ATM between a parabola & a sphere: computation

[Guy Brandenburg <gfbranden@earthlink.net>]

At the NCA/AU mirror-making class last Friday, Phil Lamphier, Jerry
Schnall, and I were talking about how to figure out how much glass
actually gets removed when moving from a sphere to a parabola. We did
some calculations, and they do not seem correct. I wonder if anybody can
point out where we went wrong. 

Here they are:

Let us assume that we have a sphere, which we can treat mathematically
here as a 2-dimensional circle, with a radius of big R. Let us also
assume that we are doing this on a Cartesian coordinate plane, and that
the origin is the 'bottom' of the mirror, which is facing UP. Thus, the
center of our circle is at (0, R) and, since it has radius R, its
equation is

(x-0)^2 + (y-R)^2 = R^2, 

or, after simplifying, we get

s^2 + y^2 - 2*y*R + R^2 = R^2 

or, simplifying some more

x^2 + y^2 - 2 y R = 0. 

I actually want to solve this for y, since I am interested in the
vertical distance between the sphere and the parabola, so I will rewrite
it as

y^2 - 2*R*y + x^2 = 0.

Now this is a quadratic, and is susceptible to solution by the quadratic
formula, the one that says that if
ax^2 + bx + c = 0, then 
x = (-b + or - sqrt (b^2 - 4*a*c))/(2*a).
And here, a = 1, b = - 2*R, and c = x^2.

so 

y = (2*r + or - sqrt (4 * R^2 - 4 * x^2) )/2, which simplifies to

y = R + or - sqrt (R^2 - x^2). This I will call the equation of the
sphere.

I maintain that the "+" solution gives you the value of y on the upper
part of the sphere, about 2 stories up, and can be ignored, so the
solution we want, down at the level of the mirror, is

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

Now on to the parabola. It is well known that, and I can derive it if
needed, that the equation of a parabola with its focus at (0, f) and its
directrix at y = - f is

y = x^2 / (4*f).

The focal point for a parabola versus our sphere is, I trust, one-half
of the radius of curvature of that circle we were talking about. Thus, f
= R/2, so if we substitute this into the prior equation, we get

y = x^2 / (2 * R). This I will call the equation of the parabola.

What is x? you may ask. It is what is commonly referred to as little r,
and is the horizontal distance from the center of the mirror to whatever
point on the mirror surface we are talking about. Remember, i have the
mirror facing up, so light would be coming from above, parallel to the
y-axis. 

If we are talking about removing glass to turn a sphere/circle into a
paraboloid/parabola, we are really talking about what is the difference
in those y values for the sphere and for the parabola. So this
difference, or d, would be simply absolute-value of ( (y of sphere)
minus (y of parabola) )

or

absval ( x^2 / (2*R) - R + sqrt (R^2 - x^2) ).

I myself have not found any errors here so far. But if I put in values
for x and R, then I get some numbers that seem generally way out of
line. For example, suppose we have an 8-inch f/6 mirror. Its radius of
curvature will be 96 inches (2 * 8 * 6), and if we let x = 3, that is to
say we are 3 inches out from the center and 1 inch away from any of the
edges, then I get the following amount of glass to be removed:

absval ( 3^2 / (2*96) - 96 + sqrt ( 96^2 - 3^2) )   , or

absval ( 9/192 - 96 + sqrt (9216 - 9) ), which is 
absval (.046875 -96 + sqrt (9207)) , or about
0.00001145 inches.

For some perspective on this, let us convert to metric. One inch is
defined as 2.54 cm, so we get about
0.000029 cm, or
0.00029 mm, or
0.29 microns or
290 nanometers, or about one wavelength of violet light (? approx), or
about 30 atoms thick.

This seems a bit high. I hadn't thought that an uncorrected 8" f/6
mirror that was perfectly spherical was that far off.

If we go right out to the edge, that is 4 inches away from the center of
the mirror, where x = 4, then I would get y(sphere) - y(parabola) to be

absval (16/192 - 96 + sqrt (96^2 - 16) ), or about

0.0000362 inches, or about
0.0000919 cm, or
0.000919 mm, or
0.919 microns, or about
920 nanometers, which is about 2 wavelengths of green or yellow light.
Again, I didn't think that a spherical 8" f/6 mirror was that far off.
That is about 92 layers of atoms.

I suppose that part of the deal is that if you raise your figure of
reference by one wavelength (say) at the center, and refer everthing
back to that, not only the center but the points I indicated, then it
would cut the error roughtly in two. And I guess that is what Texereau
and 'Tex.exe' do.

Any comments?

Guy Brandenburg

ps: Phil, I see one goof I made the other night. I forgot to double the
focal length to get the radius of curvature.