80 USE OF THE TABLES 
GREAT CIRCLE COURSE AND DISTANCE 
Like all other problems in navigation, this problem can be approached and 
solved with the same astronomical triangle; i. e., having been given two sides and 
an included angle, it is required to find the third side and one other angle. 
Let L; and , be the latitude and longitude of the point of departure and L, and 
Az be the latitude and longitude of the point of destination, respectively. Now, 
if in the astronomical triangle we make the following substitutions, we may use 
these tables with which to solve the problem: 
For ¢ substitute the difference of longitude between the two places. 
For L substitute L,; (the latitude of the point of departure). 
For d substitute L, (the latitude of the point of destination). 
Then Z will equal the initial Great Circle course and co h, will equal the Great 
Circle distance between the two points. ‘The method of computing the course and 
co h,, or 90° +h,, is given by the following rules: 
When t (diff. long.) is less than 90°, both b and Z’ have + signs. 
When ¢ (diff. long.) is greater than 90°, both b and Z’ have — signs. 
When L, and I, are in same latitude, L: is always plus. 
When IL, and I, are in different latitude, L, is always minus; combine alge- 
braically L, and 6, having regard for signs; should the result be less than 90°, give 
Z’’ the same sign, but if L.+6 is greater than 90° give Z’’ the opposite sign to 
L,+0. 
Add algebraically Z’ and Z’’, naming the initial course from the elevated pole, 
if the resultant Z has the plus sign, but name course from the depressed pole if Z 
has a minus sign. 
When L,+6 has a, plus sign, the distance is 90° —h.. 
When L,+ 6 has a minus sign, the distance is 90° + Ag. 
Problem 16.—Given two places, one in latitude 40° N., longitude 70° W., the 
other in latitude 30° S., longitude 10° W., find the Great Circle distance between 
them; also the initial course. Diff. long.—60° (H. A. between 0° and 90°). 
t_.60° L, 30° 060/0(—) 
L; 40° N.fb 30° 47°4(+) A 12595 C 178 Z’ 41°9(+) 
In+b 0 47.4(+) B 186053 D 1860 
he 0° 3575 A+B 198648 C+D 2038 Z’’ 89°5(+) 
Course=N. 131. 4 E. 
(90° —0° 35'5=D=89° 24'5=5,364.5 nautical miles. 
Problem 17.—Find the Great Circle distance and initial course between 1° N., 
122° W., and 35° N., 139° E. Diff. long. =99° (180°—99°=81°). 
t__81°|=b 83° 38.0(—) 
L,; 1°f=Lz, 35° 00.0(+) A 80302 C 5 Z’ 83°7(—) 
L.+b 48 38 (—) B 12465 D 9945 Z’’ 41.7(—) 
he—Go 477 A+B 92767 C+D 9950 Z 125.48. and W. 
or N. 54. 6° W. 
90° + 6° 47’=D=96° 47’=5,807 
nautical miles. 
Problem 18.—Find Great Circle distance and initial course between Cape Town 
34° §., 18° E., to New York 40° N., 73° W. Diff. long. =91° (H.A. between 90° 
and 180°) supplement= 89°. 
t__89° \L, 40° 0!0(—) ‘ 
L; 34 §.fo 1 28.9(—) A 25229 C 81 Z’ 1.8°(—) 
L2+6b 41 28.9(—) B 17889 D 53 
h, 21° 44.8’ A+B 43118 C+D 134 Z’’ 53.7(—) 
Course N. 55.5 W. 
90° + 21° 44.8’=111° 44.8’=6,704.8 nautical miles. 
