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    Re: Manufacture new Bygraves?
    From: Gary LaPook
    Date: 2009 Jul 10, 14:06 -0700
    It isn't a major problem but does require different procedures.
    I started this thread back in February:
    make sure you open up the first post to show quoted text.
    Most of you questions will be answered..
    Some portions of that thread.
    When declination is less than 55' on my version (less than 20' on the
    original) you can't compute "W" because you start the process with
    declination on the cotangent scale. In this case, Bygrave says to use the
    same process as when the azimuth exceeds 85º, you simply interchange
    declination and latitude and compute altitude. But Bygrave didn't tell us how
    to calculate azimuth in this case. In my testing I have found a method that
    produces quite accurate azimuths. You simply skip the computation of "W" and
    simply set "W" equal to declination. The  worst case I have found is that the
    azimuth is within 0.9º of the true azimuth but most are much closer. If the
    declination is less than one degree and the latitude is also less than one
    degree, follow this procedure and also assume a latitude equal to one degree.
    After you have computed the Az you then follow the same procedure discussed
    above for azimuths exceeding 85º by interchanging the latitude and
    declination and then computing  Hc.
    So your MHR-1 doesn't provide for cases where the declination is less than
    20', the lowest mark on the cotangent scale. Bygrave says to use the same
    procedure in this case as in the case where the azimuth is near 90?, simply
    interchange the declination and the latitude and then compute the altitude
    and this works fine and you get accurate altitudes. But the azimuth that you
    derive in this process is not the correct azimuth and is thrown away and not
    used for plotting the LOP just as in the case of azimuths near 90?. Bygrave
    gives no instruction for computing the azimuth in this case. It cannot be
    computed in the normal way since the first thing you need to to is find
    declination on the cotangent scale and values less than 20' are not on the
    Bygrave or MHR-1. This is an important special case since the sun's
    declination is in this range for several days around each equinox. I
    developed an approximation that works well giving azimuths within one degree
    of he correct value and usually much closer. I simply skip the first step, the
    derivation of "y" (Bygrave's terminology), "W" (my terminology). I simply set
    "y" equal to declination and then proceed normally.

    Hanno Ix wrote:
    Gary and Thomas:

    Could please one of you educate me/us somewhat about the "equinox problem" you discussed?


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