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

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Latitude from duration of day
From: Tony Oz
Date: 2018 Dec 17, 11:44 -0800

[Moved from "long by chron" discussion...]

I want to solve for Lat from known Dec, H and LHA for nothing serious. It came up in my conversation with a friend who is going to go on vacations in Spain. I asked him to measure the day time duration by noticing the Sunrise and Sunset, to halve the difference and to send me an SMS with that value. I was going to tell him his latitude. He did not believe me it was possible - especially by a sliderule alone - so demanded to give him a walk-through of supposed calculations. He looked-up the day-time duration for the place and date he was planning to visit, I bravely embarked in the calculation.

The original formula was:

sin(H) = sin(Lat)·sin(Dec) + cos(Lat)·cos(Dec)·cos(t)

I transformed it to:

cos(t) = (sin(H) - sin(Lat)·sin(Dec))/(cos(Lat)·cos(Dec))

Then I declared that at Sunset and Sunrise the H = 0° (which is plain WRONG!) and "simplified" the formula thus:

cos(t) = -(sin(Lat)·sin(Dec))/(cos(Lat)·cos(Dec))

From which I could get:

tg(Lat) = -cos(t)/tg(Dec)

Doing that stage with the given t = daytime/2 and DecSun for the date I gave him the answewr that was ~2° off the mark. A-ha! he cried, that's more than 200km from my planned hotel!

I did not expect that neglecting sin(H) = sin(-0°50') that actuallly is the case on sunsets/sunrises would have such an impact on accuracy of results.

So I had to take my Casio fx-85ES+ and iteratively change the Lat in the original formula untill I got the given daytime duration. My friend insists he won the argument.

Now I want to show him it is nevertheless possible to calculate his Lat from known Dec and t in one go.

I think I must use Napier's analogies for that...

So it is nothing more than a mathematical joke. :)

Warm regards,

Tony

60°N 30°E

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