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    Re: Sight Reduction Formula Question
    From: George Huxtable
    Date: 2005 Jan 8, 17:24 +0000

    Bill wrote-
    >A lot for a liberal arts major to follow.
    If Bill succeeded in avoiding any exposure to trig. in his schooldays, then
    he has a bit of catching-up to do. But it's by no means rocket-science, and
    even an intelligent liberal arts major should be able to manage it. This
    list can help, to some extent, if he explains his problems.
    > My question. Susan Howell states
    >(for the longer of the two formulas Fried gave) "if sin LHA is less than 0,
    >Zn = Z, if sin LHA is equal to or greater than 0, Zn = 360 - Z."
    >Do you perceive this as a reasonable litmus test, or is it flawed in your
    >learned opinion?  More probable, have I entirely missed the point?
    I'm not familiar with Howell's text, but she seems to have got the matter right.
    The formula that Bill, and Susan Howell, refer to is presumably-
    cos Z = (sin d - (sin L * sin H ))/(cos L * cos H)
    in which, presumably, d = declination of body, H = altitude of body, L =
    latitude of observer, and then Z = azimuth of body.
    in respect of that formula, I wrote earlier-
    "when cos Z = +.98 (for example), then Z = 11.5 degrees. Or else -11.5
    degrees (same as 348.5 deg). It's not too hard to sort out which, between
    those two, because if you sketch it out, it's obvious, if the body is
    passing to your North, that if its still to your East, then it has to be in
    the NE sector, so Z must be 11.5. So it depends on your lat, the dec of the
    body, and the sign of its its LHA, all of which are known."
    When you find an angle that has a particular cosine, using the arc-cos
    (ACS) function, (that cosine always being a number between -1 and +1), a
    calculator will always give a result that runs between 0 deg (for cos Z =
    +1), through 90 deg (for cos Z = 0), to 180 deg (for cos Z = -1).  For cos
    Z = .98, the example I chose, arc-cos of .98 will be given as +11.5
    degrees. But also, an angle of -11.5 degrees (which in terms of azimuth is
    exactly the same as +348.5 deg), has exactly the same cosine, so from that
    equation for cos Z alone, there's no telling which one is right.
    When the body is on the observer's meridian, then its LHA = 0 (same as 360)
    degrees. Its LHA always increases, at somewhere near 15 degrees per hour,
    so after meridian passage, as it travels on Westwards, the LHA is steadily
    increasing from zero, and sin (LHA) will be positive. Conversely, before
    meridian passage, as the body approaches from the East, the LHA is
    increasing toward 360 deg, in which case its sine will be negative.
    So Howell's test-
    "if sin LHA is less than 0, Zn = Z, if sin LHA is equal to or greater than
    0, Zn = 360 - Z."
    is simply saying -
    "If the body hasn't yet reached meridian passage but is still to your East,
    then sin LHA < 0, take the Azimuth as it's given by the formula, which will
    be in the range 0 to 180 deg. If meridian passage has occurred, so sin LHA
    > 0, choose instead the 360-degree complement of the calculated azimuth,
    360 - Z, which will put it into the azimuth range 180 to 360 deg.
    That fits with commonsense, if you draw some little diagrams. Agreement all
    By the way, in my earlier statement that I quoted above, I wrote-
    "So it depends on your lat, the dec of the body, and the sign of its its
    LHA, all of which are known."
    But whether you take the azimuth unchanged, or instead take its 360 deg
    complement, doesn't depend on lat and dec (as I said it did) but only on
    the LHA. Sorry about that.
    Bill thought he might have "entirely missed the point"; but he hadn't.
    contact George Huxtable by email at george@huxtable.u-net.com, by phone at
    01865 820222 (from outside UK, +44 1865 820222), or by mail at 1 Sandy
    Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.

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