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61
##### Flat Earth Projects / Re: Wiki - Moon
« on: May 21, 2019, 07:47:19 PM »
There is a very low cost and popular way to calculate Moon distance, using this recent developed technique below.

It is known that the Moon apparent size from the observer is 1.7% different when seeing it at the horizon than at Zenith.
When seeing the Moon at the horizon, the observer is farther almost the Earth radius, so the Moon will have a smaller apparent size.
The technique is to use your own smart phone over a tripod and take pictures from the Moon at horizon then at Zenith, and also note the Moon elevation (there are several smart phones app for that).

By the apparent size difference, elevation and Earth radius you can calculate its distance.

You don't need to be precise, an error of 10% or 38,000 km is accepted when calculating 380,000 km in distance, using such technique.

So, hundreds of people, not governmental agencies, could accumulate their calculations, and reach an average consensus.
If one could think that horizon atmosphere thickness could interfere in the apparent size, reaching a value closer to 384,000 km ±10%, or even ±50% will be a great result.

https://phys.org/news/2014-05-distance-moon.html

62
##### Flat Earth Community / Re: Does the lay Flat Earther understand Physics? Or are just in to conspiracies?
« on: May 21, 2019, 06:21:07 PM »

63
##### Flat Earth Projects / Re: Wiki correction: AN/DN
« on: May 21, 2019, 06:00:27 PM »
What does that have to do with the topic of the nodes?

Nodal lines are directly involved in solar (and lunar) eclipses.  Impossible to discuss one without the other.  Nothing better than eclipses to help calculate and justify the nodal lines.

64
##### Flat Earth Projects / Solar Eclipse Umbra and Penumbra sizes
« on: May 21, 2019, 05:56:51 PM »
As an interesting topic for improving general knowledge about light and shadow casting, I am starting here a discussion about how to calculate the Total Solar Eclipse Umbra and Penumbra sizes, angles of projection, etc, in order to calculate the proportional Sun and Moon sizes and respective distances.   As the umbra (the small disc of total shadow where you can not see the Sun during the eclipse) is a conical projection, it assumes a round shape if casting close to 90° angle to Earth's surface, or an ellipsoid when this angle decreases.

There are 8 variable in this project.

1.) The Sun distance
2.) The Moon distance
3.) The Sun diameter
4.) The Moon diameter
5.) The Umbra diameter if a circle
6.) The Umbra dimensions (length and width) if an elipse
7.) The Penumbra dimensions
8.) The Shadow casting Angle on Earth's surface

The Umbra and Penumbra round shape can tell us the conical result of projection, with 5 and 7 we can calculate the proportions between 3 and 4   and between 1 and 2 above.   That is pure optics and shadow casting.

A) If the Sun is bigger than the Moon (what it is), it will projects a Moon shadow in a conical form, smaller the diameter as far it gets from the Moon, until the distance is such that the cast shadow reach the apex, a very narrow point.  If nothing intercepts this cast conical shadow, it just disappears in the space, nobody sees the cone.  If anything comes inside such dark cone, it will intercepts and shows the shadow.   The apex distance from the Moon is directly proportional to the sizes Sun & Moon and the distance between them.

B) If the Sun and the Moon has the same size, it will projects a Moon shadow in a conical form with infinite apex, known as "cylinder".  The shadow diameter would be the same as the Moon and will be projected at far distances without changing size.

C) If the Sun is smaller than the Moon, it will projects a Moon shadow with an inverted cone, apex on the Sun.  The cone would increase in diameter as it distances from the Moon, blocking solar light in vast area, in such way that the entire Andromeda galaxy could be in this inverted cone shadow.

Considering history and records, close to 90° of Sun's position where Umbra happens on total eclipse, the Umbra diameter was never smaller than 120km in diameter.

Considering (A) above, the Moon must be way larger than 120km in diameter, the Sun must be way larger than the Moon and they are not close at all.  With the same measurement date for the Penumbra, one could calculate the size proportions between Sun & Moon and the proportional distances between them and Earth.

