Offline Action80

  • *
  • Posts: 2805
    • View Profile
Re: FE and ICBMs
« Reply #320 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.
« Last Edit: February 14, 2023, 02:04:05 PM by Action80 »
To be honest I am getting pretty bored of this place.

SteelyBob

Re: FE and ICBMs
« Reply #321 on: July 09, 2021, 03:39:35 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?

*

Offline AATW

  • *
  • Posts: 6488
    • View Profile
Re: FE and ICBMs
« Reply #322 on: July 09, 2021, 03:39:53 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.
Tom: "Claiming incredulity is a pretty bad argument. Calling it "insane" or "ridiculous" is not a good argument at all."

TFES Wiki Occam's Razor page, by Tom: "What's the simplest explanation; that NASA has successfully designed and invented never before seen rocket technologies from scratch which can accelerate 100 tons of matter to an escape velocity of 7 miles per second"

Offline Action80

  • *
  • Posts: 2805
    • View Profile
Re: FE and ICBMs
« Reply #323 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.
To be honest I am getting pretty bored of this place.

Offline Action80

  • *
  • Posts: 2805
    • View Profile
Re: FE and ICBMs
« Reply #324 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.
« Last Edit: July 09, 2021, 04:04:18 PM by Action80 »
To be honest I am getting pretty bored of this place.

*

Offline AATW

  • *
  • Posts: 6488
    • View Profile
Re: FE and ICBMs
« Reply #325 on: July 09, 2021, 04:09:20 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:

Quote
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)
Tom: "Claiming incredulity is a pretty bad argument. Calling it "insane" or "ridiculous" is not a good argument at all."

TFES Wiki Occam's Razor page, by Tom: "What's the simplest explanation; that NASA has successfully designed and invented never before seen rocket technologies from scratch which can accelerate 100 tons of matter to an escape velocity of 7 miles per second"

SteelyBob

Re: FE and ICBMs
« Reply #326 on: July 09, 2021, 04:28:26 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?

Offline Action80

  • *
  • Posts: 2805
    • View Profile
Re: FE and ICBMs
« Reply #327 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:

Quote
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.
« Last Edit: July 09, 2021, 05:11:01 PM by Action80 »
To be honest I am getting pretty bored of this place.

Rama Set

Re: FE and ICBMs
« Reply #328 on: July 09, 2021, 04:50:42 PM »
You also have a claim to substantiate, which you haven’t done.

*

Offline Iceman

  • *
  • Posts: 1825
  • where there's smoke there's wires
    • View Profile
Re: FE and ICBMs
« Reply #329 on: July 09, 2021, 04:52:15 PM »
Average
Velocity
Is
Meaningless
To
This
Entire
Question

Offline Action80

  • *
  • Posts: 2805
    • View Profile
Re: FE and ICBMs
« Reply #330 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.
To be honest I am getting pretty bored of this place.

Offline Action80

  • *
  • Posts: 2805
    • View Profile
Re: FE and ICBMs
« Reply #331 on: July 09, 2021, 05:07:20 PM »
You also have a claim to substantiate, which you haven’t done.
My claim is this:

The claim made that this flight as described and adhered to by proponents in this thread is bogus and the trip did not take place as offered.

The proponents of this flight being made can quite clearly and definitively close the issue, as is incumbent upon them, not me.

All they need to do is post the math proving a ballistic object traveling at a rate of 16,000km/h, at an altitude of 250km, under no propulsion, g=9.08m/s2, will gain an additional altitude of 4250km.

Simple. Go ahead. Have a go.
« Last Edit: July 09, 2021, 05:09:44 PM by Action80 »
To be honest I am getting pretty bored of this place.

Offline Action80

  • *
  • Posts: 2805
    • View Profile
Re: FE and ICBMs
« Reply #332 on: July 09, 2021, 05:08:31 PM »
Average
Velocity
Is
Meaningless
To
This
Entire
Question
d=rt is relevant to every trip taken in the history of humanity.

