Azimuth of the sun, LHA issues

Hi all, I am trying to do an Azimuth and having an issue with LHA. This is for a sunset sighting in apx. pos’n 33 54’ N 120 00’W at 19:34:00 LT on May 8th 2017 (GMT-7 for DST) 02:34:00 UT may 9 2017. When I go into the Nautical Almanac for GHA I come out with 219 22.4 then subtract my W longitude to come out with LHA of 99 22.4. This doesn’t seem right to me as it seems to indicate that the sun is past the visible horizon. Am I missing something?

First, I would tell you that an Amplitude would be better suited for sunset. Just finish working out the problem and if your azimuth is close to 270 your probably OK.

The Suns azimuth is 289.6 for your position at that time.

Salted sea, which route did you take? I am looking at a way to do it with pub 229, but it doesn’t seem right that the LHA is over 90 degrees when sun azimuth is taken right before sunset. and yeah if I had a clear horizon I would do amplitude.


Your result is right, while your reasoning is false.

Sun rise or set are not at LHA 270°/90° (with some exceptions at the equator).
That would mean the length of the day being always exactly 12 hours.

For your position, all times TZ -7:

Day length = 13h 47m
Sunrise is at 06:03 with an azimuth of 68.2°
Local noon is at 12:56 with an azimuth of 180° and a LHA of 0°
at your time, 19:34, elevation of the sun is 2.41°, azimuth is 289.6°, LHA is 6h 37m 34s
Sunset is at 19:50 with an azimuth of 292°

Ok, thanks I guess. so you are saying that LHA is correct, I guess that I am just having a hard time visualizing that but either way, if I were to solve it with pub 229, I run into issues with Next LHA and using base Dec an Lat. I know that I can just put it in the formula and out comes a Z value but I really would like to figure out why the pub 229 is not working for me.

Imagine that mid-summer you are standing at the north pole. The sun will be above the horizon through the entire 360 degree rotation of LHA.

LHA and declination are the coordinates of the GP. The calculation done in Pub 229 converts to Azimuth (Z) and altitude Hc) which is what is seen by an observer at any given point.

I never liked to use pub 229. 505 in bowditch was my jam. Always an exact answer.

thanks Kebbebc, makes sense. I understand what pub 229 is doing and how to do triple interp. just a little confused with some LHA’s above 90 , when I go to the next LHA Z value the Z jumps up a lot if it crosses the black lines in pub 229 (say LHA 101 30, Dec 17 20, Lat 33 30)

I don’t use 229 for azimuths. I just use the formula from bowditch 2. Honestly I don’t even use that anymore, I use sky mate. Unless on a crossing and super bored.

Take the GP of a body, that body will be visible (about) 90 degrees in every direction away from the GP. So if you were at the GP (the body directly overhead) you could move 90 degrees away in any direction and it would still be visible. The area where the body can be seen could be thought of as a “footprint”.

Assume for example the GP is Lat 10 N and say 120 W. Move north along the 120 W meridian, when you cross the north pole the LHA will switch from LHA=0 to LHA=180 but you will still see the body.

As far as 229, it’s nice to know how to use it but the formulas are quicker and easier.