do you understand RET? Apparently not, the first step is to agree on what RE is, and since you are not RE, yu don't get to say what it is. It is what textbooks and science teachers and diagrams all over the internet say it is today, not some obscure 19th century scientist.
Tom Bishop gets to say what FE, or more precisely, the Tom Bishop version of FE (there are many, and they conflict). So what can you say? So far, I got no map, "anomolous winds", "unknown forces". To make your idea work, you must insist that no one knows the distances between places and apparently never can.
Tom Bishop, does gps work? Does it accurately tell you and billions of others where they are all the time? Even if there are no satellites because the earth is flat and gps is faked by NASA, it still produces accurate location data (or do you think gps fails constantly?) World maps with locations of ships over all oceans shows their locations as reported by gps. This is all wrong and no one notices?
How can ships and planes have gps and not know their locations and therefor distance, even over ocean? If the distances are as advertised, why can't you make a flat map? If the distances are not as advertised, how can ships and planes arrive where they intended and when they intended?
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And as I stated, I have not driven that route and have no reason to drive it.
I am sure the people that were firing the rockets over there had plenty of opportunity to drive it plenty of times and I am sure they had distances that were accurate enough, given just how "deadly," those Scud missile attacks were. Might as well been the North Koreans firing the missiles, LOL!
So those two points were a random place in the desert in northern Iraq and Tehran - a typical Scud firing from the Iran-Iraq war.
Why would you need to have driven a journey to know the distance when you have the 'x-y' coordinates you keep telling us are so simple?
Why can't you just calculate the distance? Should be pretty easy, right?
Unless...you can't?
I could calculate the distance given those coordinates on a flat chart.
So could you, I'm sure.
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And as I stated, I have not driven that route and have no reason to drive it.
I am sure the people that were firing the rockets over there had plenty of opportunity to drive it plenty of times and I am sure they had distances that were accurate enough, given just how "deadly," those Scud missile attacks were. Might as well been the North Koreans firing the missiles, LOL!
So those two points were a random place in the desert in northern Iraq and Tehran - a typical Scud firing from the Iran-Iraq war.
Why would you need to have driven a journey to know the distance when you have the 'x-y' coordinates you keep telling us are so simple?
Why can't you just calculate the distance? Should be pretty easy, right?
Unless...you can't?
I could calculate the distance given those coordinates on a flat chart.
So could you, I'm sure.
You've got the lat/long. What more do you need? Everybody uses lat/long right?
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A globe is a map on the surface of a sphere. Whether the earth is round or flat, mathematically only one can have accurate distances, per Gauss's Remarkable Theorem. You can't represent a scaled version of a curved surface on a flat plane and preserve distance, and vice versa. If the earth is round, you can't make a flat map with constant scale and accurate distance. If the earth is flat, no globe could have accurate distances. Math fact, seen it proved, as have millions (billions?) in math classes all around the world.
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Epicycles, from the wiki page:
Epicycles worked very well and were highly accurate, because, as Fourier analysis later showed, any smooth curve can be approximated to arbitrary accuracy with a sufficient number of epicycles. However, they fell out of favour with the discovery that planetary motions were largely elliptical from a heliocentric frame of reference, which led to the discovery that gravity obeying a simple inverse square law could better explain all planetary motions.
I searched looking for contemporary astronomy using epicycles, but after many pages, I could find only Ptolemy, Copernicus, Aristotle, all history telling the story of how Kepler came up with a better explanation. Zero contemporary references to epicycles. Explained here:
https://www.spitzinc.com/blog/epicycles-and-discovering-bad-theories/
As I asked earlier, do you understand RET? Apparently not
No, actually, it is you who does not know about RET. Kepler used epicycles. This should have been a dead give away since Newton came after Kepler, and used epicycles in the previous quote I gave.
There were still epicycles in Kepler's version. He just reduced it a little more. You can see that Jupiter is on an epicycle in his version:
Dr. Samuel Herrick - https://ia802909.us.archive.org/18/items/the-foundations-of-astrodynamics/The%20Foundations%20of%20Astrodynamics.pdf
(https://i.imgur.com/4jIumNC.png)
“ Physical celestial mechanics may be said to have begun with Galileo Galilei, Isaac Newton, and the laws of force and gravitation. Astrodynamics and mathematical celestial mechanics, on the other hand, date back at least to Heracleides of Pontus in the fourth century B.C. The Greek invention of epicycles and eccentrics was developed into a system by Apollonius of Perga in the third century and Hipparchus of Alexandria in the second century B.C. It was refined and published by Ptolemy of Alexandria in the second century A.D., and came to be known as the Ptolemaic system. It is generally assumed that the epicycle was discredited by Johannes Kepler some 1500 years later, but in point of fact epicycles have persisted in astrodynamics down to the present day, and have extended their domain into other areas of science under the guise of Fourier series! ”
“ In modern perturbation theory we actually take account of the original epicyclic concept by combining several Fourier series that have arguments based upon different angular variables. ”
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And for the final freaking time, as has been demonstrated in previous threads, long distance navigation of any sort is broken down into easier to manage short trips, with frequent stops to check systems and position.
Just a simple google and this is shown to be nonsense.
https://www.travelandleisure.com/airlines-airports/longest-flights-in-the-world#:~:text=Lufthansa%20repatriated%20German%20citizens%20from,Papeete%2C%20Tahiti%2C%20to%20Paris.
Nothing, and I mean, NOTHING, has fundamentally changed concerning which routes are taken over long distances, either by sea or air, for over 100 years.
The first intercontinental flight didn't occur until 1927. The first flight over the north pole didn't occur until 1937. The first commercial polar route didn't occur until 1954.
When you are able to tell the difference between a globe and map, let us know will you?
All globes are maps, not all maps are globes.
I know self awareness is not really your thing, but I'll just leave this little nugget here.
You have no business even commenting in the thread as it is quite clear the subject matter is well over your head.
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@TomBishop
There's a really simple way to end all of this nonsense. Simply produce an accurate Flat Earth map. That's all you need to do and this all ends. It truly is just that simple.
We'll wait.
In turn, we will wait for you to produce an accurate RE map.
I like any WGS-84 chart. We used them at sea every day. They were always accurate and always based upon a spherical earth. If you can find an inaccuracy in one of them then PLEASE tell us. We need to know to enhance the safety of the ship's & crews out there every day that depend upon them. Thank You!
You have no clue whether the WGS - 84 chart was based on a spherical earth or not.
ZERO clue.
"Somebody told me they were," is a more accurate statement.
And for the final freaking time, as has been demonstrated in previous threads, long distance navigation of any sort is broken down into easier to manage short trips, with frequent stops to check systems and position.
Nothing, and I mean, NOTHING, has fundamentally changed concerning which routes are taken over long distances, either by sea or air, for over 100 years. That is fact. And none of it, I mean of NONE OF IT, requires RE to work.
Well, when WGS stands for 'World Geodetic System' and that system is based upon a globe then I could say that somebody told me that the navigational charts were based upon a global earth. Since we used those charts each and every day while at sea to navigate and we always got to our destination safely then you could say that was a satisfactory demonstration of their accuracy and effectiveness.
Your statement about 'frequent stops' is based upon ignorance, I see. Just how many international trips have you made across the earth's oceans? Do you have any navigational training? I've personally been on countless trips across all the oceans on earth. Typically we would make a trip from China to the USA and set up the entire voyage plan before even leaving the dock. That plan would be put into the ships autopilot and once we are clear of the last sea buoy at the departing port we would not stop or even slow down once until we arrived at the destination's entry buoy. This is how it works on ship's today. Everything is computerized and all the maps are electronic. International regulations still say we must have paper charts aboard (WGS-84) and our position is charted at least once an hour. I don't know which ships make frequent stops to check systems and position, but not even one ship I worked on over the last 20 years ever did that. This is from personal experience!
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Tom Bishop, do I correctly understand your position?
"Celestial navigation is allegedly based on the theory that the Round Earth is the center of the universe and that everything is revolving around it."
Where did you get that? Completely wrong. Please take a course in elementary astronomy.
It is you who needs to take some courses.
https://www.lowtechmagazine.com/2008/01/satellite-nav-1.html
"Man has navigated across the globe by means of satellites for thousands of years – however, until the mid twentieth century, these were not GPS-satellites, but stars. In reality, the sun and the stars aren’t satellites of the Earth, but celestial navigation is based on a precopernican world view (the earth was believed to be the centre of the universe)."
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Tom Bishop, do I correctly understand your position?
"Celestial navigation is allegedly based on the theory that the Round Earth is the center of the universe and that everything is revolving around it."
Where did you get that? Completely wrong. Please take a course in elementary astronomy.
It is you who needs to take some courses.
https://www.lowtechmagazine.com/2008/01/satellite-nav-1.html (https://www.lowtechmagazine.com/2008/01/satellite-nav-1.html)
"Man has navigated across the globe by means of satellites for thousands of years – however, until the mid twentieth century, these were not GPS-satellites, but stars. In reality, the sun and the stars aren’t satellites of the Earth, but celestial navigation is based on a precopernican world view (the earth was believed to be the centre of the universe)."
Please do your research before you post nonsense. Perhaps the first celestial navigators did have a pre-Copernican world view and perhaps many got lost doing so. Today a celestial navigator takes multiple sightings on some chosen heavenly bodies then uses spherical trig to break the sights down to arrive at a usable position. The assumption is always based upon the fact that the earth is a sphere. If that were not so then celestial navigation wouldn't work. It does work. I know from personal experience, so the earth must be a sphere. This has been demonstrated in the Zetetic manner.
I have worked aboard quite a few high tech computerized ships. Every ship has always had 2 sextants and a full set of paper charts aboard. This is done because it's a law. If an official in some country wants to inspect the ships sextants and you don't have them then you won't be permitted to leave the dock. If something happens to the GPS system or maybe the ship gets a serious lightning strike and knocks out all the electronics then the old fashioned sextants would be the fall back navigation system, even today. Any navigational officer must also be proficient in using a sextant to obtain a license.
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celestial navigation is based on a precopernican world view (the earth was believed to be the centre of the universe)."
And what shape was it known to be?
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RE both epicycles still being used and celestial navigation based on precopernican, straw man, Tom Bishop is misrepresenting RET. Seems like the REs should get to say what RE is. Also seems like a person should know what he is arguing against. We are here arguing with a man who says that celestial navigation is precopernican, astronomers are still using epicycles, and no one knows the distance between major cities, meaning gps doesn't work. I have a feeling FEs rely on gos for their own daily use, but disclaim it for figuring out the shape of the earth.
And yet our expectation of reasonablness and competent mental process leads us to expect to be able to explain and convince him. That a person functioning and accepted in regular life, able to read and write and do arithmetic, remember things, hold a job, have friends, in short function apparently competently in everyday life. could repeatedly reject, dismiss, delude, distort, etc in order to hold on to a belief that makes no sense is alarming and scary. It threatens our collective self image as an intelligent competent species.
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Today a celestial navigator takes multiple sightings on some chosen heavenly bodies then uses spherical trig to break the sights down to arrive at a usable position. The assumption is always based upon the fact that the earth is a sphere. If that were not so then celestial navigation wouldn't work. It does work. I know from personal experience, so the earth must be a sphere. This has been demonstrated in the Zetetic manner.
Celestial bodies and lights in the sky aren't the earth. Spherical trigonometry means that the sky is spread out around you and says nothing about the actual shape of the Earth. The bodies don't even move at a consistent speed, but speed up as they approach the horizon due to 'refraction'.
There are ways to get your latitude or longitude from the celestial bodies, such as from the angle of the Sun at noon, but all of the FE models also have a longitude and latitude.
celestial navigation is based on a precopernican world view (the earth was believed to be the centre of the universe)."
And what shape was it known to be?
The precopernican world view was a round world that required epicycles. Keep up.
RE both epicycles still being used and celestial navigation based on precopernican, straw man, Tom Bishop is misrepresenting RET. Seems like the REs should get to say what RE is.
They did.
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...but all of the FE models also have a longitude and latitude.
They do? What models are those?
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Today a celestial navigator takes multiple sightings on some chosen heavenly bodies then uses spherical trig to break the sights down to arrive at a usable position. The assumption is always based upon the fact that the earth is a sphere. If that were not so then celestial navigation wouldn't work. It does work. I know from personal experience, so the earth must be a sphere. This has been demonstrated in the Zetetic manner.
Celestial bodies and lights in the sky aren't the earth. Spherical trigonometry means that the sky is spread out around you and says nothing about the actual shape of the Earth. The bodies don't even move at a consistent speed, but speed up as they approach the horizon due to 'refraction'.
There are ways to get your latitude or longitude from the celestial bodies, such as from the angle of the Sun at noon, but all of the FE models also have a longitude and latitude.
I can see that you have NO idea of how celestial navigation really works. No, all the lights in the sky are not the earth. Those lights only provide a reference, called a zenith point, on the surface of the earth. Once you have a number of zenith points you then have to use spherical trigonometry because you will end up with a bunch of triangles that are great circles on a globe. Here is where it's important to consider the actual shape of the earth. Failure to do so will have undesirable consequences for the safety of your ship. All this is from personal experience. Please do your own research so your answers can become more meaningful. Right now your comments are mostly incorrect.
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So Tom has said:
There are ways to get your latitude or longitude from the celestial bodies, such as from the angle of the Sun at noon, but all of the FE models also have a longitude and latitude.
And Action80 said this, a little while ago in a different thread:
As if Lat/Long is not an x-y coordinate system, when it clearly is anyway.
...but now seems to be struggling to calculate the distance between a pair of lat/longs, for want, apparently, of some x-y coordinates on a 'flat chart':
I could calculate the distance given those coordinates on a flat chart.
So now is your chance to shine, FE people. Tom, Action80 or somebody...a really simple task for you, given that you agree that lat/long is a coordinate system and that all FE models have lat/long too, which can be derived from celestial navigation methods. All you have to do is demonstrate how you would calculate the distance between two lat/long pairs of coordinates, like the ones I used in my example earlier in this thread. You keep saying it's really simple to do. So do it, and show your calculations.
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The discussion is fascinating, but it is a bit bogged down in unnecessary detail.
Wiki could use an additional page to simplify matters. A list of fake jobs, based on faulty RE model. People working those jobs should realize the error of their ways and do something more productive (an additional list would be nice).
I have some suggestions to start the list of fake jobs: navigator, cartographer, astronomer, sniper, artilleryman, pilot, astronaut, teacher, physicist, rocket scientist, geologist, trucking company managers (they think they can price the jobs by mileage, lol!)... We would know that personal experience coming from these occupations should not be taken seriously.
It would not be necessary to explain time and again to some navigator that they have no idea about the shape of the earth.
But I am struggling with the other list.
Astronomers could switch to astrology.
Armies of the world should switch from artillery to something more accurate. Just imagine the resources we could save!
Teachers are not really needed, there is plenty of material on the internet to do research, each child can educate themselves. Even to become a rocket scientist, it is enough to solve quadratic equations. All one needs to know from scientific papers is to read the sentences containing words "anomalous", "wrong" and "unknown". Scientific method would be much more efficient if simplified to simple Ctrl+f searches.
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And as I stated, I have not driven that route and have no reason to drive it.
I am sure the people that were firing the rockets over there had plenty of opportunity to drive it plenty of times and I am sure they had distances that were accurate enough, given just how "deadly," those Scud missile attacks were. Might as well been the North Koreans firing the missiles, LOL!
So those two points were a random place in the desert in northern Iraq and Tehran - a typical Scud firing from the Iran-Iraq war.
Why would you need to have driven a journey to know the distance when you have the 'x-y' coordinates you keep telling us are so simple?
Why can't you just calculate the distance? Should be pretty easy, right?
Unless...you can't?
I could calculate the distance given those coordinates on a flat chart.
So could you, I'm sure.
You've got the lat/long. What more do you need? Everybody uses lat/long right?
I am sure they do.
They also use a point of origin.
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A globe is a map on the surface of a sphere. Whether the earth is round or flat, mathematically only one can have accurate distances, per Gauss's Remarkable Theorem. You can't represent a scaled version of a curved surface on a flat plane and preserve distance, and vice versa. If the earth is round, you can't make a flat map with constant scale and accurate distance. If the earth is flat, no globe could have accurate distances. Math fact, seen it proved, as have millions (billions?) in math classes all around the world.
A globe is not a map and a globe is not necessary for navigation.
End of story.
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@TomBishop
There's a really simple way to end all of this nonsense. Simply produce an accurate Flat Earth map. That's all you need to do and this all ends. It truly is just that simple.
We'll wait.
In turn, we will wait for you to produce an accurate RE map.
I like any WGS-84 chart. We used them at sea every day. They were always accurate and always based upon a spherical earth. If you can find an inaccuracy in one of them then PLEASE tell us. We need to know to enhance the safety of the ship's & crews out there every day that depend upon them. Thank You!
You have no clue whether the WGS - 84 chart was based on a spherical earth or not.
ZERO clue.
"Somebody told me they were," is a more accurate statement.
And for the final freaking time, as has been demonstrated in previous threads, long distance navigation of any sort is broken down into easier to manage short trips, with frequent stops to check systems and position.
Nothing, and I mean, NOTHING, has fundamentally changed concerning which routes are taken over long distances, either by sea or air, for over 100 years. That is fact. And none of it, I mean of NONE OF IT, requires RE to work.
Well, when WGS stands for 'World Geodetic System' and that system is based upon a globe then I could say that somebody told me that the navigational charts were based upon a global earth. Since we used those charts each and every day while at sea to navigate and we always got to our destination safely then you could say that was a satisfactory demonstration of their accuracy and effectiveness.
Your statement about 'frequent stops' is based upon ignorance, I see. Just how many international trips have you made across the earth's oceans? Do you have any navigational training? I've personally been on countless trips across all the oceans on earth. Typically we would make a trip from China to the USA and set up the entire voyage plan before even leaving the dock. That plan would be put into the ships autopilot and once we are clear of the last sea buoy at the departing port we would not stop or even slow down once until we arrived at the destination's entry buoy. This is how it works on ship's today. Everything is computerized and all the maps are electronic. International regulations still say we must have paper charts aboard (WGS-84) and our position is charted at least once an hour. I don't know which ships make frequent stops to check systems and position, but not even one ship I worked on over the last 20 years ever did that. This is from personal experience!
You are full of it.
Whether or not the stops are ACTUALLY made isn't the POINT.
Every long distance route taken by sea or air has essentially remained UNCHANGED since long distance travel commenced.
None of it was ever based on the idea of a globe.
It was always based on celestial navigation.
Ever since Googleotomy, you have been pushing this ridiculous tripe on the boards and it has always been nothing but an outright LIE of monstrous proportions.
Your schtick is old, worn out. Go lie down in the corner somewhere.
This thread is about ICBM's, which pose a SERIOUS threat to the continued existence of HUMANITY, even though NO ONE has ever seen one used, and even though they do not need RE to operate. North Korea might be able to shoot one past 100 miles any day now.
So pay attention.
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I am sure they do.
They also use a point of origin.
What, exactly, do you mean by 'point of origin'?, and how does that facilitate the calculation of a distance between two lat/long points?
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I am sure they do.
They also use a point of origin.
What, exactly, do you mean by 'point of origin'?, and how does that facilitate the calculation of a distance between two lat/long points?
Whenever you calculate distance between two point on a x/y coordinate system, there is always a point of origin.
Then it is simply the distance formula d=√((x2-x1)²+(y2-y1)²)
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So Tom has said:
There are ways to get your latitude or longitude from the celestial bodies, such as from the angle of the Sun at noon, but all of the FE models also have a longitude and latitude.
And Action80 said this, a little while ago in a different thread:
As if Lat/Long is not an x-y coordinate system, when it clearly is anyway.
...but now seems to be struggling to calculate the distance between a pair of lat/longs, for want, apparently, of some x-y coordinates on a 'flat chart':
I could calculate the distance given those coordinates on a flat chart.
So now is your chance to shine, FE people. Tom, Action80 or somebody...a really simple task for you, given that you agree that lat/long is a coordinate system and that all FE models have lat/long too, which can be derived from celestial navigation methods. All you have to do is demonstrate how you would calculate the distance between two lat/long pairs of coordinates, like the ones I used in my example earlier in this thread. You keep saying it's really simple to do. So do it, and show your calculations.
If I was struggling with it, then I would not have given the formula for how to do it.
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I am sure they do.
They also use a point of origin.
What, exactly, do you mean by 'point of origin'?, and how does that facilitate the calculation of a distance between two lat/long points?
Whenever you calculate distance between two point on a x/y coordinate system, there is always a point of origin.
Then it is simply the distance formula d=√((x2-x1)²+(y2-y1)²)
Ok...so one of the two points is your x1y1 and the other point is your x2y2
I gave you two places, one in Iraq and one in Tehran, with a lat/long for each. I don't mind which one you take as your point of origin, if that's how you like to think about it. I just want you to plug the coordinates in to your equation and work out the distance between them.
If I was struggling with it, then I would not have given the formula for how to do it.
Careful though...your formula there is for cartesian coordinates, so you'll want to convert your lat long before you plug the numbers in, or maybe just come up with a formula that works the distance out directly from the angular coordinates on, presumably, the monopole FE map. Curious to see how you go about doing that.
Looking forward to your answer!
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I am sure they do.
They also use a point of origin.
What, exactly, do you mean by 'point of origin'?, and how does that facilitate the calculation of a distance between two lat/long points?
Whenever you calculate distance between two point on a x/y coordinate system, there is always a point of origin.
Then it is simply the distance formula d=√((x2-x1)²+(y2-y1)²)
Ok...so one of the two points is your x1y1 and the other point is your x2y2
I gave you two places, one in Iraq and one in Tehran, with a lat/long for each. I don't mind which one you take as your point of origin, if that's how you like to think about it. I just want you to plug the coordinates in to your equation and work out the distance between them.
If I was struggling with it, then I would not have given the formula for how to do it.
Careful though...your formula there is for cartesian coordinates, so you'll want to convert your lat long before you plug the numbers in, or maybe just come up with a formula that works the distance out directly from the angular coordinates on, presumably, the monopole FE map. Curious to see how you go about doing that.
Looking forward to your answer!
Of course you wouldn't mind which is the point of origin, given there is no true longitudinal zero for the fake sphere.
You knew this of course.
Further proof all of it a bunch of crap.
ICBM's, as far as everyone here is concerned, are fictional.
Therefore, the OP is fictional.
It belongs in Complete Nonsense, like most of the other fictional stuff offered up by the usual crowd.
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Of course you wouldn't mind which is the point of origin, given there is no true longitudinal zero for the fake sphere.
That's completely irrelevant. Longitudinal datum is indeed entirely arbitrary - a historic debate won by the British. But that's equally true of the monopole FE map, which is, I'm assuming, your preferred model? You and Tom have both agreed that lat/long coordinates work just fine on FE, so where's the problem there?
Origins are generally arbitrary things. Your formula, using cartesian coordinates, would return the same distance between two places regardless of where the origin was placed. It could be one of the two places, or somewhere entirely different - it would all cancel out.
Your problem isn't the origin. Your problem is you've triumphantly presented a formula for cartesian coordinates but you've been given lat/long, measured in degrees. So plugging degrees into a formula designed for distances isn't going to work. You need some way of converting the two, especially since you and Tom keep saying how simple it is and how, for example, mariners have been navigating that way for ages. Well, ok...but they've been using lat/long for centuries, and very happily calculating distance between points. So you need to explain how that could be possible on a FE.
If you are going with the monopole model, what you effectively have in a lat/long is a set of polar coordinates. These are normally a distance and an angle, of course, but on the monopole FE that's kind of what you have. Longitude makes sense on either RE or FE. Latitude though can't be an angle on the monopole FE. So you need some way of converting degrees latitude to a distance from north pole - the origin of your polar coordinate system. I don't know what you want to use for this conversion because you guys never really say - I guess 1nm per minute of latitude, as per conventional navigational thinking? That would give a radial distance for the FE of 180 * 60 = 10,800 nautical miles, which is pretty close to the Wiki's estimate for the radius.
Assuming you're happy with that conversion, you should just be able to plug in any lat/long to a polar/cartesian coordinate converter and, hey presto, out will come the x/y pairings you need to plug into your formula. Then we can get some distances out of you.
With me so far?
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Of course you wouldn't mind which is the point of origin, given there is no true longitudinal zero for the fake sphere.
That's completely irrelevant. Longitudinal datum is indeed entirely arbitrary - a historic debate won by the British. But that's equally true of the monopole FE map, which is, I'm assuming, your preferred model? You and Tom have both agreed that lat/long coordinates work just fine on FE, so where's the problem there?
Origins are generally arbitrary things. Your formula, using cartesian coordinates, would return the same distance between two places regardless of where the origin was placed. It could be one of the two places, or somewhere entirely different - it would all cancel out.
Your problem isn't the origin. Your problem is you've triumphantly presented a formula for cartesian coordinates but you've been given lat/long, measured in degrees. So plugging degrees into a formula designed for distances isn't going to work. You need some way of converting the two, especially since you and Tom keep saying how simple it is and how, for example, mariners have been navigating that way for ages. Well, ok...but they've been using lat/long for centuries, and very happily calculating distance between points. So you need to explain how that could be possible on a FE.
If you are going with the monopole model, what you effectively have in a lat/long is a set of polar coordinates. These are normally a distance and an angle, of course, but on the monopole FE that's kind of what you have. Longitude makes sense on either RE or FE. Latitude though can't be an angle on the monopole FE. So you need some way of converting degrees latitude to a distance from north pole - the origin of your polar coordinate system. I don't know what you want to use for this conversion because you guys never really say - I guess 1nm per minute of latitude, as per conventional navigational thinking? That would give a radial distance for the FE of 180 * 60 = 10,800 nautical miles, which is pretty close to the Wiki's estimate for the radius.
Assuming you're happy with that conversion, you should just be able to plug in any lat/long to a polar/cartesian coordinate converter and, hey presto, out will come the x/y pairings you need to plug into your formula. Then we can get some distances out of you.
With me so far?
Essentially, being with you requires the thought the latitude/longitude system in place is based on a globe, when it isn't.
So, no.
I will never be "with you."
Because that is a flat out lie.
It's arbitrary, as you just admitted.
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Essentially, being with you requires the thought the latitude/longitude system in place is based on a globe, when it isn't.
So, no.
I will never be "with you."
Because that is a flat out lie.
It's arbitrary, as you just admitted.
All coordinate systems are to some extent arbitrary. You can have any origin you wish. As long as the coordinates all reference the same system, distances between pairs of coordinates will give the same result.
You yourself said you could work it out with your formula, but now you're saying you can't because the system is wrong. But all those mariners through the ages, diligently recording their lat/long based on celestial observations (as per Tom's statement)...were they wrong, then?
And what, exactly, is the lat/long system we should be using then? What system do you suppose our Iraqi Scud aimers were using?
And which FE model are you using?
And how big is the FE?
[edited to remove tautology!]
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Essentially, being with you requires the thought the latitude/longitude system in place is based on a globe, when it isn't.
So, no.
I will never be "with you."
Because that is a flat out lie.
It's arbitrary, as you just admitted.
All coordinate systems are to some extent arbitrary. You can have any origin you wish. As long as the coordinates all reference the same system, distances between pairs of coordinates will give the same result.
You yourself said you could work it out with your formula, but now you're saying you can't because the system is wrong. But all those mariners through the ages, diligently recording their lat/long based on celestial and solar observations (as per Tom's statement)...were they wrong, then?
And what, exactly, is the lat/long system we should be using then? What system do you suppose our Iraqi Scud aimers were using?
And which FE model are you using?
And how big is the FE?
I am certainly not stating the distance couldn't be worked out using two points.
You are not paying attention.
I am stating the two points being arbitrary would be close, but it was not used to fire those missiles which you love to use.
Like I wrote earlier, a much more likely scenario would have been actual distances obtained by close up surveillance of the land, performed by live people.
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Essentially, being with you requires the thought the latitude/longitude system in place is based on a globe, when it isn't.
You are wrong:
History of longitude
https://en.wikipedia.org/wiki/History_of_longitude
"Eratosthenes in the 3rd century BCE first proposed a system of latitude and longitude for a map of the world."
Remember, Eratosthenes was the guy who first calculated the size of the earth based upon it being spherical.
Separately, centuries later, "Ancient Hindu astronomers were aware of the method of determining longitude from lunar eclipses, assuming a spherical earth. The method is described in the Sûrya Siddhânta, a Sanskrit treatise on Indian astronomy thought to date from the late 4th century or early 5th century CE.[9]"
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Essentially, being with you requires the thought the latitude/longitude system in place is based on a globe, when it isn't.
You are wrong:
History of longitude
https://en.wikipedia.org/wiki/History_of_longitude
"Eratosthenes in the 3rd century BCE first proposed a system of latitude and longitude for a map of the world."
Remember, Eratosthenes was the guy who first calculated the size of the earth based upon it being spherical.
Separately, centuries later, "Ancient Hindu astronomers were aware of the method of determining longitude from lunar eclipses, assuming a spherical earth. The method is described in the Sûrya Siddhânta, a Sanskrit treatise on Indian astronomy thought to date from the late 4th century or early 5th century CE.[9]"
Longitude is arbitrary, meaning there is no true 0/0.
You're wrong.
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Every long distance route taken by sea or air has essentially remained UNCHANGED since long distance travel commenced.
None of it was ever based on the idea of a globe.
Wrong again. Dead reckoning and rhumb lines were the first navigational methods. Since then, for long hauls, excluding course corrections due to weather and other needs, great circle navigation is used. Look it up.
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Every long distance route taken by sea or air has essentially remained UNCHANGED since long distance travel commenced.
None of it was ever based on the idea of a globe.
Wrong again. Dead reckoning and rhumb lines were the first navigational methods. Since then, for long hauls, excluding course corrections due to weather and other needs, great circle navigation is used. Look it up.
I do not need to look it up.
Everything you mention has nothing to do with a sphere.
NOTHING.
Long distance navigation does not need a sphere to get you where you want to be.
Bye now.
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Essentially, being with you requires the thought the latitude/longitude system in place is based on a globe, when it isn't.
You are wrong:
History of longitude
https://en.wikipedia.org/wiki/History_of_longitude
"Eratosthenes in the 3rd century BCE first proposed a system of latitude and longitude for a map of the world."
Remember, Eratosthenes was the guy who first calculated the size of the earth based upon it being spherical.
Separately, centuries later, "Ancient Hindu astronomers were aware of the method of determining longitude from lunar eclipses, assuming a spherical earth. The method is described in the Sûrya Siddhânta, a Sanskrit treatise on Indian astronomy thought to date from the late 4th century or early 5th century CE.[9]"
Longitude is arbitrary, meaning there is no true 0/0.
You're wrong.
What do you mean by "true"? The Prime Meridian is used as 0 degrees longitude on pretty much every navigational map/chart in the world. And yes, it was determined by man. So what? So are the hours of a day, arbitrarily concocted to be made up of concocted and arbitrary 60 minutes for each and so on. Longitude & Latitude are used to accurately & uniformly & consistently navigate just like Hours and Minutes are used to accurately & uniformly & consistently tell time. Your "arbitrary" point means literally nothing.
"If you look at the intersection of 0 degrees latitude (known as the Equator) and 0 degrees longitude (known as the Prime Meridian) on a map, you will see that the confluence falls in the middle of the Atlantic Ocean, in the Gulf of Guinea off the coast of western Africa."
https://www.geographyrealm.com/zero-degrees-latitude-and-zero-degrees-longitude/
And you're still wrong, "Eratosthenes in the 3rd century BCE first proposed a system of latitude and longitude for a map of the world."
Remember, Eratosthenes was the guy who first calculated the size of the earth based upon it being spherical.
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Essentially, being with you requires the thought the latitude/longitude system in place is based on a globe, when it isn't.
You are wrong:
History of longitude
https://en.wikipedia.org/wiki/History_of_longitude
"Eratosthenes in the 3rd century BCE first proposed a system of latitude and longitude for a map of the world."
Remember, Eratosthenes was the guy who first calculated the size of the earth based upon it being spherical.
Separately, centuries later, "Ancient Hindu astronomers were aware of the method of determining longitude from lunar eclipses, assuming a spherical earth. The method is described in the Sûrya Siddhânta, a Sanskrit treatise on Indian astronomy thought to date from the late 4th century or early 5th century CE.[9]"
Longitude is arbitrary, meaning there is no true 0/0.
You're wrong.
What do you mean by "true"? The Prime Meridian is used as 0 degrees longitude on pretty much every navigational map/chart in the world. And yes, it was determined by man. So what? So are the hours of a day, arbitrarily concocted to be made up of concocted and arbitrary 60 minutes for each and so on. Longitude & Latitude are used to accurately & uniformly & consistently navigate just like Hours and Minutes are used to accurately & uniformly & consistently tell time. Your "arbitrary" point means literally nothing.
"If you look at the intersection of 0 degrees latitude (known as the Equator) and 0 degrees longitude (known as the Prime Meridian) on a map, you will see that the confluence falls in the middle of the Atlantic Ocean, in the Gulf of Guinea off the coast of western Africa."
https://www.geographyrealm.com/zero-degrees-latitude-and-zero-degrees-longitude/
And you're still wrong, "Eratosthenes in the 3rd century BCE first proposed a system of latitude and longitude for a map of the world."
Remember, Eratosthenes was the guy who first calculated the size of the earth based upon it being spherical.
Just throwing more meaningless words into the issue isn't going to help you here.
Eratosthenes could have done everything you want attributed to him.
Hell, he could have won an Oscar for all I care.
Doesn't matter what he proposed.
0/0, as far the earth is concerned, is arbitrary.
ICBM's, as far as anyone here is concerned, are fictional.
Therefore, the OP is a fictional pursuit and thus belongs in CN.
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Every long distance route taken by sea or air has essentially remained UNCHANGED since long distance travel commenced.
None of it was ever based on the idea of a globe.
Wrong again. Dead reckoning and rhumb lines were the first navigational methods. Since then, for long hauls, excluding course corrections due to weather and other needs, great circle navigation is used. Look it up.
I do not need to look it up.
Everything you mention has nothing to do with a sphere.
Great circle navigation certainly is all about a sphere, the shortest route on a sphere.
"A great circle, also known as an orthodrome, of a sphere is the intersection of the sphere and a plane that passes through the center point of the sphere."
https://en.wikipedia.org/wiki/Great_circle
Long distance navigation does not need a sphere to get you where you want to be.
Bye now.
Of course it does't "need it". But if you want the shortest long haul route, Great Circles win and are used by the transport of goods and humans the world over.
Bye now.
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Every long distance route taken by sea or air has essentially remained UNCHANGED since long distance travel commenced.
None of it was ever based on the idea of a globe.
Wrong again. Dead reckoning and rhumb lines were the first navigational methods. Since then, for long hauls, excluding course corrections due to weather and other needs, great circle navigation is used. Look it up.
I do not need to look it up.
Everything you mention has nothing to do with a sphere.
Great circle navigation certainly is all about a sphere, the shortest route on a sphere.
"A great circle, also known as an orthodrome, of a sphere is the intersection of the sphere and a plane that passes through the center point of the sphere."
https://en.wikipedia.org/wiki/Great_circle
Great circle navigation is just a bunch of fiction.
No way to demonstrate it is actually performed.
And, it has nothing to do with the OP.
Just more obfuscating BS peddled by your sorts to hide the original BS.
Long distance navigation does not need a sphere to get you where you want to be.
Bye now.
Of course it does't "need it". But if you want the shortest long haul route, Great Circles win and are used by the transport of goods and humans the world over.
Bye now.
Like I wrote above, just more regular BS trotted out by you and the rest.
As if people and things do not arrive when they do, regardless.
Taking a line from a flat surface and transcribing it to the surface of a sphere will usually result in a a curve.
Bye now.
-
Essentially, being with you requires the thought the latitude/longitude system in place is based on a globe, when it isn't.
You are wrong:
History of longitude
https://en.wikipedia.org/wiki/History_of_longitude
"Eratosthenes in the 3rd century BCE first proposed a system of latitude and longitude for a map of the world."
Remember, Eratosthenes was the guy who first calculated the size of the earth based upon it being spherical.
Separately, centuries later, "Ancient Hindu astronomers were aware of the method of determining longitude from lunar eclipses, assuming a spherical earth. The method is described in the Sûrya Siddhânta, a Sanskrit treatise on Indian astronomy thought to date from the late 4th century or early 5th century CE.[9]"
Longitude is arbitrary, meaning there is no true 0/0.
You're wrong.
What do you mean by "true"? The Prime Meridian is used as 0 degrees longitude on pretty much every navigational map/chart in the world. And yes, it was determined by man. So what? So are the hours of a day, arbitrarily concocted to be made up of concocted and arbitrary 60 minutes for each and so on. Longitude & Latitude are used to accurately & uniformly & consistently navigate just like Hours and Minutes are used to accurately & uniformly & consistently tell time. Your "arbitrary" point means literally nothing.
"If you look at the intersection of 0 degrees latitude (known as the Equator) and 0 degrees longitude (known as the Prime Meridian) on a map, you will see that the confluence falls in the middle of the Atlantic Ocean, in the Gulf of Guinea off the coast of western Africa."
https://www.geographyrealm.com/zero-degrees-latitude-and-zero-degrees-longitude/
And you're still wrong, "Eratosthenes in the 3rd century BCE first proposed a system of latitude and longitude for a map of the world."
Remember, Eratosthenes was the guy who first calculated the size of the earth based upon it being spherical.
Just throwing more meaningless words into the issue isn't going to help you here.
Eratosthenes could have done everything you want attributed to him.
Hell, he could have won an Oscar for all I care.
Doesn't matter what he proposed.
0/0, as far the earth is concerned, is arbitrary.
ICBM's, as far as anyone here is concerned, are fictional.
Therefore, the OP is a fictional pursuit and thus belongs in CN.
Your opinion does not matter and is irrelevant. You're still wrong.
0/0 is as arbitrary as 12 o'clock is on your watch.
ICBMs are fictional "as far as anyone here is concerned"? Wrong again. There are loads of people here who believe they, in fact, are quite real. You're fictional opinions not rooted in fact most likely belong in CN.
-
Every long distance route taken by sea or air has essentially remained UNCHANGED since long distance travel commenced.
None of it was ever based on the idea of a globe.
Wrong again. Dead reckoning and rhumb lines were the first navigational methods. Since then, for long hauls, excluding course corrections due to weather and other needs, great circle navigation is used. Look it up.
I do not need to look it up.
Everything you mention has nothing to do with a sphere.
Great circle navigation certainly is all about a sphere, the shortest route on a sphere.
"A great circle, also known as an orthodrome, of a sphere is the intersection of the sphere and a plane that passes through the center point of the sphere."
https://en.wikipedia.org/wiki/Great_circle
Great circle navigation is just a bunch of fiction.
No way to demonstrate it is actually performed.
Wrong again:
(https://i.imgur.com/2GAcDmX.png)
And, it has nothing to do with the OP.
Umm, everyone is talking about distances and routes. Now you're just scrambling.
