Maths problem

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Re: Maths problem

Post by Brian Peacock » Wed Mar 01, 2017 1:55 am

Give it to me straight then: describe the correct conception of the experiment and what it shows. Mine is laid out in the last animation - you can use it for comparison.
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Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT
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Re: Maths problem

Post by pErvinalia » Wed Mar 01, 2017 1:59 am

Oh ffs. I just told you that I've done the thought experiment for two separate constructions. Have you not read the thread? And I've already thoroughly compared what your animation represents vs the two different types of models that I have envisaged. What's the fucking point of bashing my head against the wall for another round? Go and read the thread and address all the shit you ignored the first time round.
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Re: Maths problem

Post by Brian Peacock » Wed Mar 01, 2017 2:04 am

:roll:
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Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT
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Re: Maths problem

Post by pErvinalia » Wed Mar 01, 2017 2:04 am

In micro form:
Closed clock mechanism = no time mismeasurement regardless of the train speed or angle of the clock.
Open clock mechanism = time delay due to the increased distance the sound wave has to travel to complete one "tick". Not simple geometry, as the path described by the "ping" will be a scalene triangle because the sound wave has to travel further on the way out than on the way back.
NO Doppler Effect experienced on the train, while ever an observer (or sound receiver/reflector) is stationary in relation to the train.
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Re: Maths problem

Post by Scott1328 » Wed Mar 01, 2017 2:35 am

The speed of sound is not constant. It is constant relative to the speed of its medium.

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Re: Maths problem

Post by pErvinalia » Wed Mar 01, 2017 4:08 am

yep.
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Re: Maths problem

Post by mistermack » Wed Mar 01, 2017 2:26 pm

I think everyone has missed a trick.
Sound is obviously causing problems in picturing the problem.

I suggested ages ago that you can picture it better, if you imagine the clock as a runner, who always runs at a constant speed (relative to the ground, of course) and the other parts of the clock as the emitter, the arm, and the reflector at the end of the arm.
The runner runs to the end of the arm, touches the reflector, and runs back, and moves the clock one unit each time.

The train runs at constant speed, and the only variable is the angle of the arm with respect to the track.

If you want to model the straight-ahead case, just imagine the clock offset slightly from the train, so that the runner can run in that direction.

It's exactly the same question, just easier to model.
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Re: Maths problem

Post by Scott1328 » Wed Mar 01, 2017 11:27 pm

You are modeling the problem incorrectly. Put the runner on a platform on the moving train.

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Re: Maths problem

Post by pErvinalia » Thu Mar 02, 2017 12:57 am

mistermack wrote:I think everyone has missed a trick.
I haven't missed anything. The hardest part of this is that the return journey of the ping is shorter than the outward journey (except at 90deg). So i'ts not a case of simple geometry. I've actually come up with a crazy looking equation that works it all out, although I'm assuming it's wrong like the first time I post every other equation turns out to be wrong. I'll post it and the steps I took to get it later.
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Re: Maths problem

Post by pErvinalia » Thu Mar 02, 2017 12:59 am

Scott1328 wrote:You are modeling the problem incorrectly. Put the runner on a platform on the moving train.
No, he's modelling it correctly. He's modelling it as a clock open to the surrounding (still) air. Your model here is of a clock that is closed to the surrounding air. As I said earlier, that is a trivial problem to model. Such a clock would tick at the correct time regardless of angle or the speed of the train.
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Re: Maths problem

Post by pErvinalia » Thu Mar 02, 2017 3:55 am

Ok, here's my crack at an equation to model it. See if you can make sense of the steps and diagrams to see if I did it right.

Assumptions and symbols:
vs = speed of sound = 330m/s
vt = speed of train
t = time in seconds
@ = angle of the clock to direction of travel (only focussing on 0-90deg here)
Distance between sound emitter (1) and sound receiver/reflector (2) = 1.65m
Therefore clock ticks off one one-hundredth of a second for each return journey of a sound blip. i.e. 0.01 seconds.

So here's the geometry of the clock from the reference frame of the moving train:
trainclock.jpg
Here's the geometry of the travel of the sound wave from the reference of the medium (i.e. the still air):
trainclock2.jpg
Notice the positions of the emitter/receiver as the sound wave travels through the clock.
Notice that the geometry formed by the sound "ping" is a scalene triangle. Therefore we need to convert that to two right angle triangles and do a bit of fiddling to work out the variables.

