great circle Pq continued, will cut an arc off FI=2 CP9= 2PCq= 2QCp, by Prop. 2. That is, since Fn represents the part cut off from FI, the arc Fn equals the arc lh, or <Q Ch. Q. E. D. Cor. Hence, if from the projected pole y of any circle, a perpendicular be erected to the line of measures, it will cut off a quadrant from the represen. tation of that circle. For that perpendicular will be parallel to OS, Q being at an infinite distance. Prop. 8. Let Fnk (fig. 6) be the projection of any circle FI, and p the projection of the pole P. And if Čg be the co-tangent of CAP, Cg and gВ perpendicular to the line of measures gC, and CAP be bisected by AO, and the line oB be drawn to any point B, and also PB cutting Fnk in d, then the angle goB will equal the arc Fd. For the arc FG is a quadrant, and the goA=_gpA+ Loop= (because GCA and gap are right angles) gAC+oAp=gAC+CAO =gAo. Therefore gA=go; consequently, o is the dividing centre of gB, the representation of GA, whence, by Prop. 5, <goB is the measure of gB. But, since pq represents a quadrant, p is the pole of gB, and, therefore, the great circle pdB passing through the pole of the circles, gВ and Fn, will cut off equal arcs in both; that is, Fd=gB=_goB. Q. E. D. Cor. The goB is the measure of the angle gpB. For the triangle gpB represents a triangle on the sphere, wherein the arc, which gВ represents, is eqnal to the angle which <p represents, because gp is 90°. Therefore, goB is the measure of both. N.B. The practical part, on wbich we are about to enter, depends entirely on what has been stated in the theory now finished, which being understood, no difficulty can arise. In the Gnomonical representation, the plane of projection is supposed to touch the hemisphere, which is to be projected, in its vertex, the point of contact being the centre of projection. But if it is required to make the representation npon any plane parallel to the touching plane, the process will be exactly similar, by employing a greater or lesser radius, according to the greater or lesser distance. * This is the same in effect as projecting a greater or lesser sphere upon its touching plade. When it is required to project the sphere, gnomonically, upon a given plane, it will assist the imagination, to suppose yourself placed in the centre of the sphere, with your face towards the plane, whose position is given, and from thence projecting, with the eye, the circles of the sphere upon this plane. Prop. 9. To draw a great circle, through a given point, and at a given distance from the pole of projection. With the radius of projection describe the circle ADB (fig. 7), and through the given point P, draw the right line PCA, and CE perpendicular to it; make the angle CAE equal the given distar.ce of the circle from C, and through E describe the circle EFG, and through P draw the line PK touching the circle in I, then PIK is the circle required. The same by the Plain Scale. With the tangent of the distance of the circle, from the pole of projection C, describe the circle EIF and draw PK to touch the circle, and PIK is the circle required. Prop. 10. To draw a great arcle perpendicular to a given great circle, which passes through the pole of projection, and at a given distance from the said pole. Draw the primitive ADB (fig. 7). Let CI be the given circle ; draw CL perpendicular to ci, and make the angle CLI equal the given distance, through I draw KP parallel to CL for the circle required. The same by :he Scale. In the given circle CI, set off the tangent of the given distance from C to 1; throngh I draw KP perpendicular to CI, then KP is the circle re. quirod. Prop. 11. To measure any part of a great circle, or to set off any number of degrees thereon. Let EP be the great circle (fig. 8); through C draw ID per. pendicular 1o EP, and CB parallel to it. Let EBD be a circle de scribed with the radius of projection CB, make IA=IB, then A is the dividing centre of EP; consequently, drawing AP, the CIAP is equal the measure of the given arc IP. Or, if the degrees be given, make the ZIAP equal the given dogrees, which will cut off IP, the corresponding arc. The same by the Scale: Draw ICD perpendicular to EP ; apply CI to the taugents, and set off the semi-tangent of its complement from CA, and it gives the dividing centre of EP, &c. Prop. 12. To draw a great circle to make a given angle with a given great circle, at a given point ; or to measure an angle made by two great circies. Let P be the given point (fig. 1, plate 2), and PB the given great circle. Draw through P and C the centre of projection, the line PCG, to which, from C, draw CA perpendicular, and equal to the radius of projection. Draw PA and AG perpendicular to it, at G erect BD, perpendicular to GC, cutting PB in B ; draw AO bisecting the angle CAP; then, at the point o, make BOD equal the given angle, and from D draw the line DP, then BPD is the angle required. Or, if the degrees in the angle BPD be required, from the points BD draw the lines BO, DO; and the angle BOD is the measure of BPD. Cor. It an angle be required to be made at the pole, or centre of projection, equal to a given angle, this is no more than drawing two lines from the centre, making the angle reqnired. And if one great circle be required to be drawn perpendicular to another great circle, it must be drawn through its pole. Prop. 13. To project a lesser circle parallel to the primitive. With the radius of projection AC (fig. 7), and centre C, describe the primitive circle ADB, by Cor. Prop. 3, and draw ACB and GCE perpendicular to it. Set off the circle's distance from its pole from B to H, and from H to D, and draw AFD; with the radius CE describe the circle EFG required. The sunc by the Scule. With the radius CE, equal to the tangent of the distance of the circle from its pole, describe the circle EFG, for the circle required. Prop. 14. To draw a lesser circle perpendicular to the plane of projection Through the centre of projection C (fig. 9), draw its parallel great circle Ti. At C make the angle ICN, and TCO equal the distance of the given circle, from its parallel great circle TI; make CL equal to the radius of projection, and draw LM perpendicular to CL. 