Just more obfuscating BS peddled by your sorts to hide the original BS.
Long distance navigation does not need a sphere to get you where you want to be.
Bye now.
Of course it does't "need it". But if you want the shortest long haul route, Great Circles win and are used by the transport of goods and humans the world over.
Bye now.
Like I wrote above, just more regular BS trotted out by you and the rest.
As if people and things do not arrive when they do, regardless.
Taking a line from a flat surface and transcribing it to the surface of a sphere will usually result in a a curve.
Bye now.
People and things arrive when they do regardless of the route they take. Interesting assertion.
See yah.
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Essentially, being with you requires the thought the latitude/longitude system in place is based on a globe, when it isn't.
You are wrong:
History of longitude
https://en.wikipedia.org/wiki/History_of_longitude
"Eratosthenes in the 3rd century BCE first proposed a system of latitude and longitude for a map of the world."
Remember, Eratosthenes was the guy who first calculated the size of the earth based upon it being spherical.
Separately, centuries later, "Ancient Hindu astronomers were aware of the method of determining longitude from lunar eclipses, assuming a spherical earth. The method is described in the Sûrya Siddhânta, a Sanskrit treatise on Indian astronomy thought to date from the late 4th century or early 5th century CE.[9]"
Longitude is arbitrary, meaning there is no true 0/0.
You're wrong.
What do you mean by "true"? The Prime Meridian is used as 0 degrees longitude on pretty much every navigational map/chart in the world. And yes, it was determined by man. So what? So are the hours of a day, arbitrarily concocted to be made up of concocted and arbitrary 60 minutes for each and so on. Longitude & Latitude are used to accurately & uniformly & consistently navigate just like Hours and Minutes are used to accurately & uniformly & consistently tell time. Your "arbitrary" point means literally nothing.
"If you look at the intersection of 0 degrees latitude (known as the Equator) and 0 degrees longitude (known as the Prime Meridian) on a map, you will see that the confluence falls in the middle of the Atlantic Ocean, in the Gulf of Guinea off the coast of western Africa."
https://www.geographyrealm.com/zero-degrees-latitude-and-zero-degrees-longitude/
And you're still wrong, "Eratosthenes in the 3rd century BCE first proposed a system of latitude and longitude for a map of the world."
Remember, Eratosthenes was the guy who first calculated the size of the earth based upon it being spherical.
Just throwing more meaningless words into the issue isn't going to help you here.
Eratosthenes could have done everything you want attributed to him.
Hell, he could have won an Oscar for all I care.
Doesn't matter what he proposed.
0/0, as far the earth is concerned, is arbitrary.
ICBM's, as far as anyone here is concerned, are fictional.
Therefore, the OP is a fictional pursuit and thus belongs in CN.
Your opinion does not matter and is irrelevant. You're still wrong.
0/0 is as arbitrary as 12 o'clock is on your watch.
ICBMs are fictional "as far as anyone here is concerned"? Wrong again. There are loads of people here who believe they, in fact, are quite real. You're fictional opinions not rooted in fact most likely belong in CN.
See, I am not offering opinion.
I am offering fact.
Another relevant discovery of a weak area of thought when it comes to you.
Clearly cannot cannot tell between a fact (ICBM's are currently FICTIONAL) and an opinion (Many people BELIEVE in ICBM's)
ICBM's?
Nothing offered by anyone to show they are real, except a statement of belief.
As you agree, we are done here.
-
Every long distance route taken by sea or air has essentially remained UNCHANGED since long distance travel commenced.
None of it was ever based on the idea of a globe.
Wrong again. Dead reckoning and rhumb lines were the first navigational methods. Since then, for long hauls, excluding course corrections due to weather and other needs, great circle navigation is used. Look it up.
I do not need to look it up.
Everything you mention has nothing to do with a sphere.
Great circle navigation certainly is all about a sphere, the shortest route on a sphere.
"A great circle, also known as an orthodrome, of a sphere is the intersection of the sphere and a plane that passes through the center point of the sphere."
https://en.wikipedia.org/wiki/Great_circle
Great circle navigation is just a bunch of fiction.
No way to demonstrate it is actually performed.
Wrong again:
(https://i.imgur.com/2GAcDmX.png)
And, it has nothing to do with the OP.
Umm, everyone is talking about distances and routes. Now you're just scrambling.
Just more obfuscating BS peddled by your sorts to hide the original BS.
Long distance navigation does not need a sphere to get you where you want to be.
Bye now.
Of course it does't "need it". But if you want the shortest long haul route, Great Circles win and are used by the transport of goods and humans the world over.
Bye now.
Like I wrote above, just more regular BS trotted out by you and the rest.
As if people and things do not arrive when they do, regardless.
Taking a line from a flat surface and transcribing it to the surface of a sphere will usually result in a a curve.
Bye now.
People and things arrive when they do regardless of the route they take. Interesting assertion.
See yah.
As if a web picture demonstrates actual reality.
Jesus.
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AnYtHiNg I dOnT lIkE dOeSnT eXiSt
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Every long distance route taken by sea or air has essentially remained UNCHANGED since long distance travel commenced.
None of it was ever based on the idea of a globe.
Wrong again. Dead reckoning and rhumb lines were the first navigational methods. Since then, for long hauls, excluding course corrections due to weather and other needs, great circle navigation is used. Look it up.
I do not need to look it up.
Everything you mention has nothing to do with a sphere.
Great circle navigation certainly is all about a sphere, the shortest route on a sphere.
"A great circle, also known as an orthodrome, of a sphere is the intersection of the sphere and a plane that passes through the center point of the sphere."
https://en.wikipedia.org/wiki/Great_circle
Great circle navigation is just a bunch of fiction.
No way to demonstrate it is actually performed.
Wrong again:
(https://i.imgur.com/2GAcDmX.png)
And, it has nothing to do with the OP.
Umm, everyone is talking about distances and routes. Now you're just scrambling.
Just more obfuscating BS peddled by your sorts to hide the original BS.
Long distance navigation does not need a sphere to get you where you want to be.
Bye now.
Of course it does't "need it". But if you want the shortest long haul route, Great Circles win and are used by the transport of goods and humans the world over.
Bye now.
Like I wrote above, just more regular BS trotted out by you and the rest.
As if people and things do not arrive when they do, regardless.
Taking a line from a flat surface and transcribing it to the surface of a sphere will usually result in a a curve.
Bye now.
People and things arrive when they do regardless of the route they take. Interesting assertion.
See yah.
As if a web picture demonstrates actual reality.
Jesus.
That's hilarious, a "web picture". It's a radar depiction of the route of a flight from LA to London. You can look at many, many more great circle flights and routes here (and every other flight tracking service - They all, strangely, show the same thing): https://www.flightradar24.com/-0.88,-60.28/3
Funny how those long haul non-stop flights just don't blast straight from point to point. Funny how these great circle routes look longer to the layperson. Strange that the airlines would take a visually longer route than necessary, burning more fuel, longer durations, less flights, all meaning less profit. So very strange indeed.
Oh wait, maybe you have a flat earth map that would explain this phenomenon. Can you post your flat earth map? The one you perhaps use.
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I am certainly not stating the distance couldn't be worked out using two points.
You are not paying attention.
I am stating the two points being arbitrary would be close, but it was not used to fire those missiles which you love to use.
Like I wrote earlier, a much more likely scenario would have been actual distances obtained by close up surveillance of the land, performed by live people.
But the two points aren't themselves arbitrary. They are measured from the same arbitrary datum. As the formula you yourself provided clearly shows, it matters not where that datum is, because the x and y pairs are subtracted from each other to find an x/y pair of differences, which are then squared, summed and rooted to find the hypotenuse - the distance between them. Are you suggesting that your formula would return a different result if I moved the origin for the same pair of points? Because that's basic maths, and if you're struggling with that, then we aren't going to get anywhere.
You and Tom have both acknowledged that people navigated for years using celestial nav to obtain their positions - expressed as a lat/long. You yourself described lat/long as an 'x/y coordinate system' - not strictly true, given that it's an angular system, but never mind. You said you could calculate distances between two points using your formula, but now you seem to have rolled back on that citing some very odd excuse about the arbitrary nature of longitudes - odd because of course that would also apply to the monopole FE map. Where is the zero longitude line on that?
You are refusing to engage in any discussion about how one might convert lat/long into some form of coordinate on the monopole FE map. You are refusing to discuss which FE map you believe to be correct, or what its dimensions are. This stuff should be bread and butter for an FE enthusiast, surely?
Instead, all you do is dodge questions and say everything is lies. Fine by me - I don't suppose I'll ever persuade you to change - I'm just making sure nobody else coming here is persuaded by your arguments, and your total failure to answer questions or engage in any meaningful debate is plain to see, which does rather undermine your position. Entirely up to you how you proceed.
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Essentially, being with you requires the thought the latitude/longitude system in place is based on a globe, when it isn't.
So, no.
I will never be "with you."
Because that is a flat out lie.
It's arbitrary, as you just admitted.
All coordinate systems are to some extent arbitrary. You can have any origin you wish. As long as the coordinates all reference the same system, distances between pairs of coordinates will give the same result.
You yourself said you could work it out with your formula, but now you're saying you can't because the system is wrong. But all those mariners through the ages, diligently recording their lat/long based on celestial and solar observations (as per Tom's statement)...were they wrong, then?
And what, exactly, is the lat/long system we should be using then? What system do you suppose our Iraqi Scud aimers were using?
And which FE model are you using?
And how big is the FE?
I am certainly not stating the distance couldn't be worked out using two points.
You are not paying attention.
I am stating the two points being arbitrary would be close, but it was not used to fire those missiles which you love to use.
Like I wrote earlier, a much more likely scenario would have been actual distances obtained by close up surveillance of the land, performed by live people.
I think what you actually said earlier was that they could drive a car to measure the distance. Do you think that the Iraqi army drove to Tehran before launching their Scuds, or that Hamas drove from Gaza to Tel Aviv? What other ways might live people on the ground survey the land?
Regarding the measurement between two points on Earth, the arbitrary nature of points-of-origin and what-not; why don't you nominate 2 points, around two thousand miles apart, and tell us the exact distance. You get extra points for showing us your calculations.
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Everyone, this is hilarious, you gotta read this quote from a link Tom Bishop posted saying that Kepler still used epicycles:
"Kepler accounted for the second class of deviation by his perspicuous laws of planetary motion. It is this fact that has generally been credited with the destruction of epicycles as a mechanical device."
I was going to repost the link, but too much trouble and it was a conventional science article from I think 1940 and waste of your time to read it unless you want a intro to astronomy circa 1940. Tom Bishop likes his science quotes old.
Tom Bishop, it is true there are still deviations from a perfect ellipse, but these are not epicycles, they are the calculable gravitational impact of planet's moons, nearby planets, wobble, decelleration, probably more, see a real astronomer for an accurate list. Epicycles had no known physical phenomena, just a fudge factor that made the apparent motion calculable at least pretty close, the error appraoches 0 as the number of epicycles approaches infinty, I would think.
RE the original questions do ICBMs exist and are they aimed RE or FE? I think I know the answer. A person can look at the web of facts and experts and analyze the plausibility using consistency with knwon facts and proof by contradiction to rule out wrong answers. Or one can ignore contradictions, misunderstand science, start with your conclusion and form a group willing to accept any ridiculous explanation for their treasured narrative.
Occam's razor: the simple explanation for the facts I know is that ICBMs exist and the people who aim and test them know the true shape of the earth.
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Everyone, this is hilarious, you gotta read this quote from a link Tom Bishop posted saying that Kepler still used epicycles:
"Kepler accounted for the second class of deviation by his perspicuous laws of planetary motion. It is this fact that has generally been credited with the destruction of epicycles as a mechanical device.
And sitting atop and below that quote are explanations on why that "generally credited" idea is wrong.
I was going to repost the link, but too much trouble and it was a conventional science article from I think 1940 and waste of your time to read it unless you want a intro to astronomy circa 1940. Tom Bishop likes his science quotes old.
You were literally citing the work of a scientist born in 1571. ::)
Too bad that Kepler used epicycles, contradictory to your claim. There is an image of his system with epicycles there in that link. - https://ia802909.us.archive.org/18/items/the-foundations-of-astrodynamics/The%20Foundations%20of%20Astrodynamics.pdf
Occam's razor: the simple explanation for the facts I know is that ICBMs exist and the people who aim and test them know the true shape of the earth.
I have a hard time understanding this sentence.
You know the simple explanation for the facts that ICBMs exist? How is that?
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Tom, can you help Action80 with the fiendish 'finding the distance between two lat/longs' challenge? He's still struggling.
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From the late 1940s the UK had a strategic nuclear capability in the form of manned aircraft carrying free-fall atomic and hydrogen weapons. This mirrored similar forces in perceived-allied and opposing nations (United States, USSR, France, later China). The policy was/is known as deterrence. The final iteration of this in Britain's case in the 1960s were the Vulcan and Victor bombers of the Royal Air Force carrying a short-range supersonic nuclear missile called Blue Steel.
In 1969, responsibility for Britain's nuclear deterrence was passed from the RAF to the Royal Navy, in the form of long range SLBMs to be launched by submarines; initially Polaris, currently Trident, with a range of around 7,500 miles. Blue Steel was retired, but the Vulcan and Victor continued in service for another 15 to 20 years in other roles. Britain retained a short-range tactical nuclear capability using free-fall bombs carried by Jaguars and Tornadoes until the late 1990s, when this was also retired.
Among the 5 major nuclear-strategic powers, Britain is currently alone in having all its nuclear eggs in the ICBM/SLBM basket. The USA and Russia have land-based ICBM, submarine SLBMs, and manned aircraft. France has SLBMs and manned aircraft.
Is it likely that the UK would have surrendered its nuclear deterrence to a fictional technology? Remember that the other powers are aware of the "pretense".
Why would Kim Jong Un be chasing a mythological technology to threaten the US?
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Tom, can you help Action80 with the fiendish 'finding the distance between two lat/longs' challenge? He's still struggling.
I think Tom is happy to let Action80 muddy the waters.
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As if a web picture demonstrates actual reality.
Jesus.
That's hilarious, a "web picture". It's a radar depiction of the route of a flight from LA to London. You can look at many, many more great circle flights and routes here (and every other flight tracking service - They all, strangely, show the same thing): https://www.flightradar24.com/-0.88,-60.28/3
Funny how those long haul non-stop flights just don't blast straight from point to point. Funny how these great circle routes look longer to the layperson. Strange that the airlines would take a visually longer route than necessary, burning more fuel, longer durations, less flights, all meaning less profit. So very strange indeed.
Oh wait, maybe you have a flat earth map that would explain this phenomenon. Can you post your flat earth map? The one you perhaps use.
Your claiming now that you didn't get the screenshot from the web.
Cool story bro.
Everyone knows your claim is total BS and everyone knows you want to throw all of this OFF TOPIC BS into the mix because you got nothing worthwhile to offer.
If you have any evidence to offer in regard to the actual existence of ICBM's let us know, okay.
Bye now.
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Tom, can you help Action80 with the fiendish 'finding the distance between two lat/longs' challenge? He's still struggling.
I am not struggling with it at all.
I told you how to do it.
Go ahead.
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From the late 1940s the UK had a strategic nuclear capability in the form of manned aircraft carrying free-fall atomic and hydrogen weapons. This mirrored similar forces in perceived-allied and opposing nations (United States, USSR, France, later China). The policy was/is known as deterrence. The final iteration of this in Britain's case in the 1960s were the Vulcan and Victor bombers of the Royal Air Force carrying a short-range supersonic nuclear missile called Blue Steel.
In 1969, responsibility for Britain's nuclear deterrence was passed from the RAF to the Royal Navy, in the form of long range SLBMs to be launched by submarines; initially Polaris, currently Trident, with a range of around 7,500 miles. Blue Steel was retired, but the Vulcan and Victor continued in service for another 15 to 20 years in other roles. Britain retained a short-range tactical nuclear capability using free-fall bombs carried by Jaguars and Tornadoes until the late 1990s, when this was also retired.
Among the 5 major nuclear-strategic powers, Britain is currently alone in having all its nuclear eggs in the ICBM/SLBM basket. The USA and Russia have land-based ICBM, submarine SLBMs, and manned aircraft. France has SLBMs and manned aircraft.
Is it likely that the UK would have surrendered its nuclear deterrence to a fictional technology? Remember that the other powers are aware of the "pretense".
Why would Kim Jong Un be chasing a mythological technology to threaten the US?
Why do you believe any of the crap put out by the HIGHLY RELIABLE propogandists?
Pitiful..
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I was part of it; I was in the Royal Air Force.
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I am not struggling with it at all.
I told you how to do it.
No, you didn't. You wheeled out a formula for working out the distance between cartesian coordinates and then, when I pointed out that you can't plug lat/long into that because they are measured in degrees, you went off down some weird rabbit hole about the arbitrary nature of origins, oblivious to the fact that pretty much all coordinate systems are in some way arbitrary, and it doesn't effect the distance between points. I'd be quite happy to do the maths for you, as you quite clearly can't do it yourself, but to do that I'd need you tell me a fair bit more about what size and layout you think the earth actually is - this is your model, not mine. If you can't put your support behind a map, and tell me its dimensions, we can't work out distances, can we?
Unless you want people reading this to assume that you're talking nonsense, I strongly suggest you in some way engage with the debate and put up some actual answers. Which FE map are you backing? Monopole? Something else? And how big is the map? Give us some usable dimension, like a radius, or a distance per degree of latitude, for example.
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Great circle navigation is just a bunch of fiction.
No way to demonstrate it is actually performed.
Bye now.
How about if I testified based upon daily personal experience that "Great circle navigation is used every day aboard long haul cargo ships". This is either true, or I'm a liar. Assuming that you will accuse me of lying then how could you demonstrate that? Have you ever been on the bridge of a cargo ship when the course to the next port was being planned? To call me a liar then you must have been and in that particular instance a great circle route was not planned for some particular reason. Have you witnessed more than 10 instances when a great circle course wasn't planned on a voyage of more than 5000 miles? Now assuming that happened have you been on multiple ships, multiple times, operated by different companies? Did they all refrain from planning great circle courses? If that's true then it would be interesting to know just which companies those were. Shipping is a competitive industry and knowing what the competition is doing would be an advantage. I am, however, going to speculate that you have never actually been on the bridge of a ship, or on a long haul run of any kind, and are just trolling by throwing out inciteful rhetoric without actual knowledge. You degrade the flat earth theory every time you do that and that makes you look foolish.
Bye now.
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I was part of it; I was in the Royal Air Force.
Being in the service then, you would certainly have access to the actual launch codes.
Or, you would be just as "in the know," as the rest of the regular saps who simply follow the party line.
I'm guessing #2.
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I am not struggling with it at all.
I told you how to do it.
No, you didn't. You wheeled out a formula for working out the distance between cartesian coordinates and then, when I pointed out that you can't plug lat/long into that because they are measured in degrees, you went off down some weird rabbit hole about the arbitrary nature of origins, oblivious to the fact that pretty much all coordinate systems are in some way arbitrary, and it doesn't effect the distance between points. I'd be quite happy to do the maths for you, as you quite clearly can't do it yourself, but to do that I'd need you tell me a fair bit more about what size and layout you think the earth actually is - this is your model, not mine. If you can't put your support behind a map, and tell me its dimensions, we can't work out distances, can we?
Unless you want people reading this to assume that you're talking nonsense, I strongly suggest you in some way engage with the debate and put up some actual answers. Which FE map are you backing? Monopole? Something else? And how big is the map? Give us some usable dimension, like a radius, or a distance per degree of latitude, for example.
It has nothing to do with what kind of points you offer up and you know damn well it doesn't.
Despite your objections, you know damn well it has to do with no defined 0/0.
Without a defined 0/0, the actual distance of each grid set forth by intersecting lines of lat/long, or x/y, or whatever you choose to call them, cannot be officially defined, and hence why you all struggle so mightily with issues of distance when it comes to interpreting whole world maps.
Further, when it comes to mapping specific areas, such as what you have offered up, that is a depicted as FLAT on a flat piece of paper.
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Great circle navigation is just a bunch of fiction.
No way to demonstrate it is actually performed.
Bye now.
How about if I testified based upon daily personal experience that "Great circle navigation is used every day aboard long haul cargo ships". This is either true, or I'm a liar. Assuming that you will accuse me of lying then how could you demonstrate that? Have you ever been on the bridge of a cargo ship when the course to the next port was being planned? To call me a liar then you must have been and in that particular instance a great circle route was not planned for some particular reason. Have you witnessed more than 10 instances when a great circle course wasn't planned on a voyage of more than 5000 miles? Now assuming that happened have you been on multiple ships, multiple times, operated by different companies? Did they all refrain from planning great circle courses? If that's true then it would be interesting to know just which companies those were. Shipping is a competitive industry and knowing what the competition is doing would be an advantage. I am, however, going to speculate that you have never actually been on the bridge of a ship, or on a long haul run of any kind, and are just trolling by throwing out inciteful rhetoric without actual knowledge. You degrade the flat earth theory every time you do that and that makes you look foolish.
Bye now.
I am accusing you of making an affirmative statement about something you have no independent way to demonstrate it to be true.
Not really a liar, per say.
Just a parroter of a common lie.
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Action80:
I believe what you call "HIGHLY RELIABLE propagandists" because they are HIGHLY RELIABLE.
I got on an airplane in Sydney AUS and flew to LA USA. The trip time, airliner speed and travel time matched the schedule, my phone gps matched lat/long, google maps distance matches, distance on globe with piece of string, everything matches. It is all HIGHLY RELIABLE. An ocean of things is reliable.
You never answered any of my questions. On FE, if I fly a plane from Tierra Del Fuego and keep the southern cross at 90 degrees from one side, where do I go?
On RE, you would make a circle around the south pole, gps, gyrocompass would all match, it is HIGHLY RELIABLE. You would track the latitude line you were on. Without going there, I can diagram and explain it, and that is all I am asking for from you. Even if it doesn't exist, the geometry still is consistent with itself.
Please lay out on a FE map what happens when a plane takes off from Tierra Del Fuego and keeps the southern cross 90 degrees off one side for 8000 miles. Do you have a HIGHLY RELIABLE map on which you can plot a HIGHLY RELIABLE course?
Please no picking nits on the question or denial because no personal experience. I am asking you for a consistent model of how it could be, not to prove it in person. The question is clear and simple, either you have the geometry or you don't, but somehow, I expect your answer will not be either "here is a diagram" or "I have no idea, no reasonable answer". Answer, please, don't waste time with evasion techniques.
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Why do you believe any of the crap put out by the HIGHLY RELIABLE propogandists?
Pitiful..
I took the trouble to read this reply again, and it says at lot more than the simple words on the page. "Why would I believe ..... propagandists". Your assumption is that we have all gained our knowledge second hand, or been duped in some way by "the system".
Like you, I've no idea who the people on this Forum are; their nationality, age, education or political views. I'm not claiming to be among the brightest buttons in the box but I get the impression that, like me, some of the correspondents here got themselves educated, got a job, and learned how to do something that the world needs in order to survive. We met similar people at work and exchanged views on how we thought the world worked. We traveled the world. Some of us worked in foreign countries, or with colleagues who came to our workplace from overseas and found that our education and backgrounds often coincided. We validated our education and training by going places, doing things, actually doing the job, and found that it worked.
Its equally evident that some people didn't.
Like everyone, we receive our share of propaganda and fake-news, but our experience has enabled us sort the wheat from the chaff, in most cases. The Internet is certainly an Information Superhighway, shame its not a Wisdom Superhighway.
Let me put the question back to you; why do you believe any of the crap put out by the propagandists? Who do you think are propagandists?
If I told you that I was personally a propagandist, would you believe me?
To paraphrase a well known internet sage;
"I won, you lost, I'm done here".
Bye now.
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Great circle navigation is just a bunch of fiction.
No way to demonstrate it is actually performed.
Bye now.
How about if I testified based upon daily personal experience that "Great circle navigation is used every day aboard long haul cargo ships". This is either true, or I'm a liar. Assuming that you will accuse me of lying then how could you demonstrate that? Have you ever been on the bridge of a cargo ship when the course to the next port was being planned? To call me a liar then you must have been and in that particular instance a great circle route was not planned for some particular reason. Have you witnessed more than 10 instances when a great circle course wasn't planned on a voyage of more than 5000 miles? Now assuming that happened have you been on multiple ships, multiple times, operated by different companies? Did they all refrain from planning great circle courses? If that's true then it would be interesting to know just which companies those were. Shipping is a competitive industry and knowing what the competition is doing would be an advantage. I am, however, going to speculate that you have never actually been on the bridge of a ship, or on a long haul run of any kind, and are just trolling by throwing out inciteful rhetoric without actual knowledge. You degrade the flat earth theory every time you do that and that makes you look foolish.
Bye now.
I am accusing you of making an affirmative statement about something you have no independent way to demonstrate it to be true.
Not really a liar, per say.
Just a parroter of a common lie.
My affirmative statement can be verified by YOU, if you choose to take up the challenge. Which other independent verification method would you believe? I'm not parroting anything. If I got aboard a ship as a passenger and went up to the bridge as a visitor and a ship's officer said "Yea, we will be taking a great circle route from China to the USA", and then I repeated that on here, you could say, 'parrot'. That's not the case here, I was one of the ship's officers. That fact would mean that my 'affirmative statement' was from personal experience and I would have to be lying if my ship didn't actually navigate a great circle course on a routine basis. Again, I'm saying that ships I worked aboard as an officer, routinely use great circle routes between ports every time they could, it's the shortest distance. Liar or not?
That brings up another question: what is your independent way of verifying your affirmative statement "Great circle navigation is just a bunch of fiction"?
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It has nothing to do with what kind of points you offer up and you know damn well it doesn't.
Despite your objections, you know damn well it has to do with no defined 0/0.
Without a defined 0/0, the actual distance of each grid set forth by intersecting lines of lat/long, or x/y, or whatever you choose to call them, cannot be officially defined, and hence why you all struggle so mightily with issues of distance when it comes to interpreting whole world maps.
Further, when it comes to mapping specific areas, such as what you have offered up, that is a depicted as FLAT on a flat piece of paper.
Which FE map are you backing, and what are its dimensions?
Why do you refuse to answer such a basic question?
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Action80:
I believe what you call "HIGHLY RELIABLE propagandists" because they are HIGHLY RELIABLE.
I got on an airplane in Sydney AUS and flew to LA USA. The trip time, airliner speed and travel time matched the schedule, my phone gps matched lat/long, google maps distance matches, distance on globe with piece of string, everything matches. It is all HIGHLY RELIABLE. An ocean of things is reliable.
You never answered any of my questions. On FE, if I fly a plane from Tierra Del Fuego and keep the southern cross at 90 degrees from one side, where do I go?
On RE, you would make a circle around the south pole, gps, gyrocompass would all match, it is HIGHLY RELIABLE. You would track the latitude line you were on. Without going there, I can diagram and explain it, and that is all I am asking for from you. Even if it doesn't exist, the geometry still is consistent with itself.
Please lay out on a FE map what happens when a plane takes off from Tierra Del Fuego and keeps the southern cross 90 degrees off one side for 8000 miles. Do you have a HIGHLY RELIABLE map on which you can plot a HIGHLY RELIABLE course?
Please no picking nits on the question or denial because no personal experience. I am asking you for a consistent model of how it could be, not to prove it in person. The question is clear and simple, either you have the geometry or you don't, but somehow, I expect your answer will not be either "here is a diagram" or "I have no idea, no reasonable answer". Answer, please, don't waste time with evasion techniques.
As soon as you learn to stick to one subject at a time, namely, will Kim Jong Un ever be able to fire an ICBM?
LOL!
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Kim Jong Un has fired several ICBMs.
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Kim Jong Un has fired several ICBMs.
ICBM's? Really?
Errr...no, he hasn't.
I suggest you reacquaint yourself with a dictionary and post haste.
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Google "Hwasong 14 test July 4 2017". Then come back and tell the class all about it.
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Action80:
Do ICBMs exist?
Do they hit what they aim at?
What map do they use to aim them, FE map (????) or RE globe coordinates?
My point was that an ICBM test is a test of FE/RE. If you fire an ICBM and carefully track where it hits, that track will match either FE or RE. If they know how to accurately aim, they know the actual shape of the earth.
If the earth is flat, and they know how to aim, they have a FE map. Either they know the true shape of the earth, or they don't know how to target, whether FE or RE.
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Not sure he's still here. He went to Google some stuff on Friday and, well, you know what a distraction the internet is.
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Google "Hwasong 14 test July 4 2017". Then come back and tell the class all about it.
Okay.
Reports (if they can be believed) indicate the missile may have traveled 700 miles.
Sorry, not even close to an ICBM.
I regret to inform the class you failed.
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From Wikipedia:
First test flight
The first publicly announced flight test was on 4 July 2017, to coincide with the US Independence Day celebrations. This flight had a claimed range of 933 kilometres (580 mi) eastwards into the Sea of Japan (East Sea of Korea) and reached an altitude of 2,802 kilometres (9,193,000 ft) during a 39-minute flight.[29]
This range was deliberately shortened, to avoid encroaching on other nations' territory, by 'lofting' the missile: firing it on a trajectory that was inefficiently high, rather than optimised for range. This allows the missile's performance to be tested and demonstrated, without requiring a huge test range.[29]
A prediction for the possible range, following an optimum trajectory, has been given at 6,700 kilometres (4,200 mi)[30] or as much as 10,400 kilometres (6,500 mi) not taking into account the Earth’s rotation. If true, then this brings the U.S. states of Alaska and Hawaii within the missile's range.[29]
Second test flight
Preparations for a second test flight were detected by US intelligence as early as 20 July.[28] On 28 July, the missile was fired at 11:41 p.m local time, the first time which a night time launch was carried out.[31][32] The missile was fired at a lofted trajectory with apogee of 3,700 km (2,300 mi), landing 998 km (620 mi) away with a total flight time of approximately 47 minutes. Based on the data from the test flight, if the missile were fired at the optimal efficient trajectory, it is predicted that the maximum effective range would exceed 10,000 km (6,200 mi). If factoring in the rotation of the Earth, which may provide a range boost when travelling eastward, the Hwasong-14’s coverage area would include the US West Coast, Chicago, and possibly even New York,[16] but only with a substantially reduced payload.[33]
In other words, the missiles were deliberately launched in an unusually high trajectory in order to reduce the effective range, in order to not encroach on neighboring territories, and to keep the missile within DPRK's telemetry range. Do you understand the part where it says it could reach Alaska, Hawaii, and the continental US West Coast? And New York with a reduced payload? Does this count as an ICBM in your dictionary?
The "claims" were made by DPRK's own KCNA news agency, and confirmed by US, Japanese and ROK trackers.
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From Wikipedia:
First test flight
The first publicly announced flight test was on 4 July 2017, to coincide with the US Independence Day celebrations. This flight had a claimed range of 933 kilometres (580 mi) eastwards into the Sea of Japan (East Sea of Korea) and reached an altitude of 2,802 kilometres (9,193,000 ft) during a 39-minute flight.[29]
This range was deliberately shortened, to avoid encroaching on other nations' territory, by 'lofting' the missile: firing it on a trajectory that was inefficiently high, rather than optimised for range. This allows the missile's performance to be tested and demonstrated, without requiring a huge test range.[29]
A prediction for the possible range, following an optimum trajectory, has been given at 6,700 kilometres (4,200 mi)[30] or as much as 10,400 kilometres (6,500 mi) not taking into account the Earth’s rotation. If true, then this brings the U.S. states of Alaska and Hawaii within the missile's range.[29]
Second test flight
Preparations for a second test flight were detected by US intelligence as early as 20 July.[28] On 28 July, the missile was fired at 11:41 p.m local time, the first time which a night time launch was carried out.[31][32] The missile was fired at a lofted trajectory with apogee of 3,700 km (2,300 mi), landing 998 km (620 mi) away with a total flight time of approximately 47 minutes. Based on the data from the test flight, if the missile were fired at the optimal efficient trajectory, it is predicted that the maximum effective range would exceed 10,000 km (6,200 mi). If factoring in the rotation of the Earth, which may provide a range boost when travelling eastward, the Hwasong-14’s coverage area would include the US West Coast, Chicago, and possibly even New York,[16] but only with a substantially reduced payload.[33]
In other words, the missiles were deliberately launched in an unusually high trajectory in order to reduce the effective range, in order to not encroach on neighboring territories, and to keep the missile within DPRK's telemetry range. Do you understand the part where it says it could reach Alaska, Hawaii, and the continental US West Coast? And New York with a reduced payload? Does this count as an ICBM in your dictionary?
The "claims" were made by DPRK's own KCNA news agency, and confirmed by US, Japanese and ROK trackers.
No, it does not qualify as an ICBM.
Like I wrote earlier, you and the rest wish to cosign for proven liars.
Simply contribute to the fear-mongering, terroristic warlords who lie on a daily basis in order to justify a perpetual state of war.
People who perpetuate lies in order to maintain control should be ashamed.
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So Japan, USA, ROK and North Korea agreed to tell the same lie? Love to be a fly on the wall at that meeting.
Seriously, if all references we make are "liars", we can probably agree that any further discussion of any topic is pointless. Bye.
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So Japan, USA, ROK and North Korea agreed to tell the same lie? Love to be a fly on the wall at that meeting.
Seriously, if all references we make are "liars", we can probably agree that any further discussion of any topic is pointless. Bye.
Yeah, you are right. World governments have never actually had any meetings in order to push any plans involving any sort of subterfuge and duplicity on the world populace.
https://theintercept.com/2018/02/06/lie-after-lie-what-colin-powell-knew-about-iraq-fifteen-years-ago-and-what-he-told-the-un/
https://www.independent.co.uk/news/uk/politics/chilcot-report-iraq-war-inquiry-tony-blair-george-bush-us-uk-what-happened-a7119761.html
-
It is generally accepted that Russia did fake a lot of stuff during the Cold War to pretend that they had more technological prowess than they did. During the Cold War, Russia went around parading fake ICBMs (https://apnews.com/article/1108a17523085ef3a7df9a7eea4671e5) for decades:
Moscow paraded dummy missiles
MANY OF the huge strategic missiles displayed in Red Square parades during the Soviet era were only dummies, but they scared the West into an expensive response, a Russian magazine reported yesterday.
One such fake, GR-1, an acronym for Global Missile, showed during a parade in 1965, prompted the United States to build an anti- missile defence system worth billions of dollars, said the weekly Vlast (Power). In fact, the Soviets had abandoned the GR-1 project long before the parade.
Another two mobile ballistic missiles shown in the same parade were also fakes, their test launches having been a failure, the magazine said. "Foreign military attaches were scared to death, triggering panic in Nato headquarters," it said. "A huge international uproar followed, and only those who prepared this demonstration knew they were dummies." One of the authors of the Vlast report worked as a missile engineer and said he had worked on a support system for one of the fake missiles to prevent it from bouncing on the stone-paved Red Square in Moscow. The magazine said the Soviet leader Nikita Krushchev first bluffed the West with the legend of powerful Russian missiles, saying the Soviet Union was making them "like sausage". "Such comparison sounded ambiguous for the Soviet people, because the sausage was in deficit, but it duly impressed foreigners," it said. At the time of Krushchev's comment, the Soviets had only four intercontinental ballistic missiles on duty, while the United States had 60. "The myth about the Soviet missile superiority was convenient for both the Soviet leadership and the American military industrial complex, which was getting huge contracts," the magazine said.
There is also a book about Russian Cold War fakery; Russia and the Big Red Lie.
-
As provided by Wikipedia:
"...If factoring in the rotation of the Earth, which may provide a range boost when travelling eastward, the Hwasong-14’s coverage area would include the US West Coast, Chicago, and possibly even New York,[16] but only with a substantially reduced payload.[33]
I found this part to be especially rich in asshattedness.
The fine globularists wish us to believe the rotation of the earth would not affect the positioning of a helicopter that is choosing to hover over a specific spot on the ground, but will impact the landing site of a missile that is above the earth in the same air.
What a bunch of freaking liars!
You guys really need to get a grip on all the BS.
-
A helicopter is in the atmospheric boundary layer of Earth. An ICBM is in space.
-
A helicopter is in the atmospheric boundary layer of Earth. An ICBM is in space.
LOL!
The moon is in space too, in case you forgot.
You do not even know what an ICBM is and are going to presume to now lecture about where things move and how/why the mythical G is going to affect things?
I don't think so.
-
It is generally accepted that Russia did fake a lot of stuff during the Cold War to pretend that they had more technological prowess than they did. During the Cold War, Russia went around parading fake ICBMs (https://apnews.com/article/1108a17523085ef3a7df9a7eea4671e5) for decades:
Moscow paraded dummy missiles
MANY OF the huge strategic missiles displayed in Red Square parades during the Soviet era were only dummies, but they scared the West into an expensive response, a Russian magazine reported yesterday.
One such fake, GR-1, an acronym for Global Missile, showed during a parade in 1965, prompted the United States to build an anti- missile defence system worth billions of dollars, said the weekly Vlast (Power). In fact, the Soviets had abandoned the GR-1 project long before the parade.
Another two mobile ballistic missiles shown in the same parade were also fakes, their test launches having been a failure, the magazine said. "Foreign military attaches were scared to death, triggering panic in Nato headquarters," it said. "A huge international uproar followed, and only those who prepared this demonstration knew they were dummies." One of the authors of the Vlast report worked as a missile engineer and said he had worked on a support system for one of the fake missiles to prevent it from bouncing on the stone-paved Red Square in Moscow. The magazine said the Soviet leader Nikita Krushchev first bluffed the West with the legend of powerful Russian missiles, saying the Soviet Union was making them "like sausage". "Such comparison sounded ambiguous for the Soviet people, because the sausage was in deficit, but it duly impressed foreigners," it said. At the time of Krushchev's comment, the Soviets had only four intercontinental ballistic missiles on duty, while the United States had 60. "The myth about the Soviet missile superiority was convenient for both the Soviet leadership and the American military industrial complex, which was getting huge contracts," the magazine said.
There is also a book about Russian Cold War fakery; Russia and the Big Red Lie.
So ICBMs exist. Can you confirm to @Action80 please.
-
A helicopter is in the atmospheric boundary layer of Earth. An ICBM is in space.
LOL!
The moon is in space too, in case you forgot.
And?
-
It is generally accepted that Russia did fake a lot of stuff during the Cold War to pretend that they had more technological prowess than they did. During the Cold War, Russia went around parading fake ICBMs (https://apnews.com/article/1108a17523085ef3a7df9a7eea4671e5) for decades:
Moscow paraded dummy missiles
MANY OF the huge strategic missiles displayed in Red Square parades during the Soviet era were only dummies, but they scared the West into an expensive response, a Russian magazine reported yesterday.
One such fake, GR-1, an acronym for Global Missile, showed during a parade in 1965, prompted the United States to build an anti- missile defence system worth billions of dollars, said the weekly Vlast (Power). In fact, the Soviets had abandoned the GR-1 project long before the parade.
