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    Re: Compare Methods: Lat/Lon Near Noon
    From: Antoine Couëtte
    Date: 2021 Oct 1, 12:52 -0700

    Dear Ed,

    You stated : "MOO also needs the Zn to the body".

    If I understand correctly your point, my reply would be: "yes and no", depending on the method you are using. But there will always be a final "no" in all cases in my reply.

    Referring to the enclosed document,

    (1) - If you use Option 4.0, my partial reply would be "yes" to your point. MOO needs the Zn's of the Body, at least to perform such  "preliminary and optional data processing" as described in 4.0 (see in particular 4.0.1.2). This preliminary processing involves correcting each individual height so as to cancel the effects of μDec and NS over some elapsed time from the same time origin. These effects can be adequately computed and cancelled with just only one condition : the Values of cos Zn are to stay almost equal to 1, which in fact they are (see 5.7.2.1.1). So, "yes" you need the successive Zn's, but only through their cosines. Since they all are assumed to be equal to +/-1 - which in fact they are - you actually do not explicitly use the successive Zn's values. This options enables you to process data for which UT Culmination = UT transit ( = UT LAN). But from there you are still to process some kind of quadratic curve regression.

    (2) - From this point, whether you use or not option 4.0, for your quadratic regression my full reply is : you do not need knowing the various Zn's.

    The quadratic curve your are fitting to your observations is derived from a series of (UT, Height) values, only from these.

    This Quadratic curve is here to compute H culm and UT culm, from which you derive H tran and UT tran through the appropriate Formulae. The quadratic curve does not explicitely need Zn's.

    (2.1) - If you start from "raw data" you will be using :

    (UT culm - UT tran) in seconds of time = (48 / π) * (tan Lat - tan Dec) * (μDec - NS) - (1a)

    (2.2) - Or if you start from " pre-processed data" you will be simply using :

    UT culm = UT tran 

    In all cases you are fitting you data to a quadratic curve having its symmetry axis "forced" to be exactly parallel to the heights scale axis. You just need (UT, Heights). No knowledge of Zn's is necessary.

    Hope I understood you point correctly and that it helps.

    If somebody can give his own explanations on this matter, e.g. Frank, I will certainly read them with attention.

    Best Regards,

    Antoine

    File:
    210915-Latitude-Longitude-Near-Noon-methods-comparison-.pdf
       
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