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 | |______|_________|

An analysis of this table shows that the azimuth this year (1882) increases with the latitude from 1° 28' 05" at 26° north, to 2° 3' 11" at 50° north, or 35' 06". It also shows that the azimuth of Polaris at any one point of observation decreases slightly from year to year. This is due to the increase in declination, or decrease in the star's polar distance. At 26° north latitude, this annual decrease in the azimuth is about 22", while at 50° north, it is about 30". As the variation in azimuth for each degree of latitude is small, the table is only computed for the even numbered degrees; the intermediate values being readily obtained by interpolation. We see also that an error of a few minutes of latitude will not affect the result in finding the meridian, e.g., the azimuth at 40° north latitude is 1° 43' 21", that at 41° would be 1° 44' 56", the difference (01' 35") being the correction for one degree of latitude between 40° and 41°. Or, in other words, an error of one degree in finding one's latitude would only introduce an error in the azimuth of one and a half minutes.

With ordinary care the probable error of the latitude as determined from the method already described need not exceed a few minutes, making the error in azimuth as laid off on the arc of an ordinary transit graduated to single minutes, practically zero.

REFRACTION TABLE FOR ANY ALTITUDE WITHIN THE LATITUDE OF THE UNITED STATES.

 _____________________________________________________

| | | | |

| Apparent | Refraction | Apparent | Refraction |

| Altitude. | _minus_. | Altitude. | _minus_. |

|___________|______________|___________|______________|

| | | | |

| 25° | 0° 2' 4.2" | 38° | 0° 1' 14.4" |

| 26 | 1 58.8 | 39 | 1 11.8 |

| 27 | 1 53.8 | 40 | 1 9.3 |

| 28 | 1 49.1 | 41 | 1 6.9 |

| 29 | 1 44.7 | 42 | 1 4.6 |

| 30 | 1 40.5 | 43 | 1 2.4 |

| 31 | 1 36.6 | 44 | 0 0.3 |

| 32 | 1 33.0 | 45 | 0 58.1 |

| 33 | 1 29.5 | 46 | 0 56.1 |

| 34 | 1 26.1 | 47 | 0 54.2 |

| 35 | 1 23.0 | 48 | 0 52.3 |

| 36 | 1 20.0 | 49 | 0 50.5 |

| 37 | 1 17.1 | 50 | 0 48.8 |

|___________|______________|___________|______________| 

Applications

In practice to find the true meridian, two observations must be made at intervals of six hours, or they may be made upon different nights. The first is for latitude, the second for azimuth at elongation.

To make either, the surveyor should provide himself with a good transit with vertical arc, a bull's eye, or hand lantern, plumb bobs, stakes, etc.[1] Having "set up" over the point through which it is proposed to establish the meridian, at a time when the line joining Polaris and Alioth is nearly vertical, level the telescope by means of the attached level, which should be in adjustment, set the vernier of the vertical arc at zero, and take the reading. If the pole star is about making its upper transit, it will rise gradually until reaching the meridian as it moves westward, and then as gradually descend. When near the highest part of its orbit point the telescope at the star, having an assistant to hold the "bull's eye" so as to reflect enough light down the tube from the object end to illumine the cross wires but not to obscure the star, or better, use a perforated silvered reflector, clamp the tube in this position, and as the star continues to rise keep the horizontal wire upon it by means of the tangent screw until it "rides" along this wire and finally begins to fall below it.

Take the reading of the vertical arc and the result will be the observed altitude.

[Footnote 1: A sextant and artificial horizon may be used to find the altitude of a star. In this case the observed angle must be divided by 2.]

Another Method

It is a little more accurate to find the altitude by taking the complement of the observed zenith distance, if the vertical arc has sufficient range. This is done by pointing first to Polaris when at its highest (or lowest) point, reading the vertical arc, turning the horizontal limb half way around, and the telescope over to get another reading on the star, when the difference of the two readings will be the double zenith distance, and half of this subtracted from 90° will be the required altitude. The less the time intervening between these two pointings, the more accurate the result will be.

