Fig. 42.

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 Cd. Now Cd is the cotangent of the angle Cd A, the inclination of the given circie, 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 pass 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.

Gnomonic Projection of the

If a circle be parallel to the plane of projection, and Sphere. 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 Cd is the tangent of the angle CAD, and Cf 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 fis 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 fis the nearest vertex; and GA being produced gives d the other

Fig. 43

Join AQ: then because Cm 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 ¿Q

Fig. 44.

Fig. 45.

COROLLARIES.

1. Any great circle b Qt is projected into a line of tangents to the radius SQ.

2. If the circle b C pass through the centre of projection, then the projecting point A is the dividing centre thereof, and C b is the tangent of its correspondent 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 b AD perpendicular to AO; then any straight line b Qt drawn through b will cut off the arches hl, F n 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 b Qn 1, &c. therefore n is the projection of N, and that of L; hence fn, hl, the projections of FN, HL respectively,

are equal.

PROPOSITION VII. THEOREM VII.

If Fnk, hlg, (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, o p be equal, and the perpendicular o S be erected to the line of measures " h. Then the line p n, C drawn from the poles C, p, through any point Q in the line o S, will cut off the arches Fn, hl equal to each other, and to the angle Q Cp.

The great circle A o perpendicular to the plane of the primitive is projected into the straight line o S perpendicular to gh, by Prop. i. cor. 3. Let Q be the projection of q; and since p Q, CQ are straight lines, they are therefore the representations of the arches P'q, Cq of great circles. Now since Pq C is an isosceles spherical triangle, the angles PCQ, CPQ are therefore equal; and hence the arches P q, C q produced will cut off equal arches from the given circles FI, GH, whose representations Fn, h are therefore equal: and since the angle QC p is the measure of the arch h, it is also the measure of its equal F n.

VOL. XVII. Part II.

COROLLARY.

Gnomonic Projection

of the

Hence, if from the projected pole of any circle a perpendicular be erected to the line of measures, it Sphere. 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 C g be the cotangent of CAP, and g B perpendicular to the line of measures g C, let CAP be bisected by A o, and the line o B drawn to any point B, and also p B cutting Fnk in d; then the angle go B is the measure of the arch F d.

The arch PG is a quadrant, and the angle go A= PA+ A Pg AC+0A P=g AC+CA og A o; therefore g Ago; consequently o is the dividing cen tre of g B, the representation of GA; and hence, by Prop. v. the angle go B is the measure of g B. But since pg represents a quadrant, therefore p is the pole of g B; and hence the great circle pd B passing through the pole of the circles g B and Fn will cut off equal arches in both, that is F d=g Bangle g o B.

COROLLARY.

The angle go B is the measure of the angle gp B. For the triangle g p B represents a triangle on the sphere, wherein the arch which g B represents is equal to the angle which the angle p represents; because gp is a quadrant; therefore go B 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.

centre, and P the point through which a great circle is Let ADB (fig. 46.) be the projection, C its pole or Fig. 46. 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.

Gnomonic circle PB, and C the centre of the primitive: through Projection the points P, C draw the straight line PCG; and draw of the the radius of the primitive CA perpendicular thereto; Sphere. join PA; to which draw AG perpendicular: through G draw BGD at right angles to GP, meeting PB in B ; bisect 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.

Fig. 46.

Plate

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 less 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 CR describe the circle EFG, and it is the projection required.

PROPOSITION XIII. PROBLEM V.

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

Let C (fig. 48.) be the centre of projection, and TI CCCCXLVII. a great circle parallel to the proposed less circle: at C fig. 48. make the angles ICN, TCO each equal to the distance

of the less circle from its parallel great circle TI; It CL be the radius of projection, and from the extremity L draw LM perpendicular thereto; make CV equal to LM; or GF equal to CM: then with the vertex V and *See Conic assymptotes CN, CO describe the hyperbola WVK *; or, with the focus F and CV describe the hyperbola, and it will be the perpendicular circle described.

Sections.

Fig. 49.

