:

tive.

Stereogra- for the angle AML=AKF. But since A is the pole =2NHG=NCG: hence ENC=INE+INC=NCG Stereogra

phic Pro. BLD, the chords, and consequently the arches AB +INC = a right angle ; and therefore NC is a tan- phic Projection of AL, are equal, and the arch ABG is the sum of the

gent to the primitive at N; but the arch ND is the jection of the Sphere. arches AL, BG; hence the angle AML is equal to an distance of the less circle from its nearest pole D:

the Sphere.
angle at the circumference standing upon AG, and hence NC is the tangent, and EC the secant of the
therefore equal to AGI or FGK; consequently the distance of the less circle from its pole to the radius of
angle FGK=FKG, and the side FG=FK. In like the primitive.
manner HG=HK: hence the triangles GHF, KHF
are equal, and the angle FGH=FKH=LMN.

PROPOSITION VI. THEOREM VI.
COROLLARIES.

The projection of the poles of any circle, inclined to the

primitive, are, in the line of measures, distant from 1. An angle contained by any two circles of the

the centre of the primitive, the tangent, and cotansphere is equal to the angle formed by their projec- gent, of half its inclination. tions. For the tangents to these circles on the sphere Let MN (fig. 6.) be a great circle perpendicular to Fig. 6. are projected into straight lines, which either coincide the primitive ABCD, and À the projecting point; then with, or are tangents to, their projections on the primi. P, p are the poles of MN, and of all its parallels m n,

&c. Let AP, A p meet the diameter BD in Ff, 2. An angle contained by any two circles of the which will therefore be the projected poles of MN and sphere is equal to the angle formed by the radii of their its parallels. The angle BEM is the inclination of the projections at the point of intersection.

circle MEN, and its parallels, to the primitive : and

because BC and MP are quadrants, and MC common PROPOSITION IV. THEOREM IV.

to both ; therefore PC=BM: and hence PEC is also
The centre of a projected great circle is distant from the inclination of MN and its parallels. Now EF is

the centre of the primitive ; the tangent of the incli. the tangent of EAF, or of half the angle PEC the in.
nation of the great circle to the primitive, and its ra- clination ; and E f is the tangent of the angle EAf;
dius, is the secant of its inclination.

but EA f is the complement of EAF, and Ef is the
Let MNG (fig. 4.) be the projection of a great

cotangent of balf the inclination.
circle, meeting the primitive in the extremities of the

COROLLARIES.
diameter MN, and let the diameter BD, perpendicular
to MN, meet the projection in F, G. Bisect FG in 1. The projection of that pole which is nearest to the
H, and join NH. Then, because any angle contained projecting point is without the primitive, and the pro-
by two circles of the sphere is equal to the angle formed jection of the other within.
by the radii of their projections at the point of intersec- 2. The projected centre of any circle is always be-
tion ; therefore the angle contained by the proposed tween the projection of its nearest pole and the centre of
great circle and the primitive is equal to the angle the primitive, and the projected centres of all circles
ENH, of which EH is the tangent, and NH the se- are contained between the projected poles.
cant, to the radius of the primitive.

PROPOSITION VII. THEOREM VII.
COROLLARIES.

Equal arches of any two great circles of the sphere
1. All circles which pass through the points M, N, will be intercepted between two other circles drawn
are the projections of great circles, and have their con-

on the sphere through the remote poles of those great
tres in the line BG; and all circles which pass through circles,
the points F, G, are the projections of great circles,
and have their centres in the line HI, perpendicular to

Let AGB, CFD (fig. 7.) be two great circles of the Fig. 7.
BG.

sphere, whose remote poles are E, P; through which
2. If NF, NH be continued to meet the primitive in

draw the great circle PBEC, and less circle PGE, in-
L, F; then BL is the measure of the great circle's in-
;

tersecting the great circles AGB, CFD in the points
clination to the primitive; and MT=2BL.

B, G, and D, F; then the arch BG is equal to the

arch DF.
PROPOSITION V. THEOREM V.

Because E is the pole of the circle AGB, and P the
The centre of projection of a less circle perpendicular to

pole of CFD, therefore the arches EB, PD are equal;

and since BD is common to both, hence the arch ED
the primitive, is distant from the centre of the primi- is equal to the arch PB. For the same reason, the
tive, the secant of the distance of the less circle from
its nearest pole; and the radius of projection is the

arches EF, PG are equal ; but the angle DEF is

equal to the angle BPG : bence these triangles are tangent of that distance.

equal, and therefore the arch DF is equal to the arch

BG. 5. 5.

