'^nfyp^ n^ ^"^ Digitized by the Internet Archive in 2007 with funding from IVIicrosoft Corporation http://www.archive.org/details/elementsofdescriOOwarrrich THE ELEMENTS OF DESCRIPTIVE GEOMETRY, SHADOWS AND PERSPECTIVE. WITH A BRIEF TREATMENT OF TRIHEDRALS, TRANSVERSALS, AND SPHERICAL, AXONOMETRIC. AND OBLIQUE PROJECTIONS. FOR COLLEGES AND SCIENTIFIC SCHOOLS. BY S. EDWARD WARREN, C.E., Formerly Professor in the Rensselaer Polytechnic Institute, ETC., ETC. V iW ii II A J: > UNJ\ KUslTv Ml. < ALIFUUMA NEW YORK: JOHN WILEY & SONS, PUBLISHERS, 15 AsTOR Place. 1877. Copyright, 1877, John Wiley and Sons. S. W. GREEN, PRINTER AND ELECTROTYPER, 16 and 18 Jacob Street. CONTENTS. PAGE PREFACE. DESCRIPTIVE GEOMETRY.— Definitions. General principles. Fun- damental problems 3 Preliminary 3 Definitions and first principles 4 Fundamental problems 7 A— Projections. PROBLEM I.— To make the projections of points in various positions . . 7 PROBLEM II.— To draw the projections of lines in various positions in space lo B — Tangencies. PROBLEM III.— To draw lines having various positions in the planes of projection I2 C — Intersections. PROBLEM IV. — To represent planes in various positions 13 D — Development. PROBLEM V. — To revolve a point so as to immediately show the form and extent of its path 16 Classification of surfaces and lines 17 DESCRIPTIVE GEOMETRY. BOOK I.— SURFACES OF REVOLUTION. Part I. — Ruled Surfaces. CHAPTER I. GENERAL PROBLEMS OF THE POINT, LINE, AND PLANE. A — Projections. PROBLEM VL— Having given two projections of a point and a line, to find their projections on any new planes of projection 20 1 3 \ [| y i PROBLEM IV. — To represent planes in various positions. In Space. — Planes, other than those of projection, being-, like those, supposed to be of indefinite extent, cannot, like limited magnitudes, be well represented by projections, since these would simply be in most cases the entire surface of both H and V. Thus, '\i KMN {?\. I., Fig. 8) be a portion of an unlimited plane, its projections would be the whole surface of H and V. If, however, it has a fixed position, it will intersect H and V in fixed lines which will indicate its position. These lines PQP" on H, and P'QF" on V, are called respectively the horizontal and vertical traces of the plane. Other magnitudes of indefinite extent may be simi- larly indicated. As a plane can cut a line in but one point, the traces of a plane must meet the ground line at the same point. The positions of a plane may be most simply classified as : I St. In the same direction as the ) containing it. ground line. j parallel to it. 2d. In a different direction from V perpendicular to it. the ground line. ) oblique to it. In -Fig. 8, RS and R!S' being parallel to the ground line, represent the traces of a plane in like position. \i A'B'y for example, on V, and parallel to GL, were the only trace of a plane, the latter would be parallel to H. Again, if CD, parallel to GL, and on H, were the only trace of a plane, the latter would be parallel to V. In Projectiofi. — Fig. 9 shows, pictorially, the traces PQP' and PQP" of the plane /'g^, Fig. 8, after H and V have been made to coincide. In Fig. ID, AB is the single trace, coinciding with the ground line GL, of a plane containing GL. Such a plane is not determined unless by some further condition, as that it also contains a given point, or is parallel to a given line, or makes a given angle with H or V. CD' y parallel to GL^ is 14 DESCRIPTIVE GEOMETRY. the vertical trace of a horizontal plane. EF, parallel to GL^ is the horizontal trace of a plane parallel to V. H and H\ both parallel to GL, are the traces of a plane which is paral- lel to the ground line, but oblique to H and V, and crosses WiQ first angle, as RS — R'S' , Fig. 8, crosses the third. Again : IJI' is a plane, perpendicular to the ground line, and hence to both H and V. Such a plane is often called a profile plane. KMK' is a plane perpendicular to V only and NON' is perpendicular to H only, ^he last two positions are of especially frequent use, and may be fixed in mind by noting that wJien a plane is perpendicular to either plane of projection^ its trace on the other plane of projection is perpendicular to the ground line, PQF is oblique to both H and V and to the ground line as at PQF, Fig. 8 — that is, in such a way as to make acute angles with both H and V in the first angle, on the same side of itself ; while RSR' is a plane of which the part in the first angle is situated as P"QF is in Fig. 4 — that is, so as to make an obtuse angle with V and an acute angle with H on the same side of itself— the right-hand side in Fig. 10. The angles made by the traces, on V and H respectively, with the ground line, on the same side, are just the reverse of these relative to the same planes. 15. The projecting lines of all points of a given line, evidently form a pair of projecting planes of that line, one of them perpendicular to H, and the other to V. It is thus evident that the two projections as well as the two traces of a line determine the line, since the projecting planes being erected on the projections could only intersect in the one line represented by those projections. An exception to this principle occurs when both projec- tions of the line are perpendicular to the ground line, for then the two projecting planes of the line coincide. The line must then be determined by two of its points. It is now clear that the plane, as/,;;///, PI. I., Fig. i, of the two projecting lines of any point is perpendicular to the ground line ; hence the projections of these lines, being DESCRIPTIVE GEOMETRY. 1 5 the traces of this plane, meet the ground line at the same point. Hence, after the revolution of V to coincide with H, these traces wifl form a perpendicular to the ground line, and joining the two projections,/, and//, of the same point />. (12, 1°). 16. Notation of Lines a'dd Planes— First, literal, —T\\Q pro- jections of a line, like those of a point (13), are distinguished by accents attached to the letters which denote them, and a line, like a point, is named in naming its projections. Thus ab — a'b' means the line AB in space whose horizontal and vertical projections are respectively ab and a'b\ A line may be, but not often is, designated by its traces. A plane is usually designated by lettering and naming its traces, as indicated in PI. I., Fig. 10. Second, graphical, — Invisible and auxiliary lines are dot- ted. The planes of projection being supposed to be opaque, nothing is visible out of the first angle (7). The projections of lines are inked, not according to their own visibility, but according to that of the line itself which they represent. The traces of invisible and auxiliary planes are made in broken and dotted lines, consisting of dashes alternating Avith dots. Sometimes invisible lines are distinguished from auxi- liary ones, otherwise called construction lines, by inking the projections of the former with true dots, point dots, and the latter by dash dots, or very short dashes. Construction lines may also be colored, in which case they need not be dotted. 17. Verbal 7iotation, — The terms horizontal and vertical briefly and clearly express the almost constantly recurring positions, parallel to the horizontal plane of projection, and perpendicular to the horizontal plane of projection. As the positions, parallel to the vertical plane ; perpendicular to the vertical plane, or to the ground line ; and parallel to the ground line, all occur about as frequently as the two former ones, similarly brief terms to denote these positions, l6 DESCRIPTIVE GEOMETRY. like profile plane (Prob. IV.) for the third, would greatly abridge speech in the study of descriptive geometry. The following are therefore suggested for optional use in the other cases: Cp-parallely for parallel to V, the two planes H and V, and their contents, being complementary to each other, and parallel to V being the same thing relative to V that horizontal is to H ; then perpendicular, for perpendic- ular to V, which is merely extending the established usage of perspective to the present case ; finally, bi-parallel, as clearly expressive of parallel to the ground line, that position being also parallel to both H and V. D — Development. 1 8. A fundamental operation in many practical problems of development is the revolution of a point, as shown in the following problem ; where the result, being confined to the simplest positions relative to the planes of projection, is immediately shown without construction (lo). In every revolution about an axis, we have to remember: 1°. That every point in the axis remains fixed. 2°. That every other point moves in a circle whose plane is perpendicular to the axis, whose centre is in the axis, and whose radius is the perpendicular distance of the point from the axis. 3°. That the revolving points preserve their relative position. PROBLEM V. — To revolve a point so as to immediately show the form and extent of its path. In Space. — The point must, to fulfil the requirement, re- volve in a circle parallel to a plane of projection ; hence, the axis about which it revolves must be perpendicular to the same plane. In Projection.— V\. II., Figs, ii, I2. In Fig. ii, let the vertical straight line C — CC" be a fixed axis about which a point, whose horizontal projection is a, revolves. By (i8) this point must therefore describe a horizontal circle, whose DESCRIPTIVE GEOMETRY. 1 7 horizontal projection would he an equal circle ^ and all points of this circle being at equal heights above H, must therefore (i2, 3°) be vertically projected in a straight line parallel to the ground line. Then, if the given point were in H, at aa'^ I i', 2 2', 3 3^ etc., would be projections of its succes- sive positions after revolving through the arcs a\ — a'\' , a2 — a'2\ etc. But if the point were in space, as at aa", then i i'', 2 2'\ 3 3'', etc., would be its successive positions after like revolutions in a plane at the height a'a'^ above H. Fig. 12 shows various positions of a point, aa'y in the vertical plane of projection, and of a point, a"a\ at the dis- tance, aa", in front of V, in the revolution of these points about an axis, C"C — C, which is perpendicular to V. Examples. — i'. In Fig. 11 let the given revolving point be taken succes- sively in each of the other angles besides the first. Ex. 2". Make the same changes in the position of the given point in Fig. 12. Classification of Surfaces and Lines, 19. The generation of a magnitude is its formation by the successive positions of some simpler moving magnitude. Thus the motion of a point produces a line of some form ; a moving line describes a surface of some kind. The mov- ing magnitude is called a generatrix ; any fixed magnitude employed to guide its motion is called a directrix, if it be a point or line, a director if it be a surface. The various positions of a linear generatrix are called elements of the surface which they form, and immediately successive points or elements are called consecutive, 20. Revolution and Transpositiott. — For the purposes of de- scriptive geometry, the main divisions of surfaces are con- veniently made to depend on the form or the motion of the generatrix. When the motion of any generatrix whatever is one of revolution around a fixed axis, every one of its points de- scribes a circle centred in and perpendicular to that axis (18). All surfaces so formed are thence called surfaces OF l8 DESCRIPTIVE GP:0METRY. Ri:voLUTiON, and possess the important property that their plane sections, perpendicular to the axis, are circles. When the motion of the generatrix is not one of revolu- tion, it is said to be one of transposition. Surfaces thus gen- erated are thence called surfaces of transposition y and they either possess no line which could be called an axis, or none containing the centres of a series of circular sections per- pendicular to it. 21. Ruled and double curved surfaces. — Again : all surfaces, whether of revolution or of transposition, which can be generated by the motion of a straight line, are called ruled surfaces^ since such lines can be ruled upon them. All surfaces which must be generated by a curved line are called double curved surfaces^ since, as any surface has but two dimensions, and these surfaces are straight in no direction, they are curved in the directions of both of their two dimensions^ from whatever point, and on whatever pair of tangent directions at right angles to each other these di- mensions are estimated. 22. Principal subdivisions. — Ruled SURFACES are of two kinds, plane and single-curved. A plane surface, consid- ered as a surface of revolution, is generated by the revolu- tion of a straight line, either about an axis which is perpen- dicular to it, or which is parallel to it and at an infinite dis- tance from it. . A single curved surface of revolution is generated by the revolution of a straight line about an axis, which it in- tersects obliquely, as in the cone of revolution which is that of elementary geometry ; or to which it is parallel^ as in the cylinder of revolution^ which is also that of elementary geometry; or, finally, with which it is not in the same plane. - '■ . ' In the two former cases, consecutive elements (19) are evidently in the same plane. Such surfaces being cut along any element, may be unfolded or developed upon a plane, and are therefore called developable surfaces. DESCRIPTIVE GEOMETRY. I9 In the last case, the surface formed does not possess this property, and is called a zvarped surface. 23. The foregoing are all the possible forms of ruled sur- faces of revolution^ since they exhaust all the possible relative positions of the straight lines, one of which is the genera- trix and the other the fixed axis, * The plane is evidently that particular case of the cone in which the generatrix is perpendicular to the axis, and that one of the cylinder in which the generatrix is infi- nitely distant from the axis. 24. Double curved surfaces, alike of revolution and of transposition, are either doubly convex.^ as a sphere, on which any two great circles at right angles to each other are both convex towards surrounding space, or both con- cave on the side towards their common diameter consid- ered as an axis; or are concavo-convex, as the surface of a bell in which the rim and other circular sections are con- vex towards surrounding space while the bell-shaped sec- tions are concave towards it. 25. Lines. — These, however imagined as independent of surfaces, always arise practically as the lines of contact or of intersection of surfaces. They are classed in descriptive geometry as straight or curved, and curved lines as of two kinds, curves of single curvature, or such as lie in one plane, as a circle ; and curves of double curvature, which do not. A straight line is generated by a point moving paral- lel to a fixed straight linf^^)^^ or directly towards a fixed i Al.inJRJ^lA. DESCRIPTIVE GEOMETRY. BOOK I -SURFACES OF REVOLUTION. PART I.— RULED SURFACES. CHAPTER I. GENERAL PROBLEMS OF THE POINT, LINE, AND PLANE. 26. In all the operations of descriptive geometry, lines' and planes are supposed to be of indefinite extent, so that the projections of the former and the traces of the latter can always be produced indefinitely in either direction. A — Froj ections. PROBLEM VI. — Having given two projections of a point and a lincy to Jind their projections on any new planes of pro- jection. In Space. — By (12, 3°) the heights of the projections of any fixed point upon any number of vertical planes, and meas- ured from their respective ground lines, will be equal, being all equal to the height of the given point above the horizontal plane H. Likewise, the projections of one point, DESCRIPTIVE GEOMETRY. 21 upon different planes perpendicular to the vertical plane V. will be at equal distances from the traces of these planes upon V. In Projection. — PL II., Fig. 13. Let AB — A'B' be a given line, and let its projection upon the vertical plane at MN be found. By (12) draw A A" and BB" perpendicular to MJV, and make r,A''= rA' and s,B"=sB\ and A"B" will be the projection oi AB — A'B' upon NMN' after the revolution of this plane to coincide with H. Again, let aa^ be a given point, and let its projection on a plane, XYX\ perpendicular to V, be found. Then, draw- ing a' a" perpendicular to X'Y^ and making /,«''= /^, we find a'\ the required projection of aa^ on the new plane XYX\ Examples. — 1° Let aa be in any other angle than \}^q firsts and find its pro- jection also on a new vertical plane. Ex. 2°. Let AB — A' B' be in any other position (Prob. II.), or in any other angle than theyfrj-/. Ex. 3°. Having given the traces of a line to find its projections. Ex. 4°. Through a given point, draw lines parallel to any two given lines. B — ^Tangencies. PROBLEM VII. — To draw a straight line in a plane which is given by its traces^ and a plane to contain a line zvhich is given by its traces. In Space. — We have seen (Prob. IV.) that a plane is deter- mined by its traces. It is also determined by any two in- tersecting or parallel lines, A and B ; for let it at first con- tain only ^, then by revolving about ^ as an axis, it must at some one and only one position contain B also, when it will be determined. If a line contains two points of a plane, it lies wholly in the plane ; hence a line which intersects both of any two lines in a plane lies in that plane. Again : any line which intersects a given line will deter- mine a plane containing the latter, and if the traces of the 22^ DESCRIPTIVE GEOMETRY. given line are known, lines from them to any one point (Prob. IV.) on the ground line, will be the traces of a plane containing the line. In Projection.— ^\, II., Figs. 14, 15. In both, let PQP' be the given plane. Then any line, ab — a'b'y from any point, aa', on PQ to any point, bb', on P Q, will be in PQP. Or, if the plane be given, Fig. 14, by two intersecting lines, ab — a'b' and cd — c'd\ then any line which intersects both of these — including the particular case of intersecting either and being parallel to the other — will lie in the plane of the given lines. Again, any line, cd — c'd' , drawn (12, 2°) so as to inter- sect a given line,^^ — a'b'y as at nn\ will determine a plane through ab — a'b'. Or if, as in the figures, the traces a and b' are given, any point, g, of the ground line, joined with these traces, will give the traces of a plane containing ab — a'b' . Examples. — 1°. Let the plane have any of the positions shown in Prob. IV., with any variations of the kinds of positions shown in Figs. 14 and 15. Ex. 2°. Let the traces of either the given or the required line be so chosen as to make these lines cross any of the four angles. Ex. 3°. Let each trace in succession of the line be at an infinite distance from Q. [The line will then be parallel to that trace of the plane which con- tains this trace of the Une.\ Ex. 4". Draw two lines which shall be in the same vertical plane. Ex. 5°. Draw two lines which shall both be in a plane perpendicular to V. Ex. 6°. Having given one projection of a line in a given plane, find its other projection. C — Intersections. PROBLEM VIII.-— r^ find the traces of a line given by its projections. In Space. — The required traces, being points of a given line, the projections of each of them will be on the pro- jections of the line. Each trace, being also in a plane of projection, will coincide with its projection on that plane, AVhile its other projection will be on the ground line. Com- bining these two principles, we have the following : DESCRIPTIVE GEOMETRY. 2$ Rule. To findtJie trace of a line on either plane of projection : First, Note where the projection of the line on the other plane of projection intersects the ground line. Second. Thence draw a perpendicular to the ground line, and its intersection with the projection on the pro- posed plane will be the required trace. Thus, PI. II., Fig. i6, a picture model of the problem, shows a line, AB^ in space, piercing the plane V aty, thence crossing the second angle, and piercing the back part of H at X. But, according to the rule, x , the vertical projec- tion of X is the intersection of a'b' with the ground line 6^Z, and it must be found first, when, as in practice, the line itself, AB, is not shown. Likewise, j/, found before y, is the intersection of ab and GL, In Projcctio7i. — PI. II., Figs. 17, 18. In both figures, x, the horizontal trace of the given line, ab — a'b\ and y' , its ver- tical trace, are found just as above directed and illustrated, as may be seen by inspection. Examples. — 1°. 'Find the traces of a line crossing the third angle, and of one crossing the /i7?fr/-^ angle. [Figs. 17 and 18 inverted, without change of notation, will represent these cases.] Ex. 2". Take the given portion of the line in each of the four angles suc- cessively, and then find its traces. Ex. 3°. Let the line, determined by two given points, be in a plane perpen- dicular to the ground line. [In this case, first project the line upon a new plane of projection, as in Prob. VI.] Ex. 4°. Assume pairs of traces in many different positions in any part of both planes of projection, and then construct the projections of the lines to which these traces belong. 27. Difficulties at the outset, in the solution of problems, being often caused by oversight of, or uncertainty about very elementary details, the following are, for convenience, added or repeated here : 1°. If a point is on a line, its projections will be on the projections of that line. 2°. If two planes are parallel, their traces will be parallel. 3°. If lines are parallel, their projections on the same 24 DESCRIPTIVE GEOMETRY. plane will be parallel, for their projecting planes (15) are par- allel, and hence the traces of these planes, which are the projections of the lines, will be parallel also. 4°. If a line lies in a plane, its traces will lie in those of the plane ; or if a plane contains a line, its traces will con- tain those of the line (Prob. VII.). 5°. A korizofttal line in a plane is parallel to the horizon- tal trace of the plane, and its vertical projection is therefore parallel to the ground^ line. Likewise, and with a slight change in the form of statement, a line in a plane and par- allel to its vertical trace is parallel to V, and hence its horizon- tal projectio7t is parallel to the ground line. 6°. Two straight lines can intersect only at one point, hence their projections will intersect at points on the same perpendicular to the ground line (12, 2°). If the projections intersect otherwise, the lines themselves do not intersect. See PI. II., Fig, 14, where ab — a'b' and cd — c'd\ being in the same plane, PQPy intersect at nn\ 7°. If any point, line, or figure be in one plane of projec- tion, its projection upon the other plane will be in the ground line (Prob. I.). Hence, 8°. If a plane be perpendicular to a plane of projection, its trace on that plane will contain the projections on that plane, of every thing in the given plane. 9°. If a plane contains two lines, it will contain any line which intersects both of them. Theorem I. — The projection of a right a^tgle will be a right angle when at least one of its sides is parallel to the plane of projection. For if both sides are thus parallel, their projecting planes will be perpendicular to each other; hence their traces, which are the projections of the sides of the angle, will be so also. But either of these sides, A^ being revolved about the other one, B^ as an axis, it will evidently generate a plane (23) perpendicular to that other side, B, and hence to any plane containing B. That is, the projecting planes of the two DESCRIPTIVE GEOMETRY. 2$ sides, and hence the directions of the projections of A and By remain unchanged. Theorem II. — If a line is perpendicular to a pla7tey its projec- tions will be perpendicular to the traces of the plajie. PL II., Fig. 19. Let AB be perpendicular to the plane PQP' at B, It will therefore be perpendicular to every line drawn through By and in the plane PQP' y and hence to a parallel, R (not shown) to PQ, through B. But the project- ing planes of AB and*i? are (Theor. I.) perpendicular to each other, and the horizontal projection of R is (27, 3°) parallel to PQ. Hence the horizontal projection rt:^ of ^^ is perpendicular to PQ. Likewise a'd' is perpendicular to Otherzuise, either projecting plane, as ABab, of the line ABy is perpendicular to a plane of projection, as H, and to PQP' because containing AB, a perpendicular to the latter plane. Hence it is perpendicular to the intersection, as PQy of H and PQP. Hence, for example, the traces of ABaby on H and on PQP, are both perpendicular to PQ. That is, ab is perpendicular to PQy and the like is true of a'b' and PQ. PROBLEM IX. — To construct the traces of a plane under various frequently occurring conditions. In Space. — Many particular problems are here included in this comprehensive title ; partly since the plane, as the simplest of surfaces, is auxiliary to the solution of many problems where its construction is required under various conditions ; but especially since the main feature of the solu- tion in all cases is the same, viz., the determination of two lines which shall be in the plane, and whose traces will there- fore (Prob. VII.) determine those of the plane. The principal cases, and their determining lines are as fol- lows : To construct a plane : 1°. Through three given pointSy the determining lines being then a7iy two of the three lilies connecting these points. 26 DESCRIPTIVE GEOMETRY. 2°. TJirough a given point and straight line. Here the lines are the given litte and one from any point of it to the given point, usually a parallel to the given line, through the given point. 3°. Through two given parallel or intersecting lines ; when these, or, in case of their inconvenient location, any lines intersecting both of them (Prob. VIL, 27, 9°), will be the determining lines of the plane. 4°. TJirough one given line and parallel to a^iother ; when the plane will be determined by the former line, together with a line through any point of it and parallel to* the second line. 5°. Through a given point and perpendicular to a given line ; when each of the determining lines will contain the given point and will be parallel to a trace of the plane. Hence (Theor. II.) the projection of either of these lines on the plane of projection containing such trace will be perpen- dicular to the projection of the given line on the same plane (27-3°, 5°). 6°. Through a given point and parallel to any' two given lines. The determining lines will pass through the given point and be parallel to the given lines. 7°. Through a given point and parallel to a given plane ; when the plane will be determined by lines through the given point and parallel to any two lines taken in the given plane by Prob. VII. In Projection. — The construction of cases 2°, 4°, and 5° will sufficiently illustrate all the rest, on account of the similarity of all, as stated. Case 2°. — PI. II., Fig. 20. This illustrates the particular case, frequently occurring, where the given line is one trace of the plane, and the given point is assumed on some known line of the plane. Then let Pc be the given trace, and aa' , on the known line cd — cd\ the given point. Through aa' draw ah — a'b' parallel to Pc. Its vertical trace, b' , together with d\ the vertical trace of cd^cd\ will determine the re- quired vertical trace. In reference to frequent variations occurring in practice, note that if d' were inaccessible, the vertical trace P'Q would DESCRIPTIVE GEOMETRY. 2/ join^' with the intersection Q of the horizontal trace with the ground line, but if Q and d^ were both inaccessible, we should take an auxiliary line through another point of cd — c'd and parallel either to H or to V. Case 4°, — PI. II., Fig. 21. Let ab — a'b' be the line which the required plane is to contain, and let cd — c'd' be the one to which it is to be parallel. Assume any point, pp\ on ab—a'b',2indi through it draw c^d^ — c^d^ parallel X.o cd—c'd' , Then the horizontal traces, a and ^„ of these lines, being joined, give PQ the horizontal trace of the required plane. Likewise the vertical trace FQ \oms b' and r/, the vertical traces of these lines. Two points being sufficient for each trace, Q, being common to both traces, serves as a test of the mechanical accuracy of the construction. Case 5".— PI. III., Fig. 22. Let ab—a'b' be the line to which a plane through the given point//' is to be perpen- dicular. Then pc being perpendicular to ab and p'd par- allel to GL, the line pc^p'c is a parallel to H in the required plane ; hence its vertical trace, c\ is a point of the vertical trace P'Q, perpendicular to a'b\ of the required plane. Like- wise, pd being parallel to GL, and /V perpendicular to a'b\ pd—p'd' is a parallel to V in the required plane; hence, find- ing its horizontal trace dy draw PQ through d and perpen- dicular to ab, for the horizontal trace of the plane. Remark. — The line from the given point to the intersec- tion of the plane with the given line is ihQ perpendicular from the given point to the given line. Examples. — i". Construct case 1° with variations by taking the three given points in one angle, or in two or three different angles. Ex. 2°. Construct case 1° when one of the determining lines joining two of the points is parallel to a plane of projection. Ex. 3°. Construct case 1° when the same line is parallel to the ground line. Ex. 4°. Construct case 2° in its general form, the given line and point hav- ing any position in the same or different angles. Ex. 5°. Construct case 3° when the traces of either or both of the given lines are inaccessible. Ex. 6°. Construct case 3° when either or both of the given lines are paral- lel to the ground line. Ex. 7°. Construct case 3° when the given lines intersect on the ground line. 28 DESCRIPTIVE GEOMETRY. Ex. 8°. Construct case 4" when the given lines, which in Fig. 21 both cross \)\e first angle, shall cross different angles. Ex. 9°. Construct raj^ 5° when the given line crosses either the second or the fourth angle, and when the same line is either parallel or perpendicular to a plane of projection. Ex. 10°. Construct case 5° when the given line is in a profile plane {Proh. IV.). Ex. 11°. Construct case 6°. Ex. 12°. Construct case 7". Ex. 13°. Construct a plane parallel to a given plane and at a given per- pendicular distance from it. PROBLEM X. — To construct the intersection of two planes given by their traces, A line may be cut from each given plane by an auxiliary plane. The intersection of these lines will be a point of the line of intersection of the given planes. The planes of projection may generally be the auxiliary planes, and the intersection of the given planes then joins the inter- section of their vertical traces with that of their horizontal traces. Thus, PI. III., Fig. 23, H and V being the planes of pro- jection, PQF and RSR' are two planes .whose like traces in- tersect at a and at b'. That is (27, 4°), the intersections of the like traces of two planes are the traces of the intersection of the planes. But one projection of each trace is on the ground line (27, 7°) ; hence ab and a'b' are the projectiojis of the in- tersection ab\ Hence, when the planes H and V are the auxiliary planes, we have the Rule. — Project the intersection of the traces on each plane of projection upon the ground linCy and join the points so found with the intersection of the traces on the other plane. In Projection — The general case. — PI. III., Figs. 24, 25. In both, where like letters designate like joints similarly founds though in different positions only, owing to the different po- sitions of the given planes in the two cases, let PQP and RSR' be the given planes ; then a and b' are the traces of their line of intersection, whose projections are therefore ab and d'b\ DESCRIPTIVE GEOMETRY. 29 Particular cases — the traces of the intersection inaccessible. I ° . The planes parallel to the grou nd line. — PI . III., Fig. 26. The intersection of these ^XdcaQ^MN—M' N' and XY—X'Y\ will also be parallel to the ground line. Thus, knowing its direction, one point will determine it. Take any aux- iliary plane, as NxX\ which will conveniently intersect both of the given planes. By the general case, above, this plane cuts from MN—M'N' the line Nx—7t'M', and ixomXY—X'Y the line rx — r'X' . These lines intersect at /', whose horizon- tal projection is t, whence ts — /V, parallel to the ground line, is the required intersection. A like construction, with two auxiliary planes, would have served had the planes been nearly parallel to the ground line. 2°. The planes nearly per pendictilar to the ground line. — In this case, the auxiliary planes are conveniently taken paral- lel to H or V. Thus, PI. III., Fig. 27, let PQP' and RSR' be the given planes. Then a horizontal plane, C D\ will cut the given planes in lines parallel to their horizontal traces. The vertical traces of these lines are by (27, 4°) C and D\ hence their projections are ck — Ck' and dk — D'k\ where ck and dk are parallel to RS and PQ, and they give the point, kk' , of the required intersection. * Likewise AB is the horizontal trace of an auxiliary plane parallel to V, and cutting the given planes in lines parallel to their vertical traces, and beginning at A and B. Hence, project A and B at a' and b' y draw a'h' and b'h' par- allel to SR' and QF , and meeting at h' , whence project // to h on ABy and we have hk — Ji'k' for the required intersec- tion. Had the planes been very nearly perpendicular to the ground line, we could have used, for each projection of the intersection, two auxiliary planes parallel to the ground line, and very nearly so to the plane containing the other projection of the intersection. Examples. — 1°. Construct the case just mentioned, Ex. 2°. Take planes whose intersection shall cross the third angle and the fourth angle, and find the intersection in each case. 30 DESCRIPTIVE GEOMETRY. Ex. 3°. Let one of the planes be perpendicular either to H or to V, and treat it as a new plane of projection by revolving it into the plane to which it is perpendicular, thence showing a trace of the given plane on a new plane of projection. Ex. 4°. Let the traces of the given planes on one of the planes of projection be parallel to each other. Ex. 5°. Let both planes cut the ground line at the same point. Ex. 6°. Let planes, both parallel to the ground line, be so taken as to intersect in ihe fourth angle. Ex. 7°. Let one of two given planes be parallel to the ground line, and cross either the second or fourth 2iVig\es, so that its two traces shall coincide on the paper. Ex. 8°. Find the intersection of two planes each of which is given by two lines contained in it. Ex. 9°. Find the point of intersection oi three given planes. PROBLEM XI. — To construct the intersection of a line ivith a plane. In Space, — PI. III., Fig. 28. In this picture model, H and V are the planes of projection, PQF is any given plane, and AD a given line whose projections are ad and a'd! . AD evi- dently will meet PQF by (27, 4°) in the trace upon PQP' of any plane containing AD. The projecting planes oi AD being the most convenient, Aann' represents that which projects it upon H, and whose trace on PQP is inn' , project- ed in mn and in'n' (Fig. 23). Hence, AD pierces PQP on im{ at B^ whose projections are b' and ^, on if^n' and mn. Likewise, taking the vertically projecting plane Aa'h'h, oi AD, its trace ox\. PQP \^ k'hy projected in hk and h'k'. This meets AD at B as before. Hence the Rule. — Find the intersections of t lie projections of the given linCy zuith the projections of the intersection of the given plane with either of the projecting planes of the given line. In Projection — a. The general case. — In PI. III., Fig. 29, PQP' is the given plane (differently situated from PQP' in Fig. 28, to further illustrate, as in Figs. 24 and 25, the fol- lowing of the same operations in different positions by the aid of like letters at like points), ad — a'd' is the given line, ann' is its horizontally projecting plane, intersecting the given plane (Prob. X.) in the line mn — m'n' ^ which, in turn, intersects DESCRIPTIVE GEOMETRY. 3! ad—a'd' at b'b (naming b' first because it is found first), the required intersection of ad — a'd' Avith PQP' . Likewise, the vertically projecting plane, d'h'h^ inter- sects PQP in the line whose traces are // and k' , and whose projections are hk and h'k\ giving bb\ as before, for the re- quired intersection, only that b is thus found before b' . Note. — Problems VIIL, X., and XI. have thus been emphasized hy picto- lialillustration^ and by the expression of their solution in a concise rule, because they are the ones most frequently applied in the solution of all other problems. They should therefore be made thoroughly familiar by review and varied examples before proceeding further. b. Particular cases. — 1°. The given line perpendicular either to H or V. In this case, the projection of the entire line on the plane of projection to which it is perpendicular is a point (Prob. II.) which is therefore one projection of the re- quired intersection of this line with the given plane. Hence this case is equivalent to the problem, often separately enun- ciated, ** Having one projection of a point of a given plane ^ to find the other projection of the same points In this case, any auxiliary plane containing the given line will be perpendicular to that plane of projection to which the given line is perpendicular, and will therefore be convenient. Thus, let PQP , PL IV., Fig. 30, be the given plane, and let a be the horizontal projection of a vertical line, a — a'b\ Then any line, MN^ drawn through a, will be the horizontal trace of a vertical plane, MNM\ whose trace, M'm' — MN, on PQP will contain the required intersection, c'c, oi a — a'b' with PQP. In other words, c^ is the vertical projection of that point of the plane PQP whose horizontal projection is c. Otherwise (and a little more convenient instrumentally), let/ be the horizontal projection of a vertical line whose in- tersection with PQP' is to be found. Then RSR' is a verti- cal plane containing p and parallel to PQ^ and hence cutting PQP' in a horizontal line, RS—R'r', which contains the intersection,//', of the vertical line, /—/'/, with the plane PQP. 32 DESCRIPTIVE GEOMETRY. RS could equally well have been parallel to the ground line GL, when R'r' would have been parallel to QP'. The same lines here described, only drawn in a different order, would have served to find / and a had / and c' been given. Examples. — i". Let the given line be parallel to a plane of projection. Ex. 2°. Let it be parallel to the ground line. Ex. 3°. Let the plane in Fig. 29 take the position oi FQP' in Fig. 30, the given line remaining the same. Ex. 4°. Let the given line cross any one of the four angles. Ex. 5°. Let the plane be parallel to the ground line. Ex. 6". Let the traces of the plane coincide on the paper. Ex. 7°. Let the line be in a profile plane (Prob. IV.). 2 . The plane given by other lines than its traces. — The solu- tion is the same as in Fig. 29, except that we note the inter- section of one or both auxiliary projecting planes of the given line w^ith these determining lines of the plane, rather than with the traces of the plane, in order to find the trace of this projecting plane upon the given plane. Thus, PL IV., Fig. 31, let the given plane be that of the two intersecting lines ab — a'b' and cb — c'b' ; and let RS — r's' be the line whose inter- section with this plane is required. Then, as before, RSR! is a projecting plane of the given line RS — r's\ and it inter- sects ab — a'b' at ee' , and cb — c'b' at dd! , giving ed — e'd' as the trace of this projecting plane upon the given plane. Hence, q'q, the intersection of this trace with RS — r's' , is the re- quired intersection of the line RS — r's' with the plane abc — a'b'c'. Examples. — i". Let the point bb' be in any other angle than the^n-/. Ex. 2°. Let the given line cross any other angle than \^q first. Ex. 3°. Construct a line parallel to a given line, A^ and intersecting two other given lines, C and Z>, which are not in the same plane. [Pass a plane, by Prob. IX., case ^° ^ through C and parallel to ^4, and, by Prob. XL, find where D pierces this plane. A line from the latter point, parallel to A, will be the one rec^uired.J Note. — Problems VIII. and XL will enable the student to take up perspec- tive constructions by the method of visual rays (5) according as V or some other plane is taken as the perspective plane ; also to find shadows on plane sur- faces, if the given lines be supposed to be rays of light. 1.1 \\\i A '.' V DESCRIPTIVE GEOMETRY. 