Bye.
To be honest I am getting pretty bored of this place.

*

Offline AATW

  • *
  • Posts: 6488
    • View Profile
Re: FE and ICBMs
« Reply #333 on: July 09, 2021, 05:10:35 PM »
Average
Velocity
Is
Meaningless
To
This
Entire
Question
lol, I have given up trying to explain that.
Tom: "Claiming incredulity is a pretty bad argument. Calling it "insane" or "ridiculous" is not a good argument at all."

TFES Wiki Occam's Razor page, by Tom: "What's the simplest explanation; that NASA has successfully designed and invented never before seen rocket technologies from scratch which can accelerate 100 tons of matter to an escape velocity of 7 miles per second"

*

Offline AATW

  • *
  • Posts: 6488
    • View Profile
Re: FE and ICBMs
« Reply #334 on: July 09, 2021, 05:13:48 PM »
You are the making that claim, within the parameters of the written record.
When did I make that claim?

I have repeatedly said that I don't know if it's true but have noted it's very complicated, far too complicated for you to understand - I'm basing that assertion on the fact that despite repeated explanations you don't understand much simpler concepts.

You are simply making an argument from incredulity. But if you're able to produce the maths which shows what the true final height would be, given the constantly changing value of g, then I'd like to see it.
Tom: "Claiming incredulity is a pretty bad argument. Calling it "insane" or "ridiculous" is not a good argument at all."

TFES Wiki Occam's Razor page, by Tom: "What's the simplest explanation; that NASA has successfully designed and invented never before seen rocket technologies from scratch which can accelerate 100 tons of matter to an escape velocity of 7 miles per second"

Offline Action80

  • *
  • Posts: 2805
    • View Profile
Re: FE and ICBMs
« Reply #335 on: July 09, 2021, 05:16:11 PM »
You are the making that claim, within the parameters of the written record.
When did I make that claim?

I have repeatedly said that I don't know if it's true but have noted it's very complicated, far too complicated for you to understand - I'm basing that assertion on the fact that despite repeated explanations you don't understand much simpler concepts.

You are simply making an argument from incredulity. But if you're able to produce the maths which shows what the true final height would be, given the constantly changing value of g, then I'd like to see it.
Yeah, here we go.

You do not now subscribe to the idea the flight has taken place.

Very well then.

You admit you have simply been trolling the thread.

Since you admit here you do not believe it happened, and you admit it is beyond your level of education, good bye to you.
« Last Edit: July 09, 2021, 05:21:52 PM by Action80 »
To be honest I am getting pretty bored of this place.

*

Offline AATW

  • *
  • Posts: 6488
    • View Profile
Re: FE and ICBMs
« Reply #336 on: July 09, 2021, 05:29:12 PM »
You do not now subscribe to the idea the flight has taken place.

I didn't say that either. ICBMs are definitely a thing, and I know tests of them have happened.
I don't claim to know the specifics of the flight we are talking about.
You are asserting that given a starting height and velocity when engines shut off the final height can't be a value claimed.
I would like to see your math which demonstrates that because, frankly, it's really complicated and given that you couldn't work out a simple average I struggle to believe you can work this out. Honestly, it's a bit beyond me.
So all you're left with is an argument from incredulity. I'm not making an argument at all, I'm merely pointing out that you need to understand more about all this before you make claims which you can't back up.

One thing to ponder. If d=rt then r = d/t.
We know the distance, 250km. And we know the time, 5 minutes.
So r = 250/5 = 50km/minute (note, minute, we need to multiply by 60 to get hours)

50 x 60 = 3000km/h average velocity. Not 8,000. See?

Tom: "Claiming incredulity is a pretty bad argument. Calling it "insane" or "ridiculous" is not a good argument at all."

