And, lastly, If from the point t, in which rR produced meets the axis, we take the angle CtV in proportion to the longitude of the proposed map, as MF the sine of the middle latitude is to radius, and draw the parallels and meridians as in the figure, the whole space SOQV will be the proposed part of the conical surface expanded into a plane; in which the places may now be inserted according to their known longitudes and latitudes.

Example.

V. Let Ll, the breadth of the zone, be 50°, lying between 10° and 60° north latitude; its longitude 110°, from 20° east of the Canaries to the center of the western hemisphere; comprehending the western parts of Europe and Africa, the more known parts of North America, and the ocean that separates it from the old continent.

And because Cx = Nn ⁄ Ll × MF × MT, add these three logarithms.

Log. 0.8726650 (= 50° to radius 1)-1.9408476
Log. MF (sin. 35°)-1.7585913
Log. MT (tang. 55°) 0.1547732
Take the sum-1.8542121
from log. Nn (= .6923772)-1.8403427
the remainder-1.9861306
is the logarithm of Cx. And because 1: Cx ∷ MT : xt, to this adding the log. MT 0.1547732
The sum0.1409038

is the log. of xt = 1.383260; and xR (= xr = ½ Ll) being .4363325, Rt will be 0.9469275, rt = 1.8195925. Whence having fixed upon any convenient size for our map, the center t is easily found. As, allowing an inch to a degree of a great circle, or 50 inches to the line Rr, Rt the semidiameter of the least parallel will be 54.255 inches, and that of the greatest parallel 104.255 inches.

Again, making as radius to MF so the longitude 110° to the angle StV, that angle will be 63° 5´⅗. Divide the meridians and parallels, and finish the map as usual.

Note, The log. MT being repeated in this computation with a contrary sign, we may find xt immediately by subtracting the sum of the logarithms of Ll and MF from the log. of Nn.

VI. A map drawn by this rule will have the following properties:

1. The intersections of the meridians and parallels will be rectangular.