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    Re: Sight Reduction Formula Question
    From: Bill B
    Date: 2005 Jan 8, 15:26 -0500

    Thank you George.  Your explanation helped me understand the workings of the
    formula, as opposed to plug and chug, significantly.  Thought it worked, but
    did not understand why.  Started leaning with HO229 mid 2004, and have just
    recently started playing with the various spherical trig formulas.
    
    > 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.
    
    LOL.  Actually did make it through the engineering calculus courses at
    Purdue with excellent grades, but headed for a B.S. in psychology with an
    interest in consumer research before my M.A. Different math.  Like my racing
    skis, those skills have gotten very rusty and lost their edge because of
    disuse.  I do indeed have some catching up to do.
    
    I apologize if my questions are not always clearly stated.  As they arise
    out of a combination of ignorance or confusion about what I have read, they
    are by default often confused themselves.  Fortunately, in attempting to
    state the problem clearly I find it organizes my thoughts to the point that
    I am able to solve it without the list's help in many cases.  Unfortunately,
    that leaves the list with questions that arise after I have thought myself
    into a corner and am deeply confused.
    
    Appreciate all you help,
    
    Bill
    
    > 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
    > round!
    >
    > 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.
    >
    > George.
    >
    > ================================================================
    > contact George Huxtable by email at george---.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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