# NavList:

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

**Re: Longitude by unequal altitudes - Brent**

**From:**David C

**Date:**2020 Jul 12, 22:30 -0700

Lars wrote

you should not be totally confused. The last term's numerator is clearly 620 seconds of arc, the altitude difference. In the denominator you have a constant of 2, times C which is given as 1.54 (seconds of arc)/(minutes of time)

^{2}, times (I-g) given as 77.5 minutes of time. Now 620 / (2·1.54·77.5) = 2.60 and the dimension becomes minutes of time, or 2^{m}36^{s}. I guess there is an explanation somewhere in Brent of the dimension of C as given in Table III.

Thanks. it now makes sense. In an example I have worked the correction was 3.02 minutes of time. The easiest way to convert that to min and sec is to use the sexigesimal key on the calculator which gives the answer 3 min 1 sec. What confused me was that the calculator displayed this as 3° 1' because the calculator interprets the input as 3.02°. But the input is 3.02 min of time so the answer is in min sec of time. Obvious!

To summarise the sighst, comparing the results with my GNSS position........

Noon sight in error by 0.4'.

Ex-meridian LHA 9min 28 sec am error 0.3'

Both these errors can be explained by my use of a sextant without optics.

The longitude was determind by sights at LHA 22 min am and 17 min pm. The longitude was 8.5' in error.