When imaging, it is essential to track the star field being images with very great accuracy. Both Film and CCD imagers can resolve to about 0.01 mm. Thus if the telescope has a focal length of 3000 mm the telescope has to be pointed and held to an angular accuracy of about 1 arc second. In addition to pointing the telescope to a given star in the field being imaged, the telescope has to be set up so that the star field does not rotate in the image plane. There are two ways of doing this. One is to rotate the image plane to match the rotation of the field and the other, more traditional way, is to mount the telescope axis to be parallel with that of the earth. In the latter case, it is possible to attain very excellent polar alignment, especially with a permanent pier, but very hard to attain perfect alignment. The issue of the actual motion of a star in the image plane of a polar mounted telescope is considered here.
The motion of a star in the image field depends upon three factors. One is
the amount of actual field rotation which is caused by the fact that the RA
axis of the telescope is not exactly aligned with the earth's axis. Call
this the angle of misalignment, A. This misalignment causes a rotation rate,
R. The rotation rate is 4.37E-3 radians per minute times the number of radians
of misalignment. This can be shown from a detailed study of the spherical
geometry of the celestial and observing spheres. The numbers used here
are for simplicity the worst case but with the 1/cos (declination) set to 1. The rotation will be worst at declination = 90 degrees.
The problem is that the rotation is a strong function of the pointing declination
and RA. (see another article on this web site that gives the exact formulas)
Then we have R=(4.37E-3 * A/57)/ cos (declination) Where A is in degrees of misalignment.
To simplify the calculation I assume we point well away from the pole and assume the 1/cos (declination) < say 4. Thus the examples are a bit optimistic but generally within a factor of a few and close enough to give the idea. I have ignored this factor in the examples by assuming declination = 0 and cos = 1.
The actual number of minutes of exposure possible for a given allowed amount of drift, d, is dependent upon the distance between the guide star, which is the center of rotation, and the star of concern. Call this distance, D. D is the actual distance measured at the imaging plane.
It can be found in two ways. The simplest is when the guide star and the star of concern are both on the field being imaged. If the guide star is in the center of the image for a 35 mm size image, the most distant star is in the corner which is 21.5 mm distant. For a CCD chip of 400 size, the distance from the center to the corner is 4.15 mm. If the guide star is outside the image frame its effective distance must be calculated from the angular separation of the two stars and the focal length of the telescope. For example, with a 1200 mm telescope and a separation of 1 degree, the star separation in the image plane is 1200 times 0.017 or 20.4 mm. (0.017 is the radian angle corresponding to 1 degree)
The actual number if minutes of exposure allowed to reach the limits described
is then:
Minutes of exposure = ( d/DA) * 13E3
For example with 1 degree of misalignment, an allowed 0.01 mm of star trail,
and a distance of half the diagonal of a 35 mm frame, we get:
Minutes
= 0.01/(21.5 * 1) * 13E3 = 6 minutes
For the same situation with a size 400 chip (416 or ST-7) we get:
Minutes
= 0.01/(4.15 * 1) * 13E3 = 31.3 minutes
For the same CCD chip conditions but with the guide telescope pointed 1 degree
off the axis of the center of the chip and using a 1200 mm telescope, we get,
the effective distance, D = 21 mm between the guide star and the center of the
chip. Then the center of the chip will move the specified amount of 0.01 mm
in a time given by:
Minutes = 0.01/(21 * 1) * 13E3 = 6.2 minutes
Thus we see that a significant penalty is paid by guiding on a star too far away from the center of the imaging film or chip. It is worse by a factor of 1/cos (declination) as described above. At my location (about 43 degrees) the zenith is at declination 47 degrees and the cos is about the square root of 2 or 1.4. So the rotation at the zenith for me is worse than the example by 50% or so. At some hour angles the problem is not as bad since this depends on the RA and the direction of the misalignment. So you can see that exact calculations are somewhat convoluted but the principle is to get the alignment with the pole as close as possible so as to reduce rotation as much as possible.
It has been reported to me by several persons setting up their telescopes that they are usually satisfied with polar alignment to 2 to 10 arc minutes. With a permanent pier mounting it is possible to get polar alignment to 10 arc seconds. (with care) The problem with trying to hold accurate alignment is that the compliance of the LX mount is such that an angular change in pointing introduced by a 1 kilogram load change is about 50 arc seconds. So it is important to align the telescope with a load similar to that used under actual operation.