Telescope Reviews: number of doubles per aperture

in the topic "How do I start?" one of my posts led to an off topic thread that i've moved here.

at issue was my claim that up to 40,000 double stars would be within resolvable reach of a 100mm (4") scope. a few people demurred with that estimate and asked how it was calculated.

first the answer: i calculated from the current separation the "rayleigh aperture" necessary to resolve it: D(mm) = 140/rho; this is REZ AP. (dawes would be 116/rho.) then i calculated the minimum aperture necessary to see the magnitude of the fainter star of the pair as =1.1*10^((m2-2.7)/5). (the 10% bump in the aperture is to get above the "dark sky" assumption in the formula for light grasp.) this is MAG AP. then i took the larger of the two apertures, and sorted the dataset on that value.

this ignores a third issue: the brightness necessary to stimulate foveal resolution in very close but matching pairs. thus, if you have a 12" telescope your resolution limit is 0.46 and your magnitude limit is 15.1 ... but you won't be able to separate a matched 15 mag. pair at 0.46"!

as remedy, i calculated a third aperture that would raise the fainter component to a visual brightness equal to a mag. 10 star viewed through a 12" telescope; this is TEN AP. (i find mag. 10 in my 12" is roughly where my foveal vision gives out.)

since each magnitude increase is a change of 2.5 times the luminance, the diameter of the mirror must increase by the square root of 2.5, or roughly 1.6, for each magnitude step below 10. thus, in order to observe an 11th mag. star at a brightness equal to a 10th mag. star in a 12" scope, you need a 19" aperture.

then i checked if the separation (rho) was greater than 10x the resolution limits defined by MAG AP or REZ AP (whichever aperture was larger); if it was, then the companion would be visible at the limit magnitude. if not, then i increased the aperture by adding the difference between MAG AP and either REZ PA or TEN AP multiplied by the resolution ratio (1/10 to 1).

(this may seem kludgy, but i don't believe the more "refined" nomograms or multivariate calculations are really any more accurate, without some measure of observer skill, instrument collimation, sky conditions, etc.)

here is the chart of number of resolvable pairs per millimeter of aperture in the revised calculation.

now it appears that a 4" (100mm) scope can resolve up to 13,000 double stars, and a 10" scope can resolve roughly 38,000 doubles. in fact a simple linear calculation of 130xD(mm) is a good rule of thumb.

i checked the doubles chosen by this method at different apertures, and the results seemed plausible and consistent.

someone in the previous topic mentioned that many doubles are separated by arcminutes rather than arcseconds. this led me to look at the *easiest* doubles identified by this method. here is a sampling of those with bayer/flamsteed designations:

ZET UMA (STF 1744 AC)
92 sig 1,2 TAU (STFA 11)
20,21 NU 1,2 CRB (STFA 29)
8 ALP VUL (STFA 42)
77,78 THE 1,2 TAU (STFA 10)
17 CVN (STFA 24)

anyway, this is one of the advantages of having WDS in spreadsheet format: you can really drill down on any question.