Re: ATM Optical Flat

[Aplanatic@aol.com]

Hi Richard:

> The emperor Zong of the asteroid Weird has ordered and optical flat.   I
>  thought I might be able to fill the order with a mercury dish.   The
>  asteroid is perfectly spherical and does not rotate, so a fixed mercury
>  dish would have a spherical shape.    What if I spin the dish
>  slightly?    What will be the shape of the mercury surface?  What will
>  the error curve look like?

The best you can do is cancel the second-order dependence, leaving the 
fourth-order term:

Z = (1/8)*(r^4/R^3)

where Z is the residual fourth-order error, r is the distance radial from 
mirror center, and R is the radius of curvature of the non-rotating mercury 
flat, which also happens to be the radius of the asteroid.
  
>  Yes, you predicted the next question:   what if the asteroid is
>  rotating and you know lattitude of the flat?

The rotational rate of the flat is just the asteroid's rotational rate times 
the sine of the latitude.

>  Assume that the asteroid had a spherical shape and a constant known
>  density that is the same as the earth.   Assume that the asteroid is the
>  same diameter as earth.

The earth is roughly 8,000 miles in diameter, so, R = 6500 km, approximately. 
 Assuming that Zong wants a 2 km flat to test his new ray gun, the residual 
error would be about 0.45 nm.

How fast must poor Richard rotate the 2 km flat to achieve perfect 
second-order cancellation if he lived at 33 degrees north latitude on the 
rotating earth?

Dave Rowe
Torrance, CA
Medium Format Astrophotography:
http://members.aol.com/aplanatic/photos/astro.html