#### Action80

• 1165
##### Re: FE and ICBMs
« Reply #380 on: July 16, 2021, 04:49:32 PM »
supply the average rate of velocity for the profile.

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.
« Last Edit: July 17, 2021, 05:24:43 PM by Action80 »

#### AllAroundTheWorld

• 4968
##### Re: FE and ICBMs
« Reply #381 on: July 16, 2021, 05:47:53 PM »
somehow magically continuing to accelerate without engine power now, upwards to a final velocity of 25,000 km/h!
Who is claiming that?

Quote
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:

Code: [Select]
`v0 = 16000h0 = 250g0 = 9.80665re = 6371.009hn = h0 + (v0/3600)gh = g0 * ((re*re)/((re+h0)*(re+h0)))vn = v0 - ghcounter = 1TextWindow.WriteLine("h1 = " + hn + "v1 = " + vn + " g1 = " + gh)While (vn > 0)counter = counter + 1gh = g0 * ((re*re)/((re+hn)*(re+hn)))hn = hn + (vn/3600)vn = vn - ghTextWindow.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.
"On a very clear and chilly day it is possible to see Lighthouse Beach from Lovers Point and vice versa...Upon looking into the telescope I can see children running in and out of the water, splashing and playing. I can see people sun bathing at the shore
- An excerpt from the account of the Bishop Experiment. My emphasis

#### Action80

• 1165
##### Re: FE and ICBMs
« Reply #382 on: July 17, 2021, 11:32:39 AM »
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.

#### Rama Set

• 8705
• Round and round...
##### Re: FE and ICBMs
« Reply #383 on: July 17, 2021, 01:36:07 PM »
AATW is busted because Steely Bob said something? Holy North Korea, Batman!
Th*rk is the worst person on this website.

#### AllAroundTheWorld

• 4968
##### Re: FE and ICBMs
« Reply #384 on: July 17, 2021, 01:55:55 PM »
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.
"On a very clear and chilly day it is possible to see Lighthouse Beach from Lovers Point and vice versa...Upon looking into the telescope I can see children running in and out of the water, splashing and playing. I can see people sun bathing at the shore
- An excerpt from the account of the Bishop Experiment. My emphasis

#### Action80

• 1165
##### Re: FE and ICBMs
« Reply #385 on: July 17, 2021, 02:15:44 PM »
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.

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.
« Last Edit: July 17, 2021, 04:04:51 PM by Action80 »

#### AllAroundTheWorld

• 4968
##### Re: FE and ICBMs
« Reply #386 on: July 17, 2021, 02:27:30 PM »
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.
"On a very clear and chilly day it is possible to see Lighthouse Beach from Lovers Point and vice versa...Upon looking into the telescope I can see children running in and out of the water, splashing and playing. I can see people sun bathing at the shore
- An excerpt from the account of the Bishop Experiment. My emphasis

#### Action80

• 1165
##### Re: FE and ICBMs
« Reply #387 on: July 17, 2021, 02:37:25 PM »
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.

#### Rama Set

• 8705
• Round and round...
##### Re: FE and ICBMs
« Reply #388 on: July 17, 2021, 02:38:22 PM »
Can you show us your calculation? It could put this to rest once and for all.
Th*rk is the worst person on this website.

#### Action80

• 1165
##### Re: FE and ICBMs
« Reply #389 on: July 17, 2021, 02:51:48 PM »
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.
« Last Edit: July 17, 2021, 03:12:04 PM by Action80 »

#### Rama Set

• 8705
• Round and round...
##### Re: FE and ICBMs
« Reply #390 on: July 17, 2021, 02:57:56 PM »
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.
Th*rk is the worst person on this website.

#### Action80

• 1165
##### Re: FE and ICBMs
« Reply #391 on: July 17, 2021, 03:11:31 PM »
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.
« Last Edit: July 17, 2021, 03:27:19 PM by Action80 »

#### AllAroundTheWorld

• 4968
##### Re: FE and ICBMs
« Reply #392 on: July 17, 2021, 03:41:19 PM »
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.
"On a very clear and chilly day it is possible to see Lighthouse Beach from Lovers Point and vice versa...Upon looking into the telescope I can see children running in and out of the water, splashing and playing. I can see people sun bathing at the shore
- An excerpt from the account of the Bishop Experiment. My emphasis

#### Action80

• 1165
##### Re: FE and ICBMs
« Reply #393 on: July 17, 2021, 03:55:02 PM »
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)

Quite simply, you need to go back to following your own advice which you offered to me a couple of pages back.
« Last Edit: July 17, 2021, 04:56:42 PM by Action80 »

#### stack

• 2403
##### Re: FE and ICBMs
« Reply #394 on: July 17, 2021, 05:57:14 PM »
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?

