PSN-L Email List Message

Subject: Re: Is a Lehman geometry rolling pivot inherently unstable?
From: Brett Nordgren Brett3nt@.............
Date: Sun, 06 Jul 2008 22:54:24 -0400


I think you have it.  Indeed the boom tip traces a prolate cycloid.  The=20
game, I think, is to find a way of fitting the 'best' circular arc, as=20
defined by its center and radius, to the cycloid in the narrow region of a=
few (5?) degrees plus and minus from the 'vertical' position of the=20
extension line (boom).  With trial and error I could get sub microinch=20
tracking, but haven't yet come up with a good automated error minimization=
approach.  In general, with a small ball and long boom, that should be=20
fairly easy to do.  Once you have located the center of the circular arc,=20
you can locate the plate surface to locate the rotation center where it=20
needs to be.

I'm still not sure I am visualizing the exact geometry you are looking at,=
but it's clear you're on the right track.


At 10:48 AM 7/6/2008 -0700, you wrote:
>Thanks for the trigger word, =93cycloid.=94  I had been thinking it too,=
>somehow your writing it got me thinking about a book I had stashed away,=20
>=93Technology Mathematics Handbook=94 by Jan J. Tuma.  Just the thing for a=
>discussion like this.  Our problem can be defined as class of cycloids=20
>called =93prolate cycloids.=94
>Give a circle with center C of radius R rolling on a contact line, and a=20
>Point P of K*R length (C to P), and A equals angle of CP to the normal to=
>the contact line, then the graph of P is:
>X =3D R(A =AD KsinA)   Y =3D R(1 =AD KcosA)
>This is a cycloid with loops on the end where the cusps would be if a pure=
>cycloid were graphed. (For a pure cycloid just set K=3D1)
>A point moving around a point (i.e. what we really want) is:
>X =3D RsinA         Y =3D RcosA
>What immediately comes out of this is that just simple observation of the=
>prolate cycloid curve is that the upper pivot and lower pivot are tracing=
>different curves because they are effectively 180 degrees out of phase in=
>the equation so right away balance has to be changing.  Now which way?
>I think I=92ll post this and continue with sims in Excel a bit later.  But=
>just some food for thought.  Also an important consideration is that these=
>will yield curves in the plane of the rotation, but they have to be=20
>combined in a perpendicular plane to fully establish the final effect on=20
>the bob trajectory.  I.e., the bob is a vertex on a triangle (and one not=
>necessarily a right triangle) formed by the upright, beam and suspension=
>In fact, the thought that the bob support does not have to be constrained=
>to a right triangle may provide the way out of the possible geometry=20
>problem.  More food for thought.


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