A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
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. :)