# NavList:

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

**Re: Re-doing my homework (by Bygrave)**

**From:**Paul Hirose

**Date:**2017 Feb 23, 00:18 -0800

On 2017-02-22 17:13, Tony Oz wrote: > Given: > 2016-10-31T10:11:36UTC > Lat_AP = 37°15'N; Lon_AP = 66°30'W > Regulus: Dec = 11°53'N, GHA = 40°44' I confirm the Regulus position. My variant of the Bygrave formulas, worked on a 10 inch slide rule: LHA = 40 44 - 66 30 = -25 46 = -25.77 W = arc tan(tan 11.88 / cos 25.77) = 13.13 X = 90 - 37.25 + 13.13 = 65.88 A = arc tan (tan 25.77 * cos 13.13 / cos 65.88) = 49.00 Zn = 180 - 49.00 = 131.00 (130.976 by computer) That agrees well with your stuff. > LHA = GHA + Lon_AP = 334°14' > H = 360° - LHA = 25°46'; H is OK, no rule apply. > CoLat = 90° - 37°15' = 52°45' > W = tan^-1(tan(Dec)/cos(H)) = tan^-1(tan(11°53')/cos(25°46') = 13°9' > since the names are same and LHA is 334° - the W should be added. > X = 52°45' + 13°9' = 65°54'; X is OK, no rule apply. > since X < 90° -> the Y = X == 65°54'; Y is OK, no rule apply. > Az = tan^-1(tan(H)·cos(W)/cos(Y)) = tan^-1(tan(25°46')·cos(13°9')/cos(65°54')) = 49°1' > since Lat is North, X < 90° and LHA = 334°, then > Zn = 180 - Az = 130°59' But we diverge at Hc. I multiply instead of dividing. > Hc = tan^-1(cos(Az)/tan(Y)) = tan^-1( cos(49°1')/tan(65°54')) = *16°21'* Hc = arc tan (cos 49.00 * tan 65.88) = 55.70 (55.704 by computer) Well, I actually divide by cot 65.88, that dodge being necessary with a single T scale. But it's equivalent to multiplication by the tangent.