THE PROJECTION of the sphere is a perspective representation of the circles on the surface of the sphere; and is variously denominated according to the different positions of the eye and plane of projection.

There are three principal kinds of projection; the stereographic, the orthographic, and gnomic. In the stereographic projection the eye is supposed to be placed on the surface of the sphere; in the orthographic it is supposed to be at an infinite distance; and in the gnomic projection the eye is placed at the centre of the sphere. Other kinds of projection are, the globular, Mercator's, scenographic, &c. for which see the articles GEOGRAPHY, NAVIGATION, PERSPECTIVE, &c.

Definitions.

1. The plane upon which the circles of the sphere are described, is called the plane of projection, or the primitive circle. The pole of this circle is the pole Stereographic Projection, and the place of the eye is the projecting point.

2. The line of measures of any circle of the sphere is that diameter of the primitive, produced indefinitely, which passes through the centre of the projected circle.

Axiom.

The projection, or representation of any point, is where the straight line drawn from it to the projecting point intersects the plane of projection.

SECTION I.

Of the Stereographic Projection of the Sphere.

In the stereographic projection of the sphere, the

Stereographic Projection of the Sphere, ed. The projection of the hemisphere opposite to the eye falls within the primitive, to which the projection is generally limited; it, however, may be extended to the other hemisphere, or that wherein the eye is placed, the projection of which falls without the primitive.

As all circles in this projection are projected either into circles or straight lines, which are easily described, it is therefore more generally understood, and by many preferred to the other projections.

**Proposition I. Theorem I.**

Every great circle which passes through the projecting point is projected into a straight line passing through the centre of the primitive; and every arch of it, reckoned from the other pole of the primitive, is projected into its semitangent.

Let \(ABCD\) (fig. 1.) be a great circle passing through \(A, C\), the poles of the primitive, and intersecting it in the line of common section \(BED\), \(E\) being the centre of the sphere. From \(A\), the projecting point, let there be drawn straight lines \(AP, AM, AN, AQ\), to any number of points \(P, M, N, Q\), in the circle \(ABCD\); these lines will intersect \(BED\), which is in the same plane with them. Let them meet it in the points \(p, m, n, q\); then \(p, m, n, q\), are the projections of \(P, M, N, Q\); hence the whole circle \(ABCD\) is projected into the straight line \(BED\), passing through the centre of the primitive.

Again, because the pole \(C\) is projected into \(E\), and the point \(M\) into \(m\); therefore the arch \(CM\) is projected into the straight line \(Em\), which is the semitangent of the arch \(CM\) to the radius \(AE\). In like manner, the arch \(CP\) is projected into its semitangent, \(Ep\), &c.

**Corollaries:**

1. Each of the quadrants contiguous to the projecting point is projected into an indefinite straight line, and each of those that are remote into a radius of the primitive.

2. Every small circle which passes through the projecting point is projected into that straight line which is its common section with the primitive.

3. Every straight line in the plane of the primitive, and produced indefinitely, is the projection of some circle on the sphere passing through the projecting point.

4. The projection of any point in the surface of the sphere, is distant from the centre of the primitive, by the semitangent of the distance of that point from the pole opposite to the projecting point.

**Proposition II. Theorem II.**

Every circle on the sphere which does not pass through the projecting point is projected into a circle.

If the given circle be parallel to the primitive, then a straight line drawn from the projecting point to any point in the circumference, and made to revolve about the circle, will describe the surface of a cone; which being cut by the plane of projection parallel to the base, the section will be a circle. See CONIC-Sections.

But if the circle \(MN\) (fig. 2.) be not parallel to the primitive circle \(BD\), let the great circle \(ABCD\), passing through the projecting point, cut it at right angles in the diameter \(MN\), and the primitive in the diameter \(BD\). Through \(M\), in the plane of the great circle, let \(MF\) be drawn parallel to \(BD\); let \(AM, AN\) be joined, and meet \(BD\) in \(m, n\). Then, because \(AB, AD\) are quadrants, and \(BD, MF\) parallel, the arch \(AM\) is equal to \(AF\), and the angle \(AMF\) or \(Amn\) is equal to \(ANM\).

Hence the conic surface described by the revolution of \(AM\) about the circle \(MN\) is cut by the primitive in a subcontrary position; therefore the section is in this case likewise a circle.

**Corollaries:**

1. The centres and poles of all circles parallel to the primitive have their projection in its centre.

2. The centre and poles of every circle inclined to the primitive have their projections in the line of measures.

3. All projected great circles cut the primitive in two points diametrically opposite; and every circle in the plane of projection, which passes through the extremities of a diameter of the primitive, or through the projections of two points that are diametrically opposite on the sphere, is the projection of some great circle.

4. A tangent to any circle of the sphere, which does not pass through the projecting point, is projected into a tangent to that circle's projection; also, the circular projections of tangent circles touch one another.

5. The extremities of the diameter, on the line of measures of any projected circle, are distant from the centre of the primitive by the semitangents of the least and greatest distances of the circle on the sphere, from the pole opposite to the projecting point.

6. The extremities of the diameter, on the line of measures of any projected great circle, are distant from the centre of the primitive by the tangent and cotangent of half the great circle's inclination to the primitive.

7. The radius of any projected circle is equal to half the sum, or half the difference of the semitangents of the least and greatest distances of the circle from the pole opposite to the projecting point, according as that pole is within or without the given circle.

**Proposition III. Theorem III.**

An angle formed by two tangents at the same point in the surface of the sphere, is equal to the angle formed by their projections.

Let \(FGI\) and \(GH\) (fig. 3.) be the two tangents, and \(A\) the projecting point; let the plane \(AGF\) cut the sphere in the circle \(AGL\), and the primitive in the line \(BML\). Also, let \(MN\) be the line of common section of the plane \(AGH\) with the primitive; then the angle \(FGH = LMN\). If the plane \(FGH\) be parallel to the primitive \(BLD\), the proposition is manifest. If not, through any point \(K\) in \(AG\) produced, let the plane \(FKH\), parallel to the primitive, be extended to meet \(FGH\) in the line \(FH\). Then, because the plane \(AGF\) meets the two parallel planes \(BLD, FKH\), the lines of common section \(LM, FK\) are parallel; there- for the angle \( \text{AML} = \text{AKF} \). But since \( A \) is the pole of \( BLD \), the chords, and consequently the arches \( AB \) and \( AL \), are equal, and the arch \( ABG \) is the sum of the arches \( AL \) and \( BG \); hence the angle \( \text{AML} \) is equal to an angle at the circumference standing upon \( AG \), and therefore equal to \( \text{AGI} \) or \( \text{FGK} \); consequently the angle \( \text{FGK} = \text{FKG} \), and the side \( \text{FG} = \text{FK} \). In like manner \( \text{HG} = \text{HK} \); hence the triangles \( \text{GHF}, \text{KHF} \) are equal, and the angle \( \text{FGH} = \text{FKH} = \text{LMN} \).

**Corollaries.**

1. An angle contained by any two circles of the sphere is equal to the angle formed by their projections. For the tangents to these circles on the sphere are projected into straight lines, which either coincide with, or are tangents to, their projections on the primitive.

2. An angle contained by any two circles of the sphere is equal to the angle formed by the radii of their projections at the point of intersection.

**Proposition IV. Theorem IV.**

The centre of a projected great circle is distant from the centre of the primitive; the tangent of the inclination of the great circle to the primitive, and its radius, is the secant of its inclination.

Let \( \text{MNG} \) (fig. 4.) be the projection of a great circle, meeting the primitive in the extremities of the diameter \( MN \), and let the diameter \( BD \), perpendicular to \( MN \), meet the projection in \( F, G \). Bifect \( FG \) in \( H \), and join \( NH \). Then, because any angle contained by two circles of the sphere is equal to the angle formed by the radii of their projections at the point of intersection; therefore the angle contained by the proposed great circle and the primitive is equal to the angle \( \text{ENH} \), of which \( EH \) is the tangent, and \( NH \) the secant, to the radius of the primitive.

**Corollaries.**

1. All circles which pass through the points \( M, N \), are the projections of great circles, and have their centres in the line \( BG \); and all circles which pass through the points \( F, G \), are the projections of great circles, and have their centres in the line \( HI \), perpendicular to \( BG \).

