Longitude from a noon sight

Celestial navigators, if nothing else, are opportunists. We have to be, given the variables of sea conditions, our stomachs and most importantly, cloud cover. We develop the technique of taking sights whenever we can, and become experts in squeezing all available information from those sights.
A case in point of this technique is deriving longitude from a noon sight.
Let me explain.
Let’s say we have been dead reckoning for a few days, due to heavy cloud cover. We have an idea of where we are because we have been keeping track of our course, our speed and the distance we have traveled. However we would like to be more exact, and we wait for a chance to get a sun sight. The weather finally clears enough to make it practical to take a noon meridian passage sight. Based on our DR position we calculate the time of meridian passage and prepare about 20 minutes beforehand to get our shot when the sun is at its highest. What can we learn from such a sight?
At the time of our sight we are at DR 35° 15’ N by 68° 25’ W. It is April 15, 2007. The Nautical Almanac tells us that the time of meridian passage in LMT on that day is 1200 hours. As I like to work in GMT, I need to convert the exact time of meridian passage at my longitude to GMT. To do this I check the Arc to Time conversion table in the Nautical Almanac and see that it takes the sun 4 hours 33 minutes and 40 seconds to travel 68° 25’. I add this time to the 1200 hours already noted and arrive at a time of 16 hours 33 minutes 40 seconds GMT for meridian passage.

By the time the sun has reached its highest point in the sky I check my watch and note the time of 16 hours 36 minutes 52 seconds. According to Mixter, below the 80th parallel the sun will “hang” for anywhere from two to four minutes of time. For the sake of our problem — and one of the drawbacks in obtaining an exact answer using this method — is the fact that the length of hang time can’t be factored in. Nevertheless, we can still get a great deal of useful information from this process.

We now calculate that the difference in time between our DR position and the actual position where we took our sight is 3 minutes 12 seconds. If we convert this time to arc we can establish — with some accuracy — our longitude at the time of the sight.

By entering the increments and corrections table at the rear of the Nautical Almanac we find that 3 minutes 12 seconds of time is equal to 48’ of arc. We then add that 48’ of arc to our DR longitude of 68° 25’ and get the longitude at our time of sight as 69° 13’.

We can check this answer by working the problem through using GHA. We know that by definition the meridian passage of the sun is defined as when LHA = 0°, or put another way, when the GHA is equal to the longitude of the observer.
On April 15 at 1600 the GHA of the sun is 59° 58.9’. Adding 36 minutes 52 seconds of time (time of the sight was 16:36:52 GMT) we discover that at that time the longitude was 69° 11.9’ or 69° 12’.

This confirms, to my satisfaction at least, that the method I’m suggesting is valid. The close proximity of the numbers gives me confidence that once I calculate the latitude that I can mark that spot on the chart with a square indicating an estimated position. Not as accurate as a fix but a whole lot better than a DR!

About the Writer
Contributing Editor David Berson writes the Nav Problem page in every issue of Ocean Navigator. He is also the owner and operator of Glory, an electrically powered excursion boat, in Greenport, N.Y.
Question for David? editors@oceannavigator.com
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