[SteelyBob]

Asked and answered.

You are an 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 admitted troll.

You're nothing if not persistent, I'll grant you that.

Average velocity only equals (start velocity + end velocity)/2 for a linear velocity profile, ie constant acceleration.

Why do you keep ignoring that?

Why do you keep ignoring the stunningly obvious examples we give you, that quite clearly show that you can be at 16000km/h at the end of a period of time, and average well under 8000km/h?

Like, for example, travelling at 1km/h for 4 minutes and 55 seconds, and then accelerating rapidly to 16,000km/h.

But hey, what's the point? You'll ignore the example and keep saying the same thing, right?

[Iceman]

I recommend everyone just stops indulging A80's fantasy that avg velocity during burn phase is somehow the answer to his problem with an ICBM working...

[SteelyBob]

I recommend everyone just stops indulging A80's fantasy that avg velocity during burn phase is somehow the answer to his problem with an ICBM working...

Possibly. I just see it as step one on a long road of complete, and possibly wilful, misunderstandings or misrepresentations of basic facts.

I don't think we can address anything else until we get past this one.

[Action80]

Asked and answered.

You are an 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 admitted troll.

You're nothing if not persistent, I'll grant you that.

Average velocity only equals (start velocity + end velocity)/2 for a linear velocity profile, ie constant acceleration.

Why do you keep ignoring that?

Why do you keep ignoring the stunningly obvious examples we give you, that quite clearly show that you can be at 16000km/h at the end of a period of time, and average well under 8000km/h?

Like, for example, travelling at 1km/h for 4 minutes and 55 seconds, and then accelerating rapidly to 16,000km/h.

But hey, what's the point? You'll ignore the example and keep saying the same thing, right?

I am not ignoring it.

Already admitted it is a linear method of calculation several times in the thread.

You are ignoring the fact the velocity profile in this flight took place over a path not varying to a significant degree from vertical.

Go ahead, make a horizontal line at each level of velocity achieved at any given split over the five minute time frame you put forth. Try to achieve a significant difference in the average velocity derived via calculus and those given by a linear method as I described.

[SteelyBob]

I am not ignoring it.

Already admitted it is a linear method of calculation several times in the thread.

You are ignoring the fact the velocity profile in this flight took place over a path not varying to a significant degree from vertical.

No, I completely agree with that. In fact, I'm going one further and suggesting we just model it as a vertical flight to keep things simple.

Go ahead, make a horizontal line at each level of velocity achieved at any given split over the five minute time frame you put forth. Try to achieve a significant difference in the average velocity derived via calculus and those given by a linear method as I described.

That's pretty much what I did in one of the many examples you ignored. Here it is again:

1 minute at 0km/h
1 minute at 1000km/h
1 minute at 2000km/h
1 minute at 4000km/h
1 minute at 8000km/h
1 nano second at 16000km/h

Over to you for average speed and distance travelled. Notice how different the results are from a simple average of 0 and 16000km/h

[Action80]

I am not ignoring it.

Already admitted it is a linear method of calculation several times in the thread.

You are ignoring the fact the velocity profile in this flight took place over a path not varying to a significant degree from vertical.

No, I completely agree with that. In fact, I'm going one further and suggesting we just model it as a vertical flight to keep things simple.

Go ahead, make a horizontal line at each level of velocity achieved at any given split over the five minute time frame you put forth. Try to achieve a significant difference in the average velocity derived via calculus and those given by a linear method as I described.

That's pretty much what I did in one of the many examples you ignored. Here it is again:

1 minute at 0km/h
1 minute at 1000km/h
1 minute at 2000km/h
1 minute at 4000km/h
1 minute at 8000km/h
1 nano second at 16000km/h

Over to you for average speed and distance travelled. Notice how different the results are from a simple average of 0 and 16000km/h

Oh. Remarkable you would spend any amount of time at 0 km/h, but it certainly is keeping with your desperate trolling efforts.

Quite amusing too!

Or perhaps it was the palm in your face from earlier.

Regardless.

If I travel 250km in 5 minutes, I have traveled at an average rate of travel equivalent to 3000km/h for those five minutes.

However, average velocity over a linear trajectory, which you agree is a vertical path, given the figures of 0 initial velocity and a final velocity achieved with five minutes 16,000 km/h work out to 8,000km/h and an altitude achieved of 667km.

