Scale = CD/ED
= 1/9,260,000. If CD = 6 cm then ED = 6 x 9,260,000cm = 555.6km.
555.6km = 300nm and this is the departure distance between the two meridians
10º apart.
Departure = Ch.long(')
x cos lat so cos lat =
departure/ch.long(') = 300/600 = 0.5.
0.5 is cos 60º so this is at 60º N or S.