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> From: "Bill Noyce" 

> I think the original question was, how to predict the time at which
> the sun has an azimuth of 90 or 270 degrees, given a known or assumed
> latitude.  In this case, you have the co-latitude, the co-declination
> of the sun, and the azimuth (which is a right angle) as three known
> parts of the triangle.  (The sun's declination comes from a calendar,
> plus perhaps an estimate of longitude to the nearest 15 degrees.  You
> would have to be more careful if using the moon.)
> The formulas will come out something simple like
> cos LHA = tan d / tan L  and  sin h = sin d / sin L

Interesting.  For Peter to calculate the exact moment the body is at 90/270d
he must know his location exactly.

While you state co-latitude is a known, it cannot be known exactly until the
exact time of transit is known.  However delination is a relatively slowly
changing variable, so if we can estimate the time of 90/270 to a half hour
or so we can derive a good estimate of LHA from AP Lat (LHA = tan d /tan L).

With LHA and AP lon we can estimate GHA, from which we can derive estimated
time of transit within a few minutes or better.

With time of transit, we get a closer approximation of declination and GHA.
We could use the direct computation of azimuth from the formula George
posted a while back as a sanity check.

"Tan Z = sin LHA / (cos LHA  sin lat - cos lat  tan dec)

and the rules for putting Z into the right quadrant, 0 to 360, clockwise
from 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"

If the computed azimuth is within a degree or so of 90/270d, we are pretty
much good to go as we have seen earlier. One could be off the actual time
for a given AP by several minutes or more with little damage done unless AP
latitude was horribly off.

If off more than a degree or so, make an estimate of the change in time
needed to move the azimuth to the desire angle and try again.

> I think the intent is to observe until the predicted h occurs, and
> record the resulting GMT.  From that and the matching LHA you compute
> GHA and Longitude.  Right?  And you use the graphical method to
> interpolate to hit the desired h exactly.

You would have to ask Peter about that.  My impression is once he had the
time of transit, he computed Hc.  Any difference between Hc or Ho was
plotted east/west from AP for longitude.
>
> From the modern perspective, a sight at this special time gives a LOP
> like any other; its main advantage is that the LOP runs exactly north
> and south if you're at the assumed latitude, so any error in the
> latitude has only a tiny effect on the resulting longitude.

Exactly.

Thanks

Bill B






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