# NavList:

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

**Berson's Noon Sun problem**

**From:**Frank Reed

**Date:**2013 Feb 10, 12:40 -0800

Under the Orion thread, Robin Stuart posted this problem:

"I also enjoy David Berson’s "Nav Problem" column in Ocean Navigator which contains a Celnav problem linked to an historical back story. Here’s one from last September’s edition that I’m interested in comparing answers to with list members. Calculations are done using the 2012 Nautical Almanac. It is April 12, DR position 25d15’N, 53d35’W. Height of the eye is 12 feet. What time is LAN at the DR position? A lower limb noon sight gives Hs 73d32.6’. What is Ho and the latitude?"

Well, it's nearly trivial... This is "Day One" celestial navigation, and I mean that literally.

Let's consider the part that I see as non-trivial. Why do we need to know the "time" of LAN? One purpose would be for planning. Occasionally novice navigators get confused and think they need to shoot the Sun at a particular instant of GMT. That's not so. But sometimes, if one is pressed for time, a few minutes can be saved by knowing in advance as nearly as possible the GMT when LAN is expected. But there's another reason which matters a great deal if we want a reasonably accurate latitude: the declination of the Sun at this time of year is changing at a rate of around 1' of arc per hour. The Sun is moving north at a speed of just about 1 knot. So if we enter the almanac with just the date, we might be off by as much as 24' in the Declination, which would usually be considered an unacceptable error. So when is LAN if we're in longitude 53d 35' W? Well, there are two ways to do this.

First, there's the easy method if you have a complete Nautical Almanac in front of you. Look down the column for GHA Sun on this date and find the one closest to 53 degrees. Let's see... it's just under 45 degrees at 1500 GMT (as it should be, since the Sun is moving 15 degrees per hour and it was nearly over the Prime Meridian at noon GMT) and it's just under 60 degrees at 1600 GMT. Our DR longitude is just over halfway between those two, so the GMT would be just over halfway to 1600, maybe 1535. But we don't really need that exact value since we're really interpolating between the Declination values at 1500 and 1600. They are 8d 58.7 N and 8d 59.6 N. If we interpolate halfway between and then nudge it up a little 8d 59.2 or 59.3. That's close enough. Of course, in the REAL world, it's highly unlikely that you would not have GMT on a watch in the first place, in which case you could do the interpolation of GMT in one step, ignoring the GHA.

An equivalent method for this would be to recognize that April 12 is near a zero in the equation of time so the apparent Sun nearly tracks the mean Sun. This is to noting above that the Sun was "nearly" over the Prime Meridian at 1200 GMT. Then you can just divide the DR longitude by 15 degrees per hour to get 3.57 hours GMT plus a few seconds, if you want, since the Sun is still "late" on April 12. And then, just as above, you interpolate between the values of the Declination. This is an "eyeball" interpolation that would give the same result as above unless you have an almanac table that only gives the Sun's declination once every 24 hours, in which case it's a little bit more work interpolating (but then you have the advantage of only carrying a few pages of almanac data instead of a whole volume).

Once we know the Dec, the rest is certainly trivial --so trivial that it's scarcely worth discussing. But let's do it 19th century style to make things interesting. A typical 19th century navigator would subtract the Lower Limb altitude (after correcting for IC only) from the "corrected" right angle, 89d 48'. That gives the zenith distance of the Sun's center: zd=16d 15.4'. So now we know how far away we are from the latitude where the Sun is directly overhead. The zenith distance IS the distance away from that point. The Sun, of course, is directly overhead in the latitude that is equal to the Sun's declination which we found to be 8d 59.2'. That is the DEFINITION of the Declination. Though the problem doesn't tell us which way we were facing, we can assume we're not far from the DR so the Sun is south of us. That means the zenith distance gets added onto the Sun's Declination yielding the latitude: 25d 14.6' N --very close to the DR lat. This "19th century" approach to the calculation assumes that the dip is nearly 4' (so height of eye 16 feet) and the Sun's SD is 16.0', which it nearly is on this date, and it assumes refraction is zero which it nearly is. The same calculation completed with the standard tables in the Nautical Almanac would differ by only a couple of tenths of a minute of arc.

-FER

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