(ATM) Re: Field Rotator

Bob Lombardi (blombard@bb.iu.net)
Tue, 29 Aug 1995 07:48:21 -0400 (EDT)

> ------------------------------
>
> From: PAugello@aol.com
> Date: Sat, 26 Aug 1995 19:01:04 -0400
> Subject: (ATM) Re: Field Rotator
>
> > How does a field derotator work?
>
> I'm probably not the best person to answer this, as I've never used or
> even seen one, but here goes:
>
> A (stepper?) motor is used to rotate the camera in relation to the
> optical axis and focusser. I imagine a slip-ring type of arrangement, like
> the rear cell of a S-C would be desired.
> The tricky part is calculating the rate of rotation required. I've
> thought about it a bit and understand that it would depend on your latitude
> and the DEC of the object -- at the pole, an ALT-AZ _is_ polar mounted, and
> so rotation would be zero everywhere. On the other hand, at the equator,
> objects on the Celestial Equator would have no rotation until they hit the
> zenith, and then they would need 180 deg. of rotation instantaneously, as the
> 'scope does the same thing. Then there would be no further rotation, as the
> object sets. Objects north of the CE would rotate one way, objects south
> would rotate the other.
> I haven't tried to do the math, partly because I recall an article in
> S&T describing a program used to find acceptable windows for ALT-AZ photos --
> it would calculate the total rotation, based on exposure time, location of
> the observer, LST, and RA-DEC of the object. I think the guy took into
> effect the angular size of the field and how sharp the edges needed to be,
> based upon seeing and film grain. Code was supplied, so it should be at
> S&T's WWW site. I assume that I could get the algorithms needed from looking
> at the code (yeah, right). Maybe someone else knows of a better source for
> what would be needed?
>
> Hope this is of some help...
> Peter Augello
>

The algorithms to do this are in Genet and Trueblood's first book, the paperback from Willmann Bell. The title is "Microcomputer Control of Telescopes". Don't remember the chapter, but I'm sure anyone interested can find it.

As you say, the amount of rotation depends on the apparent position in the sky, with objects going through the zenith requiring the highest slew rates. When we design tracking antennas (day job ;-), we often specify a circle around the zenith where you really can't slew at full rate. This should never have to happen for astronomical use, but if you're tracking a fast-moving satellite, it's something you need to think about.

Hope this helps, Bob

Bob Lombardi WB4EHS from Melbourne, FL >>>>> blombard@iu.net The average male bicycle rider consumes 375 calories to ride 10 miles. The average car uses 1/2 gallon of gasoline, or 18,600 calories, to go the same distance. -- from Bicycle Guide magazine, July 1994