Re: [ATM] parabolic versus spherical volume of glass

["vladimir sacek" <v959@msn.com>]

  Actually, I've sort of cut it short, because it got
  contradicting, and I wanted to figure it out.  What
  I regularly see in the literature is that the fastest 
  parabolizing method (center&edge, with the 70%
  zone least affected) produces best-fit parabola
  that focuses at the circle of least confusion of the
  initial sphere. But the shape of this parabola simply
  doesn't fit into this (logical) logic: it is deeper than
  the initial sphere, and only can be arrived at by 
  deepening the sphere most at the center, least at 
  the edge.

  The "needed" parabola shape is the one is obtained
  by taking off similar amounts of glass at both, center 
  and edge. Consequently, depth of this parabola is 
  nearly identical to that of the initial sphere. In that case,
  its paraxial radius is only slightly shorter, and its depth
  is greater than that of the sphere everywhere but at the
  edge and center. The greater depth deviation is at the
  70% zone, given by (0.7r)^4/8R^3. 

  The amount of glass removed is given by the difference
  if the blank volume before and after parabolizing. With the
  above parabola, it comes to approx. D^3/18000F^3.
  It is a very small volume: for the 16.5" f/5 it would be 
  as little as 0.002 cubic inches, or less than 1/100,000
  of an inch (less than 1/2 green/yellow wavelength) 
  averaged over the entire surface. At the edge and the 
  center, the amount of glass removed would max at 
  about 0.000016" or 0.7 wavelengths.

  That compares to 0.0184 cubic inches to remove deepening
  the center, with the depth of glass removed maxing at the 
  center (0.000215", or 10 wavelengths), with an average amount
  removed over the entire surface of about 0.000086 cubic inches,
  some nine times more than in the previous case.

  While no one can expect to execute parabolizing nearly as
  perfectly as both scenarios assume, they illustrate well how
  much of a difference parabolizing method can make.

  Vlad

   
  OK. I have just refreshed my memory on how to
  integrate volumes of revolution, like paraboloids,
  using elementary calculus. 
  (see
  http://www.mathematics-online.org/kurse/kurs9/seite51.html<http://www.mathematics-online.org/kurse/kurs9/seite51.html>
  for a little refresher.)
  I get that the volume of a paraboloid with a focal
  length of F and a sagitta (depth) of d would be

  V = 2 * pi * F * d^2

  And this works out to about 22.051 cubic inches for
  the paraboloid, as opposed to the 22.055 c.i. that I
  calculated earlier. 

  And if I redo this with the paraboloid for the f/2.25,
  twenty-inch diameter mirror with the 45" focal length,
  I get that the total volume of glass removed for the
  paraboloid is 87.815 cubic inches, as opposed to the
  87.347 in^3 for the sphere.

   As you say, Vlad, there is no practical difference.

  Guy

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