op = tan 1⁄2u; op′ = cot 1⁄2u; om = cot u; mn = cosec u cot ω.
Again, for the parallels, take Pb = Pc equal to the co-latitude, say c, of the parallel to be projected; join vb, vc cutting lr in e, d. Then ed is the diameter of the circle which is the required projection; its centre is of course the middle point of ed, and the lengths of the lines are
od = tan 1⁄2(u − c); oe = tan 1⁄2(u + c).
The line sn itself is the projection of a parallel, namely, that of which the co-latitude c = 180° − u, a parallel which passes through the point of vision.
Notwithstanding the facility of construction, the stereographic projection is not much used in map-making. It is sometimes used for maps of the hemispheres in atlases, and for star charts.
External Perspective Projection.—We now come to the general case in which the point of vision has any position outside the sphere. Let abcd (fig. 10) be the great circle section of the sphere by a plane passing through c, the central point of the portion of surface to be represented, and V the point of vision. Let pj perpendicular to Vc be the plane of representation, join mV cutting pj in f, then f is the projection of any point m in the circle abc, and ef is the representation of cm.
Let the angle com = u, Ve = k, Vo = h, ef = ρ; then, since ef: eV = mg : gV, we have ρ = k sin u/(h + cos u), which gives the law connecting a spherical distance u with its rectilinear representation ρ. The relative scale at any point in this system of projection is given by
| σ = dρ / du, σ′ = ρ / sin u, |
| σ = k (1 + h cos u) / (h + cos u)2; σ′ = k / (h + cos u), |
the former applying to measurements made in a direction which passes through the centre of the map, the latter to the transverse direction. The product σσ′ gives the exaggeration of areas. With respect to the alteration of angles we have Σ = (h + cos u) / (l + k cos u), and the greatest alteration of angle is
| sin−1 ( | h − 1 | tan2 | u | ). |
| h + 1 | 2 |