Re: ATM Herschelian, was (no subject)

["James O'Malley" <omalleyj@eclipse.net>]

On Thu, 19 Aug 1999 glheiner@pacbell.net wrote:

> 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
>

You don't need to consider it a portion of a 16", a sphere is a sphere is 
a sphere.
 
> (16^4)88.6= 5.8 million, cube root of = roughly 180 inches 
> 

so you get F 8.9 and change.

> 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 :)
> 
This needs to be ray-traced. You bring up a good point but I am not sure a
sphere has these problems. It does seem to me that you would need an
oversized primary and an aperture mask to make the primary appear circular
that far off-axis. Say a 10" F/8 primary, an 8" aperture mask 3 inches off
axis and tilted, an eyepiece 3 inches off-axis in the other direction with
the opposite tilt.

The reason I think this is true is that if you look at the way a Schimdt
camera handles off-axis rays through its small corrector, I think that we
have the same situation here.

You could leave the whole 16", make two 8" apertures in a single mask, and
have a binocular Herschelian. It would be easier than matching two
mirrors. (I'm getting out of hand now....)

jim

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