How Many Sweeps? : Solution
Glen
In absolute terms, probably the best way to calculate this is to compute the area of the sphere of the entire sky in SI steradiants (sr) - solid angles, being 4 x pi or 12.56637 sr, where 1 sr (180 / pi)^2 or 3282.8063501 sq. degrees). Area of the sphere is 4 x pi x (180/pi)^2 or 41252.96124949 sq.degrees
Convert the field area into steradiants of the individual sweep. So if the field size is as you say 15 arcmin or 0.25 deg, then its area is pi x 0.25^2 = 0.19634954 sq deg. per field (or 0.000059811497 sr, or in SI units 59.811497 /*mu*/sr. (microsteradian(s))
The number of theoretical fields to see the entire sky in the eyepiece would be 41,252.8063501/0.19634954 or 210,099.607 or 210,100 fields.
Based on shape stated by John Herschel of the 3 x 0.25 degrees, the area
of the box with two halves either end would be (0.75+0.19634954) = 0.94634954 sq.deg. per single north-south sweep.
Therefore the number of theoretical single north-south sweeps would be 41,252.8063501/0.94634954 = 43591.7 or 43592.
COMMENT:
However, the catalogued "sweeps" done by him, at least according to the catalogue, is a 3 degree height in Declination by whatever time he looked and measured the objects seen passing across the field at the sidereal rate. A formal total of 810 sweeps were done by John Herschel between 13th Feb 1823 and 22nd Jan 1838; 428 northern, 382 southern.
This was the same method as Dunlop did, but he was far less careful with his sweeps and likely never repeated them - especially near the poles. Herschel did three or four separate sweeps of the area, especially along the Milky Way, and likely did a preliminary survey checking on the Dunlop's own observations and those stars in the Paramatta Star Catalogue (PSC)
To get some realistic total, I'd guess he must of made about 340,000 to 380,000 north-south "sweeps" altogether based on various checks, rechecks and varying circumstances (like the weather, seeing, etc.) - not bad for thirteen -odd years between 1825 and 1838. These latter numbers assumes that he crosses over the sky equally at the poles as the equator, and does not account for stopping to look at interesting fields. (Note: This does not account for the other east-west sweeps done in the surveys and for computing the right ascensions for objects from the known positions of his so-called zero stars. As the 'scope was NOT equatorially mounted, so the calculating RA away from the meridian had to be calculated and compensated for - the so-called the reducing curve in right ascension for time.
Most of these calculations were done by tables (whose method incidentally was not properly understood by Dunlop - hence, the nightmare with his own positions and so many missing objects. As the observations were not duplicated, the source of the errors cannot be determined - except perhaps for simple errs like signs or wrongly interpreted written figures...)
However, if you take into account not crossing to the same patch of sky every time nearer the pole, then numbers might have been perhaps fractionally less. To decide this read his work; "XIX. Observations of Nebulae and Clusters of Stars, made at Slough, with a Twenty-feet Reflector, between the years 1825 and 1833. Philosophical Transactions of the Royal Society Appendix pg.482-493 (1833)
[Downloadable through JSTOR, if you have access to a Library that can access it, or your own account. (Else private contact me, and I'll send the actual pages specified here in pdf - if so required.)
Last edited by Enchilada; 21-01-2009 at 01:21 AM.
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