Re: FE and ICBMs
« Reply #200 on: June 15, 2021, 12:55:59 PM »
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...
"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

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Online Iceman

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Re: FE and ICBMs
« Reply #201 on: June 15, 2021, 01:01:17 PM »
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.

Offline Action80

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Re: FE and ICBMs
« Reply #202 on: June 15, 2021, 03:39:09 PM »
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.
« Last Edit: June 15, 2021, 03:43:51 PM by Action80 »

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Online stack

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Re: FE and ICBMs
« Reply #203 on: June 15, 2021, 04:59:10 PM »
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?

Re: FE and ICBMs
« Reply #204 on: June 15, 2021, 05:08:46 PM »
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?   


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Offline Rama Set

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Re: FE and ICBMs
« Reply #205 on: June 15, 2021, 06:32:34 PM »
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.
Th*rk is the worst person on this website.

Offline SteelyBob

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Re: FE and ICBMs
« Reply #206 on: June 15, 2021, 09:03:37 PM »
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.

Offline Action80

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Re: FE and ICBMs
« Reply #207 on: June 17, 2021, 06:01:20 PM »
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.

« Last Edit: June 18, 2021, 10:49:44 AM by Action80 »

Offline WTF_Seriously

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Re: FE and ICBMs
« Reply #208 on: June 17, 2021, 06:23:17 PM »
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.

Quote
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.


Quote
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.

Quote
No on, then off, then on again.
See comment immediately above.
Distance from Sydney to Perth - We don't know.
There's a mirror floating in the sky - Yup.

Re: FE and ICBMs
« Reply #209 on: June 17, 2021, 06:43:39 PM »
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 .........

Offline SteelyBob

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Re: FE and ICBMs
« Reply #210 on: June 17, 2021, 09:42:36 PM »
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?

Offline Action80

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Re: FE and ICBMs
« Reply #211 on: June 18, 2021, 10:41:02 AM »
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?

Offline Action80

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Re: FE and ICBMs
« Reply #212 on: June 18, 2021, 11:22:46 AM »
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.
« Last Edit: June 18, 2021, 12:18:36 PM by Action80 »

Offline SteelyBob

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Re: FE and ICBMs
« Reply #213 on: June 18, 2021, 12:02:31 PM »
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. 

Offline Action80

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Re: FE and ICBMs
« Reply #214 on: June 18, 2021, 12:10:47 PM »
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
« Last Edit: June 18, 2021, 12:16:46 PM by Action80 »

Offline SteelyBob

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Re: FE and ICBMs
« Reply #215 on: June 18, 2021, 12:22:58 PM »

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 if you're interested.

But don't worry about it - let's develop your point further. Where were you going with your argument?

Offline Action80

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Re: FE and ICBMs
« Reply #216 on: June 18, 2021, 12:33:26 PM »

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 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.
« Last Edit: June 18, 2021, 12:40:29 PM by Action80 »

Offline SteelyBob

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Re: FE and ICBMs
« Reply #217 on: June 18, 2021, 12:57:01 PM »

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 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.

Offline Action80

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Re: FE and ICBMs
« Reply #218 on: June 18, 2021, 01:20:14 PM »

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 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?
« Last Edit: June 18, 2021, 01:24:55 PM by Action80 »

Re: FE and ICBMs
« Reply #219 on: June 18, 2021, 01:54:30 PM »
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.