Still waiting for you to back your claim (with demonstrable math please) a ballistic object located at an altitude of 250km, subjected to g=9.08m/s² under no propulsion and guidance, can achieve an additional altitude of 4250 km.

I figured this would be easy for you to do since you already admitted the same object would fall back to earth within twelve minutes if subjected to g=9.87m/s².

[Rama Set]

Someday soon Lackey will acknowledge the gradient of gravity as altitude increases.

[SteelyBob]

Oh. Remarkable you would spend any amount of time at 0 km/h, but it certainly is keeping with your desperate trolling efforts.

Would it make you happier if the first two minutes were both at 1000km/h? It would move the answer further away from what you want it to be...entirely up to you. Telling that you haven't actually done the maths though. I wonder why?

Quite amusing too!

Or perhaps it was the palm in your face from earlier.

Regardless.

If I travel 250km in 5 minutes, I have traveled at an average rate of travel equivalent to 3000km/h for those five minutes.

However, average velocity over a linear trajectory, which you agree is a vertical path, given the figures of 0 initial velocity and a final velocity achieved with five minutes 16,000 km/h work out to 8,000km/h and an altitude achieved of 667km.

Ok, I think the issue here is that you're confusing a linear trajectory with a linear velocity profile - not the same thing. Linear trajectory, non linear velocity profile means you can't just average the start and end speed. I thought you'd agreed with that a few posts back but you now seem to have back-pedalled.

Still waiting for you to back your claim (with demonstrable math please) a ballistic object located at an altitude of 250km, subjected to g=9.08m/s² under no propulsion and guidance, can achieve an additional altitude of 4250 km.

I figured this would be easy for you to do since you already admitted the same object would fall back to earth within twelve minutes if subjected to g=9.87m/s².

We'll get to that, but let's bottom out the velocity thing first, because the second phase is even harder to grasp, and you're struggling with step one.

[stack]

Average
Velocity
Is
Meaningless
To
This
Entire
Question

I'm blindingly confused by all of this. But agree with you. Why would average velocity matter in the slightest? Isn't the only velocity of concern what it is at engine cutoff, 16000km/h? Who cares what happened between 0 and 16000, the latter is the only figure that matters.

I know this has come up a dozen times all ready, but the metaphor seems to be a bullet leaving the barrel of a gun, that velocity and a whole host of other factors, drag, trajectory, g and such, determine the distance and/or altitude traveled. I kind of think of it as the ballistic missile is fired out of a barrel that 250 km high at engine cut-off. The question becomes, just like the bullet out of the barrel, how far/high will it travel from that point on. Not what happened in the "barrel".

And like I pointed out earlier, which was considered a strawman, which it is not, is that a dozen plus parameters all enter into the calculations to figure that out. It's very, very complicated to say the least.

When the projectile hits engine cut-off at 16000km/h at an altitude of 250k and g of 9.08, having all its fuel burned off, take-off weight of the HS-15 is estimated at approximately 73 to 74t or 67000 kilos. Propellent for the 2 stages estimated to weigh 67 tons or 60000 kilos. So the projectile at cut-off now weighs 7000 kilos traveling at 16000km/h - That is your bullet velocity and some elements to be concerned with.

With that, if it continued up to let's say 500 km post cut-off, our 7000 kilo bullet would be experiencing g of now 8.43.

The question is how fast is our bullet going now?

if it continued up to let's say 1000 km post cut-off, our 7000 kilo bullet would be experiencing g of now 7.32.

The question is how fast is our bullet going now?

Let's say it kept going up to a whopping 2000 km post cut-off, our 7000 kilo bullet would be experiencing g of now 5.68.

And so on. And that's just factoring in a one, gravity, of the many elements needed to accurately calculate just how high it could go from a 16000km/h speed, 7000 kilo mass starting point.

So I agree, to even mention an average is absolutely meaningless. Is there something missing?

[SteelyBob]

Average
Velocity
Is
Meaningless
To
This
Entire
Question

I'm blindingly confused by all of this. But agree with you. Why would average velocity matter in the slightest? Isn't the only velocity of concern what it is at engine cutoff, 16000km/h? Who cares what happened between 0 and 16000, the latter is the only figure that matters.

