John:
(I don't know if each paragraph will be enclosed in asterisks. That
happened in one of my posts, but not in the other. I use an asterisk for
multiplaction and any spurious asterisk at the beginning and end of a
paragraph shouldn't be confuse with that multiplying asterisk.)
Here is the formula for h, for a given Az, Lat, and dec. ...which I said
that I'd soon post.
>From the size of this formula, it isn't surprising that others have
recommended something briefer. But this formula comes directly from the
formula for Az, from Lat, dec, & h.
This is in the form of a quadratic-formula solution. It can give 2 answers,
and I'll say something about how to choose which one is right.
What follows will be correct if I didn't make any algebraic errors.
I use these abbreviations:
Az = azimuth
Lat = latitude
dec = declination
h = hour angle (The sun's hour angle is the sundial equal-hours before or
after solar.noon).
(The Greek letters save space, when they're available, but not when they
have to be written out in Latin characters. And the above abbreviations are
clearer to people who aren't familiar with the Greek-letter symbols.)
h =
-2(tan dec * cos Lat)/tan Az plus-or-minus the square root of:
{4( tan^2 dec * cos^2 lat/tan^2 Az)
- 4(sin^2 Lat +1/tan^2 Az) * (tan^2 dec * cos^2 Lat -sin^2 Lat) }
The result of evaluating the above is divided by:
2 * (sin^2 Lat + 1/tan^2 Az)
-------------------------
Because the quadratic formula often gives 2 answers, then here are some
suggestions for choosing the right one:
If your input azimuth is east of south, then h and its sine must be
positive.
If your input azimuth is west of south, then h and its sine must be
negative.
If your input azimuth is south of the east-west line, then the cosine of h
must be greater than:
tan dec/tan Lat.
.if your input azimuth is north of the east-west lilne, the cosine of h
must be less than tan dec/tan Lat.
----------------------------
Maybe a briefer solution-fomula can be gotten by setting equal to zero, the
sun's altitude in a co-ordinate system whose "equator" is the azimuth
circle passing through your desired azimuth, and solving for the h that
would achieve that.
If that's workable, then it could have the advantage that h only appears
once in the altitude formula.
I apologize in advance for any algebraic or copying errors.
Michael Ossipoff
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