33 UNI VKt:srr V nl D-DevelopmentU (J AL i FO UN 1 A 28. As a plane cannot- be reduced to anysimpler form by transformatioHy the only problems of development here to be considered are those of transposition, the object of which is to show a line, angle, or figure in its real size by bringing it into or parallel to a plane of projection. Such problems are direct when the developments are the results required by them, and inverse when from auxiliary developments re- quired projections or other developments are found. General Methods of Solution, 29. Three geiieral methods of solution are employed in descriptive geometry, which, as they are best illustrated under the present head of development, are stated here. First, As evidently illustrated by any of the preceding problems, beginning with PL II., Fig. 20, the planes of pro- jection and the given magnitudes are both fixed ; and the required parts. are found by operating upon the given ones by auxiliary lines and planes suitably chosen for the pur- pose. • Second. The planes of projection being fixed, the given magnitudes may be brought by one or more rotations, usu- ally elementary rotations — that is, about an axis perpen- dicular either to H or V, as in Prob. V. — to such a simple position relative to the planes of projection, that the parts required there immediately appear without further con- struction. This is the method of rotatioris. Third. The given magnitudes being fixed, their projec- tions may be successively found on one or more new planes of projection as in Prob. VI., until by that means the same result is produced as by the method of rotations. This is the method of change of planes of project io7i. A comparison of the following problems with these statements will sufficiently illustrate them. 34 DESCRIPTIVE GEOMETRY. PROBLEM XII. — To find the true length of the line joining two given points. In Space. — A line may be shown in its true length by bringing it to a position in or parallel to a plane of projection. This position may be obtained by revolving the line either about an 2iX\Sy perpendicular to one plane of projection till it becomes parallel to the other one (29, Second)y or about an axis, in or parallel to a plane of projection, till the line reaches a like position. In the former methody the points of the line will move in arcs parallel to a plane of projection as in Prob. V. By the latter method, we employ the prificiple that the distance of a point in space from any line taken as an axis is the hypothenuse of a right-angled triangle, whose other two sides are the projection of this distance upo7t a plane containing the axisy and the projecting line of the point upon this plane. In Projection. — PL IV., Figs. 32, 33. First method: the axis perpendicular to H or Y« In Fig. 32, let ab — a'b' be the given line, and let it be revolved about the vertical axis at b till parallel to V. The point aa' will then describe the horizontal arc aa^ — a'a^'y having the point bb" for its centre, and limited by ba^ — b'a^'y parallel to V. Hence ba^ is par- allel to the ground line, and a^'b' is the true length of the line ab — a'b\ Note that such an axis need not intersect ab — a'b'y but may be any perpendicular to H or V. Second method: the axis in or parallel to H or V. In Fig. 33, let ab — a'b' be a given line in a plane whose horizon- tal trace is PQ {Froh. VII.) Its true length will then be shown on H by revolving it into H, about PQ as an axis. After such revolution, aa' will be found at a", on a per- pendicular ra" to PQ through a, and at a distance from PQ equal to the hypothenuse of a right-angled triangle whose base is ar, and altitude na'. Finding b" in a similar manner, a"b" is the true length oiab—a'b'. When the axis chosen is a projection of the given line, as at a'b'y Fig. 32, the plane last described becomes a pro- DESCRIPTIVE GEOMETRY. 35 jecting plane, the revolution is one of 90° and the hypothe- nuse coincides with the altitude described. Thus, on per- pendiculars to a'b' at a' and b' ^ make a'a^ = an^ and b'b^ = biUy giving aj)^ aS the true length of ab — a'b^ after revolution about its vertical projection ^'^ to coincide with V. This solution (see Prob. VI.) illustrates (29, Third), Examples. — 1°. Revolve ah — ab\ Fig. 32, about a perpendicular to V at b' till parallel to H. Ex. 2°. Revolve ab — ab\ Fig. 32, about atiy vertical axis, not intersecting ab — ab\ till the latter shall be parallel to V. Ex. 3°. Revolve ab — ab\ Fig. 32, about ab as an axis till it coincides with H. Ex. 4°. Revolve ab — a!b\ Fig. 33, aboyt the vertical trace of any oblique plane containing it, till it coincides with V. Ex. 5°. Find the true length of the perpendicular from a given point to a given plane. PROBLEM XIII. — To find the shortest distance from a given point to a given line. In Space, — By Prob. IX., case 5°, pass a plane through the given point and perpendicular to the given line. Then, by Prob. XL, find where the given line pierces this plane. The line joining this point with the given point will be the line required, and its true length may be found by the last problem. /// Projection. — PI. IV., Fig. 34. Let//' be the given point, and ab—a'b' the given line. Knowing the directions of the traces of the perpendicular plane, this plane is determined by the line pt—p't\ where p't' is perpendicular to a'b\ and // is parallel to GL, Then PQ is drawn through the trace /, perpendicular to ab, and QF through Q, perpendicular to a'b\ The perpendicular plane PQP' thus found inter- sects the given line at qq' , giving/^— /y for the required distance. Its true length is q'p^. This solution accords with (29, First), The usual solution, in which the true length is found first (28), is, to pass a plane through both the given point and the given line (Prob. IX., 2°), and then to revolve it about 36 DESCRIPTIVE GEOMETRY. one of its traces into the plane of projection containing that trace, as in PI. IV., Fig. 33, when the given point and hne being seen in their real relative position, a perpendicu- lar from the one to the other can immediately be drawn in its true length (29, Second). Examples.— 1°. Construct the solution just described. Ex. 2°. Construct Fig. 34, with/>/' in any other angle than ih^ fourth. Ex. 3°. Construct Fig. 34, with ab'—a'b' crossing any other angle than the second. PROBLEM XIV. — To find the shortest distance between two lines not in the same plane,. In Space. — The usual solution, illustrating (29, First) and readily constructed from a detailed description, is this.^' Let AB and CD be the given lines. Then — 1°. By Prob. IX., 4°, pass a plane through the line CD and parallel to the line AB. Call this plane PQF , 2°. Project the line AB upon the plane PQP' ; that is, assume any point M on AB, and by (Theor. II.) draw from it a perpendicular XY to the plane PQF ; find, by Prob. XL, the point N, where XY pierces PQP' ; and through N draw a parallel, A^B^, to ABy which will be the projection oiAB upon the plane PQP. 3°. Note the point / where the projection A^B^ of AB. upon PQP intersects CD, 4°. At / draw a parallel to XY and it will intersect AB at a point K, and IK will be the required shortest distance and common perpendicular between the two lines ; for / is where the projecting plane oi AB upon PQP cuts CD. 5°. By Prob. XII., find the true length of this common perpendicular. In constructing this description, it only remains to * In all constructions thus made from description, it is very important that the student should make the successive steps of the construction, one by one, as he reads their description ; that is, he should not attempt to imagine the whole construction from the whole description, before beginning to draw it. DESCRIPTIVE GEOMETRY. 37 denote by corresponding small letters, ab — a'b\ inm\ etc., the projections oi ^\^ry point and line above designated in space by capitals, AB^ M, etc. Otherwise^ and illustrating (29, Second^ Third) a new plane of projection, a vertical one V„ for example, may be taken through or parallel to either of the given lines, as AB. The system, composed of the two given lines, may then be revolved about any axis perpendicular to V„ till AB takes a vertical position, when the required shortest distance being horizontal will immediately appear by its projection on H. /;/ Projectiofu — PL IV., Fig. 35. Let ab — a'b' and cd—c'd be the two given lines. Make ab the ground line of a new vertical plane containing ab — a'b' , which line will therefore, by Prob. VI., appear projected upon this plane, when the latter is revolved into H, at a"b" , In like manner, cd^c'd' will appear at cd — c"d". Then revolve ab — a"b" and cd^c"d", together, about an axis perpendicular to the new vertical plane, as at m" , until ab — a"b" becomes vertical at b^ — b^'m'\ when cd — c"d" , revolving to an equal angular extent — c"in"cl' — b"m"bl' — about the same axis, will appear at mc^ — m"c^' , since mm" ^ being in the axis, remains fixed. The line ab — a'b' being now brought to the vertical position, b^ — b^"m", all perpendiculars to it will be horizontal, and by (Theor. I.) that one of them which is also perpendicular to mc^ — m"c/' will appear so in horizontal projection ; hence n.p^y perpendicular to mc^ from ^„ is the true length of the required perpendicular or shortest distance between the two lines. Its revolved new vertical projection is n^'p" (parallel to ab as a ground line), which, by counter-revolu- tion around the axis mn — m" ^ appears at np — n"p" y and thence on the primitive planes at 7ip — ;/'/'. We thus see that by the methods illustrated in Fig. 35, proceeding in an inverse order, we first find the desired true distance, and thence, by counter-revolution^ and a return to the primitive planes^ the projections of its original position, if these be desired ; while by the method of the first sqlp tion, we proceed directly by first finding the projectionsr^6f 38 DESCRIPTIVE GEOMETRY. the primitive position of the required distance, and thence its true length.* Examples. — i". Let the given lines cross different angles. Ex. 2". Let both projections of each make angles of 30"*, or less, with the ground line. Ex. 3°. Let both projections of each make angles of 60°, or more, with the ground line. Ex. 4°. Let one of the lines be the ground line. Ex. 5". Find the required shortest distance by Ex. 3, Prob. XL, 3"; the given line A being in this example perpendicular to the directions of both of the given lines C and Z>, and hence taken as the intersection of two planes each assumed by (Theor. 11.) so as to be perpendicular to one of the given lines Cand D. PROBLEM XV. — To find the true size of the angle between two given lines. In Space. — Find either trace of the plane containing the given lines (Prob. XL, 3°), then, by (Prob. XII.), revolve this plane into the plane of projection containing that trace, when the given lines will there show their real relative position. In Projection. — PI. IV., Fig. 36. Let ab — a'b' and ac — a'c' be the given lines, and hence bac — b'a'c' the projections of the required angle. By Prob. VIII., find/ and g, the hori- zontal traces of the lines, and/^ will be the like trace of their plane. Then revolve this plane about this trace until it co- incides with H, vfhQnaa' will appear at ^ on ar, the horizontal trace of the plane of its revolution (27, 4°), and at a distance, ^'V, from pq, equal to the hypothenuse of which ar and a't are the base and altitude. Since / and ^, being in the axis,- remain fixed, /^"^ is then the true size, shown on H, of the given angle. Examples. — i". Let the projections of the given angle be obtuse. Ex. 2°. Revolve the plane of the angle about its vertical trace and into V. Ex. 3°. Let the vertex of the angle be in any other angle than \h.e Jirst. * See my larger DESCRiprrvE Geometry (Mew York, 1874), where a number of j>i:oblems in which the inverse order is followed are conveniently classed under the title, ** Inverse or counter-development.** C DESCRIPTIVE GEOMETRY. 59. Ex. 4'. Let one of the given lines be parallel to the ground line. Ex. 5°. Let the two lines be the two traces of a plane. PROBLEM XVI. — To bisect a given angle in space. In Space, — Equal arcs subtend equal angles, but,^if in space, may be differently inclined to a plane of projection, and hence unequal in projection ; hence, if an angle be equally divided in space, its projections will not be so divided, unless the plane of the angle be parallel to a plane of projection, to which position it must therefore be brought by revolution about a suitable axis. In Projection, — i'^. The general case, PI. IV., Fig. 36. Here both sides of the angle bac^b'a'c' are oblique to both planes of projection. Then finding first the real size, pa"q, of the angle by the last problem, bisect it by the line a"ny and project n at n' , In counter-revolution, a" returns to aa' ^ and nn' y being in the axis, remains fixed, and na and n'a' become the projections of the bisecting line na" , 2^. Special case. — PI. V., Fig. 37. In this case, which oftener occurs in practice, one side, bc—b\ of the angle, abc — a'b'c\ is perpendicular to a plane of projection, in this case to V. Revolve the angle about be — b' as an axis till it is parallel to H, as shown by revolving any point, as aa' y of the side ab — a'b' in the arc aa^ — <^'ct! •> where b' is the centre of a'a^y till ab — a'b' becomes horizontal as at afi — a^b' , Then bisect the revolved angle ajbc by djby whose vertical pro- jection is d^b' y and in counter-revolution, afi—a^b'y returns, as shown, to ab—a'b'y and the bisecting line similarly to db — d'b' y which gives th.Q projections of the semi-angles. Examples. — 1°. Let one of the sides of the angle be vertical. Ex. 2'. Let one of the sides be parallel to the ground line. PROBLEM XVII.— r^ Ji7td the true size of the angle in^ eluded between two given planes. In Space, — Principles, leading to variously modified con- structions, are as follows (PI. V., Fig. 41): First, T\iQ 40 DESCRIPTIVE GEOMETRY. plane, mno^ of the required angle will be perpendicular to the intersection, PF ^ of the two planes. Second. Hence by (Theor. II.) the traces, as inn, of this plane will be perpen- dicular to the like projections, as Pby of the intersection ; and the sides, mo and no, of the angle will be perpendicu- lar to the intersection, PP\ of the planes. Third. As the plane mno is perpendicular to PF , it will be perpendicular to every plane through that line, and hence to its project- ing plane FbP ; but FbP being perpendicular both to H and to nino, is perpendicular to their intersection mn. Hence or, the intersection of these two planes, is perpen- dicular to PF y as being in the plane mno, and also to mn as being in the plane FbP. When one of the two given planes is a plane of projec- tion, the plane of the required angle is perpendicular to the trace of the given plane on that plane of projection, and hence to the latter plane itself. In Projection. — 1°. One of the given planes a plane of projec- tion. PI. v., Figs. 38, 39. In Fig. 38, let the angle be- tween H and the plane PQP^ be required. Then assume any auxiliary vertical plane PoF , whose horizontal trace, Po, is perpendicular to PQ. The lines Po and one in the given plane from P to P' are the sides of the required angle. Hence, by revolving this angle about Fo as an axis till it coincides with V, as indicated by the arc Pn whose centre is a, the angle is shown in its real size at F710. Fig. 39 shows the case in which the given plane PQF has the most general kind of position. Briefly, Pr, perpendicular to GL, is the horizontal, and rs\ perpendicular to FQ, is the vertical trace of the plane perpendicular both to V and PQF, and therefore containing the angle between PQF and V. Revolving this plane about Pr and into H, Ps"r shows this angle in its true size. oF'p shows the angle between PQP' and H, but does so by revolving the plane F'oF of the angle about its horizon- tal trace into H, by making op — oF and perpendicular to eP"; instead of by revolving it about its vertical trace oP\ as in Fig. 38. DESCRIPTIVE GEOMETRY. 4I 2°. The general case without auxiliary developments (29, First).— ?\. v., Fig. 40. PQP' and PRP' being the given planes whose included angle is required, Pb — P'a' is their intersection. Then assuming any point 0, on the ground line, by {Prin. Seco?id) draw on perpendicular to Pb^ and oc' perpendicular to Pa\ and noc' will be the plane of the re- quired angle. By Prob. XL, this plane cuts the intersec- tion Pb — Pa' of the two planes at dd\ which point, joined with nn' and kk'y where the plane noc' cuts the horizontal traces of the given planes, gives ndk — n'd'k' y the projections of the required angle, whose true size, shown at nDky is found precisely as in Prob. XV., Dr being the hypothenuse formed on dr and d'f as a base and altitude. 3°. With auxiliary developments. — PL V., Fig. 42. Let PQP and PRP' be the given planes. PbP' is the projecting plane of their intersection upon H, and Pb is therefore the horizontal projection of that intersection. Revolving P about b to P" , and drawing P" P' y this line is the intersec- tion PP'y in space, revolved about P'b to coincide Avith V. Then bd" ^ perpendicular to PP" (see roy Fig. 41), is the true perpendicular distance of the vertex d" of the required angle from the horizontal trace nk of its plane, 71k being perpendicular to Pb at 3. When the plane of the angle revolves about nk into H, d" will appear at dy by making bd on bP equal to bd' . Hence ndk is the required angle be- tween PQP' and PRP. Examples. — 1°. Let the traces of the given planes on one of the planes of projection be parallel. Ex. 2°. Let one of the planes be perpendicular to the ground line. Ex. 3°. Let the plane in Fig. 40, the construction remaining the same, be situated as in Fig. 42. Ex. 4°. In Fig. 42, revolve the vertical projecting plane /*<5/'' about its hori- zontal trace. Ex. 5°. Let both planes be parallel to the ground line. Ex. 6°. Let both planes be nearly perpendicular to the ground line. Ex. 7°. Find the bisecting plane of the angle between the given planes. [In Fig. 42, for example, the horizontal trace of this plane would join P with the point where the bisecting line of the angle ndk meets nk. Its vertical trace would pass through P'\ 42 DESCRIPTIVE GEOMETRY. PROBLEM XYllL—Tofind the angle made by a line with a plane. In Space. — The angle made by a line with a plane is that made by the line with its projection on the plane. But this projection, the line itself, and the perpendicular which projects any one of its points upon the plane, form a right- angled triangle (Prob. XII.), in which the angle made by the given line with the projecting perpendicular is the com- plement of the required angle, which last can thence be found without finding the projection of the line upon the plane. Otherzvise, by rotating the system of the given line and plane, till, for example, the plane should be vertical, the perpendicular to it from a point of the given line would be horizontal, and hence a convenient axis about which to again revolve the given line till it should be horizontal, when the triangle above mentioned would be immediately known in its real size. In Projection. — By the method of rotation (29; Second). PI. v.. Fig. 43. Let PQP' be the given plane, and ab — ab' the given line. Assuming any convenient plane, PRS\ perpendicular both to P'Q and V, revolve it about its trace PR as an axis into H, as indicated by the arc S' Q^ , when P'Q, revolving with PRS'j will appear at Q,P^ perpendicu- lar to GL. The point /*, being in the axis PR, remains fixed, and PQ^P^ is the given plane after revolution to a vertical position. By revolving ab — a'b' an equal angular extent about the same axis, the relative position of the line and plane* will be preserved. This is indicated by the arcs b'b^' and a'a^', both having R as their centre and being of equal angular extent with 5'2i- The horizontal projec- tions of these arcs are bb^ and aa^y parallel to GL, and giv- ing a^b^ — rt:/^/ for the revolved position of the given line. This done, the horizontal ap — a^p\ where ap is perpendicu- lar to P(2, , is the perpendicular from a^a^' to the revolved plane PQ^P^y meeting it at /, as a^b^ does at c. Hence, by making/^'' equal to the hypothenuse formed on pc and it'c' DESCRIPTIVE GEOMETRY. 43 as a base and altitude, and drawing c"a^, we have the re- quired angle, pc'^a^. By the usual solutioUy we should have found the hori- zontal traces of ab — a'b' and of a perpendicular from aa' to PQF, and should then have proceeded, as in Prob. XV., to find the angle at aa' between these two lines, which would have been, as above, the complement of the required angle. Examples. — 1°. Construct the solution just indicated. Ex. 2°. Let the plane be situated as in Fig. 39. Ex. 3°. Let the plane be parallel to the ground line. Ex. 4°. Substitute for the given plane, the planes of projection. * [Find both traces of the line, then, recalling that the angle made by a line with a plane of projection is the angle made by it with its projection on that plane, by revolving the line each way, about each trace, of each of its two pro- jecting planes, we can show its true length eight times, as well as each of the required angles four times. In Fig. 32, ^'«jV is the angle between rt<5 — a'b' and H, and that between a^b^ and a'b' the one between ab — a'b' and V. See also P'no, Fig. 38 ; oP"p and Fs"7\ Fig. 39, and P'P'b, Fig. 42, for examples of the angles made by lines with the planes of projection.] PROBLEM XIX. — To reduce an angle to the horizon. In Space. — The enunciation expresses the operation of finding the horizontal projection of an angle in space, given by the inclinations of its sides to each other and to a verti- cal line. These inclinations may be shown by auxiliary developments. In Project io7t. — PL V., Fig. 44. Let ao'o and b^o'o be the angles made by the two sides of the given angle with a vertical line 00' ^ and let b^o'b^ be the true size of the angle shown by revolving it about o'b^ into V. Then oa being the horizontal projection of o'a^ the horizontal projection, vby of o'b^ will be found by the arcs b^b with ^ as a centre, and bj) with ^ as a centre, which intersect at ^, and aob is the horizontal projection of the angle b^o'b^ . PROBLEM XX. — To construct the projections of a plane figure lying in a given plane. In Space. — Determine a sufficient number of points of the figure, given by its development, by parallels or per- 44 DESCRIPTIVE GEOMETRY. pendiculars to the traces of its plane, and transfer these points by counter-revolution to the projections of the same lines. /// Projection. — 1°. Preliminaries. PI. V., Fig. 45. Let the given figure be a circle, and let PQP' be its plane. Assuming any plane, PRR'^ perpendicular to the vertical trace P'Q^ revolve the point P of the horizontal trace, in this plane, to /*,, by making /\ the intersection of an arc of radius QP^ , equal to QP, with RR'P^, the vertical trace of the plane of the revolution. Then P'QP^ will be the plane PQP' revolved about FQ into V, and the circle with 0", assumed as a centre, will be the revolved position of the given circle. 2°. Projections of points of the circle. — Draw o"s'y for ex- ample, parallel to P,Q. Its vertical trace is s' on P'Qj and it is in space parallel to PQ ; hence, projecting / at j, its projections are s'o\ parallel to GL, and so, parallel to PQ. By counter-revolution about P'Q, 0" appears at 0' , the intersection of o"o' perpendicular to PQ, with s'o' , and thence, by projection, at o on so, or so == s'o". \\{ like man- ner, any other points may be found. Again : draw, for example, o"r" , parallel to PQ. Mak- ing Qr = Qr", we find ro, parallel to GL, for its horizontal projection ; and, projecting r at r', its vertical projection is r'o', parallel to FQ. Then o'o is found as before from 0", but on r'o' and ro. Also r'o' = r"o". Other points can be likewise found. The points aa'a" and bb'b" are the highest and lowest, and are on a"g^' perpendicular to P,Q, and whose horizon- tal projection, ag, is perpendicular to PQ, at g, found by making Qg = Qg". - . Producing 00' to k", and making Qt" — Qt, the line k"t" shows the revolved position- of a line in PQP', and whose projections are perpendicular to GL. Then drawing the tangents, Tusf'd", parallel to k"t", we find the tangents, as dd', which include between them the two projections of the circle, and, as in the previous constructions, their contacts, as d and d' , with these projections. DESCRIPTIVE GEOMETRY. 45 Examples. i°. — Let the given plane be revolved into H. Ex. 2°. Let it be in the kind of position shown at PQP\ Fig. 39. Ex. 3°. Substitute for the circle a star or other plane figure. Ex. 4°. Substitute for any plane figure a cube one face of which shall be in the given plane. Theorem III. — The projections of a circle seen obliquely are ellipses, PI. V., Fig. 45. Parallels in space are parallel in projection, and lines equally divided in space are equally divided in projection. Thus every line, as ab\ through the centre of either projection, has a pair of parallel tangents at its extremities, as at a and b\ and bisects all the chords, as e'c' y which are parallel to these tangents. One of these chords, c'e\ passes through the centre o\ and hence is a diameter. Two diameters, as a'b' and ce\ each of which, as in the original circle, bisects chords parallel to the other, are called conjugate diameters. Also, in each projection, two diameters, called the axes, are perpendicular to the parallel chords and tangents which they intersect, as seen at ab and ec. These properties be- long to the ellipse ; hence we conclude that the projections of a circle seen obliquely are ellipses. The longer and shorter axes are distinguished as re- spectively the transverse and conjugate^ or the major and minor axes. LIBR A i.' t UNIVKUsn V <»}• CALlFi^KMA 46 DESCRIPTIVE GEOMETRY, CHAPTER II. DEVELOPABLE SURFACES. Definitions, 30. Proceeding with ruled surf aces (22), we come to single curved surfaces^ of which developable surfaces, as the sim- plest, are considered first. The only developable surface of revolution is the cone^ including the cylinder of revolution as one of its extreme cases. The cone of revolution is generated by the revolu- tion of a straight line about an axis which it intersects at any angle. 31. The opposite extreme cases of the cone are, firsts that in which this angle is nothing, when the generatrix (19) is parallel to the axis, and the surface becomes a cylin- der of revolution, and, second^ that in which the same angle is a right angle, when the cone becomes a plane (23). 32. The intersection of the generatrix of a cone with its axis is called its vertex. As the generatrix is of indefinite extent in both- directions from this point, in an immaterial or purely geometrical cone, the complete conic surface consists of two equal and opposite parts, of indefinite ex- tent, on each side of the vertex. These are called the tzvo nappes, 33. The base of a cone of revolution is any plane section of it, usually understood as a circular one, at which its ele- DESCRIPTIVE GEOMETRY. 47 ments are supposed to terminate, or at which it rests upon some plane. Thus the horizontal trace of any cylinder or cone is also often called its base. Circular sections of surfaces of rev- olution generally, are called parallels, and are in planes per- pendicular to the axis of revolution. Planes containitig the axis are called meridian planes, and the sections in them, meridians, A— Projections. 34. As already seen (Theor. III.), the projections of cir- cles seen obliquely are ellipses. Such, therefore, will be one or both of the projections of the base of a cylinder or cone whose axis is oblique to either H or V, or to both. It may then evidently be convenient to be able to con- struct an ellipse when its axes, as ab and ce, PI. V., Fig. 45, are known, or when a pair of conjugate diameters are known. It will also be 'convenient to construct a tangent to an ellipse, first, at a given point, since it is evidently easier to sketch a curve through given points when the tangents at those points are known ; second, from a given exterior point, as from the projections of the vertex of a cone to those of its bases (see PI. VI., Fig. 52); third, paral- lel to a given line, as at the limiting projecting lines, dd', PI. v., Fig. 45. We will therefore begin with the two following prelimi- nary general problems upon the ellipse, embracing these particulars. PROBLEM XXI. — To construct an ellipse under various given conditions, PL VI., Figs. 46-48. 1°. Construction of the ellipse on given axes by radials from the extremities of an axis. — Fig. 46. Let 2AB and CD be the given axes. Divide AB and Be, which is equal and parallel to ^6', into the same number of equal parts, simi- larly numbered, as shown, in reckoning to or from B. 48 DESCRIPTIVE GEOMETRY. Then radials from ^Tand D, through points similarly num- bered on Be and AB^ respectively, will intersect at points, as I, 2, 3, of the curve. In like manner, each of the other three quadrants of the curve can be found. • Theory. — If CA — AB, the curve would become a circle, and by the same construction the triangles, CiDy C2D, etc., would then be right-angled at i, 2, etc., on the curve, as they should be. Fig. 46 is then simply the projection of this construction of the circle, as seen after revolution about its diameter, equal and parallel in space to AB, until the diameter perpendicular to this one is shown as at CD. 2**. Construction of the ellipse by concentric circles on the given axes.—F\g.'/^y, Let OA and OChe the given semi-axes, on which describe circles, as shown, and divide the latter into corresponding parts, as Ac and Da, by radii which may be equidistant if preferred. Then, from points of division, as b and d, on the same radius, draw, from the former, paral- lels to OA, and from the latter, parallels to OC, whose in- tersections, as at/, will be points of the required ellipse. Theory. — From the triangle, Odg, since fcl is parallel to Og, we have, fg'.bg'.'.Od : Ob iiOA: OC. The lines y^ and ^^ being called ordinates of the ellipse and circle of radius OC, respectively, the first and last couplets of the above proportion express the property of the ellipse that the ordinate to the ellipse is to the corre- sponding ordinate to the circle on its transverse axis, as the scini-conjtigate axis is to the semi-transverse axis. The curve given by the construction is thus identified as an ellipse. 3°. Construction of an ellipse on a given pair of conjugate diameters — First Method. — Fig. 48. Let ab and ^^(Theor. III.) be a pair of such diameters. Draw y^ through c and paral- lel to ab. At c erect a perpendicular, cD, to fg, make it equal to ab, and on it as a diameter describe the circle DESCRIPTIVE GEOMETRY. 49 ADBc. If now a circle be described passing through O and Oy and having its diameter on fgy right angles, fOg and fog, can be inscribed in it, and the sides of the latter will be the required axes. The centre, //, of this circle is found by drawing the line Oo, and ch perpendicular to it at its middle point. The limits, m and ;/, of the semi-axes are found by drawing Mm and Nn parallel to Oo, Making op = oju, and oq =■ on, gives the axes, having which the curve can be found as already described. Theory. — Consider the paper above ^ as a vertical plane in which lies the circle of radius OA ; and the paper below ^ as a horizontal plane on which this circle is projected by projecting lines, of which Oo is one. Then cd will be the projection of cD ; ab equal and parallel to AB ; and vio and no^ being the projections of MO and NO, are not only perpendicular to each other at the centre, 0, of the ellipse, but, like MO and NO, each bisects the chords which are parallel to the other (Theor. III.). Hence they are the axes. Second Method, — Fig. 48. At a draw a parallel, ax^ not shown, to oc, meeting y^ at x, and completing the parallelo- gram aocx. Then a construction like that of Fig. 46, by lines from c and d to like points of division on ax and ao respectively, will give points of the arc, anic, of the ellipse. Having found the entire curve in this manner, the axes can be very simply found as follows : On any diameter, as ab, describe a semicircle which will cut the corresponding half of the ellipse in a point. Lines from this point to a and b will be chords including a right angle, they being thus inscribed in a semicircle ; then lines through the centre, o, and parallel to these chords will be the required axes, by a property of the ellipse that supplementary chords — that is, chords from any point of the curve to the extremities of a diameter — are parallel to a pair of conjugate diameters, and, if perpendicular to each other, are therefore parallel to the axes. Theory. — T\iQ figure thus formed is merely the projec- tion of one like Fig. 46 upon a plane parallel to it, by means of projecting lines which are oblique to the plane of 50 DESCRIPTIVE GEOMETRY. projection, instead of being perpendicular to it, as is usu- ally the case. Examples. — 1°. In Fig. 46, find the other three quadrants of the ellipse. Ex. 2°. In Fig. 46, when conjugate diameters, instead of axes, are given. Ex. 3°. Having given the curve and a diameter, find the axes. PROBLEM XXII. — To construct tangents to a given ellipse. 1°, At a given point on the curve — First Method, — Fig. 49. Let ACB be a semi-ellipse to which a tangent is to be drawn at S. The semicircle AcB may be considered as the position, after revolution about AB and into the plane of the paper, of the circle in space whose projection is ACB, Then 5 will be found at s, and sT will be the revolved position of the required tangent. The point T being fixed because in the axis, s returns by counter-revolution to 5, and TS is the required tangent. Second Method, — Let R be the given point. Revolve it to r, and draw a radius, Os, perpendicular to Or, Then, b}^ Prob. XX., OR and OS will be a pair of semi-conjugate diameters. Hence, Re, parallel to OS, is the required tan- gent at R, 2°. TJirough a given exterior point, — Fig. 50. As in Fig. 49, let AcB represent the revolved position of the circle whose horizontal projection is ACB, and let Pbe the given point in space, in the plane of the circle. Then projecting c at c' upon the ground line GL, the arc c'C , with A' as a centre, is the vertical projection of cC, giving A'C as the vertical projection of the circle in space, and P, on A'C produced, as the vertical projection of P, since, as AB \^ perpendicular to GL, the plane of the circle is per- pendicular to V, and therefore A'C is its vertical trace. Revolving into H about AB — A' again, PP appears at pp' , and pt, tangent from p to the semicircle at /, is the re- volved position of the required tangent. In the final counter-revolution, pp' returns to PP , and t to J", and PT is the required tangent. A second tangent from P can be similarly found. DESCRIPTIVE GEOMETRY. » 5 1 3°. Parallel to a given line — First Method, — Fig. 51. Pro- ceeding by the same method as in Fig. 50, let A CB be the given ellipse, and nP the given line ; regarded as the hori- zontal projection of a circle in space and a line in its plane. Then, revolving about the axis ^^— ^', we find A'C and P as before, and, by revolving into H again, P will appear at /, and ;/, being in the axis, remains fixed, giving np for the revolved position of nP into H, and hence thy parallel to ;//, for the revolved position of the given tangent. Counter-revolving again, t returns to J", and pp' to PPy whence 77/, parallel to nP, is the required tangent. Second Method. — Draw a line, OS^ not shown, parallel to nPy meeting the ellipse at 5. By revolution, show its re- volved position, Osy as in Fig. 49, and draw Ot perpendicu- lar to Os. Then, finding T by counter-revolution, TNy parallel to OS or to nP, will be the tangent required. Examples. — 1°. In Fig. 49, take the ellipse in other positions, and apply each method separately to different quadrants of it. Ex. 2°. In Fig. 50, show the entire ellipse in a different position, change the position of P, and draw both of the tangents from it. [GL must be per- pendicular to AB."] Ex. 3°. In Fig. 51, show the entire ellipse and vary the direction of nP, 35. The perfection of the form of a curve constructed by points depends less on the number of points than on the accuracy with which they are located. In the case of an ellipse, eight points^ exactly located, four being the extremi- ties of the axeSy and one other in each quadrant y are almost always enough to enable one to sketch the curve through them. PROBLEM XXIII. — To represent a cone of given axis and diameter by the projections of its vertex y circular sectionSy and visible limits, the axis being oblique to both H and V ; also any element or point of the surface. In Space, — A cone being determined by its vertex and any plane section not containing its vertex, its projections will consist of the projections of these parts, to which are 52 DESCRIPTIVE GEOMETRY. usually added the projections on each plane of those ele- ments which form the boundaries of the convex surface, as seen in looking perpendicularly at that plane. These elements are called the extreme elements, and as all the others, and hence the projections of the given section on each plane, are included between these, their projections are tangent to those of the section. In Projection,— ?\. VI., Fig. 52. 1°. Four points of each projection of the section. Let Vo—V'o' be the axis of the cone, and ab its diameter at the point 0, This diameter is in space horizontal, and hence (Theor. I.) perpendicular to Vp, and, in vertical projection, parallel to GL, as at a'b' . That diameter at 00' which is parallel to V is shown at c'd' , equal to ab, perpendicular to V'o\ and in horizontal projection parallel to GL, as at ab. We thus get four points, aa\ bb', cc\ and dd' of the section. 2°. The shorter axis of each projection of the section. — As- sume a new vertical plane of projection, parallel to the axis of the cone, or, as at 6^,Z„ containing it. By making VV"= Vy, and perpendicular to G^L^, and 00" likewise equal to the height of ^ above GL, the new projection of the axis will be V"o"\ and n"r" — ab, and perpendicular to V"o" , through 0" , will be the new projection of that diameter of the cone which is in this vertical plane. Its horizontal projection is nr, and its vertical one, not drawn, is wV, where the heights of n' and r' above GL are equal to those of n" and r" above G,L, (Prob. VI.) As these are evidently the highest and lowest points of the section, their tangents are horizontal, and hence, in hori- zontal projection, parallel to ab, and, in vertical projection, to GL. Again : the shorter axis of the vertical projection being evidently in the plane perpendicular to V through FV, since it is perpendicular to c'd', assume G"L" , parallel to V'o' as the ground line, upon a vertical plane through cd — ^V, for example, of a new plane of projection perpendicular to V. Then 0'" will be the projection of 00' upon this plane, and making a"p"=^ ap, we get o"'a" as the projection of the given DESCRIPTIVE GEOMETRY. 53 circular section upon the new plane, and, making o"'e"= oq,, we find the radius whose vertical projection is o'e', and hori- zontal projection oe, where et= e"t". The opposite extremity of this diameter is ff. 3°. Other points. — We have now found eight points of the projections of the circular section, which, as four of them determine the axes in each projection, may serve in sketch- ing the curve. But as ihey are in this case quite ununi- formly distributed, four more may usefully be found in each projection by simply drawing in the horizontal projec- tion the diameters symmetrical with cd and ef^ and therefore equal to them, and likewise, in the vertical projection, those symmetrical with^'<^' and r'n\ giving, in all, twelve points in each projection. Finally, tangents to arbf from F, and to a'c'b'd! from V\ drawn by the eye, or found as in Fig. 50, for each projection separately, will complete the projections of the cone. In the latter case, two more points of each projection, making fourteen in all, will be known. 4". Partial plane construction of the circular section. — By first finding the axes ab and rn of the horizontal projection, and cd' and e'f of the vertical projection, the curve in each case may be completed as a plane problem by either of the constructions shown in Prob. XXL, 1°, 2°. •Also, as ab and rn are perpendicular to each other in space, each bisects chords parallel to the other, and will do the like in all projections ; hence, a'b' and r'n' are conjugate diameters, and in like manner so are cd and ef As soon, therefore, as only these diameters have been found, the re- spective projections of the curve may be completed by either method of Prob. XXL, 3°. 5°. Visibility of the curve — Elements and poiftts. — In hori- zontal projection, that portion of the curve is visible which is between the extreme elements and on the upper half of the cone, as determined from the vertical projection by in- spection. Thus, dd' being on the upper half, the full por- tion of the curve is visible, and the dotted portion invisible, in horizontal projection. 54 DESCRIPTIVE GEOMETRY. . In vertical projection, the entire circumference of the circular section would be visible if the surface of the cone were limited by it, since the vertex is further from the ob- server looking towards V than the base is. But the conic surface being produced as shown, only the front part of the circular section is visible — that is, the part through ee' , and limited by the extreme elements from V, Finally, Ve — V'e' being any element, and being careful that e and e' are the projections of the same point, kk' will be a point on that element. Examples.— 1°. Construct a larger cone and with a larger angle at its ver- tex, so as to clearly distinguish the extreme elements, tangent to the circular section, from those through ad , dd\ etc. Ex. 2°. Let the axis of the cone be parallel to H or V. Ex. 3°. Let it be nearly perpendicular to H or Y. The Co7iic Sections in General. 