Re: ATM 8" f3.8 -- Worthy project?

[Dwight Elvey <elvey@hal.com>]

"Matt Terry" <mterry@tampabay.rr.com> wrote:
> 
> 
> 
>   Hi there glass-men and women,
> 
>      A couple of quick questions:  How  many folks have made sub f/4
> smallish
>   mirrors?  A mirror class member wound up with about 3.8 on his 8" and got
>   discouraged from figuring it.  I do not often read about such speedy
>   parabaloids, though I have looked through a 41" f/3.1 at the Grand Canyon
>  SP in 1998.
>       I'd like to see if I can figure his mirror, but, are there testing
> limitations or impossible coma problems at f/3.8? Would such a mirror prove
> exponentially harder to figure than, say, the 6" f/4.25  I managed to bring
> to one-eleventh wave?  That doesn't seem likely, but there must be a reason
> such short focalities are not oft made, eh?


Hi Matt
 If you are going to used a Foucault type tester, it is
generally excepted to use a wire tester rather than a knife
edge ( although it is still not determined if the readings
that you get are in fact better). My feeling is that it
is still the best method for fast mirrors unless you
use one of the null methods or laser interferometers
( both requiring additional, expensive equipment ).
 In the wire test, you normally use a pin hole source
and a wire for the detector. The advantage is that ( assuming
the wire is small enough ) one can see both sides of the wire
at a zone, when figuring. The wire tends to make a ring
at the RoC for that zone. On fast mirrors, this ring is
quite distinct. The disadvantages are that you need a
bright point source as small or smaller than the wire
and like I said, it hasn't been shown by any of the diffraction
experts on this list as to the validity of using the center
of the wires shadow as the measurement point ( although
I'd think it was vary close to it ).
 The biggest issue for figuring will be that more glass
needs to be removed for a fast mirror. This means that
one has to have their figuring methods refined enough to
know the right thing to do for each problem.
Dwight