# 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 May 21, 09:59 -0700

John:

By definition, a body with a negative F is below your astronomical horizon, so the more obvious example would be if you're using the tables to determine a great circle route to a destination closer to your antipode than to you (i.e. further away from you than 10,800 nmi)..

As a recap, F = dec + B. The magnitudes of both dec and B range from 0° to 90°. If you're specifically looking for F < -90°, then both dec and B must be negative. Declination is treated as negative if its name is contrary to your latitude.

sin(B) = cos(lat) * cos(LHA). Since lat is always treated as positive, B will only be negative if cos(LHA) is negative (hence the rule about 90° < B < 270°). B's maximum negative magnitude occurs at cos(LHA) = -1, when LHA is 180°. We therefore need to be looking at a body relatively near lower transit. For such a body to be above your (visible) horizon it needs to have a relatively high declination, where LHA has less influence on altitude.

The maximum contrary declination you can see over your horizon is the complement of your latitude. To see a body with a large contrary declination, you need a small latitude.

To sum up, you need:

- A contrary dec
- A high dec
- A small lat

An example would be taking Polaris sights from a few miles south of the equator, probably flying at altitudes where the air is thin enough to see such a faint star so low. And your Polaris tables fell behind some shelves.

Or you're looking for a great circle route connecting Spain and New Zealand.