Re: [ATM] Polishing / Figuring Simulator

[Mark Holm <mdholm@telerama.com>]

Martin Cibulski wrote:

> Tool overhang cannot be simulated. Perhaps someone
> else has an idea to calculate the pressure distribution in this case.

I too have thought about this question of the pressure distribution
resulting from tool overhang and have not yet been successful.  It ought
to be reasonably simple once you get all the principles in a row, but I
am missing something.  I have tried to think about it from the following
assumptions.

Assume the optic and tool are circular discs.
Assume the line between the centers of the discs lies on the X axis.

1.  The top piece is not accelerating either up or down (to any serious
degree) so, neglecting force from machine components, the integral of
pressure over the contact area has to add up to the weight of the top piece.

2. Assume, on a short time scale, that the pitch behaves as a Hooke's
law elastic solid.

3. Assume that the glass or other substrate is infinitely rigid.

4. Neglect any effect of curvature (assume flat plates).

I think 2,3 and 4 combine to mean that the pressure function along the X
axis and any line parallel to the X axis is linear.  Further, that
pressure is constant over Y distance, until you get to the edge of
either disc, where it drops to zero.

If s is the overhang distance and r is the radius of the lower disc (and
the upper disc is not larger than the lower disc), then:

for s <= 0 (no overhang) the slope of the pressure function is zero,

for s = r (upper disc over edge of lower disc) the slope of the pressure
function is infinite.  (Clearly this last is impossible, but I think it
still works as a model condition.  I hope nobody actually runs their
upper disc center out to the edge of the lower disc.)

Given the slope, it ought to be not too hard to calculate the intercept
that gives the correct integral for the pressure (the weight, as noted
above.)  One could do a crude iterative approximation if not up to an
analytical solution. (I would probably go the rough and ready route.)

The question I do not know is the shape of the slope function.  A
tangent function has zero slope at zero and goes to infinity, so it
might be correct, but that is only one guess.

Another principle that might work into the calculation is that the upper
disc is not accelerating in rotation.  This means that the torques have
to balance.  The torque of the overhanging part isn't too hard to get.
The torque of the pressure distribution is the integral of the first
moment of the pressure distribution.  I think this torque condition
gives you the slope function I mentioned above, but am not 100% certain.

Another way to say it is that the combination of the pressure-area
integral = weight, and the torque integral = zero should be enough
conditions to bound and solve the problem.

Anyhow, I am not good enough at the math and physics involved to have
worked through the problem yet.   Frustratingly close, but not there.

Mark Holm
mdholm@telerama.com



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