A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Position-Finding
From: Paul Hirose
Date: 2018 Jan 7, 20:40 -0800
For a very short distance (high altitude) test of the Bygrave formulas on a slide rule, solve for the separation and position angle of two components of a double star. To make the test more extreme, let the star be Polaris, which is a triple system. Coordinates of the two main components from Wikipedia: 02h 31m 49.09s +89° 15′ 50.8″ α UMi Aa 02h 30m 41.63s +89° 15′ 38.1″ α UMi B If Aa (the star visible to the naked eye) is the "observer" and B is the "celestial body", then 89.26411 lat 89.26058 dec 0.28108° LHA W = arctan (tan 89.26058 / cos .28108) W = 89.26058 At slide rule accuracy the cosine of such a small angle is virtually 1, so W equals the declination. X = 90 - 89.26411 + 89.26058 X = 89.99647 X is in the range 0 - 180, so "altitude" is positive. Z = arc tan(tan 0.28108 * cos 89.26058 / cos 89.99647) That formula is a good lesson on techniques to extend the range of the ST scale. 1. Tan 0.28108 is not on ST, so set it as tan 2.81. 2. Cos 89.26058 (red 89.26058) may be set directly. Since red 89 = black 1, begin at black 1 and move left .26°. (A few rules include the red numbers on ST. But usually they're omitted since the numbers are obvious.) 3. The method in step 2 won't work for cos 89.99647. Therefore, delete nines to get a value you can set: 86.47. Begin at black 4 (= red 86) and move left .47° to get cos 86.47. Though it's not immediately obvious, the same principle is at work in step 1 and step 3. I.e., cos 89.99647 (red 89.99647) is equivalent to black .00353, which is outside the range of ST. By shifting the decimal point three places you get black 3.53, alias red 86.47, which is on the scale. This flexibility with the decimal point is possible on ST because the sine or tangent of a small angle is practically proportional to the angle. That's not true on the S and T scales! If the angles were literally equal to the numbers on ST, the characteristic of the tangent would be -2, i.e., the result is between .01 and .1. But if you include the decimal point adjustments in steps 1 and 3, the characteristic is 0 and so the arc tan is read on T. With the slide in its final position (index at the tangent on D), set cursor to the D index. Z = 134.26° on T at the cursor. Correct angle is 134.13. Hc = arc tan (cos 134.26 * tan 89.99647) Hc = 89.99494 The cosine can be set directly. Change the multiplication to division by cot 89.99647, which you set as red 86.47 as in the Z computation. Read 84.94, adjust to 89.99494 to account for the deleted 9s. Separation angle is the complement of that angle: .00506°. Correct angle is .005054 (18.19″) and error is .02″. It wasn't easy, but a 10 inch slide rule and the Bygrave formulas performed well in this test.