Another two mobile ballistic missiles shown in the same parade were also fakes, their test launches having been a failure, the magazine said. "Foreign military attaches were scared to death, triggering panic in Nato headquarters," it said. "A huge international uproar followed, and only those who prepared this demonstration knew they were dummies." One of the authors of the Vlast report worked as a missile engineer and said he had worked on a support system for one of the fake missiles to prevent it from bouncing on the stone-paved Red Square in Moscow. The magazine said the Soviet leader Nikita Krushchev first bluffed the West with the legend of powerful Russian missiles, saying the Soviet Union was making them "like sausage". "Such comparison sounded ambiguous for the Soviet people, because the sausage was in deficit, but it duly impressed foreigners," it said. At the time of Krushchev's comment, the Soviets had only four intercontinental ballistic missiles on duty, while the United States had 60. "The myth about the Soviet missile superiority was convenient for both the Soviet leadership and the American military industrial complex, which was getting huge contracts," the magazine said.
There is also a book about Russian Cold War fakery; Russia and the Big Red Lie.
So ICBMs exist. Can you confirm to @Action80 please.
I think the larger point Tom was making is that ICBM's do not actually exist.
Pity that went over your head or through your ears.
-
A helicopter is in the atmospheric boundary layer of Earth. An ICBM is in space.
LOL!
The moon is in space too, in case you forgot.
And?
And what?
According to you fine globularists, the moon is tidally locked.
Due to gravity.
Just give it up already.
According to your mythology, you wouldn't need to program New York as a target.
Just plot a distance say 300 miles short and let the earth do the rest.
Just sheer stupidity that allows for claims like that.
Not to mention, you are just demonstrating "I believe, I believe, I believe," on a scale not exhibited since Natalie Wood in a Miracle on 34th Street.
-
A helicopter is in the atmospheric boundary layer of Earth. An ICBM is in space.
LOL!
The moon is in space too, in case you forgot.
You do not even know what an ICBM is and are going to presume to now lecture about where things move and how/why the mythical G is going to affect things?
I don't think so.
ICBM;
M = Missile; a projectile.
B = Ballistic; not reliant on aerodynamics for its trajectory.
IC = Inter Continental; having a range typical of the distances between continents.
The same sources which told you that the initial test of Hwasong-14 had a range of 700 miles, and you trust, also say it reached an altitude of 1750 miles. Apparently, this gives it the range to hit Alaska and Hawaii. (I'm not the rocket scientist here, but you have the quadratic equations, so you do the math). Note that this was the very first test of the Hwasong 14, subsequent development and tests have improved the performance.
And you will do the math for us?
-
It is generally accepted that Russia did fake a lot of stuff during the Cold War to pretend that they had more technological prowess than they did. During the Cold War, Russia went around parading fake ICBMs (https://apnews.com/article/1108a17523085ef3a7df9a7eea4671e5) for decades:
Moscow paraded dummy missiles
MANY OF the huge strategic missiles displayed in Red Square parades during the Soviet era were only dummies, but they scared the West into an expensive response, a Russian magazine reported yesterday.
One such fake, GR-1, an acronym for Global Missile, showed during a parade in 1965, prompted the United States to build an anti- missile defence system worth billions of dollars, said the weekly Vlast (Power). In fact, the Soviets had abandoned the GR-1 project long before the parade.
Another two mobile ballistic missiles shown in the same parade were also fakes, their test launches having been a failure, the magazine said. "Foreign military attaches were scared to death, triggering panic in Nato headquarters," it said. "A huge international uproar followed, and only those who prepared this demonstration knew they were dummies." One of the authors of the Vlast report worked as a missile engineer and said he had worked on a support system for one of the fake missiles to prevent it from bouncing on the stone-paved Red Square in Moscow. The magazine said the Soviet leader Nikita Krushchev first bluffed the West with the legend of powerful Russian missiles, saying the Soviet Union was making them "like sausage". "Such comparison sounded ambiguous for the Soviet people, because the sausage was in deficit, but it duly impressed foreigners," it said. At the time of Krushchev's comment, the Soviets had only four intercontinental ballistic missiles on duty, while the United States had 60. "The myth about the Soviet missile superiority was convenient for both the Soviet leadership and the American military industrial complex, which was getting huge contracts," the magazine said.
There is also a book about Russian Cold War fakery; Russia and the Big Red Lie.
So ICBMs exist. Can you confirm to @Action80 please.
I think the larger point Tom was making is that ICBM's do not actually exist.
Pity that went over your head or through your ears.
Many. And the others?
-
A helicopter is in the atmospheric boundary layer of Earth. An ICBM is in space.
LOL!
The moon is in space too, in case you forgot.
You do not even know what an ICBM is and are going to presume to now lecture about where things move and how/why the mythical G is going to affect things?
I don't think so.
ICBM;
M = Missile; a projectile.
B = Ballistic; not reliant on aerodynamics for its trajectory.
IC = Inter Continental; having a range typical of the distances between continents.
The same sources which told you that the initial test of Hwasong-14 had a range of 700 miles, and you trust, also say it reached an altitude of 1750 miles. Apparently, this gives it the range to hit Alaska and Hawaii. (I'm not the rocket scientist here, but you have the quadratic equations, so you do the math). Note that this was the very first test of the Hwasong 14, subsequent development and tests have improved the performance.
And you will do the math for us?
Who said I trust the BS posted in that article?
I don't trust that article one bit.
At all.
Just like I do not trust you at all.
After all, you stated:
Kim Jong Un has fired several ICBMs.
So, according to you "several," constitutes one.
And that "one," didn't even travel 1000 miles (according to sources), let alone 1000's of miles.
See, when you are incapable of understanding words like, "several," and just willy-nilly toss out figures not matching your rhetoric, it becomes hard to accept.
-
It is generally accepted that Russia did fake a lot of stuff during the Cold War to pretend that they had more technological prowess than they did. During the Cold War, Russia went around parading fake ICBMs (https://apnews.com/article/1108a17523085ef3a7df9a7eea4671e5) for decades:
Moscow paraded dummy missiles
MANY OF the huge strategic missiles displayed in Red Square parades during the Soviet era were only dummies, but they scared the West into an expensive response, a Russian magazine reported yesterday.
One such fake, GR-1, an acronym for Global Missile, showed during a parade in 1965, prompted the United States to build an anti- missile defence system worth billions of dollars, said the weekly Vlast (Power). In fact, the Soviets had abandoned the GR-1 project long before the parade.
Another two mobile ballistic missiles shown in the same parade were also fakes, their test launches having been a failure, the magazine said. "Foreign military attaches were scared to death, triggering panic in Nato headquarters," it said. "A huge international uproar followed, and only those who prepared this demonstration knew they were dummies." One of the authors of the Vlast report worked as a missile engineer and said he had worked on a support system for one of the fake missiles to prevent it from bouncing on the stone-paved Red Square in Moscow. The magazine said the Soviet leader Nikita Krushchev first bluffed the West with the legend of powerful Russian missiles, saying the Soviet Union was making them "like sausage". "Such comparison sounded ambiguous for the Soviet people, because the sausage was in deficit, but it duly impressed foreigners," it said. At the time of Krushchev's comment, the Soviets had only four intercontinental ballistic missiles on duty, while the United States had 60. "The myth about the Soviet missile superiority was convenient for both the Soviet leadership and the American military industrial complex, which was getting huge contracts," the magazine said.
There is also a book about Russian Cold War fakery; Russia and the Big Red Lie.
So ICBMs exist. Can you confirm to @Action80 please.
I think the larger point Tom was making is that ICBM's do not actually exist.
Pity that went over your head or through your ears.
Many. And the others?
The others did not exist at all.
When you claim you have something, you cannot parade out nothing.
I mean, look at your replies to this post.
You are claiming you have something, but you are trotting out absolutely nothing.
Not working out too well, is it?
-
Tom made his point that ICBMs don’t exist by citing a source that said they do exist. Seems like Tom failed.
Total Lackey, meanwhile has nothing to present other than his incredulity. Real top notch work.
-
In the article they only have sources with testimonials of the fake missiles.
What a desperate position you guys are now in to have argue that known liars are truthful sometimes. ::)
-
In the article they only have sources with testimonials of the fake missiles.
What a desperate position you guys are now in to have argue that known liars are truthful sometimes. ::)
Odd projection from the guy citing sources that disprove his own claim.
-
In the article they only have sources with testimonials of the fake missiles.
What a desperate position you guys are now in to have argue that known liars are truthful sometimes. ::)
From the same AP article you cited:
"It wasn’t until 1970 that the Soviet Union reached parity with the United States in land-based intercontinental ballistic missiles, and the overall nuclear balance was attained only shortly before the 1991 Soviet collapse, Vlast said."
Seems like your statement, "In the article they only have sources with testimonials of the fake missiles." is not truthful nor accurate. That's what happens when you stop reading after you found what you thought to be a juicy cherry-picked quote.
-
In the article they only have sources with testimonials of the fake missiles.
What a desperate position you guys are now in to have argue that known liars are truthful sometimes. ::)
If ICBMs were real, what evidence would it take for you to believe it?
Would you have to actually witness a launch? Or would you need a warhead to drop on your city?
-
In the article they only have sources with testimonials of the fake missiles.
What a desperate position you guys are now in to have argue that known liars are truthful sometimes. ::)
From the same AP article you cited:
"It wasn’t until 1970 that the Soviet Union reached parity with the United States in land-based intercontinental ballistic missiles, and the overall nuclear balance was attained only shortly before the 1991 Soviet collapse, Vlast said."
Seems like your statement, "In the article they only have sources with testimonials of the fake missiles." is not truthful nor accurate. That's what happens when you stop reading after you found what you thought to be a juicy cherry-picked quote.
Ah, so you really have nothing. You are grasping at straws to justify a country who would need to lie about their ICBMs. That's not a testimonial of someone's personal knowledge of real or fake missiles. That's a brief statement of general Cold War history. The references of personal knowledge in that link says that the missiles are fake.
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They said paraded missiles were fake. You haven’t come close to showing that all the ICBMs are fake. In fact your own source disagreed with you. You are master of self-owning.
-
I didn't say that the article proved that all the ICBMs were fake. What I said about the article was as follows:
It is generally accepted that Russia did fake a lot of stuff during the Cold War to pretend that they had more technological prowess than they did. During the Cold War, Russia went around parading fake ICBMs (https://apnews.com/article/1108a17523085ef3a7df9a7eea4671e5) for decades:
You didn't even disagree with this assessment, and have accepted it, desperately moving the goal post now to demanding that the article proves that all the ICBMs are fake.
-
If that’s all your contending then you’ve waited awfully long to clarify. If that is the case, why even post it? Faking a parade float seems woefully off-topic.
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They said paraded missiles were fake. You haven’t come close to showing that all the ICBMs are fake. In fact your own source disagreed with you. You are master of self-owning.
I think the real issue is no one here has come close to demonstrating ICBM's are real, including you.
Until then, it is just a myth propagated by liars.
You have nothing but a bunch of words from known liars, which you love to repeat.
The OP needs to go to CN.
-
They said paraded missiles were fake. You haven’t come close to showing that all the ICBMs are fake. In fact your own source disagreed with you. You are master of self-owning.
I think the real issue is no one here has come close to demonstrating ICBM's are real, including you.
Until then, it is just a myth propagated by liars.
“No one at tfes.org has come close to demonstrating ICBMs are real. Until tfes.org demonstrates they are real they are a myth.”
Top notch epistemology, Total Lackey.
You have nothing but a bunch of words from known liars, which you love to repeat.
The OP needs to go to CN.
Nice opinion.
-
They said paraded missiles were fake. You haven’t come close to showing that all the ICBMs are fake. In fact your own source disagreed with you. You are master of self-owning.
I think the real issue is no one here has come close to demonstrating ICBM's are real, including you.
Until then, it is just a myth propagated by liars.
You have nothing but a bunch of words from known liars, which you love to repeat.
The OP needs to go to CN.
What about Tomahawk missiles? Are they real? Are mortars real?
At what level of technology does the weapon become fake?
-
They said paraded missiles were fake. You haven’t come close to showing that all the ICBMs are fake. In fact your own source disagreed with you. You are master of self-owning.
I think the real issue is no one here has come close to demonstrating ICBM's are real, including you.
Until then, it is just a myth propagated by liars.
“No one at tfes.org has come close to demonstrating ICBMs are real. Until tfes.org demonstrates they are real they are a myth.”
Top notch epistemology.
Actually, no one, anywhere, (that is equivalent to ZERO, by the way, in case you have learned how to count) has demonstrated ICBM's are real.
So, it is not tfes.org that has the entire burden.
But go ahead, you try.
Much like unicorns.
You have nothing but a bunch of words from known liars, which you love to repeat.
The OP needs to go to CN.
Nice opinion.
When subjects are broached that are deemed to have no factual basis or supporting evidence, those subjects belong in CN.
In this case, everything I have written is true.
Not one person has ever seen a purposefully used ICBM.
Not
One
Person
It is only alleged that ICBM's exist and those allegations comes directly from persons who have a less than stellar record of honesty.
-
They said paraded missiles were fake. You haven’t come close to showing that all the ICBMs are fake. In fact your own source disagreed with you. You are master of self-owning.
I think the real issue is no one here has come close to demonstrating ICBM's are real, including you.
Until then, it is just a myth propagated by liars.
You have nothing but a bunch of words from known liars, which you love to repeat.
The OP needs to go to CN.
What about Tomahawk missiles? Are they real? Are mortars real?
At what level of technology does the weapon become fake?
When you cannot produce evidence of use, I would suppose.
I have seen Tomahawks in use, hitting their intended targets.
I have also seen mortars in use, hitting their intended targets.
-
Actually, no one, anywhere, (that is equivalent to ZERO, by the way, in case you have learned how to count) has demonstrated ICBM's are real.
Source?
So, it is not tfes.org that has the entire burden.
But go ahead, you try.
Much like unicorns.
Unlike unicorns, there are videos of ICBMs being launched, but don't let that get in the way of a comforting narrative.
When subjects are broached that are deemed to have no factual basis or supporting evidence, those subjects belong in CN.
Fortunately, your horrendous criteria for what is true and isn't, is not used.
In this case, everything I have written is true.
Not one person has ever seen a purposefully used ICBM.
Not
One
Person
It is only alleged that ICBM's exist and those allegations comes directly from persons who have a less than stellar record of honesty.
Good thing purposely using an ICBM isn't the ultimate arbiter of their reality.
-
Actually, no one, anywhere, (that is equivalent to ZERO, by the way, in case you have learned how to count) has demonstrated ICBM's are real.
Source?
Again, you seem to be struggling.
Those alleging ICBM's are real are responsible for providing a source demonstrating ICBM's are real.
So, it is not tfes.org that has the entire burden.
But go ahead, you try.
Much like unicorns.
Unlike unicorns, there are videos of ICBMs being launched, but don't let that get in the way of a comforting narrative.
Like this, uh.
https://www.youtube.com/watch?v=9UgIXKeYimU
When subjects are broached that are deemed to have no factual basis or supporting evidence, those subjects belong in CN.
Fortunately, your horrendous criteria for what is true and isn't, is not used.
Those are not my criteria.
In this case, everything I have written is true.
Not one person has ever seen a purposefully used ICBM.
Not
One
Person
It is only alleged that ICBM's exist and those allegations comes directly from persons who have a less than stellar record of honesty.
Good thing purposely using an ICBM isn't the ultimate arbiter of their reality.
It is the only arbiter.
Until then, the best you can write is that you like to subscribe to allegations put forth by known purveyors of bullshit.
Not surprising in the least to those familiar with your MO, I am sure.
-
They said paraded missiles were fake. You haven’t come close to showing that all the ICBMs are fake. In fact your own source disagreed with you. You are master of self-owning.
I think the real issue is no one here has come close to demonstrating ICBM's are real, including you.
Until then, it is just a myth propagated by liars.
You have nothing but a bunch of words from known liars, which you love to repeat.
The OP needs to go to CN.
What about Tomahawk missiles? Are they real? Are mortars real?
At what level of technology does the weapon become fake?
When you cannot produce evidence of use, I would suppose.
I have seen Tomahawks in use, hitting their intended targets.
I have also seen mortars in use, hitting their intended targets.
I am perhaps fortunate never to have seen an actual mortar fired, or a Tomahawk hit its target. Do you have a source for their existence, or is it just your own testimony?
-
They said paraded missiles were fake. You haven’t come close to showing that all the ICBMs are fake. In fact your own source disagreed with you. You are master of self-owning.
I think the real issue is no one here has come close to demonstrating ICBM's are real, including you.
Until then, it is just a myth propagated by liars.
You have nothing but a bunch of words from known liars, which you love to repeat.
The OP needs to go to CN.
What about Tomahawk missiles? Are they real? Are mortars real?
At what level of technology does the weapon become fake?
When you cannot produce evidence of use, I would suppose.
I have seen Tomahawks in use, hitting their intended targets.
I have also seen mortars in use, hitting their intended targets.
I am perhaps fortunate never to have seen an actual mortar fired, or a Tomahawk hit its target. Do you have a source for their existence, or is it just your own testimony?
Just my own testimony.
-
It has nothing to do with what kind of points you offer up and you know damn well it doesn't.
Despite your objections, you know damn well it has to do with no defined 0/0.
Without a defined 0/0, the actual distance of each grid set forth by intersecting lines of lat/long, or x/y, or whatever you choose to call them, cannot be officially defined, and hence why you all struggle so mightily with issues of distance when it comes to interpreting whole world maps.
Further, when it comes to mapping specific areas, such as what you have offered up, that is a depicted as FLAT on a flat piece of paper.
Which FE map are you backing, and what are its dimensions?
Why do you refuse to answer such a basic question?
Action80 - still waiting for you to explain which FE map you think best represents the earth, and to give some indication of its dimensions.
Tom - still waiting for you to help Action80 out with the distance-between-two-lat/longs maths challenge.
-
When you cannot produce evidence of use, I would suppose.
I have seen Tomahawks in use, hitting their intended targets.
I have also seen mortars in use, hitting their intended targets.
So if you hadn't actually seen a tomahawk or a mortar, you wouldn't believe in those either yet they would exist. How much of reality are you missing?
-
When you cannot produce evidence of use, I would suppose.
I have seen Tomahawks in use, hitting their intended targets.
I have also seen mortars in use, hitting their intended targets.
Just my own testimony.
Why should we believe your testimony? To anyone else but you it is just unverified second hand drivel. The exact thing you have completely dismissed and ignored this entire thread and deemed as proof of nothing.
RonJ gave his own personal testimony of witnessing the test of a North Korean ICBM yet you have completely dismissed it and discounted it.
I'll give out a clue. Yes 'rocket man' can launch ICBM's. I did a little checking in my personal records that did indicate that I was aboard a ship that did transit the Sea of Japan between North Korea and Japan during the launch of Hwasong 14. That voyage was a little unique in that we had alarms going off and incoming traffic telling us that North Korea had launched a missile. This was a warning to us because what goes up has to come down. That missile was spotted on a trajectory that did pass over our ship and did come back down and landed in the sea near Japan. We were quite a distance from where the missile landed but what if something went wrong on the assent and we got hit when the missile finally landed? It was a small chance but it's something to be concerned about. So again here are some observations from first hand personal experience so I didn't have to spend any time doing Google searches.
What makes your testimony so special? Short answer is it isn't. Yours is no more proof of something as anyone else's.
In the end, Tom Bishop can sum up your entire line of debate in this thread way better than I can:
Lol "Everyone is Wrong and LiEeInG"
That is a desperate argument from a losing position. An argument from a position of strength would have positive evidence for that position.
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When you cannot produce evidence of use, I would suppose.
I have seen Tomahawks in use, hitting their intended targets.
I have also seen mortars in use, hitting their intended targets.
So if you hadn't actually seen a tomahawk or a mortar, you wouldn't believe in those either yet they would exist. How much of reality are you missing?
Perhaps you have an issue with the word "verification."
I certainly do not.
Regardless, if I am able to see something live and in living color, I am certainly in a better position to come to a personal opinion as to the existence of that something, whatever that something is.
So, in regard to your 1st statement, that is just a bunch of crap.
In regard to your 2nd statement, I am missing no more of the real existence about me and would feel rather confident stating I am experiencing more reality than you are, given your proven predilection to believe proven propaganda artists.
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When you cannot produce evidence of use, I would suppose.
I have seen Tomahawks in use, hitting their intended targets.
I have also seen mortars in use, hitting their intended targets.
Just my own testimony.
Why should we believe your testimony? To anyone else but you it is just unverified second hand drivel. The exact thing you have completely dismissed and ignored this entire thread and deemed as proof of nothing.
I never asked you to believe anything.
The MSM asks you to believe and you show your fealty and undying support for everything they put out, despite numerous proven episodes of outright lies and BS spewed on a daily basis.
RonJ gave his own personal testimony of witnessing the test of a North Korean ICBM yet you have completely dismissed it and discounted it.
What makes your testimony so special? Short answer is it isn't. Yours is no more proof of something as anyone else's.
This entire section of your post is just pure BS.
I never stated his/her testimony was bogus.
Quit lying.
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I never stated his/her testimony was bogus.
Quit lying.
You most certainly did. After being given a firsthand account, you continued with your narrative of:
Lol "Everyone is Wrong and LiEeInG"
with respect to any information regarding whether or not ICBMs exist. That's a direct assertion that RonJ's testimony was bogus.
Bye now.
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I never stated his/her testimony was bogus.
Quit lying.
You most certainly did. After being given a firsthand account, you continued with your narrative of:
Lol "Everyone is Wrong and LiEeInG"
with respect to any information regarding whether or not ICBMs exist. That's a direct assertion that RonJ's testimony was bogus.
One
more
time.
When you can come up with something that actually proves an ICBM exists, I will gladly recant.
I never saw a post from RonJ concerning the existence/use of a North Korean ICBM and if you can find a direct reply from me to RonJ concerning the existence/use of a North Korean ICBM then I will recant again.
Until then, I will leave it to your obviously challenged intellect to actually determine what the acronym "ICBM" stands for.
Bye now.
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I never stated his/her testimony was bogus.
Quit lying.
You most certainly did. After being given a firsthand account, you continued with your narrative of:
Lol "Everyone is Wrong and LiEeInG"
with respect to any information regarding whether or not ICBMs exist. That's a direct assertion that RonJ's testimony was bogus.
One
more
time.
When you can come up with something that actually proves an ICBM exists, I will gladly recant.
I never saw a post from from RonJ concerning the existence/use of a North Korean ICBM and if you can find a direct reply from me to RonJ concerning the existence/use of a North Korean ICBM then I will recant again.
Until then, I will leave it to your obviously challenged intellect to actually determine what the acronym "ICBM" stands for.
Bye now.
In an attempt to directly answer your question and provide a post, I will give the following first hand experiences:
1) While serving as a US Merchant Marine officer aboard a cargo ship transiting between the Korean Peninsula and Japan there were numerous alarms and warnings going off on the bridge of our ship.
2) Several broadcast messages to all ships at sea were received from the Japanese Coast Guard authorities regarding the immanent and later the actual launch of a North Korean ICBM.
3) Since the North Koreans don't tell everyone in advance what is happening all we can do is wait.
4) The missile launch in question was at night. Skies were clear and a missile was clearly seen traveling upwards then over the top (forward of the bow) of our ship.
5) We later got a message that the missile had landed in the sea far off the coast of Japan in the Pacific Ocean quite a distance from our ship.
We knew the danger was over at this point. Our voyage back to the USA from China was not interrupted and the North Korean ICBM test didn't cause us any difficulties.
This was my experience along with the experience of others who were performing their duties on the bridge of the ship when the event occurred.
No one is telling me what to say and/or when I could say it. This is all public record for those who wish to do their own research.
That was an interesting voyage from the start. The North Koreans were definitely saying that they could and might conduct a missile strike on Hawaii. We were also at the dock in Hono loading cargo at that time. Stress levels were a bit higher than normal. Ask anyone living in Hawaii about that period of time and they will be glad to tell you. I do believe that the tourist industry was hit some because why would someone want to be in Honolulu when 'rocket man' might get an itchy trigger finger?
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I never stated his/her testimony was bogus.
Quit lying.
You most certainly did. After being given a firsthand account, you continued with your narrative of:
Lol "Everyone is Wrong and LiEeInG"
with respect to any information regarding whether or not ICBMs exist. That's a direct assertion that RonJ's testimony was bogus.
One
more
time.
When you can come up with something that actually proves an ICBM exists, I will gladly recant.
I never saw a post from from RonJ concerning the existence/use of a North Korean ICBM and if you can find a direct reply from me to RonJ concerning the existence/use of a North Korean ICBM then I will recant again.
Until then, I will leave it to your obviously challenged intellect to actually determine what the acronym "ICBM" stands for.
Bye now.
In an attempt to directly answer your question and provide a post, I will give the following first hand experiences:
1) While serving as a US Merchant Marine officer aboard a cargo ship transiting between the Korean Peninsula and Japan there were numerous alarms and warnings going off on the bridge of our ship.
2) Several broadcast messages to all ships at sea were received from the Japanese Coast Guard authorities regarding the immanent and later the actual launch of a North Korean ICBM.
3) Since the North Koreans don't tell everyone in advance what is happening all we can do is wait.
4) The missile launch in question was at night. Skies were clear and a missile was clearly seen traveling upwards then over the top (forward of the bow) of our ship.
5) We later got a message that the missile had landed in the sea far off the coast of Japan in the Pacific Ocean quite a distance from our ship.
We knew the danger was over at this point. Our voyage back to the USA from China was not interrupted and the North Korean ICBM test didn't cause us any difficulties.
This was my experience along with the experience of others who were performing their duties on the bridge of the ship when the event occurred.
No one is telling me what to say and/or when I could say it. This is all public record for those who wish to do their own research.
That was an interesting voyage from the start. The North Koreans were definitely saying that they could and might conduct a missile strike on Hawaii. We were also at the dock in Hono loading cargo at that time. Stress levels were a bit higher than normal. Ask anyone living in Hawaii about that period of time and they will be glad to tell you. I do believe that the tourist industry was hit some because why would someone want to be in Honolulu when 'rocket man' might get an itchy trigger finger?
Interesting story.
I am sure your vessel and crew was on high alert and on the lookout.
So tell us what was it like to see a non-ICBM in action? Pretty cool, I bet.
I know I like to watch missiles being fired.
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Though its based on personal experience, looks like RonJ was able to refer to his log to refresh his memory and add a bit of gravitas to his experience. I don't know if you keep a diary @Action80, but could you provide more detail of your Tomahawk stuff, like where and when, and was it the sub-, surface- or land based version. And only if its not breaching any security protocols.
I'm still inclined to think that the whole Tomahawk thing may be a figment of the MIC pseudo-arms dealers and unicorn farmers.
Oh, and this is a nice touch: So tell us what was it like to see a non-ICBM in action? Pretty cool, I bet.
Now it looks like Ron is agreeing with you! Clever!
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Though its based on personal experience, looks like RonJ was able to refer to his log to refresh his memory and add a bit of gravitas to his experience. I don't know if you keep a diary @Action80, but could you provide more detail of your Tomahawk stuff, like where and when, and was it the sub-, surface- or land based version. And only if its not breaching any security protocols.
I'm still inclined to think that the whole Tomahawk thing may be a figment of the MIC pseudo-arms dealers and unicorn farmers.
Oh, and this is a nice touch: So tell us what was it like to see a non-ICBM in action? Pretty cool, I bet.
Now it looks like Ron is agreeing with you! Clever!
I cannot get into any particulars, except to state it was testing.
No touch to it.
He acknowledges a missile was fired, he acknowledges it landed, and he acknowledges what he witnessed did not constitute an ICBM.
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Please do your research first. Your non-ICBM statement might possibly be a bit incorrect!
If I buy a Lamborghini and only drive it to the grocery store and someone says "That's a sub-100 mph vehicle" I say "do some research".
North Korean 'rocket man' has to be a bit careful doing his tests. The exact capabilities of his missiles are not exactly public knowledge and nether are those of the missiles in USA's arsenal. It's safe to assume, from the known results of all the previous tests, that these missiles could reach the USA. That would make them an ICBM.
Who would have thought that the Japanese could launch a devastating attack on Hawaii in 1941. Do you blame Hawaiians for being cautious? Fool me once, shame on you. Fool me twice, shame on me!
-
I never stated his/her testimony was bogus.
Quit lying.
You most certainly did. After being given a firsthand account, you continued with your narrative of:
Lol "Everyone is Wrong and LiEeInG"
with respect to any information regarding whether or not ICBMs exist. That's a direct assertion that RonJ's testimony was bogus.
Bye now.
I believe we were talking about whether the Republican Legislature in the states were being nefarious about the audits. If it was found that they lied about audits, or audits in the past, that would be evidence against them. In that case we had Rama Set calling everything that disagreed with him to be wrong or a lie.
In this case we do have evidence that Russia has lied about ICBM technologies, with you summarily declaring that there is no evidence of them being nefarious about ICBMs. Quite the opposite of your portrait of the situation.
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I believe we were talking about whether the Republican Legislature in the states were being nefarious about the audits. If it was found that they lied about audits, or audits in the past, that would be evidence against them. In that case we had Rama Set calling everything that disagreed with him to be wrong or a lie.
What, exactly does this have to do anything?
In this case we do have evidence that Russia has lied about ICBM technologies, with you summarily declaring that there is no evidence of them being nefarious about ICBMs. Quite the opposite of your portrait of the situation.
Do you sell cherries? You sure do seem to do a lot of cherry picking.
You seemed to overlook this:
with respect to any information regarding whether or not ICBMs exist.
part of my post.
The nefarious activity was in effort to imply more power than may have actually been the case. The fact that ICBM technology existed wasn't in question.
-
Please do your research first. Your non-ICBM statement might possibly be a bit incorrect!
If I buy a Lamborghini and only drive it to the grocery store and someone says "That's a sub-100 mph vehicle" I say "do some research".
North Korean 'rocket man' has to be a bit careful doing his tests. The exact capabilities of his missiles are not exactly public knowledge and nether are those of the missiles in USA's arsenal. It's safe to assume, from the known results of all the previous tests, that these missiles could reach the USA. That would make them an ICBM.
Who would have thought that the Japanese could launch a devastating attack on Hawaii in 1941. Do you blame Hawaiians for being cautious? Fool me once, shame on you. Fool me twice, shame on me!
Actually, it is more safe to assume this renegade despot has no missile capable of reaching the US.
Two launches, neither coming close to defining an ICBM.
-
Total Lackey: Missiles totally work!
Also Total Lackey: Missiles don’t work!
-
Please do your research first. Your non-ICBM statement might possibly be a bit incorrect!
If I buy a Lamborghini and only drive it to the grocery store and someone says "That's a sub-100 mph vehicle" I say "do some research".
North Korean 'rocket man' has to be a bit careful doing his tests. The exact capabilities of his missiles are not exactly public knowledge and nether are those of the missiles in USA's arsenal. It's safe to assume, from the known results of all the previous tests, that these missiles could reach the USA. That would make them an ICBM.
Who would have thought that the Japanese could launch a devastating attack on Hawaii in 1941. Do you blame Hawaiians for being cautious? Fool me once, shame on you. Fool me twice, shame on me!
Actually, it is more safe to assume this renegade despot has no missile capable of reaching the US.
Two launches, neither coming close to defining an ICBM.
A wise man once showed me something about the word ASSUME. It makes an ASS out of U and ME. We try to assume nothing while at sea. King Neptune can and will come back and bite you, hard. It's much better to prepare for what the potential facts could be. You have no way of knowing what rocket man's hole card may actually be. You don't think the missiles that were actually tested were fully fueled do you? Perhaps you should just send an email saying that you are putting a bulls eye target in the middle of Jack London Square in Oakland, CA and say 'here you go, rocket man, here's your test target, give it your best shot'. That way we will both know if you have an ICBM or just a toy.
-
Please do your research first. Your non-ICBM statement might possibly be a bit incorrect!
If I buy a Lamborghini and only drive it to the grocery store and someone says "That's a sub-100 mph vehicle" I say "do some research".
North Korean 'rocket man' has to be a bit careful doing his tests. The exact capabilities of his missiles are not exactly public knowledge and nether are those of the missiles in USA's arsenal. It's safe to assume, from the known results of all the previous tests, that these missiles could reach the USA. That would make them an ICBM.
Who would have thought that the Japanese could launch a devastating attack on Hawaii in 1941. Do you blame Hawaiians for being cautious? Fool me once, shame on you. Fool me twice, shame on me!
Actually, it is more safe to assume this renegade despot has no missile capable of reaching the US.
Two launches, neither coming close to defining an ICBM.
A wise man once showed me something about the word ASSUME. It makes an ASS out of U and ME. We try to assume nothing while at sea. King Neptune can and will come back and bite you, hard. It's much better to prepare for what the potential facts could be. You have no way of knowing what rocket man's hole card may actually be. You don't think the missiles that were actually tested were fully fueled do you? Perhaps you should just send an email saying that you are putting a bulls eye target in the middle of Jack London Square in Oakland, CA and say 'here you go, rocket man, here's your test target, give it your best shot'. That way we will both know if you have an ICBM or just a toy.
The bottom line for all of this thread is this.
The only evidence that exists for ICBM's is some propaganda and gullible believers.
You know, the same type that confront Bible enthusiasts with words like, "How can you believe that garbage!?"
Guys like you and Rama.
-
The selfaware-wolf was hungry.
-
Please do your research first. Your non-ICBM statement might possibly be a bit incorrect!
If I buy a Lamborghini and only drive it to the grocery store and someone says "That's a sub-100 mph vehicle" I say "do some research".
North Korean 'rocket man' has to be a bit careful doing his tests. The exact capabilities of his missiles are not exactly public knowledge and nether are those of the missiles in USA's arsenal. It's safe to assume, from the known results of all the previous tests, that these missiles could reach the USA. That would make them an ICBM.
Who would have thought that the Japanese could launch a devastating attack on Hawaii in 1941. Do you blame Hawaiians for being cautious? Fool me once, shame on you. Fool me twice, shame on me!
Actually, it is more safe to assume this renegade despot has no missile capable of reaching the US.
Two launches, neither coming close to defining an ICBM.
A wise man once showed me something about the word ASSUME. It makes an ASS out of U and ME. We try to assume nothing while at sea. King Neptune can and will come back and bite you, hard. It's much better to prepare for what the potential facts could be. You have no way of knowing what rocket man's hole card may actually be. You don't think the missiles that were actually tested were fully fueled do you? Perhaps you should just send an email saying that you are putting a bulls eye target in the middle of Jack London Square in Oakland, CA and say 'here you go, rocket man, here's your test target, give it your best shot'. That way we will both know if you have an ICBM or just a toy.
The bottom line for all of this thread is this.
The only evidence that exists for ICBM's is some propaganda and gullible believers.
You know, the same type that confront Bible enthusiasts with words like, "How can you believe that garbage!?"
Guys like you and Rama.
I thought the bottom line of this thread was you and Tom studiously avoiding any commitment to any particular map of the FE, or any dimensions thereof. The ICBM discussion seems to be a useful distraction from that issue.
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It's like a guy comes up to you on the street. He says "give me some money, I have a gun in my pocket". You look down and see a bulge there and say "I don't believe you have a gun, I'm not gullible and it's just propaganda." Then the guy says "OK, I'll prove that I have a gun by shooting you between the eyes." What's the smart thing to do? On one side of your bet is your life and the other side is just some money. Do you feel lucky, Punk? Well do ya?
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I think its more fundamental than that Ron; before you left home, the guy placed an ad in the paper claiming to have a gun, and the local PD, the Sheriff's Dept and State cops also were on TV telling you he has a gun.
And the cops should know, because they also have guns.
-
I think it's more like so the guy pulls out the gun and fires it at the wall but since the bullet didn't hit you right between the eyes it really wasn't a bullet so the guy really doesn't have a gun like he claims because guns fire bullets.
-
Or maybe:
There's been a bunch of news stories about a sniper. The sniper has been doing some observed and documented quality tests on his rifle, scope, and bullets. There have been a few test shots and so far and all those have fallen way short of expectations. Some time passes. There's some articles in the newspapers and on TV saying that there's some rumors claiming that the sniper has been doing more research on guns & ammo lately and now has a new and improved system that is sure to be deadly and has a much longer range. Then someone steps up and makes the claim, "its all propaganda and you are all being gullible". That same person then says, "I'm so confident that I'm correct that I'm willing to have someone else stand there and be a test target". In the mean time I'll just stand behind this bulletproof wall and give the sniper the finger. When the sniper misses I get to say "told ya".
-
Please do your research first. Your non-ICBM statement might possibly be a bit incorrect!
If I buy a Lamborghini and only drive it to the grocery store and someone says "That's a sub-100 mph vehicle" I say "do some research".
North Korean 'rocket man' has to be a bit careful doing his tests. The exact capabilities of his missiles are not exactly public knowledge and nether are those of the missiles in USA's arsenal. It's safe to assume, from the known results of all the previous tests, that these missiles could reach the USA. That would make them an ICBM.
Who would have thought that the Japanese could launch a devastating attack on Hawaii in 1941. Do you blame Hawaiians for being cautious? Fool me once, shame on you. Fool me twice, shame on me!
Actually, it is more safe to assume this renegade despot has no missile capable of reaching the US.
Two launches, neither coming close to defining an ICBM.
A wise man once showed me something about the word ASSUME. It makes an ASS out of U and ME. We try to assume nothing while at sea. King Neptune can and will come back and bite you, hard. It's much better to prepare for what the potential facts could be. You have no way of knowing what rocket man's hole card may actually be. You don't think the missiles that were actually tested were fully fueled do you? Perhaps you should just send an email saying that you are putting a bulls eye target in the middle of Jack London Square in Oakland, CA and say 'here you go, rocket man, here's your test target, give it your best shot'. That way we will both know if you have an ICBM or just a toy.
The bottom line for all of this thread is this.
The only evidence that exists for ICBM's is some propaganda and gullible believers.
You know, the same type that confront Bible enthusiasts with words like, "How can you believe that garbage!?"
Guys like you and Rama.
I thought the bottom line of this thread was you and Tom studiously avoiding any commitment to any particular map of the FE, or any dimensions thereof. The ICBM discussion seems to be a useful distraction from that issue.
Having difficulty with the OP title there, Bob?
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Having difficulty with the OP title there, Bob?
I have to concur with A80. Bob, if you don't know what the topic of a discussion is, consider finding out before posting. Moreover, do not post mid-thread just to explain that you're lost. There are better places to do that.
As for other super-clever RE trolls in this thread: please don't force me to clean this up. You'll be grumpy when it happens. Get back on topic or get out of the thread.
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It's like a guy comes up to you on the street. He says "give me some money, I have a gun in my pocket". You look down and see a bulge there and say "I don't believe you have a gun, I'm not gullible and it's just propaganda." Then the guy says "OK, I'll prove that I have a gun by shooting you between the eyes." What's the smart thing to do? On one side of your bet is your life and the other side is just some money. Do you feel lucky, Punk? Well do ya?
