Re: [ATM] question about Couder Zones

[Jim Burrows <burrjaw@earthlink.net>]

At 2004-08-09 17:37 -0500, Rich Ball wrote:

>On the standard Couder mask I presume that Zone #1 is the first zone outside
>the area that would be blanked out by the secondary, perhaps 1.8 inches ID?
>..not the center of the mirror.  I note that the value of Zone #1 in
>examples I've seen is never zero.

It's best to go ahead and include the center zone in the Foucault test even 
if it's going to be inside the secondary shadow, so the test reduction 
program won't have to extrapolate (could produce even more error than the 
admittedly difficult center zone reading).  The test reduction program can 
be told the secondary diameter so it will evaluate just the illuminated 
part of the mirror.

Let H0 be the inner radius of the center zone (= 0 if the center is 
visible, like Texereau's Fig. 38 on p. 82), and H1 its outer radius.  Then 
there's three ways to compute the representative radius Hm:  Texereau's 
ave( H0, H1); Couder's RMS( H0, H1); best, Nils Olof's ave( Texereau, 
Couder).  So if H0=0, then Hm = 0.5*H1 (Texereau), 0.71*H1 (Couder), 0.6*H1 
(Nils Olof).

         -- Jim Burrows
         -- mailto://burrjaw@earthlink.net
         -- http://home.earthlink.net/~burrjaw
         -- Seattle N47.4723 W122.3662 (WGS84)  

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