2. If \( NF, NH \) be continued to meet the primitive in \( L, F \); then \( BL \) is the measure of the great circle's inclination to the primitive; and \( MT = 2BL \).

**Proposition V. Theorem V.**

The centre of projection of a less circle perpendicular to the primitive, is distant from the centre of the primitive, the secant of the distance of the less circle from its nearest pole; and the radius of projection is the tangent of that distance.

Let \( \text{MN} \) (fig. 5.) be the given less circle perpendicular to the primitive, and \( A \) the projecting point. Draw \( AM, AN \) to meet the diameter \( BD \) produced in \( G \) and \( H \); then \( GH \) is the projected diameter of the less circle; bifect \( GH \) in \( C \), and \( C \) will be its centre; join \( NE, NC \). Then because \( AE, NI \) are parallel, the angle \( \text{INE} = \text{NEA} \); but \( \text{NEA} = 2\text{NMA} = 2\text{NHG} = \text{NCG} \): hence \( \text{ENC} = \text{INE} + \text{INC} = \text{NCG} + \text{INC} \) is a right angle; and therefore \( NC \) is a tangent to the primitive at \( N \); but the arch \( ND \) is the distance of the less circle from its nearest pole \( D \); hence \( NC \) is the tangent, and \( EC \) the secant of the distance of the less circle from its pole to the radius of the primitive.

**Proposition VI. Theorem VI.**

The projection of the poles of any circle, inclined to the primitive, are, in the line of measures, distant from the centre of the primitive, the tangent, and cotangent, of half its inclination.

Let \( \text{MN} \) (fig. 6.) be a great circle perpendicular to the primitive \( ABCD \), and \( A \) the projecting point; then \( P, p \) are the poles of \( MN \), and of all its parallels \( m, n, \ldots \). Let \( AP, Ap \) meet the diameter \( BD \) in \( F, f \); which will therefore be the projected poles of \( MN \) and its parallels. The angle \( \text{BEM} \) is the inclination of the circle \( \text{MEN} \), and its parallels, to the primitive; and because \( BC \) and \( MP \) are quadrants, and \( MC \) common to both; therefore \( PC = BM \); and hence \( \text{PEC} \) is also the inclination of \( MN \) and its parallels. Now \( \text{EF} \) is the tangent of \( \text{EAF} \), or of half the angle \( \text{PEC} \) the inclination; and \( \text{E}f \) is the tangent of the angle \( \text{EAF} \); but \( \text{E}f \) is the complement of \( \text{EAF} \), hence \( \text{E}f \) is the cotangent of half the inclination.

**Corollaries.**

1. The projection of that pole which is nearest to the projecting point is without the primitive, and the projection of the other within.

2. The projected centre of any circle is always between the projection of its nearest pole and the centre of the primitive; and the projected centres of all circles are contained between their projected poles.

**Proposition VII. Theorem VII.**

Equal arches of any two great circles of the sphere will be intercepted between two other circles drawn on the sphere through the remote poles of those great circles.

Let \( \text{AGB}, \text{CFD} \) (fig. 7.) be two great circles of the sphere, whose remote poles are \( E, P \); through which draw the great circle \( \text{PREC} \), and less circle \( \text{PG} \), intersecting the great circles \( \text{AGB}, \text{CFD} \), in the points \( B, G, \) and \( D, F \); then the arch \( \text{BG} \) is equal to the arch \( \text{DF} \).

Because \( E \) is the pole of the circle \( \text{AGB} \), and \( P \) the pole of \( \text{CFD} \), therefore the arches \( \text{EB}, \text{PD} \) are equal; and since \( BD \) is common to both, hence the arch \( \text{ED} \) is equal to the arch \( \text{PB} \). For the same reason, the arches \( \text{EF}, \text{PG} \) are equal; but the angle \( \text{DEF} \) is equal to the angle \( \text{BPG} \); hence these triangles are equal, and therefore the arch \( \text{DF} \) is equal to the arch \( \text{BG} \).

**Proposition VIII. Theorem VIII.**

If from either pole of a projected great circle, two straight lines be drawn to meet the primitive and the projection, they will intercept similar arches of these circles. On the plane of projection AGB (fig. 7.) let the great circle CFD be projected into cfd, and its pole P into p; through p draw the straight lines pd, pf, then are the arches GB, cf similar.

Since pd lies both in the plane AGB and APBE, it is in their common section, and the point B is also in their common section; therefore pd passes through the point B. In like manner it may be shown that the line pf passes through G. Now the points D, F are projected into d, f; hence the arches FD, fd are similar; but GB is equal to FD, therefore the intercepted arch of the primitive GB is similar to the projected arch fd.

Corollary.

Hence, if from the angular point of a projected spherical angle two straight lines be drawn through the projected poles of the containing sides, the intercepted arch of the primitive will be the measure of the spherical angle.

Proposition IX. Problem I.

To describe the projection of a great circle through two given points in the plane of the primitive.

Let P and B be given points, and C the centre of the primitive.

1. When one point P (fig. 8.) is the centre of the primitive, a diameter drawn through the given points will be the great circle required.

2. When one point P (fig. 9.) is in the circumference of the primitive. Through P draw the diameter PD; and an oblique circle described through the three points P, B, D, will be the projection of the required great circle.

3. When the given points are neither in the centre nor circumference of the primitive. Through either of the given points P (fig. 10.) draw the diameter ED, and at right angles thereto draw the diameter FG. From F through P draw the straight line FPH, meeting the circumference in H; draw the diameter HI, and draw the straight line FIK, meeting ED produced in K; then an arch, terminated by the circumference, being described through the three points, P, B, K, will be the great circle.

Proposition X. Problem II.

To describe the representation of a great circle about any given point as a pole.

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

1. When P (fig. 8.) is in the centre of the primitive, then the primitive will be the great circle required.

2. When the pole P (fig. 11.) is in the circumference of the primitive. Through P draw the diameter PE, and the diameter AB drawn at right angles to PE will be the projected great circle required.

3. When the given pole is neither in the centre nor circumference of the primitive. Through the pole P (fig. 12.) draw the diameter AB, and draw the diameter DE perpendicular to AB; through E and P draw the straight line EPF, meeting the circumference in F. Make FG equal to FD; through E and G draw the straight line EGH, meeting the diameter AB produced if necessary in H; then from the centre H, with the radius HE, describe the oblique circle DIE, and it will be the projection of the great circle required.

Or, make DK equal to FA; join EK, which intersects the diameter AB in I; then through the three points, D, I, E, describe the oblique circle DIE.

Proposition XI. Problem III.

To find the poles of a great circle.

1. When the given great circle is the primitive, its centre is the pole.

2. To find the pole of the right circle ACB (fig. 11.). Draw the diameter PE perpendicular to the given circle AB; and its extremities P, E are the poles of the circle ACB.

3. To find the pole of the oblique circle DEF (fig. Fig. 13.). Join DF, and perpendicular thereto draw the diameter AB, cutting the given oblique circle DEF in E. Draw the straight line FEG, meeting the circumference in G. Make GI, GH, each equal to AD; then FI being joined, cuts the diameter AB in P, the lower pole; through F and H draw the straight line FH p, meeting the diameter AB produced in p, which will be the opposite or exterior pole.

Proposition XII. Problem IV.

To describe a less circle about any given point as a pole, and at any given distance from that pole.

1. When the pole of the less circle is in the centre of the primitive; then from the centre of the primitive, with the semitangent of the distance of the given circle from its pole, describe a circle, and it will be the projection of the less circle required.

2. If the given pole is in the circumference of the primitive, from C (fig. 14.), the centre of the primitive, Fig. 14. feet off CE the secant of the distance of the less circle from its pole P; then from the centre E, with the tangent of the given distance, describe a circle, and it will be the less circle required. Or, make PG, PF each equal to the chord of the distance of the less circle from its pole. Through R, G, draw the straight line BGD meeting CP produced in D; bisect GD in H, and draw HE perpendicular to GD, and meeting PD in E; then E is the centre of the less circle.

