Re: ATM between a parabola & a sphere: computation

[Jerry Hudson <hudsonjk@pacbell.net>]

I'll make an attempt to answer Guy Brandenburg's
question.

By the way, greetings to Jerry Schnall - I miss
our Friday night grinding sessions many moons ago!

Guy, your math was just fine.  It's a little
easier if you start with the formula you
derived, 

y = R - sqrt (R&2 - x^2),

and expand the sqrt using the binomial formula.
You will find you can throw away all the higher
order terms, retaining

  ~
y = (1/2) x^2 / R  +  (1/8) x^4 / R^3.

The first term on the right is your parabola, the
second term represents the deviation of your
sphere from a parabola.  For the 8 inch f:6
mirror, the second term evaluates to 3.62E-5,
(same number you got), or 1.8 wave (1 wave being 
about 2E-5 as a round figure).  Incidentally, that 
figure is in perfect agreement with Texereau's
Ch. II, eq. (7).

Now, that's exactly how much glass you would
remove if you set about parabolizing by "turning
down" the edge.  Nothing wrong with your calculation.
It would also be the amount of glass removed if
you deepened the center of a sphere, a more 
conservative approach.

The deviation just calculated can be referred to
a different sphere than the one whose curvature
you share in the middle.  Take one that is
tangent to the middle and intersects the edge of
the parabola.  Then the maximum deviation of
sphere minus parabola occurs at the .707 zone,
or 2.83 inches on your eight incher.  I'll
leave it an exercise to the reader to show that
the maximum deviation is now just 1/4 what it
was in the above calculation, or about .45 wave
for your eight incher.

This change of reference sphere is what you do
when you both turn the edge and deepen the middle,
a simpler process, almost always done on larger
optics.

- Jerry Hudson