Maths problem

[pErvinalia]

If you did that (provided the algorithm) the first time I asked then perhaps we could have avoided this confusion. From what I can tell, it seems that you are using the symmetry of the rules of geometry to more simply calculate the journey lengths. That's a great idea, if that's what you are doing. But we'll never know unless you either provide your algorithm, or at least provide a detailed description of how you are calculating the path lengths. Just saying "Symmetry for the win, derp" isn't conveying anything useful.

[Brian Peacock]

Equation are tiresome on the phone, but it ain't tricky to increment the relationships over time, work out the rate of expansion of the ping's wavefront, compare the (x, y) co-orda of reflector and/or reciever against the expanding radius of the wave, and use the principles of analytical geometry to work out the timing/distance. I'm back home on Saturday and'll pop something up then. In the meantime perhaps you can work something out for yourself - or maybe try and unmunge the JS of the third animation. It's all in there, and only really involves a bit of cosine/sine plotting and some speed * time sums.

[pErvinalia]

Equation are tiresome on the phone, but it ain't tricky to increment the relationships over time, work out the rate of expansion of the ping's wavefront, compare the (x, y) co-orda of reflector and/or reciever against the expanding radius of the wave, and use the principles of analytical geometry to work out the timing/distance. I'm back home on Saturday and'll pop something up then. In the meantime perhaps you can work something out for yourself

I suppose you missed it like you missed most of the other posts in the thread. I've already worked it all out and plotted it on a graph. I applied the symmetry of general geometry laws like you have suggested and it doesn't make it any easier. In fact it makes it conceptually harder as you have to split the problem into two physically separated sound pings. The way I've done it is the exact same geometry (actually one line in the algorithm longer) but conceptually it is one contiguous sound path.

[Brian Peacock]

Yes, that's because the point of emission and the point of reflection are fixed but you only know the point of reflection once you've incremented time from the point of emission. But it can be simplified. The only value to increment is the one for time because we know all the other starting conditions, the y-valyes of emitter and reflector never change, x-values can be solved by subtraction, and what is added to the ping's outward journey at 90 deg - n is added to its inbound journey at 90 + n so we only have to solve to 90 deg to solve the timing question for the entire thing.

[pErvinalia]

I can't make much sense of that. On the surface of it it sounds the same as what I've done. I think I'll wait till you present some equations and diagrams before I try and grok what you are doing.

[Brian Peacock]

If we say that the point of emission (where the emitter/receiver is when the 'ping' sounds) are described by the co-ordinates (Ex,Ey), that the point of reflection (where the reflector is when the 'ping' reaches it) is (Rx,Ry), and that the point of return (where the emitter/receiver is when the ping returns) is (Px,Py) then the total time (T) of each clock cycle is...

T = ( sqrt( |Rx - Ex|2 + |Ry - Ey|2 ) + sqrt( |Px - Rx|2 + |Py - Ry|2 ) ) / speed-of-sound

I'll try and go over how to determine the values of Ex, Ey, Rx, Ry, Px, and Py tomorrow.

[pErvinalia]

Yes I think the basic trig is fairly obvious. It's the "determin[ing] the values of Ex, Ey, Rx, etc..." that is the more complicated part. I assume you'll find them the same way I found them. So after all this, I assume you are doing the same thing I did. I wish you'd been clearer earlier and stopped stating strange statements that were confusing and/or flat out wrong. All you've done is make a somewhat complex problem nearly impenetrable.

[Brian Peacock]

Yeah, whatever.

[mistermack]

What's interesting is that the method of calculating the problem works exactly the same for light, as for sound. (although at much higher speeds, obviously).
In the scienceforums thread, Janus made the point that the difference between the two cases would be that a light clock would not speed up or slow down, depending on the direction you point it. He then realised that the calculation for light would be the same method, and surmised that the reason that a light clock would not be affected by the direction it's pointed would be down to length contraction.
But length contraction is a funny thing in it's own right. Going on what Wikipedia says, it's something that you would observe, if you were moving relative to the object being measured. But you wouldn't observe any length contraction, if you were co-moving with the object. ie, were in the same frame.
Einstein said :
"The question as to whether length contraction really exists or not is misleading. It doesn't "really" exist, in so far as it doesn't exist for a comoving observer; though it "really" exists, i.e. in such a way that it could be demonstrated in principle by physical means by a non-comoving observer.[20]
— Albert Einstein, 1911"

And as I said earlier in the frame, it's a result of time dilation. So there is no difference between sound and light as far as special relativity goes. It seems to work in the same way for both, if your clock is made of sound or light respectively.

[pErvinalia]

There is no difference between anything as far as special relativity goes. SR applies to everything that is moving relative to something else.

[Forty Two]

As the distance between emitter→reflector→receiver increases so the clock will slow, because sound travels at a constant in a stable medium.

Yes, I think that's a given.
But, at a constant speed, does the clock run at the same speed, no matter what angle it is pointing to?
In other words, does the sound wave cover the same distance, whatever angle you point the clock.
Will the clock ping at a different rate, if say, the clock is aligned at 20, 30, 50 degrees to the direction of travel?

It's actually not a simple thing to work out. I tried but my head hurt. I was hoping some sad mathematician could rule on it.

Since it's run on sound relying on an echo, then if you angle the clock, the echo will bounce off at an angle. Whether the sender will receive an echo depends, therefore, on the strength of the sound wave, the distance between the sender and the clock, and the angle. Provided the sending device hears an echo, then it doesn't matter what angle the clock is pointed because the sound is traveling at the same speed.

EXCEPT, that since the sound itself exerts force, the clock will get further away from the sending device if it is at 180 degrees (perpendicular to the sending device) as opposed to say at 45 degree angle, where the vector will be off slightly. I.e., if the clock is completely facing the sender source, all the force from the sound pushes the clock away a bit. If the clock is at an angle, then less forces pushes back because a bit of it is either missing the clock or pushing the clock a bit to the side.

also, you'll have to factor air density, altitude, etc.