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I guest that this formula given by Cpt Legras on this list a while ago
can be worked around to solve the problem.

Cos asd =(sin H - (sin L  sin d)) / cosL cos d

asd = semi diurnal arc (from rise to noon transit, in fact the pole angle)

H = altitude of the sun (- 0?53' flag time; -6? civil twilight; etc)
L = Latitude
d = declination (at noon)

the duration of a day is twice the asd in ? ' " , or 2 asd/ 15 in h mn sec.

If the chosen H is 0?,  a simplified formula is used whitch is :
cos asd = (- tan L  tan d)

Pierre Boucher

Jared Sherman a ?crit :
> I was in warmer climes recently and noticed "again" that traveling
> north/south, there's a big change in the amount of daylight at this time of
> year. And for some reason I thought, this should be a way that one can
> determine their latitude, no?
>
> If one times the length of the day at any given location (using whichever
> version of "sunrise" and sunset you prefer), and one knows the date relative
> to the equinox or solstice day, one should be able to calculate the latitude
> one is at, no? It seemed interesting, that with just a watch and calendar
> one should be able to get latitude this way. Since most modern watches run
> to better than a second a day, this would seem to be a reasonable navigation
> tool.
>
> Has anyone seen a formula for this, or does anyone want to give it a try?
>
>

--
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Pierre Boucher N
formation en navigation de plaisance
Ste-Therese (Quebec) Canada

la "VOILE" ... le reste n'est que du vent...
"Sail, Sail !"... the rest is only wind...
EMAIL: pboucher@lavoile.com
http://www.lavoile.com
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