# NavList:

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

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Re: Latitude from duration of day
From: Lars Bergman
Date: 2018 Dec 19, 07:30 -0800

Tony,

You can use cot(θ) = cos(t) / tan(Dec) if you like, then the final formula becomes cos(Lat-θ) = sin(H)·sin(θ) / sin(Dec). The trick is to find something that makes the parenthesis become sin(x)·cos(y)+cos(x)·sin(y) or cos(x)·cos(y)+sin(x)·sin(y), then you can replace it with sin(x+y) or cos(x-y).

These kind of transformations were very common when logarithms were used for numerical work. Then you want the right hand side to be multiplications of different trig functions. When you use logs, these multiplications become additions. And to avoid subtraction, you use the reciprocal of the trig functions in the denominator. For example, the function above becomes

cos(Lat-θ) = sin(H)·sin(θ)·csc(Dec); where cot(θ) = cos(t)·cot(Dec)

Using sin(H) = sin(Lat)·sin(Dec) + cos(Lat)·cos(Dec)·cos(t) to solve for H directly with logarithms is no nice task. A better way is to first find θ from
tan(θ) = tan(Dec)·sec(t), then H from
sin(H) = sin(Dec)·cos(Lat-θ)·csc(θ)

If you also want the azimuth, then tan(Z) = tan(t)·csc(Lat-θ)·cos(θ), with the same θ.

(With reservation for possible faults ...)
Lars 59°N 18°E

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