# NavList:

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

**Re: Averaging**

**From:**Alexandre Eremenko

**Date:**2004 Oct 21, 16:00 -0500

Thanks to all who participated in the Averging discussion. Now I can formulate the following Rules: Suppose you are averaging altitudes over an intevral of 5 minutes. 1. If the body is 20 degrees away from the meridian, the error due to nin-linearity does not exceed 0.25'. 2. If the body is 10 degrees away from the meridian, the error due to non-linearity does not exceed 0.5'. 3. If the altitude of the body is at most 70d, the error due to non-linearity does not exceed 0.5', no matter what the hour angle is. 4. If the altitude is at most 60d, the error does not exceed 0.3', no matter what the hour angle is. Thus the "worst case scenario" mentioned by Herbert Prinz is indeed the worst. Moreover, this is essentially the ONLY bad scenario for averaging: measuring a large altitude close to the meridian. Once more: DO NOT average if the altitude is more than 60d AND the body bearing is within 2 compass points of N or S. In ALL OTHER cases, averaging helps to increase precision. Few fords for "mathematically inclined" on how this was obtained. The problem is to estimate the non-linearity of altitude that comes from the formula (1) sinA=sinDsinL+cosDcosLcost, where A,D,L and t are altitude, declination, latitude and LHA respectively. It is easily seen that L and D change slowly and approximately linearly (I assume we are on a boat, not an airplane!), so t is the fastest variable in the formula. Moreover, t changes linearly for all practical purposes. So I estimate the non-linearity of A(t), assuming D and L fixed. Let t_0 be the middle of our 5-minute interval. By Taylor formula, A(t_0+h)=A(t_0)+hA'(t_0)+(h^2/2)A''(t_0)+...=linear part + the error term. We need an estimate for the error term. We have |h|<2.5 min=0.0109 rad, so h^2<0.000118 rad=0.4'. It remains to estimate the second derivative A''. To do this, one differentiates the main formula twice. (sorry, I omit the expression for the second derivative, but it is easy to derive). Now for |A|<60 deg, the expression is easy to estimate by hand. But in general, i just did it on computer, (running Matlab, but any other soft will do the job). The results are: |A"|<2.4 if 10deg