Considering (B), both Moon and Sun must have 120km in diameter in order to cast a same size shadow.  The problem here is that the Umbra diameter changes even at 90° (round Umbra), in different eclipses according to factors, one is the actual distance Moon-Earth at the moment of eclipse, another is the angle of shadow casting over Earth surface.  Considering only a closer to 90° Sun & Moon position, then both must have 120km in diameter.  The killing factor here is that on this situation the only way to have Penumbra is when the same size Sun is far away from the Moon, it can be calculated, measured, simulated on any optical lab or kitchen table.   Grab your flashlight, put it face down over a paper, using the flashlight face draw a circle with a pencil, cut this disc of paper (lets call it "blocker"), now projects the flashlight light beam 90° against the wall and use the blocker to block the light, change distances Flashlight-Wall, Flashlight-Blocker and see what happens.  No matter the distance from the wall, the blocker shadow diameter will be almost constant.   Now cut another blocker half the diameter of the first one and repeat the experience, you will notice the conical shadow projected on the wall, if the flashlight and the smaller blocker is away from the wall, there will be no casting shadow visible defined, as (A).   With the blocker same size of flashlight, the Umbra never changes size, and it changes in real life, this option (B) is eliminated.

Considering (C), Cut a blocker twice the diameter of the flashlight and play with it against the wall... it will be impossible, the cast shadow will always be larger than the blocker or the flashlight.  If flashlight and blocker are very close, there will be no light projected to the wall at all, no Penumbra. So, this option (C) is eliminated.  The only way to have a 120km round shadow Umbra (90°) and larger Penumbra for a 48km diameter Sun is for the Moon to be way from the Sun and larger.  It is easy to calculate.  With the FE Sun at altitude of 4800km and 48km in diameter, to project a Umbra shadow of 120km in diameter, the blocker (Moon) can be calculated using this formula:

SD = Sun Diameter
MD = Moon Diameter
SPD = Shadow Projected (on Earth) Diameter
H1 = Distance Sun to Cone Apex (behind/above the Sun)
H2 = Distance Sun to Moon
H3 = Distance Earth to Moon
H = H1+H2+H3

First you need to find H1

Suppose:
H2+H3 = 10
SD = 2
SPD = 8

2 = 8 * H1 / (10+H1)
2 * (10+H1) = 8 * H1
8 * H1 / 2 = 10+H1
4 * H1 - H1 = 10
H1 = 10 / 3
H1 = 3.333

Then we have the total cone H as 13.333.

Now you can calculate Moon Diameter (MD) for a certain distance (H3)

MD = SPD * (H2+H1) / H

Suppose all the numbers above the same, and H2 (Sun to Moon Distance) is 3.
MD = 8 * (3+3.333) / 13.333
MD =3.8
So, if the Sun (diameter 2) is 10 from Earth, Moon is 7 from Earth, Shadow is 8, Moon diameter must be 3.8, almost double the Sun.

If H2 = 1, MD = 8 * (1+3.333) / 13.333
MD = 2.6, still bigger than the Sun even being very much closer to the Sun.

So, for a Sun = 48km (SD=48), H2+H3 = 4800km, SPD = 120km

48 = 120 * H1 / (4800+H1)
48 * (4800+H1) = 120 * H1
120 * H1 / 48 = 4800+H1
2.5 * H1 - H1 = 4800
H1 = 4800 / 1.5 = 3200km

Now, calculate diameter of the Moon, in kilometers, by distance from Sun:

MD = 120 * (H2+3200) / 8000
MD = 0.015 * (H2 + 3200)
MD = H2*0.015 + 48

See, no matter the distance of the FE Moon from the Sun, it starts with at least 48km in diameter if very near the Sun, or 120km if very near to Earth. The Moon should increase 15 meters in diameter for every kilometer from the Sun.  The FE wiki says Sun and Moon are at the same altitude and size, in that case, a total solar eclipse would obscure the whole Earth. On that Nodal calculation of the Moon performing an angled path under and over the Sun, it becomes clear that the FE Moon would be not at the same altitude as the Sun during a total eclipse, it will be lower, so, to projects a 120km Umbra, the Moon needs to be bigger than the Sun.  If FErs publish the numbers, angle of inclination, altitude, etc, one could even calculate the Moon diameter based on the 120km Umbra and Penumbra.   It will be interesting to see those FE numbers, mostly to produce the very large penumbra that can cover diferentes continents at once.

Then you need to use the Penumbra numbers to calculate how far is the Sun from the Moon.  For example, at the maximum point of the Great American Eclipse in August 2017, the penumbral shadow spanned from Panama all the way to Greenland, covering North America, Central America, and large parts of the northern polar regions.  For such gigantic Penumbra, only option (A) is possible.  Moon is way bigger than 120km in diameter, very far away from Earth and the Sun is way bigger than the Moon, 400 times farther than the Moon.

