[ATM] ball bearing triangles

[Jay Kirkland <jaykirk2@compusmart.ab.ca>]

>Even on oddly shaped triangles, if you left the same amount of edge (say
>1/2") on the outside of the triangles beyond the the three support points,
>that wouldn't alter the CoB, would it? 
>
>Yes it would.  You have to add an amount such that the new shape has the 
>same CoB as the smaller one. 


To a tiny extent, tho. 
	Okay, one of my small  oddly shaped triangles of 1/8 stainless weighs
almost 5 oz. It has 1/2" extra material outside the balance points. That
shifts the balance point from the Plop derived toward the long apex to an
imbalance of a tiny bit more than 1/4 oz. Now the points that touch the
mirror aren't mathematical points, they're real world somethings. Say
they're 1/4" in diameter. We already know from a previous thread in May or
June that even Kriege/Berry's formula works pretty well in determining
support points, that Plop allows some leeway for good performance. 1/4"
diameter let alone the 1/2" contact points on the mirror should more than
make up for any slight shifting in balance points. What are we talking in
oz/grams per square inch here? - it's got to be infinitesimal. How can that
stress a mirror?
	I mean, I like to make my stuff as accurately as I can. Part of that
admittedly is just for ego. You want every part of your mirror cell to be
exact and symmetrical. It's just how it should be made in a perfect
universe. But I say yet again, if the parts of your mirror cell were made
with care, to reasonable tolerances, don't have like several inches of
extra 1/4"  thick material around the edges, and their sides still  need to
be counterweighted/un-weighed so they don't stress the mirror, then there
has to be something wrong with the cell's design. Verify with Plop. Verify
with Kriege/Berry, as their numbers for 18 and especially 27 point cells
should still provide very good support. Perhaps not as good as Plop,
theoretically, but still exceeding the atmosphere most of us look through
on most nights.
	Jay


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