Re: ATM Mercury flats

[Spencer Roedder <roedder@roedder.com>]

At 03:25 PM 97/05/01 -0500, you wrote:
>-- [ From: Richard Schwartz * EMC.Ver #2.5.02 ] --
...<snip 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?

I didn't send the numbers because it didn't seem to be necessary; but yes,
I did the numbers on the back of a windows calculator envelope, which is
why I said "amateur" mirrors.  For the 200-inch, the curvature comes out
more like one wave.

>
>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.
>

Good point - that the curvature of the flat, being _pure_ sphere, causes
mostly a shift in the focal length.

>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?

I'm not motivated to do the calculation.  My intuition tells me roughly one
turn every 78 minutes (the orbital period of an object at the earth's
surface).  Now, as I think about it, my intuition is strengthened by having
read somewhere that cycle time for any object moving frictionlessly on a
straight-line path from one point on the surface of the earth to another is
78 minutes.  E.g., drop an object down a hole through the earth and it will
come back 78 minutes later (ignore collisions with the hole caused by
earth's rotation).  Also e.g., drill a straight line tunnel from New York
to Chicago (not just straight on its ground track, but straight like a
needle piercing an apple), evacuate it, and put a train in it on
frictionless tracks; and gravity will take the train from one end to the
other and back in 78 minutes.  Finally, e.g., make a frictionless flat
table like those used for physics experiments, say 200 inches in diameter.
An object released at one edge will drift to the other edge and back under
the influence of gravity in 78 minutes.

>
>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]

I think you are mis-stating what you mean here.  If you can only put mass
in the mirror's plane and the earth is still there, no amount of mass will
cancel the curvature.

What fun....

--Spencer Roedder   roedder@roedder.com