# NavList:

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

**Re: NASR question on large F values**

**From:**David Iwancio

**Date:**2021 Jun 3, 02:30 -0700

John:

First off, I made a mistake earlier in describing the formula for B (I think I was confusing it with A). The real math is:

tan(B) = cos(LHA) / tan(lat)

With that out of the way, you were talking earlier about finding an F value of around -91°. Considering the sun, the greatest dec value is going to be about 20°

F = B + dec

To get F to be negative we once again consider a contrary dec.

-91° = B - 20°

So our target B is -71°.

With B between 0° and -90°, tan(B) must also be negative. Since tan(lat) will always be positive (with lat always between 0° and +90°), the sign of B is dictated by cos(LHA). This is where the rule for the sign of B being negative for LHA's between 90° and 270° comes from.

To increase the chances of a body being at least near the horizon, an LHA at or near 91° (or 269°) is probably the best bet. This is near the bottom of the columns in the tables

Glancing through the tables, for latitude 1°, B = -71°33' at LHA = 93°. Setting dec = -20° and rounding to the nearest degree, this yields F = -92° and A = 87°. H comes out to -3°, which puts the sun well below the horizon.

On the equator, if we arbitrarily declare dec to be negative, with LHA = 91° the table gives an F value of -110° and altitude of -56'. This is a possibility if your bridge wing is high enough above the waterline.