3. When the given pole is neither in the centre nor circumference of the primitive. Through P (fig. 15.), Fig. 15. the given pole, and C the centre of the primitive, draw the diameter AB, and draw the diameter DE perpendicular to AB; join EP, and produce it to meet the primitive in p; make p F, p G, each equal to the chord of the distance of the less circle from its pole; join EF which intersects the diameter AB in H; from E through G draw the straight line EGI, meeting the diameter AB produced in I; bisect HI in K: Then a circle described from the centre K, at the distance KH or KI, will be the projection of the less circle.

Proposition XIII. Problem V.

To find the poles of a given less circle.

The poles of a less circle are also those of its parallel great.

If therefore the parallel great circle be given, then its poles being found by Prob. III., will be those of the left circle. But if the parallel great circle be not given, let HMIN (fig. 15.) be the given left circle. Through its centre, and C the centre of the primitive, draw the line of measures IAHB; and draw the diameter DE perpendicular to it, also draw the straight line EHF meeting the primitive in F; make FP equal to the chord of the distance of the left circle from its pole: join EP, and its intersection P with the diameter AB is the interior pole. Draw the diameter pCL through E and L, draw ELq meeting the diameter AB produced in q; then q is the external pole. Or thus: Join EI intersecting the primitive in G; join also EH, and produce it to meet the primitive in F; bisect the arch GH in p; from E to p draw the straight line EPp, and P is the pole of the given left circle.

PROPOSITION XIV. PROBLEM VI.

To measure any arch of a great circle.

1. Arches of the primitive are measured on the line of chords. 2. Right circles are measured on the line of semitangents, beginning at the centre of the primitive. Thus, the measure of the portion AC (fig. 16.) of the right circle DE, is found by applying it to the line of semitangents. The measure of the arch DB is found by subtracting that of BC from 90°: the measure of the arch AF, lying partly on each side of the centre, is obtained by adding the measures of AC and CF. Lastly, to measure the part AB, which is neither terminated at the centre or circumference of the primitive, apply CA to the line of semitangents; then CB, and the difference between the measures of these arches, will be that of AB.

Or thus: Draw the diameter GH perpendicular to DE; then from either extremity, as D, of this diameter, draw lines through the extremities of the arch intended to be measured; and the intercepted portion of the primitive applied to the line of chords will give the measure of the required arch. Thus IK applied to the line of chords will give the measure of AR.

3. To measure an arch of an oblique circle: draw lines from its pole through the extremities of the arch to meet the primitive, then the intercepted portion of the primitive applied to the line of chords will give the measure of the arch of the oblique circle. Thus, let AB (fig. 17.), be an arch of an oblique circle to be measured, and P its pole; from P draw the lines PAD, PBE meeting the primitive in B and E; then the arch DE applied to the line of chords will give the measure of the arch of the oblique circle AB.

PROPOSITION XV. PROBLEM VII.

To measure any arch of a left circle.

Let DEG (fig. 18.) be the given left circle, and DE the arch to be measured: find its internal pole P; and describe the circle AFI parallel to the primitive, and whose distance from the projecting point may be equal to the distance of the given left circle from its pole P; then join PD, PE, which produce to meet the parallel circle in A and F. Now AF applied to a line of chords will give the measure of the arch DE of the given left circle.

PROPOSITION XVI. PROBLEM VIII.

To measure any spherical angle.

1. If the angle is at the centre of the primitive, it is measured as a plane angle. 2. When the angular point is in the circumference of the primitive; let A (fig. 19.) be the angular point, Fig. 19. and ABE an oblique circle inclined to the primitive. Through P, the pole of ABE, draw the line APp meeting the circumference in p: then the arch AFp is the measure of the angle BAD, and the arch AFp is the measure of its supplement BAF: also pF is the measure of the angle BAC, and pED that of its supplement.

3. If the angular point is neither at the centre nor circumference of the primitive. Let A (fig. 20.) be Fig. 20. the angular point, and DAH, or GAF, the angle to be measured, P the pole of the oblique circle DAF, and p the pole of GAH: then from A, through the points Pp, draw the straight lines APM, ApN, and the arch MN will be the measure of the angle DAH; and the supplement of MN will be the measure of the angle HAF or DAG.

PROPOSITION XVII. PROBLEM IX.

To draw a great circle perpendicular to a projected great circle, and through a point given in it.

Find the pole of the given circle, then a great circle described through that pole and the given point will be perpendicular to the given circle. Hence if the given circle be the primitive, then a diameter drawn through the given point will be the required perpendicular. If the given circle is a right one, draw a diameter at right angles to it; then though the extremities of this diameter and the given point describe an oblique circle, and it will be perpendicular to that given. If the given circle is inclined to the primitive, let it be represented by BAD (fig. 21.), whose pole is P, and let A be the Fig. 21. point through which the perpendicular is to be drawn: then, by Prob. I. describe a great circle through the points P and A, and it will be perpendicular to the oblique circle BAD.

PROPOSITION XVIII. PROBLEM X.

Through a point in a projected great circle, to describe another great circle to make a given angle with the former, provided the measure of the given angle is not less than the distance between the given point and circle.

Let the given circle be the primitive, and let A (fig. 19.) be the angular point. Draw the diameters AE, DF perpendicular to each other; and make the angle CAG equal to that given, or make CG equal to the tangent of the given angle; then from the centre G, with the distance GC, describe the oblique circle ABE, and it will make with the primitive an angle equal to that given.

If the given circle be a right one, let it be APB (fig. Fig. 22.) and let P be the given point. Draw the diameter GH

Stereographic Projection of the Sphere.

GH perpendicular to AB; join GP, and produce it to make Hb equal to twice Aa; and Gb being joined, the sphere intersects AB in C. Draw CD perpendicular to AB, and equal to the cotangent of the given angle to the radius PC; or make the angle CPD equal to the complement of that given: then from the centre D, with the radius DP, describe the great circle FPE, and the angle APF, or BPE, will be equal to that given.

If APB (fig. 23.) is an oblique circle. From the angular point P, draw the lines PG, PC through the centres of the primitive and given oblique circle. Through C, the centre of APB, draw GCD at right angles to PG; make the angle GPD equal to that given; and from the centre D, with the radius DP, describe the oblique circle FPE, and the angle APF, or BPE, will be equal to that proposed.

PROPOSITION XIX. PROBLEM XI.

Any great circle cutting the primitive being given, to describe another great circle which shall cut the given one in a proposed angle, and have a given arch intercepted between the primitive and given circles.

If the given circle be a right one, let it be represented by APC (fig. 24.) and at right angles thereto draw the diameter BPM; make the angle BPF equal to the complement of the given angle, and PF equal to the tangent of the given arch; and from the centre of the primitive with the secant of the same arch describe the arch Gg. Through F draw FG parallel to AC, meeting Gg in G; then from the centre G, with the tangent PF, describe an arch no, cutting APC in I, and join GI. Through G, and the centre P, draw the diameter HK; draw PL perpendicular to HK, and IL perpendicular to GI, meeting PL in L; then L will be the centre of the circle HIK, which is that required.

But if the given great circle be inclined to the primitive, let it be ADB (fig. 25.), and E its centre; make the angle BDF equal to the complement of that given, and DF equal to the tangent of the given arch, as before. From P, the centre of the primitive, with the secant of the same arch, describe the arch Gg, and from E, the centre of the oblique circle, with the extent EF, describe an arch intersecting Gg in G. Now G being determined, the remaining part of the operation is performed as before.

When the given arch exceeds 90°, the tangent and secant of its supplement are to be applied on the line DF the contrary way, or towards the right; the former construction being reckoned to the left.

PROPOSITION XX. PROBLEM XII.

Any great circle in the plane of projection being given, to describe another great circle, which shall make given angles with the primitive and given circles.

Let ADC (fig. 26.) be the given circle, and Q its pole. About P the pole of the primitive, describe an arch mn, at the distance of as many degrees as are in the angle which the required circle is to make with the primitive. About Q the pole of the circle ADC, and at a distance equal to the measure of the angle which the required circle is to make with the given circle ADC, describe an arch on, cutting mn in n. Then about n as a pole, describe the great circle EDF, cutting the primitive and given circle in E and D, and it will be the great circle required.

Scholium.