Let MN (fig. 5.) be the given less circle perpendi-
cular to the primitive, and A the projecting point.

PROPOSITION VIII. THEOREM VIII.
Draw AM, AN to meet the diameter BD produced
in G and H; then GH is the projected diameter of If from either pole of a projected great circle, two straight
the less circle : bisect GH in C, and C will be its lines be drawn to meet the primitive and the pro-
centre; join NE, NC. Then because AE, NJ are jection, they will intercept similar arches of these
parallel, the angle INE=NEA ; but NEA=2NMA circles.
Vol. XVII. Part II,

t

3 H

OB

Fig. 9.

Stcreogra

On the plane of projection AGB (fig. 7.) let the straight line EGH, meeting the diameter AB produ. Stereo phic Pro- great circle CFD be projected into cfd, and its pole P ced if necessary in H; then from the centre H, with pbic i'ro. jection of into p; through p draw the straight lines pd, pf, then the radius HE, describe the oblique circle DIE, and it jection et the Sphere. are the arches GB, f d similar.

will be the projection of the great circle required.
Since pd lies both in the plane AGB and APBE, Or, make DK equal to FA; join EK, which inter-
it is in their common section, and the point B is also sects the diameter AB in I; then through the tbree
in their common section ; therefore pd passes through points, D, 1, E, describe the oblique circle DIE.
the point B. In like manner it may be shown that
the line pf passes through G. Now the points D, F

PROPOSITION XI. PROBLEM III.
are projected into d, f: hence the arches FD, fd are
similar; but GB is equal to FD, therefore the inter-

To find the poles of a great circle.
cepted arch of the primitive GB is similar to the pro- 1. When the given great circle is the primitive, its
jected arch fd.

centre is the pole.

2. To find the pole of the right circle ACB (fig. 11.). COROLLARY.

Draw the diameter PE perpendicular to the given circle Hence, if from the angular point of a projected sphe. AB; and its extremities P, E are the poles of the circle rical angle two straight lines be drawn through the

ACB. projected poles of the containing sides, the intercepted 3. To find the pole of the oblique circle DEF (fig. Fig. 13, arch of the primitive will be the measure of the spheri

13.). Join DF, and perpendicular thereto draw the cal angle.

diameter AB, cutting the given oblique circle DEF

in E. Draw the straight line FEG, meeting the cir. PROPOSITION IX. PROBLEM I.

cumference in G. Make GI, GH, each equal to AD;

then FI being joined, cuts the diameter AB in P, the To describe the projection of a great circle through two lower pole; through F and H draw the straight line

given points in the plane of the primitive. FHp, meeting the diameter AB produced in p, which Let P and B be given points, and C the centre of

will be the opposite or exterior pole. the primitive.

PROPOSITION XII. PROBLEM IV. Fig. 8.

1. When one point P (fig. 8.) is the centre of the primitive, a diameter drawn through the given points To describe a less circle about any given point as a pole, will be the great circle required.

and at any given distance from that pole. 2. When one point P (fig. 9.) is in the circumference of the primitive. Through P draw the diameter 1. When the pole of the less circle is in the centre PD; and an oblique circle described through the three of the primitive; then from the centre of the primitive, points P, B, D, will be the projection of the required with the semitangent of the distance of the given circle

from its pole, describe a circle, and it will be the pro3. When the given points are neither in the centre jection of the less circle required.

nor circumference of the primitive. Through either of 2. If the given pole is in the circumference of the Fig. 10. the given points P (fig. 10.) draw the diameter ED, primitive, from C (fig. 14.) the centre of the primitive, Fig. 14

,

. and at right angles thereto draw the diameter FG. set off' CE the secant of the distance of the less circle From F through P draw the straight line FPH, meet- from its pole P; then from the centre E, with the taning the circumference in H: draw the dianieter HI, gent of the given distance, describe a circle, and it will and draw the straight line FIK, meeting ED produced be the less circle required. Or, make PG, PF each in K; then an arch, terminated by the circumference, equal to the chord of the distance of the less circle from being described through the three points, P, B, K, will ita pole. Through B, G draw the straight line BGD be the great circle.

meeting CP produced in D: bisect GD in H, and

draw HE perpendicular to GD, and meeting PD in PROPOSITION X. PROBLEM II.

