1461

Science & Alternative Science / Re: FE and ICBMs

« on: July 09, 2021, 05:02:57 PM »

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.

1462

Science & Alternative Science / Re: FE and ICBMs

« on: July 09, 2021, 04:47:43 PM »

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.

1463

Flat Earth Investigations / Re: Branson to go only 55 miles up !

« on: July 09, 2021, 04:02:58 PM »

As we witnessed the dome appears at 73 miles

Why do you keep posting that video and saying it's a rocket "hitting the dome"?
It's a despinning device. You can see that the rocket stops spinning, it doesn't stop or smash into pieces as it would it if was hitting something solid at that speed.

You are claiming the rocket in the video is still traveling vertical from point of origin?

1464

Science & Alternative Science / Re: FE and ICBMs

« on: July 09, 2021, 03:58:14 PM »

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.

1465

Science & Alternative Science / Re: FE and ICBMs

« on: July 09, 2021, 03:54:24 PM »

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.

1466

Science & Alternative Science / Re: FE and ICBMs

« on: July 09, 2021, 03:27:26 PM »

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.

1467

Science & Alternative Science / Re: FE and ICBMs

« on: July 09, 2021, 03:11:39 PM »

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:



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.

1468

Suggestions & Concerns / Re: Concerning all the choices in how to post

« on: July 09, 2021, 01:57:53 PM »

Thank you again.

1469

Science & Alternative Science / Re: FE and ICBMs

« on: July 09, 2021, 01:33:55 PM »

^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."

1470

Suggestions & Concerns / Concerning all the choices in how to post

« on: July 09, 2021, 12:34:41 PM »

I did a search (and perhaps it wasn't specific enough - if not I apologize) but could not find a user guide for using these choices of formatting here at TFES.

Is there a guide as to how and why one would use BBC or not use BBC, what the horizontal rule button is for, and the purpose of toggle view?

Thank you.

1471

Science & Alternative Science / Re: FE and ICBMs

« on: July 09, 2021, 11:58:12 AM »

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.

1472

Science & Alternative Science / Re: FE and ICBMs

« on: July 09, 2021, 11:20:42 AM »

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.

1473

Science & Alternative Science / Re: FE and ICBMs

« on: July 09, 2021, 10:26:47 AM »

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.

1474

Science & Alternative Science / Re: FE and ICBMs

« on: July 09, 2021, 10:25:46 AM »

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.

1475

Science & Alternative Science / Re: FE and ICBMs

« on: July 08, 2021, 06:12:12 PM »

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.

1476

Science & Alternative Science / Re: FE and ICBMs

« on: July 08, 2021, 06:08:32 PM »

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.

1477

Science & Alternative Science / Re: FE and ICBMs

« on: July 08, 2021, 05:58:23 PM »

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.

1478

Science & Alternative Science / Re: FE and ICBMs

« on: July 08, 2021, 05:36:17 PM »

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.

1479

Science & Alternative Science / Re: FE and ICBMs

« on: July 08, 2021, 04:23:55 PM »

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.

1480

Science & Alternative Science / Re: FE and ICBMs

« on: July 08, 2021, 03:56:54 PM »

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?