[ATM] Pressure per square mm

["James Lerch" <jlerch1@tampabay.rr.com>]

Greetings All,

I'm still working on the erosion simulator.  I need to solve for the pressure
(aka normal force) for each square millimeter of the contact area, and I'm not
sure how to do that, nor what 'topic' I'd search for information on the
procedure. :0

Currently I'm applying the KISS principal, with the following assumptions: (all
of which can be addressed as progress is made)

1) the contact area is flat
2) the tool and mirror won't bend / flex
3) the tool and mirror have uniform density
4) the slurry is a uniform thickness

I've created a picture that represents the problem I am trying to solve:
http://lerch.no-ip.com/atm/Pressure_per_unit_area.gif

The image shows two situations.  The top situation has a rather straight forward
solution, but things get complicated for the bottom situation.  Common sense
says that the right side of the contact area should have a higher pressure/mm
than the left side of the contact area.  I just don't know how to calculate the
distribution..

Take Care,
James Lerch
http://lerch.no-ip.com/atm (My telescope construction,testing, and coating site)

"Anything that can happen, will happen" -Stephen Pollock from:
"Particle Physics for Non-Physicists: A Tour of the Microcosmos"

" Press on: nothing in the world can take the place of perseverance.
Talent will not; nothing is more common than unsuccessful men with talent.
Genius will not; unrewarded genius is almost a proverb.
Education will not; the world is full of educated derelicts.
Persistence and determination alone are omnipotent. "
                                                           Calvin Coolidge

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