I think its more fundamental than that Ron; before you left home, the guy placed an ad in the paper claiming to have a gun, and the local PD, the Sheriff's Dept and State cops also were on TV telling you he has a gun.
And the cops should know, because they also have guns.
I think it's more like so the guy pulls out the gun and fires it at the wall but since the bullet didn't hit you right between the eyes it really wasn't a bullet so the guy really doesn't have a gun like he claims because guns fire bullets.
Or maybe:
There's been a bunch of news stories about a sniper. The sniper has been doing some observed and documented quality tests on his rifle, scope, and bullets. There have been a few test shots and so far and all those have fallen way short of expectations. Some time passes. There's some articles in the newspapers and on TV saying that there's some rumors claiming that the sniper has been doing more research on guns & ammo lately and now has a new and improved system that is sure to be deadly and has a much longer range. Then someone steps up and makes the claim, "its all propaganda and you are all being gullible". That same person then says, "I'm so confident that I'm correct that I'm willing to have someone else stand there and be a test target". In the mean time I'll just stand behind this bulletproof wall and give the sniper the finger. When the sniper misses I get to say "told ya".
Superb strawmanning harmony fellas!
Tell us, which of you is responsible to see the conductor is recompensed after the nightly performance?
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Having difficulty with the OP title there, Bob?
Well, if we're not happy talking about the shape and size of the earth, let's go back to N Korea's missile test and do some maths, shall we?
You seem reasonably happy that the reported range of the NK missile test, 950km, is accurate. You also seem content that it was indeed a ballistic missile, meaning that after a short burn its trajectory was entirely ballistic, ie dictated by gravity (if you like that sort of thing) or UA (if you don't). I'm going to go out on a limb and also assume that you also agree with with the reported flight time of 53 minutes, although do please let me know if not.
The reason the experts are coming up with ICBM ranges for the missile, despite it flying a relatively short distance, is that if something flies ballistically for that long and only travels 950km, it has to have a very steep launch angle. Indeed, the reported apogee was very high - 4500km. If it was launched at a shallower angle, it would go much further. But you don't need me to tell you that - as you keep saying, the equations are very simple. Even if you dispense with heretical things like the reduction in g as you get away from the earth, if you calculate the speed necessary at 'launch' (more accurately, the end of burn, but let's keep it simple) to fly ballistically for something like 50 minutes (the reported burn time was around 5 minutes), and then try a shallower launch angle, you'll get much bigger ranges than 950km.
If you are genuinely intellectually curious, then do the maths and let us know how you get on.
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Having difficulty with the OP title there, Bob?
Well, if we're not happy talking about the shape and size of the earth, let's go back to N Korea's missile test and do some maths, shall we?
You seem reasonably happy that the reported range of the NK missile test, 950km, is accurate. You also seem content that it was indeed a ballistic missile, meaning that after a short burn its trajectory was entirely ballistic, ie dictated by gravity (if you like that sort of thing) or UA (if you don't). I'm going to go out on a limb and also assume that you also agree with with the reported flight time of 53 minutes, although do please let me know if not.
The reason the experts are coming up with ICBM ranges for the missile, despite it flying a relatively short distance, is that if something flies ballistically for that long and only travels 950km, it has to have a very steep launch angle. Indeed, the reported apogee was very high - 4500km. If it was launched at a shallower angle, it would go much further. But you don't need me to tell you that - as you keep saying, the equations are very simple. Even if you dispense with heretical things like the reduction in g as you get away from the earth, if you calculate the speed necessary at 'launch' (more accurately, the end of burn, but let's keep it simple) to fly ballistically for something like 50 minutes (the reported burn time was around 5 minutes), and then try a shallower launch angle, you'll get much bigger ranges than 950km.
If you are genuinely intellectually curious, then do the maths and let us know how you get on.
The reason why the experts (henceforth warmongers) come up with the ICBM ranges for the missile is the perceived need to govern through means of fear, successfully pushed on the masses for millenia.
You can certainly keep mentally subjugating yourself to these clowns and asshats. I DNGAF what you do.
Plus, you want everyone here to believe this Kim Jong Un is capable of forcing missiles to achieve speeds approaching 32,000 MPH!
What a joke!
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The reason why the experts (henceforth warmongers) come up with the ICBM ranges for the missile is the perceived need to govern through means of fear, successfully pushed on the masses for millenia.
You can certainly keep mentally subjugating yourself to these clowns and asshats. I DNGAF what you do.
Plus, you want everyone here to believe this Kim Jong Un is capable of forcing missiles to achieve speeds approaching 32,000 MPH!
What a joke!
Do you accept the reported distance flown and time of flight as being accurate?
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No ICBM is reported to or required to exceed escape velocity.
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The reason why the experts (henceforth warmongers) come up with the ICBM ranges for the missile is the perceived need to govern through means of fear, successfully pushed on the masses for millenia.
You can certainly keep mentally subjugating yourself to these clowns and asshats. I DNGAF what you do.
Plus, you want everyone here to believe this Kim Jong Un is capable of forcing missiles to achieve speeds approaching 32,000 MPH!
What a joke!
Do you accept the reported distance flown and time of flight as being accurate?
No ICBM is reported to or required to exceed escape velocity.
I believe the real concern and follow up questions should be focused on the warmongering figures presented as translating into the reported "ICBM" as traveling over 32,000 mph!
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I believe the real concern and follow up questions should be focused on the warmongering figures presented as translating into the reported "ICBM" as traveling over 32,000 mph!
I may have missed something, in which case my apologies, but where did the 32,000mph figure come from?
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I believe the real concern and follow up questions should be focused on the warmongering figures presented as translating into the reported "ICBM" as traveling over 32,000 mph!
I may have missed something, in which case my apologies, but where did the 32,000mph figure come from?
Do the math.
Using the figures you provided.
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I believe the real concern and follow up questions should be focused on the warmongering figures presented as translating into the reported "ICBM" as traveling over 32,000 mph!
I may have missed something, in which case my apologies, but where did the 32,000mph figure come from?
Do the math.
Using the figures you provided.
C'mon lackey, you have to show your work if you want full marks on your homework...
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Do the math.
Using the figures you provided.
I too am keen to see how you've arrived at 32,000mph. How did you model the effect of drag on the missile? What about the reduction in mass as the fuel is consumed during the burn? I haven't had time to do the maths I'm afraid. It's not at all 'simple' - that was your word. You keep saying it's just a simple quadratic equation. The truth is far from that - you end up with a set of differential equations that can't be solved analytically, so you have to use some kind of time step solution. Tom would of course say that means rockets aren't real, given his hostility to numerical solutions of n-body problems, but that's best left to another thread.
Nevertheless, even without calculating the numbers accurately, you can presumably see that a flight time of 53 minutes for a downrange distance of just 950km is clearly not a projectile operating at maximum rage. Unless, as per my previous post, you disagree with the figures? Are you saying it wasn't a flight time of 53 minutes? Or the range wasn't as advertised?
-
The bottom line for all of this thread is this.
The only evidence that exists for ICBM's is some propaganda and gullible believers.
I've been happy to let you go on and on. Give a man enough rope kind of thing
But Jesus F*&@ing Christ.
https://www.airforcetimes.com/news/your-air-force/2021/02/24/air-force-launches-unarmed-unarmed-minuteman-3-missile-from-vandenberg-afb/
https://breakingdefense.com/2020/02/first-2020-minuteman-iii-test-launches-as-new-start-countdown-begins/
As for Dear Leader go ahead and look foolish as long as you'd like.
Bye now!
-
The bottom line for all of this thread is this.
The only evidence that exists for ICBM's is some propaganda and gullible believers.
I've been happy to let you go on and on. Give a man enough rope kind of thing
But Jesus F*&@ing Christ.
https://www.airforcetimes.com/news/your-air-force/2021/02/24/air-force-launches-unarmed-unarmed-minuteman-3-missile-from-vandenberg-afb/
https://breakingdefense.com/2020/02/first-2020-minuteman-iii-test-launches-as-new-start-countdown-begins/
As for Dear Leader go ahead and look foolish as long as you'd like.
Bye now!
Pff. You can prove anything with facts.
So what now, Lackey? CGI? We weren't there so can't know it happened? There's always some excuse...
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Do the math.
Using the figures you provided.
I too am keen to see how you've arrived at 32,000mph. How did you model the effect of drag on the missile? What about the reduction in mass as the fuel is consumed during the burn? I haven't had time to do the maths I'm afraid. It's not at all 'simple' - that was your word. You keep saying it's just a simple quadratic equation. The truth is far from that - you end up with a set of differential equations that can't be solved analytically, so you have to use some kind of time step solution. Tom would of course say that means rockets aren't real, given his hostility to numerical solutions of n-body problems, but that's best left to another thread.
Nevertheless, even without calculating the numbers accurately, you can presumably see that a flight time of 53 minutes for a downrange distance of just 950km is clearly not a projectile operating at maximum rage. Unless, as per my previous post, you disagree with the figures? Are you saying it wasn't a flight time of 53 minutes? Or the range wasn't as advertised?
Just ry and do the math using the figures you used.
See if you can come up with how I arrived at 32,000 mph.
A perfect opportunity for you to write, "WRONG and HERE is WHY!"
-
The bottom line for all of this thread is this.
The only evidence that exists for ICBM's is some propaganda and gullible believers.
I've been happy to let you go on and on. Give a man enough rope kind of thing
But Jesus F*&@ing Christ.
https://www.airforcetimes.com/news/your-air-force/2021/02/24/air-force-launches-unarmed-unarmed-minuteman-3-missile-from-vandenberg-afb/
https://breakingdefense.com/2020/02/first-2020-minuteman-iii-test-launches-as-new-start-countdown-begins/
As for Dear Leader go ahead and look foolish as long as you'd like.
Bye now!
The bottom line for all of this thread is this.
The only evidence that exists for ICBM's is some propaganda and gullible believers.
I've been happy to let you go on and on. Give a man enough rope kind of thing
But Jesus F*&@ing Christ.
https://www.airforcetimes.com/news/your-air-force/2021/02/24/air-force-launches-unarmed-unarmed-minuteman-3-missile-from-vandenberg-afb/
https://breakingdefense.com/2020/02/first-2020-minuteman-iii-test-launches-as-new-start-countdown-begins/
As for Dear Leader go ahead and look foolish as long as you'd like.
Bye now!
Pff. You can prove anything with facts.
So what now, Action80? CGI? We weren't there so can't know it happened? There's always some excuse...
So, how much was missile and how much was re-entry vehicle, I wonder.
Not an ICBM.
But, by all means, continue.
Bye now.
-
The bottom line for all of this thread is this.
The only evidence that exists for ICBM's is some propaganda and gullible believers.
I've been happy to let you go on and on. Give a man enough rope kind of thing
But Jesus F*&@ing Christ.
https://www.airforcetimes.com/news/your-air-force/2021/02/24/air-force-launches-unarmed-unarmed-minuteman-3-missile-from-vandenberg-afb/
https://breakingdefense.com/2020/02/first-2020-minuteman-iii-test-launches-as-new-start-countdown-begins/
As for Dear Leader go ahead and look foolish as long as you'd like.
Bye now!
The bottom line for all of this thread is this.
The only evidence that exists for ICBM's is some propaganda and gullible believers.
I've been happy to let you go on and on. Give a man enough rope kind of thing
But Jesus F*&@ing Christ.
https://www.airforcetimes.com/news/your-air-force/2021/02/24/air-force-launches-unarmed-unarmed-minuteman-3-missile-from-vandenberg-afb/
https://breakingdefense.com/2020/02/first-2020-minuteman-iii-test-launches-as-new-start-countdown-begins/
As for Dear Leader go ahead and look foolish as long as you'd like.
Bye now!
Pff. You can prove anything with facts.
So what now, Action80? CGI? We weren't there so can't know it happened? There's always some excuse...
So, how much was missile and how much was re-entry vehicle, I wonder.
Not an ICBM.
But, by all means, continue.
Bye now.
That's what you're going with? LMFAO.
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See if you can come up with how I arrived at 32,000 mph.
How'd you come up with 32k mph? I can't figure out where nor how you arrived at that number. Fastest known ICBM:
LGM-30 Minuteman Maximum speed:
Mach 23
(17,508 miles per hour; 28,176 kilometers per hour; 7.8267 kilometers per second)
https://en.wikipedia.org/wiki/LGM-30_Minuteman
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Just ry and do the math using the figures you used.
See if you can come up with how I arrived at 32,000 mph.
A perfect opportunity for you to write, "WRONG and HERE is WHY!"
I'm assuming you've done something like running 53 minutes, g, and 950km through a ballistic trajectory calculator or perhaps s=ut+1/2at2 and ended up with 32,000+mph, but I don't know, as you haven't shown any of your calculations. The ball is very much in your court here - you can't just pluck a number out of nowhere and then use it in your arguments - nobody is going to take you seriously unless you 'show your working', as the teachers like to say.
Like I said, I'm curious to see if you've incorporated the burn phase, change of mass and drag into your calculations. I suspect you haven't, but I can't tell because you aren't showing us. Why not just show us?
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Just try and do the math using the figures you used.
See if you can come up with how I arrived at 32,000 mph.
A perfect opportunity for you to write, "WRONG and HERE is WHY!"
I'm assuming you've done something like running 53 minutes, g, and 950km through a ballistic trajectory calculator or perhaps s=ut+1/2at2 and ended up with 32,000+mph, but I don't know, as you haven't shown any of your calculations. The ball is very much in your court here - you can't just pluck a number out of nowhere and then use it in your arguments - nobody is going to take you seriously unless you 'show your working', as the teachers like to say.
Like I said, I'm curious to see if you've incorporated the burn phase, change of mass and drag into your calculations. I suspect you haven't, but I can't tell because you aren't showing us. Why not just show us?
Whether or not you take me seriously is not the subject, nor is it any concern of mine.
I hold globular believers in very low regard, especially those incapable of performing very basic math.
So far, no one has demonstrated an ICBM even exists.
So, this entire topic belongs in CN.
How much stuff that has been thrown into the mix just on this thread is a typical display of RE-supporters.
Sane person - "You know something, not one ICBM has even been used, RE or FE. Why try to link an ICBM to the shape of the earth, if this is a fact?"
RE supporter - "Because ICBM's are REAL!"
Sane person - "OK. Show me some evidence they are real."
RE supporter - "These reports from government officials, repeated by MSM."
Sane person - "The same governments that are demonstrably lying to you each and every day and have been for millenia? The same MSM that serves as nothing but a propaganda arm for those governments?"
RE supporters - "OK, what about shipping routes and missing land area?"
It really is getting tiresome.
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Small missiles work. Large ones don’t. Total Lackey, folks. You can’t make it up.
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Sane person: Here's all the evidence of test launches of ICBMs
Action 80. Nuh uh!
It really is getting tiresome.
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The bottom line for all of this thread is this.
The only evidence that exists for ICBM's is some propaganda and gullible believers.
Have you ever seen the Eiffel Tower?
How do you know the Eiffel Tower is real?
I'm guessing you believe in the Eiffel Tower because it doesn't conflict with your worldview even though you've never seen it.
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It really is getting tiresome.
Indeed, quite tiresome. So you've decided to not show us your math as to how you arrived at 32k mph? Is there a reason you won't show us?
As well, yes, straight from the Military Industrial Complex and MSM from 1962, there has actually been a live firing of a ballistic missile with a warhead detonation - Granted, it wasn't flown intercontinentally, but it did leave and re-enter the atmosphere after traveling a 1000+ or so nautical miles, minimum-maximum altitude 98,000' - 260,000' (my bolding):
Test: Frigate Bird
Time: 23:30 6 May 1962 (GMT)
Location: Johnston Island
Test Height and Type: SLBM Airburst; 11,000 Feet
Yield: 600 kt
Device Diameter (inches): 18
Device Length (inches): 46.6
Device Weight (lb.): 717
Frigate Bird was the only US test of an operational ballistic missile with a live warhead. This test involved firing a Polaris A1 missile from a ballistic missile submarine. The missile was launched by the USS Ethan Allen (SSBN-608) at 13:18 (local) from a position 1500 nm east-northeast of Christmas Island. The re-entry vehicle (RV) and warhead flew 1020 nm downrange toward Christmas Island before re-entering the atmosphere 12.5 minutes later, and detonating in an airburst at 11,000 feet. The system tested was a combination of a Polaris A1 SLBM, and a W-47Y1 warhead in a Mk-1 RV. The Mk-1 RV had a beryllium heat-sink heat shield, and with the 717 lb warhead had a gross weight of 900 lb. The missile/RV demonstrated an accuracy on the order of 2200 yards. This warhead had a yield-to-weight ratio of 1.84 kt/kg, but the higher yield Y2 variant tested in Dominic Harlem doubled the yield and nearly doubled tht YTW ratio to 3.61 kt/kg.
The image of the Frigate Bird mushroom cloud was taken through the periscope of the USS Carbonero (SS-337) 480 nm ENE of Christmas Island. The Carbonero (along with the USS Medregal, SS-480) was within 30 miles of the burst point.
(http://nuclearweaponarchive.org/Usa/Tests/FrigateBird200c10.jpg)
http://nuclearweaponarchive.org/Usa/Tests/Dominic.html
Other than riding a missile yourself, I'm not sure what you're looking for...
(https://i.pinimg.com/originals/b3/16/b6/b316b6c63fd28a26b47289f0f260af15.gif)
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Whether or not you take me seriously is not the subject, nor is it any concern of mine.
It's not me you need to worry about - it's the undecided folks reading this. Your refusal to respond to any basic questions, like whether or not you accept the time of flight figure, or to support your apparently arbitrary figure of 32000mph, just makes it look like you're hiding something. The sum total of your arguments on this thread, and indeed others, is simply saying that stuff isn't true. That's not a debate - you need to actually engage in some kind of discussion, providing evidence and data to support your position.
I hold globular believers in very low regard, especially those incapable of performing very basic math.
You're welcome to your opinion, of course, but levelling an accusation of an inability to perform basic math when you have, on this thread, completely failed to demonstrate any math(s) skills whatsoever is somewhat hypocritical. In this thread alone you've demonstrated a lack of understanding of cartesian and polar coordinates, claimed that ballistic missiles can be aimed by means of a simple quadratic equation...although you've failed to provide that equation, and failed to comprehend that it ain't that simple. You've then plucked a random figure out of thin air, without any evidence at all, and expected us all to accept it as fact to support your argument.
There's a good reason you aren't providing your calculations, isn't there? If you're embarrassed by your mathematical skills, and you're interested in ballistic missile principles, this website has a useful estimate of the NK missile capabilities based on flight time: https://physicsfromplanetearth.wordpress.com/2017/08/18/the-range-of-north-korean-ballistic-missiles/ (https://physicsfromplanetearth.wordpress.com/2017/08/18/the-range-of-north-korean-ballistic-missiles/)
They have two pages - the second one has a more precise calculation, but that involves trusting the reported apogee height, which I'm assuming you don't.
So far, no one has demonstrated an ICBM even exists.
But you've demonstrated that you won't accept any evidence that contradicts your worldview - what's the point of engaging in debate if you aren't willing to either change your own position, or to offer up compelling evidence in order to change other peoples'? You're just sat there shouting 'it isn't true'.
So, this entire topic belongs in CN.
How much stuff that has been thrown into the mix just on this thread is a typical display of RE-supporters.
Sane person - "You know something, not one ICBM has even been used, RE or FE. Why try to link an ICBM to the shape of the earth, if this is a fact?"
RE supporter - "Because ICBM's are REAL!"
Sane person - "OK. Show me some evidence they are real."
RE supporter - "These reports from government officials, repeated by MSM."
Sane person - "The same governments that are demonstrably lying to you each and every day and have been for millenia? The same MSM that serves as nothing but a propaganda arm for those governments?"
RE supporters - "OK, what about shipping routes and missing land area?"
It really is getting tiresome.
But you yourself have willingly accepted the reported range of a ballistic missile test - you've used that same statistic to argue against RE proponents. At the above link you can see that, if you also accept the time of flight reports to be true, then that's all the information you need to work out that the range of those missiles is indeed intercontinental. You don't need anything else - no NASA, no governments...nothing. If a ballistic missile can fly for 40 or 50 minutes, then it is capable of flying many thousands of miles.
If you're genuinely curious, there's a wealth of information out there. If you're just here to cynically shout that stuff isn't true, then we can't help.
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Having difficulty with the OP title there, Bob?
Indeed, the reported apogee was very high - 4500km. (the reported burn time was around 5 minutes)
Again, I refer you to your own quote.
If you are genuinely intellectually curious, then do the maths and let us know how you get on.
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OK, those are two of the numbers that could come into play.... still waiting to see the Action80mathTM that got you your magic number
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This is like he's lost in the desert, and he wants us to find him by working out how he misread the map. It's purely conjecture, but here's some calculations:
Action80 highlighted the altitude, which corresponds to 2796 miles. If he assumed that it reached this altitude in (the highlighted) 5 minutes, at a constant velocity of 2796/5 x 60, that would give a speed of 33,500 mph, which is "over 32,000 mph".
Of course, that would mean that he thought "end of burn time" = "apogee", and that it didn't either accelerate or decelerate during its ascent, and I'm sure he wouldn't be that dumb, because he tells us that knows what the B (and all the other letters) in ICBM stands for.
However, as many correspondents have already said, we have no idea how he got where he is, so all we have is conjecture.
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Having difficulty with the OP title there, Bob?
Indeed, the reported apogee was very high - 4500km. (the reported burn time was around 5 minutes)
Again, I refer you to your own quote.
If you are genuinely intellectually curious, then do the maths and let us know how you get on.
Like I said, in the same quote you are referring to, all the maths you need is here: /https://physicsfromplanetearth.files.wordpress.com/2018/01/papertable-1-e1515874894590.png (https://forum.tfes.org/https://physicsfromplanetearth.files.wordpress.com/2018/01/papertable-1-e1515874894590.png)
It covers speed calculations for the apogee / flight time in question.
Have you read it?
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Having difficulty with the OP title there, Bob?
Indeed, the reported apogee was very high - 4500km. (the reported burn time was around 5 minutes)
Again, I refer you to your own quote.
If you are genuinely intellectually curious, then do the maths and let us know how you get on.
Like I said, in the same quote you are referring to, all the maths you need is here: /https://physicsfromplanetearth.files.wordpress.com/2018/01/papertable-1-e1515874894590.png (https://forum.tfes.org/https://physicsfromplanetearth.files.wordpress.com/2018/01/papertable-1-e1515874894590.png)
It covers speed calculations for the apogee / flight time in question.
Have you read it?
Yes.
Here is the summary:
I. 404 Not Found
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Yes.
Here is the summary:
I. 404 Not Found
Tut. The link was bad, but it didn't take too much figuring out. For the hard of thinking though, here:
https://physicsfromplanetearth.files.wordpress.com/2018/01/papertable-1-e1515874894590.png
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Yes.
Here is the summary:
I. 404 Not Found
Tut. The link was bad, but it didn't take too much figuring out. For the hard of thinking though, here:
https://physicsfromplanetearth.files.wordpress.com/2018/01/papertable-1-e1515874894590.png
Yeah, a wordpress file.
With just an analysis.
LOL!
You guys are rich!
Still waiting on the ICBM.
Still waiting on the math disproving 32,000 MPH wrong.
You got jack here and you all know it.
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Still waiting on the math disproving 32,000 MPH wrong.
Because you haven't presented your maths so we can look at what you get wrong.
Do that, then we can talk.
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Still waiting on the math disproving 32,000 MPH wrong.
Because you haven't presented your maths so we can look at what you get wrong.
Do that, then we can talk.
At this point, I will further point out that if one clicks on this link:
https://physicsfromplanetearth.files.wordpress.com/2018/01/papertable-1-e1515874894590.png
with a different browser (this is important) you will see a distinct difference in rendering, perhaps leading one to believe the handlers of SteelyBob and AATW are certainly one in the same, with instructions on the level of,"QUICK, go fix my screw up!"
RE Supporters - "We are are wondering how you got your figures. Can you show us? In the meantime, allow us to post this BS WordPress table with figures that have no explanation as to how they were derived and furthermore, no author to which it can be attributed for verification. We are sure you will not catch this."
Sane person - " You are kidding, right?"
Note differences in rendering of the webpage, one, and then note it is just a stupid WordPress document, with no mention of the author, two, and no accounting of how they arrived at their math presented in their useless table.
Clicking on AATW's, link shows the table clipped from the WordPress page, a sure sign of dishonesty.
Either way, both SteelyBob and AATW, are guilty of not providing the name of the author of this piece of trash table, something I thought TFES was keen on avoiding. If you cannot provide proper references, both of you should stop posting.
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Further, a parabola extending 4500km to the y and 950km to the x in the span of a 5 minute burn, with a total flight time of 53 minutes?
LOL!!!
-
Either way, both SteelyBob and AATW, are guilty of not providing the name of the author of this piece of trash table, something I thought TFES was keen on avoiding. If you cannot provide proper references, both of you should stop posting.
To be clear, I don't know what this link is. It was poorly formed, it was fairly obvious how to correct it but you were apparently too stupid to work it out so I helped you.
The fact you couldn't work out how to correct a simple URL does not fill me with confidence that your maths is correct, but I'm sure you're about to present it any minute now...
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Further, a parabola extending 4500km to the y and 950km to the x in the span of a 5 minute burn, with a total flight time of 53 minutes?
LOL!!!
You're almost at the 'aha!' moment you so desperately need to arrive at.
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Further, a parabola extending 4500km to the y and 950km to the x in the span of a 5 minute burn, with a total flight time of 53 minutes?
LOL!!!
You're almost at the 'aha!' moment you so desperately need to arrive at.
Already there.
Regardless of anything else related to missile functioning, d=rt remains something that cannot be explained away in this case.
It's not like you're insistent a 5 minute burn from 177,000lbf engine can elevate 159,000lbs to 4500km. You want people to believe the engine doesn't need to be operational in order to achieve such a feat (which isn't possible to begin with, as no such altitude exists).
But by all means.
Continue with your fairytale. I understand they go quite well with binkies.
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For the record, here's where the table comes from and here is the math used to populate it:
https://physicsfromplanetearth.wordpress.com/2018/01/14/the-range-of-north-korean-icbms-update/
Now why not show your math instead of dancing around every conceivable way to distract from showing your math? Put an end to the back and forth by just showing what you did. It's super weird that you won't just copy and paste your math into a post and are doing/saying so many other things except providing this one simple thing that is actually necessary. Why?
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It's not like you're insistent a 5 minute burn from 177,000lbf engine can elevate 159,000lbs to 4500km.
Correct. I don't think anyone is suggesting that. The mass at launch 159,000 lbs. However, the huge majority of that mass doesn't get anywhere near apogee because it is ....what? Here's some clues; its flammable, and there's none left after 5 minutes.
Any ideas?
Going off at a slight tangent, but lets expand our thinking a little; the standard US Army artillery piece is the M109 Howitzer. Its barrel is 6 metres long (around 20 feet), and it fires a 155 mm shell around 13 miles (21 km). I confess I don't know the answer to this myself (as I'm neither an artilleryman nor a rocket scientist), but I wonder how its burn time relates to its flight time?
-
Total Lackey’s objection still makes no sense. He admits rockets work. All an ICBM does is put all the powered acceleration at the front end of the vector.
-
For the record, here's where the table comes from and here is the math used to populate it:
https://physicsfromplanetearth.wordpress.com/2018/01/14/the-range-of-north-korean-icbms-update/
Now why not show your math instead of dancing around every conceivable way to distract from showing your math? Put an end to the back and forth by just showing what you did. It's super weird that you won't just copy and paste your math into a post and are doing/saying so many other things except providing this one simple thing that is actually necessary. Why?
Thanks for updating the link - my bad. Not sure what happened there. That was, of course, the page I was trying to refer to. Not exactly difficult to find for the intellectually curious and genuinely interested, which makes A80’s choice to distract and divert from his/her continuous failure to provide any evidence whatsoever all the more telling. Shame really - I come here to debate, and there’s not much debate being offered.
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It's not like you're insistent a 5 minute burn from 177,000lbf engine can elevate 159,000lbs to 4500km.
The mass at launch 159,000 lbs. However, the huge majority of that mass doesn't get anywhere near apogee because it is ....what? Here's some clues; its flammable, and there's none left after 5 minutes.
^Now admits the missile had no fuel at launch?
Regardless, 5 minutes of burn cannot in any form or fashion satisfy requirements of d=rt given 4500km yand 950 km x.
Burn must take place in total from time of ignition to end of burn.
No on, then off, then on again.
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I feel like playing the game this morning.
^Now admits the missile had no fuel at launch?
Evidently your reading comprehension is poor as your understanding of physics. Nowhere is it stated there is no fuel at launch. There's a huge weight of fuel at launch which decreases to zero at the end of burn which is the point of Duncan's post with regards to how much mass would be left at the end of the burn to continue to apogee.
Regardless, 5 minutes of burn cannot in any form or fashion satisfy requirements of d=rt given 4500km yand 950 km x.
Please enroll in a first year high school physics class and learn about things like acceleration and initial velocity and then reassess this and get back to us.
Burn must take place in total from time of ignition to end of burn.
Who said differently? Back to the reading comprehension again I guess.
No on, then off, then on again.
See comment immediately above.
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Thanks WTF_S;
To confirm, my contention is that the missile launches at a mass of 159,000 lbs, the majority of which is fuel, and all consumed within the initial, single, 5-minute, burn, which accelerates it on a high angle trajectory where y>x.
Total burn time = 5 minutes.
Unpowered ballistic cruise flight duration = 48 minutes.
Total flight duration from launch to impact = 53 minutes.
Anyone else having trouble with this? Happy to see if anyone can put it in simpler terms.
As for the x/y sums, still waiting .........
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It's not like you're insistent a 5 minute burn from 177,000lbf engine can elevate 159,000lbs to 4500km.
The mass at launch 159,000 lbs. However, the huge majority of that mass doesn't get anywhere near apogee because it is ....what? Here's some clues; its flammable, and there's none left after 5 minutes.
^Now admits the missile had no fuel at launch?
Regardless, 5 minutes of burn cannot in any form or fashion satisfy requirements of d=rt given 4500km yand 950 km x.
Burn must take place in total from time of ignition to end of burn.
No on, then off, then on again.
What exactly do you think are the requirements of d=rt, in the context of an accelerating object tracing an elliptical flight path?
-
It's not like you're insistent a 5 minute burn from 177,000lbf engine can elevate 159,000lbs to 4500km.
The mass at launch 159,000 lbs. However, the huge majority of that mass doesn't get anywhere near apogee because it is ....what? Here's some clues; its flammable, and there's none left after 5 minutes.
^Now admits the missile had no fuel at launch?
Regardless, 5 minutes of burn cannot in any form or fashion satisfy requirements of d=rt given 4500km yand 950 km x.
Burn must take place in total from time of ignition to end of burn.
No on, then off, then on again.
What exactly do you think are the requirements of d=rt, in the context of an accelerating object tracing an elliptical flight path?
According to the sources presented here, the parabola in this case forms the vertex between 425 and 450 km x, correct?
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Thanks WTF_S;
To confirm, my contention is that the missile launches at a mass of 159,000 lbs, the majority of which is fuel, and all consumed within the initial, single, 5-minute, burn, which accelerates it on a high angle trajectory where y>x.
You contend that 159,000 lbs consisted of a, "majority of which is fuel," based on "what exactly?"
Total burn time = 5 minutes.
Any idea on the altitude and rate of travel achieved by the time of engine cut off? It is apparent you disagree with the 4500 km achieved at that time, contending an unpowered ballistic object can continue gaining a significant amount of altitude after impetus is removed.
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It's not like you're insistent a 5 minute burn from 177,000lbf engine can elevate 159,000lbs to 4500km.
The mass at launch 159,000 lbs. However, the huge majority of that mass doesn't get anywhere near apogee because it is ....what? Here's some clues; its flammable, and there's none left after 5 minutes.
^Now admits the missile had no fuel at launch?
Regardless, 5 minutes of burn cannot in any form or fashion satisfy requirements of d=rt given 4500km yand 950 km x.
Burn must take place in total from time of ignition to end of burn.
No on, then off, then on again.
What exactly do you think are the requirements of d=rt, in the context of an accelerating object tracing an elliptical flight path?
According to the sources presented here, the parabola in this case forms the vertex between 425 and 450 km x, correct?
Not a parabola no - as the source I referred to makes clear, ballistic trajectories on a globe are actually portions of an ellipse, not a parabola. But let's go with parabola since I'm guessing you aren't to keen on the globe thing. So yes, a missile trajectory of 950km would have the 'vertex' at roughly the midpoint along the horizontal (if the world was flat) axis, so x=425km, yes.
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It's not like you're insistent a 5 minute burn from 177,000lbf engine can elevate 159,000lbs to 4500km.
The mass at launch 159,000 lbs. However, the huge majority of that mass doesn't get anywhere near apogee because it is ....what? Here's some clues; its flammable, and there's none left after 5 minutes.
^Now admits the missile had no fuel at launch?
Regardless, 5 minutes of burn cannot in any form or fashion satisfy requirements of d=rt given 4500km yand 950 km x.
Burn must take place in total from time of ignition to end of burn.
No on, then off, then on again.
What exactly do you think are the requirements of d=rt, in the context of an accelerating object tracing an elliptical flight path?
According to the sources presented here, the parabola in this case forms the vertex between 425 and 450 km x, correct?
Not a parabola no - as the source I referred to makes clear, ballistic trajectories on a globe are actually portions of an ellipse, not a parabola. But let's go with parabola since I'm guessing you aren't to keen on the globe thing. So yes, a missile trajectory of 950km would have the 'vertex' at roughly the midpoint along the horizontal (if the world was flat) axis, so x=425km, yes.
Well, if I cut an ellipse in half, what shape am I left with?
Call me suspicious, but this looks like a parabola to me:
https://en.wikipedia.org/wiki/Hwasong-15#/media/File:Trajectories_of_Hwasong-14.svg
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Well, if I cut an ellipse in half, what shape am I left with?
Call me suspicious, but this looks like a parabola to me:
https://en.wikipedia.org/wiki/Hwasong-15#/media/File:Trajectories_of_Hwasong-14.svg
They aren't the same thing - see https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwiT9_DWlaHxAhU05eAKHQmACxEQFjAHegQIHhAD&url=https%3A%2F%2Fcourses.lumenlearning.com%2Fboundless-algebra%2Fchapter%2Fintroduction-to-conic-sections%2F&usg=AOvVaw1-N5oByVcSeU7_75VNTX7G (https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwiT9_DWlaHxAhU05eAKHQmACxEQFjAHegQIHhAD&url=https%3A%2F%2Fcourses.lumenlearning.com%2Fboundless-algebra%2Fchapter%2Fintroduction-to-conic-sections%2F&usg=AOvVaw1-N5oByVcSeU7_75VNTX7G) if you're interested.
But don't worry about it - let's develop your point further. Where were you going with your argument?
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Well, if I cut an ellipse in half, what shape am I left with?
Call me suspicious, but this looks like a parabola to me:
https://en.wikipedia.org/wiki/Hwasong-15#/media/File:Trajectories_of_Hwasong-14.svg
They aren't the same thing - see https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwiT9_DWlaHxAhU05eAKHQmACxEQFjAHegQIHhAD&url=https%3A%2F%2Fcourses.lumenlearning.com%2Fboundless-algebra%2Fchapter%2Fintroduction-to-conic-sections%2F&usg=AOvVaw1-N5oByVcSeU7_75VNTX7G (https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwiT9_DWlaHxAhU05eAKHQmACxEQFjAHegQIHhAD&url=https%3A%2F%2Fcourses.lumenlearning.com%2Fboundless-algebra%2Fchapter%2Fintroduction-to-conic-sections%2F&usg=AOvVaw1-N5oByVcSeU7_75VNTX7G) if you're interested.
Interesting, even though I already understood that an ellipse and a parabola are not the same thing, hence the query as to what is left upon rendering an ellipse in two.
But don't worry about it - let's develop your point further. Where were you going with your argument?
First point.
Does this image of the flight trajectory, presented here:
https://en.wikipedia.org/wiki/Hwasong-15#/media/File:Trajectories_of_Hwasong-14.svg
depict an ellipse or a parabola.
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Well, if I cut an ellipse in half, what shape am I left with?
Call me suspicious, but this looks like a parabola to me:
https://en.wikipedia.org/wiki/Hwasong-15#/media/File:Trajectories_of_Hwasong-14.svg
They aren't the same thing - see https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwiT9_DWlaHxAhU05eAKHQmACxEQFjAHegQIHhAD&url=https%3A%2F%2Fcourses.lumenlearning.com%2Fboundless-algebra%2Fchapter%2Fintroduction-to-conic-sections%2F&usg=AOvVaw1-N5oByVcSeU7_75VNTX7G (https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwiT9_DWlaHxAhU05eAKHQmACxEQFjAHegQIHhAD&url=https%3A%2F%2Fcourses.lumenlearning.com%2Fboundless-algebra%2Fchapter%2Fintroduction-to-conic-sections%2F&usg=AOvVaw1-N5oByVcSeU7_75VNTX7G) if you're interested.
Interesting, even though I already understood that an ellipse and a parabola are not the same thing, hence the query as to what is left upon rendering an ellipse in two.
But don't worry about it - let's develop your point further. Where were you going with your argument?
First point.
Does this image of the flight trajectory, presented here:
https://en.wikipedia.org/wiki/Hwasong-15#/media/File:Trajectories_of_Hwasong-14.svg
depict an ellipse or a parabola.
That image depicts a parabola.
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Well, if I cut an ellipse in half, what shape am I left with?
Call me suspicious, but this looks like a parabola to me:
https://en.wikipedia.org/wiki/Hwasong-15#/media/File:Trajectories_of_Hwasong-14.svg
They aren't the same thing - see https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwiT9_DWlaHxAhU05eAKHQmACxEQFjAHegQIHhAD&url=https%3A%2F%2Fcourses.lumenlearning.com%2Fboundless-algebra%2Fchapter%2Fintroduction-to-conic-sections%2F&usg=AOvVaw1-N5oByVcSeU7_75VNTX7G (https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwiT9_DWlaHxAhU05eAKHQmACxEQFjAHegQIHhAD&url=https%3A%2F%2Fcourses.lumenlearning.com%2Fboundless-algebra%2Fchapter%2Fintroduction-to-conic-sections%2F&usg=AOvVaw1-N5oByVcSeU7_75VNTX7G) if you're interested.
Interesting, even though I already understood that an ellipse and a parabola are not the same thing, hence the query as to what is left upon rendering an ellipse in two.
But don't worry about it - let's develop your point further. Where were you going with your argument?
First point.
Does this image of the flight trajectory, presented here:
https://en.wikipedia.org/wiki/Hwasong-15#/media/File:Trajectories_of_Hwasong-14.svg
depict an ellipse or a parabola.
That image depicts a parabola.
Great.
So, next we have these two questions.
How many lbs of fuel contributed to the 159,000 lb weight of the missile?
