'is to be drawn. When When pm is greater than pa, the ellipse | touches the circle in two points; these points divide pe Pop, and vP, vP' cutting Ir in pp': these are the projections ellipse into two parts, one of which, being on the other side all the circles representing of the meridian plane aqr, is invisible. the of the poles, through which Stereographic Projection.-In this case the point of visiou is on the surface, and the projection is made on the plane of the great circle whose pole is V. Let kplV (fig. 12) be a great circle through the point of vision, and ors the trace of the plane of projection. Let c be the centre of a small circle whose radius is cp=cl; the straight line pl represents this small circle in orthographic projection. Fig. 12. We have first to show that the stereographic projection of the small circle pl is itself a circle; that is to say, a straight line through V, moving along the circumference of pl, traces a circle on the plane of projection ors. This line generates an oblique cone standing on a circular base, its axis being cV (since the angle pVc-angle cV); this cone is divided symmetrically by the plane of the great circle kpl, and also by the plane which passes through the axis Vc, perpendicular to the plane kpl. Now Vr.Vp, being =Vosec kVp.Vk cos kVpVo-Vk, is equal to Vs-VI; therefore the triangles Vrs, Vlp are similar, and it follows that the section of the cone by the plane rs is similar to the section by the plane pl. But the latter is a circle, hence also the projection is a circle; and since the representation of every infinitely small circle on the surface is itself a circle, it follows that in this projection the representation of small parts is (as we have before shown) strictly similar. Another inference is that the angle in which two lines on the sphere intersect is represented by the same angle in the projection. This may otherwise be proved by means of fig. 13, where Vok is the diameter of the sphere passing through the point of vision, fgh the plane of projection, kt a great circle, passing of course through V, and ouv the line of intersection of these two planes. A tangent plane to the surface at t cuts the plane of projection in the line rus perpendicular to ov; tv is a tangent to the circle kt at t, tr and ts are Fig. 13. 9 any two tangents to the surface at t. Now the angle vtu (u being the projection of t) is 90°-otV=90°-oVt-ouV-tuv, therefore tv is equal to uv; and since tes and uvs are right angles, it follows that the angles vts and vus are equal. Hence the angle ris also is equal to its projection rus; that is, any angle formed by two intersecting lines on the surface is truly represented in the stereographic projection. We have seen that the projection of any circle of the sphere is itself a circle. But in the case in which the circle to be projected passes through V, the projection becomes, for a great circle, a line through the centre of the sphere; otherwise, a line anywhere. It follows that meridians and parallels are represented in a projection on the horizon of any place by two systems of orthogonally cutting circles, one system passing through two fixed points, namely, the poles; and the projected meridians as they pass through the poles show the proper differences of longitude. To construct a stereographic projection of the sphere on the horizon of a given place. Draw the circle vlkr (fig. 14) with the diameters kv, Ir at right angles; the latter is to represent the central meridian. Take koP equal to the co-latitude of the given place, say u; draw the diameter meridians have to pass. All optanu; op'-cotu; == d k m Fig. 14. Again, for the parallels, take Pb = Pe equal to the co-latitude, say c, of the parallel to be projected; join vb, vc cutting Ir in e, d. Then ed is the diameter of the circle which is the required projection; its centre is of course the middle point of ed, and the lengths of the lines are od-tan (u-c); oe-tan (u+c). The line sn itself is the projection of a parallel, namely, that of which the co-latitude c=180°-u, a parallel which passes through the point of vision. A very interesting connexion, noted by Professor Cayley, exists between the stereographic projection of the sphere on a meridian plane (ie., when a point on the equator occupies the centre of the drawing) and the projection on the horizon of any place whatever. The very same circles that represent parallels and meri. dians in the one case represent them in the other case also. In fig. 15, abs being a projection in which an equatorial point is in the centre, draw any chord ab perpendicular to the centre meridian cos, and on ab as diameter describe a circle, when the property referred to will be observed. This smaller circle is now the stereographic projection of the sphere on the horizon of some place whose co-latitude we may call u. The radius of the first circle being unity, let ac=sinx, then by what has been proved above co=sina cotu=cosx; therefore u=x, and ac=sinu. Although the meridian circles dividing the 360° at the pole