# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: Towards a basis for Bruce Stark's Tables**

**From:**Bruce Stark

**Date:**2003 Jan 5, 17:35 EST

Fred, If I understand what you mean by "intercelestial distances" you're right: the calculation that gets comparing distances from the Almanac is the cosine-haversine formula. But you're not out of the woods yet with the formula for clearing the distance. But first, let's talk about the need to check for errors of arc and centering in your sextant. You can use my tables as long as one of the objects is the moon. Otherwise use another method of clearing. If you use one such as Duntorne or Borda, don't use the "logarithmic difference" short cut. Nothing wrong with the log-dif tables as long as one of the bodies is the moon, but if you use them for two stars you run into the same problem you would in using my method with its "Q." You could get around the problem posed by log-dif or "Q" but it would be tricky. So if you're checking your sextant with a distance from the moon use my tables; if between two other bodies, Borda, without the log-dif table, is a good choice. If you're woking Borda's with logarithms you'll want to work to six or seven decimal places. So far as I know, mine is the only rigorous method of clearing that doesn't require more than five places of decimals. For star-to-star distances above 25� you might want to use one of the approximate methods. These need only four places of decimals. Unfortunately they aren't reliable with short distances no matter how many decimals you use. I wouldn't be surprised if you can develop my formula from Borda's, or vice versa. Would be an interesting exercise. Or you can develop the formula the way I did, from the cosine-haversine and the old time-sight formula. That time-sight formula probably was used by mathematicians centuries before it was used at sea. Here it is, spelled out in a way I hope will come through un-garbled: Call the sides of spherical triangle "a," "b," and "c." Call half the sum of those three sides "s." Opposite side "a" is angle "A." Then the sine of one-half angle A is equal to the square root of: [sin(s-b)sin(s-c)]/(sin b sin c). I combined that formula with the cosine-haversine formula to get the one for the Tables. Bruce