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O.K., Now that we are all off and practicing our lunars (or, like me,
waiting for the skys to clear once more) How about I prize open another
kettle 'o fish...

How about double altitudes for latitude? I have a couple of methods from
Cotter here, one of which I can't get to work, and another which seems
to be very inaccurate.

Here is more info on what I am talking about:

O.K., you have a watch, but it is no good for measuring absolute time,
just short intervals, hence it is O.K. for lunars, but it does not keep
time well enough for the classic Sumner line approach. So, If we have a
sun altitude, and a couple of hours later another sun altitude, and we
know the time difference between the two obs, then can we compute a
latitude directly?

In theory, yes. Here is one method.

P is the north pole, X is the first obs, Y is the second obs, Z is the
Zenith.

                    .,-P
                  ,'  / \
                ,'   /    \
               Z..../       \
               |   /`---.    \
               |  /      ``.   \
               | /          `-.   \
               |/              `--. \
               Y--------------------- X


We know PX, PY and the time difference converted to arc P
So, use the law of cosines to compute XY.

Now, use the law of cosines to find angle PXY.

Now, we measured ZX, and ZY, and computed XY, so, using law of cosines,
solve for ZXY.

Now, PXZ = PXY - ZXY

Now, we know PXZ, PX, and XZ, so using law of cosines, solve for PZ

Our latitude is 90 - PZ  

So, with this, and the same lousy watch we use for lunars, we get a
latitude without having to hope for clear skies at LAN...

Comments??

Cheers!

Jeff!



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From: J.van.Puffelen@uni4nn.iaf.nl (Jan van Puffelen)
Subject: Re: double altitudes (another blast from the past...)
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>Jeff Gottfred wrote:


>Here is more info on what I am talking about:
>
>O.K., you have a watch, but it is no good for measuring absolute time,
>just short intervals, hence it is O.K. for lunars, but it does not keep
>time well enough for the classic Sumner line approach. So, If we have a
>sun altitude, and a couple of hours later another sun altitude, and we
>know the time difference between the two obs, then can we compute a
>latitude directly?
>
>In theory, yes. Here is one method.
>
>P is the north pole, X is the first obs, Y is the second obs, Z is the
>Zenith.
>
>                    .,-P
>                  ,'  / \
>                ,'   /    \
>               Z..../       \
>               |   /`---.    \
>               |  /      ``.   \
>               | /          `-.   \
>               |/              `--. \
>               Y--------------------- X
>
>
>We know PX, PY and the time difference converted to arc P
>So, use the law of cosines to compute XY.
>
>Now, use the law of cosines to find angle PXY.
>
>Now, we measured ZX, and ZY, and computed XY, so, using law of cosines,
>solve for ZXY.
>
>Now, PXZ = PXY - ZXY
>
>Now, we know PXZ, PX, and XZ, so using law of cosines, solve for PZ
>
>Our latitude is 90 - PZ  
>
>So, with this, and the same lousy watch we use for lunars, we get a
>latitude without having to hope for clear skies at LAN...
>
>Comments??
>

This method looks of little practical value at sea:

* It depends on the fact that the latitude does not chance between the two
observations. This is usually not true at sea.

* It also depends on the fact that the longitude does not change between the
two observations. After all, if the longitude changes, the LHA changes as
well. Again, this is usually not the case at sea.

* If the GMT is not known, the sun nor the moon can be used. For these two
objects the declination depends on the time. This limits the use of this
method to the brightest stars and planets.

* However, in order to measure the altitude of stars both the night sky and
the horizon must be visible at the same time. This is only the case in the
nautical twilight which is of a very limited duration (especially in the
tropics). This duration is too short for the two observations of the same
star. This makes even star sights unpractical.

The only practical application is IMHO on land, with an artificial horizon.
But then again, this method is fairly complex as opposed to the simple noon
altitude (or meridian passage). A far more simple approach would be to
determine true N or S with a magnetic compass and take the highest/lowest
altitude of a star or planet during the meridian passage. This will provide
the latitude on a very simple and accurate way.

Regards,
Jan van Puffelen
slip0039@uni4nn.iaf.nl
52d 24.5'N 4d 55'E



>Cheers!
>
>Jeff!
>
>
>
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>Do NOT send administrative requests to navigation@ronin.com. Thanks. -ben
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>

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