RE: [ATM] parabolic versus spherical volume of glass

["Donald Good" <donald.good@comcast.net>]

> -----Original Message-----
> From: atm-bounces@atmlist.net 
> [mailto:atm-bounces@atmlist.net] On Behalf Of Guy Brandenburg
> Sent: Sunday, January 02, 2005 10:31 AM
> To: atm
> Subject: [ATM] parabolic versus spherical volume of glass
> 
>...
> 
> 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).

For a volume of revolution:
V=integral(2*Pi*r*H*dr)
  r=0->R

The following equations are written in terms of focal length
(Fs, Fp) and mirror radius (R=D/2) rather than sag and RoC.
r (radial position on mirror) is the variable of integration.

For the sphere
Hs=sqrt(4*Fs^2-r^2)-sqrt(4*Fs^2-R^2) for Rs=2*Fs
Vs=-(2*Pi/3)*[sqrt(4*Fs^2-R^2)^3-8*Fs^3)]-Pi*sqrt(4*Fs^2-R^2)*R^2

Note the substitution for Rs=2Fs does not affect the accuracy and
is used to compare the sphere to a corresponding paraboloid.

For the paraboloid
Hp=(R^2-r^2)/(4*Fp)
Vp=(Pi/(8*Fp))*R^4

Note that at the center (r=0), Hs=2*Fs-sqrt(4*Fs^2-R^2) and
Hp=R^2/(4*Fp) which correspond to Guy's equations.  Also for
comparison, the following equations are evaluated:

h=2*Fs-sqrt(4Fs^2-R^2)
Vs'=(Pi/6)*h*(3*R^2+h^2)

All this has been put in a spreadsheet at:
http://www.atmlist.net/contrib/donald-dot-good-at-comcast-dot-net/index.htm

Positive values on the graph indicate adding glass onto the sphere, negative
values for polishing glass off.  Adjust the fl diff value to see the effect
on the graph.  Any linear units may be used (e.g. inches or mm).

>...
> 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. 

It is interesting to note that using the same focal length for
both sphere and paraboloid (fl diff=0) means glass has to be added.
The optimum (fl diff=-.10) shows that the center deepens by .00012"
removing .008 cu in of glass.  Hp=0.20650 and Hs=0.20638 (dif=0.00012).
Vp=22.077 and Vs=22.069 (dif=0.008).

>...
> 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. 
>...

For this 20" f/2.25, the optimum (fl diff=-.27) shows that the center
deepens by .0016" removing .165 cu in of glass.  Hp=0.5589 and 
Hs=0.5573 (dif=0.0016) at r=0.  Vp=87.793 and Vs=87.628 (dif=0.165).

I am not sure of the cause of the differences between Guy's values for
these examples and mine.  Possibly the different forms of the equations
leads to different roundoff errors (on input and output).  However, both
equations here were derived from the same source (volume of revolution)
and both use the same input variables (focal length and mirror diameter).
I agree with Guy's sphere equations, but I don't see his paraboloid
equation.  Anyway, have fun.

Clear skies,
Don


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