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Re: Figure out LAN?
From: Stan K
Date: 2014 Oct 2, 18:42 -0400
From: Stan K
Date: 2014 Oct 2, 18:42 -0400
Sam,
This is a screen shot of the Celestial Tools Noon Sight time of LAN calculation for your example. It uses the "GHA method". There are several things worth noting.
The GHA of the Sun at 16h UT of 62º35.5' agrees with the USNO site and any other reference I have, but more about this later.
The time of 165108 results in a GHA of the Sun of 75º22.7', according to the USNO site, not the 75º22.4' it is supposed to be. However, changing it by one second to 165107 gives the correct value. This is due to two things. First, the GHA difference of 12º46.9' corresponds to a time of 51m07.5999...s, which the program rules rounds to 51m08s. But the 16h GHA of the Sun is rounded to tenths of a minute. It would not have to change much for the rounding rules to result in 51m07s. It would have to be 12º46.874999..', essentially a difference of 0.025', i.e. a 16h GHA of the Sun of 62º35.525', resulting in a GHA difference of 12º46.875' (actually 12º46.874999...'). So it would appear that the rounding of the 16h GHA of the Sun to tenths of a minute results in a one second error, enough to cause the discrepancy.
But if you look at the 16h GHA of the Sun in the Nautical Almanac, it is 62º35.6', not 62º35.5'. This is due to the fact that the Almanac adjusts the GHA up in order to reduce the error caused by ignoring the v correction of the Sun, as explained in paragraph 24 on page 261 of the Almanac. By using this erroneous value, the correct UT of LAN of 165107 is obtained. Funny how a "wrong" value can give a "correct" result.
Stan
This is a screen shot of the Celestial Tools Noon Sight time of LAN calculation for your example. It uses the "GHA method". There are several things worth noting.
The GHA of the Sun at 16h UT of 62º35.5' agrees with the USNO site and any other reference I have, but more about this later.
The time of 165108 results in a GHA of the Sun of 75º22.7', according to the USNO site, not the 75º22.4' it is supposed to be. However, changing it by one second to 165107 gives the correct value. This is due to two things. First, the GHA difference of 12º46.9' corresponds to a time of 51m07.5999...s, which the program rules rounds to 51m08s. But the 16h GHA of the Sun is rounded to tenths of a minute. It would not have to change much for the rounding rules to result in 51m07s. It would have to be 12º46.874999..', essentially a difference of 0.025', i.e. a 16h GHA of the Sun of 62º35.525', resulting in a GHA difference of 12º46.875' (actually 12º46.874999...'). So it would appear that the rounding of the 16h GHA of the Sun to tenths of a minute results in a one second error, enough to cause the discrepancy.
But if you look at the 16h GHA of the Sun in the Nautical Almanac, it is 62º35.6', not 62º35.5'. This is due to the fact that the Almanac adjusts the GHA up in order to reduce the error caused by ignoring the v correction of the Sun, as explained in paragraph 24 on page 261 of the Almanac. By using this erroneous value, the correct UT of LAN of 165107 is obtained. Funny how a "wrong" value can give a "correct" result.
Stan
-----Original Message-----
From: Paul Hirose <NoReply_Hirose@navlist.net>
To: slk1000 <slk1000@aol.com>
Sent: Thu, Oct 2, 2014 1:32 am
Subject: [NavList] Re: Figure out LAN?
From: Paul Hirose <NoReply_Hirose@navlist.net>
To: slk1000 <slk1000@aol.com>
Sent: Thu, Oct 2, 2014 1:32 am
Subject: [NavList] Re: Figure out LAN?
On 2014-10-01 10:01, Samuel L wrote: > Is the following example correct in determination of LAN using GHA? > Date- October 1, 2014 > Latitude- N 41d > Longitude- W 75d 22.4min > LAN= 11:51:06.39 (12:51:06.39 DST) Doesn't look right. GHA at that time is 75°22.1′ per the USNO celestial navigation calculator. At apparent noon, GHA should match your longitude. http://aa.usno.navy.mil/data/docs/celnavtable.php But it's possible I goofed. It's time to get ready for bed so I rushed through the check. Your times are stated to excessive and unjustified precision. The arc equivalent of .01s is .15″. But the format of your longitude implies it's only known to about .1′, or 6″. Another issue is whether the time is UT1 or UTC. At .01s precision, the difference is highly significant. It may even be necessary to take polar motion (which is a few tenths of an arcsecond) into account. Such accuracy is attainable, though very unnecessary for celestial navigation. But the software I would use is on my Windows XP machine and the hour is too late to be cranking that up and fiddling with numbers. I may try it tomorrow for fun.(?)
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