What was the altitude achieved at t+5?
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Thanks WTF_S;
To confirm, my contention is that the missile launches at a mass of 159,000 lbs, the majority of which is fuel, and all consumed within the initial, single, 5-minute, burn, which accelerates it on a high angle trajectory where y>x.
You contend that 159,000 lbs consisted of a, "majority of which is fuel," based on "what exactly?"
Total burn time = 5 minutes.
Any idea on the altitude and rate of travel achieved by the time of engine cut off? It is apparent you disagree with the 4500 km achieved at that time, contending an unpowered ballistic object can continue gaining a significant amount of altitude after impetus is removed.
Majority of the mass is fuel? It's an aluminium tube with a 2000lb warhead at one end, and a rocket motor at the other. You work it out. What exactly do you think is inside the part between the motor and warhead? Its a fuel tank or, for a solid fuel motor, containment for the fuel mass.
Altitude and velocity at engine shut-down? No idea, I'm not a rocket scientist. As a layman, I couldn't be more specific than to say its high and fast; not only has it been accelerating for the last 5 minutes, but its rate of acceleration has been increasing as fuel is consumed. As an aircraft engineer, I know that its aerodynamic drag following engine shutdown will be very small-to-non-existent, due to the low-to-non-existent air density (drag being {drag-coefficient x air-density x surface-area x velocity-squared}/2). Therefore, the only braking force to its vertical velocity component is due to gravity.
Significant amount of altitude after impetus removed? You have maybe heard of the German Flak 36, the 88 mm anti aircraft gun from WW2? Its impetus was removed at an altitude of about 15 feet (the end of the barrel) and it had an effective altitude range of over 30,000 feet (around 6 miles), and that was in draggy-air.
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You contend that 159,000 lbs consisted of a, "majority of which is fuel," based on "what exactly?"
Based on the fact that it's going to be quite similar to this:
https://minutemanmissile.com/solidrocketboosters.html as well as this: https://en.wikipedia.org/wiki/LGM-30_Minuteman#:~:text=The%20Minuteman%2DIII%20has%20a,800%20ft%20(240%20m).
I'll highlight the pertinent points for the Minuteman II and III since it appears reading isn't your thang.
1st stage weight 51,251 lbs. Fuel weight 45,670 lbs.
2nd stage weight 16,057 lbs. Fuel weight 13,680lbs.
3rd stage weight 8,187 lbs. Fuel weight 7,292 lbs.
In case math ain't your deal either, that's total stage weight 75,795. Total fuel weight 66,642. With the total missile weight of 79,432 that would be 83+% fuel. I'll give you credit for knowing the definition of majority.
Any idea on the altitude and rate of travel achieved by the time of engine cut off? It is apparent you disagree with the 4500 km achieved at that time, contending an unpowered ballistic object can continue gaining a significant amount of altitude after impetus is removed.
Didn't take my advice on that high school physics class evidently. I even pointed out the important things here like initial velocity and acceleration (hint: in this case it's G). I did forget to mention how to calculate distance traveled.
Extremely curious how high you think a bullet travels when shot straight up. You do realize a bullet is an 'unpowered ballistic object' once it leaves the barrel, right?
If you'd like to continue your 'Nuh uh!' game that's great. But please, try not to embarrass yourself in the process.
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Thanks WTF_S;
To confirm, my contention is that the missile launches at a mass of 159,000 lbs, the majority of which is fuel, and all consumed within the initial, single, 5-minute, burn, which accelerates it on a high angle trajectory where y>x.
You contend that 159,000 lbs consisted of a, "majority of which is fuel," based on "what exactly?"
Total burn time = 5 minutes.
Any idea on the altitude and rate of travel achieved by the time of engine cut off? It is apparent you disagree with the 4500 km achieved at that time, contending an unpowered ballistic object can continue gaining a significant amount of altitude after impetus is removed.
Majority of the mass is fuel? It's an aluminium tube with a 2000lb warhead at one end, and a rocket motor at the other. You work it out. What exactly do you think is inside the part between the motor and warhead? Its a fuel tank or, for a solid fuel motor, containment for the fuel mass.
You have no idea then? The materials necessary would need to be of such strength as to withstand the forces sustained.
In other words, soda can aluminum would not fit the bill and not just any aluminum tube.
Plus, where does the idea of solid fuel come from?
Take a shot at a figure.
Try UDMH weight per gallon, and high tensile strength aluminum and split it all up.
Altitude and velocity at engine shut-down?
No idea.
Thanks.
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Great.
So, next we have these two questions.
How many lbs of fuel contributed to the 159,000 lb weight of the missile?
What was the altitude achieved at t+5?
Fuel mass discussed elsewhere. Height at burnout of the order of 250km or thereabouts.
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Great.
So, next we have these two questions.
How many lbs of fuel contributed to the 159,000 lb weight of the missile?
What was the altitude achieved at t+5?
Fuel mass discussed elsewhere. Height at burnout of the order of 250km or thereabouts.
Thank you.
So, travelling at just under 3000 km/h at t+5 and having attained an altitude of only 250 km, do you think the missile will continue to rise to an altitude of 4500 km in over or under 53 minutes?
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Thanks WTF_S;
To confirm, my contention is that the missile launches at a mass of 159,000 lbs, the majority of which is fuel, and all consumed within the initial, single, 5-minute, burn, which accelerates it on a high angle trajectory where y>x.
You contend that 159,000 lbs consisted of a, "majority of which is fuel," based on "what exactly?"
Total burn time = 5 minutes.
Any idea on the altitude and rate of travel achieved by the time of engine cut off? It is apparent you disagree with the 4500 km achieved at that time, contending an unpowered ballistic object can continue gaining a significant amount of altitude after impetus is removed.
Majority of the mass is fuel? It's an aluminium tube with a 2000lb warhead at one end, and a rocket motor at the other. You work it out. What exactly do you think is inside the part between the motor and warhead? Its a fuel tank or, for a solid fuel motor, containment for the fuel mass.
You have no idea then? The materials necessary would need to be of such strength as to withstand the forces sustained.
In other words, soda can aluminum would not fit the bill and not just any aluminum tube. So now you are a stress engineer as well as a mathematician. Good to know. My guess would be 2024 specification aluminium , or 7075 for high-stress areas. Maybe some titanium to cope with heat stress. But, as I said, I'm not a rocket scientist so its just a guess. As for the soda cans; no idea again. Beverage Packaging Engineer is another training course I never graduated from.
Plus, where does the idea of solid fuel come from? Many missiles use solid fuel. Just mentioned it for completeness. North Korea's SLBM in current development is speculated to have a solid fuel booster.
Take a shot at a figure. I think this is your area of expertise.
Try UDMH weight per gallon, and high tensile strength aluminum and split it all up.
Altitude and velocity at engine shut-down?
No idea.
Thanks. Some people might consider this a weakness. Whilst it may be a precedent on the Internet if I don't know I'll tell you, rather than make up a number.
Edit. Sorry, I've just seen this in your response to Bob; 3000kph at T+5? What part of left-field did that come from? That's just a bit faster than Concorde with 100 passengers. We are talking about a space-rocket FFS. Do you have your own random number generator on these posts?
Edit again. I just worked out what you did; Bob's 250 km in 5 minutes, equates to a constant 3000 kph in your brain! Brilliant. You think it got to end of burn from a standing start at a constant 3000 kph! This is absolutely mind-blowing stuff. Was that a quadratic equation by the way? Keep the posts coming, please; this is better than Big Bang Theory.
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Maybe A80 needs to ratchet up his math a bit to differential equations. You have a rocket lifting off with a specific amount of thrust. Everything could be simple with a fixed amount of acceleration over a set time and you could do some quick calculations BUT; A rocket burns fuel at an enormous rate and the bulk of the rocket's mass is from fuel so every pound of fuel burned makes it easier for the rocket to accelerate. This means that the rate of acceleration is accelerating. The increase in speed per unit time (acceleration) will be much lower right after liftoff when the rocket is heavy with fuel. Right before the rocket is empty of fuel and is much lighter, the acceleration rate peaks out. The final velocity could be quite a bit higher than you might think. Additionally as the rocket leaves the atmosphere the air drag decreases, and as the rocket gets further from the center of the earth the force of gravity decreases a bit as well. Everything comes together to allow the rocket to accelerate at an ever increasing rate until burnout.
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Thanks WTF_S;
To confirm, my contention is that the missile launches at a mass of 159,000 lbs, the majority of which is fuel, and all consumed within the initial, single, 5-minute, burn, which accelerates it on a high angle trajectory where y>x.
You contend that 159,000 lbs consisted of a, "majority of which is fuel," based on "what exactly?"
Total burn time = 5 minutes.
Any idea on the altitude and rate of travel achieved by the time of engine cut off? It is apparent you disagree with the 4500 km achieved at that time, contending an unpowered ballistic object can continue gaining a significant amount of altitude after impetus is removed.
Majority of the mass is fuel? It's an aluminium tube with a 2000lb warhead at one end, and a rocket motor at the other. You work it out. What exactly do you think is inside the part between the motor and warhead? Its a fuel tank or, for a solid fuel motor, containment for the fuel mass.
You have no idea then? The materials necessary would need to be of such strength as to withstand the forces sustained.
In other words, soda can aluminum would not fit the bill and not just any aluminum tube. So now you are a stress engineer as well as a mathematician. Good to know. My guess would be 2024 specification aluminium , or 7075 for high-stress areas. Maybe some titanium to cope with heat stress. But, as I said, I'm not a rocket scientist so its just a guess. As for the soda cans; no idea again. Beverage Packaging Engineer is another training course I never graduated from.
Plus, where does the idea of solid fuel come from? Many missiles use solid fuel. Just mentioned it for completeness. North Korea's SLBM in current development is speculated to have a solid fuel booster.
Take a shot at a figure. I think this is your area of expertise.
Try UDMH weight per gallon, and high tensile strength aluminum and split it all up.
Altitude and velocity at engine shut-down?
No idea.
Thanks. Some people might consider this a weakness. Whilst it may be a precedent on the Internet if I don't know I'll tell you, rather than make up a number.
Edit. Sorry, I've just seen this in your response to Bob; 3000kph at T+5? What part of left-field did that come from? That's just a bit faster than Concorde with 100 passengers. We are talking about a space-rocket FFS. Do you have your own random number generator on these posts?
Edit again. I just worked out what you did; Bob's 250 km in 5 minutes, equates to a constant 3000 kph in your brain! Brilliant. You think it got to end of burn from a standing start at a constant 3000 kph! This is absolutely mind-blowing stuff. Was that a quadratic equation by the way? Keep the posts coming, please; this is better than Big Bang Theory.
Are you unfamiliar with the concept of expressing average velocity over time?
And I would appreciate it if you point to where I claimed quadratic functions apply to anything else other than trajectory calculations.
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Maybe A80 needs to ratchet up his math a bit to differential equations. You have a rocket lifting off with a specific amount of thrust. Everything could be simple with a fixed amount of acceleration over a set time and you could do some quick calculations BUT; A rocket burns fuel at an enormous rate and the bulk of the rocket's mass is from fuel so every pound of fuel burned makes it easier for the rocket to accelerate. This means that the rate of acceleration is accelerating. The increase in speed per unit time (acceleration) will be much lower right after liftoff when the rocket is heavy with fuel. Right before the rocket is empty of fuel and is much lighter, the acceleration rate peaks out. The final velocity could be quite a bit higher than you might think. Additionally as the rocket leaves the atmosphere the air drag decreases, and as the rocket gets further from the center of the earth the force of gravity decreases a bit as well. Everything comes together to allow the rocket to accelerate at an ever increasing rate until burnout.
Absolutely correct.
Bringing me back to this:
" 1) While serving as a US Merchant Marine officer aboard a cargo ship transiting between the Korean Peninsula and Japan there were numerous alarms and warnings going off on the bridge of our ship.
2) Several broadcast messages to all ships at sea were received from the Japanese Coast Guard authorities regarding the immanent and later the actual launch of a North Korean ICBM.
3) Since the North Koreans don't tell everyone in advance what is happening all we can do is wait.
4) The missile launch in question was at night. Skies were clear and a missile was clearly seen traveling upwards then over the top (forward of the bow) of our ship."
So, to be perfectly clear, you saw the NK ICBM, engine ablaze, already making a clear arc across the night sky?
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Maybe A80 needs to ratchet up his math a bit to differential equations. You have a rocket lifting off with a specific amount of thrust. Everything could be simple with a fixed amount of acceleration over a set time and you could do some quick calculations BUT; A rocket burns fuel at an enormous rate and the bulk of the rocket's mass is from fuel so every pound of fuel burned makes it easier for the rocket to accelerate. This means that the rate of acceleration is accelerating. The increase in speed per unit time (acceleration) will be much lower right after liftoff when the rocket is heavy with fuel. Right before the rocket is empty of fuel and is much lighter, the acceleration rate peaks out. The final velocity could be quite a bit higher than you might think. Additionally as the rocket leaves the atmosphere the air drag decreases, and as the rocket gets further from the center of the earth the force of gravity decreases a bit as well. Everything comes together to allow the rocket to accelerate at an ever increasing rate until burnout.
Absolutely correct.
Bringing me back to this:
" 1) While serving as a US Merchant Marine officer aboard a cargo ship transiting between the Korean Peninsula and Japan there were numerous alarms and warnings going off on the bridge of our ship.
2) Several broadcast messages to all ships at sea were received from the Japanese Coast Guard authorities regarding the immanent and later the actual launch of a North Korean ICBM.
3) Since the North Koreans don't tell everyone in advance what is happening all we can do is wait.
4) The missile launch in question was at night. Skies were clear and a missile was clearly seen traveling upwards then over the top (forward of the bow) of our ship."
So, to be perfectly clear, you saw the NK ICBM, engine ablaze, already making a clear arc across the night sky?
Yes, that's what we believed we saw.
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@A80;
Are you unfamiliar with the concept of expressing average velocity over time?
Are you familiar with the concepts of momentum and kinetic energy? They are completely unrelated to average speed, so why quote that?
And I would appreciate it if you point to where I claimed quadratic functions apply to anything else other than trajectory calculations.
The point was that, over the last dozen pages, you have mentioned "quadratic equations" a gazillion times, but not actually presented one of your own. Are you familiar with the concept of irony or, like Kim's missile, has it gone clean over your head?
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While Total Lackey continues to be a contrarian let’s all remind ourselves that he agrees that rockets work. What exactly the problem with a large ballistic missile vs a smaller cruise missile is unclear.
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Are you unfamiliar with the concept of expressing average velocity over time?
Great - now we're getting somewhere, as you are at last explaining how you're getting to your erroneous statements. Average velocity would only work if the rocket's acceleration was constant. But it is far from constant. At launch, there is only a small amount of excess thrust over the weight of the rocket. But close to Hbo, the rocket is a lot lighter and yet still has the same thrust. So most of the accelerating happens in the latter stages of the burn. The Hwasong 15, for example, was estimated to have reached 7.17km/s at the end of its burn. At that velocity, it would take around 12 minutes to decelerate to 0 at 9.81ms-2
At which point you'll probably say 'so how come it flew for 50 minutes? Surely 12 x 2 = 24?'
Which then leads us to the other complicating factor that you aren't considering, which is the progressive reduction in g as you get further away from earth. At 4000km, for example, the ICBM would only experience a g of 3.7ms-2. That's why the calculations get complex very quickly - you've got variable mass and g, and a rotating planet.
Of course, I guess you disagree with the rotating round planet bit, and probably the reducing g bit as well. That's fine...but the burden then falls on you to explain what exactly did happen to the rocket if it flew for 50 odd minutes and only went 950km from the launch site, and how far it would be able to travel if it was launched at a shallower angle.
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The fact that the rocket experiences less G forces at higher altitudes kind of blows up the UA argument here doesn't it? It looks like the Wiki needs a whole lot of revisions. Maybe someone could scam up a value for the Bishop Constant.
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While Total Lackey continues to be a contrarian let’s all remind ourselves that he agrees that rockets work. What exactly the problem with a large ballistic missile vs a smaller cruise missile is unclear.
I think it's the 'intercontinental' aspect of the ICBM that he would have an issue with since he can't explain the size and positions of the major continents.
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I think the "missile" aspect is also a little challenging.
He claims some experience of witnessing a Tomahawk; "I cannot get into any particulars, except to state it was testing".
A cruise missile, of course is fundamentally different to most other guided missiles. A solid-fuel booster accelerates it to cruise speed, where a turbine engine takes over and propels it at a fairly constant sub-sonic speed for the next several hours under aerodynamic control to its target. And the cruise engine keeps running right up to impact. This was the design of the WW2 German V1, and every cruise missile since.
I'm guessing he may also have difficulty with the concept of, for instance AAMs like AIM-9 Sidewinder, AIM-7 Sparrow and early AIM-32 ASRAAM, where the solid-fuel rocket motor accelerates it rapidly to multi-mach numbers within 2 or 3 seconds, and it glides the rest of the way to its target. Judging from his posts, he seems to think that a (obviously fictional) ICBM motor should run until, at least, apogee.
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Are you unfamiliar with the concept of expressing average velocity over time?
Great - now we're getting somewhere, as you are at last explaining how you're getting to your erroneous statements. Average velocity would only work if the rocket's acceleration was constant. But it is far from constant. At launch, there is only a small amount of excess thrust over the weight of the rocket. But close to Hbo, the rocket is a lot lighter and yet still has the same thrust. So most of the accelerating happens in the latter stages of the burn. The Hwasong 15, for example, was estimated to have reached 7.17km/s at the end of its burn. At that velocity, it would take around 12 minutes to decelerate to 0 at 9.81ms-2
At which point you'll probably say 'so how come it flew for 50 minutes? Surely 12 x 2 = 24?'
Which then leads us to the other complicating factor that you aren't considering, which is the progressive reduction in g as you get further away from earth. At 4000km, for example, the ICBM would only experience a g of 3.7ms-2. That's why the calculations get complex very quickly - you've got variable mass and g, and a rotating planet.
Of course, I guess you disagree with the rotating round planet bit, and probably the reducing g bit as well. That's fine...but the burden then falls on you to explain what exactly did happen to the rocket if it flew for 50 odd minutes and only went 950km from the launch site, and how far it would be able to travel if it was launched at a shallower angle.
I completely understand what you wrote.
I also completely understand that a rocket will completely accelerate until the end of its burn, at which point it will cease to continue the process of acceleration.
However, we are still at t+5, and we are still left with 250km altitude. and we are still left with the average rate of acceleration at 3,000 km/h.
So, splitting down to minutes, t+1, t+2,etc, given, for instance, the missile could obtain 1,000 km/h at t+1, 2,000 km/h at t+2, etc., up to 5,000 km/h, in order to average 3,000 km/h. Other values are possible over the span of 5 minutes, but the sum of these values/t cannot exceed 3,000 km/h.
So, how does a missile without engine burn, as you stated, accelerate to gain 4250 km in altitude within 48 minutes, fighting against g?
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You are completely missing the point. Kinetic energy and momentum have nothing to do with average speed, and even less to do with "average acceleration". The only relevant number is its instantaneous velocity at the time of engine shutdown. Do you not understand anything about physics? If the average acceleration equals 3000 kmph (what?), do you not understand that the RATE of acceleration has been increasing exponentially for the entire 5 minute burn due to the decreasing mass of the projectile?
Sorry to appeal for assistance, but can anyone out there help him with this?
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You are completely missing the point. Kinetic energy and momentum have nothing to do with average speed, and even less to do with "average acceleration". The only relevant number is its instantaneous velocity at the time of engine shutdown. Do you not understand anything about physics? If the average acceleration equals 3000 kmph (what?), do you not understand that the RATE of acceleration has been increasing exponentially for the entire 5 minute burn due to the decreasing mass of the projectile?
Sorry to appeal for assistance, but can anyone out there help him with this?
I understand completely what you are writing.
If you are trying to claim the physics somehow does away with average rate of acceleration in this instance, however, that would be faulty.
In other words, you want to now claim that certain parameters of the entire launch, such as being in the right position at the right time, would be useless.
I understand that any set of acceleration values could be substituted at every minute, or even every second, or even every millisecond, of the 5 minute burn to an altitude of 250km, as long as the sum of the values/t = 3,000km/h.
So go ahead.
Provide a value at engine cut off that will suffice to meet the obvious requirement of 250km altitude at t+5, averaging 3,000 km/h over the 5 minute burn, and continue to accelerate the missile and cause the missile to achieve the span ranging from 250 km to 4500 km in less than 53 minutes.
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Do you mean 3,000 km/h2?
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Do you mean 3,000 km/h2?
5 minutes were spent from takeoff (0) to an altitude of 250 km.
So, no.
d=rt, requires no squaring of any values.
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Do you mean 3,000 km/h2?
5 minutes were spent from takeoff (0) to an altitude of 250 km.
So, no.
d=rt, requires no squaring of any values.
You are granted another chance and a clue:
we are still left with the average rate of acceleration at 3,000 km/h
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Action80 math : where the numbers are made up and the units dont matter.
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Do you mean 3,000 km/h2?
5 minutes were spent from takeoff (0) to an altitude of 250 km.
So, no.
d=rt, requires no squaring of any values.
You are granted another chance and a clue:
we are still left with the average rate of acceleration at 3,000 km/h
Courtesy of
The Hwasong 15, for example, was estimated to have reached 7.17km/s at the end of its burn. At that velocity, it would take around 12 minutes to decelerate to 0 at 9.81ms-2
If you see, 7.17km/s2, anywhere mentioned by him in his posts, then feel free to keep giving me passes or if not, take one for yourself.
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Oh Jesus. 7.17km/s is a velocity. As he says in his quote. They have different units than acceleration. Slow down. Take a breath.
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Oh Jesus. 7.17km/s is a velocity. As he says in his quote. They have different units than acceleration. Slow down. Take a breath.
If you understand enough to correct what it should be to your own mind, then run with it.
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Oh Jesus. 7.17km/s is a velocity. As he says in his quote. They have different units than acceleration. Slow down. Take a breath.
If you understand enough to correct what it should be to your own mind, then run with it.
This is you objectively fucking up and then doubling down on it. What will it take for you to understand that you aren’t currently knowledgeable enough to analyze this issue?
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Oh Jesus. 7.17km/s is a velocity. As he says in his quote. They have different units than acceleration. Slow down. Take a breath.
If you understand enough to correct what it should be to your own mind, then run with it.
This is you objectively fucking up and then doubling down on it. What will it take for you to understand that you aren’t currently knowledgeable enough to analyze this issue?
The rate I put forth is fine. I mistakenly put the word acceleration instead of the word velocity.
Kindly pat yourself on the back, take two if you care, my apology for using the wrong word in this case, and have a great rest of your day.
Before you go, if you can explain how a 16,038 mile per hour velocity at t+5 could possibly translate into an average rate of 3,000 mph over 5 minutes, that would be terrific.
ETA: To totally satisfy what appears to be a certain need for perfection in others, 0 - 16,038 mph translates to an acceleration of 23.9 m/s2.
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Oh Jesus. 7.17km/s is a velocity. As he says in his quote. They have different units than acceleration. Slow down. Take a breath.
If you understand enough to correct what it should be to your own mind, then run with it.
This is you objectively fucking up and then doubling down on it. What will it take for you to understand that you aren’t currently knowledgeable enough to analyze this issue?
The rate I put forth is fine. I mistakenly put the word acceleration instead of the word velocity.
Kindly pat yourself on the back, take two if you care, my apology for using the wrong word in this case, and have a great rest of your day.
I will take that as an apology for trying to impugn Steely Bob while it was in fact your error as well.
Before you go, if you can explain how a 16,038 mile per hour velocity at t+5 could possibly translate into an average rate of 3,000 mph over 5 minutes, that would be terrific.
I can't perform the calculation because it has been 25 years since I have done calculus, but as a possibility, you would want to find a plot where the jerk creates an acceleration curve that derives an average velocity of 3,000 mph. I am not sure why you would think it's impossible.
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I completely understand what you wrote.
I also completely understand that a rocket will completely accelerate until the end of its burn, at which point it will cease to continue the process of acceleration.
However, we are still at t+5, and we are still left with 250km altitude. and we are still left with the average rate of acceleration at 3,000 km/h.
So, splitting down to minutes, t+1, t+2,etc, given, for instance, the missile could obtain 1,000 km/h at t+1, 2,000 km/h at t+2, etc., up to 5,000 km/h, in order to average 3,000 km/h. Other values are possible over the span of 5 minutes, but the sum of these values/t cannot exceed 3,000 km/h.
Let's be ultra-precise with our terminology here. 'average rate of acceleration at 3000km/h' is meaningless I'm afraid. I think what you mean is 'average vertical velocity of 3000km/h', because the displacement, s, is 250km and the time, t, is 1 hour. But whilst the average velocity is absolutely 3000km/h, as I said above, there is no requirement for velocity to build in a linear fashion as you describe. Thrust is constant, mass is reducing. Acceleration is slow initially, and rapid towards the end. The final velocity is more like 25,000km/h, but that doesn't mean the average velocity isn't 3000km/h, because it builds rapidly in the latter stages. It's not a straight line.
So, how does a missile without engine burn, as you stated, accelerate to gain 4250 km in altitude within 48 minutes, fighting against g?
As I showed above. If the final velocity is around the 7km/s region, then it would take around 12 minutes to decelerate to 0 at g=9.81ms-2. But g isn't constant either, which is why it takes a lot longer.
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Oh Jesus. 7.17km/s is a velocity. As he says in his quote. They have different units than acceleration. Slow down. Take a breath.
If you understand enough to correct what it should be to your own mind, then run with it.
This is you objectively fucking up and then doubling down on it. What will it take for you to understand that you aren’t currently knowledgeable enough to analyze this issue?
The rate I put forth is fine. I mistakenly put the word acceleration instead of the word velocity.
Kindly pat yourself on the back, take two if you care, my apology for using the wrong word in this case, and have a great rest of your day.
Before you go, if you can explain how a 16,038 mile per hour velocity at t+5 could possibly translate into an average rate of 3,000 mph over 5 minutes, that would be terrific.
ETA: To totally satisfy what appears to be a certain need for perfection in others, 0 - 16,038 mph translates to an acceleration of 23.9 m/s2.
You need to understand the difference between an average and an absolute. The average family has 2.4 children. So does an average family exist?
In physics, averages might make for interesting comparison, but are no basis for calculation. The only considerations in calculating the energy state at a particular instant are its instantaneous position and velocity.
Consider this; Urbanville and Townsville and are 60 miles apart. They are served by a train which does the journey in one hour. How the f*** are you supposed to get off the train in Townsville if it is doing an average of 60 mph?
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Consider this; Urbanville and Townsville and are 60 miles apart. They are served by a train which does the journey in one hour. How the f*** are you supposed to get off the train in Townsville if it is doing an average of 60 mph?
And what if the train does the first half of the journey at 30mph? How fast does it have to go for the second half in order to travel the whole distance at an average of 60mph?
Intuitively one would thing it's 90mph, when that actual answer is it's impossible. Some of this stuff is counter-intuitive.
Frankly, Lackey, this level of maths is beyond you. And that isn't a slight, it's beyond me too. It's beyond most people. As a rocket rises the thrust remains (fairly) constant but it's burning fuel which means the acceleration is constantly increasing (because F=ma, the F of the thrust remains the same but the m constantly changes as fuel is burnt). Then you have the drag which reduces with altitude as the air gets thinner. And there's a gravitational effect too although that's probably small enough to not have to worry about it. Look at any rocket - it doesn't suddenly zoom into the air, it starts off going pretty slowly, the higher it goes and the lighter it gets the more it accelerates. The rate of acceleration is not constant, that's what makes this so complex.
TL;DR - this stuff is complicated. And it's fine to not understand it, just try not to overestimate your ability to.
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Oh Jesus. 7.17km/s is a velocity. As he says in his quote. They have different units than acceleration. Slow down. Take a breath.
If you understand enough to correct what it should be to your own mind, then run with it.
This is you objectively fucking up and then doubling down on it. What will it take for you to understand that you aren’t currently knowledgeable enough to analyze this issue?
The rate I put forth is fine. I mistakenly put the word acceleration instead of the word velocity.
Kindly pat yourself on the back, take two if you care, my apology for using the wrong word in this case, and have a great rest of your day.
I will take that as an apology for trying to impugn Steely Bob while it was in fact your error as well.
I fail to see how utilizing SteelyBob's quote is "impugning" SteelyBob.
Before you go, if you can explain how a 16,038 mile per hour velocity at t+5 could possibly translate into an average rate of 3,000 mph over 5 minutes, that would be terrific.
I can't perform the calculation because it has been 25 years since I have done calculus, but as a possibility, you would want to find a plot where the jerk creates an acceleration curve that derives an average velocity of 3,000 mph. I am not sure why you would think it's impossible.
Well, a curve in acceleration in this case would be the commencement of a deviation from the straight line trajectory to that of a parabola. Which it was already launched in order to form. So again, 16,058 mph at t+5 and starting at 0mph, cannot possibly translate to a traveled total distance of 250km, given d=rt.
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I completely understand what you wrote.
I also completely understand that a rocket will completely accelerate until the end of its burn, at which point it will cease to continue the process of acceleration.
However, we are still at t+5, and we are still left with 250km altitude. and we are still left with the average rate of acceleration at 3,000 km/h.
So, splitting down to minutes, t+1, t+2,etc, given, for instance, the missile could obtain 1,000 km/h at t+1, 2,000 km/h at t+2, etc., up to 5,000 km/h, in order to average 3,000 km/h. Other values are possible over the span of 5 minutes, but the sum of these values/t cannot exceed 3,000 km/h.
Let's be ultra-precise with our terminology here. 'average rate of acceleration at 3000km/h' is meaningless I'm afraid. I think what you mean is 'average vertical velocity of 3000km/h', because the displacement, s, is 250km and the time, t, is 1 hour. But whilst the average velocity is absolutely 3000km/h, as I said above, there is no requirement for velocity to build in a linear fashion as you describe.
Nowhere did I state that velocity is linear in this case.
I simply posted the result of d=rt, which still applies in this case. It applies because we have a given distance (by you) and we have a given time of thrust applied (also by you).
Thrust is constant, mass is reducing.
Of course.
Acceleration is slow initially, and rapid towards the end. The final velocity is more like 25,000km/h, but that doesn't mean the average velocity isn't 3000km/h, because it builds rapidly in the latter stages. It's not a straight line.
Okay, if you read the post, I gave you the 25,000km/h at t+5.
Now, starting from 0 km/h and finishing at 16,038 mph, how does that = 250km total distance traveled in 5 minutes?
So, how does a missile without engine burn, as you stated, accelerate to gain 4250 km in altitude within 48 minutes, fighting against g?
As I showed above. If the final velocity is around the 7km/s region, then it would take around 12 minutes to decelerate to 0 at g=9.81ms-2. But g isn't constant either, which is why it takes a lot longer.
I do not think you understand the implications.
A missile will most certainly gain some additional altitude without power, but certainly not 4250km, even battling 9.08g/s2.
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Oh Jesus. 7.17km/s is a velocity. As he says in his quote. They have different units than acceleration. Slow down. Take a breath.
If you understand enough to correct what it should be to your own mind, then run with it.
This is you objectively fucking up and then doubling down on it. What will it take for you to understand that you aren’t currently knowledgeable enough to analyze this issue?
The rate I put forth is fine. I mistakenly put the word acceleration instead of the word velocity.
Kindly pat yourself on the back, take two if you care, my apology for using the wrong word in this case, and have a great rest of your day.
Before you go, if you can explain how a 16,038 mile per hour velocity at t+5 could possibly translate into an average rate of 3,000 mph over 5 minutes, that would be terrific.
ETA: To totally satisfy what appears to be a certain need for perfection in others, 0 - 16,038 mph translates to an acceleration of 23.9 m/s2.
You need to understand the difference between an average and an absolute. The average family has 2.4 children. So does an average family exist?
In physics, averages might make for interesting comparison, but are no basis for calculation. The only considerations in calculating the energy state at a particular instant are its instantaneous position and velocity.
Consider this; Urbanville and Townsville and are 60 miles apart. They are served by a train which does the journey in one hour. How the f*** are you supposed to get off the train in Townsville if it is doing an average of 60 mph?
Consider this; Urbanville and Townsville and are 60 miles apart. They are served by a train which does the journey in one hour. How the f*** are you supposed to get off the train in Townsville if it is doing an average of 60 mph?
And what if the train does the first half of the journey at 30mph? How fast does it have to go for the second half in order to travel the whole distance at an average of 60mph?
Intuitively one would thing it's 90mph, when that actual answer is it's impossible. Some of this stuff is counter-intuitive.
Frankly, Lackey, this level of maths is beyond you. And that isn't a slight, it's beyond me too. It's beyond most people. As a rocket rises the thrust remains (fairly) constant but it's burning fuel which means the acceleration is constantly increasing (because F=ma, the F of the thrust remains the same but the m constantly changes as fuel is burnt). Then you have the drag which reduces with altitude as the air gets thinner. And there's a gravitational effect too although that's probably small enough to not have to worry about it. Look at any rocket - it doesn't suddenly zoom into the air, it starts off going pretty slowly, the higher it goes and the lighter it gets the more it accelerates. The rate of acceleration is not constant, that's what makes this so complex.
TL;DR - this stuff is complicated. And it's fine to not understand it, just try not to overestimate your ability to.
If I was actually making a claim for a calculation here in terms of the rocketry, you both might have a point.
However, I am simply analyzing a portion of the provided numbers.
d=rt is a given and no amount of physics or rocket surgery makes that disappear.
Neither does the misguided question of how you get off of a train at speed negate the given reality provided by SteelyBob.
The fact remains that it has been claimed that a 250km distance was traveled in 5 minutes.
That translates to an average rate of travel of 3000km/h, which is a fact.
Given a starting rate of travel of 0 mph and a final rate of travel of 16,038 mph, kindly make that fit.
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The fact remains that it has been claimed that a 250km distance was traveled in 5 minutes.
That translates to an average rate of travel of 3000 mph, which is a fact.
Given a starting rate of travel of 0 mph and a final rate of travel of 16,038 mph, kindly make that fit.
It would be a fact if you weren't mixing up your units. But OK, I've had a go.
Minute 1 - rocket accelerates from 0-600 - Average speed = 300km/h, distance travelled = 5km
Minute 2 - rocket accelerates from 600-1200 - Average speed = 900km/h, distance travelled = 15km
Minute 3 - rocket accelerates from 1200-2000- Average speed = 1600km/h, distance travelled = 26.66km
Minute 4 - rocket accelerates from 2000-4000- Average speed = 3000km/h, distance travelled = 50km
I'm going to split the last minute into two 30 second blocks as this is where the acceleration is highest
Minute 5 - first 30 seconds. Rocket accelerates from 4000-8000- Average speed = 6000km/h, distance travelled = 50km (because this is 30 seconds, not 60, so divide by 120)
Minute 5 - second 30 seconds. Rocket accelerates from 8000-16000- Average speed = 12000km/h, distance travelled = 100km
Distance travelled = 246.66km. I'd say that's close enough.
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If I was actually making a claim for a calculation here in terms of the rocketry, you both might have a point.
However, I am simply analyzing a portion of the provided numbers.
d=rt is a given and no amount of physics or rocket surgery makes that disappear.
Neither does the misguided question of how you get off of a train at speed negate the given reality provided by SteelyBob.
The fact remains that it has been claimed that a 250km distance was traveled in 5 minutes.
That translates to an average rate of travel of 3000 mph, which is a fact.
Given a starting rate of travel of 0 mph and a final rate of travel of 16,038 mph, kindly make that fit.
AATW is spot on. If you imagine plotting the velocity (y-axis) against time (x-axis) on a graph, then to find the distance travelled you need to find the area under the graph. If acceleration is constant, meaning the line on the graph is linear, then it's a straightforward calculation: s=ut + 1/2 at2, where u = initial velocity, which is zero in this case, leaving just the second term. We normally use 's', meaning displacement, instead of distance 'd' in mechanics problems, but the meaning is similar here. The problem is that 'a' is not constant in the case of our rocket, as you yourself have agreed, so we have to define a function for velocity with respect to time, and then integrate that in order to find the area under the graph - it's not a straightforward calculation, and if the function is complex, then we might even need to use numerical methods to solve it.
There is no reason why an object travelling at 16,000mph (I think our units are drifting here, but never mind - we're talking about a principle) can't average 3000mph over the previous 5 minutes - if there's a substantial upslope to the velocity profile then most of the distance covered will be in the last stages of the time period.
I'm not sure how we can make this any clearer.
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Pardon me for not reading 13 pages of replies, but coming down to the end of this, it seems to me that none of this is necessary. The target of an ICBM will be in a very different place if the earth is flat or round. All that is necessary for the question of FE vs RE is to observe that if you get the shape of the earth wrong, the missile will miss by hundreds of miles. The US has multiple test ranges and tracks missiles with radar, gps, and hydrophone grid at landing site. Russia shoots them at Siberia and can see where they landed. They want to get the CEP (circular error probability) down as low as possible (certainly less than a mile) to hit hardened targets.
The equation for a multi-stage rocket would be very complex, and I believe they have in flight guidance control of some kind. There is a gap between stages firing, the stages have different power and weight. It is not physics 101 textbook parabola. They are MIRVed and different warheads land different places.
The North Korea missiles went almost straight up. To know how far they can go if aimed more horizontally, one must know the shape of the earth.
Why do REs here go down the rabbit hole of arguing exact equations etc when there is a simple principle that will determine the point being debated? Complexity obscures, and RE should be looking for clarity and simplicity. Sometimes I think this is not about RE or FE but who has the most detailed scientific knowledge and math ability.
My point was that missiles that go a thousand miles or more will land in very different places on FE or RE. The people who test them have spent huge amounts of money tracking and instrumenting to know exactly where they are landing. Multiple countries have done this. There are 3 possibilities.
1. They know earth is round.
2. They know earth is flat.
3. All such systems have failed and none of the countries that desperately want to have their missiles hit a target have succeeded.
I do not care what the equation is. I care whether there is a giant worldwide conspiracy, required for case 2 or 3. Case 1 is ICBMs work and RE is true. Case 2 is ICBMs work and everyone involved with knows the earth is flat. Case 3 is everyone is an idiot and desperately trying to look like a genius, in which case it puzzles me that they could be so incompetent at aiming and so brilliant at faking it.
Do you think that no one knows where ICBMs will land? Because if someone knows, they know the shape of the earth. Multiple ranges and organizations in the US and more in other countries. Tracked by radar and gps and landing area instrumented with hydrophones.
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Pardon me for not reading 13 pages of replies, but coming down to the end of this, it seems to me that none of this is necessary. The target of an ICBM will be in a very different place if the earth is flat or round. All that is necessary for the question of FE vs RE is to observe that if you get the shape of the earth wrong, the missile will miss by hundreds of miles. The US has multiple test ranges and tracks missiles with radar, gps, and hydrophone grid at landing site. Russia shoots them at Siberia and can see where they landed. They want to get the CEP (circular error probability) down as low as possible (certainly less than a mile) to hit hardened targets.
