NavList:

If you know your longitude, it is possible to use a table in the Nautical Almanac to find parameters a0, a1, a2 and use them to "correct" the observed altitude Ho of Polaris for a more accurate value of your latitude:

Lat = Ho - 1 + a0 + a1 + a2

This formula has to be equivalent to the three-step calculation of Lat given below. If you have seen a development like this before, I would appreciate if the collective wisdom of the NavList members could point me to relevant references.


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We need to solve the following spherical triangle with vertices YPS:
Y: "You"  (i.e., your position)
P: "Pole" (whose altitude seen from Y we want to calculate)
S: "Star" (i.e., the GP of Polaris)


STEP 1)
Calculate the Local Hour Angle of Polaris, which is the angle at vertex P:

LHA   =   GHA Aries  +  SHA Polaris  +  Longitude

This completes the determination of the YPS triangle. The other two known quantities are:
|YS| = 90 - Ho   (from your sextant measurement)
|SP| = 90 - Dec  (Dec is the declination of Polaris, from almanac)

The YPS triangle is characterized by these two known sides and an opposite angle (LHA). This is the same arrangement encountered in solving the so-called "one-body fix," whose solution I posted earlier this year on this list. As Richard Feynman said: "The same equations have the same solutions." Hence steps 2 and 3.


STEP 2)
Calculate the second opposite angle Z at vertex Y:

sin Z  =  cos Dec  *  sin LHA  /  cos Ho

This angle can be converted into azimuth Zn of Polaris:
Zn = 360 - Z, if Z > 0 (i.e., 0 < LHA < 180)
Zn = - Z    , if Z <=0 (i.e., 180 <= LHA <= 360)

This formula for Z fails with Polaris in zenith ( cos Ho = 0 ). I am not going to worry about this singular case, since then we have Y = S and there is no triangle to solve.


STEP 3)
Calculate your latitude:

sin Lat  =  ( sin Dec * sin Ho  -  cos Dec * cos Ho * cos Z * cos LHA ) / ( 1  -  cos Dec * cos Ho * sin Z * sin LHA )

Since cos Dec ~ 0, this formula never goes bad. For Dec = 90, the formula simplifies to Lat = Ho, as expected.
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I tested this approach on the example given in the 2009 Nautical Almanac:

Input:
UT = 2009 April 21, 23h 18m 56s
Longitude = W 37 degrees 14'
Ho = 49 degrees 31.6'

Results:
Latitude = 49 degrees 54.1'
Azimuth = 359.1

which agrees perfectly with the solution printed in the almanac.


Peter Hakel