V. Let L_l_, 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 C_x_ = N_n_ ⁄ (L_l_ × 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
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Take the sum -1.8542121
from log. N_n_ (= .6923772) -1.8403427
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the remainder -1.9861306
is the logarithm of C_x_. And because 1:
C_x_ ∷ MT : _xt_, to this adding the log. MT 0.1547732
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The sum 0.1409038
is the log. of _xt_ = 1.383260; and _x_R (= _xr_ = ½ L_l_) being
.4363325, R_t_ 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 R_r_, R_t_ 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 S_t_V,
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 L_l_ and MF from the log. of N_n_.