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parallel to AB ; then, because the line GP, and the plane of projection are both perpendicular to the plane ABF, therefore, GH is both parallel to the plane of projection, and also to the line gh projected on it.
In the circle ADB, DQ?=GQH=gqh; and BP=GP=gp® ; and BP : EP, or DQ :: bp : ep or dq; and BPS : DQ® :: bp : dq; that is, gpo : gqh :: bp : dq? ; and, therefore, adgbh is an ellipsis, the major axis gh of which is the diameter of the circle. Q. E. D.
Cor. 1. Since ab is perpendicular to gh, ab is the minor axis, and is double the length of the sine of the angle ABb to the radius gp ; that is, the major axis is equal to twice the co-sine of the inclination to the radins of the . circle.
Cor. 2. The major axis is equal to twice the co-sine of its distance from its parallel great circle, for gh=ĠH=2AP- twice the sine of AK.
Cor. 3. The extremities of the conjugate axis are distant from the centre of the primitive circle, by the sines of the circles of the least and greatest distance from the axis or pole of the primitive circle. Thus, aC is the sine of AN, and IC the sine of BN.
Cor. 4. Hence it is plain, also, that the conjugate axis passes invariably through the centre C of the primitive circle, and is always in the line of measures of that circle.
N.B. Every circle in the projection represents two equal circles, which are parallel and equi-distant from the primitive circle. Every right line represents two semi-circles, one towards the eye, and the other on the opposite side. Every ellipsis represents two equal circles, only baving a contrary inclination AB, CD, the one being above the primitive circle, the other below it.
And now having shewed the principles of the theory, it will be proper to (educe some short practical rules, by way of example
Prop. 5. To project a circle parallel to the primitive.
Take the complement of the distance of the circle to be projected from the primitive circle, and set it off from A to E; and, with the centre C, and a radius, equal the perpendicular line A
ED EF, describe the circle DgG.
To do the sume by the Pluin Scale. Take the sine of the circle's distance from the pole of the primitive circle, describe the desired circle with this radius and she centre C.
Prop. 6. To project a right circle, iz one which is perpendicular 1. the plane of projection,
Through the centre C of the primitive circle draw the diameter AB, and take the dis
D tance from its parallel great circle, and set it from A to E, and from B to D, and draw ED for the
B circle required.
To do the same by the Scale. Take the sine of the distance of the circle froin its parallel great circle AB, and at that distance draw a parallel ED for the circle required.
Prop. 7. To project a small oblique circle.
Draw the line of measures AB, and take
E the nearest distance of the circle from the primitive, and set it from B to D upwards, if above the primitive, and downwards if below
A it, also take its greatest distance, and set it from A to E, and draw ED, and let fall the perpendiculars EF, DG ; and bisect FG in H, and erect the perpendicular KHI, making KH=HI= half ED; then describe an ellipsis (by the conic sections), the major axis of which is IK, aud the minor axis FG, and the curve will represent the given circle.
To do the same by the Scale. Draw the line of measures AB ; and take the sines of the least and greatest distances of the circle from the pole of the primitive, and set them off from the centre C to F aud G both ways, if the periphery encompass the pole, but the same way if it lie on one side of it; bisect FG in H, and erect HK and HI perpendicular to FG, and equal the radius of the circle given, or the sine of its distance from its own pole ; about the Axes FG, KI, describe the ellipsis, which completes it.
Prop. 8. To find the pole of a given ellipse, Through the centre of the primitive C draw the conjugate axis FG of the ellipsis; on the points F, G, erect the perpendiculars FE, GD, or set off the transverse axis IK from E to D, and bisect ED in R, and let fall RP perpendicular to AB, then P is the pole. Which may be found alsn hy the Scale.
R Thus, cake CF and CG and apply them to the
D sines, and find the corresponding degrees of the supplenients ; then take the sine of half the sum of these degrees, if F and G be both one side of C, or the sine of half the difference if they are situated on contrary sides, and set it off from the point C to the pole P
K Prob. 9. To measure an arch of a parallel cir: le, or to set off any number of degrees upon it.
With the radius of the parallel circle, and one foot of the compasses in C, describe the
9 circle Gg; draw CGB and Cgy; then Bl will measure the given arch Gg; or Gg will contain the given number of degrees set orð from B to l; hence, one being ascertained determines the other.
Prop. 10. To measure any part of a right circle.
In the right circle ED, let EA=AD; and let AB be required to be measured ; E
B make CF=AE ; and, from the centre F, KAG with the radius AB, describe the arc GI; draw CGK touching the circle in G; then I HK will be the measure of AB. For FG cqual the sine of the angle, HGK to the radius CF or AE, and BA is the same, by Corollary Proposition 3.
E found as in Proposition 9,
Cor. If the right circle passes through the centre, there is no more required than to raisc perpendicuları on it, which will cut the pri. H. mitive circle as required. Or, apply the part of the right circle to the line of sines.
Prop. 11. To set off any number of degrees upon a right circle, as DE. See the figure to proposition 10.
Draw CA perpendicular to DE, and make the angle HCK equal the degrees given ; make CF equal to the radius AE ; take FG, the nearest distance, and set it off from A to B, then AB will be equal to
E the degrees in the angle HCK as proposed.
D Another Method. On ED describe the semicircle END, then, by Prop. 11, set off NP equal to the degrees given, draw PL perpendicular to ED, then AL will contain the degrees required.
