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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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