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37. Draw from the centre to the pole P, the radius CP, and through the point r the tangent L'PF'G, to meet in G the plane of projection BD prolonged. Then, meas. of GPK = PDO = 45° + PD;

meas. of PKG = {BO + PD = 45° + †PD; therefore GPK = GKP, and PGGK: that is, the targent PG is projected into a line KG equat to it.

38. Since POE and Por are equal, the line or will not bisect ST; but KS <KT. Bisect ST in m, then ms mTradius of circle of projection; and cs, cm, CT, will be in arithmetical progression. Hence,

cm = {CT + cstan AF + § tan AE =
Siu (AF + AE)

=

sin AP

2 cos AF COS AE 2 cus § (AP + PE) COS ≥ (AP — PE) Let dcm distance of the centre m from the centre c of the plane of projection, A = PE= polar distance of the circle EF, D=AP = distance of the two poles, r = ms = MT: then

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(CTCs) = (tan AF-tan JAE)

sin (AF - AE)

=

sin A

2 cos AF COS AE COS D+COS A

[See formulæ (u), &c. chap. iv.]

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Consequently, dr:: sin D: sin A.... (9.)

From these three theorems the whole doctrine of stereographic projection may be deduced, by tracing the mutations of D and 4. The chief maxims and principles of construction may also be developed geometrically, thus:

39. Beginning with great circles, let PE 90°, and PFPE; then will EF be a diameter. Draw the right lines OTF, OES, bisect rs in r; then rst, will be the radius of the circle into which the great circle whose diameter is Er will be projected. Through o and draw oRrI. The circle described on ST will pass

through o, because EOE is a right angle. In like manner, it will pass through A, because cs is perpendicular to the middle of AO. Therefore ro = rs = rt; conseq. sor = Osr, and cro= 2 osr. cor = 90° cro

90°

90° 2 osr= - (DO BE)

= 90° — 90° + BE

=

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cor inclination

of the plane of the

F

great circle to the plane of projection. Hence, or is manifestly

sec. inclination,

and cr = tan inclination, to rad co.

Thus, with the radius sec D, and at the distance from the centre = tan », it will be easy to describe the eircle.

Again, take Ai = 2be = 2ap = 2D, and draw 01, the point of intersection r with DS will be the centre, and ro the radius.

= 90°

Also, since meas. of cor= BE, meas. of OCR = OE BE; we have COR + OCR = 90°, and conseq. CRO 90°. Hence, by drawing oRr perpendicular to FE, we find the centrer and the radius ro: also CR = sin,BE = sin D, OR = COS D.

40. or, therefore, will always make with oc an equal equal to the inclination or distance of the two poles. Let there be, then, a second circle whose inclination shall be cor'; the radius of its projection will be r'o, and the radii ro, r'o, will make at their points of intersection an angle ror' which will be their difference of inclination, or the mutual inclination of the two circles. This is a particular case of the general theorem.

41. Suppose now that AP = 90°, P will coincide with and the pole of the circle will be upon the limit of

the projection. Let BH BE A polar distance of the circle to be projected, the chord HE will be perpendicular to DB. Draw on and OES, GS will be the diameter sought. Bisect Gs in n, n will be the centre, NE = NH = ns = nG, will be the radius of the projected circle.

Meas. of EnG= 2nse = OD BE 90° BE = 90° — BCE

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CnENCE 90° ... cen = 90°... En = tan BE = tan 4, and cn = sec BE = sec A.

These values serve for all circles which have their pole on the circumference of the circle of projection, If these are great circles, then ▲ = 90°, and tan A, sec ▲ are infinite: consequently, the centres of the projections falling at an infinite distance from c, the projections themselves will be right lines passing through c, and intersecting under the angles which such circles, taken two and two, form on the sphere.

A

42. Let o be the point of observation, or place of the eye, A the pole of the projection, BDECB the plane of projection, PD an arc of a great circle which has its origin at any point whatever M" of the circle OBPE, Pt the B tangent of PD; ct will be the secant, and s the projection of P. Draw st: then from the rectilineal triangle sct we shall have,

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2cs.ct. cos sct

M'

M

N

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Substituting this in the last value of st2, we have

st2 sec2 AP + tan2 PD →→

= sec2 1AP + tan2 PD

séc2 AP + tan2 PD

Therefore st = tan PD = Pt.

2 tan AP COS DE

SID AP COS DE

2 tan AP

2 sin AP COS AP

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Consequently, the tangent pt of an arc of a great circle terminated at the plane of projection, is projected into a right line equal to it.

P

Let Pt, and Pt, two such tangents, be connected by the right line tť which will be in the plane of projection. Let st, st', be the projections of those tangents; the triangles trt', tst', are (from the above) equal in all respects: therefore the angles opposite to the common side tt' will be the same in both: conse

quently, the tangents of any two arcs terminated at the plane of projection, are projected into lines which are respectively equal to them, and which form an equal angle. Hence, two circles which intersect in P on the sphere, form on the projection an angle equal to that which they make on the sphere; because, at the point of intersection the elements of the arcs coincide with those of their tangents. Therefore, all great circles intersect mutually on the plane of projection under the same angle as on the sphere; so also do little circles which intersect at the same points, and have, by consequence, common tangents.

43. By way of showing the application of these principles, let us suppose that the eye is at the south pole of the equator. The plane of projection will then be the equator itself; the centre of the equator will represent the north pole; AP (fig. to art. 35) will be = 0; the projections of the parallels to the equator will all

have for a common centre that of the projection; and the radii of those circles will be the tangents of the halves of the polar distances. Thus,

for the polar circle ....r=tan ( 23°28′)=tan 11°44′ tropic of cancer.rtan ( 66°32′)=tan 33°16′ tropic of capricornr=tan (113°28′)=tan 56°44' tan (156° 32') = tan 78° 16′ (90°-L) = tan 45°-L or, if the lat. be south r tan (45° + 1L).

for the antarctic circle

for any latitude L.... 7tan

As for the meri

dians, whose planes all pass through the place of the eye, they all become diameters which divide the equator in its several degrees, and form at the centre of the projection angles equal to the differences of longitude. For these circles d∞, and r = (art. 38). This kind of projection, the easiest

V

d

B

2700

180

H

90°

D

YG

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of all to describe, serves very conveniently for eclipses of the sun. The meridians and the parallels are herein divided mutually into degrees: those of the parallels are equal; those of the meridians unequal; for the expression for one of their degrees is,

♪=tan (4+1°)-tan 4 =

sin 30'

cos A cos (4 + 1o).

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