It will hence be an easy matter to construct all the various spherical triangles. The reader is, however, referred to the article Spherical Trigonometry, for the method of constructing them agreeably to this projection; and also for the application to the resolution of problems of the sphere. For the method of projecting the sphere upon the plane of the meridian, and of the horizon, according to the stereographic projection, see the article Geography.

SECTION II.

Of the Orthographic Projection of the Sphere.

The orthographic projection of the sphere, is that in which the eye is placed in the axis of the plane of projection, at an infinite distance with respect to the diameter of the sphere; so that at the sphere all the visual rays are assumed parallel, and therefore perpendicular to the plane of projection.

Hence the orthographic projection of any point is where a perpendicular from that point meets the plane of projection; and the orthographic representation of any object is the figure formed by perpendiculars drawn from every point of the object to the plane of projection.

This method of projection is used in the geometrical delineation of eclipses, occultations, and transits. It is also particularly useful in various other projections, such as the analemma. See Geography, &c.

PROPOSITION I. THEOREM I.

Every straight line is projected into a straight line. If the given line be parallel to the plane of projection, it is projected into an equal straight line; but if it is inclined to the primitive, then the given straight line will be to its projection in the ratio of the radius to the cosine of inclination.

Let AB (fig. 27.) be the plane of projection, and Fig. 27. let CD be a straight line parallel thereto: from the extremities C, D of the straight line CD, draw the lines CE, DF perpendicular to AB; then by 3. of xi. of Eucl. the intersection EF, of the plane CEFD, with the plane of projection, is a straight line: and because the straight lines CD, EF are parallel, and also CE, DF; therefore by 34. of i. of Eucl. the opposite sides are equal; hence the straight line CD, and its projection EF, are equal. Again, let GH be the proposed straight line, inclined to the primitive; then the lines GE, HF being drawn perpendicular to AB, the intercepted portion EF will be the projection of GH. Through G draw GI parallel to AB, and the angle IGH will be equal to the inclination of the given line to the plane of projection. Now GH being the radius, GI, or its equal EF, will be the cosine of IGH; hence the given line GH is to its projection EF as radius to the cosine or inclination.

Corollaries.

Corollaries.

1. A straight line perpendicular to the plane of projection is projected into a point.

2. Every straight line in a plane parallel to the primitive is projected into an equal and parallel straight line.

3. A plane angle parallel to the primitive is projected into an equal angle.

4. Any plane rectilineal figure parallel to the primitive is projected into an equal and similar figure.

5. The area of any rectilineal figure is to the area of its projection as radius to the cosine of its inclination.

Proposition II. Theorem II.

Every great circle, perpendicular to the primitive, is projected into a diameter of the primitive; and every arch of it, reckoned from the pole of the primitive, is projected into its fine.

Let BFD (fig. 28.) be the primitive, and ABCD a great circle perpendicular to it, passing through its poles A, C; then the diameter BED, which is their line of common section, will be the projection of the circle ABCD. For if from any point, as G, in the circle ABC, a perpendicular GH fall upon BD, it will also be perpendicular to the plane of the primitive; therefore H is the projection of G. Hence the whole circle is projected into BD, and any arch AG into EH equal to GI its fine.

Corollaries.

1. Every arch of a great circle, reckoned from its intersection with the primitive, is projected into its verified fine.

2. Every less circle perpendicular to the primitive is projected into its line of common section with the primitive, which is also its own diameter; and every arch of the semicircle above the primitive, reckoned from the middle point, is projected into its fine.

3. Every diameter of the primitive is the projection of a great circle; and every chord the projection of a less circle.

4. A spherical angle at the pole of the primitive is projected into an equal angle.

Proposition III. Theorem III.

A circle parallel to the primitive is projected into a circle equal to itself, and concentric with the primitive.

Let the less circle FIG (fig. 29.) be parallel to the plane of the primitive BND. The straight line HE, which joins their centres, is perpendicular to the primitive; therefore E is the projection of H. Let any radii HI and IN perpendicular to the primitive be drawn. Then IN, HE being parallel, are in the same plane; therefore IH, NE, the lines of common section of the plane IE, with two parallel planes, are parallel; and the figure IHEN is a parallelogram. Hence NE = IH, and consequently FIG is projected into an equal circle KNL, whose centre is E.

Corollary.

The radius of the projection is the cosine of the distance of the parallel circle from the primitive, or the sine of its distance from the pole of the primitive.

Proposition IV. Theorem IV.

An inclined circle is projected into an ellipse, whose transverse axis is the diameter of the circle.

1. Let ELF (fig. 30.) be a great circle inclined to the primitive EBF, and EF their line of common section. From the centre C, and any other point K, in EF, let the perpendicular CB, KI be drawn in the plane of the primitive, and CL, KN, in the plane of the great circle, meeting the circumference in L, N. Let LG, ND be perpendicular to CB, KI; then G, D are the projections of L, N. And because the triangles LCG, NKD are equiangular, \( \frac{CL}{CG} = \frac{NK}{DK} \); or \( \frac{EC}{CG} = \frac{EK}{DK} \); therefore the points G, D are in the curve of an ellipse, of which EF is the transverse axis, and CG the semiconjugate axis.

Corollaries.

1. In a projected great circle, the semiconjugate axis is the cosine of the inclination of the great circle to the primitive.

2. Perpendiculars to the transverse axis intercept corresponding arches of the projection and the primitive.

3. The eccentricity of the projection is the sine of the inclination of the great circle to the primitive.

Case 2. Let AQB (fig. 31.) be a less circle, inclined to the primitive, and let the great circle LBM, perpendicular to both, intersect them in the lines AB, LM. From the centre O, and any other point N in the diameter AB, let the perpendiculars TOP, NO, be drawn in the plane of the less circle, to meet its circumference in T, P, Q. Also from the points A, N, O, B, let AG, NI, OC, BH, be drawn perpendicular to LM; and from P, Q, T, draw PE, QD, TF, perpendicular to the primitive; then G, I, C, H, E, D, F, are the projections of these points. Because OP is perpendicular to LMB, and OC, PE, being perpendicular to the primitive, are in the same plane, the plane COPE is perpendicular to LBM. But the primitive is perpendicular to LBM; therefore the common section EC is perpendicular to LBM, and to LM. Hence CP is a parallelogram, and EC = OP. In like manner, FC, DI, are proved perpendicular to LM, and equal to OT, NO. Thus ECF is a straight line, and equal to the diameter PT. Let QR, DK be parallel to AB, LM; then RO = NO = DI = KC, and PR = RT = EK = KF. But AO : CG :: NO : CI; therefore AO² : CG² :: QR² : DK²; and EC² : CG² :: EKF : DK².

Corollaries.

1. The transverse axis is to the conjugate as radius to the cosine of the circle's inclination to the primitive.

2. Half the transverse axis is the cosine of half the sum of the greatest and least distances of the less circle from the primitive.

3. The extremities of the conjugate axis are in the line of measures, distant from the centre of the primitive by the cosines of the greatest and least distances of the less circle from the primitive. 4. If from the extremities of the conjugate axis of any elliptical projection perpendiculars be drawn (in the same direction if the circle do not intersect the primitive, but if otherwise in opposite directions), they will intersect an arch of the primitive, whose chord is equal to the diameter of the circle.

**Proposition V. Theorem V.**

The projected poles of an inclined circle are in its line of measures distant from the centre of the primitive the sine of the inclination of the circle to the primitive.

Let \(ABCD\) (fig. 32.) be a great circle, perpendicular both to the primitive and the inclined circle, and intersecting them in the diameters \(AC, MN\). Then \(ABCD\) passes through the poles of the inclined circle; let these be \(P_r Q_s\); and let \(P_p, Q_q\) be perpendicular to \(AC\); \(p, q\) are the projected poles; and it is evident that \(PO = \text{sine of } BP, \text{ or } MA, \text{ the inclination.}\)

**Corollaries.**

1. The centre of the primitive, the centre of the projection, the projected poles, and the extremities of the conjugate axis, are all in one and the same straight line.

2. The distance of the centre of projection from the centre of the primitive, is to the cosine of the distance of the circle from its own pole, as the sine of the circle's inclination to the primitive is to the radius.

**Proposition VI. Problem I.**

To describe the projection of a circle perpendicular to the primitive, and whose distance from its pole is equal to a given quantity.