E; then E is the centre of the less circle.

3. When the given pole is neither in the centre nor To describe the representation of a great circle about

circumference of the primitive. Through P (fig. 15.), Fig. 19 any given point as a pole.

the given pole, and C the centre of the primitive, draw Let P be the given pole, and C the centre of the pri

the diameter AB, and draw the diameter DE perpenmitive.

dicular to AB; join EP, and produce it to meet the 1. When P (fig. 8.) is in the centre of the primitive, primitive in p; make pF, G, each equal to the chord then the primitive will be the great circle required.

of the distance of the less circle from its pole ; join EF Fig. 11, 2. When the pole P (fig. 11.) is in the circumfe.

which intersects the diameter AB in H; from E through rence of the primitive. Through P draw the diameter G draw the straight line EGI, meeting the diameter PE, and the diameter AB drawn 'at right angles to PE AB produced in I ; bisect HI in K: Then a circle dewill be the projected great circle required.

scribed from the centre K; at the distance KH or KI, 3. When the given pole is neither in the centre nor

will be the projection of the less circle. circumference of the primitive. Through the pole P Plate (fig. 12.) draw the diameter AB, and draw the diame

PROPOSITION XIII. PROBLEM V. ccccxliv. ter DE perpendicular to AB; through E and P draw

To find the poles of a given less circle. the straight line EPF, meeting the circumference in F. Make FG equal to FD; through E and G draw the The poles of a less circle are also those of its parallel 4

great circle.

ܪ .

Fig. 15

Stereogra- great circle. If therefore the parallel great circle be line of chords will give the measure of the arch DE of Stercogriphic Projection of given, then its poles being found by Prob. 111. will be the given less circle.

phic Prothe Sphere. those of the less circle. But if the parallel great circle

jection of be not given, let HMIN (fig. 15.) be the given less PROPOSITION XVI. PROBLEM VIII.

the Sphere. circle. Through its centre, and C the centre of the pri

To measure any spherical angle.
mitive, draw the line of measures IAHB; and draw the
diameter DE perpendicular to it, also draw the straight 1. If the angle is at the centre of the primitive, it is
line EHF meeting the primitive in F; make Fp equal measured as a plane angle.
to the chord of the distance of the less circle from its 2. When the angular point is in the circumference of
pole: join Ep, and its intersection P with the diame- the primitive ; let A (fig. 19.) be the angular point, Fig. 19.
ter AB is the interior pole. Draw the diameter p CL and ABE an oblique circle inclined to the primitive.
through E and L, draw ELq meeting the diameter AB Throngh P, the pole of ABE, draw the line AP p meet-

,
produced in q; then q is the external pole. Or thus : ing the circumference in p: then the arch E p

is the
Join EI intersecting the primitive in G; join also measure of the angle BAD, and the arch AF p is the
EH, and produce it to meet the primitive in F; bisect measure of its supplement BAF: also p F is the mea-
the arch GH in p; from E to p draw the straight line sure of the angle BAC, and p ED that of its supple-

IP
EP p, and P is the pole of the given less circle.

ment.

3. If the angular point is neither at the centre nor PROPOSITION XIV, PROBLEM. VI.

circumference of the primitive. Let A (fig. 20.) be Fig. 20.

the angular point, and DAH, or GAF, the angle to To measure any arch of a great circle.

be measured, P the pole of the oblique circle DAF, 1. Arches of the primitive are measured on the line and p the pole of GAH : then from A, through the of chords.

points P p, draw the straight lines APM, A p N, and 2. Right circles are measured on the line of semi- the arch MN will be the measure of the angle DAH;

tangents, beginning at the centre of the primitive. Thus, and the supplement of MN will be the measure of the
Fig. 16. the measure of the portion AC (fig. 16.) of the right angle HAF or DAG.
circle DE, is found by applying it to the line of semi-

PROPOSITION XVII. PROBLEM IX.
tangents. The measure of the arch DB is found by
subtracting that of BC from 90°: the measure of the To draw a great circle perpendicular to a projected
arch AF, lying partly on each side of the centre, is ob-

great circle, and through a point given in it.
tained by adding the measures of AC and CF. Lastly,
To measure the part AB, which is neither terminated at

