CC′ = C′U′ − CU − UU′ = tan (z + dz) − tan z − dz = tan2 z dz.
The tangents CQ, C′Q′ will intersect at q, and in the triangle CC′q the perpendicular from C on C′q is (omitting small quantities of the second order) equal to either side of the equation
tan2 z dz sin 2α = −2 tan zd α.
−tan z dz = 2 dα / sin 2α,
which is the differential equation of the meridian: the integral is tan α = ω cos z, where ω, a constant, determines a particular meridian curve. The distance of Q from the central meridian, tan z sin 2α, is equal to
| 2 tan z tan α | = | 2ω sin z | . |
| 1 + tan2 α | 1 + ω2 cos2 α |
| Fig. 21. |
At the equator this becomes simply 2ω. Let any equatorial point whose actual longitude is 2ω be represented by a point on the developed equator at the distance 2ω from the central meridian, then we have the following very simple construction (due to O’Farrell of the ordnance survey). Let P (fig. 21) be the pole, U any point in the central meridian, QUQ′ the represented parallel whose radius CU = tan z. Draw SUS′ perpendicular to the meridian through U; then to determine the point Q, whose longitude is, say, 3°, lay off US equal to half the true length of the arc of parallel on the sphere, i.e. 1° 30′ to radius sin z, and with the centre S and radius SU describe a circular arc, which will intersect the parallel in the required point Q. For if we suppose 2ω to be the longitude of the required point Q, US is by construction = ω sin z, and the angle subtended by SU at C is
| tan−1 ( | ω sin z | ) = tan−1 (ω cos z) = α, |
| tan z |
and therefore UCQ = 2α as it should be. The advantages of this method are that with a remarkably simple and convenient mode of construction we have a map in which the parallels and meridians intersect at right angles.
| Fig. 22. |