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        riddles >> medium >> Chain Watch
        
(Message started by: ThudanBlunder on Sep 5^th, 2009, 9:40am)

Title: Chain Watch
Post by ThudanBlunder on Sep 5^th, 2009, 9:40am

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?

Title: Re: Chain Watch
Post by Ronno on Sep 6^th, 2009, 12:31am

I would guess, (by the asymptote) y=k*ln(x0/x) :P

Title: Re: Chain Watch
Post by towr on Sep 6^th, 2009, 11:14am

Doesn't it depend on the speed with which the chain is pulled along the y axis?
Or am I thinking too physical here?

Title: Re: Chain Watch
Post by Grimbal on Sep 6^th, 2009, 2:23pm

... 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

Title: Re: Chain Watch
Post by towr on Sep 6^th, 2009, 2:58pm

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)

Title: Re: Chain Watch
Post by ThudanBlunder on Sep 7^th, 2009, 7:28pm

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.

Title: Re: Chain Watch
Post by Grimbal on Sep 8^th, 2009, 6:21am

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.

Title: Re: Chain Watch
Post by ThudanBlunder on Sep 9^th, 2009, 6:49am

on 09/08/09 at 06:21:03, 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.

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. :P

Title: Re: Chain Watch
Post by Grimbal on Sep 9^th, 2009, 7:40am

on 09/09/09 at 06:49:18, 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.

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.

Title: Re: Chain Watch
Post by ThudanBlunder on Sep 10^th, 2009, 9:01am

To clarify, the watch chain always forms a straight line.

Title: Re: Chain Watch
Post by ThudanBlunder on Sep 17^th, 2009, 9:28am

Gee, is everybody still semi-comatose?
[hide]
Firstly, the chain is always tangent to the curve.
Let the watch be at (x,y) and the pulling end be at (0,y_t) at time t >= 0

Then the equation of the tangent is y = x.dy/dx + c
When x = 0, y = y_t;
so c = y_t and the tangent is y = x.dy/dx + y_t

And by Pythagoras, a² = x² + (y_t - y)²

Eliminating y_t - y and taking the negative square root (as dy/dx < 0) gives dy/dx = -sqrt(a² - x²)/x
[/hide]
But before you plug this into your favourite bookmarked online integrator, how would you solve this equation for y? (Newton solved it.)

Title: Re: Chain Watch
Post by ThudanBlunder on Sep 24^th, 2009, 7:04am

Solution: [hide]y = a*sech^-1(x/a) - sqrt(a² - x²), the tractrix. [/hide]

This and the U-boat problem are from the excellent book Chases and Escapes (by Nahin), mostly available here (http://books.google.com/books?id=HeESjfM2geUC&printsec=frontcover&dq=nahin+chases#v=onepage&q=&f=false).