into equal angles must be actually the same in both systems, yet a parallel circle whose co-latitude-is e in the direct projection abs belongs in the oblique system to some other co-latitude as c'. To determine the connexion between c and c', consider the point t (not marked), in which one of the parallel circles crosses the line soc. In the direct system, being the pole, pt-1-tan (90° — c) · and in the oblique, 2 1+cot c Fig. 15. law of distance, as p=f(u). We may thus avoid the calcu- | This vanishes when h=1, that is, if the projection be stereolation of all the distances and azimuths (with reference to graphic; or for u= =0, that is, at the centre of the map. the selected centre point) of the intersections of meridians At a distance of 90° from the centre, the greatest alteration and parallels. Construct a stereographic projection of the is 90°-2 cot-1/h (See Philosoph. Mag., April 1862.) globe on the horizon of the given place; then on this pro The constants h and k can be determined, so that the jection draw concentric circles (according to the stereo- total misrepresentation, viz., graphic law) representing the loci of points whose distances from the centre are consecutively 5°, 10°, 15°, 20°, &c., up to the required limit, and a system of radial lines at intervals of 5° Then to construct any other projection,-commence by drawing concentric circles, of which the radii are M = √( ® {(~ ~ 1)2 + (o' − 1)o} sinudu, shall be a minimum, ß being the greatest.value of u, or the spherical radius of the map. On substituting the expressions for σ and of the integration,is effected without difti culty. Put 1λ= - cos B h+cos B ; v=(h−1)x, H=-(+1) log. (x+1), H'=2 (2− v + dv3). FIG. 16.-Stereographic Projection. previously calculated by the law p=f(u), for the successive values of u, 5°, 10°, 15°, 20°, &c., up to the limits as before, and & system of radial lines at intervals of 5°. This being completed, it remains to transfer the points of intersection from the stereographic to the new projection by graphic interpolation. We now come to the general case in which the point of the great circle section of the of cm. a 9 Fig. m Let the angle com-u, Ve=k, Vo=h, ef=p; then, since ef: eV mg: gV, = dh Therefore M = 4' sin2 §ẞ – H, and 1⁄2 must be determined sɔ as to make H2: H'a maximum. In any particular case this maximum can only be ascertained by trial, that is to say, log H2-log H' must be calculated for certain equidistant values of h, and then the particular value of h which corresponds to the required maximum can be obtained by interpolation. Thus we find that if it be required to make the best possible perspective representation of a hemisphere, the values of h and it are h=η47 and k=2·034; so that in this case 2.034 sin u For a map of Africa or South America, the limiting radius B we may take as 40°; then in this case 2.543 sin u P=1-625+cos u For Asia, B-54, and the distance h of the point of sight in this case is 1.61. Fig. 18 is a map of Asia having the meridians and parallels laid down on this system. Figure 19 is a perspective representation of more than a | The co-ordinates originating at the centre, take the central hemisphere, the radius B being 108°, and the distance h of the point of vision, 1.40. The co-ordinates xy of any point in this perspective may be expressed in terms of the latitude and longitude of the corresponding point on the sphere in the following manner. meridian for the axis of y and a line perpendicular to it for the axis of x. Let the latitude of the point G, which is to occupy the centre of the map, be y; if p, w be the latitude and longitude of any point P (the longitude being reckoned from the meridian of G), u the distance PG, and u the ω FIG. 19.-Twilight Projection. azimuth of P at G, then the spherical triangle whose sides for a map of Africa, which is included between latitudes 40° are 90°-y, 90°-4, and u gives these relations north and 40° south, and 40° of longitude east and west of a central meridian. Values of x and y. w=10° w=20° w=30° w=40° 00.0 =2 18.67 28.07 37.53 19.87 20:43 21.25 9.69 y= 0.00 0.00 10° 20° 2=0'00 y= 9.69 x= 0.00 y=19.43 9.32 30° 40° x= 0·00 y = 29 ·25 x= 0.00 y=39.17 19.54 Conical Development. The conical development is adapted to the construction of maps of tracts of country of no great extent in latitude but any extent in the direction of a parallel. Selecting R H 10 h Fig. 20. that PHPK, each being made equal to the arc PQ. It is clear, then, that the surface generated by HK is very nearly coincident with the surface generated by RQ when the figure rotates round ON through any angle, great or small. The approximation of the surfaces will, however, be very close only if QR is very small. Suppose, now, that the paths of H and K, as described in the revolution round ON, are actually marked on the surface of the cone, as well as the line of contact with the sphere. And further, mark the surface of the cone by the intersections with it of the meridian planes through OV at the required equal intervals. Then let the cone