A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: David Pike
Date: 2017 Nov 21, 08:43 -0800
Herman you wrote:
The formula for solving the calculated height, in the celestial triangle, is written in different order and ways.
You see Hc=arcSin((Cos L • Cos d • Cos t) + (Sin L • Sin d))
Hc=arcSin((Cos L • Cos d • Cos t) +/- (Sin L • Sin d))
Hc=arcSin((Cos L • Cos d • Cos t) ~ (Sin L • Sin d))
Hc=arcSin((Sin L • Sin d)+(Cos L • Cos d • Cos t) )
And people ad it with different rules. For instance I remember ~ says subtract the smallest value from the biggest??
Why al these differences, is it a kind of interpretation?
What is the "original" formula?
Speaking as a chap who can barely remember the rules for plane triangles these days and who never could remember anything to do with spherical triangles for more than half an hour, this is my interpretation.
What is being attempted here is to find the third side of a spherical triangle from two sides and an included angle. However, Hc, latitude, declination, and LHA are not necessarily the sides and angle in the triangle. E.g. in this case the unknown side is co-alt (90-Hc), not Hc, and t the known angle might be LHA or 360-LHA depending upon where the body is in relation to the observer. Moreover, the body might be in the same hemisphere as the observer or the opposite, and P in the PZX triangle might be the North or the South Pole, so it all gets a bit complicated. E.g. if P and Z are in one hemisphere and X is in the other, the declination side is 90+dec not co-dec. Finally, the known sides or the angle might be more that 90, so the trig functions will only be positive according to the All, Sin, Tan, Cos rule, and these latter two examples are where I would suggest the + or – comes in. If you want an original formula, I think you probably have to go back to a spherical triangle with angles A, B, and C, and sides a, b, and c, and adapt that to co-alt, co-lat, co-dec, and LHA or 360-LHA as suits your particular case.
So from the Cosine rule:
cosa=cosb.cosc + sinb.sinc.cosA
but: sinx=90-cosx and cosx=90-sinx
Hc = arcsin(coslat.cosdec.cost + sinlat.sindec)