Interesting is that if you use the RE numbers, Umbra and Penumbra fits perfectly on the solar eclipses numbers on record, calculating with Sun and Moon distances, angles of projections and ellipsoid sizes if lower than 90° projection, and lines of nodes.  The average Umbra conical shadow distance is no more than 380,000 km.

I wonder if FEs could provide some simple calculations to justify their own numbers for Sun/Moon altitude and lines of nodes, to create the same Umbra/Penumbra results.

Suppose FE Moon is farther from the North Pole than the Sun, shadow lower than 90°, and by some way it projects a shadow further yet, the ellipsoid cast shadow would be wider in the farther side from the North Pole, and narrow on the closer side.  Also, the farther side on FE will be fully projected on the ground, with good contrast, while on the RE the further side disappears due planet curvature.  The same would happen if Sun/Moon aren't well aligned to 90° and over a little different longitude.

Records show (and it is possible to calculate) the Umbra speed is around 1700km/h Eastward, the total eclipse umbra lasts no more than 7 1/2 minutes at a specific point.  A Concorde flight was able to fly under Umbra at Mach 2.5 on June 1973 for more than 74 minutes.

https://www.nature.com/articles/246072a0

MIR image shadow from the Solar Eclipse from Aug/11/1999:

From ISS on March 29, 2006:

I invite everyone to participate in this exercise of numbers.

65
##### Flat Earth Theory / Re: Wiki article of the day: LM Closer Look
« on: May 21, 2019, 01:27:15 PM »
Interesting thread, but you guys have provided zero evidence for us to consider and discuss on this matter. Are we supposed to discuss what you "think"? Try to come up with actual content to consider.

I live in Central Florida, went to NASA dozens of times.  There you can see, read, understand.  Coming from mils bground, that is exactly how things are done, brute, simple, easy, it needs to work, sometimes to work just once, it is not made for beauty. Other than that there is plenty of evidence at the Net, people who actually build such machines, engineers, diagrams, datasheets, astronauts, photos, movies, and much more.

"Evidence" ?  Interesting word coming from a person who provides NONE EVIDENCE about his own preaching.   And yes, all over the world people wipe their *** with a flimsy very fragile and thin paper, not because people love to live dangerously close to have their fingers dirty, but because 999 in a 1000 it works nicely. Even you do it, but you never requested evidence from your mother about it.  That is how science works.  You try hundreds of times, if it works repeatedly, then it is evidence enough.

I have a long list of FE evidence requests here, several of them posted without a single answer, but I will not use this post to stir the pot.

Cheers,

66
##### Flat Earth Projects / Re: Wiki correction: AN/DN
« on: May 20, 2019, 07:42:04 PM »
What force makes the FE Moon goes up and down cycle, and never colliding with the Sun?

How do you explain Eclipses of the Moon during full moon?  Another darker shadow object in middle?  If it can cast a shadow over the full moon, and based on the shadow arc angle it is much bigger than the moon, it is not transparent to cast shadow, so it blocks solar light and become illuminated, but we can not see it, it never blocked any patch of stars, is that it?

In your new explanation, the full moon becomes significantly higher than the sun.  Can you post the numbers please, so I can calculate changes in visible and measurable size?

Also, to have a complete orbit every ±28 days, it means the FE moon drags almost 51 minutes per day behind the FE sun, so, during the solar eclipse it would be casting a shadow of itself all over the earth for more than a full day/circle, since it will be aligned within 12°, almost vertical. It means, on FE the solar eclipse shadow would be visible all over the world during more than a complete circle.  That is not what happens on the real world.  RE Solar Eclipses are fast (few hours or less than that) and just over a certain part of the earth.   Care to explain the discrepancy please?

