: tive. Stereogra- for the angle AML=AKF. But since A is the pole =2NHG=NCG: hence ENC=INE+INC=NCG Stereogra phic Pro. BLD, the chords, and consequently the arches AB +INC = a right angle ; and therefore NC is a tan- phic Projection of AL, are equal, and the arch ABG is the sum of the gent to the primitive at N; but the arch ND is the jection of the Sphere. arches AL, BG; hence the angle AML is equal to an distance of the less circle from its nearest pole D: the Sphere. PROPOSITION VI. THEOREM VI. The projection of the poles of any circle, inclined to the primitive, are, in the line of measures, distant from 1. An angle contained by any two circles of the the centre of the primitive, the tangent, and cotansphere is equal to the angle formed by their projec- gent, of half its inclination. tions. For the tangents to these circles on the sphere Let MN (fig. 6.) be a great circle perpendicular to Fig. 6. are projected into straight lines, which either coincide the primitive ABCD, and À the projecting point; then with, or are tangents to, their projections on the primi. P, p are the poles of MN, and of all its parallels m n, &c. Let AP, A p meet the diameter BD in Ff, 2. An angle contained by any two circles of the which will therefore be the projected poles of MN and sphere is equal to the angle formed by the radii of their its parallels. The angle BEM is the inclination of the projections at the point of intersection. circle MEN, and its parallels, to the primitive : and because BC and MP are quadrants, and MC common PROPOSITION IV. THEOREM IV. to both ; therefore PC=BM: and hence PEC is also the centre of the primitive ; the tangent of the incli. the tangent of EAF, or of half the angle PEC the in. but EA f is the complement of EAF, and Ef is the cotangent of balf the inclination. COROLLARIES. PROPOSITION VII. THEOREM VII. Equal arches of any two great circles of the sphere on the sphere through the remote poles of those great Let AGB, CFD (fig. 7.) be two great circles of the Fig. 7. sphere, whose remote poles are E, P; through which draw the great circle PBEC, and less circle PGE, in- tersecting the great circles AGB, CFD in the points B, G, and D, F; then the arch BG is equal to the arch DF. Because E is the pole of the circle AGB, and P the pole of CFD, therefore the arches EB, PD are equal; and since BD is common to both, hence the arch ED arches EF, PG are equal ; but the angle DEF is equal to the angle BPG : bence these triangles are tangent of that distance. equal, and therefore the arch DF is equal to the arch BG. 5. 5. Let MN (fig. 5.) be the given less circle perpendi- PROPOSITION VIII. THEOREM VIII. t 3 H OB Fig. 9. Stcreogra On the plane of projection AGB (fig. 7.) let the straight line EGH, meeting the diameter AB produ. Stereo phic Pro- great circle CFD be projected into cfd, and its pole P ced if necessary in H; then from the centre H, with pbic i'ro. jection of into p; through p draw the straight lines pd, pf, then the radius HE, describe the oblique circle DIE, and it jection et the Sphere. are the arches GB, f d similar. will be the projection of the great circle required. PROPOSITION XI. PROBLEM III. To find the poles of a great circle. centre is the pole. 2. To find the pole of the right circle ACB (fig. 11.). COROLLARY. Draw the diameter PE perpendicular to the given circle Hence, if from the angular point of a projected sphe. AB; and its extremities P, E are the poles of the circle rical angle two straight lines be drawn through the ACB. projected poles of the containing sides, the intercepted 3. To find the pole of the oblique circle DEF (fig. Fig. 13, arch of the primitive will be the measure of the spheri 13.). Join DF, and perpendicular thereto draw the cal angle. diameter AB, cutting the given oblique circle DEF in E. Draw the straight line FEG, meeting the cir. PROPOSITION IX. PROBLEM I. cumference in G. Make GI, GH, each equal to AD; then FI being joined, cuts the diameter AB in P, the To describe the projection of a great circle through two lower pole; through F and H draw the straight line given points in the plane of the primitive. FHp, meeting the diameter AB produced in p, which Let P and B be given points, and C the centre of will be the opposite or exterior pole. the primitive. PROPOSITION XII. PROBLEM IV. Fig. 8. 