Working out the outward journey of the sound ping (c(out)) is relatively simple. Via Pythagoras:
c(out)2 = (vt * t + cos@ * 1.65)2 + (sin@ * 1.65)2

Note, vt * t is the distance the train travels in t seconds. And cos@ * 1.65 is the distance the sound wave has to travel in the direction of train travel between emitter/receiver 1 and 2. So total x-axis distance the sound wave needs to travel is vt * t + cos@ * 1.65

Solving the Pythagoras relationship for t gives us (via wolfphramalpha:
MSP82022a80eh5b8c01dha000067bib5c1i7gg26ef.gif
MSP82022a80eh5b8c01dha000067bib5c1i7gg26ef.gif (2.77 KiB) Viewed 2690 times
where the angle x = @

To solve, you just need to plug the values into the wolphram equation - here
So for a clock angle of 45deg and train speed of 33m/s, the time for the sound ping to travel from the emitter to the receiver is 0.00539 sec.

Now, to work out the distance of the return sound ping (c(in)), we need to get the value for a in the above diagram so we can apply Pythagoras.
a = cos@ * 1.65 - x.
x = vt * t2. I.e. velocity of the train times the amount of time it takes for the sound to travel c(in).
Note, t2sec = c/vs. i.e. t2 = c/330
Therefore a = cos@ * 1.65 - (vt * c/330)
Now we have a,b and c, we can give it the old pythag treatment.
Therefore c2 = a2 + (sin@ * 1.65)2
Therefore c2 = (cos@ * 1.65 - (vt * c/330))2 + (sin@ * 1.65)2
Note, c = 330 * t.
Therefore (330t)2 = (cos@ * 1.65 - (vt * t))2 + (sin@ * 1.65)2
Via wolframalpha that gives solves for t as follows:
MSP10100224ddae66f74ffef00004383h44cae3gc8a9.gif
MSP10100224ddae66f74ffef00004383h44cae3gc8a9.gif (2.81 KiB) Viewed 2690 times
where x = @

To solve, you just need to plug the values into the wolphram equation - here
So for a clock angle of 45deg and train speed of 33m/s, the time for the sound ping to travel from the emitter to the receiver is 0.00468 sec.

Therefore, total time for the sound ping to travel out and back is 0.00539 sec + 0.00468 sec = 0.01007 sec.

Therefore, at 45deg clock angle to train travel, and at train speed of 33m/s, the sound clock will run 0.01/0.01001 = 99.305% the speed of real time.

I need a break after that, so I'm not going to do the figures for other angles. Might do them later if no one else does them.

One point to note is that this only works for train speeds that are below a certain speed. Above a certain speed c(in) will trace a line away in the direction of train travel from the reflector (2). I haven't put any thinking power towards determining the equations for that. They won't be significantly different I guess.
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Re: Maths problem

Post by Scott1328 » Thu Mar 02, 2017 4:40 am

Stick to right angles and simplify the math
The Time between the sound clock's ping

T = h/sqrt(rs2 - rc2)

Where h is the perpendicular distance between the emitters
rs is the speed of sound
rc is the speed at which the clock is moving

When the clock is not moving the time between pings is merely the time it takes the sound to travel distance h. It is also easy to see that as the speed of the clock increases the time between pings increases.

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Re: Maths problem

Post by pErvinalia » Thu Mar 02, 2017 5:29 am

You haven't read a single thing in the thread, have you? That is patently wrong on a number of fronts, as explained throughout the thread. Ffs, am I writing in Chinese?!?
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Re: Maths problem

Post by Scott1328 » Thu Mar 02, 2017 5:46 am

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

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Re: Maths problem

Post by pErvinalia » Thu Mar 02, 2017 5:59 am

:fp:

I've explained many multiple times why you can't use simple geometry and then double it. Read the fucking thread, ffs. If you think you can do what you did then don't just provide a grade 5 level equation and expect that answers it. Work an example out and I'll show you exactly how you are wrong.

I can't keep saying it. The path traced by the sound "ping" is a scalene triangle at angles between but exclusive of 90deg and 0deg. IT'S NOT an isosceles triangle that can be bisected to make two identical right angle triangles. The path followed at low angles and low train speeds will be qualitatively as follows:
Untitled.jpg
Untitled.jpg (11.27 KiB) Viewed 2681 times
As the train speed increases and/or the angle increases, it will reach a situation where the initial emitter (i.e. the final receiver) will move past ground position of where the reflector was when it reflected the ping, before it (the final receiver) will receive the ping. In that case the scalene triangle will look qualitatively as follows:
Untitled.jpg
Please, for the love of God, can you get this into your head?
Last edited by pErvinalia on Thu Mar 02, 2017 6:16 am, edited 2 times in total.
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