sometimes known as Sanson’s, and is also sometimes incorrectly
called Flamsteed’s, is a particular case of Bonne’s in which the
selected parallel is the equator. The equator is a straight line
at right angles to the central meridian which is also a straight
line. Along the central
meridian the latitudes are
marked off at the true
rectified distances, and
from points so found the
parallels are drawn as
straight lines parallel to
the equator, and therefore
at right angles to the
central meridian. True
rectified lengths are
marked along the parallels
and through corresponding
points the meridians
are drawn. If the earth
is treated as a sphere the
meridians are clearly sine
curves, and for this reason
d’Avezac has given the
projection the name sinusoidal. But it is equally easy to plot
the spheroidal lengths. It is a very suitable projection for an
equal-area map of Africa.
Werner’s Projection.—This is another limiting case of Bonne’s equal-area projection in which the selected parallel is the pole. The parallels on paper then become incomplete circular arcs of which the pole is the centre. The central meridian is still a straight line which is cut by the parallels at true distances. The projection (after Johann Werner, 1468–1528), though interesting, is practically useless.
Polyconic Projections.
These pseudo-conical projections are valuable not so much for their intrinsic merits as for the fact that they lend themselves to tabulation. There are two forms, the simple or equidistant polyconic, and the rectangular polyconic.
The Simple Polyconic.—If a cone touches the sphere or spheroid along a parallel of latitude φ and is then unrolled, the parallel will on paper have a radius of ν cot φ, where ν is the normal terminated by the minor axis. If we imagine a series of cones, each of which touches one of a selected series of parallels, the apex of each cone will lie on the prolonged axis of the spheroid; the generators of each cone lie in meridian planes, and if each cone is unrolled and the generators in any one plane are superposed to form a straight central meridian, we obtain a projection in which the central meridian is a straight line and the parallels are circular arcs each of which has a different centre which lies on the prolongation of the central meridian, the radius of any parallel being ν cot φ.
So far the construction is the same for both forms of polyconic. In the simple polyconic the meridians are obtained by measuring outwards from the central meridian along each parallel the true lengths of the degrees of longitude. Through corresponding points so found the meridian curves are drawn. The resulting projection is accurate near the central meridian, but as this is departed from the parallels increasingly separate from each other, and the parallels and meridians (except along the equator) intersect at angles which increasingly differ from a right angle. The real merit of the projection is that each particular parallel has for every map the same absolute radius, and it is thus easy to construct tables which shall be of universal use. This is especially valuable for the projection of single sheets on comparatively large scales. A sheet of a degree square on a scale of 1:250,000 projected in this manner differs inappreciably from the same sheet projected on a better system, e.g. an orthomorphic conical projection or the conical with rectified meridians and two standard parallels; there is thus the advantage that the simple polyconic when used for single sheets and large scales is a sufficiently close approximation to the better forms of conical projection. The simple polyconic is used by the topographical section of the general staff, by the United States coast and geodetic survey and by the topographical division of the U.S. geological survey. Useful tables, based on Clarke’s spheroid of 1866, have been published by the war office and by the U.S. coast and geodetic survey.
 |
| Fig. 20. |
Rectangular Polyconic.—In this the central meridian and the parallels are drawn as in the simple polyconic, but the meridians are curves which cut the parallels at right angles.
In this case, let P (fig. 20) be the north pole, CPU the central meridian, U, U′ points in that meridian whose co-latitudes are z and z+dz, so that UU′ = dz. Make PU = z, UC = tan z, U′C′ = tan (z + dz); and with CC′ as centres describe the arcs UQ, U′Q′, which represent the parallels of co-latitude z and z + dz. Let PQQ′ be part of a meridian curve cutting the parallels at right angles. Join CQ, C′Q′; these being perpendicular to the circles will be tangents to the curve. Let UCQ = 2α, UC′Q′ = 2(α + dα), then the small angle CQC′, or the angle between the tangents at QQ′, will = 2dα. Now
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
−tan zdz = 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. |
Fig. 22 is a representation of this system of the continents of Europe and Africa, for which it is well suited. For Asia this system would not do, as in the northern latitudes, say along the parallel of 70°, the representation is much cramped.
With regard to the distortion in the map of Africa as thus constructed, consider a small square in latitude 40° and in 40° longitude east or west of the central meridian, the square being so placed as to be transformed into a rectangle. The sides, originally unity, became 0.95 and 1.13, and the area 1.08, the diagonals intersecting at 90° ± 9° 56′. In Clarke’s perspective projection a