The equation for a multi-stage rocket would be very complex, and I believe they have in flight guidance control of some kind. There is a gap between stages firing, the stages have different power and weight. It is not physics 101 textbook parabola. They are MIRVed and different warheads land different places.
The North Korea missiles went almost straight up. To know how far they can go if aimed more horizontally, one must know the shape of the earth.
Why do REs here go down the rabbit hole of arguing exact equations etc when there is a simple principle that will determine the point being debated? Complexity obscures, and RE should be looking for clarity and simplicity. Sometimes I think this is not about RE or FE but who has the most detailed scientific knowledge and math ability.
My point was that missiles that go a thousand miles or more will land in very different places on FE or RE. The people who test them have spent huge amounts of money tracking and instrumenting to know exactly where they are landing. Multiple countries have done this. There are 3 possibilities.
1. They know earth is round.
2. They know earth is flat.
3. All such systems have failed and none of the countries that desperately want to have their missiles hit a target have succeeded.
I do not care what the equation is. I care whether there is a giant worldwide conspiracy, required for case 2 or 3. Case 1 is ICBMs work and RE is true. Case 2 is ICBMs work and everyone involved with knows the earth is flat. Case 3 is everyone is an idiot and desperately trying to look like a genius, in which case it puzzles me that they could be so incompetent at aiming and so brilliant at faking it.
Do you think that no one knows where ICBMs will land? Because if someone knows, they know the shape of the earth. Multiple ranges and organizations in the US and more in other countries. Tracked by radar and gps and landing area instrumented with hydrophones.
The fact that people who are wrong about something can achieve great things does absolutely nothing to make them right on those other things.
You have to remember that correct science is not necessary, if fact it appears more and more that it is incorrect science that is necessary, for complicated things to simply just work.
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The issue is engine burn of five minutes.
The issue is distance traveled of 250km in five minutes.
Average rate of travel to achieve 250km is 3000/kmh.
AATW proposal appears to exceed that number.
Additionally, only one engine in the missile.
After burnout, how does it possibly continue to gain altitude in order to achieve an additional 4250km to apogee?
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After burnout, how does it possibly continue to gain altitude in order to achieve an additional 4250km to apogee?
Same way a baseball can fly upwards after only being hit once? It’s acceleration. Very strange that you don’t know this.
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After burnout, how does it possibly continue to gain altitude in order to achieve an additional 4250km to apogee?
Same way a baseball can fly upwards after only being hit once? It’s acceleration. Very strange that you don’t know this.
Sure I know that.
I also know it will not cover an additional 4250km, as it will start deceleration once burn out has occurred.
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The issue is engine burn of five minutes.
The issue is distance traveled of 250km in five minutes.
Average rate of travel to achieve 250km is 3000/kmh.
AATW proposal appears to exceed that number.
246.66 is less than 250, because counting.
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The issue is engine burn of five minutes.
The issue is distance traveled of 250km in five minutes.
Average rate of travel to achieve 250km is 3000/kmh.
AATW proposal appears to exceed that number.
246.66 is less than 250, because counting.
Yes.
246.66 is 3.34 less than 250.
Because it only worsens your position.
But thanks.
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Can you elaborate? You asked me to "make it fit", if you're going to quibble about a few km then I'd suggest that's just you being pretty disingenuous.
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Can you elaborate? You asked me to "make it fit", if you're going to quibble about a few km then I'd suggest that's just you being pretty disingenuous.
I am not quibbling over your numbers.
You did that.
You will see your numbers do not fit.
Analyze it.
Total distance - 246.66km
Total time traveled to attain distance - 5 min
Rate of travel = Slightly less than 3000km/h.
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*average velocity 3,000km/h. Not necessarily the velocity at t=5.
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*average velocity 3,000km/h. Not necessarily the velocity at t=5.
I understand that.
Just like an auto race, the guy could be flying well past or limping across the finish line at well below the average time necessary to have won the race.
In this case, the breakdown AATW provided surpasses the average rate of travel over that distance.
So it doesn't work.
And the missile will commence deceleration almost immediately upon burn out.
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Yes it will decelerate but all it needs is to get to the apogee of its flight path.
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^ Correct. It only needs to attain an additional 4250 km to apogee while under no continued impetus. And in accordance with RE, subjected to slightly over 9m/s2 at that altitude.
No more engine to maintain a required 8500km/h average rate of travel for let's say 30 minutes, against that immediate opposing force.
Once engine burn out occurs, a missile will commence deceleration (i.e., slow down its rate of travel).
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Yes. That’s right. Now, care to show your complete calculation showing it’s impossible?
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^ Correct. It only needs to attain an additional 4250 km to apogee while under no continued impetus. And in accordance with RE, subjected to slightly over 9m/s2 at that altitude.
No more engine to maintain a required 8500km/h average rate of travel for let's say 30 minutes, against that immediate opposing force.
Once engine burn out occurs, a missile will commence deceleration (i.e., slow down its rate of travel).
You don't appear to be reading anybody's posts. I, and several others, have patiently explained this over several posts, such as mine here - https://forum.tfes.org/index.php?topic=18167.msg240697#msg240697 (https://forum.tfes.org/index.php?topic=18167.msg240697#msg240697) - which you haven't acknowledged or engaged with in any way.
You are making repeated errors with both minor facts/numbers (such as g=9ms-2 at 4250km, whereas its actually less than half that number, as per another of my previous posts that you've ignored), and with major conceptual things, such as the fact that the average velocity over a period of time can easily be far greater than the mean of the start and end velocities if the velocity profile is non-linear.
Are you just trolling, or are you genuinely struggling with this?
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^ Correct. It only needs to attain an additional 4250 km to apogee while under no continued impetus. And in accordance with RE, subjected to slightly over 9m/s2 at that altitude.
No more engine to maintain a required 8500km/h average rate of travel for let's say 30 minutes, against that immediate opposing force.
Once engine burn out occurs, a missile will commence deceleration (i.e., slow down its rate of travel).
You don't appear to be reading anybody's posts. I, and several others, have patiently explained this over several posts, such as mine here - https://forum.tfes.org/index.php?topic=18167.msg240697#msg240697 (https://forum.tfes.org/index.php?topic=18167.msg240697#msg240697) - which you haven't acknowledged or engaged with in any way.
You are making repeated errors with both minor facts/numbers (such as g=9ms-2 at 4250km, whereas its actually less than half that number, as per another of my previous posts that you've ignored), and with major conceptual things, such as the fact that the average velocity over a period of time can easily be far greater than the mean of the start and end velocities if the velocity profile is non-linear.
Are you just trolling, or are you genuinely struggling with this?
You should read the post again.
I never claimed g=9m/s2 at 4250km.
Why are you misrepresenting my posts?
But, just in case you are not and cannot understand what the issue is:
Specifically, at 250km, g=is slightly above 9m/s2.
Now, according to you, a missile has burnt out its fuel and the engine has ceased functioning.
It immediately commences deceleration.
Yet, it needs to achieve an additional 4250km in altitude to reach apogee.
Not gonna happen.
In addition, race car velocity profiles are also decidedly non-linear. A race car can certainly be screaming full throttle or be limping across the finish line at the end of the trip. In the end, the race is decided by who completes the required miles in the least amount of time and that is determined by average speed over the entire required course. d=rt. And you still seem to think my issue is with d=rt being far greater.
You should look again.
So your argument concerning that issue is illegitimate and specious.
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You should read the post again.
I never claimed g=9m/s2 at 4250km.
Well what exactly is this sentence saying then?:
It only needs to attain an additional 4250 km to apogee while under no continued impetus. And in accordance with RE, subjected to slightly over 9m/s2 at that altitude.
There's no other altitudes mentioned in your post, or indeed the preceding one that you seemed to be responding to.
Why are you misrepresenting my posts?
I'm certainly not trying to do that - you must surely see how that might be misinterpreted? Still, you are at least acknowledging that g is, according to generally accepted scientific consensus (or 'RE', if you like) reducing all the time with increasing altitude.
But, just in case you are not and cannot understand what the issue is:
Specifically, at 250km, g=is slightly above 9m/s2.
Now, according to you, a missile has burnt out its fuel and the engine has ceased functioning.
It immediately commences deceleration.
Yep, agreed
Yet, it needs to achieve an additional 4250km in altitude to reach apogee.
Again, yes - agreed.
Not gonna happen.
Why not? It's fine to disagree, but at this point you've got to offer something up by way of calculation. I've linked you to a really good website that shows in some detail how the calculations work. It's far from easy maths, because the force due to gravity acting on the missile is reducing all the time. You could start by simplifying it - how high would the missile go if g was constant at, say 9ms-2? As maths teachers the world over like to say...show your working too, please.
In addition, race car velocity profiles are also decidedly non-linear. A race car can certainly be screaming full throttle or be limping across the finish line at the end of the trip. In the end, the race is decided by who completes the required miles in the least amount of time and that is determined by average speed over the entire required course. d=rt. And you still seem to think my issue is with d=rt being far greater.
The reason people keep calling you out on this is that the only thing that affects the missile's final trajectory from hbo onwards is its velocity at that point - 250km in this example. It's not really clear why you think a particular average speed over a distance precludes a particular velocity at the end of that distance/time. To give an example with some more down-to-earth numbers, imagine driving a Ferrari at exactly 30mph for 59 minutes and 50 secondhand then putting your foot down in the last 10 seconds. Your final speed will be very high - well over 100mph, let's say - but your average speed will be as near as doesn't matter to 30mph.
You should look again.
So your argument concerning that issue is illegitimate and specious.
If you want to progress the debate, you need to actually offer something up by way of counter-argument. You just seem to be saying 'no it isn't' to everything, without actually presenting any calculations of your own. You must have done some kind of sums to reach your conclusions. Let's see them.
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^I have offered numbers.
Your numbers.
So, if you are going to state a missile traveling at over 16,000 km/h @ an altitude of 250km, without engine power, is going to overcome g=slightly over 9m/s2 in order to travel an additional 4250km to apogee, the simple fact is we disagree. I disagree based on what you wrote earlier, when you stated a missile subject to g=9.82m/s2 would decelerate to 0 in 12 minutes if traveling the same speed.
And regardless, d=rt still applies. To travel 250km in 5 minutes results in an average rate of travel of 3000km/h. The numbers reflected do not result in that outcome.
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^I have offered numbers.
Your numbers.
So, if you are going to state a missile traveling at over 16,000 km/h @ an altitude of 250km, without engine power, is going to overcome g=slightly over 9m/s2 in order to travel an additional 4250km to apogee, the simple fact is we disagree. I disagree based on what you wrote earlier, when you stated a missile subject to g=9.82m/s2 would decelerate to 0 in 12 minutes if traveling the same speed.
Great. Now read the rest of that post - I said 'At which point you'll probably say 'so how come it flew for 50 minutes? Surely 12 x 2 = 24?'...read past that bit. As we've discussed a post or two ago, g reduces with increasing altitude, so it takes a lot longer to decelerate than 12 minutes. All covered off in that website I linked to. If you disagree, fine, but how long would it take to decelerate, and what would be the distance travelled, if g reduced in the manner generally agreed upon? You're saying it's wrong...but what's the right answer then? It can't be 12 minutes, and we know it must be bigger than that number.
And regardless, d=rt still applies. To travel 250km in 5 minutes results in an average rate of travel of 3000km/h. The numbers reflected do not result in that outcome.
Again, I'm not disputing the average velocity. But there's no reason at all why the final velocity at the end of the burn is inconsistent with that. If most of the acceleration happens in the latter stages, just like the car in my previous example, why can't you have a high end velocity but a low average?
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Great. Now read the rest of that post - I said 'At which point you'll probably say 'so how come it flew for 50 minutes? Surely 12 x 2 = 24?'...read past that bit.
Remind me to what post you are referring. Thanks.
ETA: Found it.
The Hwasong 15, for example, was estimated to have reached 7.17km/s at the end of its burn. At that velocity, it would take around 12 minutes to decelerate to 0 at 9.81ms-2
Which then leads us to the other complicating factor that you aren't considering, which is the progressive reduction in g as you get further away from earth. At 4000km, for example, the ICBM would only experience a g of 3.7ms-2. That's why the calculations get complex very quickly - you've got variable mass and g, and a rotating planet.
Of course, I guess you disagree with the rotating round planet bit, and probably the reducing g bit as well. That's fine...but the burden then falls on you to explain what exactly did happen to the rocket if it flew for 50 odd minutes and only went 950km from the launch site, and how far it would be able to travel if it was launched at a shallower angle.
Here, you want to skip to the end part where g=approx 3.7m/s2, ignoring all the values in between.
As we've discussed a post or two ago, g reduces with increasing altitude, so it takes a lot longer to decelerate than 12 minutes. All covered off in that website I linked to. If you disagree, fine, but how long would it take to decelerate, and what would be the distance travelled, if g reduced in the manner generally agreed upon? You're saying it's wrong...but what's the right answer then? It can't be 12 minutes, and we know it must be bigger than that number.
I would prefer you launder your own wash.
Neatly press, fold, and drawer, or hang in the closet as you wish.
I am telling you the missile travelling at 16,000km/h at an altitude of 250km, under no engine power, subject to g slightly above 9m/s2, will not climb an additional 4250km.
Period.
You believe in this stuff.
I do not because as I have demonstrated in numerous posts, governments lie about this type of crap ALL the time.
It is warmongering, fearmongering crap, designed to keep a terrified populace.
And regardless, d=rt still applies. To travel 250km in 5 minutes results in an average rate of travel of 3000km/h. The numbers reflected do not result in that outcome.
Again, I'm not disputing the average velocity. But there's no reason at all why the final velocity at the end of the burn is inconsistent with that. If most of the acceleration happens in the latter stages, just like the car in my previous example, why can't you have a high end velocity but a low average?
If you understand what d=rtmeans, then you know that 0-16,000km/h over the course of 5 minutes would not average out to a final distance of 250km.
It doesn't matter whether the high end acceleration takes place in front or at the end of the run.
I am giving you all your numbers.
I don't personally accept it as truth, but you should act on your own information to support your argument ICBM's exist and they need RET to operate.
Have a great day!
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Here, you want to skip to the end part where g=approx 3.7m/s2, ignoring all the values in between.
No, not at all. As I said, and as laid out nicely in the site I linked to, you have to do some calculus - some integration to find the area under the velocity time graph essentially.
As we've discussed a post or two ago, g reduces with increasing altitude, so it takes a lot longer to decelerate than 12 minutes. All covered off in that website I linked to. If you disagree, fine, but how long would it take to decelerate, and what would be the distance travelled, if g reduced in the manner generally agreed upon? You're saying it's wrong...but what's the right answer then? It can't be 12 minutes, and we know it must be bigger than that number.
I would prefer you launder your own wash.
Neatly press, fold, and drawer, or hang in the closet as you wish.
No idea what you're on about here. I've shown you calculations, websites etc. You've just said 'it isn't so', without providing any explanation.
I am telling you the missile travelling at 16,000km/h at an altitude of 250km, under no engine power, subject to g slightly above 9m/s2, will not climb an additional 4250km.
Period.
But it's not subject to g at slightly above 9ms-2, it's subject to a g profile that reduces from just above 9 down to below 4 at apogee.
You believe in this stuff.
Not so much 'believe' as 'find the evidence for it compelling'
I do not because as I have demonstrated in numerous posts, governments lie about this type of crap ALL the time.
It is warmongering, fearmongering crap, designed to keep a terrified populace.
Governments lie, but science is science.
If you understand what d=rtmeans, then you know that 0-16,000km/h over the course of 5 minutes would not average out to a final distance of 250km.
It doesn't matter whether the high end acceleration takes place in front or at the end of the run.
For simplicity, imagine a profile where its speed doubled every minute, finishing at 16,000, and to keep the maths simple, just keep the velocity flat over each minute and then step it up at the end of each minute. So zero, then 1000km/h after one minute, then 2000km/h after 2 minutes, then 4000km/h after 3 minutes, then 8000km/h after 4 minutes, and a snap to 16,000km/h at the finish line.
Total distance in that case would be 1000/60 + 2000/60 + 4000/60 + 8000/60 = 250km
Now, clearly that's an inaccurate model, because the velocity can't just instantly double, and in the rocket example the mass is never zero - it decays, presumably linearly if the thrust is constant, to whatever the zero-fuel mass of the system is. But by reducing the time step, you can increase the accuracy, and it nevertheless illustrates the point.
I really don't get how to make this any clearer, but then it's not helped by your continued refusal to actually show any maths of your own.
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^The entirety of this post just above mine here gets a big thumbs down, especially that last feeble attempt at trying to brush away the fact that 250km traveled in 5 minutes = 3000km/h average speed over those 5 minutes.
Just pitiful.
Really pitiful.
Moving on.
The missile is subject to g slightly above 9m/s2 at 250km.
It is no longer under engine power.
It will not be able to gain an additional 4250km in altitude in order to subject itself to the claimed g=3.7m/s2 at 4500km, let alone in the required time frame.
Again, you are the one claiming ICBM's are real.
I am not.
I did not start the thread and you did not either, but you are the one supporting ICBM's as reality and the fact they exist as being somehow relevant to the shape of the earth.
I am calling the entire issue of ICBM's a crock of bullshit.
Prove otherwise.
You can't.
That is why you so desperately desire for the subject to be buried now.
I have clearly shown your math to be nonsense.
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You haven’t shown anything. You’ve asserted all of your points. Showing this mathematically would be extremely compelling. We have already been presented with calculations showing it is possible and all you do is give a thumbs down. I can’t imagine a less compelling rebuttal from you other than a “nuh-uh”.
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Showing this mathematically would be extremely compelling.
I agree.
Kindly demonstrate the compelling mathematical evidence that a missile, under no engine power, traveling at a rate of 16,000km/h, at an altitude of 250km, subject to a g=slightly above 9m/s2, will continue to gain altitude to an apogee of 4500km.
By the way, how was your trip to outer space?
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Showing this mathematically would be extremely compelling.
I agree.
Kindly demonstrate the compelling mathematical evidence that a missile, under no engine power, traveling at a rate of 16,000km/h, at an altitude of 250km, subject to a g=slightly above 9m/s2, will continue to gain altitude to an apogee of 4500km.
By the way, how was your trip to outer space?
I see you are having trouble distinguishing between an assertion and showing the mathematical truth behind your claims.
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Showing this mathematically would be extremely compelling.
I agree.
Kindly demonstrate the compelling mathematical evidence that a missile, under no engine power, traveling at a rate of 16,000km/h, at an altitude of 250km, subject to a g=slightly above 9m/s2, will continue to gain altitude to an apogee of 4500km.
By the way, how was your trip to outer space?
I see you are having trouble distinguishing between an assertion and showing the mathematical truth behind your claims.
What happened to the burden being on the OP and its supporters being responsible to demonstrate the reality of their claims?
But go ahead, post the math, which you claim has already been posted in this thread, demonstrating the reality of ICBM's.
As simple as posting a link to the post.
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^The entirety of this post just above mine here gets a big thumbs down, especially that last feeble attempt at trying to brush away the fact that 250km traveled in 5 minutes = 3000km/h average speed over those 5 minutes.
Just pitiful.
Really pitiful.
Moving on.
No, not moving on. You keep saying it's wrong, but you are completely unable to back up that claim with any evidence whatsoever. Nobody is saying that 3000km/h isn't the average speed. Of course it's the average speed. And that is entirely consistent with the final velocity being 16,000km/h. The final velocity could be anything at all. As long as the velocity-time profile is such that the area under the graph is 250km, then the average speed will be 3000km/h. It could finish at 16,000km/h, or 100,000km/h, or at a standstill. The totality of the profile is all that matters.
This is somewhat painful.
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^Incorrect.
In order to to achieve the figure of 250km altitude in 5 minutes resulting in a 3000 km/h average rate of travel, the velocities contained between the 0 and the 16,000 over the five minutes would average out to 3000 km/h.
They cannot, as (final velocity + initial velocity)/2 = average velocity.
https://www.calculatorsoup.com/calculators/physics/velocity_avg.php#:~:text=The%20sum%20of%20the%20initial,u)%2C%20divided%20by%202.
One, if they would, you would have shown it by now, which is incumbent on the person making the initial claim, which you did, not me.
Two, if the missile is no longer under additional power at an altitude of 250km, and is being subjected to g=slightly over 9m/s2, you would have shown how it is possible for the missile to continue to gain 4250 km of additional altitude while decelerating, which is also your claim, not mine.
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^Incorrect.
In order to to achieve the figure of 250km altitude in 5 minutes resulting in a 3000 km/h average rate of travel, the velocities contained between the 0 and the 16,000 over the five minutes would average out to 3000 km/h.
They cannot, as (final velocity + initial velocity)/2 = average velocity.
Hallelujah…finally, some maths so we can actually understand what it is that you don’t understand.
That statement above is simply incorrect. It would work as a calculation if and only if the velocity profile was linear which, as you’ve agreed previously, it isn’t - the fuel burns down, so mass reduces, but thrust is constant.
Average velocity is not necessarily the average of the start and end velocity.
It’s easy to prove that to yourself - just consider my 30mph car example. Your calculation would have the average velocity as the average of 30 and, say, 100mph even though the car only did 100mph for a few seconds after an hour at 30mph. Average speed is clearly not 65mph, as per your maths.
Or imagine a situation where start and end velocity are the same, but there’s a burst of increased velocity in the middle somewhere. Again, it wouldn’t work.
For general cases you need the area under the velocity time graph. If the slope is linear, your equation works - the area under a triangle is half the base times the height, which is, if you think about it, exactly what your equation is. But our graph here isn’t linear - it’s exponential, so you can’t calculate that simply.
Are we finally in agreement in this point at least?
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^Incorrect.
In order to to achieve the figure of 250km altitude in 5 minutes resulting in a 3000 km/h average rate of travel, the velocities contained between the 0 and the 16,000 over the five minutes would average out to 3000 km/h.
They cannot, as (final velocity + initial velocity)/2 = average velocity.
Hallelujah…finally, some maths so we can actually understand what it is that you don’t understand.
That statement above is simply incorrect. It would work as a calculation if and only if the velocity profile was linear which, as you’ve agreed previously, it isn’t - the fuel burns down, so mass reduces, but thrust is constant.
Average velocity is not necessarily the average of the start and end velocity.
It’s easy to prove that to yourself - just consider my 30mph car example. Your calculation would have the average velocity as the average of 30 and, say, 100mph even though the car only did 100mph for a few seconds after an hour at 30mph. Average speed is clearly not 65mph, as per your maths.
Or imagine a situation where start and end velocity are the same, but there’s a burst of increased velocity in the middle somewhere. Again, it wouldn’t work.
For general cases you need the area under the velocity time graph. If the slope is linear, your equation works - the area under a triangle is half the base times the height, which is, if you think about it, exactly what your equation is. But our graph here isn’t linear - it’s exponential, so you can’t calculate that simply.
Are we finally in agreement in this point at least?
I already agree it is exponential, but I do not agree there would necessarily be a drastic difference in calculation.
According to the sources, the missile was fired in a near vertical trajectory from launch site. The attainment of 250km does not veer significantly from the vertical within the span of five minutes. 250km would, given an apogee of 4500km achieved downrange at approximately 450km, fall within a measure of approximately 50km or so. Not much of an interval to calculate and it remains very close to a linear calculation of average velocity.
0 - 16,000km/h within 5 minutes results also in a relatively straight line in what turns out to be roughly a tenth of total flight time according to your sources.
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Average velocity calculations have been accurate, but they are entirely irrelevant to the question at hand. The final velocity at the end of the burn is the only significant number here.
Talking about average is like saying that since I was 8lbs 8 oz at birth, and weight 185 now, my average weight over my lifetime would be just shy of 97 pounds.
Or like measuring a formula 1 car's lap time - you can have an average time but it's not going to tell you anything about its velocity going into a turn or at the end of a straightaway.
It's a meaningless calculation that offers nothing.
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Average velocity calculations have been accurate, but they are entirely irrelevant to the question at hand. The final velocity at the end of the burn is the only significant number here.
Talking about average is like saying that since I was 8lbs 8 oz at birth, and weight 185 now, my average weight over my lifetime would be just shy of 97 pounds.
Or like measuring a formula 1 car's lap time - you can have an average time but it's not going to tell you anything about its velocity going into a turn or at the end of a straightaway.
It's a meaningless calculation that offers nothing.
When the average velocity doesn't match the velocity profiles over the given time of flight, it is certainly relevant.
I have no doubt of your other statements, but could care less as they are in no way related to the discussion here and are meaningless and totally fraudulent analogies. Not even close to the issue here.
We are talking about achievement of a goal in a certain period of time. What it took to achieve that goal is certainly averaged out.
Since you acknowledge the average velocity calculations as correct, that means 8000km/h is just shy of 2 minutes to travel 250km, which doesn't match the 5 minute burn time, or if you wish to keep the 5 minute burn time which would almost certainly be necessary (as it stands, the remainder of the flight is at jeopardy), the missile would have been a lot higher than 250km. Which would get you closer to the 4500km required for apogee, but attainment would likely be doomed by the g-forces at the new altitude figure of 667 km, which would be around 8m/s2.
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We are talking about achievement of a goal in a certain period of time. What it took to achieve that goal is certainly averaged out.
Since you acknowledge the average velocity calculations as correct, that means 8000km/h is just shy of 2 minutes to travel 250km, which doesn't match the 5 minute burn time, or if you wish to keep the 5 minute burn time which would almost certainly be necessary (as it stands, the remainder of the flight is at jeopardy), the missile would have been a lot higher than 250km.
If it's travelled 250km in 5 minutes then the average is 3,000km/h.
It starting at 0 and ending at 16,000km/h doesn't mean the average is 8000. That's not how the calculation works. You could sit still for 4 minutes and then go at 15,000km/h for the final minute and you'd travel 250km. Your average would still be 3,000km/h.
I suggested an acceleration which would work and actually travel less than 250km, not more. You could obviously tweak it to be bang on if you really care.
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We are talking about achievement of a goal in a certain period of time. What it took to achieve that goal is certainly averaged out.
Since you acknowledge the average velocity calculations as correct, that means 8000km/h is just shy of 2 minutes to travel 250km, which doesn't match the 5 minute burn time, or if you wish to keep the 5 minute burn time which would almost certainly be necessary (as it stands, the remainder of the flight is at jeopardy), the missile would have been a lot higher than 250km.
If it's travelled 250km in 5 minutes then the average is 3,000km/h.
It starting at 0 and ending at 16,000km/h doesn't mean the average is 8000. That's not how the calculation works. You could sit still for 4 minutes and then go at 15,000km/h for the final minute and you'd travel 250km. Your average would still be 3,000km/h.
I suggested an acceleration which would work and actually travel less than 250km, not more. You could obviously tweak it to be bang on if you really care.
And I suggested a velocity profile which also (coincidentally) gave a 250km distance over 5 minutes.
But we are, I think, wasting our collective time, as A80 doesn't actually appear to be reading what we write.
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But we are, I think, wasting our collective time, as A80 doesn't actually appear to be reading what we write.
Correct. We've shown our workings, he's refused to show his.
Until he does I don't think there's much more scope for discussion.
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I already agree it is exponential, but I do not agree there would necessarily be a drastic difference in calculation.
So you agree that your 'start + end v divided by two' equation is wrong, but now we're in to a discussion about how wrong it is?
According to the sources, the missile was fired in a near vertical trajectory from launch site. The attainment of 250km does not veer significantly from the vertical within the span of five minutes. 250km would, given an apogee of 4500km achieved downrange at approximately 450km, fall within a measure of approximately 50km or so. Not much of an interval to calculate and it remains very close to a linear calculation of average velocity.
0 - 16,000km/h within 5 minutes results also in a relatively straight line in what turns out to be roughly a tenth of total flight time according to your sources.
What on earth does that bit in bold actually mean?
Why aren't you looking at either the example velocity profiles that AATW or I gave you - they both illustrate the point that you seem to be consistently missing. You're even accepting that your calculation is wrong...why not accept that the follow-on from that, ie that it's perfectly possible to accelerate to 16000km/h in 5 minutes and have an average velocity of 3000km/h over that same time period? For the - I've-lost-count-of-how-many-times time, if most of the acceleration happens in the last minute, it's not surprising at all that the average velocity is a lot less than the final velocity.
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it's perfectly possible to accelerate to 16000km/h in 5 minutes and have an average velocity of 3000km/h over that same time period?
It's not just possible, we've both given examples. And it's obvious that you can almost finish at any final velocity.
You could stay still for almost all the time and then in the last 0.000833333 seconds go at the speed of light.
You'll have travelled 250km and your average speed would have been 3,000km/h.
Obviously that's not realistic, I tried to do something more realistic.
If Lackey is struggling with the idea that there is no real link between final speed and average speed then we're a bit stuck.
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I already agree it is exponential, but I do not agree there would necessarily be a drastic difference in calculation.
So you agree that your 'start + end v divided by two' equation is wrong, but now we're in to a discussion about how wrong it is?
Evidently you are missing the point that while it is exponential, the calculation of the average velocity is such that it would not differ substantially from the result of a linear solution.
According to the sources, the missile was fired in a near vertical trajectory from launch site. The attainment of 250km does not veer significantly from the vertical within the span of five minutes. 250km would, given an apogee of 4500km achieved downrange at approximately 450km, fall within a measure of approximately 50km or so. Not much of an interval to calculate and it remains very close to a linear calculation of average velocity.
0 - 16,000km/h within 5 minutes results also in a relatively straight line in what turns out to be roughly a tenth of total flight time according to your sources.
What on earth does that bit in bold actually mean?
Why aren't you looking at either the example velocity profiles that AATW or I gave you - they both illustrate the point that you seem to be consistently missing. You're even accepting that your calculation is wrong...why not accept that the follow-on from that, ie that it's perfectly possible to accelerate to 16000km/h in 5 minutes and have an average velocity of 3000km/h over that same time period? For the - I've-lost-count-of-how-many-times time, if most of the acceleration happens in the last minute, it's not surprising at all that the average velocity is a lot less than the final velocity.
It means what I just pointed out.
You demand the area under the curve be calculated, which is fine and good, and I am stating there isn't much of a curve to be considered and the result would be much the same as a linear result.
Remember, the displacement is a total of 250km at t+5, with not much variance from the y-axis within that space. The difference in the resulting measures is not going to be that significant.
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We are talking about achievement of a goal in a certain period of time. What it took to achieve that goal is certainly averaged out.
Since you acknowledge the average velocity calculations as correct, that means 8000km/h is just shy of 2 minutes to travel 250km, which doesn't match the 5 minute burn time, or if you wish to keep the 5 minute burn time which would almost certainly be necessary (as it stands, the remainder of the flight is at jeopardy), the missile would have been a lot higher than 250km.
If it's travelled 250km in 5 minutes then the average is 3,000km/h.
It starting at 0 and ending at 16,000km/h doesn't mean the average is 8000. That's not how the calculation works. You could sit still for 4 minutes and then go at 15,000km/h for the final minute and you'd travel 250km. Your average would still be 3,000km/h.
I suggested an acceleration which would work and actually travel less than 250km, not more. You could obviously tweak it to be bang on if you really care.
Now your just throwing bogus numbers and argumentation in the mix. 15000 doesn't even get you to, what was it again? 7.17 m/s, right?
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Now your just throwing bogus numbers and argumentation in the mix. 15000 doesn't even get you to, what was it again? 7.17 m/s, right?
"you're".
And all I'm doing is explaining that final speed and average speed are not related. 15,000 is arbitrary as is the speed of light in my other example.
I was just making the point that you could travel 250km in 5 minutes and end up at pretty much any final speed.
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Now your just throwing bogus numbers and argumentation in the mix. 15000 doesn't even get you to, what was it again? 7.17 m/s, right?
"you're".
And all I'm doing is explaining that final speed and average speed are not related. 15,000 is arbitrary as is the speed of light in my other example.
I was just making the point that you could travel 250km in 5 minutes and end up at pretty much any final speed.
I am not denying that and sorry for the typo.
If you travel 250km in 5 minutes, you would certainly need to end up at some final velocity.
But given a starting velocity of 0 and a final velocity of whatever you choose it to be (in your example it was 16000km/h), that currently results in an 8000km/h average velocity.
That is a fact.
It is also a fact if the average velocity of 8000km/h is distributed over a period of time = 5 minutes, that does not = 250km traveled.
It is also a given that at 250km, RET demands g=9.08m/s2. A missile under no further engine power will certainly not continue its parabolic trajectory in order to gain an additional 4250 km in altitude while decelerating.
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If you travel 250km in 5 minutes, you would certainly need to end up at some final velocity.
Yes
But given a starting velocity of 0 and a final velocity of whatever you choose it to be (in your example it was 16000km/h), that currently results in an 8000km/h average velocity.
That is a fact.
It is literally the opposite of a fact. That only works if the acceleration was constant over the 5 minutes, which in the case of a rocket it is not.
The average velocity is simply the distance travelled divided by the time it took.
250km / 5minutes = 50 (kilometers per minute). To get to km/h multiply by the 60 minutes in an hour 50 x 60 = 3000km/h
It is also a fact if the average velocity of 8000km/h is distributed over a period of time = 5 minutes, that does not = 250km traveled.
That is a meaningless statement, you don't "distribute" an average over a period of time, the average is just what it is, and I've explained how to calculate it.
It is also a given that at 250km, RET demands g=9.08m/s2. A missile under no further engine power will certainly not continue its parabolic trajectory in order to gain an additional 4250 km in altitude while decelerating.
I am less sure about this. I found an online calculator for projectiles and that appears to be correct BUT I don't know if that was clever enough to take into account how g varies with altitude. This stuff is complicated, to be honest the maths is beyond me and, with respect, from this conversation I think it's beyond you too.
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It is literally the opposite of a fact. That only works if the acceleration was constant over the 5 minutes, which in the case of a rocket it is not.
The average velocity is simply the distance travelled divided by the time it took.
250km / 5minutes = 50 (kilometers per minute). To get to km/h multiply by the 60 minutes in an hour 50 x 60 = 3000km/h
If you think you are somehow claiming that on the one hand, I am wrong by giving the average velocity of 8000km/h derived by your figures, then only to provide the displacement/t as the correct figure, then you would need to counter this: https://www.calculatorsoup.com/calculators/physics/velocity_avg.php
Type in 0 for initial velocity and 16,000 km/h for final velocity.
See what you get.
It is also a fact if the average velocity of 8000km/h is distributed over a period of time = 5 minutes, that does not = 250km traveled.
That is a meaningless statement, you don't "distribute" an average over a period of time, the average is just what it is, and I've explained how to calculate it.
Incorrect. The average rate is indeed distributed over time to ascertain total displacement.
It is also a given that at 250km, RET demands g=9.08m/s2. A missile under no further engine power will certainly not continue its parabolic trajectory in order to gain an additional 4250 km in altitude while decelerating.
I am less sure about this. I found an online calculator for projectiles and that appears to be correct BUT I don't know if that was clever enough to take into account how g varies with altitude. This stuff is complicated, to be honest the maths is beyond me and, with respect, from this conversation I think it's beyond you too.
I appreciate your thinking and I agree that it is beyond you.
Thank you for your participation.
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If you think you are somehow claiming that on the one hand, I am wrong by giving the average velocity of 8000km/h derived by your figures, then only to provide the displacement/t as the correct figure, then you would need to counter this: https://www.calculatorsoup.com/calculators/physics/velocity_avg.php
Type in 0 for initial velocity and 16,000 km/h for final velocity.
See what you get.
I get exactly what I would expect if the acceleration is constant.
The acceleration is not constant in the scenario we are discussing. This is the thing you are continually failing to understand.
And if you don't understand that - which is the fairly simple bit - the idea that you have the ability to calculate the other stuff given the constantly changing value of g with height is a little far fetched. You need calculus for this sort of thing.
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If you think you are somehow claiming that on the one hand, I am wrong by giving the average velocity of 8000km/h derived by your figures, then only to provide the displacement/t as the correct figure, then you would need to counter this: https://www.calculatorsoup.com/calculators/physics/velocity_avg.php
Type in 0 for initial velocity and 16,000 km/h for final velocity.
See what you get.
Finally people can see where you're getting your information.
Riddle me this, Batman.
My initial velocity is 0. In one second I accelerate to 16,000 km/hr. I travel at 16,000 km/hr for 4:59 for a final velocity of 16,000 km/hr. Same numbers in your calculator.
Are you telling me my average velocity over the 5 minutes was 8,000 km/hr??????
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If you think you are somehow claiming that on the one hand, I am wrong by giving the average velocity of 8000km/h derived by your figures, then only to provide the displacement/t as the correct figure, then you would need to counter this: https://www.calculatorsoup.com/calculators/physics/velocity_avg.php
Type in 0 for initial velocity and 16,000 km/h for final velocity.
See what you get.
I get exactly what I would expect if the acceleration is constant.
The acceleration is not constant in the scenario we are discussing. This is the thing you are continually failing to understand.
And if you don't understand that - which is the fairly simple bit - the idea that you have the ability to calculate the other stuff given the constantly changing value of g with height is a little far fetched. You need calculus for this sort of thing.
We are not discussing acceleration.
And we are not discussing constant velocity either.
We are discussing exponential velocity.
You do not need calculus to determine g at altitude.
As you stated, you are clearly demonstrating the reality of the issues discussed being beyond your ability.
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As you stated, you are clearly demonstrating the reality of the issues discussed being beyond your ability.
OK. How about we both calculate the average velocity in WTF's scenario. I've already done it by the way. Do you want to have a go?
Let's see who has the better understanding, shall we?
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As you stated, you are clearly demonstrating the reality of the issues discussed being beyond your ability.
OK. How about we both calculate the average velocity in WTF's scenario. I've already done it by the way. Do you want to have a go?
Let's see who has the better understanding, shall we?
Use the calculator provided.
That is what I used.
I am not going to drift off the subject here, despite your desperate desire to do so.
As I stated earlier, and in agreement with SteelyBob, the velocity profile of a missile is not linear in form and is exponential.
However, in the case of this particular claim, the exponential velocity profile of this happens to achieve the claimed velocity over a displacement of 250km, while not drifting very far from vertical, nor does it drift very far from vertical relative to t.
Either way, the average velocity measures are not going to be that significantly different from a linear calculation.
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If the missile is no longer under additional power at an altitude of 250km, and is being subjected to g=9.08m/s2, demonstrate how it is possible for the missile to continue to gain 4250 km of additional altitude to apogee while decelerating.
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As you stated, you are clearly demonstrating the reality of the issues discussed being beyond your ability.
OK. How about we both calculate the average velocity in WTF's scenario. I've already done it by the way. Do you want to have a go?
Let's see who has the better understanding, shall we?
Use the calculator provided.
That is what I used.
I know you did. That calculator is for kids doing simple problems. It doesn't make that clear, but this is where a little knowledge is a dangerous thing.
It's obvious that you can't just take the start and end velocities and take an average. For the last time: you can only do that if the acceleration is constant.
It's obvious that in WTF's example that the average velocity must be very close to 16,000km/h because that is how fast he's going for almost the whole time.
