Re: ATM f ratio

[Anthony Stillman <atmer@flash.net>]

Frank Ward wrote:
>
>You said,".  Hence a 12.5" f/5 has a slightly larger useful
> angular field than a 10" f/4."
>
>What is it?


All,

Leave it to Frank to keep me honest.  I admit I pulled those diameters and
focal ratios out of the air and posted before checking.  Oops!  Below is
the analysis I should have done eailer.  As you can see my selection of
diameter and f/ratio was less than auspicious.  What I should have written
was a 12.5" f/5 has a slightly larger useful angular field than a 10" f/3.9
My point is still valid, only my face is egged.

Anthony


>From the "Handbook of Optics" for a parabaloidal mirror:

Spherical Abberation :     beta = 0
Sagittal coma:             beta = theta / (16*f_no^2)
Astigmatism:               beta = (L+f)^2 * theta^2 / (2*f^2 * f_no)

Where:
  beta    is the angular diameter of the blur spot in radians
  theta   is the half-field angle in radians
  f_no    is the f/ratio of the parabaloid
  f       is the focal length
  L       is the distance from the mirror to the aperture stop

The linear or physical diameter of the blur spot B is:

   B = beta * f

Adopting a photographic criteria for what is a tolerable blur spot
diameter, then for coma, theta is determined thus:

   Bmax = 0.001 inches


12.5"  f/5

f = 62.5 inches
beta = 0.001 / 62.5  = 0.000016 radians
theta = 0.000016 * 16 * 5 * 5 = 0.0064 radians


10" f/4

f = 40 inches
beta = 0.001/ 40 = 0.000025  radians
theta = 0.000025 * 16 * 4 * 4 = 0.0064 radians


10" f/3.9

f = 39 inches
beta = 0.001/ 39 = 0.000025641  radians
theta = 0.000025641 * 16 * 3.9 * 3.9 = 0.00624 radians


12.5 f/4.9  0.00627
12.5 f/5.1  0.006528