1. When one point P (fig. 8.) is the centre of the primitive, a diameter drawn through the given points To describe a less circle about any given point as a pole, will be the great circle required. and at any given distance from that pole. 2. When one point P (fig. 9.) is in the circumference of the primitive. Through P draw the diameter 1. When the pole of the less circle is in the centre PD; and an oblique circle described through the three of the primitive; then from the centre of the primitive, points P, B, D, will be the projection of the required with the semitangent of the distance of the given circle from its pole, describe a circle, and it will be the pro3. When the given points are neither in the centre jection of the less circle required. nor circumference of the primitive. Through either of 2. If the given pole is in the circumference of the Fig. 10. the given points P (fig. 10.) draw the diameter ED, primitive, from C (fig. 14.) the centre of the primitive, Fig. 14 , . and at right angles thereto draw the diameter FG. set off' CE the secant of the distance of the less circle From F through P draw the straight line FPH, meet- from its pole P; then from the centre E, with the taning the circumference in H: draw the dianieter HI, gent of the given distance, describe a circle, and it will and draw the straight line FIK, meeting ED produced be the less circle required. Or, make PG, PF each in K; then an arch, terminated by the circumference, equal to the chord of the distance of the less circle from being described through the three points, P, B, K, will ita pole. Through B, G draw the straight line BGD be the great circle. meeting CP produced in D: bisect GD in H, and draw HE perpendicular to GD, and meeting PD in PROPOSITION X. PROBLEM II. E; then E is the centre of the less circle. 3. When the given pole is neither in the centre nor To describe the representation of a great circle about circumference of the primitive. Through P (fig. 15.), Fig. 19 any given point as a pole. the given pole, and C the centre of the primitive, draw Let P be the given pole, and C the centre of the pri the diameter AB, and draw the diameter DE perpenmitive. dicular to AB; join EP, and produce it to meet the 1. When P (fig. 8.) is in the centre of the primitive, primitive in p; make pF, G, each equal to the chord then the primitive will be the great circle required. of the distance of the less circle from its pole ; join EF Fig. 11, 2. When the pole P (fig. 11.) is in the circumfe. which intersects the diameter AB in H; from E through rence of the primitive. Through P draw the diameter G draw the straight line EGI, meeting the diameter PE, and the diameter AB drawn 'at right angles to PE AB produced in I ; bisect HI in K: Then a circle dewill be the projected great circle required. scribed from the centre K; at the distance KH or KI, 3. When the given pole is neither in the centre nor will be the projection of the less circle. circumference of the primitive. Through the pole P Plate (fig. 12.) draw the diameter AB, and draw the diame PROPOSITION XIII. PROBLEM V. ccccxliv. ter DE perpendicular to AB; through E and P draw To find the poles of a given less circle. the straight line EPF, meeting the circumference in F. Make FG equal to FD; through E and G draw the The poles of a less circle are also those of its parallel 4 great great circle. ܪ . Fig. 15 Stereogra- great circle. If therefore the parallel great circle be line of chords will give the measure of the arch DE of Stercogriphic Projection of given, then its poles being found by Prob. 111. will be the given less circle. phic Prothe Sphere. those of the less circle. But if the parallel great circle jection of be not given, let HMIN (fig. 15.) be the given less PROPOSITION XVI. PROBLEM VIII. the Sphere. circle. Through its centre, and C the centre of the pri To measure any spherical angle. , is the IP ment. 3. If the angular point is neither at the centre nor PROPOSITION XIV, PROBLEM. VI. circumference of the primitive. Let A (fig. 20.) be Fig. 20. the angular point, and DAH, or GAF, the angle to To measure any arch of a great circle. be measured, P the pole of the oblique circle DAF, 1. Arches of the primitive are measured on the line and p the pole of GAH : then from A, through the of chords. points P p, draw the straight lines APM, A p N, and 2. Right circles are measured on the line of semi- the arch MN will be the measure of the angle DAH; tangents, beginning at the centre of the primitive. Thus, and the supplement of MN will be the measure of the PROPOSITION XVII. PROBLEM IX. great circle, and through a point given in it. Find the pole of the given circle, then a great circle circle