PROPOSITION XIV. PROBLEM VI.

one half of n Q: then with the vertex ƒ, and focus k, Gnomonie describe the parabola fm, for the projection of the gi- Projection ven circle FE. Sphere

PROPOSITION XV. PROBLEM VII.

To find the pole of a given projected circle. Let DMF (fig. 50.) be the given projected circle Fig. 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 bisect 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.), bisected, will give its pole in the line of measures; and in a parabola, the angle ƒ AE bisected 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.

of the

Let EP (fig. 51.) be the great circle, and IP a por- Fig. 51. tion 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 less circle, or to lay off any number of degrees on a given projected less circle.

:

plane of projection.

Draw the line of measures dp (fig. 49.); and at C, the centre of projection, draw CA perpendicular to d p, and equal to the radius of projection: with the centre A, and radius AC, describe the circle DCFG; and draw RAE 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 dƒ 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: bisect dfin 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 d Mf, 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

:

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 D n l, then H is the measure of the arch Fn.

Or let the measure be laid from H to 7, and the line DI joined will cut off F n 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 Fig. 53three sides are given: draw the radius CD (fig. 54.) 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

Fig. 55

Fig 56.

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 EF g; and when the sides GE, GF, which are substituted in place of the angles A, B, are obtuse, their supplements g E, g F 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 ACB 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

435

of the triangle; and to find the measure of the re- Gnomonic maining parts. Projection of the

Let the sides AC, CB, and the angle BAC of the Sphere. 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. 55.) 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 I 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.

and

Gnomonic and lines drawn from P to these points of division will Projection be hour lines.

of the

Sphere.

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 Ea X, or 30°. 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 equator, and is divided into hours by lines perpen- Projection dicular to it.

of the

If the plane of the dial is parallel to the equator, its Sphere. 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 15o.

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 inconveniency, M. de la Hire* proposed, that the * Hist. de eye should be placed in the axis produced at the di-Academi stance of the sine of 45° beyond the pole: In this case Royal da arches of the sphere and their projections are very near-See the ar ly proportional to each other. Hence in a map of the ticle Ge earth agreeable to this construction, the axis, instead graphy. of 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.

Seien 17

Projection

Prolate.

PRO

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.

PROJECTURE, in Architecture, the outjetting and prominency, or embossing, which the mouldings and other members have beyond the naked wall, column, &c.

PROLAPSUS, in Surgery, a prolapsion or falling out of any part of the body from its natural situation: thus we say, prolapsus intestini, "a prolapsion of the intestine," &c. See SURGERY.

PROLATE, in Geometry, an epithet applied to a spheroid produced by the revolution of a semi-ellipsis about its larger diameter. See SPHEROID.

PRO

PROLEGOMENA, in Philology, certain prepara- Prolegotory observations or discourses prefixed to a book, &c. mena containing something necessary for the reader to be apprised of, to enable him the better to understand the book, or to enter deeper into the science, &c.

PROLEPSIS, a figure in Rhetoric, by which we anticipate or prevent what might be objected by the adversary. See ORATORY, N° 80.

PROLEPTIC, an epithet applied to a periodical disease which anticipates, or whose paroxysm returns sooner and sooner every time; as is frequently the case in agues.

PROLIFER FLOS, (proles," an offspring ;" and fero, "to bear);" a prolific flower, or a flower which from its own substance produces another; a singular degree of luxuriance, to which full flowers are chiefly incident. See BOTANY.

PROLIFIC, something that has the qualities necessary for generating.

The prolific powers of some individuals among mankind are very extraordinary.-Instances have been found where children, to the number of six, seven, eight, nine, and sometimes sixteen, have been brought forth after one pregnancy. The wife of Emmanuel Gago, a labourer near Valladolid, was delivered, the 14th of June 1779, of five girls, the two first of whom were baptized the other three were born in an hour after; two of them were baptized; but the last, when it came into the world, had every appearance of death. The celebrated Tarsin was brought to bed in the seventh month of ber pregnancy, at Argenteuil near Paris, 17th July 1779, of three boys, each 14 inches and a half long, and of a girl 13 inches: they were all four baptized, but did not live 24 hours.

The

Probic