will be the axis of CD; and m, n, its poles. If EM be joined, the pole m will appear at min the primitive; but pm is the tangent of the angle Pem, or the semi-tangent of ePM, or of its equal CPF. Likewise a line drawn from E through n, the other pole will cut PF, produced beyond F, in the exterior projected pole. Now PF, produced to meet en, is evidently the tangent of the angle Nee, or the semi-tangent of nie, which is the supplement of the arc em or of the arc CF. (Z) COROLLARY. It is also obvious that (AP) the distance of the projected circle from the centre of the primitive, is the semitangent of the complement of (CF) the circle's inclination to the primitive. For Ap is the tangent of the angle AEP, which is the half of cpe, the complement of FPC. PROPOSITION IV. (Plate IV. Fig. 12.) (A) The distance of the centre of a small circle (perpendicular to the primitive) from the centre of the primitive, is equal to the secant of its distance from its own pole ; and the radius thereof is equal to the tangent of that distance. Let cd be the diameter of a small circle cutting the primitive at right angles, or, which amounts to the same thing, whose poles G and r lie in the circumference of the plane of projection. Now AB is the diameter of its representative on the plane of projection fg; bisect AB in p, and describe the small circle ACBD ; draw pc, CP, and eCB; then I say pc the radius of the small circle, is the tangent of the arc GC, the distance of the small circle from its pole g; and pp is equal to the secant of the same arc. The angle pea is equal to the angle CBA, because Ece being in a semi-circle is a right angle, consequently ECB is also a right angle; the angle EPA is a right angle, and paE=CAB, therefore the remaining angle PEA=CBA,=peB, for pe=pB; but the angle PCA is equal to PEA, because pc is equal to PE, hence PCA=Pcb, therefore the angle acB is equal to the angle pop, but ACB, is a right angle, therefore pop is a right angle, and pc a tangent to the point c, or arc ge; hence pp is the secant thereof. PROPOSITION V. (Plate IV. Fig. 13.) (B) The tangent of any arc of a great circle on the surface of the sphere, drawn from any point towards the plane of projection, will be projected into an equal straight line ;-the secant of the same arc will be represented by a line équal to itself: and any angle comprehended between two great circles, passing through the same point, on the surface of the sphere, is equal to the angle comprehended between their representatives on the plane of projection. If the sphere be cut in halves by a plane ABCD passing through its centre, and extending to an indefinite distance on all sides, in an horizontal direction; and if from any point whatever (except that point exactly perpendicular to the plane) any number of tangents be drawn towards the plane, they will all fall upon it. For, if this plane be supposed to coincide with the plane of projection, or, which is the same thing, to be perpendicular to the projecting point E, it is evident that the secant PB, of any 'arc KG, will fall wholly within the plane, and be projected into a line equal to itself; but the secant and tangent of any arc will terminate in the same point; therefore one extremity of the tangent must necessarily fall upon the plane. Let G be a point on the sphere, GB and Gc tangents to that point, meeting the plane iv B and c, and let e be the projecting point; then it is evident that GB will be projected into FB, and Gc into FC; and I say that GB=FB, GC=FC, and the angle CGB, the measure of the spherical angle igi (R.135.), is equal to the angle CFB formed on the plane of projection. For, let us be drawn parallel to AK, and let Epe be drawn through the centre P, then og=0H (Euclid 3 of III.), and the angle Egh is equal to the angle Eng. But the angle EGB is equal to the angle EHG (Euclid 32 of III.) therefore the angle EGB or FGB is equal to the angle Egh or FGH. But ng is parallel to Ak, and the angles rgh and GFK are equal to each other, being alternate angles (Euclid 29 of I.), therefore the angle FGB is equal to the angle GFB, and consequently (Euclid 6 of 1.) BG is equal to BF. Exactly in the same manner it may be proved that co= Now the triangles CGB and CFB have the two sides GB and GC, equal to the two sides rb and FC, and they stand on the same base BC, therefore (Euclid 8 of I.) the angle cab is equal to the angle CFB, which was to be demonstrated. PROPOSITION VI. (Plate IV. Fig. 11.) (C) The projected extremities of the diameter of any circle on the sphere, inclined to the plane of projection, are in that diameter of the primitive which is perpendicular to the projecting point: and distant from the centre of the primitive the semitangents of the circle's nearest and greatest distance from that pole of projection, opposite to the projecting point. =CF. For the diameter cd of a circle on the sphere inclined to the plane of projection in an angle cPf, is projected into the line ab from the projecting point at E. But AP is the semi-tangent of ec the circle's nearest distance from the pole e, and PB is the semi-tangent of ep the circle's greatest distance from the pole e. Hence the diameter as is the sum of these semi-tangents, and P the centre is half the sum. (D) COROLLARY I. The projected diameter of a small circle inclined to the plane of projections and encompassing the pole (e) of the plane of projection, is equal to the sum of the semi-tangents of that circle's greatest and least distance from the pole of projection. For such a small circle may be conceived to be parallel to the great circle, and therefore its projected diameter will fall on both sides of the pole p of the primitive. (E) COROLLARY II. The projected diameter of a small circle inclined to the plane of projection, and situated wholly on one side of the pole (e), is equal to the difference of the semi-tangents of that circle's greatest and least distance from the pole (e) of projection. For a small circle on the sphere whose diameter is mo, will be projected into mp; pp is the semi-tangent of ro, the circle's