36. A cone being often most conveniently represented by one or both of its traces (Prob. IV.), together with its ex- treme elements, the principal forms of its intersections with H and V, according to its position relative to them, are to be noted. A plane can intersect a cone in only three essentially dif- ferent ways. Thus, VAB, PL VI., Fig. 54, representing a cone whose axis, VS, and opposite elements, VA and VB, are in the plane of the paper, the intersecting plane may be supposed to be perpendicular to the paper. Then this plane may intersect the cone as follows : 1°. Parallel, as at PQ, to a plane, as MN, which contains only the vertex o( the cone. In this case, the plane cuts all the elements of the cone, and the section is an ellipse. 2°. Parallel to a tangent pla7ie, as the one of which VA is the trace. In this case the plane is parallel to the element VA only, and hence cuts all the elements but that one, and the section is called a parabola. 3°. Parallel to a plane, as the one whose trace is Vm, which contains two elements of the cone. In this case, the DESCRIPTIVE GEOMETRY. 55 plane cuts both nappes (32) of the cone, and all the elements except the two in the plane F?;/, and the curve, consisting of two opposite branches, is called a hyperbola. 37. The ellipse, Fig. 52, is a closed curve, since its plane cuts all the elements of the cone. The parabola and hyperbola are open curves, and of infinite extent, since their planes, being parallel to certain elements, intersect them only at an infinite distance from the vertex, where, since the elements diverge from the vertex, they are also at an infinite distance apart. 38. Special or elementary forms. — When the plane PQ, Fig. 54, coincides with MN, the ellipse reduces to a point. When it is perpendicular to the axis VS of the cone, the ellipse is a circle. When the plane of the parabola coincides with the tan- gent plane at VA, the parabola becomes a straight line. When the cone, by the infinite distance of its vertex, be- comes a cylinder, any intersecting plane parallel to a tan- gent plane cuts it in two parallel straight lines, which is thus another special case of the parabola. Finally, when the plane of the hyperbola coincides with that whose trace is Vm, the hyperbola becomes a pair of in- tersecting straight lines. Thus the point, straight line, parallel lines, intersecting lines, and circle, which, with figures composed of them, form the subject of elementary geometry, are simply the special and simplest forms of the conic sections. 39. Curves determined by their projections. — The projecting lines from all the points of a curve to a plane of projec- tion, form a cylindrical surface of which that curve is the directrix (19), unless the curve lies wholly in a plane which is perpendicular to that plane of projection. The trace of this cylinder upon that plane will be the projection of the given curve, and, indeed, of <^//y curve traced upon the con- vex surjface of this cylinder. But, as in the case of a point 56 , j , DESCRIPTIVE GEOMETRY. or line, the projecting cylinders erected on two given pro- jections of a curve will intersect each other only in that one curve, which is thus determined by those projections. Theorem IV. — In the ellipse , section of a cone made by a plane actually cutting all its elements^ the sum of the distances from any point of the curve to two fixed points within it is constant^ and equal to the transverse axis. Let VAB, PL VI., Fig. 54, represent a cone whose axis, F5, and opposite elements, VA and VB, are in the plane of the paper ; and let PQ be the trace of a plane perpendicu- lar to the paper and oblique to VS in the manner stated. Then the circles of centres 5 and s, in the cone's axis, and each tangent to PQ, VB, and VA, are the projections of two inscribed tangent spheres having circles of contact with the cone of which AB and ab respectively are chords. These spheres are tangent to the plane PQ at F and F^, All these points and lines, with SA, SF, and sF, being in the plane of the paper, conceive the projecting lines employed to be oblique to the paper, and the curve rpq may represent the section of the surface of the cone made by the plane PQy and A CB and acb the circles of contact of the spheres with the cone. This done, assume any point p of the curve, and through it draw the element VC of the cone, pF, and pd, which rep- resents a perpendicular from p to PQ, Then, since all tan- gents to a sphere from the same point are equal, and por- tions of elements of the cone between parallel circles are also equal, Ar-vra = Fr + rF, = Cp -^-pc — pF-^pF, also bq-\-qB — F,q + qF= Cp +pc = pF+pF, The third and fourth members of these equations being the same, their second members are equal ; then, subtracting from the latter the common part FF^, we have 2 Fr =z 2 F,q DESCRIPTIVE GEOMETRY. ■ I I ' j; 57 (JALllOKMA whence, adding FF^ again, rF^ — qF hence from the first two equations, pF+pF^ =1 Fr + Fq = rq". That is, the curve rpq is such that the sum of the distances from any one of its points to two fixed points, F and F^, Avithin it, is constant, and equal to its longest chord, rq. This property forms the usual definition of the ellipse ; rq is the transverse axis, while the fixed points F and F^ are called its foci. As the demonstration takes no account of the angle A VB at the vertex, we conclude that it is independent of it, and applies to the cylinder as a particular case of the cone, and hence that the oblique section of a cylitider of revolti- tion is an ellipse. 40. To construct an ellipse by (Theor. IV.). — Lay down rq and the points F and F^ anywhere between its extremities, but equidistant* from them. With i^and F^ successively as centres, describe arcs, first with any radms greater than Fr, and then with the remainder of rq as a radius. The inter- sections of the latter arcs with the former will give four points of the curve. These points, thus found in sets of four, will be seen to be symmetrically placed relative to rq and a perpendicular to it at its middle point. Hence the ellipse, section of the cor>e, has two axes of symmetry, and is identical with the ellipse, as defined in Prob. XX. 41. By methods* quite like those of Theorem IV., it may be shown that a parabola is a curve such that the distances of any one of its points from a fixed point within, the focus, and from a fixed line, the directrix^ are equal ; that the hy- perbola is distinguished by the property that the difference of the distances from any one of its points to two fixed points, fociy is constant and equal to the shortest chord connecting *See Olivier, Cours de Geometric Descriptive. 58 DESCRIPTIVE GEOMETRY. the two branches of the curve ; also that the ellipse and hyperbola have directrices. PROBLEM XXIV. — To represent a cone by the projectioits of its vertex, horizontal trace, and extre^ne eleme^its. In Space. — We may proceed in two ways : first, by find- ing any convenient number of the fourteen points of some circular section, by the last problem, and then finding the horizontal traces of the elements through these points ; or, second, by first finding the axes of the horizontal trace, which will usually be an ellipse (36), and then completing the curve by Prob. XXI. (1°, 2°). By an extension of (27, 4°) either trace of any element of a cylinder, or a cone, is in the like trace of that surface. In Projection. — 1°. Without the conjugate axis. Having a sufficient number of the points, aa' , bb' , cc' , etc., already de- scribed (PI. VI., Fig. 52), according to the scale of the figure or the distribution of the points, depending on the position of the cone, then, without sketching the projections of the circu- lar section through them, find the horizontal traces, as A, of Vd — V'd' (taken to represent the extreme element which is so nearly confounded with Vd'), D, of Vb — V'b', etc., and join these points, which will give the horizontal trace, ADB, of the cone, tangents to which, from F(Prob. XXII., Fig. 50), will complete the horizontal projection of the cone. Tangents as A A' to the horizontal trace, and perpen- dicular to GL, found by Prob. XXII., Fig. 51, will project this trace on the ground line at A'B', whence A'V and B'V will complete the vertical projection of the cone. Since E' , vertical projection of ?/, does not bisect ^'.5', we see that the centre of the trace is not in the axis Vu — FV^' of the cone. 2°. By the immediate construction of the axes of the trace. — The transverse axis, MR, of the trace may be found either by finding the horizontal traces M and R of the elements Vn — FV and Vr — V'r' , or by means of the auxiliary projection DESCRIPTIVE GEOMETRY. 59 V''r"n'\ in which F V and Vn" intersect G,L, in the same traces. The conjugate axis, PQ, of the trace is thus found. Pro- duce the axis V^'o" of the cone, and draw a perpendicular, NC^ to it through Oj the middle point of MRy and hence the centre of the trace of the cone ; then CN is the radius of that circular section of the cone whose plane intersects H in the line PQ, perpendicular to MH. The required conjugate axis of the trace is thus that chord of this circle which is perpen- dicular to the auxiliary vertical plane at O, The arc NS of radius CN is the revolved position of an arc of this circle, and 'OS, perpendicular to CN, is half of the desired chord ; then making OP and OQ each equal to OS, we have MR and PQ the axes of the cone's trace, from which it can be constructed by Prob. XXL, 1°, 2°; or by (40), since, as the lines from P or Q to the foci are equal to each other, and their sum equal to MR^ each is equal to MO, Hence find the foci by intersecting MR by an arc from P or Q^is a centre, and MO as a radius, and then proceed by (40). Examples. — 1°. Let VV" be in any other angle than the Jirst. Ex. 2°. Let the axis of the cone cross any other angle than ih^JlrsL Ex. 3°. Substitute a cylinder for a cone in each of the foregoing examples. Ex. 4". The axis of the cone being parallel to V, let one of the elements be parallel to H. [The horizontal trace will in this case be a parabola.] Ex. 5°. The axis being as in Ex. 4, let a horizontal plane through the vertex contain two elements. [The horizontal trace will then be a hyperbola.] PROBLEM XXV. — To constricct the projections of a cylinder of revolution having a given horizontal trace, or base ; also any clement and point of the surface. In Space. — From the symmetry of the cylinder, and its base, relative to a vertical plane through its axis, its ex- treme elements, in horizontal projection, will be tangents to the base at the extremities of its shorter axis, which is the diameter of the cylinder. Those seen in vertical projection will be parallels from the extremities of the vertical pro- jection of the base, which will be on the ground line ; and 6o DESCRIPTIVE GEOMETRY. they will be at a distance apart equal to the diameter of the cylinder. All the elements are parallel in each projection. In Projection.— V\. VI., Fig. 53. Let the ellipse ADBC be the given trace. Then, as described, ^/and DK^ tan- gent at the extremities of the shorter axis CD^ complete the horizontal projection of the cylinder. CBD is dotted, being invisible. Drawing the tangent projecting lines as EE' (Prob. XXII., 3°), E'F' is the vertical projection of the base, and O' that of its centre O. Next, describe the circle k't' of cen- tre O' and radius OC that of the cylinder, and the extreme elements in vertical projection will be tangents to this circle from E' and F' , To draw these important lines, to which all the other vertical projections of elements are parallel, some other points are provided. Thus, having, by the usual method, shown in the figure, found k\ the point of contact of a tangent from E' to the circle centred at O' , make MN, parallel to O'k', a fourth proportional, MN, to E'O', 0'k\ and any assumed distance EM. Also making E'O'' = E'O' and 0"k" equal and parallel to O'k', we have four points, k", E\ k' and N, which, if all in the same straight line, will determine it very exactly. Ff, parallel to k"E'k' ^ is then the opposite element. Finally, Hp, parallel to CI, being the horizontal projec- tion of any element, and /, of a point upon it, projecj: H at H', draw H'p' parallel to E'N, and project / upon it at /', and Hp — H'p\ and//', will be the projections Of an element and of a point upon it. Examples. — 1°. Vary the problem, beginning by drawing C7 and DK Kn the opposite directions to the present ones from Cand D. Ex. 2^. Make the application of Prob. XXI., 3°, in drawing the tangents EE' and FF . PROBLEM XXVI. — To construct the projections of a cone of revohttion having a given elliptical trace or base. In Space. — Observing from Theorem IV., Fig. 54, that a sphere inscribed in a cone of revolution and tangent to DESCRIPTIVE GEOMETRY. 6l the plane of an elliptical section, is so tangent at a focus of that section ; and that by producing or reducing FS or F^s, other cones of revolution can be found having the same elliptical section rp^, we construct the two projections of any sphere tangent at either focus of the given trace. The required cone will then be tangent to this sphere. In Projection. — PL VI., Fig. 55. Let ACBD be the given elliptical horizontal trace of a cone of revolution, with the axis AB parallel to the ground line. At either focus, as 5, F , of the trace, erect a perpendicular, and on it take any convenient radius, S' F , and with it describe the equal circles at 5 and S\ which are the projections of the tangent sphere at S, F, Then, tangents from A' and B' to the circle S' will complete the vertical projection of the cone, and, projecting V at F, on AB produced, tangents from V to circle S, which will also be tangent to the ellipse ACBD, will com- plete its horizontal projection. The sphere of radius 6"/^ may be tangent at either focus, and either above or below H, giving four cones. Examples.— i". Let AB be oblique to GL. Ex. 2°. Construct the cylinder of revolution whose horizontal trace is ACBD. [That there will be one such cylinder among the series of cones of which that in the figure is one, appears from the fact that when the variable circle S' is larger than a certain size, the tangents to it from A' and B' will converge dowmoard. There must, therefore, be one size of this circle which will make these tangents parallel. It may easily be shown by elementary geometry that the centre of this circle will be at the intersection oi F S pro- duced with the semicircle on A' B' as a diameter.] PROBLEM XXVIL — To construct the projections of a cone whose axis is parallel to the ground line, and of any of its elements, and points. In Space. — These projections will evidently be two equal isosceles triangles. The construction of any element re- quires that of any point on the circumference of the base, whose plane is perpendicular to GL. This is accomplished by revolving the base about, any suitable axis until it be- comes parallel to or coincides with a plane of projection. 62 DESCRIPTIVE GEOMETRY. In Projection.— \.^t VAD—V'B'C, PL VI., Fig. 56, be the projections of the cone. AD — A'O' is the horizontal, and OB — C'B' the vertical diameter of its base. By revolving this base about the former diameter, or any parallel to it, as an axis, the base will become parallel to H, and will then show its circular form in the horizontal projection. By revolving it about OB — CB\ or any parallel to this line, it will likewise show its circular form in vertical projection. Then let mVbe the horizontal projection of any element; revolving the base 90° to the left about AD — A' as an axis, its upper half will appear as at AB"D — A'B'", and m de- scribes the quadrant whose horizontal projection is mm" and will appear at m"m"'. By counter-revolution, this point returns in the arc m"m — m'"m\ and m' V is the vertical projection of the element inV ; but as the cone is symmet- rical relative to every plane through the axis, make A'r' = ^W, and /F' will be another element whose horizontal projection is m V. The point pp' is on the element m V—m' V . Again: let ^'F' be the given projection of an element. Revolving as before, a' will appear at a"\ whence it is pro- jected at a" , and also on the back of the cone at b" . In counter-revolution the two points a" a'" and b"a"' return to aa' and ba\ giving aVa.nd bVsiS the horizontal projections of the elements whose vertical projection is a'V\ The important elements VB—V'B\ VB—VC, VA— VA\ and VD — VA' are distinguished respectively as the highest, lowest, foremost and hindmost, the observer being understood to face the plane V. Examples. — 1°. [To become, as is very useful, familiar with revolution] revolve the base about the horizontal trace of its own plane, about the vertical trace of its own plane, and about its vertical diameter^ in finding an element one of whose projections is given. Ex. 2°. Perform the same operations when the base is the right-hand end of the cone. 42. Prob. XXIV., Fig. 52, shows how to find the size of the opening which a given cone of revolution in a given position would cover. Probs. XXV. and XXVL, Figs. 53 and 55, show how to find the size and direction of a cylin^ DESCRIPTIVE GEOMETRY. 63 der or a cone of revolution which would cover a given elliptical opening. After the full exhibition afforded by PL VI. of all the essential operations necessary in constructing the projections of single cones or cylinders of revolution, these bodies will, to avoid repeating these operations when merely illustrat- ing general principles, be simply assumed in the subsequent problems. B— Tangencies. 43. Inspection alone, PL VII., Fig. 58, is sufficient to show that a tangent plane ABC to a cone, as V—MTN, or to a cylinder, touches it all along a single element as FT", and that it contains only that element of the surface. Any line, as 'tnny or pq^ in such a plane, touches the given surface only at its point of intersection with the ele- ment of contact of the plane, and is therefore a tangent line to the surface at that point. 44. Among all the tangent lines thus lying in the tangent plane, one, VT, coincides with the element of contact of the plane with the given cylinder or cone, but it -is still accounted as only a particular case of the tangent line, since it is not a secant. That is, while other tangents as mn are tangent at c to any curve traced through c on the cone, the elemeyit through c coincides with its tangent. In other words, the tangent to a straight line is that line itself. 45. Since by {43) all the tangent lines at a given point of a cylinder or cone lie in the tangent plane along the ele- ment through that point, while any two intersecting lines determine a plane (Prob. IX.), the tangent plane to a cylinder or cone at any point is determined by a7ty two tangent lines at that pointy one of which by (44) may be the element through that point ; or, derivatively, by any lines so located as to be in the same plane with these (27, 9°). And as the element through a given point of contact is one of the tangent lines, all of which must (43) lie in the 64 DESCRIPTIVE GEOMETRY. tangent plane at that point, no plane through the vertex can be considered a tangent plane unless it contains an element ; while every tangent plane will contain the vertex since every element contains the vertex. 46. Comparing the preceding with the principle (Prob. IX.) that any plane is determined in the most elementary manner by three points, it is evident that the tangent plane to a cylinder or cone will be determined by any condition equivalent to that of three determming points, two of which are on the element of contact of the plane with the cone, and the third on any tangent at any other point on that element. The following problems will make this principle,^nd the methods of applying it, more evident. PROBLEM XXVIII. — To construct a plane tangent to a cone on a given element. In Space. — The required plane will be determined by the given element (44) and by whatever other tangent line (43) at some point of this element is made most con- venient by the position of the given cone in each case. The traces of the plane will then be determined by those of its determining lines. In Projection.— L^t the axis of the cone VAB—V'CD', PL VII., Fig. 57, be parallel to the ground line, and let Vt — V't' be the given element of contact, one projection of which is found from the other as in Prob. XXVII. The required tangent plane will then be conveniently determined by this element, with the tangent to any circu- lar section, as the base AB — C D' , at its intersection, as tt\ with this element; or, if the traces of the element are incon- veniently distant, by like tangents at any two points of Vt — r/'(43). The traces of Vt — F/ are ^ and /, which are therefore points of the horizontal and vertical traces respectively of the tangent plane. The tangent at tt' must, being in a DESCRIPTIVE GEOMETRY. 65 profile plane, be shown in revolved position in order to find its traces. Among various axes of revolution easily employed, the vertical diameter D — CD' of the base is here chosen, and the front half of the base is revolved to the right, carrying tf to t"t"\ and the vertical trace d — dD' of the plane of the base to d'' — ed'". The tangent at it' is then shown at d"'n"', a tangent at t'" , and n"n"' and d"d" are the revolved positions, and, by counter-revolution, n and d' are the primitive positions of its horizontal and vertical traces respectively. Hence nq is the horizontal and r'd' the vertical trace of the required plane PQP'y whose two traces must also meet GL at the same point Q. Examples. — 1°. Let the horizontal trace of the cone be given. Ex. 2°. Substitute a cylinder for the cone. Ex. 3°. Substitute a cylinder for the cone of Ex. 1°. Ex. 4°. Let the axis of the cone or cylinder be perpendicular to either H orV. PROBLEM XXIX. — To construct a plane parallel to a given line, and tangent to a cylinder whose axis is parallel to the ground line. In Space, — All tangent planes to a cylinder being parallel to its axis, the required planes, of which there will be two, and hence also their traces, will be parallel to the ground line. Either willy> therefore, be determined by its being parallel to a plane similarly situated and containing the given line, and by its containing a tangent line to the cylin- der parallel to some line of this auxiliary plane. In Projection,— \.Qt ABCD-E'FH'K', PI. VII., Fig. 60, be the given cylinder, and ab—a'b' the given line. The traces of this line are b and c'. Then bN and c'N', parallel to GL, are respectively the horizontal and vertical traces of the auxiliary plane, parallel to GL and containing ab—a'b'. The profile plane A QE' cuts from the auxiliary plane iW— PN' the line whose traces are Y' and N, and whose posi- tion, after revolution about the vertical trace QE', is Y'N". e^ DESCRIPTIVE GEOMETRY. The plane AQE' also cuts from the cylinder the circle AC—H'E' whose centre in the same revolution describes the horizontal arc 00"-0'0"', giving the circle of radius 0"'E"'=0'E' as the revolved position of AC—H'E'. Then the tangents, T'R" and t"r", to this circle, and parallel to Y'N", are the revolved positions of tangents to the cylinder in the two tangent planes. R" is the revolved, and R the primitive horizontal trace of T'R", found by describing the arc R"R with g as a centre. The intersection, S', of T"R" with the axis, is its vertical trace. Hence RM and S'M', parallel to GL, are the horizontal and vertical traces of one of the tangent planes. The other one may be similarly found. Examples.— 1°. The position of the cylinder remaining the same, let the plane be tangent on a given element. Ex. 2°. Let the tangent plane contain a given point in space. PROBLEM XXX. — Through a given point in space to con- struct a tangent plane to a cone whose horizontal trace is given. In Space. — Any two tangents to the cone from the given point will determine the required plane. Since this plane contains the cone's vertex (45), the line from the given point to the vertex, as PV, PL VII., Fig. 59, wiU be one of these tangents, while, for the other, a more convenient one will be the tangent, as A 7", from the horizontal trace, ^, of this one, to the horizontal trace, MTN, of the cone. Indeed, A Twill be the horizontal trace of the tangent plane. Otherwise, if any circular section of the cone be known, a tangent to it from the point where its plane cuts the tan- gent PV at the vertex will be convenient. In Projection.— Y\. VII., Fig. 61. Let VACD-V'A'B' be the cone, and aa' the point ; then aV—a'V being a line in the tangent plane, its horizontal trace, p, is a point common to the horizontal traces, PdQ and PcR, of the two planes DESCRIPTIVE GEOMETRY. 67 evidently given by the solution. Since aV—a'V does not conveniently meet V, the vertical traces of the tangent planes are determined by the points Q and R with the vertical traces, h' and k\ of the horizontal tangents, eh — e'h' and bk — ¥k\ parallel respectively to PQ and PR, and at points ee' and bb' taken at pleasure on the elements of contact yd — V'd' and Vc — FV of the planes. The vertical traces of these elements will be other points of the like traces of these planes. Examples. — i**. Let aa' be higher than the vertex. Ex. 2°. Let it be in any other angle than the first. Ex. 3°. Let the vertical trace of the cone be given. Ex. 4°. Let the projections of the cone's circular base be given. PROBLEM XXXI. — To pass a plane parallel to a given line, and tangent to a cylinder whose axis is oblique to H and V, and one of wjiose traces is given. In Space,— T\i^ two simplest determining tangents (45) would be a tangent parallel to the given line, and the ele- ment through its point of contact. But the tangent plane may be determined indirectly and more simply still as fol- lows. By Prob. IX., 4°, pass a plane through the given line and parallel to the axis of the cylinder ; since all the tangent planes are parallel to the axis. Then the tangent planes, of which there will be two, will be parallel to this one, and their traces will be tangent to the like trace of the cylinder. /;/ Projection. — W VII., Fig. 62. Let ABD—A'B' be the horizontal trace of the cylinder, and ab — a'U the given line. bda' is the plane through this line and parallel to the elements as CE, A'M' of the cylinder, since pq^p'q\ in this plane, is parallel to these elements. Then PQ, parallel to bd, and tangent to ABD at 7", is the horizontal trace of one of the required planes, and QP, parallel to da', is its vertical trace. The vertical trace of the element of contact Tr — Tr' will also be a point of the vertical trace of the plane. 6S DESCRIPTIVE GEOMETRY. Examples. — i". Find the other tangent plane to the above cylinder, paral- lel to ad — a'd\ Ex. 2°. Let the axis of the cylinder cross any other angle than the ^rsi. Ex. 3°. Let the cylinder and the given line cross different angles. PROBLEM XXXII. — To construct a plane parallel to a given line and tangent to a cone whose horizontal trace is known^ and whose axis is oblique to both H and V. In Space. — The two simplest determining- tangents (45) are a parallel L to the given line and through the cone's vertex, and a tangent to the given trace of the cone from the like trace of L. Two such tangents can generally be drawn, hence the solution gives two tangent planes. \i L coincides with an element, there will be one plane, but none, if L enters the cone. /// Projection.— "^X, VIL, Fig. 63. Let VAB—V'A'B' be the cone and ab — a'b' the given line. Vm — Vm' is the parallel to ab — a'b' through the vertex, and its traces* are m and n' , As the tangent planes contain this line, their traces contain these traces (27, 4°), hence smQ and mrS are the horizontal traces of these planes, and 71' Q and n'S are their vertical traces, which will also contain the vertical traces — not shown — of the elements of contact, Vs — FV and Vr — V'r'. Examples. — i". Let the given line cross any of the other angles. Ex. 2°. Let the axis of the cone cross any of the other angles. Ex. 3°. Let the projections of a circular section of the cone be given instead of its trace. Ex. 4°. Let the construction be made for the upper nappe (32) of a cone. 47. Problems of tangent lines to developable surfaces, or, more definitely, to curves traced on such surfaces, might seem to properly fall here under the head of tangencies, but as such curves practically arise as the intersections of such surfaces with plane or other surfaces, such problems form a portion of those under intersections. iVi?/^.— Problems XXIX.— XXXII. will enable the student here to proceed to problems of shade on cylinders and cones, and of their visible boundaries DESCRIPTIVE GEOMETRY. 69 as seen from a finite distance. Thus, if the given point in XXX. be the place of the eye, the element of contact of the plane will be the visible limit of the convex surface ; or if the given line be a ray of light, that element will be the boundary between the illuminated and unilluminated part of the surface. O — Intersections. 48. The intersections of developable surfaces, here to be considered, are of two kinds : 1°. Their intersections with //(^//^Ti". ' 2°. Their intersections with each other. The former curves, being m planes ^ are thence called plane curves. Otherwise, as the intersection of two surfaces only one of which is curved, they are also called curves of single curvature. The intersection of two curved surfaces cannot generally also be a plane curve, and is accordingly distinguished as a curve of double curvature, 49. A plane curve ^ or curve of single curvature^ is generated by a point which moves in any manner, subject only to the condition of remaining in a fixed plane. An S or an 8 are thus plane curves, as well as are circles or other simple curves. A curve of double curvature is generated by a point which moves so that no more than three consecutive positions are in the same plane. Thus, let a point, /, starting from the position 0, take the successive consecutive positions i, 2, 3, 4, 5 71, not lying in one plane. Then, as a plane can always be passed through three points, the point / will be successively in the planes 012, 123, 234 {71 — 2, n — i, ;/). 50. The line 01 may have any one of an indefinite number of directions, giving as many possible positions of the point I of a curve of double curvature. For each of these positions the line 12 may have an indefinite number of directions, and 70 DESCRIPTIVE GEOMETRY. S0 on. Hence there is an infinite variety of curves of double curvature. The same conclusion is reached from (48) by considering that there is an indefinite variety of curved surfaces, each pair of which may intersect in a great number of different ways, thus giving rise to an innumerable variety of curves, which, except in particular cases, will be of double curva- ture. 51. ^ secant is a line which intersects any given portion of a curve in two points, as at STy PI. VIII., Fig. 64. Two particular cases of the secant are of special interest. First. When, by turning the secant about either of its intersections as 7", the other intersection, 5, always continu- ing in the curve,, comes to coincide with T as at 7X, the secant is called a tangent. When tangent at an infinitely distant point of the curve, it is called an assymptote. Second. That position of the secant, as TTV, which is per- pendicular to the tangent is called a normal. 52. From (51) the tangent to a curve of single curvature may now be defined as a line lying in the plane of the curve and touching it at only one point. The tangent to a curve of dotible curvature is likewise most simply imagined as lying in the plane of the three consecu- tive points of which the middle one is the point of contact. The normal at the same point lies in the same plane with the tangent. 53. Intersections of developable surfaces with planes are all found one by one by the general principle of Prob. XI. ; that is, each point is the intersection of some element of the given curved surface with some line of the given plane which is in the same plane with the element. 54. The intersection of a plane curve, C, with any given plane, is found by finding, first, the intersection, /, of the DESCRIPTIVE GEOMETRY. 7 1 plane of the curve with the given plane ; and, second, the points in which this line, /, intersects the given curve. Examples. — 1°. Find the intersection of any curve in a horizontal plane with a given oblique plane. Ex. 2°. Let the planeof the curve be parallel to V. Ex. 3°. Let it be perpendicular to either H or V only. Ex. "4°. Let it be oblique to both H and V. 55. The intersection of a given oblique plane with a curve of double curvature (49) is found by the methods either of rotations (29, Second) or of changed planes of projection (29, Third), as follows : First. Let D be the curve, d and d' its projections, and P the plane. Then, for example, revolve both D and P about the same vertical axis, X, till /^becomes perpendicular to V. The intersection of its vertical trace with^/, vertical projection of the curve Z>, (found by revolving a sufficient number of points oiD), will then be the vertical projection, x (^ of the revolved position of the required point. The horizon- tal projection, ;ir„ of this position will be on ^„ that of the like position of the curve. By counter-revolution about JT, the real position, xx' , of the required point will be found on ^and d\ the projections oi D. Second. Otherwise: Assume a new plane, V„ for example, perpendicular to the horizontal trace of P, and find d" the new vertical projection of D upon it, by the principle of (Prob. VI.) The trace of Pupon V, will there intersect d" at a point x", auxiliary vertical projection of the required point. From this x, on d, and thence x' , on d' , can at once be found. Example.— Find the intersection of the plane PQP\ PI. VL, Fig. A, with the curve of double curvature dd' . 56. The following problems will sufficiently illustrate the construction of the intersections of planes with cylinders and cones, being those marked by a star in the following table of the various combinations ; where a right plane or cone means one that is perpendicular either to H or V. 72 DESCRIPTIVE GEOMETRY. PLANE INTERSECTIONS. Plane. Right. Oblique. Right. I 2 Obi. 3* 4* Right. 5 6* Obi. 7 S* PRO B LEM XXXIII . — To find the intersection of a plane per' pendicular to V, with a cylinder whose axis is oblique to both planes of projection, and whose horizontal trace is given. In Space, — The plane being perpendicular to V, so much of its vertical trace as lies between the extreme elements ot the cylinder is (27, 8°) the vertical projection of the entire curve ; any point of which is therefore the intersection of this projection with the vertical projection of an element of the cylinder. The horizontal projection of the point will then be on that of the same element, whose horizontal trace is on that of the cylinder. In Projection,— V\, VIII., Fig. 6^, Let ACD be the hori- zontal trace of a cylinder of revolution, whose projections by (Prob. XXV.) are ABDbd and E'C'c'e'; and X^tPQF be the given plane. Then c'e' is the vertical projection of the required intersection. Then, taking the extreme elements, and other convenient ones, as B'b\ whose horizontal projec- tion is Bb, and Pf, whose horizontal projection is Ff, project the point b' , intersection of P'Q and B'b\ upon Bb, giving b, and /', intersection of F'f and P'Q, upon Ff, giving /. Having in this way found a sufficient number of points of DESCRIPTIVE GEOMETRY. 73 the horizontal projection of the curve, connect them, as shown, observing to make the upper half of the curve full, since it is visible. This part is on elements whose horizontal traces are on the ha.U DEB of that of the cylinder. Examples.— 1°. The cylinder remaining as in Fig. 67, let the plane FQP' be vertical. Ex. 2°. Solve case (i) of the table. [See my Elem. Proj. Drawing.] Ex. 3°. Solve case (2) of the table. [See Fig. 65.] Ex. 4°. Let the cylinder be given by its axis and diameter. Ex. 5°. Substitute a cone, oblique both to H and V, for the cylinder, FQP' remaining as now. PROBLEM XXXIV.— 7^7 Jind the intersection of a plane with a cylinder y both being oblique to both planes of pro- jection. In Space. — Any auxiliary secant plane, parallel to the axis of the cylinder, will cut from the surface of the cylinder two elements, and from the given plane a straight line. Where this Hne meets those elements will be two points of the required curve. The auxiliary plane is most conveniently taken perpen- dicular to a plane of projection. The solution is thus seen to be the same as that of Prob. XI. In Projection.— Y\. VIII., Fig. 68. Let ADB be the hori- zontal trace of a cylinder of revolution, Hlhi—A'B'R', and let PQP be the given plane. Let the auxiliary planes be vertical ; they will also be parallel, since the elements of the cylinder are so, and their intersections with the given plane will therefore be parallel. MNP', where MN is parallel to //, is parallel to the auxiliary planes, hence its intersection, MN-m'P' with PQP', found by Prob. X., is parallel to those of PQP' with these planes. This done, any secant auxiliary plane, as FE, contains the elements Ff—Ff and Ee—E'e' of the cylinder, and cuts from the given plane PQP the line Ek-e'k', parallel to MN-m'P. Hence 74 DESCRIPTIVE GEOMETRY. ff and ee\ where those elements intersect this line, are the points where these elements meet PQP' , and hence are points of the required intersection. Other points being similarly found, and joined in each projection, the required curve hde—h'd'e' will be found. In the figure, the vertical projection of each point is found first, because the auxiliary planes are vertical. As in Prob. XL, were these planes perpendicular to V, the hor- izontal projection of each point would be found first. The visible half in horizontal projection is the upper one, whose points are on elements having their horizontal traces in the half HBI of the trace of the cylinder. The front half is visible in vertical projection, and crosses the elements whose horizontal traces are on the h2i\i AHB of the trace of the cylinder. Examples. — 1°. Take the auxiliary planes perpendicular to V. Ex. 2°. Let the given cutting plane have the kind of position (Prob. IV.) indicated in Fig. 71. Ex. 3°. Let the axis of the cylinder cross the first angle. Ex. 4°. Let the given trace of the cylinder be on V. Ex. 5°. Let the cylinder be given by its axis and diameter. PROBLEM XXXV.— r^ fiitd the intersection of an oblique plane with a cone whose axis is vertical. In Space. — Recollecting that a circle, if parallel to a plane of projection, is as convenient an auxiliary as a straight line, the auxiliary planes in this problem may have any position in which they will intersect the given cone, either in straight lines or circles. Then the points in which the line cut from the conic surface by any auxiliary plane meets the line cut from the given cutting plane by the same auxiliary plane, will, being thus common both to the cone and the given plane, be points of their required intersection. In Projection.~-?\. VIIL, Fig. 69. Let VABK-VA'B' be the cone, and PQP' the cutting plane, so taken that its traces coincide. As will be seen, this position of PQP' only DESCRIPTIVE GEOMETRY. 75 requires care in distinguishing its traces by their uses in the successive steps of the solution. 1°. The points on the extreme elements^ V'A' and VB' , — To find these, take the auxiliary plane ^^P which contains these elements, and, remembering that it is parallel to V, its inter- section, PF— PV, with PQP\\s parallel to the vertical trace QF, Then n'n and /j, the intersections of /V— PF with VB'— VB and VA'— VA, are the points required. As /V is in the cutting plane PQP\ and also in the same plane with the cone's axis, it intersects this axis at r\ which is there- fore the intersection oi PQF with the cone's axis. 2°. The highest and lowest points. — These are evidently on the elements lying in a plane perpendicular to the given plane, and containing the axis of the cone. VCy perpendicu- lar to PQ, is the horizontal trace of such a plane, and it con- tains the elements VL and VH, Revolving the plane VC about the cone's axis till it is parallel to V, these elements appear in vertical projection at VB' and VA' ; and the line VC, intersection of the plane VC with PQP\ appears at C"V—C"'r', since the cutting plane, as shown in (i°), inter- sects the cone's axis at r' . Hence I'" and Ji!" , intersections of C"'r' with VB' and V'A', are the revolved positions of the required highest and lowest points. By counter-revo- lution, they return to their real positions, / and //, on Or , the primitive position of C"'r'. Their horizontal projec- tions, / and //, may be found by projecting / upon VL and h' upon VH, or by projecting I'" and h'" at I" and h" , and thence revolving these, as shown, by arcs, with F as a centre, to / and //. 3°. Points on the foremost and hindmost elements. — These elements are those whose horizontal projections are VI and VK, and which are both vertically projected in Vo\ Their projections thus coincide with the traces of the auxiliary plane ImV containing them, and with om—o'm', the inter- section of this plane with the given plane PQF. Hence, revolving their plane about the cone's axis, till parallel to V, they will appear at VB—VB' and VA — VA', and om—o'm' at o"m"- o"'m"'. Hence q"'q" and k"'k" are the revolved 76 DESCRIPTIVE GEOMETRY. positions, and qq' and kk' the primitive positions of the points on the foremost and hindmost elements. 4°. Points 071 any other elements, (See Fig. 70.) — Assuming V'B\ for example, at pleasure, either as the vertical projec- tion of two elements, or as the vertical trace of their plane, VA and VB are the horizontal projections of the elements, and PB' , perpendicular to the ground line D' Q, is the hori- zontal trace of their plane. Then PQF being the given plane whose intersection with the cone VABC—V'C'D' is required, Pm—B'm' is its intersection with the auxiliary plane /!