Thus treating the earth as a sphere and applying the _Zenithal Equal-area Projection_ to the case of Africa, the central point selected being on the equator, we have, if [theta] be the spherical distance of any point from the centre, [phi], [alpha] the latitude and longitude (with reference to the centre), of this point, cos [theta] = cos [phi] cos [alpha]. If A is the azimuth of this point at the centre, tan A = sin [alpha] cot [phi]. On paper a line from the centre is drawn at an azimuth A, and the distance [theta] is represented by 2 sin ½[theta]. This makes a very good projection for a single-sheet equal-area map of Africa. The exaggeration in such systems, it is important to remember, whether of linear scale, area, or angle, is the same for a given distance from the centre, whatever be the azimuth; that is, the exaggeration is a function of the distance from the centre only. _General Theory of Conical Projections._ Meridians are represented by straight lines drawn through a point, and a difference of longitude [omega] is represented by an angle h[omega]. The parallels of latitude are circular arcs, all having as centre the point of divergence of the meridian lines. It is clear that perspective and zenithal projections are particular groups of conical projections. [Illustration: FIG. 24.] Let z be the co-latitude of a parallel, and [rho], a function of z, the radius of the circle representing this parallel. Consider the infinitely small space on the sphere contained by two consecutive meridians, the difference of whose longitude is d[mu], and two consecutive parallels whose co-latitudes are z and z + dz. The sides of this rectangle are pq = dz, pr = sin z d[mu]; in the projection p´q´r´s´ these become p´q´ = d[rho], and p´r´ = [rho]h d[mu]. The scales of the projection as compared with the sphere are p´q´/pq = d[rho]/dz = the scale of meridian measurements = [sigma], say, and p´r´/pr = [rho]h d[mu]/sin z d[mu] = [rho]h/sin z = scale of measurements perpendicular to the meridian = [sigma]´, say. Now we may make [sigma] = 1 throughout, then [rho] = z + const. This gives either the group of _conical projections with rectified meridians_, or as a particular case the _equidistant zenithal_. We may make [sigma] = [sigma]´ throughout, which is the same as requiring that at any point the scale shall be the same in all directions. This gives a group of _orthomorphic projections_. In this case d[rho]/dz = [rho]h/sin z, or d[rho]/[rho] = h dz/sin z. Integrating, [rho] = k(tan ½z)^h, (i.) where k is a constant. Now h is at our disposal and we may give it such a value that two selected parallels are of the correct lengths. Let z1, z2 be the co-latitudes of these parallels, then it is easy to show that log sin z1 - log sin z2 h = ------------------------- (ii.) log tan ½z1 - log tan ½z2