Re: ATM Herschelian, was (no subject)

[glheiner@pacbell.net]

No, I don't think that's quite right. A spherical 8" needs to be
about f/9.0 (sez Couder via Tex) to be diff limited ON axis. Also
from Couder,
minimum f/length for a spherical mirror of diameter D to stay diff
limited is f^3 = 88.6D^4 (units in inches). Let's say your scope
is "almost Herchelian" & has 1/2 it's secondary in the field. Now
you would consider your 8" as a cylinder cut out of a 16", so

(16^4)88.6= 5.8 million, cube root of = roughly 180 inches 

focal length for a 16", yielding about an f/22.5 8 incher, longer
if you want to get the secondary completely out of the field.

If that 16 is an f/4 paraboloid, cutting out your 8" section
gives you an 8" f/8 that has the coma of an f/4,  still has
1/2 the secondary in the field, & was such a total $!#@& to figure
you gave it up after your 40th figuring attempt & went Newt :)

Gary Heiner

James O'Malley wrote:
> 
> On Wed, 18 Aug 1999, Reed Alan Spaulding wrote:
> 
> >
> >
> > I'm building an eight inch scope and have been considering a modified
> > hershlinian scope design.  ............
> >
> I had read somewhere (R&vV?) that the easiest way to do this was to use a
> spherical mirror with a long enough F ratio to still be diffraction
> limited. You have to live with coma that is asymetric across the field if
> you use an off-axis portion of a parabola. I *think* the limit for an 8"
> was F/12 if you stay with a sphere.
> -------------------------------------------------------------------------------
> Jim O'Malley    |