^'F^ and aa' and bb\ where this intersection meets the assumed elements VA — V'B' and VB—V'B\ are those points of the intersection of the given cone with the given plane, which are on these elements. 5°. Points found by horizontal auxiliary planes, — Let P' R\ Fig. 69, be the vertical trace of such a plane. It cuts from the cone the circle of radius d'v\ whose horizontal projec- tion has Vd for its radius, and from the given plane PQP the horizontal line P'R'—fe, where /^ is parallel to the hori- zontal trace PQ. Then ee'y and another point, very near nn\ intersections oife—P'R! with the circle Vd—v'd\ are two points of the required intersection of PQP' with the given cone. Examples. — 1°. Complete the construction of the curve (Fig. 70) of which aa and bb' are two points. Ex. 2". Find the intersection of a right cone with a right plane (see Fig. 66). Ex. 3°. In Ex. 2 let the plane perpendicular io V be parallel to an element of the cone. Ex. 4°. Replace the cone by a pyramid. Ex. 5°. Replace the cone by any right or oblique prism. PROBLEM XXX VL — To find the intersection of a plane and a cone, both of which are oblique to both planes of projection. In Space. — As before, any point of the required intersec- tion will be where an element of the cone meets a line of the DESCRIPTIVE GEOMETRY. 77 given plane, both being in the same auxiliary plane. But as it is most convenient to take the auxiliary planes in some uniform manner, let them all contain the cone's vertex and be perpendicular either to H or V ; they will then all con- tain a line having a like position, hence their traces on the given plane will (27, 4°) all pass through the point where this line pierces that plane. In Projection,— "^X. VIIL, Fig. 71. Let VBD-VA'C be the given cone, and PQF the given plane. Let the system of auxiliary planes through VV be perpendicular to H, they will therefore intersect each other in the vertical line V-s'V, By (Prob. XL, b, 1°) this line pierces PQP' at the point F,/, found by taking any line Vx—x'y\ containing V, and in this plane, as shown by its traces, x andy, in those of the plane, either xx' or yy' being assumed. The point V^s' is then common to all the lines cut from PQP by the vertical auxiliary planes through the cone's vertex. ECV is the horizontal trace of one of these planes ; it cuts from the cone the elements EV—E'V and CV—CV, and from PQP the line ErV—s'r\ which meets these elements at // and c'c'f which are, therefore, two points of the required in- tersection. Others, as many as necessary, always taking those on the extreme elements of both projections, can be found and connected as shown. The visible points of the curve, as in Fig. 6^^ are on the visible elements of the cone. 57. Intersections of developable surfaces with each other. — These are all found on the general principle that each point of the intersection is where an element of one of the sur- faces meets one from the other surface, the two elements being in the same plane. They are sufficiently exemplified by the following problems. If the intersecting surfaces have a common axis, their intersection will be the circle perpendic- ular to that axis, generated by the point of intersection of the elements in the common meridian plane (33). yS DESCRIPTIVE GEOMETRY. PROBLEM XXXVII. — To find the intersection of two cones^ the axis of each being oblique both to H and V. « In Space. — Any plane AIV, PL IX., Fig. 72, containing elements of a cone contains its vertex. But any number of planes can be passed through two given points, hence planes can be passed through both the given vertices, Fand V, each of which will, therefore, contain elements of both cones. Such planes will all contain the line Vv joining the two vertices, and hence their horizontal traces, as lA and IB, will contain /, that of this line (27, 4°), which is, therefore, the first thing to be found. These traces will then intersect the like traces of the cones in the traces as A and 4 of the elements contained in those planes. Then, where the elements, as Av and Cv, of one cone, contained in any one of the planes, as A I, meet those I V and ^^V of the other cone, and contained in the sa7ne plane, will be points, as a, a^, c, etc., of the required intersec- tion. This solution is general for every case of pairs of bodies, each of which has a vertex. In Projection.— ?\. IX., Fig. 78. 1°. Selection of planes and construction of points. — Let VdLb- V'c'f and vADG-v'CH' be the given cones. Fol- lowing the solution in space, /is the horizontal trace of the line Vv — V'v' joining the given vertices. Hence, all lines through / and secants of the cones* bases, will be the hori- zontal traces of planes, each containing elements of both cones. Such planes will be limited, as at IbB and IfE, by those which, being tangent to one cone, cut the other, or may be tangent to both. Planes between these limits are chosen, not at random, but so as to contain the extreme elements of both cones relative to both projections, as IC contains an extreme ele- ment V'C of the vertical projection of cone vv' , and as IG contains one of its extreme elements in horizontal projection. More planes than are shown for illustration were therefore necessary in accurately finding the curve. Some of these DESCRIPTIVE GEOMETRY. 79 planes may often exactly or sensibly contain two or more extreme ijfements. Take now, for example, the plane IC. It c^its from the cone VV the elements hV—1t!V' and cV—c'V\ and from the cone vv' the elements Hv—H'v' and Cv—C'v'. The latter intersect the former in four points, as found by following them up, two by two, from the bases, at 2,2', 12,12', 6,6', and 8,8'. When, as at IE, a plane is tangent to one of the cones, it contains but one element of that cone, and therefore gives but two points, as 4,4', and 10,10', of the required intersec- tion. Any desired number of points may be similarly found. 2°. Connection of Points. — Rule. Begin at any point, and proceed from plane to plane, joining the points, till a tan- gent plane is reached. There the curve will be tangent to the elements cut from one cone by the plane which is tangent to the other. Thence the curve recrosses the elements already traversed, on the body cut by this tangent plane, through points on new elements of the other body. Thus proceed till another tangent plane is reached, and thence in like manner to the point of beginning. This is evident from Fig. 72, for as there can be no auxiliary plane exterior to the tangent one IB, that limits the curve, so that in passing from ^„ in plane lA to b, its contact with V2 in plane IB, it must thence return to plane I A, but at a point r, where a new element, Cv, of cone V, crosses the^ same element Vi of cone V, that con- tains a. Illustration. — Thus, in Fig. 78, beginning anywhere, as at 1,1', where the curve is tangent to the element Bv—B'v' of cone vv\ sketch the curve through 2,2', 3,3', and other in- termediate points not shown, to 4,4', where it is tangent to the element F^— F'/ cut from cone VV\ by the plane IE, tangent to cone vv'. Thence proceed through 5,5', 6,6', to y,f, where it is again in the tangent plane /5, but tangent to the other element, Kv—K'v', cut by that plane from the cone vv'. Thence through 8,8', 9,9', the curve, once more 80 DESCRIPTIVE GEOMETRY. in the tangent plane Zfi", and tangent at io,io' to the element Vf—V'fy returns to i,i', the starting-point chosen. 3°. Number and disposition of curves. — If both of the tan- gent planes to one body should cut the other, there would be two curves, as in Fig. 75. If one of the auxiHary planes should be tangent to both bodies, it would contain but one element of each, and would therefore give but one point, at which the curve would cross itself. Should two of the auxiliary planes each be tangent to both cones, each would give one point, making two points at which the curve would cross itself, and the curve would be found reduced to two plane curves which would there- fore generally be two ellipses. (Theor. IV.) 4°. Visibility of the curve. — In order that a point of the curve should be visible, in either projection, it must be at the intersection of elements both of which are visible in that projection. Thus the point 11, 11' is visible in hori- zontal projection, and so is the point 4,4' in vertical pro- jection. Having determined such a point, the curve is visible in both directions from it, until it reaches the near- est extreme elements found in following the curve. Thus from II, the curve is visible to 9 on vG^ and to the- point between 12 and i on Vm, 5°. Visibility of extreme elements. — This is determined by that of the elements which intersect the extreme elements. Thus Gv intersects the just invisible element dV at 5, where it enters cone VV, and emerges at 9 on the visible element Vgy and hence is visible from 9 to v. Likewise, the extreme element CV intersects the two invisible ones c'V and h'Vy and hence is invisible between the extreme elements V'n' and Vf of cone W, Examples. — 1°. Construct the problem so as to give two curves. [To cer- tainly do this, assume one cone as vv\ and its tangent planes from /, also as- sumed. Then place the vertex VV on Iv — Tv and the base of cone VV so that both of these tangent planes shall cut it.] Ex. 2°. Take the given cones so as to make a more visible curve. [For example, so that the chord ef shall be much nearer Z.] DESCRIPTIVE GEOMETRY. 8 1 Ex. 3°. Placing the ground line at right angles to its present position, make the total figure higher than it is wide, and so as to fill the plate. Ex. 4°. Let one base be within the other. Ex. 5°: Let IT be to the left of the cones and further forward from the ground line than the bases. 58. By removing the vertex of one of the cones to infinity on its axis, it becomes a cylinder as sketched in PL IX., Fig. 74. The line Vv then becomes a parallel to the axis of the cylinder, through the vertex of the cone, as shown at VI, Fig. 74. With this modification of the initial step, only, by making all the auxiliary planes contain IVv, the solution remains the same in all particulars as that of Prob. XXXVII. 59. By similarly removing the vertices of both cones, they become two cylinders as shown in PI. IX., Fig. 73. In this case, by (Prob. IX., 4°) we pass an initial auxili- ary plane, as Op, through the axis, as Om — O'm', of one cylinder, and parallel to that of the other one. The other auxiliary planes are then parallel to Op, as indicated by their parallel traces, and the solution thenceforward is in all respects the same as that of Prob. XXXVII. PROBLEM XXXVIIL— r^ find the intersection of two cylinders whose axes intersect ■ at right angles, one of them being vertical, and the other parallel to the ground liiie. In Space. — All that is peculiar to this problem, as com- pared with (59), arises from the position of the horizontal cylinder. The auxiliary planes being placed as there de- scribed, the projections of elements of that cylinder may be found as in Prob. XXVII. In Projection, — PL IX., Fig. 75. Let hr be assumed as the horizontal trace of an auxiliary plane. It cuts from the vertical cylinder two elements, horizontally projected at g and b, and from the horizontal cylinder two elements whose horizontal projection is hr. The vertical projections of 82 DESCRIPTIVE GEOMETRY. these are here shown by revolving the plane of the base po about its horizontal diameter, po—p\ till it becomes hori- zontal, when the circle pqo will be its horizontal projec- tion, and the two points r will appear on it as at r" — here supposed to be the upper one — and a point, not shown, equally to the left of r. The vertical projection of r" is r"\ and serves alike for finding by counter-revolution, as shown, r', the vertical projection of r, and k\ the vertical projec- tion of the lower point projected in r. Thence, as is evident on inspection, the auxiliary plane hr determines the four points, bit! y bb\ gt\ gg'. The construction of the highest points, as nd' ; foremost, 2iSaa' ; lowest, as nf; and hindmost, as la'y is obvious ; as is the symmetry of the curves relative to the vertical plane DOy and horizontal plane A'0\ Examples. — 1°. Construct the intersection of a cylinder and a cone for the case where there are two curves. Ex. 2°. Construct the intersection of a cylinder and a cone for the case where there is but one curve. Ex. 3". Let the axis of the cylinder be vertical, and that of the cone parallel only to H. Ex. 4°. In (3) let the axis of the cone be parallel to the ground line. Ex. 5°. Let the axis of the cone be vertical, and vary the position of the cylinder as that of the cone is in (3) and (4). Ex. 6°. Further vary (3), (4), and (5) by making either body penetrate the other, giving two curves. Ex. 7°. Construct the intersection of two cylinders, placed as in Fig. 73. Ex. 8°. Place the two cylinders, so that there shall be two curves. Ex. 9°. Vary Fig. 75 by making the axis of the horizontal cylinder there shown, oblique to H, or to V, or to both planes. Ex. 10°. Vary Fig. 75 by revolving the plane of the base po—p'c', either about its horizontal trace, or its vertical trace, or about the diameter c — v'c . Ex. 11°. In Fig. 75, let the axes, DO — A'O' and — m'Y, not intersect. PROBLEM XXXIX. — To construct the tangent line at a given point of any previous plane ^ or double-curved intersection. In Space. — 1°. The tangent line to any plane curve being in the plane of the curve (52), and the tangent line to a surface at any point being in the tangent plane at that DESCRIPTIVE GEOMETRY. 83 point (43), the required tangent line to any plane intersec- tion will be the intersection of these two planes. 2°. By the second principle just given, the tangent line at a given point of the curve of intersection of two developable surfaces will be the intersection of two planes, each of which is tangent to one of the surfaces along the element containing the given point ; as illustrated in PL IX., Fig. 74, by Qtj the tangent at /, and the intersection of the plane EQt, tangent along Et to the cylinder, with the plane FQt, tangent along Ft to the cone. In Projection. — PL VIII. By the above principles (1°), in Fig. 67 the tangent at ff to the curve bcd—c'e'y is Pf^Qf, which is the intersection of the plane PQF of the curve, and the tangent plane /!F along the element Ff—F'f containing the given point of contact ff. The point of contact being known, one trace of the tangent plane is suffi- cient. Likewise in Fig. 68, the tangent Ke — Ke', at the given point ee' of the curve fhe—f'h'e', is found as the intersection of the plane PQP' of the curve, with the tangent plane KE along the element of the given cylinder containing ee' , K being the intersection of the horizontal traces, KE and PQy its vertical projection K' is on the ground line. Also in Fig. 69, the tangent Tn — B'n' at nn' to the curve hkn — h'k'n', is the intersection of the plane PQP' of the curve, with the tangent plane BB' V along the element ^F—^'F'' containing nn' . Tangents at aa' and bb' in Fig. 70, and the tangent Kf—K'f in Fig. 71, are similarly found. 2°. PL IX., Fig. 78. The tangent MZ—M^', to the intersection of the two cones, at the point 8,8', is the inter- section of the tangent planes, HM^ to the cone vv', along the element Hv — H'v', and JiM to the cone VV , along the element /^F—//F. Examples. — 1°. Construct the tangent at any other given point in each of the foregoing figures. Ex. 2°. Construct the tangent line at the point hb\ Fig. 75. 84 DESCRIPTIVE GEOMETRY. D— Development. PROBLEM XL. — To find the true form and size of the inter- section of any cylittder or cone by a plane^ with the tange7it to the revolved curve, hi Space. — The given plane is revolved so that it shall either coincide with or be parallel to a plane of projec- tion, when the given curve will be seen in its true form and size on that plane. (See Probs. XIL and XV.) In Projection. — PI. VIIL i°. In Fig. 6j, the given plane PQF is perpendicular to V, hence if revolved into V about P'Q as an axis, any point, 2i^ff\ will be found as 2Xf"' (not shown) on a perpendicular to PQ at /', and at a distance from P'Q equal to fg. Similarly finding F'\ revolved posi- tion of P, the revolved position of the tangent Pf—Qf will be P"f"\ Or, revolving PQF about PQ into W, ff will appear at f", by making f'h—f'Q, and, P remaining fixed in the axis, Pf" is the revolved tangent. 2°. In Fig. 68, ee'y for example, revolved into V, about the vertical trace P'Q of the plane PQP of the curve ehf^e'h'f, will fall at e"^ at a distance from PQ equal to the hypothe- nuse constructed on // and cq as sides. (Prob. XII.) The tangent at ce' pierces Y in P'Q at g\ hence ^e" is its revolved position. 3°. In Fig. 71, we may proceed just as in the preceding cases, or make use of the system of lines meeting at the point VyS'. Revolving the curve ebf—e'b'f about PQ into H, this point F,/ in the plane PQP' of the curve appears at s", where s"m is the hypothenuse found from Vm and s'o as sides. Then s" is common to the revolved positions of the system of lines Vn — s'n' ^ etc., in PQP', which will therefore appear at s"n, etc. The curve and the point V,s' being on opposite sides of the axis PQ^ will so appear after revolu- tion. Thus ff appears at f", the intersection of /'?/, pro- duced, with //, perpendicular to PQ and the horizontal DESCRIPTIVE GEOMETRY. 85 projection of the arc in which //^' revolves about PQ, The horizontal trace, K, of the tangent at ff\ being in the axis, remains fixed, hence Kf" (not shown) is its revolved position. Examples.— 1°. In Fig. 67, let the axis be in PQP' 2,^^ parallel to FQ or PQ. Ex. 2°. In Fig. 68, revolve the curve about PQ, or a parallel to it, in PQP\ till in or parallel to H. Ex. 3°. Show the true size of the curve in Fig. 69. Ex. 4°. In Fig. 71, find the true size of the curve as in. preceding ex- amples ; also by using the point V,s\ but by revolution about P' Q as an axis. 60. Any two elements, either of a cylinder or cone, lie in the same plane ; then if a, b, c, d. be consecutive elements, any one, as a, may be revolved about the con- secutive one, b, till it falls into the plane of the consecutive elements b and c; next the plane of a, b, and c may be re- volved about ^into the plane of ^ and d; and so on till all the elements are brought into one plane. The surface is then said to be developed upon a plane, and as the relative position of consecutive elements is unchanged, the curved surface is merely transformed into an equivalent plane area, but not rent or destroyed. Hence, also, the relative posi- tions of lines having a fixed relation to these elements, as any curve on the surface and its tangent, will be unchanged by development. 61. Illustration, — A material cone being placed with one of its elements in a flat surface of paper, let it be rolled without slipping until the same element again coincides with the paper. The portion of paper thus traversed, if cut out, will be found to be of the form of a sector of a circle, and, if wrapped upon the cone, will exactly cover its convex surface. It is therefore the development or pattern of that surface. PROBLEM XLI . — To develop the convex surface of a cylinder, together with its intersection with a plane, or with another cylinder or a cone, and any tangent line to the curve. In Space. — The development of the convex surface of a cylinder limited by circular bases will by (60) be a rect- 86 DESCRIPTIVE GEOMETRY. angle, whose length is the circumference of the base of the cylinder, and whose altitude is the same as that of the cylinder. Whatever the position of the cylinder, the length of any portion of any element given by its projections can be found by Prob. XII. The developed tangent is readily found by means of its true length and direction. Thus, it is a hypothenuse, of which the adjacent sides are a perpendicular from any point of it to the element contaiiting the point of contact, and the portion of this element included between the tangent and this perpendicular. In Projection. — PI. IX., Figs. 75-77. In Fig. "j^, which is the development of the vertical cylinder in Fig. 75, mm^ equals the circumference of the circle whose centre is O, and inY = m'Y{¥\g. 75). Then in the respective figures, y6 and 75, m^x = may and a^x = xa' ; xy =^aby and yb = yb\ etc., giving the development of the curve of intersection a7tl—fa'd\ In Fig. 'j'jy which is the development of the horizontal cylinder in Fig. T^yDc = Dc in Fig. 75, and DD^ = the cir- cumference of the base, po — t/c\ of the horizontal cylin- der. Then, in the same figures respectively, cr = qr" y and rb = r'b'; rp = r"p and pa —p'a'; pk =pr" and ku = k'u'; kv = r"q and vf = vf\ etc., and dbauf is the development of the front half, nba — d'b'a'u'f'y of the intersection, con-, sidered as on the horizontal cylinder. The tangents at dy tty fy ^, and d^ are evidently parallel to cc^. Examples. — 1°. Construct the tangent at b (Figs. 76 and 77), development of that at bb' (Fig. 75). Ex. 2°. Develop the cylinders and intersections shown in PI. VIII., Figs. 67, 68 ; supposing them true cylinders of revolution (42). PROBLEM XLII. — To develop the convex surface of a cone of revolution with any curve on its surface y and a tangent to the curve. In Space. — The development of a cone of revolution will by (61) be the sector of a circle, whose radius is the slant DESCRIPTIVE GEOMETRY. '87 height of the cone, and whose arc equals the circumference of the cone's base. The developed tangent is found as in the last problem. In Projection. — PI. IX., Figs. 79-81. 1°. Fig. 79 is the development of Fig, 70 on PL VIII. Hence VE (Fig. 79) = V'D' (Fig. 70), and the arc EE^ equals the circumference of the base ABC of Fig. 70. The cone is here supposed to be placed with the element VE — V'E' (Fig. 70), tangent to the plane of the paper at VE in Fig. 79, and then rolled until the same element again reaches the paper at VE^^ giving on the respective figures, 79 and 70, EB = EB, BD = BB, Ad=Ad, etc. The developments, as a and d, of points, as aa' and dd\ of the intersection of the cone with the plane PQP'j are found by making Va, etc., equal to the true length of Va — Va\ This is best found by revolving Va — Va' about the cone's axis till it is parallel to V, when, as the cone is one of revolution, with its axis vertical, Va — V^a' will fall upon V'D' as at FV, by drawing a'a" parallel to GL (7), to represent the arc in which aa' revolves to a'\ The tangent de is then found by making Be (Fig. 79) =Be (Fig. 70), and drawing de. The tangent aM is similarly found. 2°. Fig. 81 sufficiently represents the method of proceed- ing to develop a cone of revolution, placed as in PI. VI., Fig. 52, and represented on PL IX. by Fig. 80, indicating the projection V'MR (Fig. 52) on a vertical plane through the cone's axis. Here VCy which bisects the angle MV'Ry is the axis of the cone, and NCr perpendicular to it is therefore a cir- cular section of the cone, half shown as such at Ncr^ by revolution about Nr into the plane of the paper. Then in the development (Fig. 81), V"N — V"N in Fig. 80; Na, ac, etc., are equal to the like arcs on Fig. 80 ; and if / (Fig. 80) is any point upon the cone's surface, its true distance from V" is Vp^y found by drawing //, parallel to Nr, as in the previous case ; hence Vp^ is laid off on the element Va, development of VA, at Vp. The elliptical base MR is similarly developed. Thus 88 DESCRIPTIVE GEOMETRY. r^R (Fig. 8 1) equals V'R on Fig. 80, r'K, = V'K, (Fig. 80), y'M (Fig. 80) would be. laid off on both VN and VN„ pro- duced. Examples. — 1°. Make the complete development, just indicated, of PI. VI., Fig. 52. Ex. 2°. Develop each of the cones in PI. IX., Fig. 78, supposing them first to have been accurately Constructed as cones of revolution. Ex. 3°. Develop the figure of Ex. 11°, Prob. XXXVIII. DESCRIPTIVE GEOMETRY, 89 J.I B K A 1; V UNlVKKsn V (>F (JALlKUliNlA CHAPTER III. WARPED SURFACES OF REVOLUTION. 62. A WARPED surface is generated by a straight line which moves so that its consecutive positions are not in the same plane. If there be a warped surface of revolution, it must be generated by the revolution of a straight line around an- other straight line not in the same plane with it ; for the only other positions which the revolving line can have with respect to the axis are, parallel to it, and intersecting it. In the former case a cylinder will be generated, and in the latter a cone, but both of these surfaces are developable (60). Theorem V. — The surface generated by the revolution of a straight line around another straight line not in the same plane with it, is warped. This theorem is proved by showing that the consecutive positions of ^the revolving line, that is, the consecutive elements of the surface, are not in the same plane. See then PI. X., Fig. 82, in which the vertical line O — O'O" is the axis, ab — a'b' the generatrix, and the shortest or perpendicular distance, Oc — c' , between the two lines is horizontal, and sweeps over the circular area of radius Oc, called the circle of the gorge. Then, first, since no point of ab — a'b' is on the axis, every point of it moves, hence consecutive positions have no com- mon point, or do not intersect. 90 DESCRIPTIVE GEOMETRY. But, second, as ab — a'b' remains perpendicular to Oc — c' while both revolve, the horizontal projections of the suc- cessive positions of ab — a'b' are tangent to the circle of radius Oc^ and hence are not parallel, and therefore are also not parallel in space. Thus the consecutive elements of the surface are not in the same plane, and therefore the surface composed of them is warped. Theorem VI. — The warped surface of revolution has two sets of element Sy such that at every point of the surface two ele- ments can be drawn, one of each set. This theorem is evident (see PL X., Fig. 85) from the symmetry of the two lines ab — a'b' and ab — a"b" , relative to the plane containing the axis O — 1i!'h' and the common per- pendicular, Oq — q', to it and to both of these lines. For by reason of this symmetry, any two points, as aa' and bb" , one on each element, and in the same plane perpendicular to the axis, are equidistant from the axis, and will therefore describe the same horizontal circle ; thus the series of circles described by all the points of ab — a'b' is identical with that described by all the points of ab — a"b". Hence the surfaces generated by these lines are identical. The two sets of elements are distinguished as those of the first and second generations. Again : from any point, as n on ab, draw a tangent nr. Then nr = nq. But each of these represents two lines, one above and one below the plane Oqr, and all equally inclined to it. Hence, taking, for example, the pair that ascends from q and r to ;/, they, being equal, because equal in horizontal projection and in inclination to H, really intersect at a point whose projection is 71. But they are of different gene- rations, for my ascending from r, is an element of the gene- ration to which ab — a'b' belongs, while qn, ascending from q, is an element of the generation to which ab — a"b" belongs. Similar reasoning applies to n when taken anywhere on any position, either of ab — a'b' or of ab — a"b". Hence, at DESCRIPTIVE GEOMETRY. QI every point of the surface two elements may be drawn^ one of each generation. 63. It follows, from the last theorem, and considering all the consecutive positions of n on any one element, that, in the warped surface of revolution, any element of 07ie genera- tion intersects all those of the other generation. 64. A line is fixed in position when it intersects three given straight lines not in the same plane, and contains a given point on one of them. For let Ay By and C be the three given lines, and a the given point on A. The plane containing the point a and line B is then fixed, and not con- taining the line C will intersect it at some point c. The line ac will then be determined by the two points a and c and will intersect the three given lines. A — Projections. PROBLEM XLIII. — To construct the projections of the warped surface of revolution. In Space — I. General Construction. Since each element of one generation intersects all those of the other (63) it will intersect any three of them. If then we have any three lines whose perpendicular distances from a fourth, taken as an axis, are equal and in the same plane ; and which are equally inclined in the same sense to that plane, they will be elements of one generation of the required surface, and will therefore be directrices from which, by (Theor. VI.), any number of elements of the other generation can be found. II. Special Constructions. — In these, having given the axis as vertical, and an element parallel to V, we know at once the radius of the gorge, and the inclination of all the ele- ments to H, and thence can find the horizontal trace of the surface, and any number of elements, the horizontal pro- 92 DESCRIPTIVE GEOMETRY. jections of the latter being (Theor. V.) tangent to the like projection of the gorge. In Projection. — PI. X., Fig. 85. Let the surface be given by its axis O — h"h', and the element ab — a'b'^ parallel to V. Then at once ab — a"b" is the element of the other genera- tion at qq'. Every tangent to the circle Oq is likewise the horizontal projection of two elements, one of each genera- tion. Thus, hk is the horizontal projection of the pair of elements whose vertical projections are h'k' and h!'k". See also ef—e'f and ef—e"f", which intersect at 00' ; ef—e"f" and eg—e"f", which intersect at ee" , etc. Visibility. — In horizontal projection, the inner side of that part of the surface which is above the gorge is visible, and in vertical projection, the part which is in front of the meridian plane ey is visible. But the symmetry of the ele- ments, relative to both of these limits, makes every line, which represents elements, visible throughout. Thus hp is visible as the horizontal projection of h"p' and kp as that of k'p\ Likewise h"s" is visible, considered as the vertical projection of //j, and s"k" is visible as that of si, whose pro- jecting plane upon V also contains hk — h'^k". Each line of the vertical projection is thus also the ver- tical projection of two elements. Example. — Assume n at any point of any other element as efy and e at any point not on ey and on any other horizontal circle except the gorge, and draw the elements containing them. PROBLEM XLIV.— 71? represent the vertical projection of the warped surf ace of revolution by its meridian parallel toM , In Space. — 1°. Pairs of elements, symmetrical with re- spect to any meridian plane, intersect in that meridian plane, and hence at points of the meridian in that plane (33). 2°. Points assumed on any element, and revolved around the axis of the surface, until they fall upon the same meri- dian plane, wilLalso give the meridian curve in that plane. Hence — DESCRIPTIVE GEOMETRY. 93 In Projection. — 1°. By symmetrical elements. PI. X., Fig. 85. eOy is the horizontal trace of the meridian plane parallel to V and containing the required meridian curve. Then pairs of elements, symmetrical relative to this plane, meet as at Ay s and e, horizontal projections of points of the curve, whose vertical projections, A' , s' and s" ^ e' and e" , etc., are on the vertical projection, e"s"A's'e'y of the curve. As this curve is evidently convex towards the axis h"h' of the surface, while no points of the surface in the meridian plane eOy can be exterior to the curve, the points e" , s" ^ etc., are points of contact of the vertical projections of ele- ments which are thus tangents to that of the meridian. 2°. By horizontal circles. — PL X., Fig. 83. The surface being given by its axis O — O'O" and the elements ab — a'b' and ab — a"b'\ assume points at pleasure as r/, ee\ ee" ^ bb\ bb'\ etc., and revolve them about O — O'O", in the arcs rn — r'n'j ef—e'f, etc., till they reach the meridian plane mOn^ when they will be points nn' , ff\ gg\ etc., of the meridian curve gfng—g"n'g'. Points of any other element, as p on cd, will give other points as ^/ of the meridian curve. Theorem VII. — The meridian curve of the warped stir face of revolution is a hyperbola. First. From PI. X., Fig. 85, the symmetry of the ele- ments, in vertical projection, relative both to A'B' and h'h" y and the construction of Fig. 83, alike show that the meridian curve consists of tzvo equal and opposite branches^ having tivo rectangular axes of symmetry, and convex towards one of them. Second. In Fig. 8$, ab and mu, being parallel to the meridian plane, meet in it, at points of the meridian, only at an infinite distance from the gorge, but as the vertical pro- jections of the elements are tangent to that of the meridian curve, a'b' and a"b" are tangent to the latter only at an infinite distance from q' . Likewise in Fig.- 83, the further from rr' the point bb" is taken, the more nearly the arc bg 94 DESCRIPTIVE GEOMETRY. becomes a perpendicular to V. This occurs only when rb — r'b" (or r'b') is infinite, when b"g' vanishes, and thus r'b' and r'b" become tangents at mfinity to the meridian curve, that is asymptotes to it (51). All these properties show the meridian curve to be a hyperbola ; but — TJiird, See Fig. 84. Let AB—A'B' be the gorge of a warped surface of revolution. Then, first, the sphere generated by the circle of the gorge is unique, being the only one having a great circle of contact Avith the surface ; second, the asymptotes, AB — a'b' and AB — e'h^ projections of ab on the meridian plane AB, are a unique pair of tan- gents, being the only ones which (Fig. 85) represent ele- ments parallel to V ; third, hence the tangents, as p' F' and ^fy at the intersections of the asymptotes with this sphere, are unique, and hence their intersections F' and f with the axis A'B\ which is a unique line relative to the meridian curve, are a pair of unique points on that axis, equidistant from the centre O' . The hyperbola has one and only one such pair of unique points on its transverse axis, viz., its foci ; moreover, drawing the semicircle on a'F\ we find a second right angle Fq'a' having its vertex on A'p'B', This agrees first with the property of the hyperboloid (Theor. VI.) that, at any point a' of an element a'O' , one and only one other element a'q' can be drawn, and, second, with the property of the hyperbola that the locus of the intersection of perpendiculars from the foci upon the tan- gents are on the circle described on the transverse axis. Hence we conclude that the unique points F' and f are the foci of the meridian curve, which is thus further shown to be a hyperbola. 65. The warped surface of revolution having thus a hyperbola for its meridian curve, is called a hyperboloid of revolution of two united nappes, to distinguish it from that of two separate nappes, which is generated by the revolution of a hyperbola about the axis as m'n' (Fig. 83) which inter- sects it. The nappes are the equal and opposite halves of the DESCRIPTIVE GEOMETRY. 95 surface which meet at the gorge as the two nappes of a cone do at its vertex. 66, The hyperboloid of revolution of two united nappes, also distinguished as the warped hyperboloid of revolu- tion, since that of two separate nappes is obviously a dou- ble-curved surface, may, from all that has now been shown, be generated in three ways : jfirstj by the revolution of a straight line about an axis not in the same plane with it ; second^ by the motion of a straight line on three fixed straight lines at equal perpendicular distances in the same plane, from a fourth line ; third, by the revolution of a hyperbola around its conjugate axis. 6^, By similar reasoning to that in Theor. VII. {Second, Fig. 83), it may be shown that any plane parallel to the vertical planes at dg and ab and between them cuts the hyperboloid in a hyperbola whose transverse axis is horu zontal, while if such a plane is in front of ab, that is, exterior to the gorge, it will cut the hyperboloid in a hyperbola each of whose branches is wholly on one nappe of the sur- face, and whose transverse axis is therefore vertical. PROBLEM XLV. — Having given either projection of a point on a warped hyperboloid of revolution, to find its other pro- jection. In Space. — The two projections of any point of the hyper- boloid are on those of any line of the surface containing the point, the element and the parallel being the most con- venient ones. In Projection. — PI. X., Fig. 83, / being the given hori- zontal projection of a point, draw pd tangent to the gorge at s for the horizontal projection of the element, and the cir- cle of radius Op for that of the /^r^//^/ containing it. Then p' , now supposed to be on the upper nappe, will be found either on d's', the vertical projection of the element ds, or 96 DESCRIPTIVE GEOMETRY. on q'p\ that of the parallel pp"q. The latter is found by projecting p" upon a'b\ or q upon the meridian d'ni'q' and drawing q'p"'p'' If /' were given,/'/ would be the vertical projection of the parallel, and a tangent p's'd! from / to the meridian, and found by any known method, would be that of the ele- ment containing p'p\ whence p would be on the horizontal projection qp"p of the parallel, or on dsp that of the element. B — Tangencies. Theorem VIII. — Every tangent plane to a warped hyperboloid of revolution is also a secant plane, containing two elements, and is tangent at their point of intersection. As in (45), the tangent plane to any surface at any point contains all the tangent lines to the surface at that point ; any two of which will determine the plane. Then (see PL X., Fig. 83) the tangent plane at ff, for example, may be conveniently determined by the tangent to the meridian g'n'g" 2itff, and the tangent to the circular parallel ef—e'f at ff. But the former curve is convex towards the axis O — O'O" , and the latter is concave towards it, hence the tangent to the first is on the left, or inner side of the sur- face 2itff, and the tangent to the second is on the right, or outer side of the surface at ff. These tangents being thus on opposite sides of the surface, the tangent plane deter- mined by them is necessarily also a secant plane, and evi- dently has only a point of contact with the surface. But there must somewhere be limits between those tangents at ff which, like ft—f, are exterior to the sur- face, and those which, like fk', are tangent to its interior side. These limits must necessarily lie in the surface, and being straight, must be elements of the surface. The two ele- DESCRIPTIVE GEOMETRY. 97 ments at any point (Theor. VI.) are thus merely that pair of tangents which coincide' with the surface, which agrees with (44). Hence the two elements, one of each generation, at any point, considered as a pair of tangents at that point, determine most simply the tangent plane at that point. PROBLEM XL VI. — To construct the tangent plane at a given point of the warped hyperboloid of revolution. In Space, — By Theor. VI 1 1., the required tangent plane will be determined by the two elements through the given point, and hence (27, 4°) its traces will be determined by their traces. In Projection, — PL X., Fig. 86. Let tt' be the given point. Then tangents to the gorge, // and tq^ will be the horizontal projections of the determining elements, from which their vertical projections, t'p' and t'q\ are found, and thence their traces, as r and u on W, and P" on V. Then PruQ is the horizontal, and QP" the vertical trace of the required plane. The tangent plane at tt^ is seen to be also a secant plane, since its horizontal trace cuts that of the hyperboloid at T and u. It is also perpendicular to the meridian plane Ot, whose horizontal trace Ot is perpendicular to PQ, This is evi- dent from the symmetry of the elements tp and tq relative to the vertical meridian plane Ot, Since the tangent plane to the warped hyperboloid has only a point of contact (Theor. VIII.) while three points determine a plane, a tangent plane to it may contain a given line, or be parallel to a given plane. Examples. — 1°. Take //' anywhere on the upper nappe of the surface. Ex. 2°. Take //' on the meridian plane O—O'O", Ex. 3°. Construct the tangent plane at a given point, by means of the tangent to the parallel at that point, and the meridian plane through the same point. 98 DESCRIPTIVE GEOMETRY. PROBLEM XLVII. — To construct a plane, tangent to a warped hyperboloid of revolution, and containing a given line. In Space. — Since every tangent plane to the hyperboloid contains two elements of the surface, every other line in this plane must intersect those elements, and hence the surface, unless the line happens to contain the point of con- tact, when it would be a tangent. Now, let a given line L intersect the surface at two points / and p^, and let G and g be the two elements (Theor. VI.) at/, and G^ and^i those at /„ 6^ and G^ being of the same generation. Now, as G and^j intersect both each other (63) and Z, and as g and G^ do the same, two and only two tangent planes are determined by the given line L and the elements at/ and/,. In Projection. — PL X., Fig. 86. Let mh — m'h' be the given line, the hyperboloid being shown by its axis O — O'O" , and an element ab — a'b\ and hence also by its gorge dpq, and horizontal trace abr. The vertical plane through mh — m'h' cuts the hyper- boloid in the hyperbola kcd — k'c'd!, found by projecting up k, c and d upon the vertical projections a'b", d'c' and o'p' of the parallels containing these points. The given line pierces the hyperboloid at its intersection ee' with this curve. Then tangents to the gorge from e will be the horizontal projections of the two elements, each one of which, with mh — m'h' y will determine one of the tangent planes. Thus the element ^/ gives the horizontal trace/, and that of mh — m'h' is /, hence fi is the horizontal trace of one of the tangent planes. The vertical trace of this plane passes through n that of mh — m'h' and jf, that of a parallel to fQi at any point as hh' of mh — m'h'. A tangent to the gorge from g would be the horizontal projection of the other element of the hyperboloid con- tained in this plane, and would intersect ef in the horizon- tal projection of the point of contact. It would also intersect mh in the horizontal projection of the second point in which DESCRIPTIVE GEOMETRY. 