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The position of objects on the sky changes in right ascension direction at an angular rate of 15 degrees per hour times cos(declination). How long does it take the sky to rotate by one field width, as a function of declination? It takes 3.1/15 = 0.2 hours = 12 minutes on the equator, and so t(dec)=cos(dec)*12 minutes elsewhere. So if we were scanning along the meridian we would have return to a given declination at an interval of cos(dec)*12 minutes, to get full coverage at minimum airmass for each declination band. At 50 seconds per field (average) in 12 minutes we would cover 3.1*12*(60/50)=45 degrees of declination.
One potential approach:
determine the rank-ordered priority of all fields on the meridian, or at that night's minimum airmass if they don't transit, in each passband, for different potential values of seeing.
reject the fields that never appear in the top 1000~1000. These have such low priority we'd never get to them in a single night.
For each parametric value of seeing, compute the sequence of observations that maximizes the merit function, including the slew overhead contribution.
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