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    Shadwell and least squares
    From: Lars Bergman
    Date: 2012 Jan 18, 04:09 -0800

    In Shadwell's "Notes on the Management of Chronometers" from 1861 there is described, in chapter V, a method of how to rate chronometers by the combination of severel error observations.


    Shadwell writes (page 82) "In order to enable us to treat the observations for rate in a systematic manner, and to deal with them in accordance with mathematical principles, it is necessary to assume two postulates: first, that equal confidence can be placed in all the observations (...), and secondly, that if any change of rate is taking place, and the rate be not constant, then that the change of the rate is progressing uniformly and in proportion to the time."

    He then makes a derivation of the method, and on page 85 writes that "is is advisable to combine the equations so as to give the most probable values of the unknown quantities, and the Method of Least Squares, well known to astronomers, seems best adapted for that purpose."

    Then Shadwell shows how to derive the different sums that are needed and finally how to calculate two parameters, x and y, where x represents the rate at the time of the first observation, and x+y represents the rate at the time of the last observation. He then gives several examples of the method.

    Can anyone explain which "least square" Shadwell is obtaining with his method?

    By postulating that the rate is linearly changing with time, then I think the observed chronometer errors have to be modelled as a second order function of time, as the rate is the derivate of the error with respect to time.

    Then to find a least squares solution of the errors observed, you need to handle sums containing time up to the power of four.

    If you go directly to make a least square solution of the (somehow calculated) rates, then you will manage with time to the power of two.

    Shadwell uses sums up to the power of three.

    I don't understand the method as described by Shadwell and would appreciate any help on the subject.

    Lars, 59N 18E

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