99 that line cuts the hyperboloid, and thus would give no additional plane. Examples. i°. Find the other tangent plane containing mh—m'h, and both projections of the point of contact of each of them. Ex. 2°. Construct a tangent plane to a warped hyperboloid, and parallel to a given plane. C — Intersections. PROBLEM XLVIII. — To find the intersection of a plane ayid zvarped hyperboloid of revolution^ and a tangent to the curve at a given point. In Space. — 1°. By auxiliary planes. Intersect the given plane by any plane which will intersect the hyperboloid either in an element or a parallel ; then where either of the latter intersects the line cut from the given plane by the auxiliary one, will be points of the required curve. 2°. By auxiliary cones. — Any line of the given plane, drawn from the intersection of the plane with the axis of the hyperboloid, will, by revolution around that axis, gene- rate a concentric cone, cutting the hyperboloid in two parallels, and the given plane in two straight lines, ele- ments of the cone, one of which will be the assumed one. These lines and parallels all being on the same cone must intersect. But the former being in the given plane and the latter in the hyperboloid, these intersections must be points of the required intersection of the plane and hyper- boloid. In Projection. — PL X., Fig. 87. Let the hyperboloid be given by its axis O—O'O", and the element ab^-o'b', and let PQP' be the given plane. Then— 1°. By auxiliary planes. Assume an)^ horizontal plane as R'S'. It cuts from the plane PQP a line qh—RS\ paral- lel to QP, and whose vertical trace is R' . It also cuts from the hyperboloid the parallel KS', one point of which is s' , lOO DESCRIPTIVE GEOMETRY. on o'b'y which, projected upon ab, gives j, and thence the horizontal projection, with radius Os^ of the parallel. Then the line hq — RS' cuts the parallel shv — RS' in the two points hh' and vv' of the required curve. Any desired number of other points may be found in the same manner. 2°. By auxiliary cones. We shall mention here only the one which contains the highest and lowest points. The generating line of this cone {hi Space 2°) is that line of the given plane which is cut from it by the meridian plane perpendicular to it, since this line evidently contains those points. The cone, being tangent to the given plane, will contain no other line of it, and hence will give only the highest and lowest points, if any. The given plane PQP' cuts the axis 0—0' O" at Oy (XI. by 1°), hence Oc — r'c', where Oc is perpendicular to PQ^ is the generatrix of the cone, and the circle of radius Oc is the cone's base. One point of each of the parallels in which this cone cuts the hyperboloid, if at all, is where it cuts the given element ab — o'b' , To find the latter points, pass a plane through ab — o'b' and the vertex Oy of the cone, and it will cut from the cone those elements which will meet ab — &b' in a point of each parallel. This plane is determined (Prob. IX., 2°) by ab — o'b' and by a parallel to it, Od — r'd' from the cone's vertex. Hence db is its horizontal trace, but it does not cut circle Oc, the base of the cone, and hence contains no elements of the cone ; which shows that in the figure this cone does not intersect the hyperboloid. Had it done so, the construction would have been completed as follows. 1st. Note the intersection of the trace db with the cone- base of radius Oc. 2d. Join these points with O, and note where they (pro- duced if necessary) meet ab, 3d. Describe arcs, with (9 as a centre, through the latter points, and their intersections with Oc will be the highest and lowest points (attending carefully to the positions in space so as to note the proper intersections, which will be DESCRIPTIVE GEOMETRY. lOI indicated by their vertical projections, or by other points of the curve). In the figure, a hyperbolic intersection is shown for variety, an elliptical one being usually represented. The transverse axis of the horizontal projection will bisect perpendicularly any chord of the curve drawn paral- lel to the infinite or conjugate axis Oc. By an analogy which holds good, between the hyperbolas in Figs, 83 and 87, as the asymptotes in Fig. 83 are parallel to the elements con- tained in a tangent plane parallel to the plane of the hyper- bola, the like is true in Fig. 87. 3°. The tangent line, — Let vv' be the point of contact. Then, by (Prob. XLVL), draw the tangent plane at vv' , and its intersection with PQP will, by the principle of (Prob. XXXIX.), which is general, be the tangent. Examples. — 1°. Representing the hyperboloid as in PI. X., Fig. 85, find its intersection with any plane, by constructing by (Prob. XI.) the intersections of its elements with that plane. Ex. 2°. In Fig. 87, assume PQ.P' in such a position as to give an elliptical intersection, and make the construction indicated above for finding its highest and lowest points. Ex. 3°. Make the intersection, as Oz — rV, of any other meridian plane, Ozy with PQP\ the generatrix of a concentric cone, cutting the hyperboloid in two parallels which will meet the two elements (of which Oz — r'z' is one) cut by PQP' from this cone in four points of the curve. 6Z. If the hyperbola g"n'g\ PI. X., Fig. 83, and its asymp- totes, a'b' and a"b" , be simultaneously revolved about the axis O — O'O" ^ the former will generate a hyperboloid, and the latter a cone whose vertex is Oy, and which will be tangent to the hyperboloid on the circles at an infinite dis- tance from the gorge m'n' , and generated by the points of contact of the asymptotes with the hyperbola. This cone is therefore called the conic asymptote, or asymptote cone of the hyperboloid. 69. The radius of the gorge of a warped hyperboloid and the inclination of its elements to the plane of the gorge are both arbitrary. If the former be reduced to zero, the I02 DESCRIPTIVE GEOMETRY. hyperboloid becomes a cone ; if the latter be made 90°, the hyperboloid becomes a cylinder. The cone and cylinder are thus only particular cases of the hyperboloid. Again, the common meridian plane of the hyperboloid and its asymptote cone {6y) intersects the former in a hy- perbola and the latter in a pair of intersecting straight lines, which is a particular case of the hyperbola. Any plane parallel to this meridian plane intersects both bodies in hy- perbolas. The tangent plane to the asymptote cone (Fig. 83) contains the pair of parallel elements of the hyperboloid, of which ab — a'b' is one, but a straight line, and a pair of parallels are each particular cases of the parabola. Also planes parallel to the gorge cut the two bodies in circles which are particular cases of the ellipse. All these facts lead us to infer the probability of what is really true,* that every plane section of a warped hyperboloid of revolution is a co7tic section of the same kind as that cut from the asymptote cone by the same plane, 70. Since all ruled surfaces are composed of rectilinear elements, the general problem of the intersection of the warped hyperboloid of revolution with cylinders and cones, or with other like hyperboloids, reduces to finding the in- tersection of a straight line with the hyperboloid ; f but many special cases, as in the following examples, can be easily solved. - v Examples. — 1°. Find the intersection of a cylinder or of a cone with a hyperboloid when their axes are parallel. Ex. 2°. Find the intersection of a cone and hyperboloid when the vertex of the former is in the axis of the latter. Ex. 3° Find the intersection of a cylinder and hyperboloid when their axes are not in the same plane, while the directions of these axes are perpen- dicular to each other. 71. Even the general case, the axes of the two given sur- faces having any position not in the same plane, may be *See my larger Descriptive Geometry, Theorem XLII. f See my larger Descriptive Geometry, Problem C. DESCRIPTIVE GEOMETRY. IO3 solved tentatively by assuming elements of either genera- tion of the hyperboloid, which it is supposed will contain points of the intersection, and by then passing auxiliary planes through these elements and the vertex of the cone (Prob. IX., 2°), or parallel to the axis of the cylinder (Prob. IX., 4°). Elements, one on each given surface, and thus in the same plane, will meet in points of the required intersection of the hyperboloid with the cylinder or cone. Example. — Find the intersection, by the last article, of a warped hy- perboloid of revolution, with either a cylinder or a cone whose axis has any position not in the same plane with that of the hyperboloid. D— Development. Example. — To find the true size of the intersection of a plane and a warped hyperboloid of revolution^ and the corre- sponding position of a given tangent to the curve, [See Prob. XL.] LI Ji II \\l\ UNJVKKSITY 01 (JALIFOIINIA. PART II.— DOUBLE-CURVED SURFACES. 72. Double-curved surfaces of revolution (24) are those which can be generated only by the revolution of a curved line of some kind, about an axis, and hence on which no straight line can be drawn. The generatrix may always be a plane curve, for sup- pose at first that the surface has been generated by a curve of double curvature, its meridian section would necessarily generate the same surface. 73. The most important distinctions among double- curved surfaces, relative to the purposes of Descriptive Geometry, are — First, Their generation by circles, or by other curves : a distinction of importance, owing to the ease of drawing tangents perpendicular to a radius. Second, That of being doubly convex, or concavo- convex. A sphere is an example of a doubly convex surface, any two plane sections whatever taken at right angles to each other through the centre being both convex towards sur- rounding space. A bell is an example of a concavo-convex surface, its horizontal sections, as it hangs at rest, being convex, while its meridian sections are concave towards surrounding space. DESCRIPTIVE GEOMETRY. IO5 74. The kinds of double-curved surfaces of revolution indicated will be sufficiently illustrated by such as can be generated by the revolution of some conic section about its own axis, or an axis in its own plane. These surfaces are, the sphere, the oblong ellipsoid (prolate spheroid), generated by the revolution of an ellipse about its transverse axis ; the flattened ellipsoid (oblate spheroid), generated by the revolution of an ellipse about its conju- gate axis ; the paraboloid, generated by the revolution of a parabola about its axis ; and the hyperboloid of separate nappes, generated by the revolution of a hyperbola about its transverse axis. Also the annular torus, generated by the revolution of a circle about an axis exterior to it but in its own plane (73). Two important general properties of the surfaces gene- rated by the revolution of a conic section about either of its axes are, that all their plane sections, and their curves of contact with circumscribed tangent cylinders or cones, are conic sections. A— Projections. 75. In representing singly any double-curved surface of revolution, it is usual to assume its axis of revolution as vertical. Its parallels will then be in horizontal, and its meridians in vertical planes (33). When such a surface is one of a group of objects vari- ously placed, it may be necessary to represent it with its axis oblique to one or both of the planes of projection. PROBLEM XLIX. — To represent the projections of any double-curved surface of revolution. In Space. — The required projections will consist of those of the apparent contours, that is, of the curves of contact of two circumscribing tangent cylinders, the elements of one I06 DESCRIPTIVE GEOMETRY. of which are perpendicillar to H, while those of the other are perpendicular to V. Or, if, as in the case of the para- boloid and the hyperboloid, the surface is unlimited in the direction of its axis of revolution, its projection on H, the axis being vertical, will be indicated simply by any one of its parallels. In Projectio7u — PL XL In Figs. 90 and 91, the circles with O and O' as a centre represent a sphere, whose pro- jections will always be equal circles. Fig. 92 represents an oblong ellipsoid, its horizontal pro- jection consisting of its greatest parallel, and its vertical one of that meridian which is parallel to V. In Fig. 88 is represented an annular torus, generated by the circle, of centre ff and radius fh—fh', in revolving about the axis O — 0'0'\ The portion generated by the semicircle vi'h'n' is doubly convex. That generated by m'q'n' is concavo-convex. The circle of radius Oq — O'g^ is the gorge. The generating circle may be tangent to the axis, thus giving the limit at which the torus ceases to be annular ; or it may intersect it. Fig. 89 represents an oblong ellipsoid with its axis first oblique to H only, and then to both H and V. We construct first the ellipse on the given axes A'B' and CD', Then the horizontal projection is the base of a vertical circumscribed tangent cylinder shown by the vertical tangents at c' and d'. The curve of contact of this cylinder being an ellipse, c'O'd! is its vertical pro- jection; hence OE =^ OF — 0'A\ and the horizontal pro- jection is the ellipse on the axes EF and cd. Next, suppose the ellipsoid to be revolved about a ver- tical axis through its centre till the vertical plane on cd, as shown at c"d", makes any angle as nO"c" with its former position. Then, first, tangents perpendicular to V, as at e" and f'\ will represent the cylinder which projects the ellip- soid upon V ; and, second, no point of the surface having changed its height, the new vertical projection will still be included between the horizontal tangent planes a' a'" and DESCRIPTIVE GEOMETRY. 10/ h'b'". Hence, projecting a' at ^, make 0"a" = Oa and pro- ject a'^ at a'" and find b"^ likewise. Also construct cf by making ce = c"e" , project e and / to c' and /', and thence over to c'" and f"\ the vertical projections of e" and /^'. The ellipse a'"f"'b"'e"' can be sketched with tolerable ac- curacy through these points of contact with its circum- scribed tangents, or additional points can be found by draw- ing tangents perpendicular to V, to other sections parallel and similar to c'd\ PROBLEM L. — Having one projection of a poiftt tipon a double -curved surface, to find its other projection. In Space, — Every plane section of a sphere being a circle, assume any circular section containing the point and per- pendicular to H or V, so that by revolution it may be made » parallel to V or H, respectively ; preferring for convenience a great circle, since this may by revolution be made to coincide with a projection of the sphere. Or choose a section parallel to H or V. In case of other surfaces, take either the' meridian or the parallel through the given point. In all cases, the other projection of the point will be on the other projection of the section containing it. In Projection, — PI. XL i°. The point on a sphere. Fig. 91. Let / be the given projection, then pr is the horizontal pro- jection of an arc through / and parallel to V, Supposing/ to be on the upper half of the sphere, r'p' is the vertical pro- jection of this arc, and/' is found by projecting/ upon it. A similar construction in an inverse order serves to find q' when q is given. 2°. The point on any non-spherical surface, — Fig. 93. Let T be the given projection of a point on the upper half of the torus, and let (97" be the meridian containing it. Revolve (9 r till parallel to V at 0T'\ when T" will be projected at T'", on the revolved meridian which coincides with that of the vertical projection of the torus. In counter-revolution. I08 DESCRIPTIVE GEOMETRY. T"T"' returns to Thy the parallel t"T—T"'T', on which, therefore, T is projected at T', It is now evident that the same construction lines, omit- ting OT, which is unnecessary, represent the solution by means of a parallel, instead of a meridian through the point. Examples. — 1°. In Fig. 91, assume a vertical great circle through / to find /'. Ex. 2°. Assume a great circle perpendicular to V through p' , to find /. Ex. 3°. In Fig. 93 let //' be given, and find in various ways the horizontal projections of the four points which it represents, once by revolving the plane — O'O" about its horizontal trace into H- B — Tangencies. "j^. Since a double-curved surface contains no straight line, a tangent plane can have only ti point of contact with it. A plane being determined by three points, a tangent plane to a double-curved surface will be determined by any three points, so conditioned as to be in such a plane, as, for example, a given point of contact and a point on each of any two tangents at that point (45), or by a required point of con- tact and any two exterior points, and hence by the line con- taining the latter. 77. As a second exterior point may have any position in respect to one first assumed, and remaining fixed, an infinite number of tangent planes may be drawn through the for- mer — that is, through one exterior point, and tangent to a double-curved surface. These will all evidently be tangent to a cone whose vertex is the fixed point and whose elements are the tangents from that point to the given surface. Again : a pair of exterior points may evidently be chosen in an infinite number of ways, such that the line joining them shall be parallel to a given line. Hence an infinite number of planes may be drawn parallel to a given line. These will all be tangent to a cylinder, all of whose ele- ments are tangent to the given surface and parallel to the given line. DESCRIPTIVE GEOMETRY. IO9 78. If any two lines, one of them at least curved, are tangent to each other at some point, they will, by simul- taneous revolution around any one axis, generate two sur- faces of revolution having a common axis, and their point of contact, remaining such, will describe a circle of con- tact, common to both surfaces. That is, if two surfaces of revolution Jiaving a common axis are tangent to each others they will have a circle of contact whose plane is perpendicular , to the common axis (57). When one of the two surfaces is a cylinder, it will be tangent to the double-curved surface on its greatest or least parallel. A cone may be tangent on any parallel. When the parallel reaches a vertex of the surface, it becomes a point, and the tangent cone becomes a tangent plane at that point. 79. Having a plane, tangent to a double-curved surface at a given point, any single-curved surface, developable or warped, may be placed tangent to this plane so that its element of contact with it shall pass through the given point of contact. A single-curved surface will thus have only a point of contact with a double-curved surface. 80. The tangent plane to a surface of revolution is perpen- dicular to the meridian plane through the point of contact ; for it contains the tangent to the parallel at the pomt of con- tact, and it is sufficiently evident from Fig. 92, for example, that the tangent at nn\ for example, to the parallel snr is horizontal, and perpendicular to the meridian plane On^ which is vertical. See also (Prob. XLVI.) PROBLEM LI. — To construct a plane tangent to an annular torus at a given point of contact. In Space, — The surface being generated by a circle, the distinction in (73) may be applied, and the plane found in part by the principle that it is perpendicular to the radius of the generatmg circle containing the point of contact. no DESCRIPTIVE GEOMETRY. Also the surface being one of revolution, the plane may be found on the principle of (80) as well as on the general principle of {y6). In Projection. — PL XL, Fig. 88. 1°. Special method appli- cable with a circular generatrix, — Let tt' be the given point of contact, either of its projections being found by (Prob. L.) when the other is given. ff\ the centre of the generating circle qfh — q'm'h\ appears at gg\ when revolved into the meridian plane Ot of the point of contact. Then gt—g't' is the radius of this point, and as the tangent plane is perpen- dicular to it, its traces will be perpendicular to the pro- jections of the radius. The directions of these traces thus being known, it is only necessary to find one point of each. Revolving the plane Ot into H, gg' appears atg^\ and tt^ at t" (Prob. XIL), and^'V is the revolved radius of contact. Then f'^f perpendicular to g"t" ^ is the revolved intersection of the tangent plane Avith the plane Ot, hence k, its horizon- tal trace, is a point of the horizontal trace RS^ perpendicu- lar to gt^ of the tangent plane. Its vertical trace, r'e\ is perpendicular to g't\ through r\ the vertical trace of the tangent tr — t'r' to the parallel through tt' , 2°. Geiieral method^ applicable to any surface of revolution. The solution can be understood by inspection of the construction, fully given, of the plane PQP\ tangent at TT and determined by the tangents to the parallel and the meridian at that point. If TT were on the inner or concavo-convex portion of the torus (73) the tangent plane would (Theor. VI 1 1.) be found to be also a secant plane. Two tangent planes, perpendicular to the axis, will each have a circle of contact with the torus. Examples. — 1°. Take the point ti' on the concavo-convex portion of the torus, retaining the method there shown. Ex. 2". Take the point it' on the concavo-convex portion of the torus, re- taining the method shown at TT. 81. Solutions of problems of tangent planes, in which an intermediate auxiliary surface is employed so that the DESCRIPTIVE GEOMETRY. Ill required tangent plane is tangent both to it and to the given surface, are indirect. Otherwise the solution is direct. Such intermediate surfaces may by the principle of {jf) be circumscribed tangent surfaces, or by that of (80) may simply have a common ^xis with the given surface. PROBLEM LII. — To construct a tangent plane to a sphere^ and through a given linCj by the direct method. In Space, — The points of contact of the two planes, which can obviously be drawn, will be those of a pair of tangents, from the point of intersection of the given line with a plane perpendicular to it through the centre of the sphere, and to the great circle contained in this plane. This plane, since the directions of its traces are known (Theor. II.), will most » conveniently be determined by lines through the centre of the sphere and parallel to those traces. In Projection. — PI. XL, Fig. 90. Let 00' be the centre of the sphere, and AB=A'B' the given line. Oa — O'a' and Ob—0'b\\\\iQYQ (7Vis perpendicular to A'B\ and Ob to AB^ are by (Theor. II.) lines of the plane perpendicular to AB—A'B' through 00\ By (Prob. XI., b, 2°) AB—A'B' pierces this plane, thus given, at cc'. Revolving the same plane about the horizontal Ob as an axis, and into the horizontal plane 0'b\ cc' will fall at C, by making bC equal to the hypothenuse b"'c' constructed as shown on b"'yy = be and c'y. Then Ct" and CT' are the tangents to the great circle of the sphere contained in the perpendicular plane, after a like revolution, and t" and T" are therefore the revolved points of contact of the required tangent planes. In counter-revolution, N, on the axis, remains fixed, and C returns to r, giving cNt for the horizontal projection of the primitive position of Ct" , on which t" is found by coun- ter-revolution at /, the intersection of t"t, the arc of counter- revolution, with ct. Then projecting iVat N' on the plane O'b', gives c'N', the vertical projection of cN^ on which t is projected at t'. 112 DESCRIPTIVE GEOMETRY. TT may be similarly found, except that, as in the figure, CT" does not always conveniently intersect the axis Ob. In this case, draw any transversal m"n'\ intersecting Ct" and CT\ and the axis bO as at q. In counter-revolution, m" .returns to m, on ct ; by drawing in!'m perpendicular to Ob, q remains fixed, giving mq^ to which n" similarly returns at n^ whence it is vertically projected on m'q' at -n! , Then en — c'n' is the tangent to which T" returns at 7", by coun- ter-revolution, in the arc T"T^ perpendicular to bO, and is thence projected at T. PQF is the required tangent plane at tt'. Its horizontal and vertical traces are respectively perpendicular to Ot and O't' the radii of the point of contact (Theor. II.), and con- tain the like traces ^ and B', oiAB—A'B', Examples. — 1°. Let the centre of the sphere be in H. Ex. 2°. Let it be in the ground line. Ex. 3°. Let the given line cross any other angle than the second. Ex. 4°. Let it be on the right of the sphere. PROBLEM LIII. — To construct a plane tangent to a sphere and through a given line, by means of an auxiliary tangent cone. In Space. — Make any point of the given line the vertex of a cone, tangent to the sphere on a circle of contact. Tangents to this circle from the traces of the given line on the plane of the circle will (Prob. XXX.) be the traces, on that plane, of tangent planes to the cone, through the given line. But by {jf) these planes will also be tangent to the sphere, and hence will be the required ones. In Projection.— ?\. XL, Fig. 91. In order that one pro- jection of the base of the auxiliary cone may be a straight line, which will evidently be most convenient, its plane must be perpendicular to a plane of projection, and the axis of the cone must consequently be parallel to the same plane. Hence let the vertex of the cone be the trace of the given line,^^ — A'B' oxi a plane parallel to H or to V through 00\ DESCRIPTIVE GEOMETRY, II3 the centre of the given sphere. In the figure, aa' is the trace of the given line on the former plane. Then tangents ac and ad, from a to the horizontal projection of the sphere, give acd for the horizontal projection of the auxiliary tangent cone, and cd th^it of its base. CC is the trace of AB — A'B' on the vertical plane, cd, of this base. Revolving this plane about the diameter cd as an axis till horizontal, the cone's base appears as the circle on cdy and CC at C\ by making CC equal to Cd\ Then (T'V and C^fz'^ are the revolved positions of the traces of the tangent planes oil the plane, cdy of the cone's base. In counter-revolution, /'' and ?i'\ the revolved points of contact, return in vertical arcs projected in t'^f and n"7i^ perpendicular to the axis cd, to t and ;/, whence their verti- cal projections, f and ;/, will be found on perpendiculars to GL from t and n, and at distances /V above, and n^k' below the horizontal plane O^a^ equal to ^'V and n''7t. Having now the points of contact //' and nn' of the required planes, the traces of the latter can be found as in Probs. LI. or LII. Examples. — 1°. Let the centre of the sphere be in H. Ex. 2°. Let it be in GL. Ex. 3". Let AB — A'B' cross any other than the first angle. Ex. 4°. Let AB—A'B' be to the left of the sphere. 82. By (80) if two surfaces of revolution have a common axis, without being tangent to each other, and if a plane be tangent to both of them, its points of contact with both will be in the same meridian plane, and a common tangent line to the two meridian curves contained in this plane would be a line of the common tangent plane. But the latter plane could not contain this line and another one arbitrarily chosen, since, unless the two Imes happened to intersect, they would be equivalent to four points, while a plane is determined by three. Hence if a plane is to be drawn through a given line and tangent to a surface of revolution by means of an exterior auxili- ary concentric surface, that surface must be generated by the 114 DESCRIPTIVE GEOMETRY. given line. Then this line, with the point of contact on the given surface, will be equivalent to three determining points of the tangent plane. It is further evident that the given line must not intersect the given surface. Finally, as this method requires the given surface to have only a point of contact with the tangent plane, it applies as well to the hyperboloid of revolution (Prob. XL VI I.) as to a double-curved surface of revolution. PROBLEM LIV. — To construct a plane through a given line and tangent to a double-curved surface of revolution^ by means of a concentric auxiliary surface, • In Space, — By (82) the concentric surface must be gener- ated by the given line, and hence will be a warped hyper- boloid of revolution (Theor. VII.). The common tangent to the two meridian curves will give one point of each of the parallels, one on each surface, containing the points of contact of the tangent planes with the two surfaces. But the point of contact on the hyperboloid is also on the given line (Theor. VI 1 1.), and that on the given surface is on the same meridian plane with the former one (80), which deter- mmes both of them. In Projection, — PL XL, Fig. 93. Let the given surface be the annular torus, generated by the revolution of the circle ab—a'b' about the axis 0^0' O" ; and \^\.AB—A'B' be the given line. Revolving a few points as NN',ss',BB\ etc., oiAB—A'B' into the common meridian plane OB^ we find the meridian curve B's'^n'b' of the hyperboloid, observing that as ON is perpendicular to AB — A'B\ the point NN' generates the gorge, and hence determines the axis N'n' of the meridian hyperbola. Also b' and t'" are found as symmetrical with B' and s" relative to N'n\ Next, the common tangent f"T"\ being a line of the common tangent plane, determines a point of the parallel DESCRIPTIVE GfeOMETRY. II5 containing the point of contact on each surface. Then revolving t"'t" about O—O'O" to ft on AB—A'B' {t not shown), we have the point of contact on the hyperboloid. But both points of contact are on the same meridian plane (80); hence draw tO, and revolve T"'T" likewise upon it at TTy which is the point of contact of the required tangent plane to the torus. The traces of this plane can then be found as in Prob. LI. As four common tangents can be drawn to the hyper- bola, and to the complete meridian of the torus, which em- braces the two opposite circles of centres rr' and 00', four tangent planes can be drawn which will fulfil the required conditions. Examples.— I ". Find any one additional tangent plane in the above problem. Ex. 2°. Apply the method of solution to any other double-curved surface of revolution, or to the warped hyperboloid. PROBLEM LV. — To construct a plane through a point in space ^ and tangent to a double-curved surface of revolution^ on a given section of the surface. In Space. — Considering only parallels and meridians, if the section is a parallel, the required plane will be tangent to the cone whose base is that parallel (Prob. XXX. and Tf). If it is a meridiany then by (80) the plane yill contain a per- pendicular to the plane of that meridian from the given point. In Projection. — PI. XL, Fig. 92. Let the ellipsoid, whose centre is 00' , be the given surface, and A A' the given point. 1°. A parallel section. — Let snr — ;;^V be the given paral- lel, then Vm! is an element of the cone having this parallel for its base, and V'A' — OA, the line joining the vertex with the given point (Prob. XXX.) pierces the plane n'm' of the base at p'p, whence tangents from / at s and r (found by a circle on pO) are the traces of the tangent planes from //' to the cone ; and ss' and rr* are the intersections of their ele- Il6 DESCRIPTIVE GEOMETRY. ments of contact on the cone, with the circle of contact m'n' of the cone and ellipsoid, and hence are the required points of contact. The cone becomes a cylinder at the greatest parallel D'E', and A/ and Ad are the horizontal traces of vertical tangent planes to it, and hence to the ellipsoid at ff and dd\ 2°, A meridian section, — Let Oa be the plane of such a section. Aa is a perpendicular to it from A, and a!' a!" is the position of a after revolving about O — O'V into the meridian plane OA" parallel to V. Then tangents, as a"'b"', to the ellipse h'D'E\ from a"'y are the revolved positions of the traces upon Oa of the tangent planes required, and b"'b" is the point of contact of one of them. This point, in coun- ter-revolution, follows the parallel through it to hb' , the point of contact of one of the two required tangent planes which can be drawn to the meridian in Oa, In like manner, cc' is the contact of one of the tangent planes to the meridian whose plane contains AA' ; and, finally, tangents^'// and A't' are the vertical traces of tangent planes, perpendicular to V, through AA', and hence give points of contact, h'h and fty on the meridian OA" parallel to V. Examples. — i". Find the traces of the tangent planes whose points of con- tact have now been found. Ex. 2°. Let the given parallel be below D'E\ Ex. 3°. Let A A' be below D' E and on the right of the ellipsoid. Note, — Problehis (LII.-LIV.) will enable the student to ass at once to the construction oi curves of shade on double- curved surfaces, in shades and shadows (Prob. LV.) and to that of curves of apparent contour of the same surfaces in perspective. C — Intersections. 83. A given double-curved surface may be required to intersect another given double-curved surface, or any of the three kinds of ruled surface : plane, developable, or warped. DESCRIPTIVE GEOMETRY. II7 The subject of intersections is thus here made complete, and admits the following general statement of the method of solution for all cases. Choose such auxiliary surfaces, in respect both to form and position, as will intersect each of the given surfaces in its simplest, or most easily used, sections. Then, where the sections, cut from the two given surfaces by the same auxiliary surface, intersect each other, will be points com- mon to both given surfaces — that is, points of their inter- section. The following problems will fully illustrate this general statement. PROBLEM LVI. — To find the intersection of a double-ctirved surface of revolution zuith a plane. In Space. — The best auxiliary surfaces for this case are planes. However placed, they will cut the given plane in straight lines, and they may be so taken as to cut the double-curved surface either in parallels or meridians, which are the two simplest ways. That is : the intersections of any parallel or meridian of the given curved surface with the line cut from the given plane by the plane of that meridian or parallel, will be points of the required intersection. The meridian plane perpendicular to the given cutting plane is a plane of symmetry of this plane and of the given surface, and hence of their intersection. I. In Projection. — The intersection of a flattened ellipsoid (oblate spheroid) by a plane. PL XII., Fig. 95. I ° . The method of meridians. The ellipsoid is represented by the projection of the circle OA^ its greatest parallel, and of the ellipse AB—A'B'CD\ its meridian parallel to V. PQF is the given plane. Oh, perpendicular to the horizontal trace PQy is the like trace of the meridian plane of symmetry, and it cuts PQP Il8 DESCRIPTIVE GEOMETRY. in the line Oah-^p'a'h\ whose vertical projection is deter- mined by a\ vertical projection of a, and p' that of the point where PQP cuts the axis O — CD' of revolution, and founds as shown, by (Prob. XL, b^ i°). Revolve the meridian plane Oh about O — CD' till it becomes parallel to V. This meridian will then coincide with OA—A'B'CD', and the line Oa—p'a'h' will appear at Ok"—p'a"'h"', giving h'" and I" as the revolved positions of the highest and lowest of the required points. Their primi- tive positions, hh' and //, can then be found in either of the two ways already shown for the like points in (Prob. XXXV., 2°). This method can be applied with any other meridian plane, but requires a like revolution and counter-revolu- tion, in every case except that of the meridian OA^ parallel to V. The plane OA cuts PQF in the line Of—p'f, where p'f is parallel to FQ, and gives at once k'k and g'g {g not shown) as the points in the plane OA, 2°. The method of parallels. — This is most convenient for finding additional points as they are found without the pro- cess of revolution and counter-revolution. Thus, assuming any horizontal auxiliary plane as c'd' between h' and /', it cuts from the ellipsoid the parallel of radius Ob, = c'b', and from PQP' the horizontal c'd' — de {de parallel to PQ) which intersects the circle Ob — c'b' at two points of the required curve, one of which is e, thence ver- tically projected at e'. Other points may be similarly found. To avoid confus- ing the figure, the curve, one half of which passes through hkegl and h'k'e'g'l', is not shown. II,— The plane sections of the annular torus. — PI. XII., Fig. 94. These, all found in the way just described, are of a great variety of forms, some of the general character- istics of which may be determined in advance by observ- ing the relation of the line /, intersection of the given cut- ting plane with the meridian plane of symmetry, to the meridian M, contained in the latter plane. This line / may have numerous different tangent or secant positions, DESCRIPTIVE GEOMETRY. II9 as ijy xyy hly etc., relative to M, each of which will indicate a new form of the intersection, while an exterior position, as mtty shows that the given plane does not intersect the torus. A plane, like PQP\ PL XII., Fig. 94, tangent at two points, //' and TT\ is called a bi-tangent plane. Theorem IX. — The section of an annular torus y made by a bi- tangent plane y consists of a pair of circles whose diameter is that of the generating circle plus that of the gorge, PI. XII., Fig. 94. — Draw the radius A'E' of the generat- ing circle to the extremity £' of the diameter of the parallel containing any point, as uu' y of the curve, also the radius O'E' of the torus, to the same point E', Suppose the curve padcy found as in the last problem, to be revolved about the line BD — PQ as an axis, till parallel to V. Then uu' will fall at u" on the perpendicular u'u" to FQ, at a distance u"u' from PQy equal to that of u from BD. Draw O's perpen- dicular to E'A' produced; O'x" perpendicular to PQ; E'F and u'v perpendicular to O'A'; Tx" perpendicular to O'O"; and finally draw ti"x". In the figure, uv happens to be very nearly tangent to the circle ^'Z>', because PQ makes very nearly an angle of 45° with GL. The triangles A'E'F and A'O's are similar, and so are (9V2/ and O'A'T, These pairs give respectively the pro- portions, A'E' : A'O' :: E'F: O's and A'T : A'O' '.: u'v : O'u' whence, by comparing terms, we see that 0's=0'u' also O'u" z= O'E' ; hence the triangles O'u'u" and O'E's are equal, the angle O'E' A' is equal to the angle 0'u"u'y and hence to u"0'x" . Comparing the triangles 0'u"x" and OF'A'y we thus find equal angles included between equal sides, since O'x" = A'T = A'E', Hence the remaining 120 DESCRIPTIVE GEOMETRY. sides u"x" and O'A' are equal. But O'A' is constant for all positions of tin' , hence the different positions of u" , revolved positions of different points as uu^ on the curve, are at a constant distance from the fixed point x" , and thus form a circle y of a radius equal to 0'A\ There are evidently two such circles in the plane PQF , and another like pair in a plane, tangent at t" . The horizontal projections of the former are ellipses, in each of which the transverse axis, as pd, equals the sum of the diameters of the gorge and generating circle, and the conjugate axis, as ae^ is the horizontal projection of a'e\ But Oa = Oe= O'A' — O" B\ the radius of the highest and lowest parallels. But O'A' - \ I'D' = i pd. That is, Oa=:^ pdy hence (9 is a focus of the ellipse pade, and hence also of the symmetrical one qbc. The revolution of the tangent BD—P'Q about 0—0' O" generates a bi-tangent cone, having the parallels through T and t' for its circles of contact, one on each nappe. Every tan'gent plane to this cone cuts the torus in a pair of circles equal to those just described. The annular torus has therefore three sets of pairs of circular sections, ist. Its parallels. 2d. Its meridians. 3d. Those in tangent planes to the bi-tangent cone. All the circles in each of the two latter sets are equal. Example. —Construct a series of figures of intersections of an annular torus, with cutting planes whose intersections with the meridian plane perpen- dicular to it shall have every possible variety of position relative to the meridian in the latter plane. When PQP\ Fig. 94, revolving about Od—0' becomes so nearly vertical that T,t and find the traces of a plane which is given by two lines, neither of which intersects the planes of projection within convenient limits. In Space. — Let A and B be the two lines. Take any point J/ on A and any point iVon -5, then the projections of the line MN joining these points will be mn — m'n\ and the traces of this line will be points of the respective traces of the plane. Two such lines, each intersecting both of the given lines, will determine the required plane, and evident- ly in a way agreeing with the definition of the plane as a surface of transposition (85). See also (27, 9°). In Projection. — Examples. — 1°. Solve the problem when the plane is regarded as a surface of transposition. Ex. 2°. Solve the problem when it is regarded as a cylinder. Ex. 3**. Solve the problem when it is regarded as a cone. C— Intersections. 