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Kieran Kelly wrote-

>I sent the post below recently as a way of showing people how land explorers
>calculated compass variation in the days before modern maps. I was intrigued
>about the Gregory method shown below and went hunting for the maths.
>Surprisingly I found the answer in a Bowditch of 1977 Vol 1 as follows:
>
>Hav(180dd-Z)= sech secL cosS cos(S-p)
>
>Where
>Z = Azimuth angle (not Azimuth)
>h = altitude
>L = latitude
>p = polar distance (visible pole)
>S = 1/2 (h+L+p)
>
>I am not a mathematician but it appears he was trying to solve for Z - the
>Azimuth Angle - and compare it to the compass to get the difference between
>true and magnetic north. In this case Gregory was using the south visible
>pole so substituted 360dd in his formula for 180dd. Note that this is not a
>universal formula because it relies on the restricted navigational triangle
>based on the visible celestial pole.

===============

This has turned out to be a rather interesting question.

What Kieran has quoted is an expression for the azimuth of a body, to
compare with its compass bearing and so obtain the compass error. If
there's no deviation (which should be the case for a land explorer, unless
he has just discovered an iron mine) the compass error should equal the
magnetic variation.

Kieran informed me that the expression was to be found on page 562 of
Bowditch vol 1 1977. It reads-

hav (180-Z) = sec h sec L cos S cos (S - p)

I have quite a collection of navigational texts, but in none of them can I
find any referece to that same expression for azimuth. Although Bowditch is
usually rather good at explaining or deriving the expressions he uses, in
this case he doesn't seem to do so (or I can't find it).

I think it's beyond my limited trigonometrical powers to derive that
expression from first principles, but others may wish to pick up that
challenge.

Note that on page 564 the same expression is used again, under "Soule and
Dreisonstok", except that both sides have been inverted; otherwise, it's
EXACTLY the same, as follows-

1 / (hav (180 - Z)) = sec S sec (S-p) / (sec L sec h)   , remembering sec =
1 / cos.
However, there's a strange comment goes with it, as follows-

"The "azimuth" determined in this way is the direction of the line of
position (Z?90deg) rather than that of the celestial body"

If that is true, then it should apply just the same to the expression on
page 562, but there's no such comment on that page. Alternatively, it
shouldn't apply to either. Which, I wonder?

George.

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