Re: ATM Mercury flats

[Richard Schwartz <richas@idt.net>]

-- [ From: Richard Schwartz * EMC.Ver #2.5.02 ] --


-------- REPLY, Original message follows --------

> Date: Thursday, 01-May-97 10:47 AM
> 
> From: Spencer Roedder          \ Internet:    (roedder@roedder.com)
> To:   Lawrence D. Lopez        \ Internet:    (lopez@mv.mv.com)
> cc:   ATM Mailing List         \ Internet:    (atm@shore.net)
> 
> Subject: Re: ATM Mercury flats
> 
> At 10:59 AM 97/05/01 -0400, you wrote:
> >Does anyone have a handle on the effect of the
> >earth's gravity on the figure of a mercury flat?
> >
> >I'm just curious.
> >
> >Larry
> 
> Yeah.  It makes it flat.
> 
> --Spencer Roedder   roedder@roedder.com
> 
> P.S.  Guessing that you really meant a different question:  Of course that
> "flat" is really a section of a sphere of 4000 mile radius, just like a
piece
> of the sea surface.  For mirror diameters in the amateur range, this
curvature
> is way too small to matter.
> 
> 

-------- REPLY, End of original message --------

Once again, I am disappointed in a blanket pronouncement without any numbers
to back it up.  How many times does the usefulness of s=(r^2)/(2*R) need to
be preached?

The largest amateur telescope I know of has an aperture of 72".  To avoid
edge effects, it should be collimated on a flat of radius 80".  So r=40"   R
=4000 * 5280 * 12.

Working the formula gives a saggita of 3.1 micro-inches, or about 1/6 wave
of typical "light".  The mercury flat would be convex.

I believe that even this can be tuned out by a small shift of focus, but I
have not worked out the tiny remaining spherical abberation.  If the flat
surface is approximately spherical, the resulting mirror will be
approximately elliptical with one focal point just barely outside the
parabolic focus, and another focal point at the center of the earth.   I
personally would be delighted with such a mirror.

I don't know if such a mirror would be any good for viewing devils, like in
Disney's "Black Hole" movie.  Perhaps Art Bell could discuss that some time.

The next assignment is for the student: how fast would the mercury flat have
to rotate to make it perfectly FLAT?

And finally, for the stuent, of course, what distribution of mass around the
edge might cancel the curvature?  (what mass and radius of the ring in the
mirror's plane?)  When you do this, what is the equation for the mirror's
surface? [solution within reach of any sophomore in physics/calculus]