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Messages - SteelyBob

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1
Flat Earth Investigations / Re: Cavendish experiment
« on: July 28, 2021, 09:01:17 PM »
The feather thing is an analogy. It’s not literally what they are trying to do. Clinging on to that phrase is starting to look a little desperate, frankly.

But if you want to play with the analogy…several different teams of scientists have all tried to weigh the feather, so to speak. They did it using all kinds of different methods - maybe different old scales, to stretch it further, in a variety of wind conditions.

And they all came up with numbers within 400ppm of each other. Most people would conclude that the number must be, give it take a few hundred ppm, the weight of the feather. You are saying the lack of better (non specific - it’s always non specific around here, isn’t it?!) precision means the feather doesn’t exist, and/or the scientists don’t understand the feather.


2
Flat Earth Investigations / Re: Cavendish experiment
« on: July 28, 2021, 09:13:16 AM »
I didn’t ask you whether or not they understand it. I asked you what they were measuring. Given that different methods are all returning a number for G that is remarkably consistent, considering the challenge, we must surely conclude that G is indeed a thing, and it clearly relates to a force between bodies of mass.

Your argument seems to be that a lack of precision (although you don’t specify what degree of precision would cause you to change your mind) indicates that the experiments aren’t measuring anything. That is an absurd argument.

3
Flat Earth Investigations / Re: Cavendish experiment
« on: July 27, 2021, 11:04:33 PM »
It's not remarkably similar. The physicist Terence Quinn above says that the range undermines their science of the metrology of the small forces.

https://www.nature.com/articles/nphys3651.pdf?proof=t

"Who needs a more accurate numerical value of G (the current recommended value is 6.67408 ± 0.00031 × 10−11 kg−1 m3 s−2)? The short answer is, nobody, for the moment, but being apparently unable to converge on a value for G undermines our confidence in the metrology of small forces."

"Despite intensified efforts, measurements of the gravitational constant continue to fail to converge, as Terry Quinn explains."

I'll repeat my question (again): what then, exactly, were they measuring, if not G?

4
Flat Earth Investigations / Re: Cavendish experiment
« on: July 27, 2021, 10:16:08 PM »
It seems like you don't believe there exists corruption in science community. The popular view of science community is no way to guarantee it is truthful.

Nothing I said in my post supports your contention. Scientists exhibit all of the failings of other human beings - of course they do. I simply pointed out that lots of different scientific teams all measured something, using different methods, and came up with remarkably similar numbers.

I'll repeat my question: what then, exactly, were they measuring, if not G?

5
Flat Earth Theory / Re: Why do objects fall at dofferent speeds?
« on: July 20, 2021, 06:55:51 AM »

Drop an object with an initial velocity of 0 from 1000 feet and it will hit the ground in 8 seconds with gravity.  An upward accelerating earth with an initial velocity of (conservatively) 1000 mph will cover the same 1000 feet and meet a stationary object in .6768 seconds.

Rate of acceleration is meaningless if you don’t consider initial velocity.

But if you drop an object with initial velocity zero, then that initial velocity is with respect to the earth's surface, so both the object and the earth (and indeed the atmosphere) are moving at the same velocity. From the moment it leaves your hands, the earth and atmosphere would continue accelerating, but the object would not, other than a small acceleration caused by the atmosphere pushing on it.

From your perspective as the dropper of the ball, the behaviour of the ball would be identical under UA and the more conventional model.

There is plenty to disagree with regarding UA - the lack of an energy source, the uniform value of g you would observe compared to the variations that are actually measured, the lack of an explanation for how satellites stay where they are...it's a long list - but this ain't it.


6
Science & Alternative Science / Re: FE and ICBMs
« on: July 10, 2021, 10:52:57 AM »
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.

7
Science & Alternative Science / Re: FE and ICBMs
« on: July 10, 2021, 08:21:07 AM »
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 Hbo 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.

8
Science & Alternative Science / Re: FE and ICBMs
« on: July 09, 2021, 09:17:38 PM »

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.

9
Science & Alternative Science / Re: FE and ICBMs
« on: July 09, 2021, 06:46:37 PM »

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

10
Science & Alternative Science / Re: FE and ICBMs
« on: July 09, 2021, 06:20:28 PM »
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.