Prop. 12. To measure an arc of an ellipsis, or to set off any number of degrees upon it. About AR, the transverse axis of the
B elipsis, describe a circle ABR; erect the perpendiculars BED, KFI,on AR ; then BK is the measure of EF, or EF is the representation of the arc BK, which may be measured, or any degrees set off upon it, as in Prop. 9.
These problems are all evident from the three first propositions, and need no other demonstration. If the sphere be projected on any plane, parallel to the primitive circle, the projection will be the very same, for being effected by parallel lines, which are always at the same distance, the same figure 0: representation will be invariably formed.
projected into a right line.
For, all lines, drawn from the projecting point to this c'rele, pass
through its intersection with the plane of projection, which is a night line.
Cor. 1. A great circle passing through the poles of the primitive circle, is projected into a right line passing through the centre.
Cor. 2. Any circle passing throngh the projecting point, is projected into a right line perpendicular to the line of measures, and distant from the celltre, the semi-tangent of its nearest distance from the pole opposite to the projecting point.
Prop. 2. Every circle that does not pass through the projecting point, is projected into a circle.
Case 1. Let the circle EF be parallel to the priinitive BD, lines drawn to all points of it, from the projecting point A, will form a conic
B! surface, which, being cut parallel to the base, by the plane BD, the section GH, into which EF is projected, will be a circle ; by the conic
a sections. Case 2. Let BH be the line of mea
EM sures to the circle EF; draw FK parallel to BD, then will the arc AK equal AF, and therefore the angle AFK or AHG
I is equal to AEF; wherefore, in the triangles AEF and AGH, the anzles at E and H are equal, and the angle A common ; therefore, the angles at F and G are equal.
Whence the cone of rays AEF, the base of which EF is a circle, is cut by a sub-contrary section, by the plane of projection BD; and, therefore, by the conic sections, the section GH, which is the projection of the circle EF, will also be a circle.
Cor. When AF is equal to AG, the circle EF is projected into a circle equal t, itself.
For then the similar triangles AHG and AEF will also be equal, and GH equal to EF.
Prop. 3. Any point on the surface of the sphere is projected into a point, distant from the centre, the semi-tangent of its distance from the pole opposite to the projecting point.
Thus the point. E, in the last figure, is projected into G, and Finto H; and CG is the semi-tangent of EM, and CH of MF.
Cor. 1. A great circular perpendicnlar to the primitive, is projected into a line of semi-tangents passing throagh the centre, and produced ad infinitum.
For MF is projected into CH its semi-tangent, and FM into the semi. tangent CG.
Cor. 2. Any arc EM of a great circle, perpendicular to the primitive, is projected into its semi-tangent. Thus, EM is projected into GC.
Cor. 3. Any arc EMF of a great circle is projected into the sum of its semitangents, of its greatest and least distances from the opposite pole M, if it lie on both sides of M; or the difference of tae semi-tangents when all on one side.
Prop. 4. The angle made by two circles on the surface of the
sphere, is equal to that made by their representations upon the plane of projection.
Let the angle BPK be projected ; through the angular point P and the
H centre C draw the plane of a great circle PED, perpendicular to the
E plane of projection EFG. Let a plane
D PHG touch the sphere in P; then, since the circle EPD is perpendicular both to this plane and to the plane of projection, it is perpendicular to their
A. intersection GH.
But the angles made by circles are the same as those made by their tangents; therefore, in the plane PGH, draw the tangents PH, PF, PG, to the arcs PB, PD, PK, and they will be projected into the lines pH, pF, pK; then the < CPA will = 2 HpG.
For the angle CPF is a right angle = CPA + CAP; therefore, taking away the equal angles CPA and CAP, and 2 pPE = CpA or PpF; consequently, pF = PF. Therefore, in the right-angled triangles PFG and PFG there are two sides equal, and the included angle is a right angle; therefore, the hypothenuse PG=pG; and, for the same reason, in the right-angled triangles PFH and PFH, PH=pH. Lastly, in the triangles PHG and pHG, all the sides are respectively equal ; and, therefore ZP=lp. Q. E. D.
Cor. The angle made by two circles on the sphere, is equal to the angle made by the raclii of their projection at the point of intersection. For the angle made by two circles on a plane, is the same with that made by their radii drawn to the point of intersection.
Prop. 5. The centre of a lesser circle projected, perpendicular to the primitive, is in the line of measures, distant from the centre of the primitive, the secant of the lesser circle's distance from its own pole, and its radius is the tangent of that distance.
Let A be the projecting point, EF the circle to be projected, GH the
E projected diameter. From the centre C and D draw CF, DF; and the tri- K
IBD angles CFI, DFI, are right-angled at I; then ZIFC= _ FCA=2 Z FEA, or 2 Z FEG = 2 Z FHG = 2 FDG. Therefore, iFC+IFD=FDG +IFD equal a right-angle; that is, CFD is a right-angle, and the line CD is the secant of BF, and the radius FD is the tangent of it. Q. E. D.
Cor. If these circles be actually described, it is plain the radius FD is a tangent to the primitive at F, where the lesser circle cuts it.
Prop. 6. The centre of projection of a great circle is in the line of measures, distant from the centre of the primitive the tangent of its inclination to the primitive; and its radius is the secant of its inclination.