greatest distance from the pole; pm the semi-tangent of em, the circle's least distance from the pole; and the difference between these semi-tangents is mp. (F) COROLLARY III. The points where the projected diameter of a great circle terminates, are distant from the centre of the primitive, the tangent and co-tangent of half the complement of the circle's inclination to the primitive. For ce is the complement of rc the circle's inclination to the primitive, and ap is the tangent of the half of ce, or the angle AEP. And, because AEB is a right angle, PEB is the complement thereof; but PB is the tangent of PEB, that is, the tangent of the complement of AEP, or tho co-tangent of 3 ce. PROPOSITON VII. (Plate V. Fig. 5.) (G) In any projected great circle, inclined to the primitive, the radius of the projected circle, is to the radius of the primitive, as the distance of the projected pole from the centre of the projected circle, is to the distance of the projected pole from the centre of the primitive. Let Eaeec be the projection of a great circle cutting the primitive EnFeG in E and e, and the line FG produced, in A and c; and let p be the centre of the primitive, and o the centre of the projected circle; p the internal, and r the external pole of the *projected circle. I say Join Eo, ep, and Er; then the angle Peo is the measure of the circle's inclination to the primitive: for po is the tangent of PEO to the radius of the sphere, and it is likewise the tangent of the circle's inclination to the primitive. (X. 154.) The angle Pep is the measure of half the circle's inclination to the primitive (Y. 154.), therefore ep bisects the angle peo. And, because Ep is perpendicular to Er (for p and r are the two poles), the angle rep is equal to the angle pek, being each of them a right angle; also pep and peo have been shewn to be equal, therefore the angle rep is equal to the angle oek. Consequently Ep makes equal angles internally, and IEK equal angles externally with the sides EP and Eo of the triangle PEO. Therefore (by Dr. Simson's or Keith's Euclid 3 of VI. and Proposition A.) EO : EP::po: pp; and, EO : EP::ro : rp. (H) COROLLARY I. The line ro, comprehended between the centre of the projected circle and its external pole, is harmonically* divided by the internal pole of the projected circle, and the centre of the primitive. For EO : EP::po: pp And eo : EP::r0 : TP Therefore, ro : rp.po: pp. (I) COROLLARY II. Every projected great circle, making an angle with the primitive, is greater than the primitive. For the radius Eo of the projected great circle EACBC, is the secant of that circle's inclination to the primitive (X. 154.); and therefore is always greater than EP, the radius of the primitive. PROPOSITION VIII. (Plate V. Fig. 6.) (K) If through the internal projected pole of a great circle, and the extremities of an arc of that projected circle, straight lines be drawn to cut the primitive ; the intercepted arcs of the projected circle and the primitive will be similar. Let Eres represent the primitive circle, and Elaehf the projection of a great circle inclined to the primitive; P the pole of the primitive, and p the internal pole of the projected great circle; draw pa, pl, pk, and pg, then will the arc GK be similar to the arc AL. For, it is obvious that if a straight line be drawn from a point within a circle to meet the circumference, the angle formed at the circumference between this line and the radius, will be less When a straight line is divided into three such parts, that the whole is to the first part, as the third part is to the second; it is said to be divided in harmonical ratio. than a right angle: therefore in the triangles Pkp, plo, the angles pkp and plo are each less than a right angle. The angle PPK is equal to the angle ipo (Euclid 15 of I.), and the sides about the remaining angles at p and o are proportional. For OL : PK::op: pp (G. 157.) Hence PK : pp:: 0L : op, therefore the triangles ppk and pol are equiangular and similar (Euclid 7 of VI.), and consequently the angle app is equal to the angle pol But equal angles at the centres of circles are subtended by similar circumferences, therefore the arc Gk is similar to the arc AL; and whatever part of the circumference of the projected circle al is, the same part of the circumference of the primitive circle will gk be. For ol : PK::the arc Al : the arc GK. (L) COROLLARY I. If straight lines be drawn from the projected pole of a great circle (towards the pole of the primitive ) to cut the projected great circle and the primitive; the arc of the primitive intercepted between these lines, will be the measure of an arc on the sphere, represented by that part of the projected great cirele intercepted between these lines. Let PF, ph, and pe be drawn; FH will be the measure of a part of a great circle on the sphere represented by AL; and Fe will be the measure of a part of a great circle on the sphere represented by AE ; for, join oE and pm, then ol : PK:: arc AL : arc GK (K. 158.), and OE : Pm::arc AE : arc ym. (K.158.) But oESOL and Pm=PK, therefore arc AL : arcgK::arc AE : arc Gm. Hence, whatever part of the great circle Eachfis represented by AE, the same part of the primitive gm must represent; but AE is the projection of a quadrant of a circle on the sphere, having the same radius as the primitive, therefore it must be measured by FE, and not Gm; for the same reason AL must be measured by FH and not by GK. (M) COROLLARY II. Hence it follows, that all great circles inclined to the primitive, have equal arcs on the sphere, represented by unequal arcs on the plane of projection. PROPOSITION IX. Problem. (Plate IV. Fig. 14.) (N) To find the pole of any great circle. I. The pole of the primitive aEbD is the centre P.(O. 152.) II. Let ab be a right circle. Through p the centre of the primitive draw DE perpendicular to as, then D and E are the poles of ab. III. Let cae be an oblique circle. Through the centre of the primitive draw cpe, and mpn at right angles to it, cutting the oblique circle in o; draw ev through o, and makes vw an ar |