Re: Planar Linkage Problem

From: Narasimham G.L. (mathma18_at_hotmail.com)
Date: 10/04/04 

Date: 4 Oct 2004 01:49:33 -0700

Honda_Kiai@hotmail.com (Ken Honda) wrote in message news:<7156fcf8.0410031830.258fc691@posting.google.com>...
> Given a planar linkage consisting of four bars of unit length (in
> which bars are permitted to cross) of the form given below:
>
> x---x---x---x
>
> compute the expected distance between endpoints when the angle between
> each bar is chosen randomly.
>
> It seems to me that one could compute the manifold arising from the
> configuration space of such an object, and a point on this manifold
> corresponds to a configuration of this planar linkage.

Yes, but for two given end points the middle link is free to rotate,
as one degree of freedom is still available, it is not deterministic.

> ..Then, integrate over the manifold.

Why bring in integration? If rods are constant length and only
included angle varies, trig is OK.

Label joints as : O---1---2---P

Fix the first fulcrum O, call angles at the second and third joints
1,2 as th and gm. By Cosine law in Trigonometry, RADIUS^2 at tip of
last point P is something like 4 s^2 + 4*s*sin (gm+th/2) +1, s = sin
(th/2).(But please check this). It must have maximum value 3 when bars
are straight for the configuration in your sketch and as also as per
Triangle inequality.

Now if RADIUS and th are given, gm can be calculated. When there are
two roots, you have two possible configurations that can be sketched.
If RADIUS alone is given, the it is a function of two parameters th
and gm. Bring in some criterion like middle rod parallel to OP. (Or,
centre link middle point is also middle of OP). Then find roots by
iteration...

Or, did you mean that integration will give you a procedure where you
need not bother about which of the two/several roots is to be taken?
There is also a vector/complex variable method to do this...

 HTH.

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