11
Science & Alternative Science / Re: FE and ICBMs
« on: July 09, 2021, 06:08:12 PM »

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?

12
Science & Alternative Science / Re: FE and ICBMs
« 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?

13
Science & Alternative Science / Re: FE and ICBMs
« 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?

14
Science & Alternative Science / Re: FE and ICBMs
« on: July 08, 2021, 08:22:27 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.

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.

15
Science & Alternative Science / Re: FE and ICBMs
« on: July 08, 2021, 02:39:04 PM »
I already agree it is exponential, but I do not agree there would necessarily be a drastic difference in calculation.

So you agree that your 'start + end v divided by two' equation is wrong, but now we're in to a discussion about how wrong it is?

According to the sources, the missile was fired in a near vertical trajectory from launch site. The attainment of 250km does not veer significantly from the vertical within the span of five minutes. 250km would, given an apogee of 4500km achieved downrange at approximately 450km, fall within a measure of approximately 50km or so. Not much of an interval to calculate and it remains very close to a linear calculation of average velocity.

0 - 16,000km/h within 5 minutes results also in a relatively straight line in what turns out to be roughly a tenth of total flight time according to your sources.

What on earth does that bit in bold actually mean?

Why aren't you looking at either the example velocity profiles that AATW or I gave you - they both illustrate the point that you seem to be consistently missing. You're even accepting that your calculation is wrong...why not accept that the follow-on from that, ie that it's perfectly possible to accelerate to 16000km/h in 5 minutes and have an average velocity of 3000km/h over that same time period? For the - I've-lost-count-of-how-many-times time, if most of the acceleration happens in the last minute, it's not surprising at all that the average velocity is a lot less than the final velocity.

16
Science & Alternative Science / Re: FE and ICBMs
« on: July 08, 2021, 02:31:36 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.

And I suggested a velocity profile which also (coincidentally) gave a 250km distance over 5 minutes.

But we are, I think, wasting our collective time, as A80 doesn't actually appear to be reading what we write.

17
Science & Alternative Science / Re: FE and ICBMs
« on: July 07, 2021, 06:56:50 PM »
^Incorrect.

In order to to achieve the figure of 250km altitude in 5 minutes resulting in a 3000 km/h average rate of travel, the velocities contained between the 0 and the 16,000 over the five minutes would average out to 3000 km/h.
They cannot, as (final velocity + initial velocity)/2 = average velocity.

Hallelujah…finally, some maths so we can actually understand what it is that you don’t understand.

That statement above is simply incorrect. It would work as a calculation if and only if the velocity profile was linear which, as you’ve agreed previously, it isn’t - the fuel burns down, so mass reduces, but thrust is constant.

Average velocity is not necessarily the average of the start and end velocity.

It’s easy to prove that to yourself - just consider my 30mph car example. Your calculation would have the average velocity as the average of 30 and, say, 100mph even though the car only did 100mph for a few seconds after an hour at 30mph. Average speed is clearly not 65mph, as per your maths.

Or imagine a situation where start and end velocity are the same, but there’s a burst of increased velocity in the middle somewhere. Again, it wouldn’t work.

For general cases you need the area under the velocity time graph. If the slope is linear, your equation works - the area under a triangle is half the base times the height, which is, if you think about it, exactly what your equation is. But our graph here isn’t linear - it’s exponential, so you can’t calculate that simply.

Are we finally in agreement in this point at least?

18
Science & Alternative Science / Re: FE and ICBMs
« on: July 07, 2021, 02:10:00 PM »
^The entirety of this post just above mine here gets a big thumbs down, especially that last feeble attempt at trying to brush away the fact that 250km traveled in 5 minutes = 3000km/h average speed over those 5 minutes.

Just pitiful.

Really pitiful.

Moving on.

No, not moving on. You keep saying it's wrong, but you are completely unable to back up that claim with any evidence whatsoever. Nobody is saying that 3000km/h isn't the average speed. Of course it's the average speed. And that is entirely consistent with the final velocity being 16,000km/h. The final velocity could be anything at all. As long as the velocity-time profile is such that the area under the graph is 250km, then the average speed will be 3000km/h. It could finish at 16,000km/h, or 100,000km/h, or at a standstill. The totality of the profile is all that matters.

