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Subject: Re: Is a Lehman geometry rolling pivot inherently unstable?
From: Charles Patton charles.r.patton@........
Date: Mon, 07 Jul 2008 00:02:39 -0700


Brett, Chris,
Thanks to Chris for tossing the wrench in the works -- :-) -- I have now 
spent the day trying to resurrect trig math I havenít done seriously for 
35 years.  I think I have the formula for Chrisís bottom pivot, but Iím 
still psyching out the upper pivot, so I havenít started sims yet.  What 
I think at this point is that as the gate swings the lower point 
trajectory tightens up (radius decreases) with swing while the upper 
pivot flattens out (radius increases) which I would assume leads to 
unstability Ė the bob going lower as it swings.  But this seems to fly 
in the face of Chrisís success in long period Lehman. The answer may lie 
in the combination in that the gate is twisting as it swings so the 
vertical position of the bob would play an important part of the 
stability.  There has to be some point that the beam twists about, and 
if the bob is mounted above or below this point the twist could 
compensate or increase the trajectory error of the bob.

The morning is spoken for, so Iíll try to get back on the problem in the 
afternoon.

Now Iíll take a big leap of faith and put forth the formulas I think 
describe the lower pivot.  Assume a cylinder of radius R with a beam of 
length K having a flat face resting on the cylinder.  Angle T is the 
angle of the contact point of cylinder/face (beam angle).  The angle of 
a line thru the center C of the cylinder and plumb bob P is equal to 
angle T minus angle B.  Line G = C to P.  All angles are in radians.
Then:
B = T(R/K)
G = (R+K)/cosB
		x & y are referenced from C
x =  G(sin(T-B))
y =  G(cos(T-B))

If thereís interest, I can try to do a standard proof deriving the 
above.  I didnít get to the formulas above with a step-by-step written 
proof, so it very well could be flawed.  Good for discussion though.

Anyway, later.
Chas.
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