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. _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. [Illustration: FIG. 20.] 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 = 2[alpha], UC´Q´ = 2([alpha] + d[alpha]), then the small angle CQC´, or the angle between the tangents at QQ´, will = 2d[alpha]. Now 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 2[alpha] = -2 tan zd [alpha]. -tan z dz = 2d[alpha] / sin 2[alpha], which is the differential equation of the meridian: the integral is tan [alpha] = [omega] cos z, where [omega], a constant, determines a particular meridian curve. The distance of Q from the central meridian, tan z sin 2[alpha], is equal to 2 tan z tan [alpha] 2[omega] sin z ------------------- = ------------------------- 1 + tan² [alpha] 1 + [omega]² cos² [alpha] [Illustration: FIG. 21.] At the equator this becomes simply 2[omega]. Let any equatorial point whose actual longitude is 2[omega] be represented by a point on the developed equator at the distance 2[omega] 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 2[omega] to be the longitude of the required point Q, US is by construction = [omega] sin z, and the angle subtended by SU at C is /[omega] sin z\ tan^(-1) ( ------------- ) = tan^(-1) ([omega] cos z) = [alpha], \ tan z /