TFES Wiki Occam's Razor page, by Tom: "What's the simplest explanation; that NASA has successfully designed and invented never before seen rocket technologies from scratch which can accelerate 100 tons of matter to an escape velocity of 7 miles per second"

Offline Action80

  • *
  • Posts: 2805
    • View Profile
Re: FE and ICBMs
« Reply #337 on: July 09, 2021, 05:43:31 PM »
You do not now subscribe to the idea the flight has taken place.

I didn't say that either. ICBMs are definitely a thing, and I know tests of them have happened.
I don't claim to know the specifics of the flight we are talking about.
You are asserting that given a starting height and velocity when engines shut off the final height can't be a value claimed.
I would like to see your math which demonstrates that because, frankly, it's really complicated and given that you couldn't work out a simple average I struggle to believe you can work this out. Honestly, it's a bit beyond me.
So all you're left with is an argument from incredulity. I'm not making an argument at all, I'm merely pointing out that you need to understand more about all this before you make claims which you can't back up.

One thing to ponder. If d=rt then r = d/t.
We know the distance, 250km. And we know the time, 5 minutes.
So r = 250/5 = 50km/minute (note, minute, we need to multiply by 60 to get hours)

50 x 60 = 3000km/h average velocity. Not 8,000. See?
Yeah, which was the figure I offered at the onset. It is apparent for anyone reading the thread, I was the first one to offer d=rt in this case is equivalent to 3,000km/h.

Which, when compared to your figures, blow your only figures really offered here (0 to 16,000km/h over the course of that five minutes travel to reach 250 km in altitude) out of the water.

You see, the average velocity of the figures you provided (0-16,000km/h) results in the 8,000km/h, hence not possibly matching d=rt.

Good bye.

You already admitted to trolling the thread.

I suggest you stop.
« Last Edit: July 09, 2021, 05:49:52 PM by Action80 »
To be honest I am getting pretty bored of this place.

*

Offline AATW

  • *
  • Posts: 6488
    • View Profile
Re: FE and ICBMs
« Reply #338 on: July 09, 2021, 05:52:44 PM »
Which, when compared to your figures, blow your only figures really offered here (0 to 16,000km/h over the course of that five minutes travel to reach 250 km in altitude) out of the water.
Do they, though?
I broke down that suggested acceleration pretty carefully to show how it could happen.
If you think I’ve made a mistake then can you show where?

But as I and others have explained, the average velocity is irrelevant here. All that matters is the height and velocity at the time of engine shut down.
Tom: "Claiming incredulity is a pretty bad argument. Calling it "insane" or "ridiculous" is not a good argument at all."

TFES Wiki Occam's Razor page, by Tom: "What's the simplest explanation; that NASA has successfully designed and invented never before seen rocket technologies from scratch which can accelerate 100 tons of matter to an escape velocity of 7 miles per second"

Offline Action80

  • *
  • Posts: 2805
    • View Profile
Re: FE and ICBMs
« Reply #339 on: July 09, 2021, 05:59:27 PM »
Which, when compared to your figures, blow your only figures really offered here (0 to 16,000km/h over the course of that five minutes travel to reach 250 km in altitude) out of the water.
Do they, though?
I broke down that suggested acceleration pretty carefully to show how it could happen.
If you think I’ve made a mistake then can you show where?

But as I and others have explained, the average velocity is irrelevant here. All that matters is the height and velocity at the time of engine shut down.
Asked and answered.

You are a self-admitted troll and are continuing to do so here.

Average velocity = (final velocity + initial velocity)/2.

Since we are discussing a portion of a trip, velocity can be considered equivalent to rate of travel.

d=rt has been relevant to all trips taken in the history of humanity.

The figures you provided, 0 - 16000 km/h in the span of 5 minutes, results in an average rate of travel of 8,000 km/h, resulting in an altitude of 667 km.

Bye to the self-admitted troll, as demonstrated here:
I don't claim to know the specifics of the flight we are talking about.
« Last Edit: July 09, 2021, 06:05:14 PM by Action80 »
To be honest I am getting pretty bored of this place.