Find the pole of the given circle, then a great circle
the centre or circumference of the primitive, apply CA described through that pole and the given point will be
to the line of semitangents; then CB, and the difference perpendicular to the given circle. Hence if the given
between the measures of these arches, will be that of

circle be the primitive, then a diameter drawn through AB.

the given point will be the required perpendicular. If Or thus : Draw the diameter GH perpendicular to

the given circle is a right one, draw a diameter at right DE; then from either extremity, as D), of this diame- angles to it; then through the extremities of this diater, draw lines through the extremities of the arch in- meter and the given point describe an oblique circle, tended to be measured ; and the intercepted portion of and it will be perpendicular to that given. If the given the primitive applied to the line of chords will give the

circle is inclined to the primitive, let it be represented measure of the required arch. Thus IK applied to the by BAD (fig. 21.), whose pole is P, and let A be the

Fig. 21. line of chords will give the measure of AB.

point through which the perpendicular is to be drawn :.
3. To measure an arch of an oblique circle : draw then, by Prob. I. describe a great circle through the
lines from its pole through the extremities of the arch points P and A, and it will be perpendicular to the ob-
to meet the primitive, then the intercepted portion of lique circle BAD.
the primitive applied to the line of chords will give the
measure of the arch of the oblique circle. Thus, let

PROPOSITION XVIII. PROBLEM X.
AB (fig. 17.), be an arch of an oblique circle to be Through a point in a projected great circle, to describe
measured, and P its pole; from P draw the lines PAD,

another great circle to make a given angle with the
PBE meeting the primitive in B and E; then the arch

former, provided the measure of the given angle is DE applied to the line of chords will give the measure

not less than the distance between the given point and of the arch of the oblique circle AB.

circle.
PROPOSITION XV. PROBLEM VII.

Let the given circle he the primitive, and let A (fig.
To measure arch of a less circle.

19.) be the angular point. Draw the diameter A E, DF
any

perpendicular to each other; and make the angle CAG Fig. 19.

Let DEG (fig. 18.) be the given less circle, and equal to that given, or make CG equal to the tangent DE the arch to be measured : find its internal pole P; of the given angle; then from the centre G, with the and describe the circle AFI parallel to the primitive, distance GC, describe the oblique circle ABE, and it and whose distance from the projecting point may be will make with the primitive an angle equal to that equal to the distance of the given less circle from its given. pole P: then join PD, PE, wbich produce to meet If the given circle be a right one, let it be APB (fig. Fig. 22. the parallel circle in A and F. Now AF applied to a 22.) and let P be the given point. Draw the diameter

GH

a

Fig 17.

3 H 2

Stereogra- GH perpendicular to AB; join GP, and produce it to about n as a pole, describe the great circle EDF, cut- Stereogra

:

n phic Pro- a; make H b equal to twice A a: and Gb being joined ting the primitive and given circle in E and D, and it phic Pro jection of intersects AB in C. Draw CD perpendicular to AB, will be the great circle required.

jection of the Sphere and equal to the cotangent of the given angle to the

the Spbar radius PC; or make the angle CPD equal to the com

SCHOLIUM. plement of that given : then from the centre D, with the radius DP, describe the great circle FPE, and the It will bence be an easy matter to construct all the

angle APF, or BPE, will be equal to that given. various spherical triangles. The reader is, however, Fig. 23. If

IF APB (fig. 23.) is an oblique circle. From the referred to the article Spherical Taigonometry, for
angular point P, draw the lines PG, PC through the the method of constructing them agreeably to this pro-
centres of the primitive and given oblique circle. jection ; and also for the application to the resolution of
Through C, the centre of APB, draw GCD at right problems of the sphere. For the method of projecting
angles to PG; make the angle GPD equal to that the sphere upon the plane of the meridian, and of the
given ; and from the centre D, with the radius DP, horizon, according to the stereographic projection, see
describe the oblique circle FPE, and the angle APF, the article GEOGRAPHY.
or BPE, will be equal to that proposed.