be cut along a generating line and opened out into a plane, and we shall have a representation as in fig. 21 of the spherical surface contained between the latitudes of Q and R. The parallels here are represented by concentric circles, the meridians by lines drawn through the common centre of the circles at equal angular intervals. Taking the radius of the sphere as unity, and being the latitude of P, we see that VP=cot 4, and if w be the difference of longitude between two meridians, the corresponding length of the arc Pp is w cos p. The angle between these meridians themselves is w sin p. P Fig. 21. Suppose, now, we require to construct a map on this principle for a tract of country extending from latitude -m to +m, and covering a breadth of longitude of 2n, m and n being expressed in degrees. In fig. 21 let HKkh be the quadrilateral formed by the extreme lines, so that HK-hk-2m; then the angle HVh is 2n sin o expressed in degrees. Now, taking the length of a degree as the unit, VP 57.296 cot 4, and VH = 57 296 cot - m. It may be convenient in the first instance to calculate the chords Hh, Kk, and thus construct the rectilinear quadrilateral HKkh, The lengths of these chords are = of the centre meridian and centre parallel a line perpendicular to the meridian and therefore touching the parallel. Let the coordinate x be measured from the centre along this line, and y perpendicular to it. Then the coordinates of a point whose longitude measured from the centre meridian is w are x= cot & sin (∞ sin ), y=2cot & sin3(w sin 4)=x tan }(∞ sin 4), the radius of the sphere being the unit; if a degree be the unit, these must be multiplied by 57.296. The great defect of this projection is the exaggeration of the lengths of parallels towards either the northern or southern limits of the map. Various have been the devices to remedy this defect, and amongst these the following is a system very much adopted. Having subdivided the central meridian and drawn through the points of division the parallels precisely as described above, then the true lengths of degrees are Fig. 22. set off along each parallel; the meridians, which in this case become curved lines, are drawn through the corresponding points of the parallels (fig. 22). This system is that which was adopted in 1803 by the "Dépôt de la Guerre" for the map of France, and is there known by the title "Projection de Bonne." It is that on which the Ordnance Survey map of Scotland on the scale of one inch to a mile is constructed, and it is frequently met with in ordinary atlases. It is ill-adapted for countries having great extent in longitude, as the intersections of the meridians and parallels become very oblique-as will be seen on examining the map of Asia in most atlases. If . be taken as the latitude of the centre parallel, and co-ordinates be measured from the intersection of this parallel with the central meridian, as in the case of the conical projection, then, ifp be the radius of the parallel of latitude 4, we have pcot .+.-4. Also, if S be a point on this parallel whose co-ordinates are x, y, so that VS=p, and be the angle VS makes with the central meridian, then p✪ = w cos &; and x=psine, y=còt. -p cos 8. Now, if we form the differential coefficients of x and y with respect to and w, the latitude and longitude of S, we get which indeed might have been more easily obtained. In the case of Asia, the middle latitude 4.40°, and the exHh=2(57 296 cot p-m) sin (n sin ø), treme northern latitude is 70°. Also the map extends 90° Kk=2(57 296 cot p+m) sin (n sin p), of longitude from the central meridian; hence, at the northand the distance between them is 2m cos (n sin 4). The west and north-east corners of the map the angles of interinclined sides of this trapezoid will then meet in a section of meridians and parallels are 90° ± 33°54'. But point at V, whose distance from P and p must corre- for comparatively small tracts of country, as France or Scotspond with the calculated length of VP. Now with land, this projection is very suitable. this centre V describe the circular arcs representing the parallels through H, K, P. Also if the parallels are to be drawn at every degree of latitude, divide HK into 2m equal parts, and through each point of division describe a circular arc from the centre V. Then divide Pp into 2n equal parts, and draw the meridian lines through each of these points of division and the centre V. If the centre V be inconveniently far off, it may be necessary to construct the centre parallel by points, that is, by calculating the coordinates of the various points of division. For this purpose, draw through the intersection Another modification of the conical projection consists in taking, not a tangent cone, but a cone which, having its vertex in the axis of revolution produced, intersects the sphere in two parallels,-these parallels being approximately midway between the centre parallel of the country and the extreme parallels. By this means part of the error is thrown on the centre parallel which is no longer