# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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From: Guy Schwartz
Date: 2006 May 5, 20:37 -0500

Thank you, Thank you, Thank you.
Your instructions are wonderful, fortunately I pressed the correct button on
my Casio fx 260
The missing step for me was the reverse sin key which on this calculator is
shift sin or sin-1.
All this calculating power is a device that was less than \$10.00 WOW.
If you have ever were or are a teacher you know about the AH HA moment when
the student gets it.
Thank you for the AH HA moment.

New question, what is the name of the inverse sin function is excel?

Have you have ever seen the Horatio Hornblower series? There is a scene
where Horatio and others
are sitting on the deck of the ship learning celestial navigation using a
slate and chalk.

Thanks again 8-).
Guy

----- Original Message -----
From: <Fr75300346@aol.com>
To: <NavList@fer3.com>
Sent: Thursday, May 04, 2006 11:12 PM
reduct...

>
> "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
>
> 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
>
> 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
>
>
> >
>
>
> --
> No virus found in this incoming message.
> Checked by AVG Free Edition.
> Version: 7.1.392 / Virus Database: 268.5.5/333 - Release Date: 5/5/2006
>
>

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