# NavList:

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

**Re: Interpolation of Meridional Part Table**

**From:**George Huxtable

**Date:**2009 Mar 26, 14:47 -0000

The plot thickens- Andres calculates, from the expression he gives in [7784] WGS84 a = 6378137 [m] = = 6378137/1852 [nm] f = 1.0/298.257223563 The result is: MP( 45.000000 ) = 3019.058271357112100000 The expression Andres quoted used eccentricity (epsilon) rather than the flattening f that he tells us is f = 1.0/298.257223563 for WGS84. However, we can deduce one from the other using the expression given in Meeus, in which eccentricity = square root of ((2 * f) - (f * f)) for which I have taken the result to be 0.08181919, exactly in keeping with Earle's figure. =========================== On the other hand, I have tried to do exactly the same thing with my pocket calculator and with the following values, taken from Earle- E= 0.08181919 A = 3437.7468, the Earth's equatorial radius in geodetic miles. Latitude L= 45 and here's a transcription of the equation I've used, now corrected, which runs on a 20-year-old Casio programmable calculator- M=A*LN(TAN(45+.5*ABSL)*((1-E*SINABSL)/(1+E*SINABSL))^(E/2)) and that gives me M(45�) = 3013.648, which conforms with Bowditch, and not (quite) with Andres.. Why do Andres and I differ (slightly), and why do I agree with Bowditch, when Andres and I are trying to calculate the same thing with the same quantities, using what appears to be the same expression? You can see that I've rejoined the nit-picking purists, as my natural home. George. contact George Huxtable, at george@hux.me.uk or at +44 1865 820222 (from UK, 01865 820222) or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK. --~--~---------~--~----~------------~-------~--~----~ Navigation List archive: www.fer3.com/arc To post, email NavList@fer3.com To unsubscribe, email NavList-unsubscribe@fer3.com -~----------~----~----~----~------~----~------~--~---