represented by its true length, but is made too small, while the parallels forming the intersections of the cone are truly represented in length. The exact position of these particular parallels may be PHP C m ence of longitude of which is du, and two consecutive = = = prhpdp pr sin udμ' whence d= =h du sin u and integrating, p=k(tan 1)*, Fig. 24. T determined so as to give, upon the whole, the least amount | sphere contained by two consecutive meridians the differ of exaggeration for the entire map. This idea of a cutting cone seems to have originated1 with the celebrated Gerard Mercator, who in 1554 made a map of Europe on this principle, selecting for the parallels of intersection those of 40° and 60°. The same system was adopted in 1745 by Delisle for the construction of a map of Russia. Euler in the Acta Acad. Imp. Petrop., 1778, has discussed this projection and determined the conditions under which the errors at the northern extremity, at the centre, and at the southern extremity of a map so constructed shall be severally equal. Let c, c' be the co-latitudes of the extreme northern and southern parallels, y, y' those of two intermediate parallels, which are to be truly represented in the projection. Let OC', Om' (fig. 23) be two consecutive meridians, as represented in the developed cone; the difference of longitude being w, let the G angle at O be hw. The degrees along the meri- c' dian being represented by their proper lengths, CC'-c'-c; and P corresponding to the pole, let Fig. 23. OP=2, then OC=z+c; and so for G, G', C. The true lengths of G'n' and Gn, namely, w siny and w sin y, are equal to the represented lengths, namely, hw (z+y) and ha (z+y) respectively, whence y and y' are known when h and z are known. Comparing now the represented with the true lengths of parallel at the extremities and at the centre, if e be the common error that is to be allowed, then e=hw(z+c) - w sin c, •= − h@(x+jc++c′)+w sin }(c+c), G n' m' The difference of the first and third gives h, and then subtracting the second from the mean of the first and third, we get where the constant h is to be determined according to the where a is the radius of the sphere. It is a minimum re-when u = = cos-1 h. This minimum should occur in the vicinity of the central parallel of the map; if u be the colatitude of this parallel, we may put z+1(c+c)= {(c' − c) cot }(d−c) tan }(c+c). Thus z being known, the common centre of the circles presenting the parallels is given. The value of h is given by the equation h(cc) sin c' - sin and γ and y can be easily computed. But there is no necessity for doing this as we may construct the angles at O, which representing a difference of longitude o are in reality equal to hw. For instance, to construct a map of Asia on this system, having divided the central meridian into equal spaces for degrees, z must be calculated. Here we have c= 20°, 80°, whence z +50° 15° tan 50° cot 15° 66° 7. Hence in this case the centre of the circles is 16° 7 beyond the north pole; also h='6138, so that a difference of longitude of 5° is represented at O by an angle of 3° 4' 9". The degrees of longitude in the parallel of 70° are in this map represented too large in the ratio of 1.150:1; those in the mid-latitude of 40° are too small in the ratio of 0.933: 1; and those in 10° latitude are too large in the ratio of 1.05 to 1. Gauss's Projection = 0 c' = may be considered as another variation of the conical system of development. Meridians are represented by lines drawn through a point, and a difference of longitude is represented by an angle ho, as in the preceding case. The parallels of latitude are circular arcs, all having as centre the point of divergence of the meridian lines, and the law of their formation is such that the representations of all small parts of the surface shall be precisely similar to the parts so represented. Let u be the co-latitude of a parallel, and p, a function of u, the radius of the circle representing this parallel. Consider the infinitely small space on the See page 178 of Traité des Projections des Cartes Géographiques, by A. Germain, Paris, an admirable and exhaustive essay. See also the or entitled Coup d'oeil historique sur la Projection des Cartes de Géographie, by M. d'Arozac, Parig 1863. Having drawn a line representing the central meridian, and selected a point on it as the centre of the concentric circles, let arcs be described with the above radii as parallels. For meridians, in this system a difference of longitude of 10° is represented by an angle of two-thirds that amount, or 6° 40'. The chord of this angle on the parallel of 10°, whose radius is 92.801, is easily found to be 10-792. Now stepping this quantity with a pair of compasses along the parallel, we have merely to draw lines through each of the points so found and the common centre of circles. The points of division of the parallel may be checked by taking the chord of 20°, or rather of 13° 20′, 2 Beiträge zum Gebrauche der Mathematik und deren Anwendung, vol. iii. p. 55, Berlin, 1772. |