I know this has come up a dozen times all ready, but the metaphor seems to be a bullet leaving the barrel of a gun, that velocity and a whole host of other factors, drag, trajectory, g and such, determine the distance and/or altitude traveled. I kind of think of it as the ballistic missile is fired out of a barrel that 250 km high at engine cut-off. The question becomes, just like the bullet out of the barrel, how far/high will it travel from that point on. Not what happened in the "barrel".

And like I pointed out earlier, which was considered a strawman, which it is not, is that a dozen plus parameters all enter into the calculations to figure that out. It's very, very complicated to say the least.

When the projectile hits engine cut-off at 16000km/h at an altitude of 250k and g of 9.08, having all its fuel burned off, take-off weight of the HS-15 is estimated at approximately 73 to 74t or 67000 kilos. Propellent for the 2 stages estimated to weigh 67 tons or 60000 kilos. So the projectile at cut-off now weighs 7000 kilos traveling at 16000km/h - That is your bullet velocity and some elements to be concerned with.

With that, if it continued up to let's say 500 km post cut-off, our 7000 kilo bullet would be experiencing g of now 8.43.

The question is how fast is our bullet going now?

if it continued up to let's say 1000 km post cut-off, our 7000 kilo bullet would be experiencing g of now 7.32.

The question is how fast is our bullet going now?

Let's say it kept going up to a whopping 2000 km post cut-off, our 7000 kilo bullet would be experiencing g of now 5.68.

And so on. And that's just factoring in a one, gravity, of the many elements needed to accurately calculate just how high it could go from a 16000km/h speed, 7000 kilo mass starting point.

So I agree, to even mention an average is absolutely meaningless. Is there something missing?

A80 is disputing the possibility of achieving the velocity quoted in the timeframe mentioned at the height given. He is suggesting, oddly, that the average speed of 3000km/h means the velocity at H_bo cannot be 16000.

Part two of his dispute seems to be that, even if you could achieve 16000km/h, you wouldn’t then achieve the apogee quoted.

I’m shooting at part one at the moment. We’ll move on to part two in due course.

[AATW]

Right. And my point over the last couple of pages is that Lackey doesn’t even understand Part 1. He doesn’t understand how to calculate averages or the limitations of the calculator he found - which scenarios it can be used in and when it can’t be used.

Given that, it seems unlikely he is able to do the maths to back up his assertion about the second part which is orders of magnitude more complex. Which is no crime, this stuff is really complicated. I'm just pointing out that all he’s doing is making an argument from incredulity.

[SteelyBob]

Right. And my point over the last couple of pages is that Lackey doesn’t even understand Part 1. He doesn’t understand how to calculate averages or the limitations of the calculator he found - which scenarios it can be used in and when it can’t be used.

Given that, it seems unlikely he is able to do the maths to back up his assertion about the second part which is orders of magnitude more complex. Which is no crime, this stuff is really complicated. I'm just pointing out that all he’s doing is making an argument from incredulity.

Yep.

[Action80]

When you travel 250 km in 5 minutes, according to the figures provided, you will have achieved an average rate of travel of 3000km/h. Pleaase note that rate of travel = velocity.

If you have an initial velocity of 0 km/h and your final velocity after 5 minutes is 16,000 km/h, then your average velocity is 8000 km/h.

There is no dispute here.

Anyone here who claims the scenario provided by AATW is possible is gaslighting, plain, pure, and simple.

Further, an object traveling at a velocity of 16,000km/h at a height of 250 km, is at that moment, subjected to g=9.08m/s². Given those figures, that object will fall to the earth in just over 17 minutes.

This is a fact.

No way this missile flight took place at all in the form or complete fashion expressed on Wikipedia and other sources.

Next.

[AATW]

When you travel 250 km in 5 minutes, according to the figures provided, you will have achieved an average rate of travel of 3000km/h. Pleaase note that rate of travel = velocity.

Yes

If you have an initial velocity of 0 km/h and your final velocity after 5 minutes is 16,000 km/h, then your average velocity is 8000 km/h.

No.

Or rather "it depends". That is true if and only if your rate of acceleration during those 5 minutes is constant

Further, an object traveling at a velocity of 16,000km/h at a height of 250 km, is at that moment, subjected to g=9.08m/s². Given those figures, that object will fall to the earth in just over 17 minutes.

