To Convert Arc into Distance.-The log. of the difference of any latitude or longitude expressed in seconds, less the log. taken out from the table (at the approximate latitude), equals the log. of the required distance in miles and decimals.

TO CONVERT A DISTANCE OBLIQUE TO THE MERIDIAN INTO ARCS OF LAT. AND LONG.

Let

48. When the distance is neither on the meridian nor at right angles thereto, it is necessary to use log. N from the table, and the exact azimuth of the line in question. A be a station in latitude 32° 26', and B another station 25 700 miles from A on a true bearing of 53° 10′ 14′′ thenFormulæ for short distances.

(1) LOG. DIFF. LONG. = log. dist.log. N, + log. sin. az. + log. sec. lat. (B approx.)

(2) LOG. DIFF. AZIMUTH

==

log. dist.log. N, + log. sin. az. + log. sec. lat. B, + log. sin. mid. lat.; equal to log. diff. lon. + log. sin. mid. lat.

(3) LOG. DIFF. OF LAT. =

log. dist.log. M, + log. cos. az

(log. dist. + log. sin. az. + log. N x 2 + log. sin. 1" + log. tan lat. B. approx.);

NOTE.-In formula (3) log. M. should be taken out for mid. lat. approx. The first three terms give approx. diff. of lat., which should be taken out and applied to lat. of A, so as to write down the lat. of B. The remainder of the formula gives a correction to be then in every case subtracted from lat. of B.

Combining the above formulæ, difference of longitude, azimuth, and latitude may be calculated with comparatively few figures by adding up and down from the middle, as in the following example, in which azimuth means an angle

D

from the meridian either north or south, such that its sine represents difference of longitude and cosine difference of

RIGOROUS METHOD OF FINDING DIFFERENCES OF LONGITUDE, AZIMUTH, AND LATITUDE, WHEN THE STATIONS ARE FAR APART, BY SPHERICAL TRIGONOMETRY.

49. Given: The latitude of A, the azimuth of B observed at A, and the distance A B. To compute differences of azimuth, longitude, and latitude. Ist-Convert the distance

A B into arc by dividing it by the normal previously multiplied by the sine of one second. 2nd-With the arc just found as one side of a spherical triangle, the co-latitude of A for another side, and the azimuth, reckoned for this purpose always from the elevated pole, as the contained angle, find the other two angles, which will respectively be the back azimuth and the difference of longitude. 3rdAdd together the log. sine of half the difference of the

azimuths, the log. cosecant of half their sum, the log. of the distance between the stations, and the log. ar. co. of meridian radius of curvature, previously multiplied by the sine of one second. The sum is the log. of the number of seconds in the difference of latitude between A and B, nearly.

The whole of the formula is as follows:

12

Logarithms to four places being sufficient for latter part.

Half back az. - half diff. of long. log.tan. 9.8731772 = 36° 45′ 2′′

Approx. diff. of lat... 0° 23' 00" 3 1380" 3.... 3'1399461

Distance in arc2

..log. 7.4832

..log. 9.9722

The correct difference of lat. is therefore 0° 23' 00":377

GEODETIC CALCULATION OF LATITUDE, LONGITUDE, AND AZIMUTH, BY CLARK'S SPHERICAL EXCESS METHOD.

50. Given the distance from Adelaide, A, to Melbourne, B, as 408-6617 miles, and the true bearing, 121° 12' 49" 4, equal azimuth 58° 47′ 10′′ 6 from south; the latitude of A, 34° 55′ 33′′, and the difference of longitude between the two places, 6° 23′ 27′′-6. It is required to find the difference of latitude, longitude, and azimuth, using values for the spheroid given in the geodetic table of constant logarithms, page 45.

These results give 37° 49′ 53′′ for the latitude of Melbourne; 6° 23′ 27′′-6 for the difference of longitude; 117° 25' 8" for the back azimuth and 297° 25' 8" for the true bearing from Melbourne. By spherical trigonometry treating the earth as a sphere the azimuths would be 58° 40′ 38′′ and 117° 31′ 43′′ respectively, giving true bearings of 121° 19′ 20′′ and 297° 31′ 43′′, which are about 6 minutes in error, and give a distance half a mile too short, as shown in the next example.

This method is not exact for long distances, but is admirably suited for use between stations visible from each other.

Note. When calculating with bearings in a reverse direction, the spherical excess and drop of curve change their sign.

CALCULATION OF TRUE BEARINGS AND DISTANCE

BETWEEN ADELAIDE AND MELBOURNE BY SPHERI-
CAL TRIGONOMETRY, TREATING THE EARTH AS A
SPHERE.

51. Given Latitude Adelaide, 34° 55′ 33′′; latitude Melbourne, 37° 49′ 53′′ 5; and difference of longitude, 6° 23' 27" 5.