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    Re: Sight Reduction Formula Question
    From: Bill B
    Date: 2005 Jan 7, 23:10 -0500

    George
    
    A lot for a liberal arts major to follow.  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?
    
    Thanks
    
    Bill
    
    
    
    > From George.
    >
    > No, not just a 180 deg ambiguity. It's far more awkward than that. A 180
    > degree ambiguity would be rather easy to allow for.
    >
    > If you derive Z from its cosine, then 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.
    >
    > In the convention I use, North is positive, and West is positive, so hour
    > angles are always increasing, and Az is measured clockwise from North
    > (which differs from the convention used by many astronomers).
    >
    > For the formula that derives Z from its cosine, the ambiguity problem is
    > easy to resolve, by inspection. However, another problem remains: that near
    > azimuths of O deg (North) and 180 deg (South), the cosine changes hardly at
    > all for quite large changes of azimuth. So in order to obtain the azimuth
    > with any accuracy when near due North or due South, you need to know its
    > cosine VERY precisely indeed. That implies that the altitude H, which was
    > calculated at an earlier stage, also needs to have been obtained to a high
    > accuracy. Otherwise you might find that cos Z might turn out to be greater
    > than 1, which would be embarrassing. Fortunately Z is seldom required to
    > great accuracy.
    >
    > Now consider the other formula, deriving Z from its sine. If, in this case,
    > that formula has given us sin Z = +0.98, what do we do then? Well, one
    > solution is that Z = 88.5 deg, which might well be the right answer. On the
    > other hand, sin 101.5 deg is also +0.98, so that's an equally valid answer.
    > It could be either, then. How do we discover which is the one we want? In
    > this case, there's no simple way, that I know of, to resolve the ambiguity
    > by using inspection and logic to determine whether the wanted result is
    > just North of East or just South of East.
    >
    > You will find the advice, in some textbooks, "just take a compass bearing
    > on the body, and it should be easy to distinguish which one is right".
    > Well, perhaps that's reasonable, in that widely-spread example, but say the
    > two possible angles were closer, say 88 deg and 92 deg. Would you be sure
    > which was the right one then? And anyway, if you're prepared to accept a
    > compass-bearing for azimuth, why are you bothering to calculate it?
    >
    > No, deriving azimuth from its sine is the worst possible option. In
    > addition, it has the same problem of inaccuracy as the cos Z method, but
    > this time for azimuths that are near East and West.
    >
    > There's a third option, that for some reason doesn't find its way into many
    > textbooks. Get the azimuth from its tan!
    >
    > This formula is-
    >
    > Tan Z = sin (hour-angle) / (cos (hour-angle) sin lat - cos lat tan dec)
    >
    > and the rules for putting Z into the right quadrant, 0 to 360, clockwise
    > fron North, are-
    >
    > If tan Z was negative, add 180 deg to Z.
    > If hour-angle was less than 180 deg, add another 180 deg to Z.
    >
    > You can see from this that the calculated altitude doesn't even enter into
    > the formula above, so there's no longer a requirement to calculate altitude
    > at all, if only an azimuth was going to be needed.
    >
    > With a programmable calculator or computer which can derive polar
    > coordinates using the "POL" or "ATAN2" function, it's even simpler, as that
    > function is designed to put the result into the correct quadrant
    > automatically.
    >
    > For example, in Casio Basic,
    > use X = POL ((Tan dec cos lat - cos (hour-angle) sin lat), - sin (hour-angle))
    >
    > and then the azimuth will appear in variable Y, as an angle measured
    > clockwise from North between -180 and + 180 deg. You can put this into 0 to
    > 360 notation, if you need to, by simply adding 360 deg if it's negative.
    >
    > But don't try simplifying out that expression above for X by cancelling any
    > minus-signs, or you will spoil it.
    >
    > Using the tan Z expression, then tan Z is changing quickly with azimuth at
    > all angles, so there are no "dead-spots" with low accuracy, as exist in the
    > other methods. I don't understand why it's not used more widely.
    >
    > George.
    >
    >
    >> --- Fred Hebard  wrote:
    >>> I believe Turner gives the reason for the longer
    >>> formula and also shows
    >>> the shorter one, saying the shorter one can be
    >>> inaccurate in some cases.
    >>> Fred
    >>>
    >>> On Jan 7, 2005, at 10:24 AM, Fried Squash wrote:
    >>>> In _Celestial for the Cruising Navigator_ by Merle
    >>>> Turner, the navigational triangle azimuth formula
    >>> is
    >>>> given as:
    >>>>
    >>>> cos Z = (sin d - (sin L * sin H ))/(cos L * cos H)
    >>>>
    >>>> but Dutton's Navigation and Piloting gives the
    >>> much
    >>>> simpler:
    >>>>
    >>>> sin Z = ( cos d * sin t ) / cos H
    >>>>
    >>>> They both seem to work.  But there must be some
    >>> reason
    >>>> that the longer one is better, right? Works in
    >>> more
    
    
    

       
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