be the primitive, then a diameter drawn through AB. the given point will be the required perpendicular. If Or thus : Draw the diameter GH perpendicular to the given circle is a right one, draw a diameter at right DE; then from either extremity, as D), of this diame- angles to it; then through the extremities of this diater, draw lines through the extremities of the arch in- meter and the given point describe an oblique circle, tended to be measured ; and the intercepted portion of and it will be perpendicular to that given. If the given the primitive applied to the line of chords will give the circle is inclined to the primitive, let it be represented measure of the required arch. Thus IK applied to the by BAD (fig. 21.), whose pole is P, and let A be the Fig. 21. line of chords will give the measure of AB. point through which the perpendicular is to be drawn :. PROPOSITION XVIII. PROBLEM X. another great circle to make a given angle with the former, provided the measure of the given angle is DE applied to the line of chords will give the measure not less than the distance between the given point and of the arch of the oblique circle AB. circle. Let the given circle he the primitive, and let A (fig. 19.) be the angular point. Draw the diameter A E, DF perpendicular to each other; and make the angle CAG Fig. 19. Let DEG (fig. 18.) be the given less circle, and equal to that given, or make CG equal to the tangent DE the arch to be measured : find its internal pole P; of the given angle; then from the centre G, with the and describe the circle AFI parallel to the primitive, distance GC, describe the oblique circle ABE, and it and whose distance from the projecting point may be will make with the primitive an angle equal to that equal to the distance of the given less circle from its given. pole P: then join PD, PE, wbich produce to meet If the given circle be a right one, let it be APB (fig. Fig. 22. the parallel circle in A and F. Now AF applied to a 22.) and let P be the given point. Draw the diameter GH a Fig 17. 3 H 2 Stereogra- GH perpendicular to AB; join GP, and produce it to about n as a pole, describe the great circle EDF, cut- Stereogra : n phic Pro- a; make H b equal to twice A a: and Gb being joined ting the primitive and given circle in E and D, and it phic Pro jection of intersects AB in C. Draw CD perpendicular to AB, will be the great circle required. jection of the Sphere and equal to the cotangent of the given angle to the the Spbar radius PC; or make the angle CPD equal to the com SCHOLIUM. plement of that given : then from the centre D, with the radius DP, describe the great circle FPE, and the It will bence be an easy matter to construct all the angle APF, or BPE, will be equal to that given. various spherical triangles. The reader is, however, Fig. 23. If IF APB (fig. 23.) is an oblique circle. From the referred to the article Spherical Taigonometry, for SECTION II. describe another great circle wbich shall cut the jection, at an infinite distance with respect to the dia. Plate If the given circle be a right one, let it be represent meter of the sphere; so that at the sphere all the visual ccccxLved by APC (fig. 24.); and at right angles thereto draw rays are assumed parallel, and therefore perpendicular fig. 24. the diameter BPM; make the angle BPF equal to the to the plane of projection. complement of the given angle, and PF equal to the Hence the orthographic projection of any point is tangent of the given arch ; and from the centre of the where a perpendicular from that point meets the plane primitive with the secant of the same arch describe the of projection ; and the orthographic representation of arch Gg. Through F draw FG parallel to AC, meet any object is the figure formed by perpendiculars drawn ing Gg in G; then from the centre G, with the from every point of the object to the plane of protangent PF, describe an arch no, cutting APC in 1, jection. and join GI. Through G, and the centre P, draw the This method of projection is used in the geometrical diameter HK; draw PL perpendicular to HK, and delineation of eclipses, occultations, and transits. It is IL perpendicular to GI, meeting PL in L; then L also particularly useful in various other projections, such will be the centre of the circle HIK, which is that re as the analemma. See GEOGRAPHY, &c. PROPOSITION I. THEOREM I. mitive, let it be ADB (fig. 25.), and E its centre : Let AB (fig. 27.) be the plane of projection, and Fig. 27 Eucl. the intersection EF, of