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 eD the circle's greatest distance from the pole e. Hence the diameter AB is the sum of these semi-tangents, and 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 мo, will be projected into mp; pp is the semi-tangent of to, the circle's greatest distance from the pole; pm the semi-tangent of eм, 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 Fc 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 the co-tangent of 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.

A

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

EO EP: popp; or, EO: EP¦¦ro ; rp.

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 rEK 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: pr

And EO: EP::ro: rp

Therefore, ro re::po: pr.

(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.

pro

Let EreG represent the primitive circle, and ELAehf the jection 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, pê, 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 Lpo (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::OL: op, therefore the triangles PPK and poL are equiangular and similar (Euclid 7 of VI.), and consequently the angle KPP 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 circle 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 oɛ: Pm::arc AE : arc Gm. (K.158.) But OE OL and PmPK, therefore arc AL: arc GK::arc AE:

arc Gm.

Hence, whatever part of the great circle Eaehf is 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.(0.152.) II. Let aв be a right circle. Through P the centre of the primitive draw DE perpendicular to aв, then D and E are the poles of ав.

III. Let cae be an oblique circle. Through the centre of the primitive draw cre, and men at right angles to it, cutting the oblique circle in o; draw ev through o, and makes vw an arc of 90 degrees; join ew, then p is the pole.

Or, the semi-tangent* of the complement of po set from P to p, will give the pole. (Y. 154.)

The pole of a small circle is the same with the pole of its parallel great circle.

(O) The truth of this proposition is shewn at Y. 154.; for on, the measure of the angle c, is the complement of po, and the semi-tangent of on is applied from P to p.

(P) PROPOSITION X. Problem. (Plate IV. Fig. 14.)

Through any given point in the circumference of the primitive circle, to describe a great circle making any given angle with the primitive.

Let EaDB be the primitive circle, and c the given point.

Through c draw CPe, and men at right angles to it; make the angle Pcr equal to the given angle, and draw cr; with r as a centre, and distance rc, describe the circle cae, then ACB will be equal to the given angle.

Or, Make Pr equal to the tangent of the given angle to the radius CP; or make cr the secant thereof.

Or thus: Set off the semi-tangent of the complement of the given angle from P to o, and through the three points coe describe a circle.

(Q) These constructions are obvious from X. 154. and Z. 155.

(R) PROPOSITION XI. Problem. (Plate IV. Fig. 15.)

Through a given point ▲ in a right circle aв, to describe a great circle cae, making an angle at A, equal to a given number of degrees.

Draw DE at right angles to aв; find the centre s of a circle which will pass through the three points DAE. Through s draw dsm parallel to DE; with the centre A, and a radius equal to PB, describe an arc ww, set off bo from a scale of chords equal to the complement of the given angle a, and draw aod; then d is the centre of the circle cae required.

(S) The truth of this construction may be shewn thus: s is the centre of the circle DAE, and since all great circles cut each other at a semi-circle's distance (C. 133.), all circles passing through a must necessarily pass through the point &, and therefore their centres must be somewhere in the line dm. Now dae is a right angle, therefore sae=CAB (N. 135.) taken from dae, leaves das the complement of the angle CAB, agreeably to the construction.

*The common plain scale generally contains a line of semi-tangents, marked S. T. See Book I. Chap. IV. page 16,

(T) PROPOSITION xii. Problem. (Plate IV. Fig. 16.) Through a given point c, within the primitive, to draw a great circle Ace making a given angle BAC with the primitive.

From the centre P of the primitive, with the tangent of the given angle as a radius, describe an arc; and from the point c, with the secant of the same angle as a radius, cross it in o. The point o is the centre, and oc is the radius of the circle Ace required.

(U) The truth of this construction is shewn at X. 154.

PROPOSITION XIII. Problem. (Plate IV. Fig. 17.)

(W) Through two given points within the primitive, to draw a great circle.

Let o and m be the two points. Through one of the points, as o, and the centre of the primitive, draw AB of an unlimited length towards B, and draw CD at right angles to AB. Join co, and produce it till it cuts the primitive in v, through v draw the diameter vw, meeting the circumference in w, and from c through w, draw cwE, meeting AB in E. Then a circle described through the three points o, m, E, will be the great circle required.

(X) The angle vcw being in a semi-circle is a right angle (Euclid 31 of III.), therefore the distances PO, and PE, measured on the line of semi-tangents are together equal to 180 degrees, consequently the point E is diametrically opposite to o, and OmE is a great circle.'

PROPOSITION XIV.

Problem. (Plate IV. Fig. 18.)

(Y) Through a given point, in any projected great circle, to draw another great circle perpendicular to the given one.

General Rule. Find the pole of the given circle (N. 159.) and through that pole and the given point draw a great circle (W. 161.) For (I. 134.) if one great circle pass through the pole of another, it cuts it at right angles.

Or, I. The pole of the primitive is in the centre, therefore a diameter of the primitive always cuts its circumference at right angles.

II. To draw an oblique circle perpendicular to an oblique circle. Let BAD be the given oblique circle and a the given point, find p the pole of the oblique circle BAD (N. 159.), and through the points p, and A, draw the great circle ECAp. (W. 161.) III. To draw an oblique circle perpendicular to a right circle. Let HG be the right circle and A the given point, draw DB at right angles to HG, then B, and D, are the poles of HG;

M