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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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Re: Latitude from duration of day
From: Frank Reed
Date: 2018 Dec 18, 10:02 -0800

Tony,
There are a number of problems with getting latitude by the length of the day. Some of these are practical. Some of them are fundamental. The calculation itself is really the last thing that you should fuss over. And in fact, the solution that you eventually discovered, solving by iteration (that's "high-low" educated guessing, for anyone reading along who doesn't quite catch the meaning), is a 100% valid solution to the problem, so you have indeed solved the math puzzle. You ended up with a solution that is not "closed form" but the vast majority of things we can try to solve for in the real world fall into that category. It's no flaw. Iterative solutions are often the best solutions to real-world puzzles.

Some of the problems:
First, let's consider your textbook version of the puzzle. You looked up the times of sunrise and sunset in a table. These are good starting points but they are the result of a standardized computation that represents an idealized observer. That idealized observer is at sea level at the location in question with height of eye above the "water" equal to zero and no obstructing hills. We can look up the times of sunrise and sunset in Denver in the US or Madrid in Spain, and they are calculated for that idealized sea level observer (somehow with eyes right on the surface of the water). But no real observer would ever agree with those times except by chance. Times of sunrise and sunset for inland observers are arbitrary things, calculated for reasons of civil law and for classification of meteorological observations, but otherwise they do not correspond to anything real. Even if you're standing on a promontory on Ibiza overlooking the sea, your times will not match. If you're 100 feet above sea level, that adds an additional 10' of dip, and (ignoring secondary corrections) that will change the times by over 50 seconds in that latitude.

Also, it's relatively rare to find oneself in a location where you can observe both sunrise and sunset on the same day. Indeed, if your friend is in Spain, and doesn't travel more than ten miles during the day, and is able to see both sunrise and sunset, then we can probably make some good guesses about location just from the possibility of the observation. Maybe in the islands: Mallorca, Menorca, Ibiza (or somewhere in the Canaries ...if they count). Or at a few places on the southern coast of Spain.

Next, consider that the duration of the day has relatively "low resolution" for latitude. It's an interesting issue: how good can it get? On the equinoxes, duration of daylight is about twelve hours everywhere on Earth at all latitudes. Latitude is not resolved at all! How good is the resolution two weeks away from equinoxes? And two months?

Next, the Sun's dec changes so this problem is not as symmetrical as it seems. The sunrise and sunset hour angles are not equal. We could correct for this, but it's yet another thing to worry about. Note that the time of year you have chosen (this week -- solstice in a few days) mostly removes this concern for now.

Finally, an observational issue: it's really rather difficult to observe the exact moment of sunrise or sunset. There's no sudden instant when the upper limb bursts above the horizon, at least not in normal conditions. You could time it to within about 15 seconds on most dates, but don't count on it. And as always, there is uncertainty from anomalous refraction and dip at the horizon. Figure another 5 to 15 seconds there. If your timing is out by 30 seconds, how much does that affect the latitude, assuming everything else works perfectly.

Back to the calculation. How could you at least do that part right? You'll need to include dip of the horizon, and you might include changes in temperature from sunrise to sunset (that thicker air at sunrise refracts quite a bit more, but then again the water temp won't change much), and you'll need to include the changing dec of the Sun. How can you do all that? You could generate a relatively complicated set of equations, but computation is cheap. So get out the calculator and iterate. Just as you did already. You can go from the textbook scenario to the real world by sticking with iteration. It's unlikely that you'll ever get a satisfactory "closed form" equation (that's a vague expression for a "nice" equation) for the real-world problem. But you can get an answer in short order by iteration. The real world likes iterative solutions!

Yet another approach: don't observe actual sunrise and sunset, but instead time the Sun's arrival at the true horizon. This is similar to the advice given for compass checks by amplitudes. Then, if handled correctly, the very simple equation that you derived actually applies. This approach also has the advantage that it can be used inland in areas where there are no hills around the horizon. The downside to this idea is that you'll need some tool to estimate the apparent altitudes that correspond to the Sun being on the true horizon. It doesn't have to be a sextant, but that would give the best results.

Frank Reed

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