NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Azimith forumula for Great Circle sailings- having problems
From: Gary LaPook
Date: 2014 Dec 23, 15:54 -0800
From: Samuel L <NoReply_SamuelL@fer3.com>
To: garylapook@pacbell.net
Sent: Tuesday, December 23, 2014 2:51 PM
Subject: [NavList] Azimith forumula for Great Circle sailings- having problems
From: Gary LaPook
Date: 2014 Dec 23, 15:54 -0800
How do you get LHA's accurate to 0.1 of a second of arc when the Nautical Almanac only gives the GHA to a 6 second precision (0.1 minute)? Somehow you are getting your LHA's 60 times more accurate than the Naval Observatory, how do you do it, I would like to do the same?
Same with your destination latitude, how do you find that to a
precision of 0.1 of a second which is only 10 feet on the ground, do you have super accurate charts that allow you to read out the latitude to within 10 feet? Even GPS only claims 10 meter accuracy 95% of the time which would be about 0.3 seconds of latitude so you are getting accuracies three times better than even GPS, how are you able to do that?
gl
From: Samuel L <NoReply_SamuelL@fer3.com>
To: garylapook@pacbell.net
Sent: Tuesday, December 23, 2014 2:51 PM
Subject: [NavList] Azimith forumula for Great Circle sailings- having problems
I use the following formula to obtain Azimith-
Z = tan-1 (sin (LHA) / (cos (LHA) x sin(AP Latitude) - cos(AP Latitude) x tan(Declination))
If LHA is greater than 180 it's treated as a negative quantity.
If the Azimith angle as calculated is negative, add 180 to it.
If the Azimith angle as calculated is negative, add 180 to it.
In one Great Circle sailing probelm I'm working on to determine Z
the answer is negative unless the LHA is entered into the formula as a negative number. The answer should be about 3d 55min. If I add 180 to to the negative answer the wrong Z is obtained.
tan-1((sin(LHA) / (cos(Present Latitude) x tan(Destination Latitude) – (sin(Present Latitude) x cos(LHA))
Here's the figures;
LHA= 0d 9m 40.9sec
Present Latitude- N 39d 58m
Destination Latitude= N 41d 43m 31.8sec
The answer I get is -3d 55min (notice the negative sign.
It would be nice to have an Azimith formula that's independent of Hc and provides the correct figure regardless of LHA greater or less than 180d.
How do I solve this problem?
Thanks,
Sam
Lohengrin
