Worked ExampleScenarioSomewhere in the South Atlantic, with a deduced position of 33° S and 16° East, I shoot three stars: Sirius, Canopus and Betelgeuse.
I have worked out a combined Index Error and Dip correction. I now apply this for each sight, along with Refraction (extracted from the Almanac) , to get an Observed Altitude.
At this point I extract, from the nautical almanac, the GHA
And I add the fractions for minutes and seconds for each star - from the nautical almanac.
The next thing to do would be to convert my Observed Altitude of each star to True Zenith Distance by subtracting the Observed Altitude from 90°.
These are the actual distances I am from the sub-stellar points of the three observed stars. Now entering the Nautical Almanac at the correct place I extract from it the SHA - that's Sidereal Hour Angle - and declination for the three stars. They happen to be:
From this point on I am only going to work the Sirius starsight. But each star must be worked through exactly the same as Sirius. We now enter the murky world of spherical trigonometry. We have to solve for the triangle APB where A is the latitude and longitude of the star and P is the pole.
 The leg PA is easy, it is obtained from the declination which is S16 41.8 + 90°. Or in decimal the angular distance of PA is 106.69666. We also know AB - which is the true zenith distance. What we are now basically going to do is use our deduced latitude and longitude and work out by how much it is in error.
 To obtain our polar angle, P, we do the following - we take the GHA Aries and add the Sidereal Hour Angle (SHA) for Sirius to it. This will give us the GHA of the star. In this case it is 299.85333°. Which is how far WEST Sirius is from Greenwich. To find out how far EAST it is we subtract from 360 and get 60.14667°. One final step is to get our assumed Local Hour Angle (LHA) for the star and as we have assumed we are 16° East we subtract this figure from the GHA*. This give us 44.14667°.
 PB is our assumed latitude, 33° plus 90° = 123°. Now we know that Hav(AB) = Hav(P) x Sin(PA) x Sin(PB) + Hav(PA~PB). Plugging the numbers into our equation it looks like this: Hav(AB) = Hav(44.14667) x Sin(106.69666) x Sin(123) + Hav(123 - 106.69666) or Hav(AB) = 0.1412205 x 0.9578393 x 0.8386706 + 0.0201055 or Hav(AB) = 0.1134441 + 0.0201055 or Hav(AB) = 0.1335496 or AB = 42.86987 or AB = 42° 52.2' The figure 42° 52.2' is our Calculated Zenith Distance or CZD. However we know that our TZD (true zenith distance) is 42° 36.6' and therefore our assumed or deduced position must be in error by 15.6' or 15.6 nautical miles (52.2 - 36.6). We also know that that distance must be towards the star as it is positive.
 In the same way we can now solve for angle B and we find this to be: Hav(B) = (Hav(PA) - Hav(PB~AB))/Sin(PB) x Sin(AB) Hav(B) = (Hav(106.69666) - Hav(123-42.86987))/Sin(123) x Sin(42.86987) Hav(B) = (0.6436524 - 0.4142944)/(0.8386706 x 0.6803237) Hav(B) = 0.2293579/0.5705774 Hav(B) = 0.4019751 B = 78.693942° True.
 And our solution table now looks like this.
Once we have calculated the other two sights ...
We can now draw, to a suitable scale, a chart laying off the bearings on it.
We would lay off the bearings from our deduced or assumed position - 16 East and 33 South at 78.7° for Sirius, 140.7° for Canopus and 41.1° for Betelgeuse. These are the red lines. We would measure, using our scale, the distances of 15.6, 25.6, and 31.2 nautical miles along each. If the figure we obtained was positive - as in Sirius and Betelgeuse - we would measure towards the star. If negative, as in the case of Canopus, away from it. In other words on a reciprocal bearing to 140.7deg; or 320.7°. At right angles to our bearing lines we draw the arc of the position circle - using a straight line as the curvature is so slight because the circles are so large. Those are the green lines - each one named after the star at the circle's centre. Where the lines meet - and they seldom meet as neatly as in the above drawing which is not to any accurate scale - there we are! In this case 32° 26' South and 16° and 09' East.
 Should the deduced postion be badly in error these calculations will not be good enough. We will have to take the postion indicated by working out the three stars as our new deduced or assumed postion and rework all the starsights again. However in practice the deduced postion is usually fairly close and the slight error is of little significance.
 And we can now transfer our rough work to a proper chart. Of course there is more to it than that - like compensating for the time it took between each sextant sight and the movement of the vessel - if we were on one - but these are merely minor refinements. With practice a set of sights and their working-out can be achieved in 15 minutes - with the warm glow of accomplishment that one can confidently point to a spot on a chart or a map and say: "That's where I am".
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- Greenwich Hour
Angle of Aries - for 1800 UT. The WHAT? Aries, represented by a set of ram's
horns is a point in the sky where the ecliptic meets the equator and from
which the positions of all heavenly bodies are referenced.