86. Lines intersecting given lines which lie on the same plane are called transversals relative to the latter. The fol- lowing principles and problems will partly illustrate the connection of descriptive geometry — that is, the method of projections — with the subject of transversals. 128 DESCRIPTIVE GEOMETRY. PROBLEM LXllL— To find the intersection of two planes, each of which is given by its horizontal trace and one point. In Space. — Any plane containing the given points and intersecting the given traces will cut each plane in a line, joining the given point of that plane with the point cut from its trace. The intersection of these lines will be a point of the intersection of the planes. In Projection. — PL XIII., Fig. 98. Using the peculiar notation of Olivier* once for illustration : Let H^ and H^ be the given traces of the planes P and Q. Let ^, a^ be the projections of the given point a of the plane P, and ^, b^ those of b^ that of the plane Q, * Designed to be completely systematic and self-explaining, to facilitate thus the reading of projections. Symbol. Meaning. I. — On the primitive H andM. a!"— horizontal projection of point a. a> = vertical " «. •< .« A^ = horizontal " " line A. A* = vertical " ** •• H^ — horizontal trace of plane P, F^ = vertical •• " " " Common System. ad = projections of a point A. ab — a'3'= projections of a line AB. PQF — traces of a plane. a'\a"', etc., or ai ai, etc., = the projections of a point A either on new planes or after a rotation. PxQxPx', or P"Q'P"\ etc., = the like for planes. II. — On new planes of projection. a*' = projection of the point a (a* a>) on a plane perpendicular to V and taken as a new plane H. a«' = projection of the point a (a* a') on a new vertical plane. A^\ A''' = projections of a line .-4 on a new H or V, respectively. I/'P = the trace of the plane P on a new plane H perpendicular to V. V^ = the trace of the plane P on a new plane V. III. — A/ter rotation about an axis. a\, «'» = the projections of the same point a as before, after rotation about an axis perpendicular to either H or V. ^'*,^'* = the projections of the line A, after a like rotation. //^, V^ = the traces of the plane P, after a like rotation. Various causes, among them the large number of slightly different symbols, the difficulty of using them on crowded diagrams, and the necessary recourse to numbers instead of letters in large and complicated figures, have prevented this notation from coming into general use. DESCRIPTIVE GEOMETRY. 1 29 Draw the line D — that is, the line whose projections are jy^ jy — and find its horizontal trace n^. Then any line, //^^, through n^y and intersecting H^ and H^^ will be the hori- zontal trace of a suitable auxiliary plane X. The plane X cuts from the plane P the line B (B^R), and from Q the line C. These lines intersect at/, a point of the required inter- section // also Illy the intersection of the given traces, is another point of/.* 87. We know that the intersection of two planes is a straight line passing through the intersections of their like traces, and that the last problem gives a correct construc- tion of that line. That is, regarding now the horizontal projection alone, the point /* will always be on the line /* drawn through ;;/, however the line H^ is drawn. That is, we have the following : Theorem I. — Of transversals. — If any three lines, as Z)*, H^y H^, cut each other two and two, and if we have three points, n^, a^, d^ on one of them, i^, and then draw from one of* these points, as ;/, any number of transversals, as //^, cutting the lines H^ and H^, in points which shall be joined with d" and U" respectively by straight lines, the intersec- tions of these last lines with each other will, together with m^, fall on a straight line /\. , In like manner, Z^\ 7/^, P might have been the given lines, and d" the origin of the transversals, and a like series of final intersections would have fallen on a line H^ contain- ing m^. Or D"y H^y P being given, and I^ taken as the origin of transversals, a like series of intersections would have de- termined the line H^, PROBLEM LXIV.— 7;? find the intersection of tivo planes whose traces do not meet within the limits of the figure. In Space. — Intersect the given planes by auxiliary planes * Observe that in this notation, a point, line, or plane is named by the single letter which indicates it in space, and which with the indices h and v indicates the projections of the two former, or with ^and V, indicates the traces of the latter. I30 DESCRIPTIVE GEOMETRY. parallel to the ground line, then the pair of lines cut from the given planes by any one of the auxiliary planes will meet at a point of the required intersection. hi Projection. — PL XIII., Fig. 99. Returning to the usual notation, let PQP' and RSR' be the two given planes, and let TU and PR be the traces of an auxiliary plane taken nearly parallel to Y as well as parallel to GL, Then Tp and Ur^ the horizontal projections of its intersections with the given planes, meet at «, a point of the horizontal projection of the given planes. Another such point, m, is found by means of a similar plane, VWy PR' ; then mn is the horizontal projection of the required intersection. Its ver- tical projection could be similarly found by similar auxiliary planes making a small angle with H. 88. As in Prob. LXIIl., we know that the last construc- tion correctly gives the intersection of two planes, and that this line is straight and passes through the intersection of QP and SR. Examining then the horizontal projection only of Fig. 99, as a plane figure, we deduce the following : Theorem II. — Of transversals. — If we have three lines, as PQ, QSy SRy which intersect in three points, and two points, as / and r, on one of them, QS ; and if we draw a series of transversals, as TW and VlVy parallel to QS, and draw lines from their intersections with PQ to /, and from those with RS to r, the pairs of lines, as Tp and Ur, thus found, will meet on a straight line mn containing the inter- section of PQ and RS. This theorem is in fact that case of the last one (87), in which n^, Fig. 98, is at infinity on the line Z>*, in which case the different positions of //^ would be parallel to JD^. Returning to the figure in space, this case is that in which the points a and d being at equal heights above H, the line D (/>*, i?) would be horizontal, which would carry «* to infinity, and would make the solution the same as that of Fig. 99, except that each auxiliary plane, instead of being given by its traces, would be given by two lines, one of DESCRIPTIVE GEOMETRY. I3I which would be its horizontal trace, while the other would contain the two given points of equal altitude. 89. From Theorems (I. and II.) the following reciprocal one is now derived : Theorem III. — Of transversals. — If there be four straight lines, D, P, Q, /, Fig. 98, three of which meet at a point, and each cuts a fourth one, and if we join any number of points on one of the former, as /, with two points, as a and b, on the fourth, Z>, these lines from a will cut P, and those from b will cut Q, so that lines joining the pairs of points, as c and d, thus found on P and Q, will all meet at one point of D, as ;2*, Fig. 98, or be parallel to it, as in Fig. 99. 90. If, in PI. XIII., Fig. 99, the auxiliary planes had been parallel to each other, Vp would have been parallel to Tp, and Wr to Ur, but m would evidently still have remained on the line inn. That is : Theorem IV. — Of transversals. — If two given lines, as PQ and RS, Fig. 99, be cut by a series of parallel transversals, as 05, TU, WVj and if through the intersections of these parallels with the given lines we draw two series of trans- versals, as Tp and parallels to it from Q and V; and Ur and parallels to it from 5 and W; the pairs of lines, as Tp and Ury thus drawn will intersect on a straight line, as mn, which contains the intersection of the given lines QP and SR, PROBLEM LXV. — Having given two lines and a point, to con- struct a third line through this point so that the three lines shall all meet in one point. PI. XIII., Fig. TOO. Let AC?Lnd EF be the given lines, and in the given point. Draw any line, AD, and then Amb and aniB. Draw Bb and note its intersection D with AD. Then draw DC, cB, and Cb ; the intersection of the last two will give n, which, with in, will determine the required line. Passing from this plane construction to its original in space, AC and EF may be the horizontal traces of two 132 DESCRIPTIVE GEOMETRY. planes whose intersection is required, and which are given, as in Prob. LXIL, by these traces and the points B of the plane E and b of the plane A. Then D being the horizontal trace of the line Bb in space, any transversals, as DA and DC^ will represent the horizontal traces of auxiliary planes, the former cutting from the two planes the lines Ab and aB, whose intersection is one point of that of the given planes, and the latter, DCy cutting them in the lines Cb and cB, which give another point, n^ of the intersection inn^ which being thus evidently the required intersection, necessarily contains the point of concurrence of -^ 6* and EF, 91. The last problem permits the enunciation of Theorem V. — Of transversals, — Having any two lines, AC3,nd EF, and a point, Z>, in their plane, if any number of transversals be drawn from D to the given lines, also diagonals of the quadrilaterals, as ABab, thus formed, the intersections, as m and ;/, of these diagonals and that of the given lines will all be on the same line L. These theorems of transversals show how plane and solid geometry mutually assist each other, being derived from constructions in space, while in the explanation of the last problem we return from a plane construction to one in space. PROBLEM LXVI. — Through a given point, to draw a line that shall intersect two given lines. First solution — In Space,-— Y'void where the plane contain- ing the point and one of the given lines cuts the other given line. The line through this and the given point will be in this plane, and will therefore evidently fulfil the required conditions. In Projection, — PL XIII., Fig. 102. Let pp' be the given point, and AB—A'B' and CD— CD' the given lines. The auxiliary plane contains //' and CD — CD\2iS is shown by its being determined by the lines pC—p'C and pb^p'b', both of which intersect CD — CD', This plane intersects DESCRIPTIVE GEOMETRY. 1 33 AB—A'B' at mm' (Prob. XL, by 2°), hence mpn—m'p'n' is the required line. Second solution — In, Space. — This problem can also be solved by two auxiliary planes, each of which contains the given point and one of the given lines. The intersection of these two planes will be the required line, and the construc- tion is evidently the same as that for finding where a given line pierces a plane which is given by a point and a line. In Projection. — PI. XIII., Fig. loi. Considering the con- struction from the point of view last named, we will examine the peculiar case in which the two traces of the given plane coincide, as at PQP\ and in which also the two projections of the given line coincide in the line OB, Then applying (Prob. XL, a) this line pierces that plane at a point whose two projections coincide in in. A line whose projections thus coincide is in the bisecting plane of the second and fourth angles, hence in is also in that plane. Therefore if a series of such lines as those whose projections coincide in OC, OD^ etc., be passed through O where the given line OB meets the ground line, they, and consequently their intersections with PQP ^ all found by Prob. XL, will be in the same bisecting plane. But these points are also in the plane PQP\ hence in the intersection of these two planes. That is, they are all in the same straight line. 92. The last problem, looking upon its construction, Fig. loi, as a plane figure, gives the following: Theorem VI. — Of transversals. — If two lines, GA and DAy intersect at a point A, and if from a point O on GA a series of lines, OB^ OC^ OD, be drawn, cutting the other line, DAy as at By Cy Dy and if from the latter points per- pendiculars, BEy CFy etc., be drawn to GAy and if Oo be drawn, also perpendicular to GAy the lines oEy oFy oG will cut OBy OCy OD in points ;«, ;2,/, which, with Ay are on the same straight line. The perpendiculars to GA at Ey Fy (9, G may have any other direction if only they remain parallel. 134 DESCRIPTIVE GEOMETRY. CHAPTER II. DEVELOPABLE SURFACES. Defiiiitions^ etc. 93. There are two kinds of developable surfaces : 1°. Those in which any two elements are in the same plane. 2°. Those in which only consecutive ones are so situ- ated. Cylinders and cones form the first of these two kinds of developable surfaces. 94. A cone of transposition — that is, a cone defined in the most general manner — is the surface formed by joining a fixed point with all the points of any fixed curve of any kind whatever, and is therefore generated by a line, mov- ing, so as always to contain a fixed point, the vertex, and upon a fixed curved directrix. Any two elements are there- fore in the same plane. Wh^n the vertex is at infinity, the elements become parallel and the surface becomes a cylinder. When the linear directrix is straight, the cone of trans- position becomes a plane by the above definition, as the cone of revolution does by (22). See Prob. LXIII., where the intersection of the given planes is really found exactly as in Prob. XXXVII. 95. The oblique cone, and cylinder, with a circular base, are types of the entire class of cones and cylinders of trans- position (20) just defined. DESCRIPTIVE GEOMETRY. I35 PL XIII., Fig. 103, shows, at VAB~VCD\ an oblique cone with a circular base, more briefly termed a scalene coney in contrast with the cone of revolution Vgd — V'r'm! , Its section perpendicular to the axis is an ellipse. 96. The sub-contrary section, — PL XIIL, Fig. 103. VO^ the line from the vertex to the centre of the circular base of a scalene cone, is its axis. A plane perpendicular the base and containing the axis is called a principal plane, and con- tains evidently the longest and shortest elements, or those respectively of least and greatest inclination to H. Neglecting now the horizontal projection, suppose VCD' to be a scalene cone with circular base whose prin- cipal plane is parallel to V. Then Xx parallel to H will be a circular section. Now let Vyhe a, section such that the angles at V and Xy and hence all the angles of the triangles XVjs, and xjyz, are equal. Then X3 \ yz w Yz \ xz hence Xz xxz =z Yz X yz. But the section Xx is a circle, hence Xz x xz equals the square of the ordinate at z ; but this ordinate is common to the two sections, and hence also equals Yz xyz. Hence the section Yy is also a circle, and relative to Xx is called a sub-contrary section, 97. The second kind of developable surfaces embraces in- numerable surfaces formed as follows : In PI. XIIL, Fig. 104, let bd represent a line intersect- ing eg at dy not in the plane PQy but in space. Likewise let eg intersect hj at gy let hj intersect kl at /, etc. Also let acfi .... be a curve tangent to these lines as at r, /", /, etc. This curve will be of double curvature (49), since the lines bdy egy etc., are not all in one plane. Now, when these lines become consecutive, the intersec- tions dy gy etc., will be confounded with the points of con- tact Cy fy etc., and the lines will constitute a surface whose elements are all tangent at consecutive points of a curve of double curvature. 136 DESCRIPTIVE GEOMETRY. 98. This surface is developable since its consecutive elements intersect^ though not all at the same point as in a cone. This may be more explicitly shown as follows : bd may be re- volved about eg as an axis into the plane egh^ this plane may then be revolved about hj as an axis till it coincides with the plane hjk^ thus bringing the lines bd, eg, hj, kl, all into the same plane. This process, equally applicable when these lines become consecutive, being continued indefinitely, all the elements will be brought into one plane without changing the relative position of consecutive ones. That is, the surface is developable. 99. There being an infinite number of different curves of double curvature (50), there will be an infinite variety of developable surfaces of the kind just described. The directrix acfi . . . being a line, from every point of which two elements diverge in opposite directions, it is analogous to the vertex of a cone, first in being the limit be- tween the lower nappe^ acfiabeh, and the upper one acfipqr; and, second, in being the place where the surface is most contracted, or retreats furthest into itself. Hence ac/i ... is called the regressive, or re-entrant edge. It is also called the cuspidal edge because any section of the surface in a plane cutting acfi . . . is of the pointed form CfC^ called a cusp ; Cf being on the lower and C^f on the upper nappe. The numerous class of surfaces just described, being of infrequent practical use, is sufficiently illustrated by a sin- gle example, beginning with a description of its curved directrix. 100. The helix may be simply defined in two ways, either of which is an obvious consequence of the other.. 1°. The helix is generated by a point which has at once a uniform angular motion around a fixed axis, and a uniform rectilinear motion parallel to it. Hence — 2°. The helix lies on a cylinder of revolution whose axis is that of the helix, and crosses its elements at a constant angle, which is acute. DESCRIPTIVE GEOMETRY. 1 37 Hence also, as the elements of a cylinder are parallel both before and after development, the development of a helix upon that of the cylinder containing it is straight. loi. In PI. XIII. , Fig. 104, suppose acfm to be a helix lying on a vertical cylinder of revolution, and that ih is a tangent to it at any point i. Then by (52) the angle made by ih with the horizontal plane PQ, is the angle made at i by the helix with that plane. But by (100, 2°) this angle is • uniform, hence all the tangents to a helix so placed^ and the helix itself^ make a const aitt angle with the horizontal plane. 102. Hence (100) the rectilinear development of the helix coincides with its tangent, and the two paths from i to PQ, ica the helical one, and ih its tangent, both being of the same constant inclination to PQy are equal. That is — The helical arc and its tangent^ included between any two parallel planes perpendicular to its axisy are equal. A — Projections. PROBLEM LXVII. — To construct the projections of a helix and of its tangent. In Space. — Both parts of the construction follow directly from the previous definitions. In Projectio7i. — PL XIII. , Figs. 105, 106. Let the point 00' in H ascend a distance o',8^ = O'O" while making a complete revolution around the axis — O'O" . Both motions being uniform, divide the circle 08, representing the angular motion, into any convenient number of equal parts, and the corresponding ascent O'O" into the same number of equal parts, and draw horizontal lines through the latter points of division. Then beginning with 00' the point i is projected at i' in the horizontal plane through ^,, the point 2 at 2' on the next equidistant line, 3 at 3' on a^ 3', etc. Otherwise : making ^4, Fig. 106, equal to the semicircle 024, Fig. 105, and AZ equal to half of O'O", then (100) the straight line 4,8 will be the development of the half turn of 138 DESCRIPTIVE GEOMETRY. the helix from 4,4' to 8,8', Fig. 105. Then divide 4,8 into as many equal parts as there are on the semicircle 468, and draw horizontal lines 55', 66\ etc., through the points of division, which will meet the perpendiculars to GL^ similarly numbered in Fig. 105, at the points 5',6', etc., of the vertical projection of the helix. The tangent. — By (loi) the horizontal projections of equal distances on the helix and its tangent will be equal. Hence if TT be a given point of contact, make Tt equal to the arc 7b, and t will be the horizontal trace of the tangent at TT . Hence / is vertically projected at t' and Tt and Tt' are the projections of the tangent at TT' , 103. The developable helicoid is generated by a straight line which moves so as to be always tangent to a helix, and hence so as to make a constant angle with any plane perpen- dicular to the axis of the helix. PROBLEM LXVIIL— r^ construct the projections of a de- velopable helicoidy its axis being vertical. In Space. — First case. If the surface be given by its helical directrix, its elements can be constructed as tangents to that curve, as in the last problem. Second case. If the surface be given by its axis and one element, any number of additional elements can be found on the principle that their inclination to H is constant, and that, therefore, the portions of them included between any two horizontal planes will be equal, as well as equidistant from the axis. In Projection. — PI. XIV., Fig. 107. First. Let acg— a'c'g'Q be the given helical directrix^ constructed as in the last problem. Then, as in the same problem, construct any desired number of tangents to this helix, and they will col- lectively constitute the projections of the surface. Thus making dT^ tangent at d and equal to the arc dba^ projecting T upon the ground line at T' , and-drawing T'd' , we have TdD^ — T'd'D^ an element of the surface. DESCRIPTIVE GEOMETRY. 1 39 Second. Let — O'Q! be the axis of the surface, and aQ' — a'Q" the portion of an dement included between H and the horizontal plane QQ!". Then draw various radii Ob—oJ)\ Oc — o^c\ etc. ; also bJ)B,, perpendicular to Ob at b, and equal to aQ\ with bb^ equal to the arc ba; c,cC^ perpen- dicular to Oc at c, and equal to aQ' with cc^ equal to the arc ca, etc. Finally, project a at a' and Q' at Q" ; b, at b,' and B, at ^/, etc., and aQ'—a'Q"\ b,B~-b,'B,\ c,C,—c^C,\ etc., will be elements, identical with those before found at the same points aa'y bb\ etc. The projections of the helical re-entrant edge can^ then be sketched, tangent to the successive ele- ments. The curves ab.c^T, and Q'B^C^ G^a, found by join- ing the traces of the equal elements just found, on the planes H and QQ" respectively, are the traces or bases of the helicoid on these planes. Visibility, — In horizontal projection. Q"a is visible. Sup- posing the figure wholly in the first angle, B^b„ for example, is visible from B^ to b, and thence invisible till it reappears from b^ to aQ", for below bb' this element is on the lower nappe (99) of the surface, and is therefore concealed by the portion of the upper nappe which is shown. In vertical projection. Situated as the surface is, the inner surface of the upper nappe is seen, except where composed of the parts of elements upward from their point of contact with gha—g'h'Q. Also elements of the lower nappe whose horizontal traces accordingly are on ab^c^ 7", are visible in vertical projection from the ground line up to their ap- parent intersection with a'Q!'\ until we pass the point of contact of a tangent to the base, perpendicular to the ground line. Examples. — i". Construct the figure when turned 90° about its axis o—aq. Ex. 2°. Construct the figure when likewise turned 180°. [In these ex- amples certain elements, readily determined by inspection, will be visible throughout. The elements generally will be seen from their visible ex- tremity, readily found, to their point of contact with the helix.] 140 DESCRIPTIVE GEOMETRY. Examination of the horizontal trace of the helicoid. • 104. Involutes, Evolutes, — When a straight line rolls tangentially upon any curve whatever, any one point of the rolling line will generate a curve which is called an involute of the given curve. Thus in PI. XIV., Fig. 107, when the tangent aQ" ^ considered as a line in H, becomes succes- sively tangent at b, c, etc., and all the intermediate consecu- tive points, the point a will describe the involute a b^c^^ etc., and the point Q' will describe the involute Q'B^C^D^, etc. The fixed curve on which a given straight line rolls is called an evolute^ relative to its involute. 105. Tangents, Normals, — The point r,, for example, is, at the instant of passing that point, describing a circular arc about c the corresponding point of contact of c^c with the circle Oa, hence a tangent at c^ will be perpendicular to cc^. That is, the tangent to the involute at any point is perpendicular to the position of the generating line at that point. But the perpendicular as cc^ to a tangent at its point of contact is a normal. That is, the normal to the involute is a tangent to the evolute, 106. Involute of the helix, — The tangential rolling of aQ" — a's upon the circle Oa, and of aQ" — a'Q"'' upon the helix abc, are simultaneous, being effected by the rolling of the triangle a'sQ" upon the vertical cylinder whose base is abeg. Hence ^, as the point common to a' Q" rolling on the helix, and aQ' rolling on the circle abeg, describes the same curve ab^c^ . . . which is thus an involute of the helix as well as of its horizontal projection abeg. B— Tangencies. 107. Tangent planes to cylinders and cones of trans- position exist under the same conditions (46), and are con- structed in the same way that has already been shown (Prob. XXVIII., etc.) DESCRIPTIVE GEOMETRY. I4I The helicoid above described being developable, tangent planes may be drawn to it under the same conditions as for cylinders and cones, as will next be shown. PROBLEM LXIX. — To construct a tangent plane along a given element of a developable helicoid. In Space, — The required plane will be determined by any two tangent lines in it, one of which may be the given element (44). In Projection,— ¥\. XIV., Fig. 107. Let Td—Td'hQ the given element. Knowing the base ab^c^ ... of the surface, TS tangent to it at T, and hence perpendicular to Td (105), is at once the horizontal trace of the tangent plane, as well as one of the tangents which determine that plane. The vertical trace of the plane is then determined by the point 5, and by the vertical trace of Td — T'd\ or of any tangent parallel to TS, as ij—i'j', whose vertical trace, j'y gives Sj' that of the plane. Example. — Construct the tangent plane on a given element of the develop- able helicoid situated as in the examples to the last problem. PROBLEM LXX. — To construct a tangent plane to a de- velopable helicoid, through a given exterior point. In Space, — Make the given point the vertex of a cone of revolution having the same declivity that elements of the helicoid have. Then a common tangent to the bases of the two surfaces will be the horizontal trace of the required plane. Its vertical trace can then be determined as in the last problem. In Projection, — The construction is left as an example. PROBLEM LXXI. — To construct a tangent plane to a develops able helicoid, and parallel to a given line. In Space, — Make any point of the given line the vertex of an auxiliary cone of revolution as in the last problem. 142 DESCRIPTIVE GEOMETRY. Then the required plane will be parallel to one through the given line and tangent to this cone. In Projection.— F\. XIV., Fig. 107. Let kl-k'l' be the given line. At any point kk' of this line, draw km — k'm' parallel to aQ" — a'Q'"y and draw the circle of radius km, which will be the base of the auxiliary cone employed. Then lo, tangent to the circle km, from / the horizontal trace of kl — k'l'y will be the horizontal trace of the plane parallel to the required one ; hence C/F parallel to lo and tangent to the base ab,c^ of the helicoid is the horizontal trace of the required plane. Its vertical trace can then be found as Sj' was, or may be drawn parallel to the vertical trace, also easily found, of the auxiliary plane lo. 108. Declivity of the developable helicoid, — The element of contact of any tangent plane to this surface, being perpen- dicular to its horizontal trace, is a line of declivity (Prob. VII.) of the plane. But all the elements, and hence all the tangent planes, have equal declivities ; hence the surface is one of the class called surfaces of uniform declivity. 109. Other examples of this class of surfaces are those in which the horizontal traces of the tangent planes, all having equal declivities, are tangent to some given curve in the horizontal plane. Thus an embankment of uniform slope surrounding an elliptical or other pond would be a surface of uniform declivity, and so, on a smaller scale, would be the bevelled edge of an oval picture-frame. no. Cone direct or. -^"W hen every element of a surface is parallel, each to some element of a certain cone, the latter is called the cone director of the surface. When, moreover, the tangent plane on any element of the surface is parallel to that on the parallel element of the cone director the sur- face is developable. For, see PI. XIII., Fig. 104, if at any one point lines be drawn parallel to be, ef hi, etc., they will form a pyramid having faces, which, with the planes containing them, are parallel to the faces bdc, egh, etc., of the develop- DESCRIPTIVE GEOMETRY. I43 able polyedron and their planes, Fig. 104. Then as these faces become infinitely many and small, they become ele- ments of a cone director, parallel to corresponding ones of the developable surface, with parallel tangent planes along the parallel elements of the two surfaces. The cone director of a surface of uniform declivity is evidently one of revolution. C—Intersections. Developable Helicoids. 111. The general, or re-entrant developable surface just illustrated by the developable helicoid, is easily represented by its elements. Its intersection by any plane is, therefore, readily found by constructing the intersections of its ele- ments with that plane by Prob. XL 112. The intersection of the developable helicoid (PL XIV., Fig. 107) with any plane perpendicular to its axis, is a pair of involutes of the circle abfg, one on each nappe, and beginning at the intersection of the plane with the helical directrix. 113. The intersection of the developable helicoid with a cylinder having the same axis as the helicoid, consists of two helices, one on each nappe, and projected on H, the common axis being vertical, in a circle concertric with the horizontal projection of the helical directrix. 1 14. The intersection of any form of the general develop- able surface with any cone is found by means of auxiliary planes, each of which contains an element of the developable surface and the vertex, and hence, generally, an element also of the cone, whence the rest of the solution is obvious. 115. The intersection of the surface with any cylinder requires the use of auxiliary planes, each of which shall 144 DESCRIPTIVE GEOMETRY. contain an element of the surface, and shall be parallel to the axis of the cylinder (Prob. IX., 4°). Examples. — 1°. Find the intersection of the developable helicoid with any plane oblique to its axis. Ex. 2°. Find the intersection of the same surface with any cylinder. Ex. 3°. Find the intersection of the same surface with any cone. Ex. 4°.' In each of the last three examples, construct a tangent line at a given point of the intersection. General View of the Intersections of Cones — Infinite Branches, 116. The manner of locating the auxiliary planes in each of the three general cases — two cylinders (59), two cones (Prob. XXXVIL), and a cylinder and a cone (58), and the details of the construction of the points of intersection (57), being essentially the same in all cases ^ the following remarks will be a sufficient guide to the construction of every variety of form which the general problem can present. 117. I. Varieties of intersection depending on the relation of the tangent auxiliary planes, •1°. When both of the auxiliary planes tangent to one body cut the other body, the intersection of the two bodies will consist of two separate curves, one of entrance and one of departure. 2°. When one tangent plane to each body cuts the other body, there will be one curve. See PL IX., Fig. 78. 3°. When there is one common tangent plane, the curve will cross itself once, 4°. When there are two common tangent planes, the curve will cross itself twice, and will consist of a pair of conic sections whenever any plane section of each body is a conic section. 118. II. Varieties of intersection depending on the relation of either cone, V, to a cone C coinposed of elements through the vertex of V and parallel to those of the other cone, V^. This test cone, C, is precisely analogous in its uses to the DESCRIPTIVE GEOMETRY. I45 plane through the vertex, whose position determined the character of the pa,r aWel plane intersections (36). 1°. When the parallel cone, (7, has onfy its vertex in com- mon with the given cone F, the two given cones can have no element of either parallel to elements of the other ; hence their intersection can have no infinite branch, and conse- quently will be a closed curve. This, though of double cur- vature (49), yet being analogous to the closed plane inter- section, may be distinguished as belonging to the class of elliptic intersections. 2°. When the parallel cone, C, is tangent to the same given cone F, there will be a pair of parallel elements, e and ^„ on Fand V^ respectively, and the intersection will con- sist of one infinite branch having no asymptote (51), since the tangent planes along e and e^ being parallel, their inter- section, which (Prob. XXXIX.) is the tangent to the inter- section of the cones at the point where the elements e and r, meet, is wholly at infinity. The curve in this case, therefore, belongs to the class of parabolic intersections. 3°. When the parallel cone, C, intersects cone Fin two elements, there will be two pairs of parallel elements, as in PL XIV., Fig. Ill, and two asymptotes (51), and the curve will have an infinite branch with asymptotes, belonging therefore, by analogy, to the class of hyperbolic intersections. 4°. When one of the cones becomes a cylinder, the con- dition for the existence of an infinite branch in the inter- section is, that one element of the cone must be parallel to those of the cylinder, and hence must coincide with the auxiliary line through the vertex of the cone parallel to the axis of the cylinder (58). Then, as will be evident on making the construction, that one of the auxiliary planes, all passing through this line (58), which is tangent to the cone, will either not cut the cylinder or will be tangent to it, or will cut it in two elements. In the first case, it will give no points, being outside of the limiting planes (Prob. XXXVII. , 3°) which include the intersection. In \hQ second 146 DESCRIPTIVE GEOMETRY. case, there will be one infinite branch, and in the thirds two of them. After the construction already given (Probs. XXXVI I., etc.) the foregoing principles are sufficiently illustrated in the next problem. PROBLEM LXXII. — To find the intersection of two cones when the curve has infinite branches. In Space. — Adjust the form or position of the given cones until by applying (118) it is found that there will be infinite branches, and then proceed as in Prob. XXXVII. In Projection.— ?\. XIV., Fig. 1 11 . Let vABk—v'A ' B' and VCbD — V'Ciy be the given cones, vq — v'q' , parallel to the axis VQ — VQ y is the axis, and vr — v'r' (where v'r' is paral- lel to VD') an extreme element of the auxiliary cone at w' and similar and parallel to the cone VV , Its base arh in- tersects AdB^ that of cone vv\ at a and h^ which shows that two elements of the cone VV are parallel to the elements av and hv of the cone w' . Then find X^ the horizontal trace of the line Vv — FV, common to all the auxiliary planes (Prob. XXXVII., In Space)^ and Xad 2Lad Xhk are the hori- zontal traces of the planes of pairs of parallel elements on the two cones. That is, cV—c'V is parallel to av — a'v\ and jV—j'V'\,ohv—hv". These planes also contain the elements whose horizon- tal traces are b and d in the plane Xd, and i and k in the plane Xk^ and therefore, as usual, each gives four points of the curve. Thus va and Vb give e ; vd and Vb give ff; vd and Vc give gg'; and finally va and Vc, the parallel ele- ments, intersect only at infinity in each direction on the in- finite branch of the required curve. Likewise the plane Xk gives the points whose horizontal projections are s, y^ /, and the point at infinity, intersection oihv — h'v' SindjV—fV in both directions. Number and character of the curves. — As both of the aux- iliary planes, Xx and Xz, tangent to cone VV y are secant to cone vv'y there will be two curves, one at which the smaller DESCRIPTIVE GEOMETRY. I47 cone VV enters, and one at which it leaves the larger cone vv'. One of these being composed of points at which all the elements of cone VV\ beginning at w, intersect those of cone vv' met in proceeding from zto x and back again, is a closed curve. The other, embracing the intersections of all the elements of cone VV with those of vv' from vtou and back, includes the intersections of the parallel elements, and is therefore infinite. Other relative positions of the cones might have made both curves infinite. The tangents to the points at infinity, — These are found as in any other case. Thus the asymptote (118, 3°) tm — t'm\ parallel to the parallel elements av — a'v' and Vc— V'c' , is the intersection of the tangent planes at and ct along these ele- ments. The other asymptote Tn is similarly found. When the parallel elements are in a limiting plane, as Xx, the element of contact of that plane itself becomes the asymptote. Connection of points, — (Prob. XXXVII., 2°.) The only peculiarity under this head is the manner of proceeding on passing a plane, as Xd, containing parallel elements. The parallel elements va — v'a' and Vc — V'c' must be considered as meeting indifferently in either of their two opposite di- rections, from V towards a or the reverse ; yet, as a point of the curve, and because two straight lines can only meet in one point, these two infinitely distant points must be con- sidered as one point. This is clearly seen by taking aux- iliary planes very near to Xd, but on opposite sides of it. Call F, and K, such planes, and let F, be behind Xd—AhdX is, on the same side with Xx. The point ^„ then determined in F„ by the elements near a and r, will be very far behind Xd. Next Xd itself will give the point x^ at infinity. Then in F„ the elements near and in front of a and c will inter- sect at jTa very far in front of Xd. This shows that the curve passing successively, without reversing its direction, through j^„ ;tr„ j„ the point x^ must be considered indiffer- ently as infinitely far before and behind Xd, Examples. — 1°. Make Xd coincide with Xx by changing the vertices so that uv and p V shall be parallel. 148 DESCRIPTIVE GEOMETRY. Ex. 2*. Let the given cones be so placed that one only of the tangent auxiliary planes to each shall be secant to the other. Ex. 3°. Place the cones so that the auxiliary cone shall be tangent to one of the given cones. Ex. 4°. Replace one of the cones by a cylinder as in (58) and find the curve. [Devote a whole plate to Prob. LXXIL, and to each of tUese examples, making a complete study.] D — Development. 1 19. The direct development of a cone of transposition consists in developing an inscribed pyramid of many sides. This method is inexact, both in principle and mechanically, and the construction is therefore postponed until, by means of a combination of developable and double-curved surfaces, a better practical solution shall be found. PROBLEM LXXIIL— r^ develop the convex surface of a cylinder of transposition whose elements are oblique to both planes of projection. In Space. — The solution embraces these three distinct operations. 1°. Find the intersection of the cylinder with a plane perpendicular to its axis by (Prob. XXXIV.), since only such a section will develop into a straight line showing the circumference of the right section. 2°. Find the true size of this right section by revolving its plane into or parallel to H or V as in (Prob. XL.) 3°. Find the true lengths of the segments of elements between the right section and the base or any other given section, as in (Prob. XIL) In Projection. — 1°. Lay off on a straight line a distance equal to the circumference of the true size of the right section, dividing the latter for this purpose into a sufficient number of parts, a, b, c, etc. 2°. Find, as described, the true lengths of the elements DESCRIPTIVE GEOMETRY. I49 through these points of division, and make them all per- pendicular to the developed right section at the corre- sponding points ^„ b^y ^„ etc., and connect their extremities, which will give the developed circumference of the base or other limiting section chosen (33). As the construction requires no new operations, it is left as an example. PROBLEM LXXIV.— r^ develop a helix by means of its curvature. In Space. — Since both of the component motions of the generatrix of a helix are uniform, the rate of curvature of the cur^e must be so also, hence when this rate, as shown by the radius of curvature, is known, the helix can be de- veloped as a circular arc. In Projection.