67
##### Flat Earth Theory / Re: Wiki article of the day: LM Closer Look
« on: May 20, 2019, 07:19:31 PM »
As an extra information, the LEM was not made to stand life support, it was just some kind of extra light vehicle to take astronauts from the orbiter to the lunar surface and back.  As a matter of fact it was very fragile, you could rupture its walls with a rock trow.  It doesn't needed to be solid or resistant, other than the structure to hold the fuel tanks and the rockets.  It has no sits or chairs, the astronauts stay standing while landing or launching.  Everything was done thinking in light weight, less fuel overall.  Each extra gram on LEM would need more fuel, would need more fuel on Saturn V, the weight multiplication was huge.  Every time one of the engineering groups could reduce few grams without compromising everything, it was commemorated. The clothes, boots, tools, cameras, films, food, should be lighter, smaller.  The astronaut gloves were made several times, each time lighter and more functional.  The boots dole was special made polymer, lighter and stronger.  Lots of things were left behind on launch back to the orbiter, just to have less weight possible for the rockets and fuel.  Everything was meticulously calculated and planned. Thousands of engineers and technicians worked very hard, thousands of hours, to reach that final possible solution.  Some people can't even understand how their fridge works.

68
##### Flat Earth Community / Re: Does the lay Flat Earther understand Physics? Or are just in to conspiracies?
« on: May 16, 2019, 10:42:13 PM »
Questions 261, 273 and 299 are absurdly pure ignorance, no answer.   There are few continents in this world, each one has its own great mountain, some very well known; Everest, K2, Namcha.  The Aconcagua was even depicted on a Disney movie about South America, the small animated airplane trying to overcome it.  Anyone that paid attention in science class remember the name Anemometer.  Christiaan Barnard the first heart transplant surgeon, had plenty of publishing on the media, impossible no one knows his name.  Just one in four knew the name "odometer" for the car's distance (not only mileage as stated, jeez...).   If some math or science questions would be there, a disaster. The test say "general information"... hmm.

69
##### Flat Earth Theory / Re: Universal Acceleration - how does the ISS maintain an orbit?
« on: May 16, 2019, 09:40:26 PM »
Also, pay attention that ISS orbit is not circular around the North Pole, there is an orbital inclination of 51.6°, each orbit takes around 92 minutes, depending on ISS altitude.  So, how and why the ISS altitude changes its orbital time?   And, what kind of force or energy makes each orbit to advance its non circular path over the same place on FE ?  On RE it is very much simple to explain, indeed, not needing any extra energy to advance its position each orbit.

This off the center path on FE would put the orbital pivot closer to equator line, but then this circling would cover much more degrees when closer to North Pole than when closer to the edge (ice wall).  What causes this totally not circular (more an egg shape path) motion on FE ?

Observe on the second image above, how easy is to perform such orbital inclined path, just apply some angle and voilá, it will automatically advance its position above ground for each orbit, even considering the path to be a straight line over the globe, with just few corrections for altitude and direction now and then, just to keep going on the planned path.  This is why FErs want terrible to dismiss any possibility to admit the existence of satellites, it is incredible difficult (to impossible) to explain them over FE.

Below a drawing of what ISS orbit would be (without the position advancement) over FE

70
##### Flat Earth Theory / Re: Gravity is not constant so why is acceleration?
« on: May 16, 2019, 08:49:12 PM »
Air pressure does not only affect scales trivially. See the following illustration and text:

https://www.artofmanliness.com/articles/fair-or-foul-how-to-use-a-barometer/

Quote

Air pressure decreases as altitude increases.

Atmospheric pressure — or barometric pressure — is simply the weight of the air at ground level. It’s a little easier to understand when you think about the concept of water pressure first. As you get deeper in water, the pressure increases. This is because as you descend, the built up weight of the water on top of you increases. In 1 foot of water, you have the weight of that foot of water pressing down on you. In 2 feet of water, you have the weight of an extra foot of water pressing on you. It’s quite logical, really.

Tom, I think you understand very well the difference between absolute and differential pressure gauges, how the absolute needs calibration (zero tara) when moved to a difference place, etc.    A Bourdon tube needs needle zero adjustment for that particular location.  A differential gauge needs not adjustment whatsoever, only the factory relationship to the needle movement to differential pressure, thus, calibration.  Most spring based mechanical scales are considered differential, being "gravity" one of the sides of the measurement element.  The force necessary to move the spring will not change based on air pressure, the mass doesn't change eight, but over a mountain the gravity will change, also changing the weight you can measure of such mass.

A regular barometer (Aneroid, Torricelly, open tube, etc) is an absolute gauge that measures the inner pressure and atmospheric pressure.

Sorry, your statement is not really correct and carries a huge misconception.  When you descend into a deeper water, the pressure of the water is not OVER you, is all around you, it does NOT make you weight heavier over any fixed regular scale, as a matter of fact, you would be lighter, your body will be less dense than the water. This is why diver workers need to use extra weight on their suits to counter effect buoyancy, in order to reach deeper water.