Having now found the altitude, correct it for refraction by subtracting from it the amount opposite the observed altitude, as given in the refraction table, and the result will be the latitude. The observer must now wait about six hours until the star is at its western elongation, or may postpone further operations for some subsequent night. In the meantime he will take from the azimuth table the amount given for his date and latitude, now determined, and if his observation is to be made on the western elongation, he may turn it off on his instrument, so that when moved to zero, after the observation, the telescope will be brought into the meridian or turned to the right, and a stake set by means of a lantern or plummet lamp.

Another Method 360 14b

It is, of course, unnecessary to make this correction at the time of observation, for the angle between any terrestrial object and the star may be read and the correction for the azimuth of the star applied at the surveyor's convenience. It is always well to check the accuracy of the work by an observation upon the other elongation before putting in permanent meridian marks, and care should be taken that they are not placed near any local attractions. The meridian having been established, the magnetic variation or declination may readily be found by setting an instrument on the meridian and noting its bearing as given by the needle. If, for example, it should be north 5° east, the variation is west, because the north end of the needle is west of the meridian, and vice versa.

Local time may also be readily found by observing the instant when the sun's center[1] crosses the line, and correcting it for the equation of time as given above--the result is the true or mean solar time. This, compared with the clock, will show the error of the latter, and by taking the difference between the local lime of this and any other place, the difference of longitude is determined in hours, which can readily be reduced to degrees by multiplying by fifteen, as 1 h. = 15°.

[Footnote 1: To obtain this time by observation, note the instant of first contact of the sun's limb, and also of last contact of same, and take the mean.]

APPROXIMATE EQUATION OF TIME.

 _______________________

| | |

| Date. | Minutes. |

|__________|____________|

| | |

| Jan. 1 | 4 |

| 3 | 5 |

| 5 | 6 |

| 7 | 7 |

| 9 | 8 |

| 12 | 9 |

| 15 | 10 |

| 18 | 11 |

| 21 | 12 |

| 25 | 13 |

| 31 | 14 |

| Feb. 10 | 15 |

| 21 | 14 | Clock

| 27 | 13 | faster

| M'ch 4 | 12 | than

| 8 | 11 | sun.

| 12 | 10 |

| 15 | 9 |

| 19 | 8 |

| 22 | 7 |

| 25 | 6 |

| 28 | 5 |

| April 1 | 4 |

| 4 | 3 |

| 7 | 2 |

| 11 | 1 |

| 15 | 0 |

| |------------|

| 19 | 1 |

| 24 | 2 |

| 30 | 3 |

| May 13 | 4 | Clock

| 29 | 3 | slower.

| June 5 | 2 |

| 10 | 1 |

| 15 | 0 |

| |------------|

| 20 | 1 |

| 25 | 2 |

| 29 | 3 |

| July 5 | 4 |

| 11 | 5 |

| 28 | 6 | Clock

| Aug. 9 | 5 | faster.

| 15 | 4 |

| 20 | 3 |

| 24 | 2 |

| 28 | 1 |

| 31 | 0 |

| |------------|

| Sept. 3 | 1 |

| 6 | 2 |

| 9 | 3 |

| 12 | 4 |

| 15 | 5 |

| 18 | 6 |

| 21 | 7 |

| 24 | 8 |

| 27 | 9 |

| 30 | 10 |

| Oct. 3 | 11 |

| 6 | 12 |

| 10 | 13 |

| 14 | 14 |

| 19 | 15 |

| 27 | 16 | Clock

| Nov. 15 | 15 | slower.

| 20 | 14 |

| 24 | 13 |

| 27 | 12 |

| 30 | 11 |

| Dec. 2 | 10 |

| 5 | 9 |

| 7 | 8 |

| 9 | 7 |

| 11 | 6 |

| 13 | 5 |

| 16 | 4 |

| 18 | 3 |

| 20 | 2 |

| 22 | 1 |

| 24 | 0 |

| |------------|

| 26 | 1 |

| 28 | 2 | Clock

| 30 | 3 | faster.

|__________|____________|