# 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:**Peter Hakel

**Date:**2014 Dec 24, 01:37 +0000

Sam,

You should add 360 (not 180) to your negative result. For your example, you can use the spreadsheet sailings.xls with the following inputs:

B2: 39 58/60

C2: =9/60+40.9/3600

E2: =41+43/60+31.8/3600

F2: 0

The answer is in the yellow cell C6. It differs from your negative result by exactly 360 degrees. If you want to dig deeper, look at the formula in cell F21 (from which cell C6 gets its value) and go from there.

Peter Hakel

You should add 360 (not 180) to your negative result. For your example, you can use the spreadsheet sailings.xls with the following inputs:

B2: 39 58/60

C2: =9/60+40.9/3600

E2: =41+43/60+31.8/3600

F2: 0

The answer is in the yellow cell C6. It differs from your negative result by exactly 360 degrees. If you want to dig deeper, look at the formula in cell F21 (from which cell C6 gets its value) and go from there.

Peter Hakel

**From:**Samuel L <NoReply_SamuelL@fer3.com>

**To:**pmh099---.com

**Sent:**Tuesday, December 23, 2014 3:51 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