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Slew times.
Oct 19 2013, CWS.
angular distance moved is limited by both the maximum angular rate and the maximum angular acceleration. If we imagine the max angular rate is 3 deg/s and max angular acceleration is 3 deg/s/s, then to move one field width requires a slew of 3 degrees in angle (gives small overlap). For no coast phase this takes a time given by t=2*sqrt(2*1.5deg/alpha)=2 seconds. The maximum angular rate achieved is omega=alpha(1)=3 deg/sec. So for these parameters for any slew larger than a field width, we are angular-rate-limited, and the slew requires a time t_slew~2+dtheta/3 seconds. IF we can slew during readout, the overhead between images separated by an angle theta then is approximately (2+theta/3) seconds.
Arguably the system operation is optimized if the slew time is exactly equal to the (unavoidable) 2 second readout time. This suggests that we should tile the sky with overlapping observations that slew by half a field width between successive exposures.
Operating on the meridian in this mode, with 15 second exposures we'll assume it would take 20 seconds total on average, per exposure. It would take 24*(3.1/360)~12 minutes for the sky to move by one field width, on the equator. In the course of 12 minutes we can acquire 12minutes*3images/min=36 images on the meridian. At this half-overlap rate we could cover 36*3.1/2~56 degrees.
Sky rotation.
The position of objects on the sky changes in right ascension direction at an angular rate of 15 degrees per hour times cos(declination).
Some references
LSST science book
lucent paper, 1965, on TSP
SPIE_2006
Hubble Space Telescope scheduler
genetic_edge_recombination_operator
genetic_algorithms_review
kubanek_MS_thesis
genetic_alg_scheduler_SPIE2012
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