— ^QQ first PL XIV., Fig. 108. Let €e\ff, 7in' be three equidistant points on a helix, and let the plane jbf be perpendicular to both H and V. Suppose a circle to be passed through these three points, e'fn' is on its ver- tical projection, and its diameter, vertically projected at/', is the hypothenuse of a right-angled triangle in which fn — fn' is one side. Drawing the chord en — e'n' y denoting the points in space by the capitals of the same letters used on the figure, and calling D the diameter of the circle EFN^ we have (FN)' = Dxfo. Likewise in the triangle fnd^ calling fd = 2r, (fn y = 2r X fo. Divide these equations member by member, and let n'e'c' =z (3j and -i— = — cos^fi ~~ 2r Now as N successively approaches and finally coincides with Fy the angle (i approximates and at last equals a, the I50 DESCRIPTIVE GEOMETRY. angle made by the tangent at F with H, and at the same time D approaches and finally becomes 2/? the diameter of the osculatory circle containing E, F and N when these points become consecutive, and whose radius is therefore the radius of curvature of the helix ; hence, finally, cos a This value is easily constructed as follows: Fig. 107. Lay off Q"'p^ Oc = r; draw pq perpendicular to QQ!" ; draw qr perpendicular to a'Q", and Q^'r will be the value of /o. For r = Q'"P = Q"'q cos a, or Q'^q = ^ but 0-/ : G-^ : : Q-^q : Q"'r hence Q"'r = p = — 5— Therefore, finally, an arc of radius Q"^r and length equal to Q'^V will be the development of the helix acg — a'c'g'Q\ Fig. 107, by means of its curvature. PROBLEM LXXV.—To develop so much of the surface of a developable helicoid as lies between two planes perpendicular to its axis. In Space, — The helix is developed by means of its curva- ture, since the elements* tangent to it in space must, to pre- serve their relative position, be tangent to its equivalent plane curve (60) of the same curvature (98). Also the re- quired portions of the elements being equaU will be so in development, and of their true length. In Projection. — To develop the helicoid. Fig. 107. PL XIV., Fig., 109. Describe a circle Oc of a radius equal to Q!"r, Fig. 107. On it lay off the arc acgq, equal to Q"'a'y Fig. 107, which will, by the last problem, be the proper DESCRIPTIVE GEOMETRY. I5I development of the helix in Fig. 107. At the points a, b^ Cy d, etc., of this arc, corresponding to aa\ bb\ cc\ etc.. Fig. 107, draw the tangents aQy afiQ^y <^/0a» ^tc, each equal to Q!"a'y the true length of an element, and placed as in Fig. 107 — that is, so that the same point of the helix and of a tan- gent shall coincide in space and in development. Then will a^b = Arcab ; a^c = abc ; Q^c = qfdCy etc., making the de- velopments of the involute bases, involutes of the developed helix beginning at a and q. The shortest distance between two points in a plane being straight, the development of the shortest path on the heli- coid between any two of its points will be a straight line joining the developments of those points. From this de- velopment, the projections of the same line can easily be constructed. 120. Examination of consecutive elements. — Several exam- ples of the apparent contradictions that arise whenever an infinity is involved have already occurred : The ques- tion whether a tangent, as requiring two points to determine ity shall be defined as containing two consecutive points, or, not being a secant, as containing only one point of a curve (52) : That of the parabola as expanding in width to infinity, yet as necessarily cutting the elements consecutive to the one to which it is parallel on the cone containing it (36, 2°) : That of connecting the points of infinite branches of intersections where one point of the curve is, in con- struction, either of two points at an infinite distance apart (Prob. LXXII.) 121. A new and interesting illustration of the last article occurs in connection with the developable helicoid. De- fined as the limit reached when the edges of a develop- able polyedron, PI. XIII., Fig. 104, become consecutive, it is plainly developable. Yet tangents to a curve of double curvature (97) are not in the same plane. But see now PI. XIV., Fig. no, where ab — a'b' is a curve of double curvature placed in the simplest position — that 152 DESCRIPTIVE GEOMETRY. is, SO that the tangents at ^^'and bb' shall be parallel to H, while one of them, that at aa' , is perpendicular to V. The line n — a!n' is the common perpendicular, or shortest distance between these tangents. Then, drawing the chord ab — a'b\ we have ' : ' a'n' = b'c' = a'c' . tang, b'a'c' = ad. tang, bad, tang, b'a'c' , Now as the point bb', continuing on the curve, ap- proaches aa' y the plane H must rotate about the tangent ad — a' in order to remain parallel to the tangents at aa' and the moving point bb\ and the limit of its successive posi- tions will be the osculatory plane containing the three con- secutive points of which aa' is the middle one. Also the plane aab' of the tangent at aa' and the point bb' will, in the same approach of b to a, have the same limit. Conse- quently the angle b'a'c' and the distance a'n' , and hence all the factors of the above equation, become infinitely small together. Therefore a'n' becomes an infinitesimal of the third order — that is, as each of the factors is then an infi- nitely small fraction of any finite unit, their product is an infinitesimal part of an infinitesimal. Hence when infinitesi- mals of different orders occur together, the higher can be neglected in comparison with the lower, but not with re- spect to each other ; as ordinary ones are, as compared with finite quantities. Accordingly, we may say, for all pur- poses involving finite quantities, that consecutive tangents to a curve of double curvature do intersect. DESCRIPTIVE GEOMETRY. 1 53 J.IBR AU Y > UNlVKKSn V <>}^ OAl.lI^X)i;MA. CHAPTER III. WARPED SURFACES OF TRANSPOSITION. Principles. 122. Warped surfaces of transposition are of endless variety, depending on the conditions to which the motion of the generatrix is subject. But these conditions may all be reduced to the two following. A warped surface may be generated — 1°. By a straight line, (7, moving upon three fixed lines Cy C„ C^; either of them straight or curved. 2°. By a straight line, (7, moving on two fixed lines and parallel to a cone, called the cone director, 123. The first method. — If any point c of a curve C be made the vertex of a cone whose directrix is C^y this cone will, in general, cut any portion of C^ in only one point, as c^. Then the line cc^y resting on C, 6',, C^j will be fixed. But proceeding likewise with a point c\ consecutive with c on the curve C, the new element ^V/, consecutive with cc^, could not generally be in the same plane with cc„^ when the three curves (7, C^y C^ were arbitrarily chosen. Hence, A warped sicrf ace may be ge7terated by a line moving on three given fixed lines. Supposing then any warped surface to be given, gen- erated under any conditions whatever. Any three curves traced upon it may be considered as the curves Cy 6^,, C^y bywhichy as directrices y it might have been generated. If the three fixed lines are straight, the cone described will become a plane (64). 154 DESCRIPTIVE GEOMETRY. 124. The second method. — If through any point/ in space, lines be drawn parallel to all the elements of a warped sur- face, they will constitute a cone to which all the elements will be parallel, and which is called the cone director. Now, if any point c, of any curve C, be made the vertex of a cone V similar to a fixed given cone V^, the cone V will in general cut any assumed curve C^ in only one point c,. Hence the line cc^, resting on the curves C and (7„ and parallel to some element of the cone V^, is fixed. As before, the consecutive line c'c^ could not generally be in the same plane with cc^. Hence — A warped surface may be generated by the second method ; andf however actually generated^ in any given case, any two curves, C and C^, traced upon it, and a cone having elements parallel to its elements, may be taken as its two directrices and cone director. 125. The simplest case under the first method, is when all the three directrices are straight, and in particular at equal perpendicular distances, taken in one plane, from a fourth line. This giwQSihQ warped hyper boloid of revolution. (Prob. XLIII.) The simplest case under the second method is when both directrices are straight, and the cone director reduces to a plane, called the plane director. This gives the surface called the hyperbolic paraboloid. 126. We shall illustrate warped surfaces of transposition by the following examples, taking those which differ most from each other, and are of most frequent use : I. Hyperbolic paraboloids. 2. Elliptic hyperboloids. 3. Helicoids. 4. Conoids. 5. Warped arches. 6. General warped surfaces. For convenience of comparison, the first five of these are here defined together. 127. 1°. In th.Q hyperbolic paraboloid, the generatrix moves upon two fixed straight lines, its directrices, and is always parallel to a fixed plane, \\.s plane director. DESCRIPTIVE GEOMETRY. 1 55 2°. The elliptic hyperboloid differs from that of revolution (^£) in that its plane sections, perpendicular to the axis, are similar ellipses. 3°. The warped helicoid is generated by a straight line moving upon a helix and its axis as directrices, and making a constant angle with the axis. 4°. The conoid has two directrices and a plane director, but one of the directrices is curved. 5°. The warped arch has three directrices, viz., two circles in parallel planes, and a straight line, which, together with the line joining the centres of the circles, is in a plane per- pendicular to those of the circles. Theorem X. — The hyperbolic paraboloid is doubly ruled, or has two sets of elements, PI. XIV., Fig. 112. Let Ka and Mb be two lines not in the same plane, and let ab, DE, KM, be three positions of a line moving on Ka and Mb, and always parallel to the plane HP, The moving line will then generate a hyperbolic paraboloid. Now, let F be a plane, parallel to Ka and Mb, and cutting ab, DE, KM at c, F, N, and let Mbm and Kak be planes parallel to V, through the lines Mb and Ka. These three planes will then cut HP in the parallel lines GL, bin and ak. Again : drawing Ff and Nn perpendicular to GL, pass planes through Z^/^and iy and through iTA^and Nn, They will cut HP in the lines df and kn, parallel to DF and KN, and the planes Mbm and Kak in the lines Ee and Dd equal and parallel to Ff, and Mm and Kk equal and parallel to Nn, We shall now show that NFc is a straight line. The triangles Kak and Mbm give : ad',ak :: Dd: Kk (i) and be : bm w Ee : Mm (2) But by reason of the equality of the third and fourth terms ad\ ak : : be -, bm (3) 156 DESCRIPTIVE GEOMETRY. which shows that aCy df, kn all meet in one point, hence also ad\ ak : : be \ bm : : cf\cn (4) and as //" = Dd or Ee, and Nn = Kk or Mm. cf '.en :: Ff -. Nn (5) which shows that NFc is a straight line, and hence that the same surface can be generated, either by the motion of ae, parallel to HP, and upon Ka and Mby or by the motion of No parallel to F, and upon NK and FD. That is, the hyperbolic paraboloid is doubly ruled, as stated. 128. Consequences of the double generation. — The hyperbolic paraboloid is seen by the last theorem to have two plane directors, one for each set of elements. These two sets are distinguished as those of the first and second generation. More- over, it is now evident that any element of either generation inter- sects all those of the other, and hence that the directrices are no peculiar lines, but that any two elements of either generation may be taken as the directrices of the other. When the plane directors are perpendicular to each other, the surface is called right, or isosceles ; otherwise, it is called scalene. 129. Proportional division of elements. — Substituting aD and aK, bE and bM, ^/^and cN ior ad, ak, etc., in the proportion (4) of Theor. X., we see that the elements of one generation divide those of the other, and hence the directrices, pro- portionally. Also by drawing perpendiculars to GL from a and b, passing planes through them and aK and bM respec- tively, and passing other planes through FD and KN paral- lel to HP, we can directly demonstrate as before, that abibc iiDEiEF i: KMiMN. 130. Vertex. Axis. — As the elements parallel to Fmake larger angles with HP as we proceed from Ka towards V, DESCRIPTIVE GEOMETRY. 1 5/ there will be one of them whose direction is perpendicular to that of GL. Also as indicated by ac and kn, some element between ac and KN will be perpendicular to the direction of GL. The intersection, v, of these elements, one of each generation, is called the vertex of the surface. A line through v, parallel to GL the intersection of the plane directors, is called the axis of the surface. A — Projections. PROBLEM LXXVI.— r^? construct the projections of a hyper- bolic paraboloid, having given the directrices and plane director of one generation. In Space. — The line joining the points in which the plane director cuts the directrices will be one element. The line joining the points in which a parallel plane cuts the same directrices will be a second element. Having thus two elements, apply the principle of (129), and lay off on each directrix parts equal to those inter- cepted .on it between the parallel planes, and the lines joining the points of division will be elements. Or, for better accuracy, let the parallel planes be as far apart as possible, and divide the space between them on each direc- trix into the same number of equal parts. To construct elements of the second generation. — Pass a plane by (Prob. IX., 4°) through one of the given directrices and parallel to the other. This will be the plane director of the second generation, and any two of the elements before con- structed will be its directrices ; and we can then proceed as in constructing the elements of the first generation. In Projection.— ?\. XIV., Fig. 113. Let AB—A'B' and CD — CD' be the directrices, and PQF the plane director. By (Prob. XL, a) the plane PQF cuts AB—A'B' at 00', and CD — CD' at 00' . Then 00 — o'o' is an element of the re- quired surface. RS, parallel to PQ, is the horizontal trace of a plane parallel to PQF as shown by making ci — c'\' and 158 DESCRIPTIVE GEOMETRY. gd—g'd' parallel to ab — a'b' and ef—e'f'y these being the intersections of these planes with the auxiliary vertical planes containing the directrices. Thus the plane RS cuts the directrices at i,i' and dd\ giving \d—\'d' for a second element. Finally, lay off on AB — A'B' spaces 12 — \'2\ etc., each equal to 01 — o'l', and on CD — CD', spaces dl—d'l', Iq — l'q\ etc., each equal to od—o'd\ and the lines, as 3^ — 3'/, joining the like points, will be other elements, any number of which can thus be constructed. Examples. — 1°. Vary the figure, to become familiar with the surface by its projections, by taking numerous and widely different positions of the direc- trices and plane director. Ex. 2°. Suppose both plane directors to be vertical. That is, assume the directrices so that elements of each generation shall be parallel in horizontal projection. Ex. 3°. In PI. XIV., Fig. 113, construct an element of the second genera- tion. [The above solution in space being only a repetition of Prob. IX., 4°, and of the above operations, taking any two of the elements there found as directrices, the construction is made an example.] PROBLEM LXXVII. — Having one projection of a pohit on the surface of a hyperbolic paraboloid^ to find its other pro- jection. In Space, — Pass a plane through that projecting line (Prob. XL, by i°) of the point, which determines the given projection, and find the intersection of this plane with the surface. The intersection of this line with the given pro- jecting line will be the point whose required projection will thence be known. In Projection. — PI. XIV., Fig. 113. Let / be the given projection of a point on the paraboloid. A line pk in any convenient direction will then be the horizontal trace of a vertical plane containing the projecting line whose projec- tion is /. This plane intersects the elements of the para- boloid, giving the curve knr — k'nW ^ on which / is vertically projected at/', the point required. Example. — Let />' be given, and find/. DESCRIPTIVE GEOMETRY. 1 59 131. Four straight lines intersecting at four points, not in the same plane, form a warped quadrilateraL The opposite sides of this quadrilateral, as A'B' and C'D\ PL XIV., Fig 114 (the vertical projection), are pairs of ele- ments, one pair of the first, and the other of the second generation of a hyperbolic paraboloid, for through A'B\ for example, a plane can be passed parallel to CD\ which will be the plane director of these elements. Likewise, a plane can be passed through A'C, parallel to B'D', which will be the plane director of these elements. PRO BLE M LXX VI 1 1 . — Having given a hyperbolic paraboloid by a pair of elements of each generation^ and having also a point of the surface ^ to construct an element of either genera- tion through that point. In Space. — Since all the elements of the same generation are parallel to the same plane, a pair of lines through the given point, parallel to elements of the first generation, will determine a plane through that point and parallel to the plane director of that generation. Hence the intersection of this plane with the elements of the other generation will be points of an element of the first generation. In Projection.— Y\, XIV., Fig. 114. Let AB—A'B' and CD — CD' be elements of the first generation, and AC — A'C and BD — B'D' those of the other, and let //' be the given point. In practical cases, pp' may have been found as in the last problem, more elements of either generation being given than are here shown. Through //' draw pa—p'a' and pd—p'd', respectively parallel to AB—A'B' and CD— CD', and they will deter- mine a plane containing an element through//'. By (Prob. XI., b, 2) this plane cuts AC— A'C at qq' , and BD—B'D' at rr' , hence qr — q'r' is an element of the first generation, through //'. Likewise, pc—p'c' parallel to AC— A'C, and pb—p'b'^ l60 DESCRIPTIVE GEOMETRY. parallel to BD — B'D\ determine the plane containing the element ns — n's' of the second generation, through pp' , 132. A straight line, moving upon three straight lines which are parallel to a plane, but not to each other, gener- ates a hyperbolic paraboloid. For, let PQ, PI. XV., Fig. 117, be the plane, and cdy efy gh, the three given lines, and let ga and hb be any two positions of the line which moves upon them. Then cd^ ef, gh are evidently three elements of the hyperbolic paraboloid of which ag and bh are directrices and PQ the plane director ; whence (Theor. X.) ag and bh are elements of the other generation of the same surface. PROBLEM LXXIX.— r^ construct the projections of a warped elliptical hyperboloid. In Space. — Parallels of a hyperboloid of revolution (33) appearing in horizontal projection as concentric circles, while those of the elliptical hyperboloid appear as similar and similarly situated concentric ellipses, the appropriate construction of the latter is by (Prob. XXL, 2°) as oblique projections of the former. In Projection. — PL XV., Fig. 116. The surface is suffi- ciently illustrated by the figure, which shows the quarter below the gorge a'b' , and in front of a vertical plane AB^ through the axis O — C'O^ of the surface. Let AB and 2CO be the axes of the elliptical base ACB — A'B\ and ab — a'b' the transverse axis of the gorge. Oc, the semi-conjugate axis of the gorge, is then a fourth proportional to OA^ OC, and Oa^ found by drawing a line ac parallel to A C. As a pair of tangents to the gorge issue in horizontal projection from every point of any other parallel alike for the circular and the elliptic hyperboloid, we construct the ellipse ACB by (Prob. XXL, 2°), and /, g, C, h, etc., will then be points symmetrically distributed relative to AB and 2OC, and from which the horizontal projections of elements DESCRIPTIVE GEOMETRY. l6l fm and fk^ ga and gB, etc., will issue, as shown, tangent to the horizontal projection of the gorge. The vertical projections of elements. — First method. By the method of (Prob. XXI I.), find the exact points of contact ;/, r, etc., of elements, mC^ Ah^ etc., and project them as at n' r\ etc. Also project A, f, g, Cy etc., at A', /', g', C, etc. Then the two elements issuing from A A' will be vertically projected in A'r\ because they are symmetrical relative to the vertical plane on AB. The two issuing from ff will be projected in fn' and/C^', the first in fn'^ because the two issuing from mm' and coinciding in horizontal projection with/;;^ and Cm are, as there seen, tangent to the gorge at n and at a point symmetrical with it relative to AB, The two issuing from gg' are vertically projected at g'a' and g's', and so on. Second method. — ^^Draw a line A" B" equal to A'B' and symmetrical with it relative to a'b\ and project A.f^g, C, h, etc., upon it at A",f',g", etc., as well as upon A'B', as be- fore done. Then join the same letters, one on A'B' and one on A"B" , that are joined in horizontal projection. That is, draw A'h" and A"h' , whose intersection will be the point r'r of the gorge, etc. In the figure, hb is perpendicular to AB because Ob is half of OBy while HB is an arc of 60°. Any other propor- tions and divisions of AHB might have been chosen. Examples. — 1°. Make the projections of the entire surface. Ex. 2°. Make the projections of the entire surface with AB perpendicular to the ground line. PROBLEM LXXX. — To construct the projections of an oblique helicoid. In Space. — The construction is founded immediately on the definition, according to which the rectilinear elements join points of equal division on the helical directrix with corresponding points of equal division on the axial direc- trix. l62 DESCRIPTIVE GEOMETRY. In Projection, — PL XV., Fig. 115. Let 0369 — o'3'6'9', con- structed as in (Prob. LXVIL), be the helical directrix, and O — O'O" the axis of an oblique helicoid, and let oO — do" , making any given angle, o'd'O' ^ with the axis, be the initial element, parallel to V. Then set off on the axis from o" spaces, equal to those laid off from O' in constructing the helix, to mark the uniform ascent of all points of the gen- eratrix. The horizontal projections of elements are then radial lines Oo, (9i, etc., of the circle 0369, and their vertical projections are oV, I'l", 2^2'^, etc. Contour and visibility, — The apparent contour in vertical projection is a curve — not shown — tangent to the successive vertical projections of elements. In horizontal projection, every radial represents the visible parts of one or more elements. Representing the surface as a warped polyedron, as in the figure, each element beginning at a visible point of it, in H readily seen by inspection, continues visible upward, as shown in the figure (supposing elements in the fourth angle visible at w!n' and n'a\ till it apparently intersects the next, or disappears at its apparent intersection with the apparent contour. 133. The oblique helicoid has a cone director (122) whose axis is that of the helicoid, and whose elements make the same angle with that axis that is made by those of the heli- coid. When the angle just mentioned is one of 90°, the heli- coid is called a right helicoidy and evidently has a plane director perpendicular to its axis. Besides these helicoids, there is one of a more general form, in which the axis is replaced by a cylinder, separate from that containing the helical directrix, and to which the elements are tangent, while also resting upon the helix. In this case these two cylinders have a common axis, but may be of any form, of revolution or not. DESCRIPTIVE GEOMETRY. 163 PROBLEM LXXXI.^ To construct the projections of a conoid. In Space. — In practice, the plane director may generally be made a plane of projection, when the elements will be immediately determined by their projections parallel to the ground line, on the other plane of projection, which will then show where they intersect the given directrices. In Projection. — PI. XV., Fig. 119. Let H be the plane director, A—A'B' the rectilinear, and CD—C'ED' the curved directrix (127, 4°). Parallels to the ground line CF\ as G'h\ Lk\ etc., are then the vertical projections of elements, whose horizontal projections are found by projecting G , L\ etc., at G^ Z, etc., and drawing GAy LA, etc. Examples. — 1°. Let the rectilinear directrix be oblique to the plane - director. Ex. 2°. Let the directrices and the plane director have any oblique posi- tions in space. [Take any three or more lines in the plane director by Prob. VII., and draw parallels to them from any point, /, on the curved directrix. Then find by (Prob. LXXXIV.) where the plane thus determined cuts the straight directrix, and the line from this point to the assumed point p will be an element.] 134. Right conoid. — When the rectilinear directrix of a conoid is perpendicular to the plane director, the surface is called a right conoid^ and this directrix, the line of striction^ as containing the shortest distances between the elements. In fact, every warped surface has such a line, but it is generally curved. The conoid is a very comprehensive form of warped surface, and includes some of the cases of other warped sur- faces. Thus a right helicoid is also a form of right conoid by the definition of the latter. PROBLEM LXXXII.— Tb construct the projections of a warped arch, also called the Cow's Horn. In Space.— The plane to which the curved directrices are both parallel (127, 5°) may be taken as a plane of projection, 164 DESCRIPTIVE GEOMETRY. when their intersections with any plane containing the straight directrix will be immediately obvious, and will be the points which determine the element in such a plane. In Projection. — PL XV., Fig. 121. Let the semicircles AB—A'D'B' and CE'—CD'E\ this one in V and the other parallel to it, be the curved directrices, and let OF—F\ in H and perpendicular to V, be the straight directrix. Then, at once, any lines, as Fa'c' and F'fe'y radiating from F\ and intersecting both semicircles, are vertical projections of elements. Their horizontal projections are found as at caO by projecting c' at c and a' at a' on the horizontal projection of the proper semicircle and joining the points so found. The element at U is horizontal ; those to the right of it, as ef—e'ff meet OF behind V, while those to the left of it, as ac — ^V, meet OF in front of V. . When the distance ^'C' becomes nothing, the surface be- comes a cylinder. When it equals CE' by the coincidence oi E' with A\ the surface reduces to two oblique cones tan- gent to each other on the line AA\ Examples. — 1°. Let OF, still in H, be oblique to V. [The traces of planes, as OFm' , upon the parallel planes AB and C D\ instead of coinciding as in Fm\ will be parallels from the then separate vertical projections of D and /".] Ex. 2°. Draw the figure with A' to the left of C. 135. The most general surface of which the warped arch used in practice is a specific example, would be one having any three directrices, one of them straight, and in any posi- tion. 136. The foregoing are all the warped surfaces which are of any considerable use in the arts, and nearly all which have received specific names and investigation. One, termed the cylindroidy and which may serve as the surface of a descend- ing arch connecting, transversely, two parallel arched pas- sages on different levels, is thus derived. Imagine a cylin- der whose elements are parallel to the ground line. Intersect it obliquely by two vertical planes lying on opposite sides of the plane of right section, as at Cm and Cpy relative to DESCRIPTIVE GEOMETRY. * 165 COy in PL XV., Fig. 116. Elevate one of the oblique sec- tions, Af keeping- it in its own plane and so that all its points shall ascend equally to its new position A\ the other section B remaining fixed. Then the lines joining the same points of A' and B^ that were joined on A and B by the elements of the cylinder, will still be parallel to V as a plane director, but will be elements of a warped surface called a cylindroid. General warped surfaces not having specific names, and of either of the kinds mentioned in (123, 124) are of occasional use. The following examples illustrate the methods of con- struction employed in such cases, as well as the character of such warped surfaces in general. PROBLEM LXXXIIL— r^ construct elements of a general warped surface having three given directrices. In Space, — The construction directly illustrates the defi- nition of such surfaces (123). In Projection.— V\, XV., Fig. 122. Let AB—A'B', CD—CD\ EF—E'F' be the three directrices. On any one of them, as AB — A' B' ^ assume any point as mm! , and join it by straight lines with points as aa' ^ bb\ cc\ taken on either of the other directrices. These lines form a conic surface which intersects the cylinder which projects the remaining directrix EF — E'F' on H in a curve whose hori- zontal projection, def coincides with EF, and whose verti- cal projection, d'ef\ is found by projecting d, e, f upon m'a\ m'b' , m'c' , This curve intersects EF—E'F' at nn' , which is thus the point where the conic surface cuts EF—E'F. Hence the line mn — m'n' by construction intersects the three directrices, and is evidently the only one through mm' that does so ; it is therefore an element of the required warped surface. Other elements can be similarly found. Examples.— 1°. Let the given point mm' be on either of the other directrices. Ex. 2°. Let one or more of the directrices be in other angles than the/rj/. l66 DESCRIPTIVE GEOMETRY. PROBLEM LXXXIV.— T;? construct elements of a general warped surface having two directrices and a plane director. In Space. — Any element may either be made parallel to a given line in the plane director^ or through 2^ given point on one of the directrices. In the former case^ it will lie on the surface of a cylin- der whose elements are parallel to the given line, and which has one of the directrices for its base.. In the latter case, the element sought will lie in a plane parallel to the plane director, through the given point. A determining point of the element will be where the cylinder in the first case, and the plane in the second, cuts the other directrix. In Projection.— ?\. XV., Fig. 123. Let AB—A'B' and CD— CD' be the directrices, and MN—M'N' the plane director of a warped surface. Then for the first case, ab — a'b' being a given line in the plane director, draw the elements ad — a'd'y be — b'e' y cf—c'f of the auxiliary cylinder parallel to it, and having AB — A'B' for its directrix (94), and find, just as in the last problem, the point nn' where this cylinder cuts the other directrix. Then mn — m'n' , parallel to ab — a'b' , is the element required. For the second case, had the element been required to pass through aa^ for example, we should first have taken several other lines besides ab — a'b' in the plane director, and should then have drawn parallels to them through aa'. These lines would have formed a plane parallel to MN—M'N' and containing aa\ Its intersection with CD — CD' , found exactly as in the two preceding construc- tions, would give the point which, joined with aa' , would be the element required. Examples. — 1°. Make the construction just described. Ex. 2°. Vary the position of the plane director. B— Tangencies. 137. Every tangent plane to a warped surf ace is i7i general a secant plane also. This arises from the property, more or DESCRIPTIVE GEOMETRY. 1 6/ less obvious on inspection of any of the warped surfaces already shown, that any two sections made by secant planes at right angles to each other, through any point, /, of the surface, will generally be curves convex in opposite direc- tions, as shown on the hyperboloid of revolution (Theor. VIII.). Hence the tangents to these curves at p must be on opposite sides of the surface, and hence the tangent plane determined by them must also be a secant plane. 138. Every tangent plane to a warped surface is in general tangent only at a point of contact. For as the elements of a warped surface are never, except in the case of the two sets on the hyperbolic paraboloid, parallel to more than one plane, and are generally parallel to none, a plane passed through any element will, in general, be oblique to all the others, and will therefore cut them all in points forming a curve which will intersect the given element at some point c. Then this element, and this curve, being considered as two sections of the surface at r, the element which is its own tangent (44) and the tangent to the curve at c determine the plane which is tangent to the surface at c and there only. This plane is thus, as well as in (137), shown to be also a secant plane. 139. It follows, from the last article, that every plane con- taining an element of a warped surface is a tangent plane to the surface at some point of that element — viz., at the point where the element intersects the curve cut from the surface by the same plane. Also that as the plane revolves about this element as an axis, its point of contact with the surface shifts along the element. The exception to the two preceding articles is that of a tangent plane parallel to the plane director, when the latter exists. Such a tangent plane has an element instead of a point of contact, and is not also a secant plane. This is clearly illustrated by PI. XV., Fig. 119, where there may evidently be a tangent plane along the whole ex- tent of that element, EA — E'A', of the conoid, which is 1 68 DESCRIPTIVE GEOMETRY. furthest from the plane director, and it is not a secant plane, since there 'are no two sections through any point on AE — A'E'y such that one would be convex in an opposite direction to the convexity of the other. 140. Raccordment, — The mutual tangency of two warped surfaces which are in contact along a common element, is often distinguished as their raccordment. They raccord along this element. When two warped surfaces have a common tangent plane at each of three points on an element common to both, they are tan- gent at all points of that element. For (see PI. XV., Fig. 120) let P, Q, R be the points of contact of such common tan- gent planes on the common element PR, Then any secant plane at each of these points will cut from the two warped surfaces a pair of curves tangent to each other, as Pa and Pb at P; Qc and Qd at Q ; Re and Rf at R. Now, as three directrices determine a warped surface, while the two curves of each pair have no points in common but P, Q, R, their points of contact ; if PR moves on Pa, Pc, Pe, it will generate a warped surface everywhere separate, except on PRy from that generated by the motion of PR on Pb, Pd, Pf as directrices. Hence, as the two are tangent, and not secant at P, Q, Ry they will be so at every point of PR, 141. When two warped surfaces have a common plane director, common tangent planes at two points on a com- mon element are sufficient to determine the surfaces as mutually tangent all along that element, since two direc- trices and a plane director determine a warped surface, and the reasoning is otherwise the same as in the last article. 142. Any number either of warped hyperboloids or of hyper- bolic paraboloids may raccord with any given warped sur- face, along any given element of it. This property follows from (Theor. VI.) and (132). For (see PL XV., Fig. 120) the three pairs of curves at P, Q, R may be taken in three parallel planes, or they may not. DESCRIPTIVE GEOMETRY. 169 143. If they are so taken, their common tangent lines at P, Q, R will be parallel to one plane, and the motion of PR on these three tangents will generate a hyperbolic parabo- loid (132), which will raccord with both of the given warped surfaces along PR, See Fig. 118, where the direc- trices AB, CD, EF, and hence their tangents, are all paral- lel to V, hence a raccording hyperbolic paraboloid on BF will be generated by the motion of BF upon these tan- gents. 144. If the three pairs of curves are not in three parallel planes, neither will their common tangent lines be so, and the raccording surface generated by the motion of PR upon these tangents will (Theor. VI.) be a warped hyperboloid. For three straight lines must, either two or all of them, be parallel to some one plane, while the hyperbolic paraboloid and the warped hyperboloid are the only warped surfaces, all of whose directrices are straight. 145. Moreover, in the former case, an infinite number of sets of parallel planes through P, Q, R may be taken, each of which (143) will contain a pair of curves tangent to each other at those points, and consequently having common tangent lines at those points. One of these sets may be perpendicular to the plane director, to which the corre- sponding positions of the generatrix PR remain parallel in moving on these three tangents. That is, one of all the hy- perbolic paraboloids of raccordment may have plane direc- tors perpendicular to each other, and thus be (128) a right hyperbolic paraboloid. 146. Owing to the ease w'ith which planes can be drawn, tangent to the warped hyperboloid (Prob. XLVI.) and in the same manner to the hyperbolic paraboloid, tangent planes to other warped surfaces are sometimes most con- veniently drawn by substituting for them one of these as a raccording surface along the element containing the given point of contact. I/O DESCRIPTIVE GEOMETRY. 147. Since the tangent plane to a warped surface has generally (138) only a point of contact, the principle of (45) applies, and any two of all the tangent lines at the point of contact determine the tangent plane at that point. In the two warped surfaces of double generation, the hyperboloid and hyperbolic paraboloid, the two elements at the given point, may be taken as these determining tangents (Theor. VIII.). In other warped surfaces, the single element at that point, with any other tangent, will determine the tangent plane (138). PROBLEM LXXXV.— 7b construct the tangent plane at a given point of a hyperbolic paraboloid. In Space. — Construct an element of each generation through the given point ; their plane will be the tangent plane required. In Projection. — PL XIV., Figs. 113, 114. h^ipf on either figure be the given point. Then, remembering (128) that the directrices of either generation, as AB — A'B' and CD — CD\ Fig. 113, are elements of the other generation, construct by (Prob. LXXVIII.) an element of each generation through pp', and the plane of these two lines (147) will be the re- quired plane. Examples.— 1°. Complete the construction for PI. XIV., Fig. 113. Ex. 2°. Complete the construction for PI. XIV., Fig. 114, Ex. 3°. Let H be the plane director, and let the directrices be parallel to V, and construct a tangent plane at a given point of the surface thus given. PROBLEM LXXXVI. — To construct the tangent plane at a given point of an elliptical hyperboloid. In Space. — The plane is determined as in (Prob. XLVL). In Projection. — PL XV., Fig. 116. Let //' be the given point. Then two tangents, in and /j, from / to the horizontal projection of the gorge, will be the horizontal projections of DESCRIPTIVE GEOMETRY. I7I the determining elements (Prob. XXII., 2°), whose vertical projections are thence found as in (Prob. LXXIX.). The traces of the required plane are thence readily found. PROBLEM "LXXXVll,— 'To construct the tangent plane at a given point of a warped surface by the direct method — that isy without an auxiliary r according surface, hi Space. — The test of convenient applicability for this method will be the easy construction of a curved section through the given point, to which a tangent line can easily be drawn, which (138), with the element through the given point, will determine the required plane. The helicoid ful- fils this test. Then — In Projection. — PL XV., Fig. 115. Let 2,2' be the given point. From the definition (127, 3°) of the helicoid, every point of its generatrix describes a helix ; and (100) 0369 — o'3'6V is the helix containing the given point. By (Prob. LXVII.) 2P—2'p' is the tangent to this helix at 2,3', which, with the element b2h—'h'h' at 2,2', determines the tangent plane Pgi^ at the same point. If either or both of these lines are inconveniently sit- uated, apply (27, 9°) as in every such case, and employ any one or more lines, each intersecting both of these primary ones, in finding PQP . If the given point of contact were on the axis as at 0,2'\ the tangent plane would be vertical, for the helix described by such a point is the axis, and hence its tangent is (44) also the axis. Thus bhh' is the tangent plane at O2" . The plane PQP' makes a larger angle with H than the element Ob — 2"b' does, since the perpendicular from O2" to PQ (Prob. VII.) is shorter than Ob—2"b', But as the ascent of every helix in one revolution about the axis is uniform, while their radii vary from zero to infinity, the angle made by a tangent plane with H will vary from 90°, as just shown, 172 DESCRIPTIVE GEOMETRY. to the angle made by the elements with H, when the point of contact is at an infinite distance from the axis. Examples. — 1°. Construct the tangent plane at 3,3'. Ex. 2°. Construct the tangent plane at ee\ and at W on tf'-t'f'. PROBLEM LXXXVIII.— 21? construct a tangent plane to a warped surface indirectly — tJiat isy by means of an auxiliary r according surface. In Space, — This method usually consists in finding an ele- ment of each generation of the auxiliary hyperbolic para- boloid, which is tangent to the given surface along the ele- ment containing the given point of contact. The latter element is one of the two required ones, which together determine the tangent plane required. /;/ Projection. — This method is appropriately illustrated in connection with the conoid and the warped arch surfaces, one of which has, and the other has not, a plane director. 