'Set off LM from C to V, or CM from C to F; then, through the vertex V, between the assymptotes CN and CO, describe che hyperbola WVK. Or, to the focus F, and semi-axis major CV, describe the hyperbola ; which will represent the circle required. To do the same by finding Points. Through the centre of projection C, draw the line of measures CF, and TCI perpendicular to it, draw any number of right lines CV, DE, GH, IK, &c. and PQ, RS, TW, &c. perpendicular to TI ; and, by Prop. 11, make CV, DE, GH, &c. each equal to the distance of the given circle from its parallel great circle. Then all the points, W, S, Q, V, E, H, K, &c.joined by a regular curve, will be the representation of the circle required. Or tlns, make the angle iak equal the distance of the given circle, from its parallel great circle; then, through the centre of projection C, draw the great circle TCI parallel to the circle given, npon which erect the perpendicular CA equal the radius of projection. Also draw any number of right lines CV, DE, GH, IK, &c. perpendicular to TI ; then take each of the distances froin A to C, D, G, I, &c. and set them off from a to c, &, d, i, &c, and to ui draw the perpendiculars co, de, gh, ik, &c. and make CV, DE, GH, IK, &c. respectively equal to co, de, gh, ik, &c. which will give the points V, E, H, K, &c.; after ihe same manner find on the other side the points Q, s, w, &c. then, through all these points, w, s, Q, V, E, H, K, &c. draw a regular curve, which will be au liyperbola, representing the cirole given. Prop. 15. To project a lesser oblique circle. Draw the line of measures dp(fig. 10), and at C, the centre of projection, draw CA1 to dp, and equal the radius of projection ; with the centre A describe the circle DCFG, and draw RAE parallel to dp; then take the least and greatest distances of the circle from the pole of projection, and set them off from C to D, and for the circle DF; and from A, the projecting point, draw AFf, and ADd, then df will be the major axis of the ellipsis. But if D fall beyond the line RE, as at G, then draw a line from G backward through A to D, and then df is the major axis of an hyperbola. But if the point D fall in the line RE, as at E, then the line AE meets the line of measures no where, and the projection of E is at an infinite distance, and then the circle will be projected into a parabola, having its vertex at the point f. Lastly, bisect df in H the centre, and for the ellipses, take half the difference of the lines Ad, As, and set it off from H to K for the focus. But, for the hyperbola, take balf the sum of Ad, Af, and set it off from H to the focus k of the hyperbola. Then, with the major axis df, and focus K or k, describe the ellipses dMf, or the hyperbola fm, for the projection of the given circle But, for the parabola, make EQ=Ff, and draw fn 1 AQ, and set off 2nQ from f to K the focus ; then, with the vertex f, and focus k, describe the parabola fm, for the projection of the given circle FE. To do the same by finding Points. Through the centre of projection C (fig. 11), draw the line of measures CF, passing through the pole P, if P be given; but, if not, find it by setting off CP equal the distance of that pole from the centre of projection, by Prop. 11, then set off PD, PF, equal to the given distance from its pole, by Prop. 11. Through P draw a suficient number of right lines, La, Mú, Nr, Oo, Rr, Ss, &c. which will all represent great circles. Find the dividing centres of each of these lives, and, by Prop. 11, set them off upon each of them from P, the given distance of the circle from its pole, as PL, PX, PM, Pu, &c, and through all the points L, M, N, O, R, &c. draw a curve line, for the circle required. Or thus, draw the line of measures PCG, and by Prop. 11, make CG egnal the distance of the parallel great circle from the pole of projection, and draw AGK perpendicular to it, which will represeut a great circle, te pole of which is P. Draw any number of right lines through P to AK, as AP, BP, HP, &c. ; and, by Prop. 11, set off from AK the parts AL, BM, HO, &c. each equal to the distance of the circle from its parallel great circle. Then all the points L, M, D, 0, &c. being joined by a regular curve, will represent the parallel circle required. Or thus, through the centre of projection C, draw the line of measures DCF, and the radius of projection CW perpendicular to it, and AGK1 to GC for a great circle, the pole of which is P. Draw up=WP, and want to it; draw any number of right lines AP, BP, GP, &c. and make pg, ph, pa, &c. equal PG, PB, PA, &c.; also make the pul and pux equal the distance of the circle from its pole P, or uwl equal the distance from its parallel great circle; and, upon PG, PB, PA, &c. make PD, PM, PL, &c. equal pd, pm, pl, &c. respectively, Or maké GD, BM, AL, &c. equal gd, bm, al, &c. After the same manner, find the points O, R, &c. and through all the poiuts R, O, D, M, L, &c. draw a regular curve making no angles, and it will represent the parallel circle required. Likewise, when any line ap cuts wx, that distance from p will give the point n, or is equal to P, and so of any other of the lines bp, 6P, &c. Prop. 16. To find the pole of any circle DMF. From the centre of projection C (fig. 12), draw the radius of projection CA perpendicular to the line of measures DF; and to A, the projecting point, draw DA, FA, and bisect the angle DAF by the line AP, then P is the pole. But if the curve be an hyperbola, as fm, (fig. 10,) dA must be produced, and the angle fAG must be bisected. And in a parabola, where the poiut d is at an infinite distance, bisect the angle fAE. Or thus, draw CA perpendicular to DC, draw DA, and make the angie DAP to equal the distance of the circle from its pole, and it will give the pole P. Prop. 17. To measure any arc of a lesser circle, or to set off any number of degrees thereon., Let Fn (fig. 13) be the given circle. From the centre of projec.tion C, draw CA perpendicular to the line of measures GH. TO P the pole of the given circle draw AP and AO bisecting the angle CAP. And draw AD1 to AO. Describe the circle GIH, by Prop. |