From: "Dwight K. Elvey" 
To: atm{at}shore.net
Reply-To: "Dwight K. Elvey" 


Hi
 I haven't been following this thread so I hope this
is new information. Field rotation of Alt-Az is not constant with time but
is cyclic over a 24 hour time.
 There was some talk about this on sci.astro.amateur.
I described some thought problems to demonstrate this.
 Lets assume that we have a telescope siting right on
the equator and we track a star that goes right overhead. To track this
star, we just run the Alt at a constant speed. We don't need any field
rotation.
 Now, lets assume that the star we want to track is 1 degree
south. We know that we will now have to use both Alt and Az drives. We know
that this will cause field rotation as well but lets look at how it is
effected.
 As the star comes up from the horizon, there is little
rotation since it is almost on the same track as the other star was. As the
star just reaches over head, the Az drive has to track quite fast, while
the Alt drive now changes direction. As you can see, the field rotation has
suddenly increased in speed to keep up with the fast rate of the Az drive.
 This was an exagerated case but demonstrates that field
rotation isn't constant and one needs to know the rates and positions of
the Alt and Az drives, as well as the location to determine the specific
rate at that instant of time.
 I haven't worked out the formula for this but I hope this
will keep others from thinking it might be something as simple as the sine
of the angle or something. Dwight


>From: mdholm{at}telerama.com
>
>
>Tom,
>
>I just (literally) did some hand waving and convinced myself that you would
get
>less field rotation with first axis east-west rather than north-south. It
>probably depends on latitude.  Below 45 degrees this may be true.  Above 45
>degrees the north-south orientation would yield less rotation.  At the
equator,
>a north-south, alt-alt is an equatorial.  At the poles, an alt-az is an
>equatorial.
>
>No orientation other than equatorial gives no rotation.
>
>Sorry, I don't have the equations for that transformation.
>
>Mark Holm
>mdholm{at}telerama.com
>
>

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