Let \(PAPB\) (fig. 33.) be the primitive circle, and \(P, p\) the poles of the right circle to be projected. Then if the circle to be projected is a great circle, draw the diameter \(AB\) at right angles to the axis \(PP\), and it will be that required. But if the required projection is that of a less circle, make \(PE, PF\) each equal to the chord of the distance of the less circle from its pole; join \(EF\), and it will be the projection of the less circle required.

**Proposition VII. Problem II.**

Through a given point in the plane of the primitive to describe the projection of a great circle, having a given inclination to the primitive.

1. When the given inclination is equal to a right angle, a straight line drawn through the centre of the primitive, and the given point, will be the projection required.

2. When the given inclination is less than a right angle, and the given point in the circumference of the primitive. Let \(R\) (fig. 34.) be a point given in the circumference of the primitive, through which it is required to draw the projection of a great circle, inclined to the primitive in an angle measured by the arch \(QP\) of the primitive.

Through the given point \(R\) draw the diameter \(RCS\), and draw \(GCg\) at right angles to it. Make the arch \(GV\) of the primitive equal to \(QP\), and draw \(VA\) at right angles to \(GC\); and in \(Gg\), towards the opposite parts of \(C\), take \(CB\) equal to \(AC\); then, with the greater axis \(RS\), and less axis \(AB\), describe an ellipse, and it will be the projection of the oblique circle required.

3. When the distance of the given point from the primitive is equal to the cosine of the given inclination.

Every thing remaining as in the preceding case; let \(A\) be the given point, and \(AC\) the cosine of an arch \(GV\), equal to the given arch \(QP\); then drawing the diameter \(RCS\) at right angles to \(ACB\), the ellipse described with the given axis \(RS, AB\) will be the projection of the inclined circle.

4. When the distance of the given point from the centre of the primitive is less than the semidiameter of the primitive, but greater than the cosine of the given inclination.

Let \(D\) be the given point, through which draw the diameter \(ICi\); and at the point \(D\) draw \(DL\) perpendicular to \(DC\) meeting the primitive in \(L\); also draw \(LK\), making with \(LD\) the angle \(DLK\) equal to the complement of the given inclination. Let \(LK\) meet \(DC\) in \(K\); then will \(DK\) be less than \(DC\). On \(DC\) as diameter describe a circle, and make \(DH\) equal to the sphere, \(DK\); through \(H\) draw a diameter of the primitive \(RCS\), and describe an ellipse through the points \(R, D, S\), and it will be the projection of the inclined circle.

**Proposition VIII. Problem III.**

Through two given points in the plane of the primitive to describe the projection of a great circle.

1. If the two given points and the centre of the primitive be in the same straight line, then a diameter of the primitive being drawn through these points will be the projection of the great circle required.

2. When the two given points are not in the same straight line with the centre of the primitive; and one of them is in the circumference of the primitive.

Let \(DR\) (fig. 34.) be the two given points, of which \(R\) is in the circumference of the primitive. Draw the diameters \(RCS\), and \(GCg\), \(FDH\) perpendicular to it, meeting the primitive in \(GgF\). Divide \(GCgC\), in \(A, B\), in the same proportion as \(FH\) is divided in \(D\); and describe the ellipse whose axes are \(RS, AB\), and centre \(C\); and it will be the projection required.

3. When the given points are within the primitive, and not in the same straight line with its centre.

Let \(D, E\) (fig. 35.) be the two given points; through \(C\) the centre of the primitive draw the straight lines \(ID, KEi\); draw \(DL\) perpendicular to \(Ii\), and \(EO\) perpendicular to \(Kk\), meeting the primitive in \(L, O\). Through \(E\), and towards the same parts of \(C\), draw \(EP\) parallel to \(DC\), and in magnitude a fourth proportional to \(LD, DC, OE\). Draw the diameter \(CP\) meeting the primitive in \(R, S\), and describe an ellipse through the points \(D\) and \(R, S\), and it will also pass through \(E\). This ellipse will be the projection of the proposed inclined circle.

**Proposition IX. Problem IV.**

To describe the projection of a less circle parallel to the primitive, its distance from the pole of the primitive being given.

From the pole of the primitive, with the fine of the given distance of the circle from its pole, describe a circle, and it will be the projection of the given left circle.

PROPOSITION X. PROBLEM V.

About a given point as a projected pole to describe the projection of an inclined circle, whose distance from its pole is given.

Let P (fig. 36.) be the given projected pole, through which draw the diameter Gg, and draw the diameter Hb perpendicular thereto. From P draw PL perpendicular to GP meeting the circumference in L; through which draw the diameter Ll. Make LT, LK each equal to the chord of the diameter of the left circle from its pole, and join TK, which intersects Ll, in Q. From the points T, Q, K, draw the lines FA, QS, KB, perpendicular to Gg; and make OR, OS, each equal to QT, or QK. Then an ellipse described through the points A, S, B, R will be the projection of the proposed left circle.

PROPOSITION XI. PROBLEM VI.

To find the poles of a given projected circle.

1. If the projected circle be parallel to the primitive, the centre of the primitive will be its pole.

2. If the circle be perpendicular to the primitive, then the extremities of a diameter of the primitive drawn at right angles to the straight line representing the projected circle, will be the poles of that circle.

3. When the projected circle is inclined to the primitive.

Let ARBS (fig. 36, 37.) be the elliptical projection of any oblique circle; through the centre of which, and C the centre of the primitive, draw the line of measures CBA, meeting the ellipse in B, A; and the primitive in G, g. Draw CH, BK, AT perpendicular to Gg, meeting the primitive in H, K, T. Bifect the arch KT in L, and draw LP perpendicular to Gg; then P will be the projected pole of the circle, of which ARBS is the projection.

PROPOSITION XII. PROBLEM VII.

To measure any portion of a projected circle, and conversely.

1. When the given projection is that of a great circle.

Let ADBE (fig. 38.) be the given great circle, either perpendicular or inclined to the primitive, of which the portion DE is to be measured, and let Mm be the line of measures of the given circle. Through the points D, E, draw the lines EG, DF parallel to Mm; and the arch FG of the primitive will be the measure of the arch DE of the great circle, and conversely.

2. When the projection is that of a left circle parallel to the primitive.

Let DE (fig. 39.) be the portion to be measured, of the left circle DEH parallel to the primitive. From the centre C draw the lines CD, CE, and produce them to meet the primitive in the points B, F. Then the intercepted portion BF of the primitive will be the measure of the given arch DE of the left circle DEH.

3. If the given left circle, of which an arch is to be measured, is perpendicular to the primitive.

Let ADEB (fig. 40.) be the left circle, of which Fig. 40. the measure of the arch DE is required. Through C, the centre of the primitive, draw the line of measures Mm, and from the intersection O of the given right circle, and the line of measures, with the radius OA, or OB, describe the semicircle AFGB; through the points D, E, draw the lines DF, EG parallel to the line of measures, and the arch FG will be the measure of DE, to the radius AO. In order to find a similar arch in the circumference of the primitive, join OF, OG, and at the centre C of the primitive, make the angle mCH equal to FOG, and the arch mH to the radius Cm will be the measure of the arch DE.

4. When the great projection is of a left circle inclined to the primitive.

Let RDS (fig. 41.) be the projection of a left circle inclined to the primitive, and DE a portion of that circle to be measured. Through O the centre of the projected circle, and C the centre of the primitive, draw the line of measures Mm; and from the centre O, with the radius OR, or OS, describe the semicircle RGFS; through the points D, E, draw the lines DF, EG parallel to the line of measures, and FG will be the measure of the arch DE to the radius OR, or OS. Join OF, OG, and make the angle mCH equal to FOG, and the arch mH of the primitive will be the measure of the arch DE of the inclined circle RDS.

The converse of this proposition, namely, to cut off an arch from a given projected circle equal to a given arch of the primitive, is obvious.

The above operation would be greatly shortened by using the line of lines in the sector.

It seems unnecessary to insist farther on this projection, especially as the reader will see the application of it to the projection of the sphere on the planes of the Meridian, Equator, and Horizon, in the article Geography; and to the delineation of Eclipses in the article Astronomy. The Analemma, Plate CCXXXV. in the article Geography, is also according to this projection; and the method of applying it to the solution of astronomical problems is there exemplified.