So to calculate the average you have to split it into sections (which is what I did when I outlined how a rocket could end up at 16,000km/h and go approximately 250km/h)
In WTF's example there are only 2 sections - the first second where he accelerates and then the rest of the time when he's going at 16,000km/h.
You need to calculate the distance of each part, add them up and then you can calculate the average velocity:
1 second
0-16,000 - average = 8000km/h (you CAN do that here because it's a constant acceleration)
Distance = speed * time
= 8000 * 1/3600 = 2.22222km [divide by 3600 so the units of hours match]
4m 59 seconds = 299 seconds
Distance = 16,000 * 299 / 3600 = 1328.88888km
Total distance = 2.22222 + 1328.88888 = 1331.11111
Velocity = distance / time = 1331.11111 / 5 * 60 = 15973.333km/h [times by 60 to get back to hours].
See? Not that hard. And here's the point. The fact you didn't understand that should tell you that you do not understand any of this as well as you think you do.
And when you get into variations of g with height you're into a whole other level of complication.
You do not have the understanding to discuss this. Which isn't a crime, but in this thread you are just being a breathing Dunning-Kruger curve. My friendly advice is to recognise your own limitations in understanding before talking so confidently on things you don't know much about.
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As you stated, you are clearly demonstrating the reality of the issues discussed being beyond your ability.
OK. How about we both calculate the average velocity in WTF's scenario. I've already done it by the way. Do you want to have a go?
Let's see who has the better understanding, shall we?
Use the calculator provided.
That is what I used.
I am not going to drift off the subject here, despite your desperate desire to do so.
As I stated earlier, and in agreement with SteelyBob, the velocity profile of a missile is not linear in form and is exponential.
However, in the case of this particular claim, the exponential velocity profile of this happens to achieve the claimed velocity over a displacement of 250km, while not drifting very far from vertical, nor does it drift very far from vertical relative to t.
Either way, the average velocity measures are not going to be that significantly different from a linear calculation.
You now appear to be throwing in the horizontal aspect of the missile’s trajectory and muddling that with the vertical velocity profile. That has nothing to do with the issue we are discussing here. For the sake of argument, everything we have discussed would be entirely valid for a missile going vertically straight up. The exponential velocity growth, the decreasing g profile as altitude increases…just keep it simple and vertical until you’ve grasped this basic concept.
Why don’t you have a go at calculating the average speed of a velocity profile like the one I suggested earlier. Go ahead…what would the average velocity of this profile be? :
1 minute stationary
1 minute at 1000km/h
1 minute at 2000km/h
1 minute at 4000km/h
1 minute at 8000km/h
0 minutes at 16000km/h
Go ahead - do the maths. Then repeat, but reduce the time step to 30 seconds, with speeds altered accordingly but still following the same exponential growth. Then do it again at 15 seconds, 7 seconds, 3 seconds….now you’re getting close to a numerical solution of a simple integral.
Then we can talk about how to calculate the velocity profile, and how to find the area under the velocity time graph.
Then we can develop a function for g with increasing altitude, and use it to develop another velocity profile for the missile as it decelerates. Then we can integrate that one too.
Then we can throw in the x axis, and start discussing elliptical versus parabolic trajectories and the effect of the curvature of the earth.
But I think we’re a long way off that, don’t you? Given that you seem to be massively struggling with the whole 250km thing.
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As you stated, you are clearly demonstrating the reality of the issues discussed being beyond your ability.
OK. How about we both calculate the average velocity in WTF's scenario. I've already done it by the way. Do you want to have a go?
Let's see who has the better understanding, shall we?
Use the calculator provided.
That is what I used.
I am not going to drift off the subject here, despite your desperate desire to do so.
As I stated earlier, and in agreement with SteelyBob, the velocity profile of a missile is not linear in form and is exponential.
However, in the case of this particular claim, the exponential velocity profile of this happens to achieve the claimed velocity over a displacement of 250km, while not drifting very far from vertical, nor does it drift very far from vertical relative to t.
Either way, the average velocity measures are not going to be that significantly different from a linear calculation.
You now appear to be throwing in the horizontal aspect of the missile’s trajectory and muddling that with the vertical velocity profile. That has nothing to do with the issue we are discussing here. For the sake of argument, everything we have discussed would be entirely valid for a missile going vertically straight up. The exponential velocity growth, the decreasing g profile as altitude increases…just keep it simple and vertical until you’ve grasped this basic concept.
Why don’t you have a go at calculating the average speed of a velocity profile like the one I suggested earlier. Go ahead…what would the average velocity of this profile be? :
1 minute stationary
1 minute at 1000km/h
1 minute at 2000km/h
1 minute at 4000km/h
1 minute at 8000km/h
0 minutes at 16000km/h
Go ahead - do the maths. Then repeat, but reduce the time step to 30 seconds, with speeds altered accordingly but still following the same exponential growth. Then do it again at 15 seconds, 7 seconds, 3 seconds….now you’re getting close to a numerical solution of a simple integral.
Then we can talk about how to calculate the velocity profile, and how to find the area under the velocity time graph.
Then we can develop a function for g with increasing altitude, and use it to develop another velocity profile for the missile as it decelerates. Then we can integrate that one too.
Then we can throw in the x axis, and start discussing elliptical versus parabolic trajectories and the effect of the curvature of the earth.
But I think we’re a long way off that, don’t you? Given that you seem to be massively struggling with the whole 250km thing.
There is no muddying the vertical velocity profile with the trajectory profile.
As stated, the trajectory profile, according to you, achieved a 250km altitude, at a velocity of 16,000 km/h.
The trajectory profile is such that it does not vary too widely from vertical, so little in fact, that any derived measure using calculus to gain the result will not vary too far from the result using a linear method.
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but in this thread you are just being a breathing Dunning-Kruger curve.
My friendly advice is to heed your own advice and bow out.
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but in this thread you are just being a breathing Dunning-Kruger curve.
My friendly advice is to heed your own advice and bow out.
I think I will, because I don't feel I have the ability to do the calculations given the varying 'g' with altitude. As I've said, this stuff is complex.
And, with respect, you definitely don't have the ability. The fact that you couldn't calculate the average speed in WTF's scenario demonstrates that.
If an object is accelerates to 16,000km/h in 1 second and then goes at that speed for 4m 59s then clearly the average speed over the 5 minutes can't be 8,000km/h.
That's like saying that the average of a sequence of numbers like:
0 | 16,000 | 16,000 | 16,000 | 16,000 | 16,000 | 16,000 | 16,000
is 8000 simply because the first number is 0 and the last is 16,000. Of course it isn't. That is not how you calculate averages.
So your simple calculator won't work in that scenario, and the fact you didn't understand that or know how to calculate the actual average shows that you don't have the ability to do these even more complicated calculations.
So you are basically reduced to making an argument from incredulity. That's all you've got.
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but in this thread you are just being a breathing Dunning-Kruger curve.
My friendly advice is to heed your own advice and bow out.
I think I will, because I don't feel I have the ability to do the calculations given the varying 'g' with altitude.
And, with respect, you definitely don't have the ability.
a=GM/r2
Add the additional altitude to the r value.
See ya.
PS: Quit claiming these calculators online are not acceptable because they do not provide the answers you want them to have. People use these calculators all the time to perform university classwork, which is accepted in the university.
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but in this thread you are just being a breathing Dunning-Kruger curve.
My friendly advice is to heed your own advice and bow out.
I think I will, because I don't feel I have the ability to do the calculations given the varying 'g' with altitude.
And, with respect, you definitely don't have the ability.
a=GM/r2
Add the additional altitude to the r value.
See ya.
You've simply presented a formula which I guess you looked up just like you looked up the calculator which you didn't understand.
Can you work it through and work out the maximum height of the missile?
I suspect not but prove me wrong, let's see your workings.
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PS: Quit claiming these calculators online are not acceptable because they do not provide the answers you want them to have. People use these calculators all the time to perform university classwork, which is accepted in the university.
Like any calculator, you have to understand how to use them and in which scenarios they apply and which ones they do not.
You have demonstrated that you do not understand this even when I have explained it multiple times.
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but in this thread you are just being a breathing Dunning-Kruger curve.
My friendly advice is to heed your own advice and bow out.
I think I will, because I don't feel I have the ability to do the calculations given the varying 'g' with altitude.
And, with respect, you definitely don't have the ability.
a=GM/r2
Add the additional altitude to the r value.
See ya.
You've simply presented a formula which I guess you looked up just like you looked up the calculator which you didn't understand.
Can you work it through and work out the maximum height of the missile?
I suspect not but prove me wrong, let's see your workings.
I certainly didn't look to you for the formula, but along with what has proven to be the rest of your nonsense, it seems to now be a prerequisite for people who you deem to be worthy to contribute that they be born with formulas already included if they choose to contribute here in order to be valid. Never mind, they need to consume formula prior to understanding language.
Nonetheless, check out the figures I already provided at 250km, which happens to be 9.08m/s2 and at 667km, which happens to be around 8m/s2.
According to RET:
G = 6.67408 * 1011
M = 5.972 * 1024
r = 6.371 * 102
At 250km, you add 2.5 *105 to the radius, resulting in 6.621 *106.
6.67408 * 1011 * 5.972 * 1024
(6.621 *106)2
I can't post the picture of the calculator I used to find the result of this equation here, but doing so would probably only result in you claiming I do not know how to use that calculator either or even worse (God forbid), I didn't make the calculator myself, therefore it isn't valid.
You are dismissed now.
Have a nice day.
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I certainly didn't look to you for the formula, but along with what has proven to be the rest of your nonsense, it seems to now be a prerequisite for people who you deem to be worthy to contribute they be born with formulas they choose to contribute here in order to be valid. Never mind, they need to consume formula prior to understanding language.
You need to be able to understand how to use formulas and calculators. You demonstrated above that you do not.
And you can say that it has been "proven to be the rest of your nonsense" as much as you like, that proof seems to only exist in your own mind.
You showed above that you can't work out averages correctly because you didn't understand in which circumstances some calculator you found online applied and when it doesn't apply.
I'm sorry if you didn't understand my explanation but I don't think I can make it any clearer. You not understanding this doesn't make you right.
At 250km, you add 2.5 *105 to the radius, resulting in 6.621 *106.
6.67408 * 1011 * 5.972 * 1024
(6.621 *106)2
Well colour me surprised but it seems in this instance you have used a calculator correctly. Well done you.
But what makes calculating the maximum height so complicated is that as the height increases g continues to decrease.
So that has to be taken into account - as has been explained to you, you need calculus for this sort of thing and it's complicated.
Given that you don't understand averages, I'd gently suggest this is above your level of understanding and math.
But if you want to have a go and show you're workings then I'm happy to have a look
(Spoiler: You won't)
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^Translation of above post for those in need.
"I am relatively peeved you demonstrated you do know how to use calculators and that you have blown a tremendous hole in our fictional working. How dare you!"
To summarize again for those unfortunate enough to be relegated to sifting through all the attempts to bury the obviously correct math:
Average velocity (linear) is calculated by (final velocity+ initial velocity)/2 as demonstrated here: https://www.calculatorsoup.com/calculators/physics/velocity_avg.php
It has already been admitted the calculation I provided is correct.
At issue is whether the trajectory profile and the velocity profile of the November 2017 Hwasong-15 missile are such that they vary so far from vertical as to fundamentally affect the results if they were measured using calculus to derive average velocity.
They will not.
The other fundamental question is this: "If the missile is no longer under additional power at an altitude of 250km, and is being subjected to g=slightly over 9m/s2, how it is possible for the missile to continue to gain 4250 km of additional altitude while decelerating."
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Average velocity (linear) is calculated by (final velocity+ initial velocity)/2 as demonstrated here: https://www.calculatorsoup.com/calculators/physics/velocity_avg.php
That is true if - and try and stay with me here - if and only if the acceleration is at a constant rate between the initial velocity and final velocity.
That is not the case with WTF's example. That is not the case with rockets.
I'm sorry you don't understand that but you are the only person in this thread who is confused about this.
The other fundamental question is this: "If the missile is no longer under additional power at an altitude of 250km, and is being subjected to g=slightly over 9m/s2, how it is possible for the missile to continue to gain 4250 km of additional altitude while decelerating."
I actually don't know if that is possible, the math is beyond me and thus it's definitely beyond you.
I had a look at an online calculator too and it appeared you'd need to be going a lot faster than that.
BUT that calculator did not take into account the effect of g decreasing with altitude. The difference between us is I understood that the calculator wasn't appropriate for answering this question.
It's taking the variations in g with altitude into account which makes this so complex.
So all you're left with is an argument from incredulity. Weak.
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Average velocity (linear) is calculated by (final velocity+ initial velocity)/2 as demonstrated here: https://www.calculatorsoup.com/calculators/physics/velocity_avg.php
I'm going to paraphrase a comment made to me in another thread.
To be frank, if you believe the above applies to rocket trajectory you're not qualified to be having this discussion.
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It has already been admitted the calculation I provided is correct.
I'm not so sure of that at all. This rocket trajectory estimation business is way more complicated than what you lay out.
This from an MIT Lab notes called, "Trajectory Calculation - Lab 2 Lecture Notes"
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwiklKfV-NXxAhUKDzQIHclzAj4QFjAMegQIAhAD&url=https%3A%2F%2Fweb.mit.edu%2F16.unified%2Fwww%2FFALL%2Fsystems%2FLab_Notes%2Ftraj.pdf&usg=AOvVaw1JqxPzrCnEHqwBJZyr-9ys
Here's all the stuff you have to take into consideration when it comes to calculating rocket trajectories and such. It's not just a simple average of velocity equals altitude/distance. It's way, way more complicated:
(https://i.imgur.com/mFPhAxa.png)
The lab notes go on to bring all of these factors to the fore. Through a slew of daunting equations. You're factoring like two or three parameters out of the 16+ that need to be taken into account. Just try and bend your mind around fuel mass flow rate, for one, and your head will explode.
So no, you are no where near qualified to say your calculations show that ICBM's aren't viable (neither am I), but there are people that are able to show that they are viable.
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It has already been admitted the calculation I provided is correct.
I'm not so sure of that at all. This rocket trajectory estimation business is way more complicated than what you lay out.
This from an MIT Lab notes called, "Trajectory Calculation - Lab 2 Lecture Notes"
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwiklKfV-NXxAhUKDzQIHclzAj4QFjAMegQIAhAD&url=https%3A%2F%2Fweb.mit.edu%2F16.unified%2Fwww%2FFALL%2Fsystems%2FLab_Notes%2Ftraj.pdf&usg=AOvVaw1JqxPzrCnEHqwBJZyr-9ys
Here's all the stuff you have to take into consideration when it comes to calculating rocket trajectories and such. It's not just a simple average of velocity equals altitude/distance. It's way, way more complicated:
(https://i.imgur.com/mFPhAxa.png)
The lab notes go on to bring all of these factors to the fore. Through a slew of daunting equations. You're factoring like two or three parameters out of the 16+ that need to be taken into account. Just try and bend your mind around fuel mass flow rate, for one, and your head will explode.
So no, you are no where near qualified to say your calculations show that ICBM's aren't viable (neither am I), but there are people that are able to show that they are viable.
Nice strawman.
I was referring to the calculation regarding average velocity I provided.
Since the flight in question has already been taken and the flight path along with trajectory has already been established for the record, no need to go back and calculate a trajectory.
The claims (according to RE adherents here) is the trajectory was such as to not vary from vertical to a wide degree.
Using calculus to determine average velocity over the given profile would not yield a significantly different result from a linear calculation.
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Average velocity (linear) is calculated by (final velocity+ initial velocity)/2 as demonstrated here: https://www.calculatorsoup.com/calculators/physics/velocity_avg.php
I'm going to paraphrase a comment made to me in another thread.
To be frank, if you believe the above applies to rocket trajectory you're not qualified to be having this discussion.
I believe the above calculator performs the function of calculating average velocity very well.
I have also stated it is a linear function given.
I have also stated the Hwasong-15 trajectory in November 0f 2017 did not vary so far from vertical at 250km that a result using calculus would differ substantially from the linear result.
That is true.
It is also true that an object at 250km, with no active propulsion, will not be able to gain an additional 4250km in altitude, as RE is claiming here.
To be frank, if you believe I made any claim the calculator above applies to rocket trajectory you're not qualified to be having any discussion.
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At issue is whether the trajectory profile and the velocity profile of the November 2017 Hwasong-15 missile are such that they vary so far from vertical as to fundamentally affect the results if they were measured using calculus to derive average velocity.
No, that's not at issue at all. The need to use calculus to derive the average velocity / distance travelled etc has nothing to do with the horizontal component of the trajectory. It would be equally true if the rocket was fired purely vertically straight up. The problem, yet again, is that the rocket's mass changes all the time as the fuel is burnt. This means the 'm' in f=ma is changing with respect to time, which gives you an exponential shape on the velocity time curve as the acceleration increases over time - or in other words, the rate of change of velocity is in itself also changing. You can't just plug that in to a simple average of two numbers calculation. To find the area under the graph (which is the distance travelled, as distance = velocity x time) you need to create a velocity function as a function of time and then integrate it with respect to time. Happy to do that for you in an example if you want, but not entirely convinced that you'd read it, as you don't seem to be reading anything else that I post. Can you assure me that you'll digest it and respond?
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I believe the above calculator performs the function of calculating average velocity very well.
IF the acceleration is constant from the starting velocity to the final velocity.
That is not the case with rockets and it's not the case with the example WTF gave.
This is all a diversion, it actually doesn't matter what the average velocity is. What matters it the height and velocity at the time of engine shutdown.
THAT is what determines how high it will go.
But the answer is complicated for reasons which have been explained to you.
The average speed doesn't matter, but what does matter is that you don't understand how to calculate it. The fact you can't do that relatively simple math shows that you are nowhere near qualified to do the far more complex calculations needed to work out how high the rocket will go given the starting conditions.
So all you're left with is an argument from incredulity.
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At issue is whether the trajectory profile and the velocity profile of the November 2017 Hwasong-15 missile are such that they vary so far from vertical as to fundamentally affect the results if they were measured using calculus to derive average velocity.
No, that's not at issue at all. The need to use calculus to derive the average velocity / distance travelled etc has nothing to do with the horizontal component of the trajectory. It would be equally true if the rocket was fired purely vertically straight up. The problem, yet again, is that the rocket's mass changes all the time as the fuel is burnt. This means the 'm' in f=ma is changing with respect to time, which gives you an exponential shape on the velocity time curve as the acceleration increases over time - or in other words, the rate of change of velocity is in itself also changing. You can't just plug that in to a simple average of two numbers calculation. To find the area under the graph (which is the distance travelled, as distance = velocity x time) you need to create a velocity function as a function of time and then integrate it with respect to time. Happy to do that for you in an example if you want, but not entirely convinced that you'd read it, as you don't seem to be reading anything else that I post. Can you assure me that you'll digest it and respond?
Look, I know the rocket travels up at an exponential rate.
Not arguing with that.
The fact the rocket burns fuel and loses mass in its flight has nothing to do with measuring the average velocity of the figures given.
Since you were earlier trying to brush off the fact the missile didn't vary far from vertical with an argument offering an example of using calculus to measure the area found under a triangle or trapezoid (which I naturally ignored as we are dealing with a curved trajectory in the instance), in order to find average velocity, what are you going to do now?
It is this simple.
The trajectory of the Hwasong -15 missile in November of 2017 was such that at an altitude of 250km, it varied from vertical of launch point 0 to a point no more than 50 km down range, more than likely near 30km.
Go ahead and apply your calculus to determine average velocity of that profile and state the measure.
I will tell you right now the result would not differ significantly from the one derived using the linear calculator provided.
In addition, an object under no propulsion at an altitude of 250km, experiencing g=9.08m/s2, will not, under any circumstance, gain an additional 4250km of altitude.
Have a great day.
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I believe the above calculator performs the function of calculating average velocity very well.
IF the acceleration is constant from the starting velocity to the final velocity.
That is not the case with rockets and it's not the case with the example WTF gave.
This is all a diversion, it actually doesn't matter what the average velocity is. What matters it the height and velocity at the time of engine shutdown.
THAT is what determines how high it will go.
But the answer is complicated for reasons which have been explained to you.
The average speed doesn't matter, but what does matter is that you don't understand how to calculate it. The fact you can't do that relatively simple math shows that you are nowhere near qualified to do the far more complex calculations needed to work out how high the rocket will go given the starting conditions.
So all you're left with is an argument from incredulity.
This from the same guy who tried to make up numbers so it would fit an average speed of 3000km/h for an object to travel 250km in 5 minutes.
You should honestly take your own advice and stop posting in this thread.
In addition, an object under no propulsion at an altitude of 250km, experiencing g=9.08m/s2, will not, under any circumstance, gain an additional 4250km of altitude.
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This from the same guy who tried to make up numbers so it would fit an average speed of 3000km/h for an object to travel 250km in 5 minutes.
Yeah. You couldn't get through your head how it was possible. So I showed one way it's possible. No need to thank me.
But, again, the average velocity isn't the issue here, the final velocity is.
But the fact you don't know how to calculate the average velocity is telling, it shows that you are not qualified to make statements like:
an object under no propulsion at an altitude of 250km, experiencing g=9.08m/s2, will not, under any circumstance, gain an additional 4250km of altitude.
Can you show the math which proves that - given that g decreases with altitude which makes these calculations extremely complex.
You need to account for the constantly changing value of g as the rocket ascends. If you think you can do that then great, let's see your working
(Spoiler - you can't)
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Look, I know the rocket travels up at an exponential rate.
Not arguing with that.
The fact the rocket burns fuel and loses mass in its flight has nothing to do with measuring the average velocity of the figures given.
Since you were earlier trying to brush off the fact the missile didn't vary far from vertical with an argument offering an example of using calculus to measure the area found under a triangle or trapezoid (which I naturally ignored as we are dealing with a curved trajectory in the instance), in order to find average velocity, what are you going to do now?
It is this simple.
The trajectory of the Hwasong -15 missile in November of 2017 was such that at an altitude of 250km, it varied from vertical of launch point 0 to a point no more than 50 km down range, more than likely near 30km.
Go ahead and apply your calculus to determine average velocity of that profile and state the measure.
I will tell you right now the result would not differ significantly from the one derived using the linear calculator provided.
In addition, an object under no propulsion at an altitude of 250km, experiencing g=9.08m/s2, will not, under any circumstance, gain an additional 4250km of altitude.
Have a great day.
Facepalm.
I wasn't talking about the shape of the flightpath, I was talking about the shape of the velocity - time graph. That, is, fire the rocket straight up (so there is only one component of velocity to worry about) and plot its velocity (y-axis of the graph) against time (x-axis of the graph).
To calculate distance travelled you need the area under the graph. That is the velocity multiplied by the time, for every infinitesimally small chunk of time. For a simple profile, it's easy maths. If velocity is constant over a set time, then it's just the area of a rectangle - the velocity multiplied by the time. If it's a linear acceleration from zero, it's now a straight sloping line - a triangular shape. So the distance travelled is 1/2 x base x height, or in other words, the starting velocity (zero) plus the end velocity all divided by two, multiplied by the time. That's what your average velocity equation was doing.
But if the velocity profile is a more complex shape, and because of the changing mass and hence variable acceleration it absolutely is (a progressively steeper up-sloping curve in this case) then to find the area under the graph you have to do the calculus I described.
Do you now understand why the horizontal component is completely irrelevant?
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This from the same guy who tried to make up numbers so it would fit an average speed of 3000km/h for an object to travel 250km in 5 minutes.
Yeah. You couldn't get through your head how it was possible. So I showed one way it's possible. No need to thank me.
But, again, the average velocity isn't the issue here, the final velocity is.
But the fact you don't know how to calculate the average velocity is telling, it shows that you are not qualified to make statements like:
an object under no propulsion at an altitude of 250km, experiencing g=9.08m/s2, will not, under any circumstance, gain an additional 4250km of altitude.
Can you show the math which proves that - given that g decreases with altitude which makes these calculations extremely complex.
You need to account for the constantly changing value of g as the rocket ascends. If you think you can do that then great, let's see your working
(Spoiler - you can't)
Spoiler - I am not making the claim the flight occurred.
You are the one making that claim, within the parameters of the written record.
You can't prove it.
Have a good day.
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You also have a claim to substantiate, which you haven’t done.
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Look, I know the rocket travels up at an exponential rate.
Not arguing with that.
The fact the rocket burns fuel and loses mass in its flight has nothing to do with measuring the average velocity of the figures given.
Since you were earlier trying to brush off the fact the missile didn't vary far from vertical with an argument offering an example of using calculus to measure the area found under a triangle or trapezoid (which I naturally ignored as we are dealing with a curved trajectory in the instance), in order to find average velocity, what are you going to do now?
It is this simple.
The trajectory of the Hwasong -15 missile in November of 2017 was such that at an altitude of 250km, it varied from vertical of launch point 0 to a point no more than 50 km down range, more than likely near 30km.
Go ahead and apply your calculus to determine average velocity of that profile and state the measure.
I will tell you right now the result would not differ significantly from the one derived using the linear calculator provided.
In addition, an object under no propulsion at an altitude of 250km, experiencing g=9.08m/s2, will not, under any circumstance, gain an additional 4250km of altitude.
Have a great day.
Facepalm.
I wasn't talking about the shape of the flightpath, I was talking about the shape of the velocity - time graph. That, is, fire the rocket straight up (so there is only one component of velocity to worry about) and plot its velocity (y-axis of the graph) against time (x-axis of the graph).
To calculate distance travelled you need the area under the graph. That is the velocity multiplied by the time, for every infinitesimally small chunk of time. For a simple profile, it's easy maths. If velocity is constant over a set time, then it's just the area of a rectangle - the velocity multiplied by the time. If it's a linear acceleration from zero, it's now a straight sloping line - a triangular shape. So the distance travelled is 1/2 x base x height, or in other words, the starting velocity (zero) plus the end velocity all divided by two, multiplied by the time. That's what your average velocity equation was doing.
But if the velocity profile is a more complex shape, and because of the changing mass and hence variable acceleration it absolutely is (a progressively steeper up-sloping curve in this case) then to find the area under the graph you have to do the calculus I described.
Do you now understand why the horizontal component is completely irrelevant?
For the final time, I have attempted to maintain a pretty decent level of decorum in this thread.
Your little comments like "facepalm" are not appreciated, nor are they necessary.
Your continued objections that average velocity, d=rt, and all the other things I have pointed out that do not happen to match the narrative of this fairy tale, and suddenly could not or do not matter is simply related to your recognition they do not match the narrative. If I travel 250km in 5 minutes and I do not GAF if I am traveling in a circle or if I am traveling in a straight line or if I am traveling in one direction and then suddenly veer off to the left or right, at the end of the trip I have averaged a rate of travel equivalent to 3000km/h over the course of that trip.
Period. End of sentence.
For purposes of this discussion, that rate of travel is equivalent to VELOCITY since we are discussing a scalar quantity.
The numbers you want to claim for the rocket do not fit the requirements.
So, every time you guys want to tag up and bury these facts, do not forget I will be here, pointing out the claimants have not provided the math.
In case you forgot, you (along with the rest) are the claimants.
Instead of providing the math (which I am the only one to do so far, truth be told) all you guys have done is say, "nuh uh."
Go ahead and post your figures for an average velocity of a projectile that travels 250km in 5 minutes.
Post your math demonstrating a ballistic object traveling at a velocity of 16,000 km/h, at an altitude of 250km, experiencing g= 9.08m/s2 under no further propulsion, will gain an additional altitude of 4250 km to apogee.
It is that simple.
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You also have a claim to substantiate, which you haven’t done.
My claim is this:
The claim made that this flight as described and adhered to by proponents in this thread is bogus and the trip did not take place as offered.
The proponents of this flight being made can quite clearly and definitively close the issue, as is incumbent upon them, not me.
All they need to do is post the math proving a ballistic object traveling at a rate of 16,000km/h, at an altitude of 250km, under no propulsion, g=9.08m/s2, will gain an additional altitude of 4250km.
Simple. Go ahead. Have a go.
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d=rt is relevant to every trip taken in the history of humanity.
Bye.
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lol, I have given up trying to explain that.
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You are the making that claim, within the parameters of the written record.
When did I make that claim?
I have repeatedly said that I don't know if it's true but have noted it's very complicated, far too complicated for you to understand - I'm basing that assertion on the fact that despite repeated explanations you don't understand much simpler concepts.
You are simply making an argument from incredulity. But if you're able to produce the maths which shows what the true final height would be, given the constantly changing value of g, then I'd like to see it.
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You are the making that claim, within the parameters of the written record.
When did I make that claim?
I have repeatedly said that I don't know if it's true but have noted it's very complicated, far too complicated for you to understand - I'm basing that assertion on the fact that despite repeated explanations you don't understand much simpler concepts.
You are simply making an argument from incredulity. But if you're able to produce the maths which shows what the true final height would be, given the constantly changing value of g, then I'd like to see it.
Yeah, here we go.
You do not now subscribe to the idea the flight has taken place.
Very well then.
You admit you have simply been trolling the thread.
Since you admit here you do not believe it happened, and you admit it is beyond your level of education, good bye to you.
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You do not now subscribe to the idea the flight has taken place.
I didn't say that either. ICBMs are definitely a thing, and I know tests of them have happened.
I don't claim to know the specifics of the flight we are talking about.
You are asserting that given a starting height and velocity when engines shut off the final height can't be a value claimed.
I would like to see your math which demonstrates that because, frankly, it's really complicated and given that you couldn't work out a simple average I struggle to believe you can work this out. Honestly, it's a bit beyond me.
So all you're left with is an argument from incredulity. I'm not making an argument at all, I'm merely pointing out that you need to understand more about all this before you make claims which you can't back up.
One thing to ponder. If d=rt then r = d/t.
We know the distance, 250km. And we know the time, 5 minutes.
So r = 250/5 = 50km/minute (note, minute, we need to multiply by 60 to get hours)
50 x 60 = 3000km/h average velocity. Not 8,000. See?
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You do not now subscribe to the idea the flight has taken place.
I didn't say that either. ICBMs are definitely a thing, and I know tests of them have happened.
I don't claim to know the specifics of the flight we are talking about.
You are asserting that given a starting height and velocity when engines shut off the final height can't be a value claimed.
I would like to see your math which demonstrates that because, frankly, it's really complicated and given that you couldn't work out a simple average I struggle to believe you can work this out. Honestly, it's a bit beyond me.
So all you're left with is an argument from incredulity. I'm not making an argument at all, I'm merely pointing out that you need to understand more about all this before you make claims which you can't back up.
One thing to ponder. If d=rt then r = d/t.
We know the distance, 250km. And we know the time, 5 minutes.
So r = 250/5 = 50km/minute (note, minute, we need to multiply by 60 to get hours)
50 x 60 = 3000km/h average velocity. Not 8,000. See?
Yeah, which was the figure I offered at the onset. It is apparent for anyone reading the thread, I was the first one to offer d=rt in this case is equivalent to 3,000km/h.
Which, when compared to your figures, blow your only figures really offered here (0 to 16,000km/h over the course of that five minutes travel to reach 250 km in altitude) out of the water.
You see, the average velocity of the figures you provided (0-16,000km/h) results in the 8,000km/h, hence not possibly matching d=rt.
Good bye.
You already admitted to trolling the thread.
I suggest you stop.
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Which, when compared to your figures, blow your only figures really offered here (0 to 16,000km/h over the course of that five minutes travel to reach 250 km in altitude) out of the water.
Do they, though?
I broke down that suggested acceleration pretty carefully to show how it could happen.
If you think I’ve made a mistake then can you show where?
But as I and others have explained, the average velocity is irrelevant here. All that matters is the height and velocity at the time of engine shut down.
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Which, when compared to your figures, blow your only figures really offered here (0 to 16,000km/h over the course of that five minutes travel to reach 250 km in altitude) out of the water.
Do they, though?
I broke down that suggested acceleration pretty carefully to show how it could happen.
If you think I’ve made a mistake then can you show where?
But as I and others have explained, the average velocity is irrelevant here. All that matters is the height and velocity at the time of engine shut down.
Asked and answered.
You are a self-admitted troll and are continuing to do so here.
Average velocity = (final velocity + initial velocity)/2.
Since we are discussing a portion of a trip, velocity can be considered equivalent to rate of travel.
d=rt has been relevant to all trips taken in the history of humanity.
The figures you provided, 0 - 16000 km/h in the span of 5 minutes, results in an average rate of travel of 8,000 km/h, resulting in an altitude of 667 km.
Bye to the self-admitted troll, as demonstrated here:
I don't claim to know the specifics of the flight we are talking about.
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Asked and answered.
You are an admitted troll and are continuing to do so here.
Average velocity = (final velocity + initial velocity)/2.
Since we are discussing a portion of a trip, velocity can be considered equivalent to rate of travel.
d=rt has been relevant to all trips taken in the history of humanity.
The figures you provided, 0 - 16000 km/h in the span of 5 minutes, results in an average rate of travel of 8,000 km/h, resulting in an altitude of 667 km.
Bye to the admitted troll.
You're nothing if not persistent, I'll grant you that.
Average velocity only equals (start velocity + end velocity)/2 for a linear velocity profile, ie constant acceleration.
Why do you keep ignoring that?
Why do you keep ignoring the stunningly obvious examples we give you, that quite clearly show that you can be at 16000km/h at the end of a period of time, and average well under 8000km/h?
Like, for example, travelling at 1km/h for 4 minutes and 55 seconds, and then accelerating rapidly to 16,000km/h.
But hey, what's the point? You'll ignore the example and keep saying the same thing, right?
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I recommend everyone just stops indulging A80's fantasy that avg velocity during burn phase is somehow the answer to his problem with an ICBM working...
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I recommend everyone just stops indulging A80's fantasy that avg velocity during burn phase is somehow the answer to his problem with an ICBM working...
Possibly. I just see it as step one on a long road of complete, and possibly wilful, misunderstandings or misrepresentations of basic facts.
I don't think we can address anything else until we get past this one.
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Asked and answered.
You are an admitted troll and are continuing to do so here.
Average velocity = (final velocity + initial velocity)/2.
Since we are discussing a portion of a trip, velocity can be considered equivalent to rate of travel.
d=rt has been relevant to all trips taken in the history of humanity.
The figures you provided, 0 - 16000 km/h in the span of 5 minutes, results in an average rate of travel of 8,000 km/h, resulting in an altitude of 667 km.
Bye to the admitted troll.
You're nothing if not persistent, I'll grant you that.
Average velocity only equals (start velocity + end velocity)/2 for a linear velocity profile, ie constant acceleration.
Why do you keep ignoring that?
Why do you keep ignoring the stunningly obvious examples we give you, that quite clearly show that you can be at 16000km/h at the end of a period of time, and average well under 8000km/h?
Like, for example, travelling at 1km/h for 4 minutes and 55 seconds, and then accelerating rapidly to 16,000km/h.
But hey, what's the point? You'll ignore the example and keep saying the same thing, right?
I am not ignoring it.
Already admitted it is a linear method of calculation several times in the thread.
You are ignoring the fact the velocity profile in this flight took place over a path not varying to a significant degree from vertical.
Go ahead, make a horizontal line at each level of velocity achieved at any given split over the five minute time frame you put forth. Try to achieve a significant difference in the average velocity derived via calculus and those given by a linear method as I described.
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I am not ignoring it.
Already admitted it is a linear method of calculation several times in the thread.
You are ignoring the fact the velocity profile in this flight took place over a path not varying to a significant degree from vertical.
No, I completely agree with that. In fact, I'm going one further and suggesting we just model it as a vertical flight to keep things simple.
Go ahead, make a horizontal line at each level of velocity achieved at any given split over the five minute time frame you put forth. Try to achieve a significant difference in the average velocity derived via calculus and those given by a linear method as I described.
That's pretty much what I did in one of the many examples you ignored. Here it is again:
1 minute at 0km/h
1 minute at 1000km/h
1 minute at 2000km/h
1 minute at 4000km/h
1 minute at 8000km/h
1 nano second at 16000km/h
Over to you for average speed and distance travelled. Notice how different the results are from a simple average of 0 and 16000km/h
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I am not ignoring it.
Already admitted it is a linear method of calculation several times in the thread.
You are ignoring the fact the velocity profile in this flight took place over a path not varying to a significant degree from vertical.
No, I completely agree with that. In fact, I'm going one further and suggesting we just model it as a vertical flight to keep things simple.
Go ahead, make a horizontal line at each level of velocity achieved at any given split over the five minute time frame you put forth. Try to achieve a significant difference in the average velocity derived via calculus and those given by a linear method as I described.
That's pretty much what I did in one of the many examples you ignored. Here it is again:
1 minute at 0km/h
1 minute at 1000km/h
1 minute at 2000km/h
1 minute at 4000km/h
1 minute at 8000km/h
1 nano second at 16000km/h
Over to you for average speed and distance travelled. Notice how different the results are from a simple average of 0 and 16000km/h
Oh. Remarkable you would spend any amount of time at 0 km/h, but it certainly is keeping with your desperate trolling efforts.
Quite amusing too!
Or perhaps it was the palm in your face from earlier.
Regardless.
If I travel 250km in 5 minutes, I have traveled at an average rate of travel equivalent to 3000km/h for those five minutes.
However, average velocity over a linear trajectory, which you agree is a vertical path, given the figures of 0 initial velocity and a final velocity achieved with five minutes 16,000 km/h work out to 8,000km/h and an altitude achieved of 667km.
Still waiting for you to back your claim (with demonstrable math please) a ballistic object located at an altitude of 250km, subjected to g=9.08m/s2 under no propulsion and guidance, can achieve an additional altitude of 4250 km.
I figured this would be easy for you to do since you already admitted the same object would fall back to earth within twelve minutes if subjected to g=9.87m/s2.
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Someday soon Lackey will acknowledge the gradient of gravity as altitude increases.
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Oh. Remarkable you would spend any amount of time at 0 km/h, but it certainly is keeping with your desperate trolling efforts.
Would it make you happier if the first two minutes were both at 1000km/h? It would move the answer further away from what you want it to be...entirely up to you. Telling that you haven't actually done the maths though. I wonder why?
Quite amusing too!
Or perhaps it was the palm in your face from earlier.
Regardless.
If I travel 250km in 5 minutes, I have traveled at an average rate of travel equivalent to 3000km/h for those five minutes.
However, average velocity over a linear trajectory, which you agree is a vertical path, given the figures of 0 initial velocity and a final velocity achieved with five minutes 16,000 km/h work out to 8,000km/h and an altitude achieved of 667km.
Ok, I think the issue here is that you're confusing a linear trajectory with a linear velocity profile - not the same thing. Linear trajectory, non linear velocity profile means you can't just average the start and end speed. I thought you'd agreed with that a few posts back but you now seem to have back-pedalled.
Still waiting for you to back your claim (with demonstrable math please) a ballistic object located at an altitude of 250km, subjected to g=9.08m/s2 under no propulsion and guidance, can achieve an additional altitude of 4250 km.
