Let the latitudes, on the same side of the equator, be 10° and 60°; then the middle latitude and its complement are 35° and 55°, and half the difference of the latitudes is 25°: and the difference of longitude being 110°, the operation will stand as below.

Log. 6600´ (in 110°) 3.8195439
Constant log. -4.1015105
-1.9210544
Log. sin. 35° -1.7585913
S = -1.6796457
Again T= 1.4281480
m= .4513202
Log. (T + m) (= 1.8794682) 0.2740350
Log. (T - m) (= 0.9768278) -1.9898180
Log.0.2842170 = D = -1.4536500
S - D (= log. tangent 59° 16´) = 0.2259957

agreeing to a minute with the solution by a table of meridional parts.

Example 2.

The rest remaining, let the difference of longitude be only 40°; then

Log. 2400´ (in 40°) 3.3802112
Constant log. -4.1015105
-1.4817217
Log. sin. 35° -1.7585913
S = -1.2403130
D (as before) = -1.4536500
S - D (= log. tang. 31° 27´ ½) -1.7866630

Example 3.

Let the difference of longitude be 40°; but the latitudes 56° and 80°;

And log. 2400´
+ log. constant		 = -1.4817217
Log. sin. 68° = -1.9671659
S = -1.4488876
T (tang. 22°) = .4040262
m = .2109980
Log. (T + m) (= .6150242) -1.7888921
Log. (T - m) (= .1830282) -1.2625181
Log. 0.5263740 = D = -1.7212944
S - D (= log. tangent 28° 6´) = -1.7275932

wanting of the true answer no more than 1° 4´.