From: Brett Nordgren Brett3nt@.............

Date: Sun, 06 Jul 2008 22:54:24 -0400

Charles, 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= =20 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= =20 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,= =20 but it's clear you're on the right track. Regards, Brett At 10:48 AM 7/6/2008 -0700, you wrote: >Brett, >Thanks for the trigger word, =93cycloid.=94 I had been thinking it too,= but=20 >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= =20 >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= =20 >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= =20 >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= =20 >prolate cycloid curve is that the upper pivot and lower pivot are tracing= =20 >different curves because they are effectively 180 degrees out of phase in= =20 >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= =20 >just some food for thought. Also an important consideration is that these= =20 >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= =20 >necessarily a right triangle) formed by the upright, beam and suspension= wire. > >In fact, the thought that the bob support does not have to be constrained= =20 >to a right triangle may provide the way out of the possible geometry=20 >problem. More food for thought. >Regards, >Chas. __________________________________________________________ Public Seismic Network Mailing List (PSN-L)