the plane CĚFD, with the plane of projection, is a straight line: and because PROPOSITION XX. PROBLEM XII. the straight lines CD, EF are parallel, and also CE, Any great circle in the plane of projection being given, DF; therefore by 34. of i. of Eucl. the opposite sides to describe another great circle, which shall make gi are equal ; hence the straight line CD, and its projecven angles with the primitive and given circles. tiou EF, are equal. Again, let GH be the proposed straight line, inclined to the primitive; then the lines Let ADC (fig. 26.) be the given circle, and Q its GE, HF being drawn perpendicular to AB, the interpole. About P the pole of the primitive, describe an cepted portion EF will be the projection of GH. arch m n, at the distance of as many degrees as are in the Through G draw GI parallel to AB, and the angle angle which the required circle is to make with the pri. IGH will be equal to the inclination of the given line mitive. About Q the pole of the circle ADC, and to the plane of projection. Now GH being the radius, at a distance equal to the measure of the angle which GI, or its equal EF, will be the cosine of IGH; hence the required circle is to make with the given circle the given line GH is to its projection EF as radius to ADC, describe an arch on, cutting mn in n. Then the cosine or inclination. Fig. 20. COROLLARIES. 25 429 Orthogra tance of the parallel circle from the primitive, or the OrthograCOROLLARIES. phic pro sine of its distance from the pole of the primitive. phic pro. jection of 1. A straight line perpendicular to the plane of pro jection of the Sphere. jection is projected into a point. PROPOSITION IV. THEOREM IV. the Sphere. 2. Every straight line in a plane parallel to the pri- An inclined circle is projected into an ellipse, wbose mitive is projected into an equal and parallel straight transverse axis is the diameter of the circle. line. 3. A plane angle parallel to the primitive is projected 1. Let ELF (fig. 30.) be a great circle inclined to Fig. 30. into an eqnal angle. the primitive EBF, and EF their line of common sec4. Any plane rectilineal figure parallel to the primi- tion. From the centre C, and any other point K, in tive is projected into an equal and similar figure. EF, let the perpendicular CB, KI be drawn in the 5. The area of any rectilineal figure is to the area plane of the primitive, and CL, KN, in the plane of Let LG, ND be perpendicular to CB, KI; then G, D are the projections of L, N. And because the tri angles LCG, NKD are equiangular, CL: CGʻ:: NK*: projected into a diameter of the primitive; and every points G, D are in the curve of an ellipse, of which axis. Fig. 28. Let BFD (fig. 28.) be the primitive, and ABCD COROLLARIES. a great circle perpendicular to it, passing through its 1. In a projected great circle, the semiconjugate axis is the cosine of the inclination of the great circle to the primitive. 2. Perpendiculars the transverse axis intercept also be perpendicular to the plane of the primitive: corresponding arches of the projection and the primitherefore H is the projection of G. Hence the whole 3. The eccentricity of the projection is the sine of circle is projected into BD, and any arch AG into EH the inclination of the great circle to the primitive. equal to GI its sine. Case 2. Let AQB (hig. 31.) be a less circle, incli- Fig. 31. COROLLARIÉS. ned to the primitive, and let the great circle LBM, per pendicular to both, intersect them in the lines AB, LM. 1. Every arch of a great circle, reckoned from its in From the centre 0, and any other point N in the diatersection with the primitive, is projected into its versed meter AB, let the perpendiculars TOP, NQ, be drawn in the plane of the less circle, to meet its circumference in T, P, Q. Also from the points A, N, O, B, let every and from P, Q, T, draw PE, QD, TF, perpendicular to the primitive ; then G, I, C, H, E, D, F, are the , dicular to LBM ; therefore the common section EC is perpendicular to LBM, and to LM. Hence CP is PROPOSITION III. THEOREM III. parallelogram, and EC=OP. In like manner, FC, DI, are proved perpendicular to LM, and equal to OT, NQ. Thus ECF is a straight line, and equal to LM; then RO=NQ=DI=KC, and PR XRT-EK : COROLLARIÉS. 3. The extremities of the conjugate axis are in the line of measures, distant from the centre of the primi tive by the cosines of the greatest and least distances of The radius of the projection is the cosine of the dis- the less circle from the primitive. a Fig. 29 |