SECTION III.

Of the Gnomonic Projection of the Sphere.

In this projection the eye is in the centre of the sphere, and the plane of projection touches the sphere in a given point parallel to a given circle. It is named gnomonic, on account of its being the foundation of dialling: the plane of projection may also represent the plane of a dial, whose centre being the projected pole, the semiaxis of the sphere will be the style or gnomon of the dial.

As the projection of great circles is represented by straight lines, and left circles parallel to the plane of projection are projected into concentric circles; therefore many problems of the sphere are very easily resolved. Other problems, however, become more intricate on account of some of the circles being projected into ellipses, parabolas, and hyperbolas.

PROPOSITION PROPOSITION I. THEOREM I.

Every great circle is projected into a straight line perpendicular to the line of measures; and whose distance from the circle is equal to the cotangent of its inclination, or to the tangent of its nearest distance from the pole of the projection.

Let BAD (fig. 42.) be the given circle, and let the circle CBED be perpendicular to BAD, and to the plane of projection; whose intersection CF with this last plane will be the line of measures. Now since the circle CBED is perpendicular both to the given circle BAD and to the plane of projection, the common section of the two last planes produced will therefore be perpendicular to the plane of the circle CBED produced, and consequently to the line of measures; hence the given circle will be projected into that section; that is, into a straight line passing through d, perpendicular to C d. Now C d is the cotangent of the angle C d A, the inclination of the given circle, or the tangent of the arch CD to the radius AC.

COROLLARIES.

1. A great circle perpendicular to the plane of projection is projected into a straight line passing through the centre of projection: and any arch is projected into its correspondent tangent.

2. Any point, as D, or the pole of any circle, is projected into a point d, whose distance from the pole of projection is equal to the tangent of that distance.

3. If two great circles be perpendicular to each other, and one of them passes through the pole of projection, they will be projected into two straight lines perpendicular to each other.

4. Hence if a great circle be perpendicular to several other great circles, and its representation passes through the centre of projection; then all these circles will be represented by lines parallel to one another, and perpendicular to the line of measures, for representation of that first circle.

PROPOSITION II. THEOREM II.

If two great circles intersect in the pole of projection, their representations will make an angle at the centre of the plane of projection, equal to the angle made by these circles on the sphere.

For since both these circles are perpendicular to the plane of projection, the angle made by their intersections with this plane is the same as the angle made by these circles.

PROPOSITION III. THEOREM III.

Any less circle parallel to the plane of projection is projected into a circle whose centre is the pole of projection, and its radius is equal to the tangent of the distance of the circle from the pole of projection.

Let the circle PI (fig. 42.) be parallel to the plane GF, then the equal arches PC, CI are projected into the equal tangents GC, CH; and therefore C the point of contact and pole of the circle PI and of the projection, is the centre of the representation G, H.

COROLLARY.

If a circle be parallel to the plane of projection, and 45 degrees from the pole, it is projected into a circle equal to a great circle of the sphere; and therefore may be considered as the primitive circle, and its radius the radius of projection.

PROPOSITION IV. THEOREM IV.

A less circle not parallel to the plane of projection is projected into a conic section, whose transverse axis is in the line of measures; and the distance of its nearest vertex from the centre of the plane of projection is equal to the tangent of its nearest distance from the pole of projection; and the distance of the other vertex is equal to the tangent of the greatest distance.

Any less circle is the base of a cone whose vertex is at A (fig. 43.) and this cone being produced, its intersection with the plane of projection will be a conic section. Thus the cone DAF, having the circle DF for its base, being produced, will be cut by the plane of projection in an ellipse whose transverse diameter is df; and C d is the tangent of the angle CAD, and C f the tangent of CAF. In like manner, the cone AFE, having the side AE parallel to the line of measures df, being cut by the plane of projection, the section will be a parabola, of which f is the nearest vertex, and the point into which E is projected is at an infinite distance. Also the cone AFG, whose base is the circle FG, being cut by the plane of projection, the section will be a hyperbola; of which f is the nearest vertex; and GA being produced gives d the other vertex.

COROLLARIES.

1. A less circle will be projected into an ellipse, a parabola, or hyperbola, according as the distance of its most remote point is less, equal to, or greater than, 90 degrees.

2. If H be the centre, and K k, l the focus of the ellipse, hyperbola, or parabola; then HK = \(\frac{A d - A f}{2}\) for the ellipse; \(H k = \frac{A d + A f}{2}\) for the hyperbola; and \(f n\) being drawn perpendicular to AE \(f l = \frac{n E + F f}{2}\) for the parabola.

PROPOSITION V. THEOREM V.

Let the plane TW (fig. 44.) be perpendicular to the plane of projection TV, and BCD a great circle of the sphere in the plane TW. Let the great circle BED be projected into the straight line b e k. Draw CQS perpendicular to b k, and C m parallel to it and equal to CA, and make QS equal to Q m; then any angle QS t is the measure of the arch Q t of the projected circle.

Join AQ; then because C m is equal to CA, the angle QC m equal to QCA, each being a right angle, and the side QC common to both triangles; therefore Q m, or its equal QS, is equal QA. Again, since the plane ACQ is perpendicular to the plane TV, and b Q to the intersection CQ; therefore bQ is perpendicular both to AQ and QS; hence, since AQ and QS are equal, all the angles at S cut the line bQ in the same points as the equal angles at A. But by the angles at A the circle BED is projected into the line bQ. Therefore the angles at S are the measures of the parts of the projected circle bQ; and S is the dividing centre thereof.

**Corollaries.**

1. Any great circle bQt is projected into a line of tangents to the radius SQ. 2. If the circle bC passes through the centre of projection, then the projecting point A is the dividing centre thereof, and Cb is the tangent of its corresponding arch CB to CA the radius of projection.

**Proposition VI. Theorem VI.**

Let the parallel circle GLH (fig. 44.) be as far from the pole of projection C as the circle FNI is from its pole; and let the distance of the poles C, P be bisected by the radius AO; and draw bAD perpendicular to AO; then any straight line bQt drawn through b will cut off the arches hI, Fn equal to each other in the representations of these equal circles in the plane of projection.

Let the projections of the less circles be described. Then, because BD is perpendicular to AO, the arches BO, DO are equal; but since the less circles are equally distant each from its respective pole, therefore the arches FO, OH are equal; and hence the arch BF is equal to the arch DH. For the same reason the arches BN, DL are equal; and the angle FBN is equal to the angle LDH; therefore, on the sphere, the arches FN, HL are equal. And since the great circle BNLD is projected into the straight line bQnI, &c., therefore n is the projection of N, and I that of L; hence fn, hl, the projections of FN, HL respectively, are equal.

**Proposition VII. Theorem VII.**

If Fnk, hlq, (fig. 45) be the projections of two equal circles, whereof one 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 C, p be divided in o, so that the degrees in Co, op be equal, and the perpendicular oS be erected to the line of measures gh. Then the line pn, Cl drawn from the poles C, p, through any point Q in the line oS, will cut off the arches Fn, hl equal to each other, and to the angle QCp.

The great circle Ao perpendicular to the plane of the primitive is projected into the straight line oS perpendicular to gh, by Prop. i, cor. 3. Let Q be the projection of q; and since pq, CQ are straight lines, they are therefore the representations of the arches pq, Cq of great circles. Now since pqC is an isosceles spherical triangle, the angles PCQ, CPQ are therefore equal; and hence the arches pq, Cq produced will cut off equal arches from the given circles FI, GH, whose representations Fn, hl are therefore equal; and since the angle QCp is the measure of the arch hl, it is also the measure of its equal Fn.

**Corollary.**

Hence, if from the projected pole of any circle a perpendicular be erected to the line of measures, it will cut off a quadrant from the representation of that circle.

**Proposition VIII. Theorem VIII.**

Let Fnk (fig. 45.) be the projection of any circle FI, Fig. 45; and p the projection of its pole P. If Cg be the cotangent of CAP, and gB perpendicular to the line of measures gC, let CAP be bisected by Ao, and the line oB drawn to any point B, and also pB cutting Fnk in d; then the angle goB is the measure of the arch Fd.