This is somewhat painful.

19
Science & Alternative Science / Re: FE and ICBMs
« on: July 07, 2021, 12:21:43 PM »

Here, you want to skip to the end part where g=approx 3.7m/s2, ignoring all the values in between.

No, not at all. As I said, and as laid out nicely in the site I linked to, you have to do some calculus - some integration to find the area under the velocity time graph essentially.

As we've discussed a post or two ago, g reduces with increasing altitude, so it takes a lot longer to decelerate than 12 minutes. All covered off in that website I linked to. If you disagree, fine, but how long would it take to decelerate, and what would be the distance travelled, if g reduced in the manner generally agreed upon? You're saying it's wrong...but what's the right answer then? It can't be 12 minutes, and we know it must be bigger than that number.
I would prefer you launder your own wash.

Neatly press, fold, and drawer, or hang in the closet as you wish.

No idea what you're on about here. I've shown you calculations, websites etc. You've just said 'it isn't so', without providing any explanation.

I am telling you the missile travelling at 16,000km/h at an altitude of 250km, under no engine power, subject to g slightly above 9m/s2, will not climb an additional 4250km.

Period.
But it's not subject to g at slightly above 9ms-2, it's subject to a g profile that reduces from just above 9 down to below 4 at apogee.

You believe in this stuff.

Not so much 'believe' as 'find the evidence for it compelling'

I do not because as I have demonstrated in numerous posts, governments lie about this type of crap ALL the time.

It is warmongering, fearmongering crap, designed to keep a terrified populace. 

Governments lie, but science is science.

If you understand what d=rtmeans, then you know that 0-16,000km/h over the course of 5 minutes would not average out to a final distance of 250km.

It doesn't matter whether the high end acceleration takes place in front or at the end of the run.

For simplicity, imagine a profile where its speed doubled every minute, finishing at 16,000, and to keep the maths simple, just keep the velocity flat over each minute and then step it up at the end of each minute. So zero, then 1000km/h after one minute, then 2000km/h after 2 minutes, then 4000km/h after 3 minutes, then 8000km/h after 4 minutes, and a snap to 16,000km/h at the finish line.
Total distance in that case would be 1000/60 + 2000/60 + 4000/60 + 8000/60 = 250km

Now, clearly that's an inaccurate model, because the velocity can't just instantly double, and in the rocket example the mass is never zero - it decays, presumably linearly if the thrust is constant, to whatever the zero-fuel mass of the system is. But by reducing the time step, you can increase the accuracy, and it nevertheless illustrates the point.

I really don't get how to make this any clearer, but then it's not helped by your continued refusal to actually show any maths of your own.

20
Science & Alternative Science / Re: FE and ICBMs
« on: July 07, 2021, 10:41:48 AM »
^I have offered numbers.

Your numbers.

So, if you are going to state a missile traveling at over 16,000 km/h @ an altitude of 250km, without engine power, is going to overcome g=slightly over 9m/s2 in order to travel an additional 4250km to apogee, the simple fact is we disagree. I disagree based on what you wrote earlier, when you stated a missile subject to g=9.82m/s2 would decelerate to 0 in 12 minutes if traveling the same speed.

Great. Now read the rest of that post - I said 'At which point you'll probably say 'so how come it flew for 50 minutes? Surely 12 x 2 = 24?'...read past that bit. As we've discussed a post or two ago, g reduces with increasing altitude, so it takes a lot longer to decelerate than 12 minutes. All covered off in that website I linked to. If you disagree, fine, but how long would it take to decelerate, and what would be the distance travelled, if g reduced in the manner generally agreed upon? You're saying it's wrong...but what's the right answer then? It can't be 12 minutes, and we know it must be bigger than that number. 

And regardless, d=rt still applies. To travel 250km in 5 minutes results in an average rate of travel of 3000km/h. The numbers reflected do not result in that outcome.

Again, I'm not disputing the average velocity. But there's no reason at all why the final velocity at the end of the burn is inconsistent with that. If most of the acceleration happens in the latter stages, just like the car in my previous example, why can't you have a high end velocity but a low average?

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