How Many Sweeps?

[glenc]

John Herschel swept the sky with his 18.5" aperture 20' long reflector.
He moved the telescope back and forth (north-south) in sweeps three degrees long. The eyepiece field was 15'.
How many thousand sweeps did it take to cover the entire sky? (Assume a sweep is 3.0 x 0.25 degrees)

[Terry B]

A very large number. Also when he found something, to stop and write down the description and position would have taken quite a while.

[glenc]

I think the answer is 55,000 sweeps. Is that correct?
That is a big effort pushing a 20 foot scope back and forth that many times!

[leon]

Haven't the slighest Idea Glen, :shrug: but it did pay off for him in the end ;)

Leon :thumbsup:

[Enchilada]

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.)

[Enchilada]

I should also comment... This same method of doing sweeps could be automated. It probably could be done by some real budding amateur with a small alt-azimuth refractor telescope with a 1 to 2-degree field, CCD camera with a really decent automated guide system with programmable up and down slewing along the north-south meridian - with a guiding system good enough for about 120 seconds.
It might be a bit of a bugger of a job to organise, but once figured out, you could very easily produce your very own southern sky star atlas / deep-sky atlas, within a year and a bit. Observations could only be over a dozen good nights in an entire year - whose only conclusion would be based naturally on the weather and the quality of the site.
Observations could be made along the meridian, with exposures of 1 to 2 minutes a pop (multi-wavelength would even be possible.) With so many IIS imagers here, hell, two or three similar telescopes scattered in differing latitudes might even be able to do it in short time, whose results
Image calibration could be done by a bright computer whizz :nerd: basically using only the free C.D.S.'s ALADIN, who could even write simple program software for automated reductions, and epoch adjustment for 2000.0, or any current (2010) or future epoch I.e. 2025 or 2050.
Best of all you could produce a photographic atlas / star catalogue, probably better than Sky Catalogue 2000.0 or Uranometria 2000.0, and would not have to curtail copyright limitations - only having to acknowledge the CDS software. (Once completed you could then do variable star observations or double stars measures forever and a day.)
:2thumbs:

Ummmm... Almost very tempted to do this myself...

"The Enchilada Star Atlas : 2010" (ESA) ??

....we have the technology and the manpower folks! :fishing:

NOTE: Had Dunlop ever seen this, I think he would have either had a "fit of the vapours" or would need a few stiff drams of the ol' Scottish whiskey!! :scared:

[glenc]

Thanks Enchilada, that was a good analysis. My calculation:
Radius in degrees = 360 divided by (2 times pi) and sweeps = 4 times pi times (57.29.. squared) divided by 0.75 = 55,004.
(I didn't worry about the halves each end.)