This page provides the derivation of the polar alignment error formula
TXF/M=23636.23
where T is integration time in seconds, X is polar alignment error radius in arc-minutes, F is objective focal length in mm, and M is the width of the pixel in microns.
We start with three premises:
1) a ploar alignment error of X arc-minutes results in an error trace circle with a radius of X arc-minutes in one sidereal day.
2) the angle of sky given in arc-seconds covered by one pixel is given by
R=(206265*M)/(1000*F)
3) A displacement of 1 half pixel represents the maximum desired error. Beyond this point, a star looses it's roundness.
Given 1,2,3 above we can determine the maximum integration time T in seconds given a polar alignment error radius, a focal length, and a pixel size. The maximum time is, of course, the time it takes to traverse R/2.
Therefore:
If the error radius of X arc-minutes traces a path which is
eq0) 2*pi*X
arc minutes in length over a siderial day (approximately 24 hours for convenience), we can say:
a) 1 arc-second=1/60 arc-minute and
b) 1 day=24*3600 seconds.
Applying this to eq0 we get:
eq1) 2*60*pi*X/(24*3600) reducing to
eq2) pi*X/720
arc-seconds per second velocity due to error X.
If T represents integration time in seconds, then we can say
eq3) T*pi*X/720
represents error displacement over time T due to error X.
We know the maximum displacement we can tolerate to be R/2 from the third premise, and from this we can say:
eq4) T*pi*X/720=R/2=206265M/(2*1000F)
Reducing eq4 we get: (note: we substitute 3.1415927 for pi)
eq5) TFX/M=23636.23
with minor alteration of eq5 we get:
eq6) T=23636.23M/FX
eq7) X=23636.23M/FT
and so forth.
I derived the above equation while on vacation in Sedona Arizona, so if there are errors, I'll just have to blame the red rocks and the spirits therein. I am, however interested in your comments. Send them to srobinso@!mindspring.com