Unfortunately your posted drawing is not trying to represent anything else than atmospheric pressure gauge readings using an absolute air pressure sensor (barometric measurement), nothing to do with mass measurements unde different gravity.   The same changes in different mass by gravity could be exercised on Earth or the Moon (no atmosphere).

Also, your image posted comes from this website http://www.flightlearnings.com/2011/10/23/atmosphere-atmospheric-pressure/ related to atmospheric pressure, so, why you use this to try to make a reference to different mass measurement for different altitudes?  What you thought is the other way around, if only thinking about air pressure interfering with weight over a scale, more pressure around you including underwater would make you lighter, not heavier, by buoyancy effect.

71
##### Flat Earth Theory / Re: Total Eclipse July 02 2019
« on: May 16, 2019, 08:06:16 PM »
Here the animation of what will be the July 2nd 2019 solar eclipse... large penumbra, small umbra.
Watch and learn

72
##### Flat Earth Community / Re: Does the lay Flat Earther understand Physics? Or are just in to conspiracies?
« on: May 16, 2019, 07:54:26 PM »
Most of those things are facts taught in elementary school

Not really.   Even so, everybody at some point of academic life did in fact learn how to calculate hypotenuse from a rectangle triangle, how many from those really remember that?  Some even did learn how to calculate integrals, some became scientists or teachers, not everyone.  Most of people just went to school because there is a normal path, not because they wanted to learn. Example, all US teens at some point did learn few years of Spanish at High School, ask people around you (not Hispanic descendants, or even them) to say "this is a wonderful sunny morning" in Spanish.  All adult Americans that I know and already had a conversation about different languages, remember how to say "holla" in Spanish, nothing else.  Most of them had no interest to learn it, in their brain this learning had no future use, so they just learn enough to have the tests grades, nothing else, and immediately forgot about it.  I would guess that more than 70% of the average people hate math, biology and sciences, and now as adults none of those remember how to calculate the volume of a sphere or the factorial of 4, or the sulfuric acid chemical composition.  People are not the same, they may looks like the same, their brain, memory and inner working differ very much, and that makes a huge difference.  Square root of 3 is a simple test to ask around, make your own statistics.  Don't even try to ask about the speed of light, that would be overboard for most people.

73
##### Flat Earth Community / Re: Does the lay Flat Earther understand Physics? Or are just in to conspiracies?
« on: May 16, 2019, 05:15:15 PM »
It is statistical that 97 in 100 people have no idea about how an airplane or a bird flies.
More than 95 in 100 doesn't know the cycle of rain.
More than 95 still thinking plants and trees are the responsible for the oxygen on the planet.
More than 98 have no idea why it is hot in the summer and cold in the winter.
Less than 1% can tell you by memory the names of the chemical elements Na, Ag, Au, Cu, S.
Less than 1% heard any of the following expressions; "NPN, If-Then-Else, Sin(x), nanosecond, 50ml, v=d/t"
Less than 1% can understand how to calculate with 9.8m/s².
Less than 1% can tell you the 5 decimal digits of PI number.
Less than 5% have no idea what PI is used for.
Less than 1% can tell you the value of hypotenuse of a isosceles right triangle with sides equal to 1
Less than 1% can tell you the square root of 2 or 3
For lots of people "tangent" may be a color
Lots of people still believing in ghosts, spirits and supernatural
Less than 1% can tell you how many cubic cm into a 1 cubic meter
Very few can calculate the area of a circle
Less than 1% knows how may pieces are in a chess game board.
Less than 1% knows what Tegucigalpa is and where it can be found, not knowing if it is a city, a lake or a fruit's name.
A very little percentage of the world's population ever entered into an airplane, much less traveled to another country.
Less than 1% knows how an air conditioned, fridge or freezer unit works.
Less than 2% knows the function of pancreas in the human body.
Very few people will know where Hydrocarbons can be found.
Very few use 10 fingers on a computer keyboard.
Very few read at least one book per year.

This is why we, humans, institucionalize schools and teachers, research and scientific development, trying to perpetuate knowledge and even improve it.
This is why there are scientists, universities, technology development, health and food improvement.

The stupid and ignorants are easy pray for any wrong information and manipulation, bad guys knows it, and use it for their own gain and profit.
It is no surprise why most of the world's population still poor, hungry and sick.

