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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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Time sights for Jim T
From: Frank Reed CT
Date: 2004 Sep 26, 19:47 EDT
Jim T, you wrote:
"They do not go on to explain how to take the result of the lunar and derive longitude."

It's pretty easy, Jim. The lunar gives you Greenwich time. The time sight gives you local time. The difference is the longitude (divide by 15 to convert from degrees to hours).

A time sight literally turns a sextant into a sundial. A sundial displays local hour angle as time directly. A sextant yields local hour angle indirectly by a calculation from the Sun's measured altitude (corrected for dip, semi-diameter, and refraction in the usual manner).

And:
"1. How was it done in Nelson's time?  (We are getting bits of that
information now, in other replies of this thread)."

There were many different methods. But they are nothing but trig identities applied to the basic equation (below) in order to facilitate logarithmic computation on paper.

And you wrote:
"2. How might a backyard or small boat lunartic do it today with a handheld
calculator, following on Frank Reed's "Easy Lunars" method?"

It's really very easy, Jim. And honestly, it's something that any celestial navigator who has understood the basic celestial triangle should be able to do himself. Maybe you're missing the point that the Sun's local hour angle *IS* the local apparent time. For example, if it's afternoon, and the Sun's LHA is 45 degrees, then the time is 3:00PM. That's all. They're the same thing in different units. The Sun's Local Hour Angle is the Local Apparent Time.

You know how to calculate LHAs generally, right? For any object, the simplest equation for the LHA is
cos(LHA) = (sin(alt) - sin(dec)*sin(Lat)) / (cos(dec)*cos(Lat))
where alt is the Sun's corrected altitude, dec is the declination (based on the best available estimate of Greenwich time), and Lat is the observer's DR latitude. It's easy to work this up on a hand-held calculator and even with logarithms, it's not terribly difficult. The rest of this post is just details. This short equation for cos(LHA) is the key to the process.

For your amusement, here is a method given in Bowditch in the 19th century for performing the same calculation:
First, calculate the Sun's polar distance: pdist = 90 - dec.
Next take half the sum of the altitude, the latitude, and the polar distance. Call that the "half-sum". From the half-sum subtract the altitude. Call that the "remainder". To the constant log 9.6990, add the following four logarithms: logsec(Lat), logcsc(pdist), lcos(half-sum), logsin(remainder). That sum is equal to the logsin(LHA). So we look up the sum in the logsin column and take out the corresponding angle (many tables included this in time units, but if not you convert to time by dividing by 15 degrees). [this procedure may sound strangely familiar to people who have been looking at the lunar methods of Mendoza Rios, Bowditch, Arnold, etc. It's got nothing to do with lunars specifically. It's just a trig identity which makes log calculations somewhat less prone to error].

The procedures above give us local apparent time. If you're using lunar distance tables based on Greenwich Apparent Time (such as are available on my web site today, or as were published in the almanacs before 1834), then you don't have any further steps except to calculate the difference between the two times as noted at the top of this message. On the other hand, if your lunars are based on Greenwich Mean Time (as the published tables were after 1834; my web site tables can also be generated for GMT), then you have to add one more step. You have to correct the calculated apparent time for the "equation of time". The equation of time is tabulated on every page of the Nautical Almanac. When the value in the almanac (modern almanac) is shaded, add the value of the equation of time to the calculated local apparent time. When unshaded, subtract. You can skip all of this equation of time junk by using lunar distances precalculated in Greenwich Apparent Time.

What if you can't see the Sun? It's not too difficult to get a local apparent time from an altitude of a star (and you can easily figure out how to do that yourself by thinking about the meaning of the GHA numbers in the almanacs), but it's largely irrelevant historically. If you're doing lunars at night, chances are good that you're keeping LAT on an ordinary watch.

Again, as noted in a previous post, the local apparent time derived from a time sight has to be updated for motion of the vessel if the time is being carried on a watch. If you're chasing the Sun (traveling west) then your watch (set to local time at a more easterly longitude) will display a time that is progressively ahead of the local time at your current longitude.

Frank R
[ ] Mystic, Connecticut
[X] Chicago, Illinois Browse Files

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