These two co-ordinates, the altitude and azimuth, will determine the position of any object with reference to the observer's place. The latter's position is usually given by his latitude and longitude referred to the equator and some standard meridian as co-ordinates.

The latitude being the angular distance north or south of the equator, and the longitude east or west of the assumed meridian.

We are now prepared to prove that the altitude of the pole is equal to the latitude of the place of observation.

Let H P Z Q¹, etc., Fig. 2, represent a meridian section of the sphere, in which P is the north pole and Z the place of observation, then H H¹ will be the horizon, Q Q¹ the equator, H P will be the altitude of P, and Q¹ Z the latitude of Z. These two arcs are equal, for H C Z = P C Q¹ = 90°, and if from these equal quadrants the common angle P C Z be subtracted, the remainders H C P and Z C Q¹, will be equal.

To determine the altitude of the pole, or, in other words, the latitude of the place.

Observe the altitude of the pole star when on the meridian, either above or below the pole, and from this observed altitude corrected for refraction, subtract the distance of the star from the pole, or its polar distance, if it was an upper transit, or add it if a lower. The result will be the required latitude with sufficient accuracy for ordinary purposes.

The time of the star's being on the meridian can be determined with sufficient accuracy by a mere inspection of the heavens. The refraction is always negative, and may be taken from the table appended by looking up the amount set opposite the observed altitude. Thus, if the observer's altitude should be 40° 39' the nearest refraction 01' 07", should be subtracted from 40° 37' 00", leaving 40° 37' 53" for the latitude.

TO FIND THE AZIMUTH OF POLARIS.

As we have shown the azimuth of Polaris to be a function of the latitude, and as the latitude is now known, we may proceed to find the required azimuth. For this purpose we have a right-angled spherical triangle, Z S P, Fig. 4, in which Z is the place of observation, P the north pole, and S is Polaris. In this triangle we have given the polar distance, P S = 10° 19' 13"; the angle at S = 90°; and the distance Z P, being the complement of the latitude as found above, or 90°--L. Substituting these in the formula for the azimuth, we will have sin. Z = sin. P S / sin P Z or sin. of Polar distance / sin. of co-latitude, from which, by assuming different values for the co-latitude, we compute the following table:

AZIMUTH TABLE FOR POINTS BETWEEN 26° and 50° N. LAT.
LATTITUDES
___________________________________________________________________
| | | | | | | |
| Year | 26° | 28° | 30° | 32° | 34° | 36° |
|______|_________|__________|_________|_________|_________|_________|
| | | | | | | |
| | ° ' " | ° ' " | ° ' " | ° ' " | ° ' " | ° ' " |
| 1882 | 1 28 05 | 1 29 40 | 1 31 25 | 1 33 22 | 1 35 30 | 1 37 52 |
| 1883 | 1 27 45 | 1 29 20 | 1 31 04 | 1 33 00 | 1 35 08 | 1 37 30 |
| 1884 | 1 27 23 | 1 28 57 | 1 30 41 | 1 32 37 | 1 34 45 | 1 37 05 |
| 1885 | 1 27 01 | 1 28 35½ | 1 30 19 | 1 32 14 | 1 34 22 | 1 36 41 |
| 1886 | 1 26 39 | 1 28 13 | 1 29 56 | 1 31 51 | 1 33 57 | 1 36 17 |
|______|_________|__________|_________|_________|_________|_________|
| | | | | | | |
| Year | 38° | 40° | 42° | 44° | 46° | 48° |
|______|_________|__________|_________|_________|_________|_________|
| | | | | | | |
| | ° ' " | ° ' " | ° ' " | ° ' " | ° ' " | ° ' " |
| 1882 | 1 40 29 | 1 43 21 | 1 46 33 | 1 50 05 | 1 53 59 | 1 58 20 |
| 1883 | 1 40 07 | 1 42 58 | 1 46 08 | 1 49 39 | 1 53 34 | 1 57 53 |
| 1884 | 1 39 40 | 1 42 31 | 1 45 41 | 1 49 11 | 1 53 05 | 1 57 23 |
| 1885 | 1 39 16 | 1 42 07 | 1 45 16 | 1 48 45 | 1 52 37 | 1 56 54 |
| 1886 | 1 38 51 | 1 41 41 | 1 44 49 | 1 48 17 | 1 52 09 | 1 56 24 |
|______|_________|__________|_________|_________|_________|_________|
| | |
| Year | 50° |
|______|_________|
| | |
| | ° ' " |
| 1882 | 2 03 11 |
| 1883 | 2 02 42 |
| 1884 | 2 02 11 |
| 1885 | 2 01 42 |
| 1886 | 2 01 11 |
|______|_________|