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Re: Star - Star Observations
From: Frank Reed
Date: 2010 Mar 9, 18:27 -0800

As George says, you should do this as a spherical trig problem if you're not doing so already, to cover the general case. By the way, be sure to use positions for the stars that are current for the given date within a few weeks. You mentioned that you're getting star positions from your Skyscout. Have you checked to see if those are updated for annual aberration? If they're not, they will be rather inaccurate for this task in some cases at some times of year. You could always get your position data, as a check, from my online nautical almanac which corrects for aberration and proper motion, too.

To your question, in general, refraction is a tide that lifts all boats :-). Refraction pushes all of the stars upwards, away from the horizon, compressing the constellations towards the zenith. And since the refraction is very nearly proportional to the tangent of the zenith distance for altitudes above 15 degrees, and since the tangent function is nearly equal to the angle for fairly small angles, the effect of refraction for altitudes above about 45 degrees is simple proportional compression. ANY DISTANCE BETWEEN ANY PAIR OF STARS is compressed by very nearly the same fractional amount. All you have to do is multiply the observed distance by 1.00034 to compare it with the calculated distance when both bodies are above 45 degrees in altitude. OR to make things even simpler, take the observed distance in degrees and divide by fifty. The result is the number of minutes of arc to add to the observed distance (if the observed distance is 15 degrees, add 0.3 minutes... if it's 40 degrees, add 0.8 minutes). Then compare with the true distance. It's another one of those cases where a spherical trigonometry problem is radically simplified in a special case. Note that this doesn't help you with angles larger than 90 degrees since you can't get both objects about 45 degrees altitude. In that case, you'll need objects specially aligned or you'll need a full spherical triangle solution. Also, it's a bit more limiting than it may appear since the portion of the sky above 45 degrees is not half of the sky. It's only about 35% of the "usable" sky (where I am rather arbitrarily counting the area above 10 degrees altitude as "usable" for observations of this sort).

Here's another trick: suppose you want to simplify the math and work only with star pairs that are inline (and not necessarily both above 45 degrees). They would be either exactly in the same azimuth or in azimuths separated by 180 degrees. Now you could guess this by eyeball. That's probably good enough if you're patient. But there's a clever geometric trick that you can use. If those two stars are in the same azimuth, then the great circle that contains them is vertical. This implies that the *pole* of that great circle is right on the horizon. So for any pair of stars, you find the pole of the great circle that passes through them, then you treat that as an "imaginary star" and work out its rising time. That's when you observe. The stars in that pair are guaranteed to be vertical when that "imaginary star" is on the horizon. Clever, huh? But I can't take credit for this trick. This was actually the basis for that idea that Lord Ellenborough introduced before the UK House of Lords in November of 1902 only to have it shot down by the First Lord of the Admiralty, who at least admitted he didn't know any better. Interestingly enough, Lord Ellenborough's idea was in fact published just a year later in very nearly the form he intended. I wouldn't be at all surprised if he hired the computers and had the tables published partially at his own expense. They were published under the title "Stars and Sextants: star distance tables for facilitating the use of Lord Ellenborough's method of correcting the centring and total errors of sextants at sea." Apparently they were available for a little over a decade starting in 1903 though I don't know how often they were updated. These tables are online:
Of course, they would have to be re-calculated to be used today, but the principle is valid. You'll note that they give examples for "Zenith Between Stars" and "Both Stars on Same Side of Zenith"

Just to repeat, from my own experience, using star-star distances to test a sextant's arc error is worthwhile but somewhat limited in accuracy unless a high-power telescope is used. You could certainly detect arc errors larger than one arcminute, and probably as small as a third of an arcminute with careful trials and a 7x scope, but I doubt better than that. I haven't tried this in over a year, and it might be interesting to see if I can get better results using the Ellenborough system of restricting observations to bodies in the same vertical circle.

-FER

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