The arch pg is a quadrant, and the angle goA = PAo + Ap = gAC + oA = gAC + CAo = goAo; therefore goA = go; consequently o is the dividing centre of gB, the representation of GA; and hence, by Prop. v. the angle goB is the measure of gB. But since pg represents a quadrant, therefore p is the pole of gB; and hence the great circle pdB passing through the pole of the circles gB and Fn will cut off equal arches in both, that is Fd = gB = angle goB.

**Corollary.**

The angle goB is the measure of the angle gpB. For the triangle gpB represents a triangle on the sphere, wherein the arch which gB represents is equal to the angle which the angle gp represents; because gp is a quadrant; therefore goB is the measure of both.

**Proposition IX. Theorem I.**

To draw a great circle through a given point, and whose distance from the pole of projection is equal to a given quantity.

Let ADB (fig. 46.) be the projection, C its pole or Fig. 46. centre, and P the point through which a great circle is to be drawn: through the points P, C draw the straight line PCA, and draw CE perpendicular to it: make the angle CAE equal to the given distance of the circle from the pole of projection C; and from the centre C, with the radius CE, describe the circle EFG: through P draw the straight line PIK, touching the circle EFG in I, and it will be the projection of the great circle required.

**Proposition X. Problem II.**

To draw a great circle perpendicular to a great circle which passes through the pole of projection, and at a given distance from that pole.

Let ADB (fig. 46.) be the primitive, and CI the given circle: draw CL perpendicular to CI, and make the angle CLI equal to the given distance: then the straight line KP, drawn through I parallel to CL, will be the required projection.

**Proposition XI. Problem III.**

At a given point in a projected great circle, to draw another great circle to make a given angle with the former; and, conversely, to measure the angle contained between two great circles.

Let P (fig. 47.) be the given point in the given great Fig. 47. circle. Gnomonic circle PB, and C the centre of the primitive: through the points P, C draw the straight line PCG; and draw the radius of the primitive CA perpendicular thereto; join PA; to which draw AG perpendicular: through G draw BGD at right angles to GP, meeting PB in B; bifeet the angle CAP by the straight line AO; join BO, and make the angle BOD equal to that given; then DP being joined, the angle BPD will be that required.

If the measure of the angle BPD be required, from the points B, D draw the lines BO, DO, and the angle BOD is the measure of BPD.

**Proposition XII. Problem IV.**

To describe the projection of a left circle parallel to the plane of projection, and at a given distance from its pole.

Let ADB (fig. 46.) be the primitive, and C its centre: set the distance of the circle from its pole, from B to H, and from H to D; and draw the straight line AED, intersecting CE perpendicular to BC, in the point E: with the radius CE describe the circle EFG, and it is the projection required.

**Proposition XIII. Problem V.**

To draw a left circle perpendicular to the plane of projection.

Let C (fig. 48.) be the centre of projection, and TI a great circle parallel to the proposed left circle: at C make the angles ICN, TCO each equal to the distance of the left circle from its parallel great circle TI; let CL be the radius of projection, and from the extremity L draw LM perpendicular thereto; make CV equal to LM; or CF equal to CM: then with the vertex V and asymptotes CN, CO describe the hyperbola WVK †; or, with the focus F and CV describe the hyperbola, and it will be the perpendicular circle described.

**Proposition XIV. Problem VI.**

To describe the projection of a left circle inclined to the plane of projection.

Draw the line of measures dP (fig. 49.) and at C, the centre of projection, draw CA perpendicular to dP, and equal to the radius of projection: with the centre A, and radius AC, describe the circle DCFG; and draw RA parallel to dP; then take the greatest and least distances of the circle from the pole of projection, and set them from C to D and F respectively, for the circle DF; and from A, the projecting point, draw the straight lines AFF, and ADd; then df will be the transverse axis of the ellipse; but if D fall beyond the line RE, as at G, then from G draw the line GADd, and df is the transverse axis of an hyperbola: and if the point D fall in the line RE, as at E, then the line AE will not meet the line of measures, and the circle will be projected into a parabola whose vertex is f: bifeet df in H, the centre, and for the ellipse take half the difference of the lines Ad, Af, which laid from H will give K the focus: for the hyperbola, half the sum of Ad, Af being laid from H, will give k its focus: then with the transverse axis df, and focus K, or k, describe the ellipse dMf, or hyperbola fm, which will be the projection of the inclined circle: for the parabola, make EQ equal to Ff, and draw fn perpendicular to AQ, and make fk equal to one half of nQ: then with the vertex f, and focus k, describe the parabola fm, for the projection of the given circle FE.

**Proposition XV. Problem VII.**

To find the pole of a given projected circle.

Let DMF (fig. 50.) be the given projected circle Fig. 50. whose line of measures is DF, and C the centre of projection; from C draw the radius of projection CA, perpendicular to the line of measures, and A will be the projecting point: join AD, AF, and bifeet the angle DAF by the straight line AP; hence P is the pole. If the given projection be an hyperbola, the angle fAG (fig. 49.), bifeeted, will give its pole in the line of measures; and in a parabola, the angle fAE bifeeted will give its pole.

**Proposition XVI. Problem VIII.**

To measure any portion of a projected great circle, or to lay off any number of degrees thereon.

Let EP (fig. 51.) be the great circle, and IP a portion thereof to be measured: draw ICD perpendicular to IP; let C be the centre, and CB the radius of projection, with which describe the circle EBD; make IA equal to IB; then A is the dividing centre of EP; hence AP being joined, the angle IAP is the measure of the arch IP.

Or, if IAP be made equal to any given angle, then IP is the correspondent arch of the projection.

**Proposition XVII. Problem IX.**

To measure any arch of a projected left circle, or to lay off any number of degrees on a given projected left circle.

Let Fn (fig. 52.) be the given left circle, and P its pole: from the centre of projection C draw CA perpendicular to the line of measures GH, and equal to the radius of projection; join AP, and bifeet the angle CAP by the straight line AO, to which draw AD perpendicular: describe the circle G/H, as far distant from the pole of projection C as the given circle is from its pole P; and through any given point n, in the projected circle Fn, draw DnI, then HI is the measure of the arch Fn.

Or let the measure be laid from H to I, and the line D joined will cut off Fn equal thereto.

**Proposition XVIII. Problem X.**

To describe the gnomonic projection of a spherical triangle, when three sides are given; and to find the measures of either of its angles.

Let ABC (fig. 53.) be a spherical triangle whose three sides are given: draw the radius CD (fig. 54.) perpendicular to the diameter of the primitive EF; and at the point D make the angles CDA, CDG, ADI, equal respectively to the sides AC, BC, AB, of the spherical triangle ABC (fig. 53.), the lines DA, DG intersecting the diameter EF, produced if necessary in the points A and G: make DI equal to DG; then from the centre C, with the radius CG, describe an arch; and from A, with the distance AI, describe another arch, intersecting the former in B; join AB, CB, and ACB will be the projection of the spherical triangle (fig. 53.); and the rectilineal angle ACB is the measure of the spherical angle ACB (fig. 53.). PROPOSITION XIX. PROBLEM XI.

The three angles of a spherical triangle being given, to project it, and to find the measures of the sides.

Let \(ABC\) (fig. 55.) be the spherical triangle of which the angles are given; construct another spherical triangle \(EFG\), whose sides are the supplements of the given angles of the triangle \(ABC\); and with the sides of this supplemental triangle describe the gnomonic projection, &c., as before.

It may be observed, that the supplemental triangle \(EFG\) has also a supplemental part \(EFg\); and when the sides \(GE, GF\), which are substituted in place of the angles \(A, B\), are obtuse, their supplements \(gE, gF\) are to be used in the gnomonic projection of the triangle.

PROPOSITION XX. PROBLEM XII.

Given two sides, and the included angle of a spherical triangle, to describe the gnomonic projection of that triangle, and to find the measures of the other parts.

Let the sides \(AC, CB\), and the angle \(ACB\) (fig. 53.), be given; make the angles \(CDA, CDG\) (fig. 56.) equal respectively to the sides \(AC, CB\) (fig. 53.); also make the angle \(ACB\) (fig. 56.) equal to the spherical angle \(ACB\) (fig. 53.), and \(CB\) equal to \(CG\), and \(ABC\) will be the projection of the spherical triangle.