SECTION II.
PROPOSITION XIX. PROBLEM XI.
Any great circle cutting the primitive being given, to Of the Orthographic Projection of the Sphere.

describe another great circle wbich shall cut the
given one in a proposed angle, and bave a given arch The orthographic projection of the sphere, is that in
intercepted between the primitive and given circles. which the eye is placed in the axis of the plane of pro-

jection, at an infinite distance with respect to the dia. Plate If the given circle be a right one, let it be represent

meter of the sphere; so that at the sphere all the visual ccccxLved by APC (fig. 24.); and at right angles thereto draw

rays are assumed parallel, and therefore perpendicular fig. 24. the diameter BPM; make the angle BPF equal to the

to the plane of projection. complement of the given angle, and PF equal to the

Hence the orthographic projection of any point is tangent of the given arch ; and from the centre of the

where a perpendicular from that point meets the plane primitive with the secant of the same arch describe the

of projection ; and the orthographic representation of arch Gg. Through F draw FG parallel to AC, meet

any object is the figure formed by perpendiculars drawn ing Gg in G; then from the centre G, with the

from every point of the object to the plane of protangent PF, describe an arch no, cutting APC in 1,

jection. and join GI. Through G, and the centre P, draw the

This method of projection is used in the geometrical diameter HK; draw PL perpendicular to HK, and

delineation of eclipses, occultations, and transits. It is IL perpendicular to GI, meeting PL in L; then L

also particularly useful in various other projections, such will be the centre of the circle HIK, which is that re

as the analemma. See GEOGRAPHY, &c.
quired.
But if the given great circle be inclined to the pri-

PROPOSITION I. THEOREM I.
Fig. 25.

mitive, let it be ADB (fig. 25.), and E its centre :
make the angle BDF equal to the complement of that Every straight line is projected into a straight line. If
given, and DF equal to the tangent of the given arch, the given line be parallel to the plane of projection,
as before. From P, the centre of the primitive, with it is projected into an equal straight line; but if it
the secant of the same arch, describe the arch Gg, and is inclined to the primitive, then the given straight
from E, the centre of the oblique circle, with the ex- line will be to its projection in the ratio of the radius
tent EF, describe an arch intersecting Gg in G. Now to the cosine of inclination.
G being determined, the remaining part of the opera-
tion is performed as before.

Let AB (fig. 27.) be the plane of projection, and Fig. 27
When the given arch exceeds 90°, the tangent and let CD be a straight line parallel thereto: from the ex-
secant of its supplement are to be applied on the line tremities C, D of the straight line CD, draw the lines
DF the contrary way, or towards the right; the former CE, DF perpendicular to AB; then by 3. of xi. of
construction being reckoned to the left.

Eucl. the intersection EF, of the plane CĚFD, with

the plane of projection, is a straight line: and because PROPOSITION XX. PROBLEM XII.

the straight lines CD, EF are parallel, and also CE, Any great circle in the plane of projection being given, DF; therefore by 34. of i. of Eucl. the opposite sides to describe another great circle, which shall make gi

are equal ; hence the straight line CD, and its projecven angles with the primitive and given circles.

tiou EF, are equal. Again, let GH be the proposed

straight line, inclined to the primitive; then the lines Let ADC (fig. 26.) be the given circle, and Q its GE, HF being drawn perpendicular to AB, the interpole. About P the pole of the primitive, describe an cepted portion EF will be the projection of GH. arch m n, at the distance of as many degrees as are in the Through G draw GI parallel to AB, and the angle angle which the required circle is to make with the pri. IGH will be equal to the inclination of the given line mitive. About Q the pole of the circle ADC, and to the plane of projection. Now GH being the radius, at a distance equal to the measure of the angle which GI, or its equal EF, will be the cosine of IGH;

hence the required circle is to make with the given circle the given line GH is to its projection EF as radius to ADC, describe an arch on, cutting mn in n. Then the cosine or inclination.

Fig. 20.

COROLLARIES.

25

429 Orthogra

tance of the parallel circle from the primitive, or the OrthograCOROLLARIES. phic pro

sine of its distance from the pole of the primitive. phic pro. jection of 1. A straight line perpendicular to the plane of pro

jection of the Sphere. jection is projected into a point.

PROPOSITION IV. THEOREM IV. the Sphere. 2. Every straight line in a plane parallel to the pri- An inclined circle is projected into an ellipse, wbose mitive is projected into an equal and parallel straight

transverse axis is the diameter of the circle. line.

3. A plane angle parallel to the primitive is projected 1. Let ELF (fig. 30.) be a great circle inclined to Fig. 30. into an eqnal angle.

the primitive EBF, and EF their line of common sec4. Any plane rectilineal figure parallel to the primi- tion. From the centre C, and any other point K, in tive is projected into an equal and similar figure. EF, let the perpendicular CB, KI be drawn in the

5. The area of any rectilineal figure is to the area plane of the primitive, and CL, KN, in the plane of
of its projection as radius to the cosine of its inclina- the great circle, meeting the circumference in L, N.
tion.

