In this projection, the eye, situated at the centre of the sphere, throws every point of its surface upon that of a tangent or secant cone, the axis of which is coincident with the axis of the sphere. Let a sphere and cone be generated by the revolution of the arc, NnsES, and the straight line, vns, intersecting each other in n, s, about the common axis, vN0,0,OS, O being the centre of the sphere, N and S the poles, E a point of the equator, and no„, so, perpendiculars upon NS. 1°. For the magnitude of a parallel of latitude, putting OE=Os = On = ON = OS = radius = 1, arcA, length of an arc of A° in the = parallel through n, E n arc Ado. through s, arcA = do. through E; Fig. 131. (600) 20. For the radii, vn, vs, with which to describe the sectoral surface, snNns,S, the development of the conical surface upon a plane, we have, 3o. To find the number of degrees in the angle, svs, corresponding to a difference of latitude of A°; with the trigonometrical radius, vr = 1, describing the arc, rr2, there results, Now, to make the plane of projection approximate the spherical surface as much as may be conveniently done, take n and s so that these and the middle point of the map shall divide its meridian equally,, or as nearly so as the degrees of latitude or their aliquot parts will permit. Calculate the radii, vn, vs (601), and find the degrees, v° (602), corresponding to the degrees of longitude, A°, to be embraced by the map; from a scale of the required magnitude, lay off the meridian, NS vs - vn, and produce it in v, about which as a centre, and with the radii, vn, vs, taken from the same scale, describe the arcs, nNn, sSs,, and, having made the arc, rr2 = v°, and drawn vns, vn2s2, divide the arc, nNn, into A parts; through these points of division and corresponding ones in NS and NS produced, draw the remaining meridians and parallels of latitude. = 4 To find the half chords, cn, cs, and their altitudes, Nc, Sc. nc : nv = sino sin90°, Forms (603), (604), (605), enable us to make the construction with accuracy when the point, v, is so distant as not to admit of employing the radius, vn, by finding the successive values of v° (602) for A = 1o, 2o, 3o, .... Scholium I. This projection is well adapted to the construction of maps of moderate extent from north to south; and when a greater number of degrees of latitude are to be embraced, it will also become quite accurate by regarding the conical surface as composed of several placed end to end, joining each other in parallels little distant, as 10, 20, or 30'. Scholium II. When the points, n, s, are situated at equal distances on opposite sides of the equator, E, the cone becomes a cylinder; for we have, 11 = -12 .. (601), radius vn = and (605), NS2 sinl Scholium III. When the points, n, s, are taken on opposite sides of the pole, N, and at equal and moderate distances, the projection becomes a circular plane, well calculated to represent the circumpolar regions, either of the earth or the heavens. EXERCISES. 1°. Construct a map of the sphere as it would appear to an observer elevated to the zenith of lat. 43° N., lon. 78° W. 2o. Project the sphere as, supposing it transparent, its opposite surface would appear on the rational horizon to an eye situated in the surface of the place just given. 3°. Make a conical projection of the State of New York and the countries extending a degree or two on the north and south, assuming the centre of the map near Syracuse, and calculating the radii to each degree of latitude. 4°. Prove that the section made by a plane through the vertex of a right circular cone, is an isosceles triangle, passing into a straight line by a revolution of the plane. 5°. Show that if a plane cut a right circular cone parallel to its side, the section will be a parabola, vanishing in a straight line. 6°. The truncation of a right circular cone will be found to be an ellipse, passing on the one side into a circle and a point, and on the other into a parabola. 7°. If a right circular cone be cut by a plane not passing through the vertex, not truncating it, nor parallel to its side, the section will be a hyperbola, whose limits will be a straight line or parabola. BOOK THIRD. NAVIGATION. SECTION FIRST. Problem of the Course. The earth is an oblate spheroid, formed by the revolution of an ellipse about its minor axis, the polar diameter being 7899 170 and the equatorial 7925 648 English miles; but, for the purposes of Navigation, it may be regarded as a sphere with a radius of 3437.75 geographical miles or minutes. If a ship could be conveniently guided in the arc of a great circle, which would be the most direct path, nothing would be easier than, by Spherical Trigonometry, to determine her course from oue point on the earth's surface to another, or, having given the point of departure and the distance run, to find her place. But as, in practice, the vessel is directed by the magnetic needle so as to cut all the meridians at a constant angle, the path actually described is a curve of double curvature, denominated the Rhumb Line, or Loxodromic Curve. PROPOSITION I To find the difference of latitude. Let a ship sail from X, to X and describe the rhumb z, intersecting the meridians x1, x, x +h, at the constant angle c = course. P is the pole, 1 the polar distance of X1, ≈ that of X, and y (x1,x) = the difference of longitude; k, h, i, are the vanishing increments of y, z, z, and j is a corresponding vanishing arc described about X Fig. 132. 292 DIFFERENCE OF LATITUDE AND LONGITUDE. P with the radius, x, which may therefore be regarded as the arc of a great circle perpendicular to x + h. It is required to find the function x = pz. = [ + ] = sin(90° – c) sin90° constant, = COSC X1=Xz_0= constant; or Difference of Latitude = Distance X cosine of Course. (606) PROPOSITION II. To find the difference of Longitude. Let y = 4x be the function required; there results, But, in order to adapt this form to the purposes of computation, we must convert the Napierian into common logarithms, and adopt, as is customary, for the unit of measure the nautical mile or minute instead of the earth's radius, which will be done by multiplying by 1 M = 230258509 and 3437'75; and there results, |