| |
[index]
[month]
[prev]
[next]
Subject: [M]: Re: mapug-digest V1 #3780 Field Rotation
From: John Oliver
Reply To: mapug@shore.net
Date: Sun Sep 23 04:01:42 2001
|
While you're here, how about checking out the
Astronomy Book
List ? |
mapug-digest wrote:
> mapug-digest Saturday, September 22 2001 Volume 01 : Number 3780
...
> Date: Sat, 22 Sep 2001 18:54:08 -0400
> Subject: [M]: Calculation of Field Rotation
>
> Rick Foster wrote:
>
> Snip ---
>
> I found a formula off of Doc G's site to calculate field rotation, but I
> need some help with the calculation. From a reference of Meeus "Astronomical
> Algorithms" Chapter 13, a formula is given:
> Angular rate of rotation = (a constant) X cos (azimuth angle) / cos =
> (altitude angle).
> Further it is said that the "a constant" for latitude 43 degrees =
> (assuming this is Doc G's latitude) is 11.1. It states that the constant is the
> angular rate of rotation of the earth times the cosine (local latitude).
> A table is given that shows the rotation rate as compared to the normal
> drift rate (11.1 ? and this seems to also be the "a constant") for a number of
> altitudes and azimuths.
> I cannot seem to get 11.1 by taking the cos of 43. It seems I am missing
> the "angular rate of the rotation of the earth"?
> I am at latitude 32 degrees, can someone help me with the calculation to
> get my constant rate (drift rate)?
> I am attempting to calculate a similar table to let me know the rotation
> rates at the various altitudes to determine better targets for alt/az
> imaging.
> Very much appreciated,=20
> Rick in Tucson=20
>
> - ------------------------------
>
> My reply:
>
> You must be misstating the equation, because at an "altitude angle of 90
> degrees" you would be dividing by zero. This is unacceptable in a valid
> mathematical equation.
>
> ------------------------------
a) angular rotation rate of Earth=15 deg/hour; cos(43)=0.731; so result
is 15 * .731 = 11.0 (Doc must be at 42 deg 15 min). For you,
cos(32)=0.848 so the constant is 15*.848=12.72
b) Yes, at the pole of any coordinate system, the rotation rate to
follow an object across that pole becomes infinite. Imagine you are
altAz tracking an object that will pass 1 arcminute from the zenith.
Just before it reaches the zenith you will be pointing very close to 0
degree azimuth, moving only in altitude. As the object passes the
zenith you will shift over to a fixed altitude of 89 deg 59 min but the
azimuth will have to go from 0 deg to 180 deg in just a few seconds. In
theory, the az rotation rate will have to be infinite for an object
passing exactly through the zenith. All altAz scopes have a "dead zone"
at the zenith. They stop observing before this dead zone, rotate
around, and start again when they pick up the object emerging from the
zone. Equatorial scopes have the same dead zone but astronomical
objects do not track across the pole. Anyone who has tried to move an
equatorial scope around while centering Polaris is aware of how slow the
scope moves in RA, even at max slew rate.
--
John Oliver
Department of Astronomy
University of Florida
View index by [date] [author] [subject]
Previous message: Re: [M]: Re: A new idea., Bill Arnett
Next message: [M]: Angle for equitorial mount, Gordon Lupien Jr.
|
|