Prop. 5. Let the plane TW (fig. 5) be perpendicular to the piane of projection TV, and BCD a great circle of the sphere in the plane TW, and let the great circle BED be projected into the right line bek. Draw CQS 1 bk, and Cm || to it, and equal to CA, and make QS= Qm; then any angle Qst=Qt.

Suppose the hypothenuse AQ to be drawn, then, since the plane ACQ is perpendicular to the plane Tv, and BQ is I to the intersection CQ, therefore bQ is perpendicular to the plane ACQ, and, consequently, IQ is 1 to the hypothepuse AQ; but AQ=Qm=Qs, and Qs is also I to lQ. Therefore, all angles made at S cut the line IQ in the same point as the angles made at A ; but, by the angles at A, the circle BED is projected into the line 6Q. Therefore the angles at s, are the measures of the parts of the projected circle bQ, and S is the dividing centre thereof. Q. E. D.

Cor. Any great circle iQb is projected into a line of tangents to the radius SQ.

For Qt is the tangent of the angle QSt to the radius QS or Qm.

Cor. 2. If the circle 6C pass through the centre of the projection, then A, the projecting point, is the dividing centre thereof, and Co is the tangent of its corresponding arc CB, to CA the radius of projection.

Prop. 6. Let the parallel circle GEH (fig. 5) be as far froin the pole of projection C, as the circle FKI is fronı its pole P, and let the distance of the poles C and P be bisected by the radius AO; and draw bAD perpendicular to AO; then any right line bek drawn through l, will cut off the arcs hl=Fn, and ge=kf, supposing f the other vertes, in the representation of these equal circles in the plane of projection.

For, let G, E, R, L, H, N, R, K, I, be respectively projected into the points g, e, r, l, h, 1, 1, k, f. Then, since in the sphere the arc BF=DH, and the arc BG=DI, and the great circle BEKD makes the angles at B and D equal, and is projected into a right line as bl; therefore the triangular figures BFN and DHL are similar and equal ; and, likewise, BGE and DIK are similar and equal, and LH is equal to NF, and KI equal to EG; whence it is evident, that their projections lh=nF, and kf=ge. Q. E. D.

Prop. 7. If hlg and Fnk (fig. 6) be the projections of two equal circles, one of which is as far from its pole P, as the other from its pole C, which is the centre of projection ; and, if the distance of the projected poles Cp be divided in o; so that the degrees in Co, op, be equal, and the perpendicular os be erected the line of measures gh, then the lines pn, Cl, drawn from the poles Cp, through any point Q, in the line os, will cut off the arc Fn=2QCp.

For, drawing the great circle GPI, in a plane perpendicular to the plane of projection ; the great circle AO perpendicular to CP is projected into os, by Prop. i, Cor. 3. Now let Q be the projection of 4, and since pQ and CQ are lines, they will represent the great circles Pq, Cq.

But the spherical triangle PqC is an isocelis triangle, therefore the angles at P and C are equal. But, because P is the pole of FI, the

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great circle Pq continued, will cut an arc off FI=2 CP9= 2PCq= ¿QCp, 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 p of any circlo, a perpendicular be erected to the line of measures, it will cut off a quadrant from the representation 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 Cg be the co-tangent of CAP, and gВ perpendicular to the line of measures gC, and CAP be bisected by AO, and the line oB be drawn to any poivt 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=4gpA+ LoAp= (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 gВ. 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=4 goB. Q. E. D.

Cor. The goB is the measure of the angle gpB. For the triangle gpB represents a triangle ou the spliere, 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 which 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 snpposed 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.

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Prop. 10. To draw a great circle perpendicular to a given great circle, which passes through the pole of projection, and at a given dis tance from the said pole.

Draw the primitive ADB (fig. 7). Let CI be the given circle ; draw CL perpendicular to Cl, 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 ; through I draw KP perpendicular to CI, then KP is the circle re. quired.

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 IAP is equal the measure of the given arc IP.

Or, if the degrees be given, make the IAP equal the given degrees, 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 , 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 reqnired, from the points BD draw the lives 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.


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