I figured this would be easy for you to do since you already admitted the same object would fall back to earth within twelve minutes if subjected to g=9.87m/s2.
We'll get to that, but let's bottom out the velocity thing first, because the second phase is even harder to grasp, and you're struggling with step one.
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I'm blindingly confused by all of this. But agree with you. Why would average velocity matter in the slightest? Isn't the only velocity of concern what it is at engine cutoff, 16000km/h? Who cares what happened between 0 and 16000, the latter is the only figure that matters.
I know this has come up a dozen times all ready, but the metaphor seems to be a bullet leaving the barrel of a gun, that velocity and a whole host of other factors, drag, trajectory, g and such, determine the distance and/or altitude traveled. I kind of think of it as the ballistic missile is fired out of a barrel that 250 km high at engine cut-off. The question becomes, just like the bullet out of the barrel, how far/high will it travel from that point on. Not what happened in the "barrel".
And like I pointed out earlier, which was considered a strawman, which it is not, is that a dozen plus parameters all enter into the calculations to figure that out. It's very, very complicated to say the least.
When the projectile hits engine cut-off at 16000km/h at an altitude of 250k and g of 9.08, having all its fuel burned off, take-off weight of the HS-15 is estimated at approximately 73 to 74t or 67000 kilos. Propellent for the 2 stages estimated to weigh 67 tons or 60000 kilos. So the projectile at cut-off now weighs 7000 kilos traveling at 16000km/h - That is your bullet velocity and some elements to be concerned with.
With that, if it continued up to let's say 500 km post cut-off, our 7000 kilo bullet would be experiencing g of now 8.43.
The question is how fast is our bullet going now?
if it continued up to let's say 1000 km post cut-off, our 7000 kilo bullet would be experiencing g of now 7.32.
The question is how fast is our bullet going now?
Let's say it kept going up to a whopping 2000 km post cut-off, our 7000 kilo bullet would be experiencing g of now 5.68.
And so on. And that's just factoring in a one, gravity, of the many elements needed to accurately calculate just how high it could go from a 16000km/h speed, 7000 kilo mass starting point.
So I agree, to even mention an average is absolutely meaningless. Is there something missing?
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I'm blindingly confused by all of this. But agree with you. Why would average velocity matter in the slightest? Isn't the only velocity of concern what it is at engine cutoff, 16000km/h? Who cares what happened between 0 and 16000, the latter is the only figure that matters.
I know this has come up a dozen times all ready, but the metaphor seems to be a bullet leaving the barrel of a gun, that velocity and a whole host of other factors, drag, trajectory, g and such, determine the distance and/or altitude traveled. I kind of think of it as the ballistic missile is fired out of a barrel that 250 km high at engine cut-off. The question becomes, just like the bullet out of the barrel, how far/high will it travel from that point on. Not what happened in the "barrel".
And like I pointed out earlier, which was considered a strawman, which it is not, is that a dozen plus parameters all enter into the calculations to figure that out. It's very, very complicated to say the least.
When the projectile hits engine cut-off at 16000km/h at an altitude of 250k and g of 9.08, having all its fuel burned off, take-off weight of the HS-15 is estimated at approximately 73 to 74t or 67000 kilos. Propellent for the 2 stages estimated to weigh 67 tons or 60000 kilos. So the projectile at cut-off now weighs 7000 kilos traveling at 16000km/h - That is your bullet velocity and some elements to be concerned with.
With that, if it continued up to let's say 500 km post cut-off, our 7000 kilo bullet would be experiencing g of now 8.43.
The question is how fast is our bullet going now?
if it continued up to let's say 1000 km post cut-off, our 7000 kilo bullet would be experiencing g of now 7.32.
The question is how fast is our bullet going now?
Let's say it kept going up to a whopping 2000 km post cut-off, our 7000 kilo bullet would be experiencing g of now 5.68.
And so on. And that's just factoring in a one, gravity, of the many elements needed to accurately calculate just how high it could go from a 16000km/h speed, 7000 kilo mass starting point.
So I agree, to even mention an average is absolutely meaningless. Is there something missing?
A80 is disputing the possibility of achieving the velocity quoted in the timeframe mentioned at the height given. He is suggesting, oddly, that the average speed of 3000km/h means the velocity at Hbo cannot be 16000.
Part two of his dispute seems to be that, even if you could achieve 16000km/h, you wouldn’t then achieve the apogee quoted.
I’m shooting at part one at the moment. We’ll move on to part two in due course.
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Right. And my point over the last couple of pages is that Lackey doesn’t even understand Part 1. He doesn’t understand how to calculate averages or the limitations of the calculator he found - which scenarios it can be used in and when it can’t be used.
Given that, it seems unlikely he is able to do the maths to back up his assertion about the second part which is orders of magnitude more complex. Which is no crime, this stuff is really complicated. I'm just pointing out that all he’s doing is making an argument from incredulity.
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Right. And my point over the last couple of pages is that Lackey doesn’t even understand Part 1. He doesn’t understand how to calculate averages or the limitations of the calculator he found - which scenarios it can be used in and when it can’t be used.
Given that, it seems unlikely he is able to do the maths to back up his assertion about the second part which is orders of magnitude more complex. Which is no crime, this stuff is really complicated. I'm just pointing out that all he’s doing is making an argument from incredulity.
Yep.
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When you travel 250 km in 5 minutes, according to the figures provided, you will have achieved an average rate of travel of 3000km/h. Pleaase note that rate of travel = velocity.
If you have an initial velocity of 0 km/h and your final velocity after 5 minutes is 16,000 km/h, then your average velocity is 8000 km/h.
There is no dispute here.
Anyone here who claims the scenario provided by AATW is possible is gaslighting, plain, pure, and simple.
Further, an object traveling at a velocity of 16,000km/h at a height of 250 km, is at that moment, subjected to g=9.08m/s2. Given those figures, that object will fall to the earth in just over 17 minutes.
This is a fact.
No way this missile flight took place at all in the form or complete fashion expressed on Wikipedia and other sources.
Next.
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When you travel 250 km in 5 minutes, according to the figures provided, you will have achieved an average rate of travel of 3000km/h. Pleaase note that rate of travel = velocity.
Yes
If you have an initial velocity of 0 km/h and your final velocity after 5 minutes is 16,000 km/h, then your average velocity is 8000 km/h.
No.
Or rather "it depends". That is true if and only if your rate of acceleration during those 5 minutes is constant
Further, an object traveling at a velocity of 16,000km/h at a height of 250 km, is at that moment, subjected to g=9.08m/s2. Given those figures, that object will fall to the earth in just over 17 minutes.
This is a fact.
Please show the math which demonstrates this fact. Note that as the object continues to ascend the value of g will continue to reduce.
If you just use a flat rate of 9.08 then you're going to get the wrong answer. I don't know how to do the math to work this out and, with respect, given your second statement above, I don't believe you do either.
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When you travel 250 km in 5 minutes, according to the figures provided, you will have achieved an average rate of travel of 3000km/h. Pleaase note that rate of travel = velocity.
Yes
If you have an initial velocity of 0 km/h and your final velocity after 5 minutes is 16,000 km/h, then your average velocity is 8000 km/h.
No.
Or rather "it depends". That is true if and only if your rate of acceleration during those 5 minutes is constant
This is demonstrably false!
Acceleration for a race car is certainly not constant, yet average velocity over the course (also not linear) is expressed as defining the winner.
We are a discussing average velocity not average acceleration or even average speed. Acceleration is simply the variation of speed over time, or in this case: dV/dt where d = change
Take your extreme BS out of here.
Further, an object traveling at a velocity of 16,000km/h at a height of 250 km, is at that moment, subjected to g=9.08m/s2. Given those figures, that object will fall to the earth in just over 17 minutes.
This is a fact.
Please show the math which demonstrates this fact. Note that as the object continues to ascend the value of g will continue to reduce.
If you just use a flat rate of 9.08 then you're going to get the wrong answer. I don't know how to do the math to work this out and, with respect, given your second statement above, I don't believe you do either.
It is not a flat rate of 9.08.
I have given the number, which accounts for ascent and velocity.
Since you are trying to claim acceleration can possibly impact the calculation of average velocity, I see no need to further engage you on this subject.
You are proving to be nothing but a troll.
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You can describe the distance travelled in a given time frame for any moving body (whether its accelerating or not) by multiplying its avg velocity by the time. No one disputed this.
But that figure offers zero predictive powers for the instantaneous velocity at any time. For this discussion, AVERAGE VELOCITY FOR THE BURN PHASE IS A MEANINGLESS VALUE.
Please stop repeating this nonsense.
RIP to what started as an interesting thread.
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AVERAGE VELOCITY FOR THE BURN PHASE IS A MEANINGLESS VALUE.
Correct. But I’d suggest that Lackey’s confusion about this and inability to calculate that average is meaningful and telling.
It shows that the rest of his argument is simply one from incredulity.
In an attempt to get this back on track, do you know how to calculate the maximum height, given the known altitude and velocity at the time of engine shut down? Given the constantly varying value of g we are in the realms of calculus and differential equations, I fear, which is a bit beyond me.
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Further, an object traveling at a velocity of 16,000km/h at a height of 250 km, is at that moment, subjected to g=9.08m/s2. Given those figures, that object will fall to the earth in just over 17 minutes.
This is a fact.
How did you calculate the 17 minutes after cut-off bit? It seems like a fairly precise number you came up with.
According to your calculations, what was the altitude of the rocket after 17 minutes?
Did you factor in the mass of the rocket after cut-off?
Did you factor into the 17 minutes what the g was 1 minute after cut-off, 2 minutes after cut-off, etc, up to 17?
Did you factor in the density of the atmosphere, drag, 1 minute after cut-off, 2 minutes after cut-off, up to 17?
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Acceleration for a race car is certainly not constant, yet average velocity over the course (also not linear) is expressed as defining the winner.
Yes, it's almost as if they can take the total approximate distance the car has driven, and then divide that number by the amount of time it took the driver to run the course. That is, notably, different than how you're coming up with the (wrong) average velocity in your own example. Do you see the difference?
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An interesting real-world example to share.
Aerie Luyendyk averaged 236.986mph (381.391km/h) in qualifying for the Indy 500 in 1996, with his four-lap time of 2m31.908s still unmatched today.
That car, like all race cars, starts out at 0km/h. By lackey-math, that means it must have finished the race at 762.79km/h in order to achieve the record-holding average velocity of 381.391km/h. So lackey, can you show us evidence that IndyCars regularly finish races at velocities greater than 700km/h? That seems like a hard thing to just take on faith.
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You can describe the distance travelled in a given time frame for any moving body (whether its accelerating or not) by multiplying its avg velocity by the time. No one disputed this.
But that figure offers zero predictive powers for the instantaneous velocity at any time. For this discussion, AVERAGE VELOCITY FOR THE BURN PHASE IS A MEANINGLESS VALUE.
Please stop repeating this nonsense.
RIP to what started as an interesting thread.
Considering you are the one offering nonsense (a strawman, in other words) that has nothing to do with the incorrect figures offered by AATW (i.e., why a calculation of average velocity over the first five minutes of the flight clearly demonstrating a final velocity of 16,000 km/h at t+5 is not possible), then I will return the favor and ask you to stop repeating the strawman.
AVERAGE VELOCITY FOR THE BURN PHASE IS A MEANINGLESS VALUE.
Correct.
Incorrect.
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Acceleration for a race car is certainly not constant, yet average velocity over the course (also not linear) is expressed as defining the winner.
Yes, it's almost as if they can take the total approximate distance the car has driven, and then divide that number by the amount of time it took the driver to run the course. That is, notably, different than how you're coming up with the (wrong) average velocity in your own example. Do you see the difference?
Of course. That is why I was first in this thread to offer the d=rt solution, using five minutes for t and 250km for d, equaling 3000km/h. As far as your objection to the other calculation of average velocity = final velocity+initial velocity/2, well, you will just need to take that up with physicists and the other "sciency" dudes. Cause it is just as legit and it clearly demonstrates AATW's figures to be totally bogus.
An interesting real-world example to share.
Aerie Luyendyk averaged 236.986mph (381.391km/h) in qualifying for the Indy 500 in 1996, with his four-lap time of 2m31.908s still unmatched today.
That car, like all race cars, starts out at 0km/h. By lackey-math, that means it must have finished the race at 762.79km/h in order to achieve the record-holding average velocity of 381.391km/h. So lackey, can you show us evidence that IndyCars regularly finish races at velocities greater than 700km/h? That seems like a hard thing to just take on faith.
Since you are demonstrating extreme ignorance concerning how an average velocity of a four lap qualifying velocity, or even an average velocity of a whole race, is actually calculated, I doubt you should remain in the conversation any longer.
I will offer you a clue. The average velocity calculation for the four lap qualifier does not, should not, or would not ever include a value of 0 in its calculation. Of course, the average velocity over the entire race would include some 0 velocity figures (dependent on pit stops, wrecks, etc.), but even that calculation would not start with a 0 velocity figure.
Amazingly, your strawmanning was all accomplished much faster than the rocket we are discussing.
boydster, once you come to grips with all this, then come back. Until then, have a great day.
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The car is starting at 0km/h. What speed is it going when it crosses the finish line? I'd like to plug those 2 values into your calculator and see what it comes up with for average velocity. It'll definitely be very accurate, right?
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The car is starting at 0km/h. What speed is it going when it crosses the finish line? I'd like to plug those 2 values into your calculator and see what it comes up with for average velocity. It'll definitely be very accurate, right?
The car is not starting at 0 km/h in either four lap qualifying or the actual race.
Have you ever watched auto racing?
Go home, boydster. I think timmy is asking for your presence elsewhere.
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A Formula 1 race starts when the red lights go out, not when the car starts moving. If a car is moving when the red lights go out, it is disqualified.
Its velocity at the start is zero.
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A Formula 1 race starts when the red lights go out, not when the car starts moving. If a car is moving when the red lights go out, it is disqualified.
Its velocity at the start is zero.
I guess once it's finished the race and comes to a halt then its velocity at the end is also 0.
Therefore it's average velocity throughout the race must be 0, according to Lackey's calculator.
I wonder if anyone can spot the flaw in that argument.
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A Formula 1 race starts when the red lights go out, not when the car starts moving. If a car is moving when the red lights go out, it is disqualified.
Its velocity at the start is zero.
That is correct for Formula One, which was not the example discussed prior to this.
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A Formula 1 race starts when the red lights go out, not when the car starts moving. If a car is moving when the red lights go out, it is disqualified.
Its velocity at the start is zero.
I guess once it's finished the race and comes to a halt then its velocity at the end is also 0.
Therefore it's average velocity throughout the race must be 0, according to Lackey's calculator.
I wonder if anyone can spot the flaw in that argument.
Yeah, I am sure they can spot the flaw.
As usual, it is your gaslighting the issue.
The actual end of the race is what is called crossing the FINISH LINE with the required amount of displacement "under your belt" so to speak.
The official declaration of the end of the race does not include the process of bringing the car or other cars to a halt.
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As far as your objection to the other calculation of average velocity = final velocity+initial velocity/2, well, you will just need to take that up with physicists and the other "sciency" dudes. Cause it is just as legit and it clearly demonstrates AATW's figures to be totally bogus.
It is legit IF the rate of acceleration is constant. Which is not the case with rockets or in the examples which have been worked through with you.
Is your entire problem with all of this that you work out the average velocity one way and get 3,000 and you do it another way and get 8,000 and you don't understand why you get two different figures? Reason is what I keep telling you. The 8,000 figure is only valid if the rate of acceleration is constant. It is not constant for rockets, which is how a rocket could end up going 250km in 5 minutes and end up with a final velocity of 16,000. I literally broke that down minute by minute with an example of how that could happen.
But none of this really matters because...
AVERAGE VELOCITY FOR THE BURN PHASE IS A MEANINGLESS VALUE.
Your entire response to that was "Incorrect". Why is it incorrect?
IF the rocket is going at 16,000km/h and is at 250km altitude when the engines shut down then that's all you need to know to calculate the maximum altitude it will reach (well, you need to know the mass and radius of earth too so you can take into account how g varies as the rocket ascends - that is the complex part). The average velocity or acceleration before the point of engine shut down has absolutely no bearing on the answer.
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A Formula 1 race starts when the red lights go out, not when the car starts moving. If a car is moving when the red lights go out, it is disqualified.
Its velocity at the start is zero.
That is correct for Formula One, which was not the example discussed prior to this.
Fine. I'm not a fan of racing, I'll concede I didn't know IndyCar begins with the cars moving. That doesn't matter much, we could just as easily talk about a race where the cars start at 0mph and finish at however fast they are going when they cross the finish line - I guess that'd be Formula 1, based on DD's post? The point of the thought experiment is to show that your method of finding an average velocity is hopelessly flawed.
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Two identical cars sit on a flat road. One is travelling at a constant speed of 8000 km/h. As soon at it passes car 2, car 2 hits the gas and continues accelerating. Both cars travel for 5 minutes, car 1 at a constant 8000km/h, car 2 accelrating to a final velocity of 16000 km/h, before both take their feet off the gas at the same time and just coast, allowing friction to gradually slow them down.
They both have the same avg velocity for the 5 minute trip, but if we want to know if one of them will travel further before it comes to a comlete stop,
Average
velocity
doesnt
matter.
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As far as your objection to the other calculation of average velocity = final velocity+initial velocity/2, well, you will just need to take that up with physicists and the other "sciency" dudes. Cause it is just as legit and it clearly demonstrates AATW's figures to be totally bogus.
It is legit IF the rate of acceleration is constant. Which is not the case with rockets or in the examples which have been worked through with you.
Is your entire problem with all of this that you work out the average velocity one way and get 3,000 and you do it another way and get 8,000 and you don't understand why you get two different figures? Reason is what I keep telling you. The 8,000 figure is only valid if the rate of acceleration is constant. It is not constant for rockets, which is how a rocket could end up going 250km in 5 minutes and end up with a final velocity of 16,000. I literally broke that down minute by minute with an example of how that could happen.
But none of this really matters because...
AVERAGE VELOCITY FOR THE BURN PHASE IS A MEANINGLESS VALUE.
Your entire response to that was "Incorrect". Why is it incorrect?
IF the rocket is going at 16,000km/h and is at 250km altitude when the engines shut down then that's all you need to know to calculate the maximum altitude it will reach (well, you need to know the mass and radius of earth too so you can take into account how g varies as the rocket ascends - that is the complex part). The average velocity or acceleration before the point of engine shut down has absolutely no bearing on the answer.
For the final time, any figures relating to acceleration do not interject themselves into calculations regarding average velocity.
You contributions and assertions they somehow do are patently false.
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For the final time, any figures relating to acceleration do not interject themselves into calculations regarding average velocity.
You contributions and assertions they somehow do are patently false.
http://hyperphysics.phy-astr.gsu.edu/hbase/avari.html
Note that even in this one-dimensional case for non-constant acceleration, the average velocity is not equal to the average of the initial and final velocities.
The average velocity expression from the constant acceleration equations works only for constant acceleration where the graph of velocity as a function of time is a straight line, the average being the midpoint of that line over the chosen time interval
Apology accepted.
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For the final time, any figures relating to acceleration do not interject themselves into calculations regarding average velocity.
You contributions and assertions they somehow do are patently false.
http://hyperphysics.phy-astr.gsu.edu/hbase/avari.html
Note that even in this one-dimensional case for non-constant acceleration, the average velocity is not equal to the average of the initial and final velocities.
The average velocity expression from the constant acceleration equations works only for constant acceleration where the graph of velocity as a function of time is a straight line, the average being the midpoint of that line over the chosen time interval
Apology accepted.
And as previously acknowledged elsewhere in this thread, the velocity profile in the case of this particular, given a time frame of five minute burn, varies so little from the vertical, an average velocity figure derived via calculus would not fit a 5 minute flight achieving 250km.
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velocity profile
You keep using that word. I do not think it means what you think it means.
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velocity profile
You keep using that word. I do not think it means what you think it means.
And you keep using "that word," when you call out what is, quite obviously, TWO WORDS.
Have a nice day.
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velocity profile
You keep using that word. I do not think it means what you think it means.
Inconceivable!
Anyway...the average velocity or the "velocity profile" are completely irrelevant here. The only things that matter are the altitude and velocity at the time of engine shutdown, and the strength of earth's gravitational field. With those things you can work out the maximum altitude of the rocket. And by "you" I mean someone who is better at math than me. Over to you, Lackey. I'm looking forward to seeing your workings.
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And you keep using "that word," when you call out what is, quite obviously, TWO WORDS.
Have a nice day.
Since we're not advancing the argument at this stage, I think you quite obviously mean "..... what are, quite obviously, TWO WORDS".
Anywho; Has everyone heard of "muzzle velocity". Its the key parameter in the ballistics of firearms and artillery. Nothing to do with rates of acceleration. Nothing to do with average speed in the barrel.
Muzzle velocity. The instantaneous speed at the end of burn.
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Yeah, speed at the end of the burn, which happens to be in the scenario discussed here, t+5.
Working with a displacement of 250km at t+5.
Yielding, according to d=rt, an average velocity of 3,000 km/h.
If a man made rocket could even travel at a velocity of 16,000 km/h (I doubt it could), and did reach that speed at the very end of 5 minutes of burn, it blows either the distance traveled figure or the yielded rate of velocity figure out of the water and puts a nail in the coffin of this BS fiction pushed by warmongers.
No wonder you guys don't come up with your supposed saving grace of calculus to supply the average rate of velocity for the profile.
You know it is closer to the 8000 km/h average figure yielded by linear calculation.
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supply the average rate of velocity for the profile.
I beg your pardon?
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supply the average rate of velocity for the profile.
I beg your pardon?
Quite, I'm sure.
Anyway, back to the subject.
Here we have the great fraud, claiming a velocity of 16,000 km/h at end of burn, somehow magically continuing to accelerate without engine power now, upwards to a final velocity of 25,000 km/h!
Even the great DD, proponent and lover of ballistics, will admit a ballistic object will not gain velocity without a source of continued propulsion.
Once the propulsion source is gone, deceleration takes place. Velocity reduces.
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somehow magically continuing to accelerate without engine power now, upwards to a final velocity of 25,000 km/h!
Who is claiming that? ???
Once the propulsion source is gone, deceleration takes place. Velocity reduces.
Yes. But, g decreases with altitude. So as the rocket ascends so the value of g gets lower. You can't use a fixed value of g to calculate its maximum height. The only way I could figure out how to simulate this was to do it in time steps:
So
Initial velocity v0 = 16000
Initial height h0 = 250km
g0 = 9.8 (g at sea level, roughly)
re = 6731 (radius of earth)
You can then iterate, let's say in second steps
Height after the first second:
h1 = h0 + (v0/3600)
The velocity after the first second - reduce the velocity by the gravity at h0, I'll call this gh0:
v1 = v0 - gh0
where gh0 = g0 x ((re x re) / (re + h0) x (re + h0))
Then you iterate. The next second:
h2 = h1 + v1/3600
v2 = v1 - gh1 (gh1 has to be recalculated)
And so on.
I wrote a little basic program to do this in second increments. It's just finished running.
The maximum height would be 9,831km
Here's the code and the site I wrote and ran it in:
v0 = 16000
h0 = 250
g0 = 9.80665
re = 6371.009
hn = h0 + (v0/3600)
gh = g0 * ((re*re)/((re+h0)*(re+h0)))
vn = v0 - gh
counter = 1
TextWindow.WriteLine("h1 = " + hn + "v1 = " + vn + " g1 = " + gh)
While (vn > 0)
counter = counter + 1
gh = g0 * ((re*re)/((re+hn)*(re+hn)))
hn = hn + (vn/3600)
vn = vn - gh
TextWindow.WriteLine("h" + counter + " = " + hn + "v" + counter + " = " + vn + " g" + counter + " = " + gh)
EndWhile
https://superbasic-v2.azurewebsites.net/
Now this is all very simplistic, the smaller the time interval you use the better approximation you'll get. And this assumes the rocket is heading straight up, which is not the case for ICBMs. Also neglects wind resistance of course. I'm sure there are differential equations which give you the right answer but that's a bit beyond me.
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somehow magically continuing to accelerate without engine power now, upwards to a final velocity of 25,000 km/h!
Who is claiming that? ???
Acceleration is slow initially, and rapid towards the end. The final velocity is more like 25,000km/h,
Yeah, you are busted.
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AATW is busted because Steely Bob said something? Holy North Korea, Batman!
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AATW is busted because Steely Bob said something? Holy North Korea, Batman!
Well, this is what Steely Bob actually said.
Lackey “accidentally” left out the context:
But whilst the average velocity is absolutely 3000km/h, as I said above, there is no requirement for velocity to build in a linear fashion as you describe. Thrust is constant, mass is reducing. Acceleration is slow initially, and rapid towards the end. The final velocity is more like 25,000km/h, but that doesn't mean the average velocity isn't 3000km/h, because it builds rapidly in the latter stages. It's not a straight line.
My emphasis. But it’s clear from the above that Bob is not claiming any acceleration after the engines have shut off. He’s talking about the way the acceleration is not linear, as I am. Which renders Lackey’s 8,000 figure incorrect. The correct value is 3,000 but is also meaningless.
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AATW is busted because Steely Bob said something? Holy North Korea, Batman!
Well, this is what Steely Bob actually said.
The final velocity is more like 25,000km/h
But it’s clear from the above that Bob is not claiming any acceleration after the engines have shut off.
I struck out the incorrect claim you made in your post.
SteelyBob did claim the rocket would continue accelerating, to a final velocity of 25,000km/h.
Because at a velocity of 16,000 km/h, at 250km, experiencing g=9.08m/s2, and decelerating, the rocket will reach 0 in 17 min.
Will not travel the additional 4250 km to apogee given these conditions.
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Because at a velocity of 16,000 km/h, at 250km, experiencing g=9.08m/s2, and decelerating, the rocket will reach 0 in 17 min.
Will not travel the additional 4250 km to apogee given these conditions.
Are you accounting for the fact that as the rocket continues to ascend the value of g will continue to get lower?
If not then you’re doing it wrong. Doing so is complicated, I suggested an iterative way above which I think approximates it, but I simplified and assumed the rocket is going straight up. Feel free to show your workings.
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Because at a velocity of 16,000 km/h, at 250km, experiencing g=9.08m/s2, and decelerating, the rocket will reach 0 in 17 min.
Will not travel the additional 4250 km to apogee given these conditions.
Are you accounting for the fact that as the rocket continues to ascend the value of g will continue to get lower?
If not then you’re doing it wrong. Doing so is complicated, I suggested an iterative way above which I think approximates it, but I simplified and assumed the rocket is going straight up. Feel free to show your workings.
First, let me point out again, I do not believe any of the claims made by RE in regard to g or this rocket.
Second, I am using the figures provided by RE and their proponents here.
Third, I took into account decreasing g and what you left out, the deceleration.
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Can you show us your calculation? It could put this to rest once and for all.
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Can you show us your calculation? It could put this to rest once and for all.
Given your willingness to accept any numbers trotted out by SteelyBob or the rest, I will wait to do so until I witness equitable demands made by you upon the others.
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Can you show us your calculation? It could put this to rest once and for all.
Given your willingness to accept any numbers trotted out by SteelyBob or the rest, I will wait to do so until I witness equitable demands made you upon the others.
Hey, if you can’t do it, no skin off my back. I’d just be surprised you can do calculus, since you seem convinced this is an algebra problem.
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Can you show us your calculation? It could put this to rest once and for all.
Given your willingness to accept any numbers trotted out by SteelyBob or the rest, I will wait to do so until I witness equitable demands made you upon the others.
Hey, if you can’t do it, no skin off my back. I’d just be surprised you can do calculus, since you seem convinced this is an algebra problem.
Hey, I already did it and presented my numbers.
No skin off my back if you desire to continue in your intellectually and philosophically disingenuous demands of one side and not the other.
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Can you show us your calculation? It could put this to rest once and for all.
Obviously he can’t. He had to use an online calculator to effectively work out the average of 2 numbers, and didn’t understand why that wasn’t appropriate for this problem.
So it seems implausible that he’s able to do the far more complex calculations involved here. I’m going to guess he’s used another online calculator for the second problem but I doubt that took into account how g continues to vary as the missile continues to ascend. As you’ve said, you need calculus for this sort of thing. I think my iterative approach is a reasonable approximation but I may have made a mistake and I made some simplifying assumptions.
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Can you show us your calculation? It could put this to rest once and for all.
Obviously he can’t. He had to use an online calculator to effectively work out the average of 2 numbers, and didn’t understand why that wasn’t appropriate for this problem.
^This is a flat out lie.
I was the first here to point out the average velocity of the rocket to an altitude of 250km in a time frame of 5 minutes, using d=rt, comes out to 3,000 km/h.
I also acknowledged on numerous occasions, that (final velocity + initial velocity)/2 is indeed a linear calculation.
I then pointed out that an average velocity calculation derived by calculus would not help you or the rest of the RE crowd here. In fact, it would only further serve to prove your bogus flight times and velocities provided at t+1, t+2, etc., as the bs they are.
I may have made a mistake.
Nope, you just posted a lie, as the record of the thread demonstrates.
In your follow up BS iteration program, you claim it demonstrates a final altitude of nearly 10,000km!
The maximum height would be 9,831km
As a reminder, the rocket we are discussing is claimed to have achieved an altitude of only 4,500 km.
I will also point out you set g = 9.8m/s2, which is the value at sea level, not at 250km.
Initial height h0 = 250km
g0 = 9.8 (g at sea level, roughly)
In total, your contributions to this thread have been entirely worthless.
Quite simply, you need to go back to following your own advice which you offered to me a couple of pages back.
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Because at a velocity of 16,000 km/h, at 250km, experiencing g=9.08m/s2, and decelerating, the rocket will reach 0 in 17 min.
Will not travel the additional 4250 km to apogee given these conditions.
According to your calculations, what was the rockets altitude after 17 minutes?
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I was the first here to point out the average velocity of the rocket to an altitude of 250km in a time frame of 5 minutes, using d=rt, comes out to 3,000 km/h.
OK. I mean, that's trivial - distance travelled divided by the time taken = average velocity.
I go 60km in 2 hours, my average velocity must have been 60/2 = 30.
So fine, you got that right but that's hardly rocket science, pun intended.
I also acknowledged on numerous occasions, that (final velocity + initial velocity)/2 is indeed a linear calculation.
OK, good. We are getting somewhere. But why then do you think that's appropriate for a rocket which as it burns through the fuel the acceleration will keep increasing till the engine shuts off. F=ma, the engine will keep producing the same"F" but the "m" keeps decreasing so the "a" will keep increasing. It's not a linear acceleration.
In your follow up BS iteration program, you claim it demonstrates a final altitude of nearly 10,000km!
Well yes. Because, as I said, I made a simplifying instruction of the rocket going straight up. Obviously an object going straight up at a certain velocity will end up going a lot higher than one going at a, say, 45 degree angle at that velocity.
Just calling it "BS" without explanation is meaningless and doesn't advance the discussion. If you think my method or code is flawed then explain why.
I will also point out you set g = 9.8m/s2, which is the value at sea level, not at 250km.
Yes. Correctly. Because when I calculate the value of gh at each step - the g at a certain altitude - I use g0 as the starting point.
In this line:
gh = g0 * ((re*re)/((re+hn)*(re+hn)))
As I said, it's possible I made a mistake in my code, but that isn't it.
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In your follow up BS iteration program, you claim it demonstrates a final altitude of nearly 10,000km!
Well yes.
And that is over double the claimed apogee of the rocket in question.
Just calling it "BS" without explanation is meaningless and doesn't advance the discussion.
I already wrote why it is BS.
It is BS because of two reasons, one being the rocket in question only achieved an apogee of 4500km, the other being a classic example of your worn and tired schtick - an interjection of meaningless dribble in order to detract from your prior failures in this thread.
I will also point out you set g = 9.8m/s2, which is the value at sea level, not at 250km.
Yes. Correctly. Because when I calculate the value of gh at each step - the g at a certain altitude - I use g0 as the starting point.
In this line:
gh = g0 * ((re*re)/((re+hn)*(re+hn)))
As I said, it's possible I made a mistake in my code, but that isn't it.
h0 = 250km cannot possibly be g=9.8m/s2, according to RE.
You are obviously trolling.
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And that is over double the claimed apogee of the rocket in question.
You accuse me of trolling, but you deliberately left out the bit where I explained why it was a lot higher than 4500km.
My initial code made a lot of simplifying assumptions, one of which was the rocket going straight up. If you want to maximise the height then you aim straight up. If you want to maximise the range then you aim at 45 degrees.
I have now amended the code which allows you to set an angle - the code is below. I ran the code with a few different angles, here are the results for an angle of 55 degrees.
The range is 14,597km, the maximum height is 4483km. The flight time is around 1 hour 35 minutes
Here's a graph of the flight which I made from the results:
(https://i.ibb.co/kmpyqTC/Missile.jpg)
Just to clear up one thing:
h0 = 250km cannot possibly be g=9.8m/s2, according to RE.
g0 is 9.8. I'm working out the values of g at different heights based on that value. It's outlined in this calculator:
https://www.vcalc.com/wiki/KurtHeckman/Gravity+Acceleration+by+Altitude
Here's the code. If you want to amend the angle then simply change the 55 in the first line.
This version splits the 16,000km/h into the vertical and horizontal parts based on the angle, like this:
(https://i.ibb.co/3TYL5Np/Velocity.jpg)
Only the vertical part is subject to gravity.
I've made the simplifying assumption that the horizontal velocity remains constant, i.e. I've made no allowance for wind resistance but given that the missile spends most of its time above much of the atmosphere I think that's OK.
angle = 55 * Math.pi/180
v0 = 16000
vhor = v0 * Math.Cos(angle)
vvert = v0 * Math.Sin(angle)
h0 = 250 * Math.Sin(angle)
range = 250 * Math.Cos(angle)
g0 = 9.80665
re = 6371.009
hn = h0 + (vvert/3600)
range = range + (vhor/3600)
gh = g0 * ((re*re)/((re+h0)*(re+h0)))
vvert = vvert - gh
hmax = hn
counter = 1
TextWindow.WriteLine(Math.Round(hn) + " ," + Math.Round(vvert) + " ," + gh)
While (hn > 0)
counter = counter + 1
gh = g0 * ((re*re)/((re+hn)*(re+hn)))
hn = hn + (vvert/3600)
if (hn > hmax) then
hmax = hn
endif
vvert = vvert - gh
range = range + (vhor/3600)
TextWindow.WriteLine(Math.Round(hn) + " ," + Math.Round(vvert) + " ," + gh)
EndWhile
TextWindow.WriteLine("hmax = " + hmax)
TextWindow.WriteLine("range = " + range)
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OK. Full disclosure. I made a mistake above. I adjusted the value of 'g' correctly, BUT when I updated the vertical velocity I mixed up my units - I was taking g at m/s from the velocity at km/h >o<
So...I've fixed my code and...
If the rocket is fired straight up, has a velocity of 16,000km/h (4444.444m/s) and a height of 250km at engine shut down then the maximum height, according to my code, will be: 1259km if you ignore the varying value of g with altitude
I'm pretty happy with that, because if you use this calculator you get a similar value
https://www.omnicalculator.com/physics/trajectory-projectile-motion
(about 1266km, but you can only put in 89 as the maximum angle and because I'm using an iterative approach I'd expect some discrepancy)
But if you take into account the variations in g then the maximum height will be 1553km
Here's my amended code and the site you can run it in. I've added some comments and cleaned it up a bit. This has been quite good fun, you know. Haven't written any code for years.
https://superbasic-v2.azurewebsites.net/
'Set up Initial Conditions - angle in degrees (90 is straight up)
'v0 is iniial velocity in meters/second
'dist is the distance in km travelled before engine shut down
angle = 90
v0 = 4444.44444
dist = 250
'Convert angle to radians
angle = Math.GetRadians(angle)
'work out vertical and horizontal parts of velocity and initial distance
vhor = v0 * Math.Cos(angle)
vvert = v0 * Math.Sin(angle)
hn = (dist * 1000) * Math.Sin(angle)
range = (dist * 1000) * Math.Cos(angle)
'Gravity at sea level and radius of earth
g0 = 9.80665
re = 6371009
'set the maximum height to the initial height
hmax = hn
seconds = 0
While (hn > 0)
seconds = seconds + 1
'Calculate the value of the g at the rocket's height
gh = g0 * ((re*re)/((re+hn)*(re+hn)))
'Add the vertical velocity to the height and the horizontal velocity to the range
hn = hn + vvert
range = range + vhor
'Update the maximum height if necessary
if (hn > hmax) then
hmax = hn
endif
'Adjust the vertical velocity by the value of g at this height - change gh to g0 to ignore these variations
vvert = vvert - gh
TextWindow.WriteLine(Math.Round(hn) + " ," + Math.Round(vvert) + " ," + gh)
EndWhile
TextWindow.WriteLine("hmax = " + hmax)
TextWindow.WriteLine("range = " + range)
TextWindow.WriteLine("time = " + seconds)
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Interesting data from Blue Origin today and how it may relate to some of the calculations here. Rough estimate of telemetry:
Blue Origin
MECO:
Time = T+ 02:21
Speed = 2,178 MPH
Altitude = 188,044 FT
G = 9.63
(https://i.imgur.com/2FaeBqs.png)
APOGEE:
Time = T+ 04:03
Speed = 000 MPH
Altitude = 351,210 FT
G = 9.48
(https://i.imgur.com/FwVH4NO.png)
Duration after MECO to Apogee: 01:42
Altitude gained after MECO: 161,748 FT
So after MECO, Blue Origin, unpowered, almost doubled its altitude. Interesting.
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I'm not sure those figures are right. I plugged them in to my calculator ©
And got 248,219ft if you don't take variations in G into account and 250,776ft if you do.
Now, it could be that my calculator is still wrong, but if you use these figures:
1878mph and 229,738ft - which is the point at which Zero G occurs, and thus is when I believe all engines must have shut off,
then my calculator gives you 352,310ft for apogee, which is pretty much spot on
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I'm not sure those figures are right. I plugged them in to my calculator ©
And got 248,219ft if you don't take variations in G into account and 250,776ft if you do.
Now, it could be that my calculator is still wrong, but if you use these figures:
1878mph and 229,738ft - which is the point at which Zero G occurs, and thus is when I believe all engines must have shut off,
then my calculator gives you 352,310ft for apogee, which is pretty much spot on
I'm not sure the G figures are correct. I used some online calculator for those. As for the other figures, I just pulled them from the telemetry displayed on the video feed. The bar on the left highlights each event as it occurs. For instance, when the rocket hits MECO, the word "MECO" brightens up. Same for "Separation/0 G". But I'm not really sure exactly when the latter occurred. But the latter lights up right about at:
T+ 02:35
Speed = 1873 MPH
Altitude = 230,234 FT
Approximately 14 seconds after MECO.
Again, not a super accurate assessment, just a ballpark. But still interesting that the capsule continues up another 160,000 FT after engine cut-off. It's basically just a bullet at that point.