Lots of people do not have a minimum idea where North direction is, or even read a map on paper.  Do you really expect them to have a 3D mental mechanic visualization capability to understand the universe, gravity, elliptic orbit, shadow projections, photons and radiation path and refraction, stars, space and time, etc?   They have no idea even what "elliptic" means.  They can not understand.

For those people, if you promise them "a diploma, or making part of a group" without needing to think, learn, use a calculator, study, etc, just by following you, and listening to you saying "believe me, I know it", they will follow you easy, they will pile up to share your "knowledge without efforts".  That is how humanity is behaving for thousands of years.   The worst atractors are the ones without any logical of scientific explanations, you just need to believe, and that's it.  On purpose they don't explain by numbers just to have an easy escape route in case of any contesting.  If you state your home has 2000 square feet facing north, it is better to have it right, somebody may prove you wrong and a liar.  But if you say "my home is big, facing the wind"... you can always get rid of someone trying to prove you wrong.  A nasty trick used by many crooks.

One of the best things I learn in life was that people are not the same.  If you learn something a little more than your neighbor, congratulations, but you will be along with few ones in this universe.  You can try to teach your neighbor, but don't expect that he will be capable to understand a single word you say, or worse, he may not even want to hear you.  For lots of people, the act of "thinking" hurts, it is difficult, tiresome, even boring.

Science and experience proves that to build a bridge or a building you need to do a lot of calculations, it could not be done only by guessing and believing.

74
##### Flat Earth Theory / Re: Total Eclipse July 02 2019
« on: May 16, 2019, 04:49:41 PM »
Wow... the Oppolzer northern hemisphere 1900-1918 eclipses map posted on FE wiki Eclipses, as a reference to promote FE, had a hidden twin map for the South Pole (Antarctic continent) as center.  That is fantastic.  So, FE now improved, it is double-sided.  See how Australia's shape is more real. I wonder which way UA pushes it.

Calculating here how Sun and Moon, both at 4800km altitude and 48km in diameter could promote the strange total eclipse path on  11/13/2012, changing more than 30° of latitude in a matter of few hours, considering only a small 12° of circling difference (longitude) between Sun and Moon in 24 hours See, same altitude, can only promote a straight down vertical shadow, total or partial, never angled.    July/22/2028 will have more than 40 degrees of latitude change.  FE behavior is amazing.  Someone may say the Moon is way down below the Sun, with a chaotic circling path (we don't observe that in the real world), but both Sun and Moon being 48km in diameter the projected umbra shadow will never be wider then 48km, the minimum noted all times was never smaller than 120km.

A lot of things don't add up. FErs scientists and high knowledge specialists need urgent to define and post the right numbers.

75
##### Flat Earth Theory / Total Eclipse July 02 2019
« on: May 14, 2019, 05:48:11 PM »
Can any FEr demonstrate how the Total Solar Eclipse of July 2nd 2019 is predicted under FE map and conditions?
How the FE Moon comes under the FE Sun on that particular path?

76
##### Flat Earth Theory / Re: How does the Sun Create Energy on a Flat Earth?
« on: May 14, 2019, 03:16:06 PM »
Any spectrograph measurement, using diffraction gratings (>2000gr/mm), shows exactly the radiation solar spectrum, no matter if you think FE or RE, the Sun doesn't change.  The analysis points to Hydrogen fusion into Helium, releasing energy, no matter what.  So, FE Sun is a Hydrogen Fusion Reactor, but considering its tinny small size of only 30km in diameter, it is totally impossible to accumulate enough gas to promote such pressure and temperature necessary to ignite the fusion process. Not even considering that gravity doesn't exist in the FE world, any gas would disperse in vaccuum.   In the real universe, not even Jupiter could do it, with a diameter of 142984km, 4766 times larger diameter than the 30km FE Sun) composed with 90% hydrogen, mass equivalent to 318 Earths.  The smallest Red Dwarf star, is 80 times bigger than Jupiter, so it needs to pack another 79 Jupiters into the actual one for it to have a narrow chance to ignite fusion and become a star.  We are talking about 381000 times larger than the FE Sun.

