Re: [ATM] [atm_free] RE: data and musings on thin mirror

["Jerry" <wa4guu@verizon.net>]

Well, it seems it should work.... maybe.

It might just be that f3.5 is hard to judge. I'm thinking smoother curve and
slower would be easier estimate correction. With the fast mirror the grating
has to be moved quite a distance from the COC to get the edge of the mirror
on the same side of the COC as the center of the mirror. Then there are a
bunch of lines in the middle.

On the last post I decided there were 3 1/2 bands at the edge but then used
3 for the calculation because the edge was a bit difficult to judge with the
cluster of zones near the edge. The ratio was real close to 2 to 1, and it
being an estimate and simpler to figure in the head, I rounded to 2 to 1 and
the result was .075". If I had gone with the 3 1/2 bands I would have been
closer to the .060 that I got with the simulated ronchi. Something like
.063. So really, the simulated and real match almost exactly to my eye as
for judging correction. I like the 3 1/2 number better than 3. So about 10%
correction.

That is all assuming that a 100 line per inch grating would show 6 lines
with the grating .15 inch from the COC on a 16 inch diameter sphere of 110.5
inch radius.

Within a small margin of error by my still highly suspect reasoning, the
cone of light from a 16" mirror of 110.5 inch radius should be about .145
inch diameter 1 inch from the COC. If that is correct then 1 inch from the
COC the cone would be intersecting 14 1/2 lines of a 100 lpi grating. So at
.15 inch from the COC it would intersect 2 lines. To get the 6 lines I
figure the grating would be a little more than .4 inch from COC and
depending which of my previous line width ratios, would mean somewhere
between .17 to .2 inch or 30 to 35% correction.


Are Foucault test numbers available?  


Jerry





    

-----Original Message-----
From: Stephen Koehler 

Jerry,

> Now I estimate how many bands could fit on the mirror if they were all the
> size of the three bands in the center spherical area. I would say 6.

I have never seen this before, but it makes sense.  This is a very
useful test!  Thanks for explaining it.

I think the test for basic correction can be done without a need to
--
Steve Koehler

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