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In 17872 Robert showed one way to compute a least square solution using Shadwell's data. Since I haven't used Matlab in some 15 years it took me some time to penetrate the code, but now I believe I got hold of the essentials. Still, however, I don't understand how the "corrns" are calculated. Anyway, Robert shows that his solution has a smaller sum of squares than Shadwell's.

As I understand it, Robert works, like Shadwell, with the chronometer error differences and the times between observations. Then Robert models the error as

error(p) = a*p + (b/(2*m))*p^2

where p is the time variable (=2, 5, 6, 9, 11) and m is the last day (=11). Furthermore, a=4.2269 and b=-0.3072.

The rate is then calculated as d/dp of the error, i.e.

rate(p) = a + b*p/m

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My own approach has been to work directly with the calender date as time variable and the actual recorded chronometer errors (disregarding the constant hour and minute error). Thus I get six pairs of time and error. Both Robert and Shadwell use only five pairs, but I don't understand why they have omitted the sixth pair, in their case the origin.

Then I model the error as

error(d) = a + b*d + c*d^2

where d is the date (in May). Making a least square fit I get
a=-7.3221, b=4.1150 and c=-0.0045.

Comparing the predicted errors with the actual errors, I get a sum of squares that is less than Robert's.

Then the rate is found as

rate(d) = b + 2*c*d

The predicted rates for 3, 5, 8, 9, 12, 14 May then become

4.0880, 4.0701, 4.0431, 4.0341, 4.0071, 3.9891

Curiously (or maybe not?) the sum of my predicted rates equals the sum of Shadwell's predicted rates, but the difference between mine and Shadwell's numbers is different from day to day.

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Thanks Robert! I have learned a lot by trying to understand your code. Now I am able to more efficently calculate least squares by using matrixes directly, something that I mastered many years ago but have been since long forgotten.

The mystery remains unsolved however, what was the theory behind Shadwell's method?

Lars, 59N 18E
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