Let LG, ND be perpendicular to CB, KI; then G,
PROPOSITION II. THEOREM II.

D are the projections of L, N. And because the tri

angles LCG, NKD are equiangular, CL: CGʻ:: NK*:
Every great circle, perpendicular to the primitive, is DK; or EC: CG* :: EKF : DK : therefore the

projected into a diameter of the primitive; and every points G, D are in the curve of an ellipse, of which
arch of it, reckoned from the pole of the primitive, EF is the transverse axis, and CG the semiconjugate
is projected into its sine.

axis. Fig. 28. Let BFD (fig. 28.) be the primitive, and ABCD

COROLLARIES. a great circle perpendicular to it, passing through its

1. In a projected great circle, the semiconjugate axis
poles A, C; then the diameter BED, wbich is their

is the cosine of the inclination of the great circle to the
line of common section, will be the projection of the
circle ABCD. For if from any point, as G, in the

primitive.
circle ABC, a perpendicular GH fall upon BD, it will

2. Perpendiculars the transverse axis intercept also be perpendicular to the plane of the primitive: corresponding arches of the projection and the primitherefore H is the projection of G. Hence the whole

3. The eccentricity of the projection is the sine of circle is projected into BD, and any arch AG into EH

the inclination of the great circle to the primitive. equal to GI its sine.

Case 2. Let AQB (hig. 31.) be a less circle, incli- Fig. 31. COROLLARIÉS.

ned to the primitive, and let the great circle LBM, per

pendicular to both, intersect them in the lines AB, LM. 1. Every arch of a great circle, reckoned from its in

From the centre 0, and any other point N in the diatersection with the primitive, is projected into its versed

meter AB, let the perpendiculars TOP, NQ, be drawn
sine.

in the plane of the less circle, to meet its circumference
2. Every less circle perpendicular to the primitive is
projected into its line of common section with the pri- AG, 'NI, OC, BH, be drawn perpendicular to LM;

in T, P, Q. Also from the points A, N, O, B, let
mitive, which is also its own diameter: and

every
arch

and from P, Q, T, draw PE, QD, TF, perpendicular
of the semieircle above the primitive, reckoned from the

to the primitive ; then G, I, C, H, E, D, F, are the
middle point, is projected into its sine.
of a great circle ; and every chord the projection of a primitive, are in the same plane, the plane COPE is
3. Every diameter of the primitive is the projection projections of these points. Because OP is perpendicu-

,
less circle.
4. A spherical angle at the pole of the primitive is perpendicular to LBM. But the primitive is perpen-

dicular to LBM ; therefore the common section EC is
projected into an equal angle.

perpendicular to LBM, and to LM. Hence CP is PROPOSITION III. THEOREM III.

parallelogram, and EC=OP. In like manner, FC,

DI, are proved perpendicular to LM, and equal to
A circle parallel to the primitive is projected into a
circle equal to itself, and concentric with the primio the diameter PT. Let QR, DK be parallel to AB,

OT, NQ. Thus ECF is a straight line, and equal to
tive.

LM; then RO=NQ=DI=KC, and PR XRT-EK
Let the less circle FIG (fig. 29.) be parallel to the XKF. But AO : CG :: NO : CI ; therefore AOS:
plane of the primitive BND. The straight line HE, CGʻ :: QR : DK'; and EC : CG? :: EKF:

:
which joins their centres, is perpendicular to the primi- DK.
tive; therefore E is the projection of H. Let any radii

COROLLARIÉS.
HI and IN perpendicular to the primitive be drawn.
Then IN, HE being parallel, are in the same plane ; 1. The transverse axis is to the conjugate as radius to
therefore IH, NE, the lines of common section of the the cosine of the circle's inclination to the primitive.
plane IE, with two parallel planes, are parallel; and 2. Half the transverse axis is the cosine of half the
the figure IHEN is a parallelogram. Hence NE= sum of the greatest and least distances of the less circle
IH, and consequently FIG is projected into an equal from the primitive.
circle KNL, whose centre is E.

3. The extremities of the conjugate axis are in the
COROLLARY.

line of measures, distant from the centre of the primi

tive by the cosines of the greatest and least distances of The radius of the projection is the cosine of the dis- the less circle from the primitive.

a

Fig. 29