# NavList:

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

**Re: Lecky's mistake**

**From:**Lars Bergman

**Date:**2020 Nov 1, 07:49 -0800

Dave, you asked "Which antilog of 8.230947 is 59minutes 58 seconds?".

It is log hav 59^{m}58^{s}. The formula he use is

hav t = sec φ · csc p · cos s · sin(s-h)

where t is hour angle, φ is latitude, p is polar distance (or co-declination), h is altitude and s = (h+φ+p)/2. The "remainder" is (s-h). The formula could be derived from the ususal PZX triangle altitude formula:

sin h = sin φ · cos p + cos φ · sin p · cos t

Then cos t = (sin h - sin φ · cos p) / (cos φ · sin p) . Subtract each side of this equation from unity:

1 - cos t = (1 - (sin h - sin φ · cos p)) / (cos φ · sin p) = (cos φ · sin p - sin h + sin φ · cos p) / (cos φ · sin p) = (sin(φ+p) - sin h) · sec φ · csc p

Now,

sin(x+y) = sin x · cos y + cos x · sin y

sin(x-y) = sin x · cos y - cos x · sin y

From this follows that sin(x+y) - sin(x-y) = 2·cos x · sin y

We have

x+y = φ+p

x-y = h

Hence x = (φ+p+h)/2 and y = (φ+p-h)/2 = (φ+p+h)/2 - h. Thus we get sin(φ+p) - sin h = 2 · cos((φ+p+h)/2) · sin((φ+p+h)/2 - h) and the final formula becomes

(1 - cos t) / 2 = cos((φ+p+h)/2) · sin((φ+p+h)/2 - h)· sec φ · csc p, or

hav t = sec φ · csc p · cos s · sin(s-h), where s = (h+φ+p)/2. This form is suitable for logarithmic use.

Lars