To find the measure of the side \(AB\): from \(C\) draw \(CL\) perpendicular to \(AB\), and \(CM\) parallel thereto, meeting the circumference of the primitive in \(M\); make \(LN\) equal to \(LM\); join \(AN, BN\), and the angle \(ANB\) will be the measure of the side \(AB\).

To find the measure of either of the spherical angles, as \(BAC\): from \(D\) draw \(DK\) perpendicular to \(AD\), and make \(KH\) equal to \(KD\); from \(K\) draw \(KI\) perpendicular to \(CK\), and let \(AB\) produced meet \(KI\) in \(I\), and join \(HI\); then the rectilineal angle \(KHI\) is the measure of the spherical angle \(BAC\). By proceeding in a similar manner, the measure of the other angle will be found.

PROPOSITION XXI. PROBLEM XIII.

Two angles and the intermediate side given, to describe the gnomonic projection of the triangle; and to find the measures of the remaining parts.

Let the angles \(CAB, ACB\), and the side \(AC\) of the spherical triangle \(ABC\) (fig. 53.), be given; make the angle \(CDA\) (fig. 56.) equal to the measure of the given side \(AC\) (fig. 53.); and the angle \(ACB\) (fig. 56.) equal to the angle \(ACB\) (fig. 53.); produce \(AC\) to \(H\), draw \(DK\) perpendicular to \(AD\), and make \(KH\) equal to \(KD\); draw \(KI\) perpendicular to \(CK\), and make the angle \(KHI\) equal to the spherical angle \(CAB\): from \(I\), the intersection of \(KI, HI\), to \(A\) draw \(IA\), and let it intersect \(CB\) in \(B\), and \(ABC\) will be the gnomonic projection of the spherical triangle \(ACB\) (fig. 53.). The unknown parts of this triangle may be measured by last problem.

PROPOSITION XXII. PROBLEM XIV.

Two sides of a spherical triangle, and an angle opposite to one of them given, to describe the projection of the triangle; and to find the measure of the remaining parts.

Let the sides \(AC, CB\), and the angle \(BAC\) of the spherical triangle \(ABC\) (fig. 53.) be given; make the angles \(CDA, CDG\) (fig. 56.) equal respectively to the measures of the given sides \(AC, BC\); draw \(DK\) perpendicular to \(AD\), make \(KH\) equal to \(DK\), and the angle \(KHI\) equal to the given spherical angle \(BAC\); draw the perpendicular \(KI\), meeting \(HI\) in \(I\); join \(AI\); and from the centre \(C\), with the distance \(CG\), describe the arch \(GB\), meeting \(AI\) in \(B\); join \(CB\), and \(ABC\) will be the rectilineal projection of the spherical triangle \(ABC\) (fig. 53.) and the measures of the unknown parts of the triangle may be found as before.

PROPOSITION XXIII. PROBLEM XV.

Given two angles, and a side opposite to one of them, to describe the gnomonic projection of the triangle, and to find the measures of the other parts.

Let the angles \(A, B\), and the side \(BC\) of the triangle \(ABC\) (fig. 53.) be given; let the supplemental triangle \(EFE\) be formed, in which the angles \(E, F, G\), are the supplements of the sides \(BC, CA, AB\), respectively, and the sides \(EF, FG, GE\), the supplements of the angles \(C, A, B\). Now at the centre \(C\) (fig. 56.) make the angles \(CDA, CDK\) equal to the measures of the sides \(GE, GF\) respectively, being the supplements of the angles \(B\) and \(A\); and let the lines \(DA, DK\) intersect the diameter of the primitive \(EF\) in the points \(A\) and \(K\); draw \(DG\) perpendicular to \(AD\), make \(GH\) equal to \(DG\), and at the point \(H\) make the angle \(GHI\) equal to the angle \(E\), or to its supplement; and let \(EI\), perpendicular to \(CH\), meet \(HI\) in \(I\), and join \(AI\); then from the centre \(C\), with the distance \(CG\), describe an arch intersecting \(AI\) in \(B\); join \(CB\), and \(ABC\) will be the gnomonic projection of the given triangle \(ABC\) (fig. 55.): the supplement of the angle \(ACB\) (fig. 56.) is the measure of the side \(AB\), (fig. 55.); the measures of the other parts are found as before.

It has already been observed, that this method of projection has, for the most part, been applied to dialling only. However, from the preceding propositions, it appears that all the common problems of the sphere may be more easily resolved by this than by either of the preceding methods of projection; and the facility with which these problems are resolved by this method has given it the preference in dialling. It may not perhaps be amiss, in this place, to give a brief illustration of it in this particular branch of science.

In an horizontal dial, the centre of projection \(Z\) (fig. 57.) represents the zenith of the place for which Fig. 57. the dial is to be constructed; \(ZA\) the perpendicular height of the style; the angle \(ZPA\), equal to the given latitude, determines the distance \(ZP\) of the zenith from the pole; and \(AP\) the edge of the style, which by its shadow gives the hour: the angle \(ZAP\), equal also to the latitude, gives the distance of the equator \(EQ\) from the zenith: let \(Ea\) be equal to \(EA\), and \(a\) will be the dividing point of the equator. Hence if the angles \(EaI, EaII, \&c.\) \(EaXI, EaX, \&c.\) be made equal to \(15^\circ, 30^\circ, \&c.\) the equator will be divided into hours;

If the dial is either vertical, or inclined to the horizon, then the point Z will be the zenith of that place whose horizon is parallel to the plane of the dial; ZE will be that latitude of the place; and the hours on the former dial will now be changed into others, by a quantity equal to the difference of longitude between the given place and that for which the dial is to be constructed. Thus, if it is noon when the shadow of the style falls on the line PX, then the difference of meridians is the angle EAX, or 32°. Hence, when a dial is to be constructed upon a given plane, either perpendicular or inclined to the horizon, the declination and inclination of that place must be previously found.

In an erect direct south dial, its zenith Z is the south point of the horizon, ZP is the distance of this point from the pole, and ZE its distance from the equator. If the dial is directed to the north, Z represents the north point of the horizon; PZ the distance of Z from the pole under the horizon; and ZE the elevation of the equator above the horizon.

If the dial is an erect east or west dial, the zenith Z is the east or west points of the horizon accordingly, and the pole P is at an infinite distance, for the angle ZAP is a right angle; and therefore the line AP will not meet the meridian PZ. The line ZA produces Gnomonic the equator, and is divided into hours by lines perpendicular to it.

If the plane of the dial is parallel to the equator, its zenith Z coincides with one of the poles of the equator P; and hence the hour lines of this dial are formed by drawing lines from the point Z, containing angles equal to 15°.

In the preceding methods of projection of the sphere, equal portions of a great circle on the sphere are represented by unequal portions in the plane of projection, and this inequality increases with the distance from the centre of projection. Hence, in projections of the earth, those places towards the circumference of the projection are very much distorted. In order to avoid this inconvenience, M. de la Hire * proposed, that the * Hiff. de eye should be placed in the axis produced at the di- l'Academie stance of the fine of 45° beyond the pole: In this case Royal des arches of the sphere and their projections are very near. See the ar- ly proportional to each other. Hence in a map of the tile Geo- earth agreeable to this construction, the axis, instead of graphy being divided into a line of semitangents, is divided equally, in like manner as the circumference. The map of the world is constructed agreeable to this method of projection.

PROJECTION, in Perspective, denotes the appearance, or representation of an object on the perspective plane.

The projection of a point is a point through which an optic ray passes from the objective point through the plane to the eye; or it is the point wherein the plane cuts the optic ray.

And hence may be easily conceived what is meant by the projection of a line, a plane, or a solid.

PROJECTION, in Alchemy, the casting of a certain imaginary powder, called powder of projection, into a crucible, or other vessel, full of some prepared metal, or other matter; which is to be hereby presently transmuted into gold.

Powder of Projection, or of the philosophers stone, is a powder supposed to have the virtue of changing any quantity of an imperfect metal, as copper or lead, into a more perfect one, as silver or gold, by the admixture of a little quantity thereof.

The mark to which alchemists directed all their endeavours, was to discover this powder of projection. See Philosophers Stone, and Chemistry, History of.