Re: [ATM] parabolic versus spherical volume of glass

[Woodchuck <djv@bedford.net>]

On Sun, 2 Jan 2005, Guy Brandenburg wrote:

> 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.

That would be me.

> The writer was indeed correct in citing the two
> formulas for calculating sagittae: r^2 / 2R for a

Amazing, even a pea-brained marmot can get it right sometimes.

> 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

Just integrate a volume element, using cylindrical coordinates.
I agree, not a fit task for 2 Jan.

> Well, I'm slowly working on a 16.5" f/5 mirror, and I

> 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

Right, the original poster was talking about an F:2 or F:1
mirror of diameter around 20.

For a 20 in diameter, F:2, R=80 in  we find a sagitta difference of
0.6250 - 0.6275 = -.0025.

Certainly this *might* be polished out with CeO or even rouge, but
it seems to me to be pushing it to expect a smooth figure to result.
But I think a bigger objection would be the length of time required.
Since I've not personally figured such a mirror, my estimate of what
constitutes tedium is very loose.  The problem of figuring would
probably deter me from even attempting it, especially the question
of testing.

To approximate the volume of glass, calculate the volume of a
cylinder of radius r, height of 1/2 ds, i.e. 1/4*PI*r*r*ds
In this example, 1/4*3*10*10*.0025 = .19 cu in.  Of course this
assumes that the optician goes from sphere to paraboloid without
touching the very edge, and gets it right the first time. (This
approximation errs on the low, optimistic side, I think.)

> 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.

Sounds like a lot of glass to me, but de gustibus non disputandam.

> 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.

:-)  Approximations *should* be made in this sort of calculation.
An exact answer would not change anything.

> Anybody care to comment?
>
> Guy

I'm happy to.

Dave

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