NavList:

"Would you please be able to give me a sample  problem and work through the
answer so that I can repeat the steps and know  I'm getting to the correct
answer."

Here's one. Suppose our assumped  position, or AP, is 72d 00'W, 35d 15'N.
Further suppose that we find in the  Nautical Almanac that the Sun's GHA is 117d
00'W and its Dec is 4d 24'S. That  means that our AP is in the North Atlantic
somewhere off Cape Hatteras and the  Sun is directly overhead in the Pacific a
few hundred miles south of the equator  (which means that the date must be
somewhat before the spring equinox or  somewhat after the fall equinox since the
Sun is directly over the equator on  those dates). The difference in the
longitudes is 45d 00' exactly (and since the  Sun is west of us, that means that
our Local Apparent Time is 15:00:00 --just a  check on things). The difference
in longitudes (dLon) is the same thing as the  LHA of the body. A body on the
observer's meridian has an LHA equal to zero  which is the same thing as
saying that the observer has the same longitude as  the spot where the object is
directly overhead.

Now let's work out the  altitude Hc from this information. The equation we'll
use  is:
sin(Hc)=sin(dec)*sin(lat)+cos(dec)*cos(lat)*cos(LHA)
We have to  convert the angles to decimal degrees and use the right signs (+
for N, - for  S):
sin(Hc)=sin(-4.40)*sin(35.25) + cos(-4.40)*cos(35.25)*cos(45.00).
You  can do the products first, but you will find that your calculator does
this  automatically (in other words, if you key in 2*3+4*5, you should get 26
when you  press "equals"). Let's do it in pieces anyway:
sin(Hc)= -0.044278 +  0.575751
or
sin(Hc)= 0.531473
Now we push the inverse sine key(s) and  get
Hc=32.105 deg
or Hc=32d 06.3'.

And that sounds about right to  me --the Sun should be at about that level in
mid-March from that location at 3  in the afternoon. So if at that moment
(when the Sun's GHA and Dec had the  values quoted above) I had measured the
Sun's altitude and after correcting it  for dip, refraction, etc., I find that the
altitude is 32d 16.3', then I must be  on a line of position which is 10
miles away from my assumed position. In which  direction? Well, after I calculated
my azimuth, I can plot that direction on a  chart, and I know that I must be
closer to that place where the Sun is straight  up because I see it higher
than the calculated  altitude.

-FER
http://www.HistoricalAtlas.com/lunars 


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