Page:Encyclopædia Britannica, Ninth Edition, v. 10.djvu/223

Rh isi.—irHi3.i.i'rie.iL.] may be made to serve the important purpose of enabling one to trace on the ordinary Mercator’s chart the track of a great circle joining any two places, and of indicating at the same time the distance of the two places. For this F$_ / /Z’'‘ /4 ./K Fig. 27. purpose the two charts must be on the same scale, one of them being on tracing paper or tracing linen. The trans- parent chart being placed over the other, the equator in the ordinary chart must coincide with the initial meridian in the meridian Mercator. Retaining this relative position, let the upper chart be moved until the two points (the pro- jection of the great circle joining which is required) on the ordinary Mercator are found to lie on a great circle of the meridian Mercator. The curvatures of the meridians and parallels in the meridian Mercator are expressed by very simple formulae. Let .1‘, 3/be the coordinates, measured from the pole along and perpendicular to the initial meridian, of any point S of the representation,—.7c corresponding to an arc of the sphere = ac, and 3/ to an are 17 which is on the sphere the distance of S from the initial meridian. Then if ac’, 3/’ be the centre of curvature of the parallel at S, ac", 3/" the centre of curvature of the meridian at S, 1 a-’:.v—tan:c, :e”=.1:+ , tan :1: y’=y-sinn, y”:y— .1. S1111] The corresponding radii of curvature are sin B+ cos .r, where‘ B is the spherical radius of the small circle, and 1 + sin x sin y, where y is the longitude of the great circle, counted from the initial meridian. Polyconic Development. Imagine a hollow globe formed of a mere surface of paper, to be cut by a system of parallel planes along equidistant GEOGRAPHY 209 parallels of latitude ; let also one meridian be cut through, from north pole to south pole, 180°. In this state let the whole be opened out into a plane from the meridian exactly opposite to the one cut through, and the previously spherical surface is converted into a number of strips of paper, each of which is part of a circular belt, with the ex- ception of the equator, which will be straight. All points which lay on the parallel whose co—latitude is we now lie on an arc of a circle whose radius is tan u and length 212-sinu ; moreover, the centres of these arcs lie in the same straight line, which is the central meridian produced. The parallels being now deﬁned, we must deﬁne meridians. These may be formed by laying off on each parallel the degrees of longitude according to their true lengths, which is the system 0' adopted in the maps of the United C States Coast Survey. Or we may take for meridians that system of lines which cuts the parallels at right angles, forming the rectangular poly- P conic system. In this case, let P (fig. 28) be the north pole, CPU the central meridian, U, U’ points in that meridian whose co-latitudes are ac and u+du, so that UU’:du. Make PU :u, UC:tan u, U’C’:tan (u+du); and with CC’ as centres describe the arcs F. r ‘,8 UQ, U’Q', which represent the parallels of '5' “ ' co-latitude u and u+clw. 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:°¢, UC’Q’:2(¢+d¢), then the small angle CQC', or the angle between the tangents at QQ’, will:‘.2d¢. Now CC’:C’U' —— CU — UU'=tan (zc +dw) - tan u — duztan “’u, du; and in the triangle CC’Q the perpendicular from C on C’Q’ is equal to either side of the equation tanﬂu. the sin 2:1): —tan u (143. 2(l¢ EEK ’ Lo, L: Q, — tan to (Zn: which is the diﬂ'erential equation of the meridian: the integral is tan :1): (0 cos it, where w, a constant, determines a particular meridian curve. The distance of Q from the central meridian, tan u sin 2:1), is equal to i_ tan u tap _¢ _ 2w sin it 1 + tan‘-‘¢ 1 + w‘-’ cos‘-‘u. At the equator this becomes simply 20.». Let any equatorial point whose actual longitude is 2:.» be represented by a point on the de- veloped equator at the distance 2w fi'oin the central meridian, then we have the following very simple con- struction (due to Mr O’Farrell of the Ordnance Survey Ofiiee). Let P (ﬁg. 29) be the pole, U any point in the central meridian, QUQ’ the re- presented parallel whose radius CU:tan 25. Draw SUS’ perpen- dicular to the meridian through U; then to determine the point Q, whose longitude is, say, 3°, lay oil" US equal to half the true length of the arc of parallel on the sphere, F13" 29- 'z'.c., 1° 30' to radius sin u, and with the centre S and radius SU describe a circular are, which will intersect the parallel in the required point Q. For if we suppose 2m to be the longitude of the required point Q, US is by C01)Stl‘uCti01)=w sin 21, and the angle subtended by SU at C is w simo tan ‘it and therefore UCQ:2¢, as it should be. The advantages of this method are that with a rcniarlc-ably simple and convenient mode ot construction we have a map in which the parallels and meridians intersect at right angles. The following table contains the lengths of the radii for describing parallels, and also the lengths of degrees of loii- gitude for every 5° of latitude,—the radius of the sphere being 57'296. tan‘1 ):tan‘1(weosz:¢, X.—27