# NavList:

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

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Re: Luni-Solar Distance
From: Lars Bergman
Date: 2010 Oct 28, 05:02 -0700

George wrote
"However, sometimes a lunar is taken for a different purpose than to
establish an immediate longitude. Sometimes, it might just be to set, or
correct the error of the chronometer, and it might not be thought
necessary, in mid-ocean, to make such a time-sight for local time, there
and then. Can the altitudes be successfully calculated under those
circumstances, I wonder?"

I cannot see that the altitudes can be calculated if you only know your latitude but don't know neither time nor longitude. Thus the sun time sight is essential to get knowledge of your local time. Also, with a fix from two (or more) stars you can find LHA Aries (corresponding to local siderial time) and latitude. Reducing these two local times to a common instant - the time of the lunar observation - makes it possible to estimate GMT.

As LHA Sun - LHA Aries = GHA Sun - GHA Aries you can then find an approximate GMT by looking for this difference in the Almanac.

Now to the example with the observer at the equator. Let us assume that he, from a previous (or later) observation, knows his LMT at the time of the lunar observation. According to the pre-requisites his LMT-clock then shows 00h00m00s 26th of March at the lunar. But, as the observer's DR longitude is 1 degree east, he calculates the GMT to 25d23h56m00s, which information, together with the DR longitude, is used for the altitude calculations. His calculated altitude becomes 86d21.6m and 56d39.1m for moon and Regulus, respectively. Correcting 'backwards' for semidiameter, parallax and refraction the apparent altitude becomes 86d02.5m and 56d39.7m respectively. These apparent altitudes together with the given lunar distance gives a cleared lunar of 37d01.5m, corresponding to a GMT of 26d00h00m00s. The observer now knows the correct GMT, and also finds his longitude as being zero, because LMT=GMT.

Using a DR longitude of 6d east instead, just to test, the used GMT becomes 25d23h36m00s and the calculated altitudes 86d31.6m and 56d39.9m, the apparent altitudes 86d49.9m and 56d40.5m. The cleared lunar will still be the same as above. (I haven't used Frank's machine for these calculations.)

Why does it work? Although you do take out GHA and DEC from the Almanac for a GMT that is in error by several minutes of time, the largest error, that of GHA, is almost completely cancelled out by your (errouneous) DR longitude, so the resulting LHA is pretty near the truth. There will be a slight residual error because GHA Moon and GHA Aries do not change at exactly 15 degrees per hour. And there will be a slight error in the moon's declination.

Lars
59N 18E

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