Want to make a comparison?  Think about a 1/2" (12.7mm) small glass marble as being the FE Sun, now, 12.7mm x 381000 = 3.024 miles, that is the equivalent diameter of more than 650 city blocks put together to form a 3 miles diameter circular area (do you want me to post the calculation?), or the equivalent to 519,841,729 US school buses piled into a huge ball, and that is the smallest Red Dwarf known to be able to ignite.  Do you really think a tinny glass marble 1/2" in diameter will ignite fusion?

US total yellow school buses in 2015/2016: 474194.  It would be necessary 1096 times the entire US school bus fleet to build such 3 miles diameter ball, just to make a small Red Dwarf ignite as a star, when compared to the size of FE Sun as a small 1/2" marble.   Think again.
https://files.schoolbusfleet.com/stats/SBFFB18StateByState.pdf
A regular US school bus is 2.6m wide, 13.7m long, 3.2m high, 114m3
A 3 miles diameter ball has a volume (V=1.33*PI*R*R*R) 5.92 E+10m3

Below the solar spectrograph, with the black absortion lines showing its radiation and gases composition.

77
##### Flat Earth Theory / Re: Scientific proof??
« on: May 13, 2019, 07:51:03 PM »
With so many online applications and software helping to locate the next visible satellite at evening time on your exact latitude/longitude, is there anyone in this world that never saw a cruising satellite reflecting solar light, yet???  Start looking up.

78
##### Flat Earth Theory / Re: How is it possible to see the sun rise or set?
« on: May 13, 2019, 07:32:23 PM »
Sorry Reer, you are wrong, about the Sun's altitude and viewing angle.

The FE statement for the Sun is 30 km in diameter, 3000 km in altitude.  I made no calculations whatsoever, but it seems FErs use this altitude because it is the only possibility to flat a sphere with a very far away Sun with parallel rays and have the same shadows based on the oblate spheroid model.  Also, the diameter is purely based on apparent size of view (angular size).

Then, based on your assumption, the Sun being over the ICE wall (worst case) and the observer being also over the 180° opposite ICE wall, the rectangle triangle would have a base of 20000 km and the vertical of 3000 km, what gives a (atan(3/20)) of 8.53 degrees.  This would be the lowest inclination (altitude) the Sun would appear anywhere over the FE for an observer.  Anywhere the observer or the Sun moves, the altitude will increase.

The best possible analogy for what is 8.5° of altitude, is looking to your home front door from the curb across the street.   A regular door is about 80 inches tall, a regular city street is about 30 ft wide plus 15 ft from the curb to the door, total 45ft = 540 inches.  It would be atan(80/540) = 8.4°.

So, just walk to the curb across the street and look back to the top of your home front door, that is the lowest altitude the Sun would be anywhere over FE.

Now, thinking about apparent size.  If the FE Sun right over you will have "x" view diameter, and it is 3000 km of altitude, on that viewing experience Ice wall to Ice wall, the hypotenuse will be sqr(3000²+20000²) = 20223 km, the delta size = 3000/20223 = 0.15 or 15%.  Suppose the apparent Sun size right over you is around a US Quarter Coin, at that longest distance it will be the size of your shirt button.  That is big enough to be completely visible and shinning bright on the sky, mostly considering that (according to FE wiki) the Sun is a globe spinning, shinning in all directions, not only as a disc spotting light down, as it was said before.   Notice that according to this size and altitude, vanishing point does not make it disappear at all. It would looks like a street lamp at 150ft (50m) away.

So, where is the night sun?

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##### Flat Earth Theory / Re: Theory/Model Request
« on: May 13, 2019, 04:00:25 PM »
Oh so are you assuming that the sun is just a big lamp or what?

Hmmm, I guess it is not a natural fusion floating device generating so much energy in a flimsy 30km diameter thing.  Even Saturn is not big and dense enough to ignite fusion.  If yes, we need to find out how it works so we could create few more on the earth's surface... can you imagine the free energy?   30km is my daily comute distance, pretty small.  It is not fission device, at such distance we would be all cooked by radiation.

80
##### Flat Earth Theory / Re: Angle and Length of a pole's shadow
« on: May 13, 2019, 03:54:33 PM »
Curious as to why you chose 4pm instead of 3pm?  3pm would have a 45° elevation angle just like 9am does, but 4pm will not.

45° is already there at 9pm, why repeat?
Now, 4pm was chosen exactly for the little offset on the shadow and angle, as a "control experience" and being high enough in the sky to suffer minuscule refraction on the atmosphere.

I wonder if Tom Bishop would have time to help with the numbers.

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