This is a fact.

Please show the math which demonstrates this fact. Note that as the object continues to ascend the value of g will continue to reduce.
If you just use a flat rate of 9.08 then you're going to get the wrong answer. I don't know how to do the math to work this out and, with respect, given your second statement above, I don't believe you do either.

[Action80]

When you travel 250 km in 5 minutes, according to the figures provided, you will have achieved an average rate of travel of 3000km/h. Pleaase note that rate of travel = velocity.

Yes

If you have an initial velocity of 0 km/h and your final velocity after 5 minutes is 16,000 km/h, then your average velocity is 8000 km/h.

No.

Or rather "it depends". That is true if and only if your rate of acceleration during those 5 minutes is constant

This is demonstrably false!

Acceleration for a race car is certainly not constant, yet average velocity over the course (also not linear) is expressed as defining the winner.

We are a discussing average velocity not average acceleration or even average speed. Acceleration is simply the variation of speed over time, or in this case: dV/dt where d = change

Take your extreme BS out of here.

Further, an object traveling at a velocity of 16,000km/h at a height of 250 km, is at that moment, subjected to g=9.08m/s². Given those figures, that object will fall to the earth in just over 17 minutes.

This is a fact.

Please show the math which demonstrates this fact. Note that as the object continues to ascend the value of g will continue to reduce.
If you just use a flat rate of 9.08 then you're going to get the wrong answer. I don't know how to do the math to work this out and, with respect, given your second statement above, I don't believe you do either.

It is not a flat rate of 9.08.

I have given the number, which accounts for ascent and velocity.

Since you are trying to claim acceleration can possibly impact the calculation of average velocity, I see no need to further engage you on this subject.

You are proving to be nothing but a troll.

[Iceman]

You can describe the distance travelled in a given time frame for any moving body (whether its accelerating or not) by multiplying its avg velocity by the time. No one disputed this.

But that figure offers zero predictive powers for the instantaneous velocity at any time. For this discussion, AVERAGE VELOCITY FOR THE BURN PHASE IS A MEANINGLESS VALUE.

Please stop repeating this nonsense.

RIP to what started as an interesting thread.

[AATW]

AVERAGE VELOCITY FOR THE BURN PHASE IS A MEANINGLESS VALUE.

Correct. But I’d suggest that Lackey’s confusion about this and inability to calculate that average is meaningful and telling.
It shows that the rest of his argument is simply one from incredulity.

In an attempt to get this back on track, do you know how to calculate the maximum height, given the known altitude and velocity at the time of engine shut down? Given the constantly varying value of g we are in the realms of calculus and differential equations, I fear, which is a bit beyond me.

[stack]

Further, an object traveling at a velocity of 16,000km/h at a height of 250 km, is at that moment, subjected to g=9.08m/s². Given those figures, that object will fall to the earth in just over 17 minutes.

This is a fact.

How did you calculate the 17 minutes after cut-off bit? It seems like a fairly precise number you came up with.

According to your calculations, what was the altitude of the rocket after 17 minutes?

Did you factor in the mass of the rocket after cut-off?
Did you factor into the 17 minutes what the g was 1 minute after cut-off, 2 minutes after cut-off, etc, up to 17?
Did you factor in the density of the atmosphere, drag, 1 minute after cut-off, 2 minutes after cut-off, up to 17?

[Clyde Frog]

Acceleration for a race car is certainly not constant, yet average velocity over the course (also not linear) is expressed as defining the winner.

Yes, it's almost as if they can take the total approximate distance the car has driven, and then divide that number by the amount of time it took the driver to run the course. That is, notably, different than how you're coming up with the (wrong) average velocity in your own example. Do you see the difference?

[Clyde Frog]

An interesting real-world example to share.

Aerie Luyendyk averaged 236.986mph (381.391km/h) in qualifying for the Indy 500 in 1996, with his four-lap time of 2m31.908s still unmatched today.

That car, like all race cars, starts out at 0km/h. By lackey-math, that means it must have finished the race at 762.79km/h in order to achieve the record-holding average velocity of 381.391km/h. So lackey, can you show us evidence that IndyCars regularly finish races at velocities greater than 700km/h? That seems like a hard thing to just take on faith.