#### AllAroundTheWorld

• 4968
##### Re: FE and ICBMs
« Reply #395 on: July 17, 2021, 09:54:35 PM »
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.

Quote
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.

Quote
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.

Quote
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.
"On a very clear and chilly day it is possible to see Lighthouse Beach from Lovers Point and vice versa...Upon looking into the telescope I can see children running in and out of the water, splashing and playing. I can see people sun bathing at the shore
- An excerpt from the account of the Bishop Experiment. My emphasis

#### Action80

• 1165
##### Re: FE and ICBMs
« Reply #396 on: July 17, 2021, 10:47:57 PM »
Quote
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.

Quote
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.
« Last Edit: July 17, 2021, 10:50:19 PM by Action80 »

#### AllAroundTheWorld

• 4968
##### Re: FE and ICBMs
« Reply #397 on: July 18, 2021, 07:59:56 PM »
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:

Just to clear up one thing:

Quote
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:

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.

Code: [Select]
`angle = 55 * Math.pi/180 v0 = 16000vhor = v0 * Math.Cos(angle)vvert = v0 * Math.Sin(angle)h0 = 250 * Math.Sin(angle)range = 250 * Math.Cos(angle)g0 = 9.80665re = 6371.009hn = h0 + (vvert/3600)range = range + (vhor/3600)gh = g0 * ((re*re)/((re+h0)*(re+h0)))vvert = vvert - ghhmax = hncounter = 1TextWindow.WriteLine(Math.Round(hn) + " ," + Math.Round(vvert) + " ," + gh)While (hn > 0)counter = counter + 1gh = g0 * ((re*re)/((re+hn)*(re+hn)))hn = hn + (vvert/3600)if (hn > hmax) then     hmax = hn endif vvert = vvert - ghrange = range + (vhor/3600)TextWindow.WriteLine(Math.Round(hn) + " ," + Math.Round(vvert) + " ," + gh)EndWhileTextWindow.WriteLine("hmax = " + hmax)TextWindow.WriteLine("range = " + range)`
« Last Edit: July 18, 2021, 08:12:23 PM by AllAroundTheWorld »
"On a very clear and chilly day it is possible to see Lighthouse Beach from Lovers Point and vice versa...Upon looking into the telescope I can see children running in and out of the water, splashing and playing. I can see people sun bathing at the shore
- An excerpt from the account of the Bishop Experiment. My emphasis

#### AllAroundTheWorld

• 4968
##### Re: FE and ICBMs
« Reply #398 on: July 20, 2021, 04:59:08 PM »
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
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/

Code: [Select]
`'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 downangle = 90v0 = 4444.44444dist = 250'Convert angle to radiansangle = Math.GetRadians(angle)'work out vertical and horizontal parts of velocity and initial distancevhor = 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 earthg0 = 9.80665re = 6371009'set the maximum height to the initial heighthmax = hnseconds = 0While (hn > 0)seconds = seconds + 1'Calculate the value of the g at the rocket's heightgh = g0 * ((re*re)/((re+hn)*(re+hn)))'Add the vertical velocity to the height and the horizontal velocity to the rangehn = hn + vvertrange = range + vhor'Update the maximum height if necessaryif (hn > hmax) then    hmax = hnendif'Adjust the vertical velocity by the value of g at this height - change gh to g0 to ignore these variationsvvert = vvert - ghTextWindow.WriteLine(Math.Round(hn) + " ," + Math.Round(vvert) + " ," + gh)EndWhileTextWindow.WriteLine("hmax = " + hmax)TextWindow.WriteLine("range = " + range)TextWindow.WriteLine("time = " + seconds)`
"On a very clear and chilly day it is possible to see Lighthouse Beach from Lovers Point and vice versa...Upon looking into the telescope I can see children running in and out of the water, splashing and playing. I can see people sun bathing at the shore
- An excerpt from the account of the Bishop Experiment. My emphasis

#### stack

• 2403
##### Re: FE and ICBMs
« Reply #399 on: July 20, 2021, 10:44:47 PM »
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

APOGEE:
Time = T+ 04:03
Speed = 000 MPH
Altitude = 351,210 FT
G = 9.48

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.