Charles,

It was +/- 0.00043" over 10 degrees.

A quick and dirty attempt to plot out the locus for your rolling ball case 
using a CAD program gives some interesting results.  They suggest that a 
rolling pivot may be generating a perfectly circular locus for the end (or 
any other point) of the beam.  I seem to remember seeing a proof of that 
somewhere, though I'm not sure.  The way the locus radius varies with the 
length of the beam is somewhat interesting.  Obviously for a very long beam 
the radius becomes large, but if you choose a point on the beam close to 
the center of the pivot ball, the radius also becomes large.  So for any 
selected pivot ball diameter, there should be a beam length which has a 
minimum radius for its locus.....interesting.

I tried a 1"dia ball with a 5" long beam, and over a motion of +/- 5 
degrees, the end traced a circle of radius of 6.0494935"  +/- 
0.0000001"  which was the limit of the accuracy of my CAD program.  That 
suggests that an analytical approach is worth looking into, to prove that 
it is indeed a perfect circle (if it is) and to come up with an expression 
to locate the center.  In the extreme example above, I think if you located 
the plate approximately 0.5494935" forward from a point directly under the 
upper wire attachment you would obtain the circular motion you were looking 
for.

I am now starting to think that it's not going to be a geometrically 
perfect circle.  If you try a 1/2" long beam with a 1" dia ball, you get a 
cycloid.

More later,
Brett

At 10:49 PM 7/5/2008 -0700, you wrote:
>Brett,
>It would seem me, too, that the figure-8 pivot is very similar to a ball 
>on a plate except that the figure-8 pivot would have about twice the 
>effect due to the point of contact moving along a second curved 
>surface  adding to the effective pivot point movement.  One quick 
>observation -- you say, "...traced out a circle, accurate to +/- 0.00043" 
>or 43 microinches per degree."
But 0.00043" is 430 microinches, not 43 microinches and in my brief 
conjecture, 100 microinches is enough to lead to failure of the swing 
trajectory in a 20 second period Lehman.
Regards,
Chas.


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