# NavList:

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

**Re: Lunars: Jupiter's BIG.**

**From:**Fred Hebard

**Date:**2003 Dec 22, 21:36 -0500

Herbert, I hope I get my terminology correct in the following. On thinking on this a bit more, I believe that to _rate_ a chronometer one needs at least three lunars spread over at least three days. In that procedure, one is looking for the rate of change of chronometer time relative to astronomic time, as well as the absolute departure. Then one would regress clock time on astronomic time and look for significant departures from 1.0 in the slope. This would not be possible over a series of sights _within_ one lunar, such as over a period of one hour, because the clock error would be negligible on any chronometer worth winding. To uncover the absolute departure of chronometer time from astronomic time in one lunar, the mean time of a set of observations could be compared to the mean astronomic time. I don't think a regression would be necessary, because, again, over such a short period, the rate of change of chronometer time compared to astronomic time should equal one, at least to any obtainable level of precision, given a usable chronometer. With more than one lunar, the absolute departure of chronometer time from astronomic time would be a time function of the slope and intercept of the regression of chronometer time on astronomic time. This of course is assuming a straight line relationship between the two. I would imagine that people then found that including temperature as a covariable could account for most significant discrepancies. I am not sure how this second equation would be formulated. I would be delighted to know. So a regression of distance on clock time is not needed to reduce the observations from a single lunar. It can be one relatively simple method (using a computer) of drawing a line through the observations to check for outliers and to assess how scattered the data are. If a point falls conveniently close to the line near the middle of the observations, it would be a reasonable one to clear. Thanks, Fred > >> Fred, >> >> It would seem that you are applying the least square fit incorrectly. >> To apply >> it to the plot of distance versus time is to shoot with guns at >> sparrows. You >> can have this cheaper by averaging the time, averaging the distances, >> and then >> proceed as if the averaged values where the actual observation - just >> as you >> are suggesting. This is what was normally done, and what is feasible >> to the >> navigator without electronic tools. To use this technique does not >> prevent you >> from plotting the distances versus time to find outlyers. As long as >> you have >> the individual distances, that is, because from repeating instruments >> (such as >> the Borda circle), one does not get them. >> >> However, if you really must apply a least square fit to the given >> data, (and I >> could not argue with you if you claim that this is the only rigorous >> treatment >> of any set of more than 2 observations), the only correct method is >> to solve >> for the watch error for which the sum of the square of the distance >> residuals >> becomes a minimum. Besides from being correct, it has the additional >> benefit >> that you can combine observations to different objects, in particular >> observations on either side of the moon, which helps to cancel certain >> instrument and observation errors. >> >> It goes without saying that this has no practical value on the boat. >> But it >> could have been used this way in surveying, right before the >> telegraph was >> introduced. >> >> Herbert Prinz >> >> Fred Hebard wrote: >> >>> Interestingly, I usually plot the raw distance against the time of >>> observation and use the least squares fit to pick out a point for >>> reduction. >> >>> I think perhaps the old method of using the mean of the observations >>> would be better than using a line of best fit, although plotting the >>> data instantly tells one how good they are. Using the mean, the time >>> would have been out by 8 seconds, about 2 minutes of longitude. >> >