Re: [ATM] astigmatism as seen via ronchi?

["Arjan te Marvelde" <Arjan.te.Marvelde@hetnet.nl>]

Derivation of CoG, let's see:

To start with, the sagitta s can be derived with pythagoras and resolving 
the roots of the quadratic equation:
  s = R - sqrt(R^2 - h^2)
where R is radius of curvature and h is radius of mirror.

The volume of the sphere cap is:
  V = pi * s^2 * (R - s/3)

The assumption is that the CoG of this spherical cap is at the sag where the 
volume is half.
Sagitta t for this half volume cap can be found by resolving:
  2 * pi * t^2 * (R - t/3) = pi * s^2 * (R - s/3)

When R >> s,
  t = s/sqrt(2)  (or 0.7*s)

However, we should not take the CoG of the cap, but that of the remaining 
glass. Location would be around 0.3*s from center thickness (C), and hence 
the CoG of the mirror is at approximately:
  CoG = (C + 0.3*s)/2, or alternatively CoG = (E - 0.7*s)/2

This answer has been suggested by Dale a.o., however with a small error.

.. Arjan


> The expression as derived by Jim Burrows is:
> (E^3-C^3)/3*(E^2-C^2) where E is the edge thickness and C is the center
> thickness (E minus sagitta).
> Don't try it on a flat mirror, though ;-) See
> http://web.telia.com/~u41105032/sling/sling.htm
> near the end (this gives a quite different value from Dale's expression -
> with due respect I trust Jim's more).

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