[ATM] parabolic versus spherical volume of glass

[Guy Brandenburg <gfbrandenburg@yahoo.com>]

I forget who wrote it, but someone was saying that
with any significant sized mirror, a parabolic shape
has to be fine-ground in, because the volume of glass
that has to be removed to make it parabolic from a
spheroid is just too great; the writer challenged
anybody to figure it out for themselves.

The writer was indeed correct in citing the two
formulas for calculating sagittae: r^2 / 2R for a
parabola and R - sqrt(R^2 - r^2), where r is the
radius of the piece of glass (half the diameter of the
mirror) and R is the radius of the circle of curvature
(or twice the focal distance). I got stuck trying to
use my rusty calculus skills in calculating from
scratch the volume of a paraboloid and a 'cap' of a
sphere, so I took the easy way out and just looked up
the formula for the volume the cap on a sphere, i.e.
the part of a sphere that is cut off by a plane. If
the sphere has a radius of R, and the 'cap' has a
sagitta  or height of h, and the circle that is formec
by the intersection of the plane and the sphere has a
radius of r, then the webpage
http://mathforum.org/dr.math/faq/formulas/faq.sphere.html
claims that the volume of the 'cap' is (Pi/6)*(3*r^2 +
h^2)*h.

Well, I'm slowly working on a 16.5" f/5 mirror, and I
did just those calculations, and I don't think that
for this mirror that fine grinding into a paraboloid
is necessary or even wise, because the difference in
volume is about 14 thousandths of a cubic inch; it
should not be hard to remove that with
polishing/figuring alone. I get that the sagitta is
.20638 for a sphere, and about .20625 for a parabola,
a difference that my sphereometer cannot distinguish.
I find that I would need to remove about 22.055 in^3
for a sphere, and 22.069 cubic inches for a
paraboloid. 

OK, maybe my mirror isn't 'serious' enough, so I
decided to calculate for a 20-inch diameter mirror
with a 90 inch ROC; the two sagittae are about 0.5556"
and 0.5573". My sphereometer MIGHT be able to
distinguish those. MAYBE. When I calculate the
difference in volume between those two spheres, Iget
87.347 cu.in vs 87.631 cubic inches, or abbout a
quarter of a cubic inch to be removed by polishing. I
still think that's doable.

Of course, what I ahve NOT done is actually calculate
the volume of a PARABOLOID versus a cap of a sphere.
Brain too tired on January 2, and plus I should be
back grading papers.

Anybody care to comment?

Guy

Anybody care to comment?

=====
Guy  Brandenburg
Washington, DC
My home page:
http://home.earthlink.net/~gfbranden/GFB_Home_Page.html
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