Maths problem

[Scott1328]

I didn't dispute your equation. I restricted attention to a simple case to keep the math easy, I further simplified by using two emitters. When the emitters are perpendicular to the motion of the clock you have a right triangle for the outbound journey and a right triangle for the inbound. The triangles are congruent and therefore only one needs to be considered. This was merely to confirm the OPs point and to quantify the slowing of the pings

So fuck off

[pErvinalia]

Well it certainly appeared as if you were disputing my equation given your comments to "stick to right angles and simplify the maths" and "learn pythagoras" followed directly on from mine. And you didn't say you were only applying it to the single case where the emitters were perpendicular to the motion of the clock. Did you forget or are you just inventing this now to cover a mistake? And not to mention that your example had been performed and answered multiple times in the thread by both Brian and I. So why the hell were you doing it again? We already knew the case at 90deg was patently simple.

[Scott1328]

Because I skipped over all your toxic shit with Brian and wrote that after visualizing the problem for 5 minutes. But I can always count on you to leap straight to hostile defensiveness. So again fuck you.

[pErvinalia]

Ok. If you say so. I still suspect nonsense given your previous post on the topic was modelling the problem wrong. In any case, both Brian and I answered the case of 90deg well before Brian stopped reading critiques of problems with his model (what you refer to as "toxic shit").

[pErvinalia]

I will grant, though, that your solution for 90deg is far simpler and cleaner than mine from earlier in the thread.

Anyway, here's some data and a graph. Note, the trend line is just an auto generated one in excel.

Untitled.jpg (63.31 KiB) Viewed 2480 times

So you can see that as the angle of the clock increases away from train travel, the clock speeds up relative to real time. This should answer MM's original question.

[Brian Peacock]

Get a grip. And learn pythagoras. Also the math is easier so I used two emitters. Doesnt change the result though.

He won't be told. The system is symmetrical - geometry for the win.

[pErvinalia]

Jesus, Brian. You're the new 42. I've explicitly shown how it isn't symmetrical. Scott agrees with me. It's only symmetrical at 90deg to clock/train travel. Read the thread and be open to learning something.

[Brian Peacock]

The distance the ping travels from emitter to a reflector at 90 degrees + n is the same as the return distance from a reflector at 90 degrees - n back to the emitter.

The distance the ping travels from emitter to a reflector at 90 degrees - n is the same as the return distance from a reflector at 90 degree + n back to the emitter.

The distance the ping travels from emitter to a reflector at 90 degrees or from the reflector at 90 degree back the to emiittrr is the same.

When the angle from the emitter to the reflector is 90 degrees +/- n the angle of the ping's return is 90 degrees +/- n + 180.

The emitter and the reflector cover the same distance in the same time in the same direction.

The speed of sound is constant.

The system is symmetrical - still a win for geometry I think.

[pErvinalia]

The distance the ping travels from emitter to a reflector at 90 degrees + n is the same as the return distance from a reflector at 90 degrees - n back to the emitter.

The distance the ping travels from emitter to a reflector at 90 degrees - n is the same as the return distance from a reflector at 90 degree + n back to the emitter.

Wrong. The train/clock is moving relative to the sound waves.

The distance the ping travels from emitter to a reflector at 90 degrees or from the reflector at 90 degree back the to emiittrr is the same.

yep. This is the only point that the model is symmetrical.

When the angle from the emitter to the reflector is 90 degrees +/- n the angle of the ping's return is 90 degrees +/- n + 180.

Wrong. See my post where I calculate the difference in the outgoing and return pings. The reason it is different is the same reason it was the last 5 times I explained it to you. It has to travel further on the way out (if the clock is between 0 and 90deg) than on the way back, because the clock is moving relative to the medium in which the sound wave is moving. So on the way out it has to travel further as the receiver is moving away from the sound wave, and on the way back it has to travel less as the receiver (the initial emitter) is moving towards the sound wave.

The system is symmetrical - still a win for geometry I think.

Wrong. If only you would actually do the sums yourself and/or read over mine where I've done it, you would see how this is wrong. For some reason, though, you don't appear to be willing to accept that you might have got this wrong. It's just a repeat of your Doppler balls up.

[pErvinalia]

Here, I did some hokey diagrams to help.

Clock is moving relative to the background (i.e. the stationary air, and therefore the sound wave). As the receiver is moving away from the soundwave, you can see how the distance the sound wave has to travel is greater than if the clock/train was stationary.

In the next diagram the sound wave hits the receiver and is reflected back towards the emitter. But as the emitter is moving towards the sound wave, you can see that the distance the sound wave has to travel back is shorter. Indeed, without checking, I'm pretty sure this is what your second animation showed. So I don't know what's happened between when you wrote that and now to change your mind on how the clock moves relative to the sound wave.

[pErvinalia]

I just had a look at your second animation and it's certainly showing shorter return times than outward times. So you are at least qualitatively right with it. I was wrong when I suggested you might not be adding/subtracting the relative speed of the train/clock. You clearly are somewhere there. Whether it is quantitatively right, who knows? I asked you previously to tease out the equations you used to generate that, but you ignored that request. Can you do it now?

[Brian Peacock]

You should look at the third animation. http://www.rationalia.com/forum/viewtop ... 0#p1698500

[pErvinalia]

Same thing. So the question remains - what happened between then and now? Why is it you now think that the sound travels the same distance outbound as inbound, when clearly from your own animation it doesn't?

And for the third time, can you provide the algorithm you used to generate these animations?

[Brian Peacock]

I don't 'now think that the sound travels the same distance outbound and inbound' - what I said was that the sound travels the same total distance (outbound+inbound) for a reflector at 90 degrees + n as for a reflector at 90 degrees - n (where n is the same value in each case!), and that for a reflector at 90 degrees - n the return distance from reflector to receiver (inbound) is the same distance as from emitter to reflector (outbound) at 90 degrees + n, and vice versa.

[pErvinalia]

Ok, I see what your saying. The symmetry I've been talking about is that the outbound journey is the same distance as the inbound journey as happens at 90deg. Hence why pages ago I talked about the concept of bisecting an isosceles triangle into two identical right angle triangles. Additionally, if the unit is closed, then that symmetry applies to all angles, although it's irrelevant as the sound wave travels at the same speed regardless of angle and train speed, so no trig is needed.

And for the fourth time, can you provide the algorithm you used in your animations??