# NavList:

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

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Re: John Karl's Celestial Navigation in the GPS Age
From: Robert Bernecky
Date: 2018 May 1, 06:31 -0700

Hopefully Sean's post answered your questions, but Karl apparently does not use the interpolation tables that Sean references.

The tables show the solution (Hc and Z) for whole number declination values of the body.  In the attached, we see Hc for declination of 18 and 19:

18   37° 40.2'   -47.6'

19   36°  52.6

We want the Hc solution for a declination of 18° 08.5' which will be between the two values showed above.  Notice the -47.6' entry is the difference between the two Hc values:  37° 40.2' - 47.6' = 36° 52.6'  and it means that when the declination 18° changes by 1 full degree (or 60 minutes), the Hc solution changes by -47.6'.  For example, if the declination were 18° 30' (exactly halfway between 18 and 19) we would expect the Hc solution to be halfway between the two values, or 1/2 * -47.6= -23.8' different from 37° 40.2'

In this example, dec 18° 08.5',  we are a fraction 8.5/60 of the way between 18 and 19, and thus the Hc solution will be 8.5/60 * -47.6 different. Karl shows this calculation in the lower right box, but he uses 8 rather than 9 (I would round 8.5 to 9) and -47 rather than -48 (again, I would round 47.6 to 48). But using all the digits, we get an "adjustment" of  8.5/60 * -47.6= -6.7 that we apply to 37° 40.2' - 06.7' = 37° 33.5'  which we could round to 37° 34'.

As Sean explains, the tables have an interpolation table (Karl does not discuss them!) that works out the -6.7 correction, so the user does not need to explicitly work out 8.5/60 * -47.6, which is a pain, back in the day, with no electronic calculator.

File:
pub229_Example.pdf
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