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Topic: Chain Watch (Read 1017 times) |
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ThudnBlunder
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A chain watch is lying on a table-top with the chain of length a pulled taut. The watch is originally on the x-axis and the other end of the chain is at the origin. The end at the origin is now pulled along the y-axis so that the chain remains taut. What is the equation of the watch's path?
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Ronno
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Re: Chain Watch
« Reply #1 on: Sep 6th, 2009, 12:31am » |
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I would guess, (by the asymptote) y=k*ln(x0/x) 
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towr
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Re: Chain Watch
« Reply #2 on: Sep 6th, 2009, 11:14am » |
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Doesn't it depend on the speed with which the chain is pulled along the y axis?
Or am I thinking too physical here?
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Grimbal
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Re: Chain Watch
« Reply #3 on: Sep 6th, 2009, 2:23pm » |
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... or whether there is any friction. I believe that if the friction is zero, and the chain is of zero weight, then the path of the watch would be a cycloid.
http://mathworld.wolfram.com/Cycloid.html
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towr
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Re: Chain Watch
« Reply #4 on: Sep 6th, 2009, 2:58pm » |
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If it's accelerate along the y-axis (and no friction), it'd be more like an (accelerating) pendulum swing. (With friction there'd be damping)
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ThudnBlunder
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Re: Chain Watch
« Reply #5 on: Sep 7th, 2009, 7:28pm » |
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I thought it was obvious that this is purely a curve of pursuit problem. 
The only assumption you need to make is that all the necessary information is contained in the problem statement.
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| « Last Edit: Sep 8th, 2009, 12:20am by ThudnBlunder » |
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Grimbal
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Re: Chain Watch
« Reply #6 on: Sep 8th, 2009, 6:21am » |
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I don't agree. In the pursuit problem the pursuer is running at a constant speed, while in the chain watch problem it is the distance between pursuer and target that is constant.
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ThudnBlunder
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Re: Chain Watch
« Reply #7 on: Sep 9th, 2009, 6:49am » |
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on Sep 8th, 2009, 6:21am, Grimbal wrote:| I don't agree. In the pursuit problem the pursuer is running at a constant speed , while in the chain watch problem it is the distance between pursuer and target that is constant. |
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There are many pursuit problems. Perhaps you are thinking of Bouguer's original pure pursuit problem, aka courbe de chien, curva di caccia, hundekurve, and verfolgungskurve.
But if 'pursue' means chasing with the intention of catching, then I agree a better name would be 'following curve' or 'tailing curve'.
Anyway, this problem predates Bouguer.
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| « Last Edit: Sep 10th, 2009, 8:59am by ThudnBlunder » |
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Grimbal
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Re: Chain Watch
« Reply #8 on: Sep 9th, 2009, 7:40am » |
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on Sep 9th, 2009, 6:49am, ThudanBlunder wrote:
There are many pursuit problems. Perhaps you are thinking of Bouguer's original problem, aka as courbe de chien, curva di caccia, hundekurve, and verfolgungskurve. |
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In fact, I was thinking of a fox chasing a chicken that is running along a wall. I don't know when and where I read about it. Yes, it is basically the dog chasing path.
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| « Last Edit: Sep 9th, 2009, 7:41am by Grimbal » |
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ThudnBlunder
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Re: Chain Watch
« Reply #9 on: Sep 10th, 2009, 9:01am » |
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To clarify, the watch chain always forms a straight line.
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| « Last Edit: Sep 10th, 2009, 10:29pm by ThudnBlunder » |
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ThudnBlunder
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Re: Chain Watch
« Reply #10 on: Sep 17th, 2009, 9:28am » |
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Gee, is everybody still semi-comatose?
Firstly, the chain is always tangent to the curve.
Let the watch be at (x,y) and the pulling end be at (0,yt) at time t >= 0
Then the equation of the tangent is y = x.dy/dx + c
When x = 0, y = yt;
so c = yt and the tangent is y = x.dy/dx + yt
And by Pythagoras, a2 = x2 + (yt - y)2
Eliminating yt - y and taking the negative square root (as dy/dx < 0) gives dy/dx = -sqrt(a2 - x2)/x
But before you plug this into your favourite bookmarked online integrator, how would you solve this equation for y? (Newton solved it.)
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| « Last Edit: Sep 21st, 2009, 5:56am by ThudnBlunder » |
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ThudnBlunder
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Re: Chain Watch
« Reply #11 on: Sep 24th, 2009, 7:04am » |
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Solution: y = a*sech-1(x/a) - sqrt(a2 - x2), the tractrix.
This and the U-boat problem are from the excellent book Chases and Escapes (by Nahin), mostly available here.
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