1°. The tangent plane to a conoid. — PI. XV., Fig. 119. Let gg' be the given point of contact. AG — Ji!G is the element of the conoid through that point. Then GF—G'F and A — A'B', the tangents, parallel to V at GG' and A,h\ are directrices, and H is the plane director, of a hyperbolic paraboloid tangent to the conoid along AG — h'G', For, as is evident by inspection of the vertical projection, ^6^ — h'G\ in moving on these directrices, can divide them propor- tionally (129) only by continuing parallel to H. Hence AF—B'F' is the horizontal trace, as well as an element of the auxiliary paraboloid, and consequently ^—^y^, parallel to V, and having its horizontal trace in AF, is an element of the second generation tangent at^', counting AG — h'G' as one of the first generation. Hence the plane PQP of these two elements, PQ being parallel to ^6^ (27, 5°), is the de- sired tangent plane at gg', 2°. The tangent plane to the warped arch. — PI. XV., Fig. 121. Let //' be the given point. Oc — F'c\ and OF—F'y the one an element, and the other a directrix of the surface, are two DESCRIPTIVE GEOMETRY. I73 sections of it at O^P ; which, being rectilinear, are their own tangents ; therefore OFc', the plane of these two lines, is the tangent plane at the point 0,F' . Hence mOk — ni'Fk' , in this plane, with ha — h'a' and 7iF—n'c\ all parallel to V, are three tangents, serving (132) as directrices of the auxiliary tangent paraboloid generated by Oc — Fc' in moving upon them. To find, next, another element of the same generation with Oc — Fc\ pass a plane (64, 123) through the directrix mk — in'k\ and a point as hh' of the directrix ha — h'a\ The horizontal trace of this plane is Ohy and its vertical trace ^/ is parallel to m'k\ This trace git' cuts the third directrix, at n'n, gWmg n'h'k' — nhk as the element desired. Hence p'r'—pr^ parallel to V, is an element at //' of the same gen- eration as the directrices (128) ; and PQP\ which is the plane of the elements OC — Fc' and/^— /'/, ^^pp\ of the auxiliary paraboloid, is the required tangent plane at//'. 148. The tayigent to a general warped surface ^ as at nn' y in PL XV., Fig. 122, is determined, as in the previous cases, by the element through that point, together with the tangent at nil' to any curved section, as EF—E'F, of the surface through that point. There is this difference however: the curved section is of unknown properties, and therefore the tangent to it can only be located by the eye, unless some special construction independent of these properties can be made. 149. The following is such a construction, partly shown. Assume points on EF, for example, as /, ^, d, etc., on each side of ;/, and draw secants tif, ne, nd, etc. Note the points, /, on nf, e^ on ne, etc., produced on the same side of ;?, where these secants intersect a circle of any convenient radius, and centre at n. Lay off on f^nf and in the same direction from/j that f is from n^ a distance /,/, equal to the chord nf Do the, like for each secant, and a curve /j^X .... will be formed, which will cross the arc whose centre is /^ at a point ;/i, where w,« is the limit between secants on the one 174 DESCRIPTIVE GEOMETRY. and the other side of the point «, hence n^n is the tangent line at n. The curve f^e^d^ .... is called a trial curve, or *' curve of error." The method by means of such curves is practi- cally as exact as a geometrical construction, and sometimes more convenient. C — Intersections. 150. The intersection of a warped surface with any plane is found by constructing the intersections of its elements with that plane by Prob. XL, and then joining the points thus ascertained. If the warped surface has a plane director, the auxiliary planes should be parallel to that plane. Each of them will cut the given plane in a right line, and the Avarped surface in one or more elements whose intersection with that line will be points of the required curve. 151. The intersection of a warped surface with any ruled surface^ other than a plane, is found by noting the intersec- tion of pairs of elements, one on each surface and in the same plane. Thus, if one of the surfaces be a cone^ each auxiliary plane may contain an element of the given warped surface, and the vertex of the cone (Prob. IX., 2°). If one of the given surfaces be a cylinder, each auxiliary plane may contain an element of the warped surface and be parallel to the axis of the cylinder (Prob. IX., 4°). 152. If both of the given surfaces be warped.the simplest solution for each case must be sought. Thus, if both sur- faces have the same plane director, any auxiliary plane parallel to it will contain an element of each surface, whose intersection will be a point of the intersection of the two surfaces. But, in general, it may be necessary to pass a DESCRIPTIVE GEOMETRY. 1 75 plane through each of a number of elements of one of the warped surfaces, and thence to note where each element pierces the curve cut from the other warped surface by this plane. 153. The plane sections, not straight, of a hyperbolic paraboloid, 2,rQ parabolas, when made by planes parallel to the intersection of the plane directors, and hyperbolas in all other cases. 154. The intersection of a helicoid with any cylinder of revolution having the same axis as the helicoid is a helix. Theorem XI. — The intersection of a right conoid having an elliptical directrix, with any plane parallel to that directrix, is an ellipse, . In PI. XV., Fig. 1 19, CD — CE'D' is a semicircle, and the directrix of the conoid there shown, and cd—c'e'd' is the section made by a plane parallel to CD, Then by similar triangles, eg : EG :: ed : ED or by substitution of equal values for these terms, oY : O'G' : : r'd : R'D' and as R'D'=R'E', the last proportion expresses the pro- perty of the ellipse that the ordinate of the ellipse is to the corresponding ordinate of the circle, as the semi-conju- gate axis is to the semi-transverse axis. The curve cd — c'e'd' is therefore an ellipse. Theorem XII. — The intersection of an oblique helicoid by a plane perpendicular to its axis, is a spiral of Archimedes, In PL XV., Fig. 115, revolve all the elements except oO — do" about O — O'O" until their horizontal projections coincide with o(9. They will then all be parallel to V, and, 176 DESCRIPTIVE GEOMETRY. as they make a constant angle with the axis, these revolved positions will be parallel to each other, in space, and hence in vertical projection also. Then, as the intersections of their vertical projections with O'O" are equidistant, those with the ground line will be so also, making the distances of the latter points from O' in arithmetical progression. Hence returning to their original positions, the distances of o, ^, b, c, etc., from O are in arithmetical progression. But the circular distances 01, 02,03, etc., are in like progression. Hence the base oabc ... of the helicoid, whose plane H is perpendicular to the axis, is a spiral of Archimedes ; for the definition of this curve is that its successive radial distances OOy Oay etc., from a fixed point (9, and its corresponding angular distances 01, 02, etc., from a fixed line Oo are in arithmetical progression. Continuing the process on the opposite side of OOy the spiral terminates at Oy tangent to that element which pierces H at 00\ 155. The last theorem affords a construction of the plane problem : To draw a tangent to a spiral of Archimedes at a given point. Let dd' be the point. Construct an arc of the helix whose horizontal projection is the circle of radius Od and which lies on a helicoid formed by drawing lines by (Theor. XII.) from o, ^, by Cy etc., to meet O — O'O" at any con- stant angle. Then the horizontal trace of a tangent plane at dd' to this helicoid will be the tangent to the spiral at dd\ Example. — Make the construction here indicated. 156. Having a warped surface ABEFy PL XV., Fig. 118, if any tangent (raccording) hyperbolic paraboloid, as BFkgy be revolved 90° about its element of contact BFy it will be- come a normal hyperbolic paraboloid to the surface along the same element. That one of the tangent hyperbolic parabo- loids, one of whose plane directors is perpendicular to BFy is a right one (128), H, to which BF is parallel, being its other plane director. By revolving such a tangent parabo- DESCRIPTIVE GEOMETRY. 1 77 loid 90°, it becomes that one of the normal paraboloids, also a right one, whose elements^ all perpendicular to BF^ are normals to the given surface. In other normal hyperbolic paraboloids at BF^ it would only be tangents at points on BF to sections of them, made by planes perpendicular to BF, that would be normals to the given surface at those points. 157. If from a point in space tangents be drawn to the curves cut from a warped surface by a series of planes containing the given point, they will be elements of a cone, whose vertex is the given point, and which is tangent to the warped surface on the curve connecting the points of contact of the tangents. 158. If a warped surface be intersected by a series of planes, each parallel to a given line, tangents to the result- ing curves, drawn parallel to that line will form a cylin- der, tangent to the given surface and parallel to the given line, and whose curve of contact with the warped surface will contain the points of contact of the parallel tangents. 159. Any plane tangent to the cone (157) or to the cylin- der (158) will be tangent to the warped surface to which these are tangent. This principle in connection with {j6) indicates the solution of the problem: to draw a plane through a given line and tangent to a warped surface. PART II.— DOUBLE-CURVED SURFACES OF TRANSPOSITION. 1 60. Double-curved surfaces of transposition are of innumer- able varieties. Among them we will only mention the ellipsoid of three unequal axeSy PL XVI., Fig. 124, the elliptic paraboloid^ the elliptic hyperboloid of separate nappes, and the serpentine. The elliptic paraboloids and hyperboloids have elliptical sections perpendicular to the principal axis, in place of cir- cular ones as in the analogous surfaces of revolution. The serpentine is a double-curved helicoid, being gene- rated by a sphere whose centre moves on a helix. That is, the surface of the serpentine encloses tangentially all the positions of the moving sphere. Any surface thus formed is called the envelope of all the positions of the generating surface. A cone of revolution, for example, is the envelope of all its tangent planes — that is, of all the positions of a plane which revolves about an axis with which it makes a constant angle at a fixed point A— Projections. PROBLEM LXXXIX.-T;? construct the projections of a serpentine. In Space. — Imagine every point of a helix to be the cen- tre of a position of the generating sphere. The two pro- DESCRIPTIVE GEOMETRY. 1 79 jections of each position will be equal circles. Curves tan- gent to the projections of all the spheres will be the pro- jections of the serpentine. In Projection,— Y\. XVI., Fig. 130. Let heh—a'b'e'k' be the helix on which moves the sphere whose radius is hm—h'm\ The equal circles, shown only in vertical pro- jection, represent positions of this sphere, their centres be- ing at the points aa!, bb\ cc\ etc., of the helix. The horizontal projection of the serpentine consists then of the concentric circles Oo and Om^ equidistant from Ohy that of the helix, by the distance hm — koy the radius of the sphere. Its vertical projection consists of the curves tan- gent to the circles at a\ b\ c\ etc. The curve, ascending from l\ suddenly stops at n\ descends to r\ and again ascends towards 0\ in order to be continuous and yet tangent to all the circles. The like takes place at u' and / and illustrates the occur- rence, sometimes, of a cusp (99) in the projections of the com- plete contour of a surface. The horizontal projection of the point I' of the apparent contour is /, on //, a parallel to GL, Other points of the horizontal projection of the apparent contour, seen in fac- ing V, may be similarly found. B—Taugencies. PROBLEM XC. — To construct a plane through a given line and tangent to an ellipsoid of three unequal axes. In Space, — If any two points of the given line be made the vertices of two tangent cones, each circumscribing the ellipsoid, each cone will, by the known properties of the surface, have an ellipse of contact with the ellipsoid. Every plane, tangent to either cone, will also be tangent to the ellipsoid on its curve of contact with that cone. The two points of intersection of the two ellipses of contact will, l80 DESCRIPTIVE GEOMETRY. therefore, be the points of contact of the two planes, each of which will be tangent to both the cones, and hence will con- tain the given line. In Projection,— V\. XVI., Fig. 124. Let the ellipses whose centres are O and 0\ and whose longer axes, only , are equal, be the projections of the given ellipsoid, OC — O' being its intermediate, and O — O'D' its shortest semi-axis ; and let Vv — V'v' be the given line. For convenience in making the planes of the ellipses of contact perpendicular to the planes of projection, take the vertices of the auxiliary tangent cones at VV and vv\ where the given line is cut by planes through 00^ 2ind parallel respectively to V and H. Then vcd is the horizontal projection of one of these cones, and Va'd^ is the vertical projection of the other one, the omitted projections being unnecessary. Also cd is the horizontal projection of the curve of contact of the former cone, and cRS' is its plane ; and a'd' is the vertical projection of the curve of contact of the latter cone, and b'NM is its plane. The intersection, cR — b'N' (Prob. X.) of these two planes intersects the surface of the ellipsoid, at its two points of intersection with either of the curves of contact. In applying this solution to the sphere, these points are easily found, first in revolved position, the curves of contact being circles, but here it is equally convenient to find both projections of one of the curves of contact. Accordingly, projecting c and d at c' and d' , and e at e' and e"y we have four points of the vertical projection of the ellipse cd. Then ff'y projected from / the middle of cdy bisects c'd' and is an axis of this projection, limited by making o'f ^o'f'y a fourth proportional to Oy, O'D', fd{Oy, parallel toed). Having now its axes, the ellipse c'f'd'f can be easily found, and p'p and q'q, its intersections with the ellipse a'b' , are then the points of contact of the required planes. Finally, taking the point //', Vp — V'p' is the element of contact of the tangent plane at that point with cone VV, and vp — v'p' its contact with cone vv'. Hence the traces, k of the former, and r' of the latter, and also the traces h and DESCRIPTIVE GEOMETRY. l8l ^ of the given line, are points in the traces kh and r'^ of the tangent plane, PQF , 2Xpp\ The tangent plane at qq' may be similarly found. i6i. The above solution is here applied to the general ellipsoid, instead of to the sphere, better to illustrate its utility. The method of Prob. LIII. would here have re- quired the construction of tangents to an ellipse ; while in that of Prob. LI I. the plane of the ellipse of contact of a cylinder parallel to the given line would no longer have been perpendicular to the given line. Examples. — 1°. Construct the tangent plane at qq\ Ex. 2°. Operate so as to find / and q before p' and q. Ex. 3°. Apply the method of this problem to a sphere whose centre is in the ground line. .162. The curve of contact of the serpentine, PL XVI., Fig. 130, with one of its internally tangent spheres, will, in general, not differ sensibly from that great circle of the sphere which is perpendicular to the tangent to the helical direc- trix at the centre of the sphere. The tangent plane to the serpentine at any point of this great circle will then be determined by the tangent to that circle at that point, and the tangent to the helix containing the same point. C — Intersections. 163. The ellipsoid of three unequal axes (PI. XVI., Fig. 124) has two sets of circular sections. Those of each set lie in parallel planes, perpendicular to that of the longest and shortest axes, the central one of these planes for each set being determined by the intermediate axis, and a diameter equal to it. Thus, the planes containing the axis lOC — O' and the diameters 2(9V, or 26^ V, each cut the ellipsoid in circles; and by the known properties of the surface, all planes 1 82 DESCRIPTIVE GEOMETRY. parallel to these will also cut the ellipsoid in circles. The centres of all the circles of each set are in the diameter which is conjugate (Theor. III.) to that, as 2.0'x' y or 20'x", which determines the positions of the planes of these circles. 164. Let two cylinders of revolution, C and 6'^, have a common element of contact, E, and let one of them, t7„ be fixed, while the other, (7, rolls upon it. The original ele- ment of contact, Ey will then describe a species of surface of transposition, which will be a cylinder, being composed of the various parallel positions of E, The right section of this cylinder is the curve generated by the point of contact, e, of the right sections, c and ^„ of C and C^y as c rolls on c^. This curve is called an epicycloid, and if c^ be straight, C^ then being a plane, the curve is a cycloid. PL XVI., Fig. 129, shows the epicycloid, o«a;<5^«/, generated by the point of contact, o, of circles O and Ao\y as circle O rolls to the right from o to 4. The construction directly embodies the definition. Thus set off from o equal spaces on both circles ; draw arcs with 0, centre of the fixed circle as a centre, through the points on circle (9, and through O ; note (9„ O^y etc., the intersection of radii through Q and the points I, 2, 3, 4, on the fixed circle, with the arc 00^; draw circles, equal to circle (9, with (9„ O^y etc., as centres, and their intersections with the arcs through the corresponding points, I, 2, 3, on circle O, will be the points, a, by Cy etc., of the epicycloid. 165. Let Cand C^ be two cones of revolution, having a common vertex, F, and element of contact, E, and let B and B^ be the bases, or right sections, which are tangent to each other at any point/, on E. This supposed, let C roll upon (7„ which remains fixed. The point /, remaining a point of By rolling on B^y will generate a kind of epicycloid, which, being everywhere equidistant from F, lies on the surface of a sphere whose centre is F, and radius Vp. The curve is, therefore, called DESCRIPTIVE GEOMETRY. 1 83 a Spherical epicycloid. It is the intersection of the derivative or secondary cone, C^^ generated by the element E, with the sphere of radius Vp, PROBLEM XCI. — To construct the projections of a spherical epicycloid, and of a tangent line to it at any point. In Space, — The generatrix/ (165) of the curve is at any moment in the circumference of the base B of the rolling cone C, and also in the common meridian plane of the two cones through the common element of. contact at the same moment, and its position is constructed according to these principles. In Projection.-^i°, PL XVL, Fig. 128. Let FF be the common vertex of the fixed cone Vfm — VaV and the rolling cone V'aK\ and let / be the original point of contact of their bases. Revolve the base aK' about the common tangent at a^ into H, at afK", and then lay off from «, on both circles, equal spaces ab, aj?^y etc., one of which shall exactly divide af Then /, is the point that was originally at the point of contact /; hence project /, at //, and counter-revolve it to FF\ which is the point of the epicycloid reached by / when a is the point of contact of the cones* bases. 2°4 To find other points, we may proceed in either of two ways : First, To find the point of the epicycloid, corresponding, for instance, to V'Vc as the common meridian plane of the two cones, take V Vc as a new vertical plane of projection, and then proceed with that plane just as with V in finding the point FF\ Second, Imagine Wc revolved about WtiW it coincides with V, find DJDl, exactly as FF' was found, since d^ is the third point from a^ as /and /, are from c and c^, and then counter-revolve V Va back to its original position, V Vc, when Dfil revolving through an equal angle, will appear 1 84 DESCRIPTIVE GEOMETRY, as at DD'y the point reached by / (/,) when c and c^ are in contact. 3°. The tangent line^ at any point of the epicycloid, will be the intersection of the tangent plane at the given point to the sphere containing the epicycloid, with the tangent plane to the sphere whose centre is the corresponding point of contact of the cones' bases, and whose radius is the line from that point to the given point. The latter sphere is properly chosen, since the genera- trix is at each instant evidently describing a momentary arc, for that instant only, about the point of contact of the cones' bases for the same instant. Examples. — 1°. Construct the tangent line just described. Ex. 2°. Find a point of the epicycloid corresponding to any position as V Vm of the common meridian plane of contact. Ex. 3°. Construct the tangent at a given point on the epicycloid, PI. XVI., Fig. 129. [The tangent at c, for example, is perpendicular to the line cr.'l PROBLEM XCII. — To find the intersection of a cone and sphere y the centre of the sphere being at the vertex of the concj . together with a tangent to the curve. In Space. — The auxiliary planes employed in the solution may either be perpendicular to a plane of projection and through the cone's vertex, or parallel to a plane of projec- tion. In the former case, they will cut elements fro|n the cone, and great circles from the sphere. In the latter case, they will cut sections similar to the base from the cone, and small circles from the sphere. In each case, the line cut from the cone will intersect that cut from the sphere by the same plane, in points common to both surfaces— that is, in points of their intersection. In Projection.— ?\. XVI., Fig. 125. Let VFBN—V'A'B' be the cone, oblique with a circular base. The arc S'S" with V for its centre sufficiently represents the sphere. 1°. The highest and lowest points. — These are on the long- est and shortest elements, VH and VK^ in the vertical plane DESCRIPTIVE GEOMETRY. 1 85 VH containing the axis VO — V0\ Revolving the plane VH about a vertical axis at V till parallel to V, the element VH—VH' will appear at VH"—V'H"'. The similarly re- volved vertical great circle of the sphere contained in the plane F//" will intersect VH"—V'H"'2X b"\ the revolved position of the highest point. In counter-revolution, this point returns by the horizontal arc vertically projected in b'"b' to b' on VH\ whence it is projected down to b on VH, Otherwise, b"^ might have been projected on VH" and the counter-revolution" shown first in horizontal projection by an arc with F as a centre, giving b first. The lowest point may be similarly found. 2°. The points on the right and left elements, — These ele- ments are horizontally projected in VA and VB {AB paral. lei to GL). The points on them are found as just explained, and as shown at aa'a'" the point on VA — V'A', 3°. The points on elements parallel to V. — These elements are VC—V'C and VD—V'D', and the points on them, r^' and dd', are immediately found without revolution, S'S" being itself that great circle which is in their plane. 4°. The points on the extreme elements seen on H. — These ele- ments are horizontally projected in VE and FF, and the points on them (not shown) are found by revolving them till parallel to V, as before. 5°. Points in horizontal auxiliary planes, — m'v' is such a plane. It cuts from the cone a circle of centre oo\ on the axis VO — V'0\ and radius o'q' ; and from the sphere a circle of radius m's' — Vs, The horizontal projections of these circles intersect at r and n, whose vertical projections r' and 71' are on m'v' . This method obviously applies only to points between the highest and lowest. At the latter, the circles would be tangent to each other, but the assumed plane could only be accidentally at the proper height to contain them. 6°. The tangent to the intersection, — This, at any point, as nn\ is the intersection of two planes, one tangent to the sphere at that point, the other to the cone, along the ele- 1 86 DESCRIPTIVE GEOMETRY. ment through the same point. The former plane is found as in (Prob. LL), the latter as in (Prob. XXVIIL). Thus V,x, the intersection of a tangent (not drawn) to S'S'^ at ss\ with the line VV\ is the vertex of a cone tangent to the sphere on the horizontal circle m's'. Then xn! — Vn is the element of this cone containing nn\ and t't is its hori- zontal trace. Hence TV, perpendicular to Vuy is the hori- zontal trace of the tangent plane to the sphere at nn\ But NT is the like trace of the tangent plane to the given cone on the element VnN — FV, hence Tn^ the horizontal projec- tion of the intersection of these two planes, is the horizontal, and Tn' is the vertical projection of the required tangent. Examples. — i". Find all the points described but not shown on the figure. Ex. 2". Find the points in an auxiliary plane perpendicular to V. ^—Development. PROBLEM XCIII. — To develop the surface of an oblique cone. In Space, — All points of the intersection of the cone and sphere, Prob. XCIL, being on the surface of a sphere, are equidistant from its centre ; this curve will therefore, in the development of the cone, appear as a circular arc with a radius equal to that of the sphere. It only remains to determine the length of this arc as a means of locating the developed elements. The lengths of these can then be found as in (Prob. XII.). To find the length of the intersection of the cone and sphere, we develop either of its projecting cylinders. In Projection. — First. PI. XVI., Fig. 126, represents the development of the cylinder whose base is the horizontal projection, nhcrday Fig. 125, of the intersection of the cone and sphere, and whose elements, nn\ bb\ etc., are the heights, n'n", b'b''^ etc., of the points of the curve, above H. Then nbcrdUy Fig. 125, develops into the straight line nn^^ DESCRIPTIVE GEOMETRY. 1 8/ Fig. 126, and n'b'a'n^ is the true length of the intersection itself. Second, Fig. 127. With Fas a centre and F5', Fig. 125, as a radius, describe the arc «'«/, and make it, and its divi- sions, equal to 7t'n^ and its divisions in Fig. 126, and in like order of parts. Then n'nl, Fig. 127, is the development of the intersection of the cone and sphere, as seen on the developed cone. Lines from F through n' , b\ etc., will then be developed elements, where VH=: V'H'", Fig. 125, like- wise VA — VA'", and VC and VD respectively equal V'C and VD' in Fig. 125, etc., giving VNHDN' for the required development of the cone in Fig. 125. To develop the tangents Tn and 27V, Fig. 125, draw «T perpendicular to VN, and limit it by an arc from iVas a centre with radius iVT= NT, Fig. 125. ^^ i ii \{ \ \i V SHADES AND SHADOWS. First Principles. i66. An opaque body, B, PL XVII., Fig. i, being exposed to light which proceeds in any direction, as Rr, will be partly illuminated and partly dark. The unilluminated portion CdT is called the shade of the body, and the boundary CPTO between this portion and that which receives light, is the line or curve of shade. 167. The portion of space from which light is excluded by the opaque body just supposed, is the shadow in space of that body. When any other surface Q intercepts this shadow, the portion of it, cpto^ which is thus deprived of light by the opaque body, is the shadow of that body on that surface, and the boundary, which determines this shadow, is called the line or curve of shadow, 168. Any pointy P, of the curve of shade of a body is the point of contact of a ray of light tangent at that point to the body, for any othei- ray, not tangent to the body, would either intersect it as Mn, at n, in a point therefore illu- minated, or would be wholly exterior to it as at Rr, SHADES AND SHADOWS. 189 169. Neither of the two rays just mentioned could give any point of the curve of shadow of the opaque body ; the exterior ray, because it nowhere touches the body, nor the secant ray, because it would be stopped by the body. Hence any point of the shadow of a body B^ upon another surface Q, is the intersection of that surface with a ray tangent to that body. That is, the curve of shadow ^ as cpto^ of a body is the shadow of its curve of shade ^ as CPTO, Hence in the construction of shades and shadows, the shades (166) must be found first. 170. Light proceeds from a luminous point in every direction, in straight lines called rays. The surface of rays radiating from such a point, and enclosing an opaque body tangentially, will be a coftCy when the luminous point is at a finite distance from the body, and a cylinder, CcTt, Fig. i, when the rays are parallel, as they are when proceeding from a body at a vast distance, as the sun, to the points of any terrestrial object. 171. The character of Shades and Shadows as an appli- cation of Descriptive Geometry is now apparent. The curve of shade of any body is the curve of contact of that body with a circumscribed tangent cylinder or cone^ according respectively as the source of light is very remote or quite near. The curve of shadow of a body is the curve of intersection of the cylinder or cone just mentioned with whatever sur- face receives the shadow. The terms cylinder and cone are here used in the widest sense as meaning respectively surfaces composed of parallel or of converging elements. In the case of any plane-sided opaque body, as a cube, the tangent cylinder and cone of rays would become respectively an enclosing prism and pyramid of rays, and the line of shade would consist of the sum of those edges of the body which separated its illumi- nated from its unilluminated faces. 100 SHADES AND SHADOWS. 172. The manner of finding either the contact or the intersection of a straight line, or ray, element of the tangent surface ot rays, with a given surface, depends largely on the nature of that surface. Hence problems of shades and shadows are naturally classified like those of descriptive geo^netry ; giving shades and shadows on plane, developable, warped, double- curved surfaces. 173. Finally, in order that the reality indicated in PL XVII., Fig. I, may be exactly shown on paper, the source of lights the body casting the shadow, and the surface receiving it, must all be given by their projections. The source of light, if near, will be given by its projections ; if remote, by the two projections of any ray, to which all the others will be parallel. 174. In making drawings for practical purposes, the light is usually taken in one fixed direction viz., so that its projections make angles of 45° with the ground line, as at A,a'—A'a\ PI. XVII., Fig. 2. The corresponding direction of the light itself is evi- dently that of the diagonal of a cube, one of whose faces coincides with H, and the other with V, the diagonals of those faces being the projections of the ray, and all converg- ing towards the right to one point on the ground line. The ray itself thus makes an angle of 35° 16' with H and with V. In general problems, to illustrate principles, the light is taken in any direction that is most convenient. A— Shades aud Shadows on Planes. PROBLEM I. — To find the shades and shadows on a rHU angular block having a panel in front and a tablet on top ; all its edges being parallel or perpendicular to H and V. In Space. — The edges of shade (166) in all such cases, and in many others, can be determined by inspection. SHADES AND 6HAD0WS. I9I The shadow of such a body consists of the shadows of its limiting points and edges. The shadow of a point upon a plane is where the ray through that point pierces that plane. The shadow of a straight line upon a plane is the trace upon that plane of a plane of rays containing the line. In Projection. — Let PL XVII., Fig. 2, represent the block with its panel and tablet. As a temporary aid to the learner while becoming fami- liar with shadow construction, we will, in a few of the earlier problems, denote points and their shadows by the same let- ter — the former by a capital, the latter by a small letter. Aa — A'a! taken at 45° (174) being the projections of a ray of light, the edges of shade (166) are evidently the ver- tical ones at A and (7, and the horizontal ones, AB — A' and BC — A'C of the top of the block ; together with the cor- responding edges of the tablet, with H—KH' and HI—HT of the panel. 1°. To find the shadow of a point upon H. — het AA\ the front, right, upper corner, be the point. Through this point draw the ray Aa—A'a\ which (D. G., Prob. VIII.) pierces H at ^, which is therefore the shadow oi AA'. 2°. To find the shadow of a point tipon V. — Let -5,^' be the point through which draw the ray Bb — A'b' ^ which pierces Y at b\ which is therefore the shado\^ oi B^A' on V. 3°. To find the shadow of a vertical line on H. — Let A — D'A' be the line. The point AD' being in H, is its own shadow at ^, and rt; is the shadow (1°) of ^^'. Hence ^^ is the shadow oi A — DA' , Likewise Cc is the shadow of a part of C—R'C on H. a-^ 4°, To find the shadow on V of a line perpendicular to it. — Let the portion BA^ — A' of AB be the line, having drawn the ray a'A^ which determin-es the point A^ of which a' is the shadow. Then b' being the shadow of B,A' on V, a'b' is the shadow of A,B—A' on V. 5°. To find the shadow on H of a line parallel to it. — Let AB — A' be the line. Then a being the shadow of -^^' (1°), 192 SHADES AND SHADOWS. and h^ that oi B,A' on H, supposing V for the moment to be transparent, ab^, obviously equal and parallel to AB, is the shadow oiAB—A' on H. 6°. To find the shadow on M of a line parallel to it. — Let C—R!C be the line. Then (2°) c' being the shadow of CC on V, and r' that of C,R! on the lower part of V, supposing H transparent, c'r', evidently equal and parallel to C — R!C\ is the shadow on V. Likewise b'c' is the shadow oi BC — A'C on V. The shadows, Ee of E—ME'; cf oi EF—E' ; fg of FG-.E'G', and gG of G—G'N\ all on the top of the block ; likewise h'l' of a part oiH'K\ and h'f of a part of HT on the panel, being found in the same manner as the preced- ing, need no further explanation. If we revolve the vertical plane of rays on A a about its horizontal trace Aa into H, the line A — £>'A^ will appear as a perpendicular AA'^ (not shown) to Aay at Ay and equal to A'D'; A' 'a will then show the true length of the ray Aa — A'a\ and AaA" will be the angle made by the ray with H. Examples. — 1°. Let the block be so far from V that its shadow will be wholly on H. Ex. 2°. Let it be so far above H that its shadow will be wholly on V. Ex. 3". Let the tablet be 9n the front, and the panel in the top. Ex. 4°. Find the angle made by the light with V. Ex. 5". Let the tablet on top be high enough to cast its shadow partly on H and V. Ex. 6". Let an opening of the size of the panel extend through the block from front to back. Ex. 7°. Let there be tablets on the two ends of the block. 175. Useful elementary principles derived from the last problem. 1°. The shadow of a vertical line on H is in the direction of the horizontal projection of the light ; for it is the horizontal trace of the vertical plane of rays through that line. 2°. Likewise the shadow on V of a perpendicular to it, is in the direction of the vertical projection of the light. SHADES AND SHADOWS. I93 3®. The shadow of a line on a plane parallel to it is equal and parallel to the line. 4°. Parallel lines will have parallel shadows on the same, or on parallel planes. Also the shadows of the same line on parallel planes are parallel. 5°. Whenever the direction of a shadow is known, one point of it will locate it. Otherwise two will be necessary. 6°. When the shadow of a line falls partly on each of two intersecting planes, the point on the intersection is common to both shadows (see a' ^ Fig. 2), for the two traces {aa! and A' a' of the plane of rays through the line AB — A') necessarily meet that intersection at the same point. PROBLEM II. — To find the shades and shadows of a stotte cross. In Space, — The edges of shade (166) can still be found mostly by inspection, or certainly by means of planes of rays through the vertical edges. Some of the edges of the cross being oblique to the planes which receive their shadows, we must apply (175, 5°). In Projection,— ^\, XVII., Fig. 3. 1°. The Shades.— i:\iQ visible edges of shade are those that are made heavy. Under the theory, however, that these edges of shade are, strictly, the elements of contact of planes of rays with the minute quarter cylinders, which, instead of sharp mathe- matical lines, really form the edges of the body, such ele- ments, when situated as at G, would be invisible in facing V. The vertical edge at G, and other similar ones, are there- fore inked lightly. The vertical edges at D and G, for example, are cer- tainly known to be lines of shade by means of the vertical planes of rays, Dd and Gg, containing them, since these show the vertical faces on KD and KG to be in the light, and those on FG and FD to be in the dark. 2°. The Shadoivs, — Beginning with AA' , its shadow, 194 SHADES AND SHADOWS. Aa — A'a^ being the direction of the rays, is a (Prob. I., i°). That of AyB' is ^, similarly found, whence ad is the shadow, on H, of ^ — ^'^'(175, 1°). Likewise that of CC is ^, also found by making dcy from d, already found, equal and parallel to ^46^ (175, 3°). Thence r/^ equal and parallel to CH is the shadow of CH — CH', interrupted at /and 7 by the shadows, Dl and Gj\ of the lower parts of the vertical edges ^ at D and G. Gl is the shadow of a part of G — G'N' on the top, HF—H"N', of the horizontal arm. hJi" is the shadow on H of a part of the vertical edge, H — H' H" , The last four shadows, with de^ that of the upper portion of the vertical edge D — D'E\ are all parallel to ab (175, 1°, 4°). Returning to ^, the shadow ad, oi AD — A'D', on H, is equal and parallel to that line. Entering V, we find / the shadow of EE' (Prob I., 2°), whence e'e, parallel to D' E' , is the shadow on V of the upper part of that line. Then/', being the shadow of FF' , gives //' that of DF — E'F ; and f'g', similarly found, as shown, isthatofi^G^ — F'G'. The remaining points of shadow on V, which can be similarly found, are invisible. Finally, the portion KO — K'O' casts a shadow on the vertical face on KD, beginning at KK\ which is its own shadow on that surface, and ending at Dyo\ the shadow of 00' on the vertical edge at D, Examples. — 1°. Turn the cross horizontally till the facQ at ATXs in the dark. Ex. 2°. Turn it till HC is parallel to V. Ex. 3°. Keeping it as it is, change the direction of the light. PROBLEM in. — To find the shadow of a turnstile on H and on a vertical plane oblique to Y. In Space. — The new given plane being vertical, its hori- zontal trace may be considered as the ground line, and the operations will be precisely similar to those of the last two problems. SHADES AND SHADOWS. I95 /// Projection,— V\. XVII., Fig. 4. The construction being fully shown, the explanation is left to the student. Examples. — i'. Removing the fence, find the shadow of the turnstile wholly on H. Ex. 2°. Placing the turnstile near V, find its shadow on H and V. Ex. 3°. Draw the figure inverted, but with AB and the rays then inclined to the right of ^, C, etc., which will suit the enunciation : To find the shadow of the axis, perpendicular to V, and arms, in a vertical plane, upon a roof slope AB — A' B' perpendicular to V. PROBLEM IV. — To find the shadow cast on a flight of steps by one of its piers. In Space. — The new features in this problem ^lvq— first, lines of shade oblique to both planes of projection; second, narrowly limited surfaces receiving the shadow ; third, an auxiliary elevation on a plane perpendicular to the ground line. The second feature will serve to show the convenience in such cases of finding points of shadow in surfaces pro- duced, but that the real portion of the shadow is only on the real portion of the surface. The third will illustrate that, in general, any two of three projections will be suffi- cient for the purpose of finding the shadows on all. ht Projection. — 1°. Preliminary. — PI. XVII., Fig. 5. This is a flight of three steps, distinguished by numbering their tops i, 2, 3. The pier at the left has a level top, CDK—CK'—C'D", a slope, AC—A'C'K'—A"C", and a front, AB—H'B'A'—A"B", also sloping. The side eleva- tion, B"D"y is on the plane TUT' , where the hidden profile of the steps is dotted. Thus ^4' and ^'' are on the same horizontal line. Also like points on the side elevation and the plan are equidistant from the trace TU, and from the ground line H'U, respectively. Thus B' B — UB" , etc. The edges of shade of the pier which cast shadows on the steps, and on the ground in front of step i, are AB—A'B'—A"B", AC—A'C—A"C", CD—C—C"D", 196 SHADES AND SHADOWS. The light is taken in the conventional direction of 45° in projection. Thus Aa — A' a' — A" a" are the three projections (174) of one ray, to which all others are parallel. 2°. The shadows of ^^—^'^'.— Beginning at BB', this point being in H, is its own shadow on H. The shadow of A A' on H is aa'a", where the ray Aa — A' a' — A" a" pierces H. Then Bn, so much of ^^ as is in front of step i, is the shadow cast by a part of AB — A'B' on the ground — that is, on H. The shadow oi AA'A" on the front of step i is a^a^a^\ Here {In Space j Third) a^ can be found, either by projecting up a^ upon A'a\ or by projecting over a I' upon A' a'. Then, projecting n at n\ gives n'al the shadow of a part of AB—A'B' on the front of step i. 3°. The remaining shadows on step i. — The ray Aa — A' a' pierces the plane A'c^ of the top of step i at ^/, which, pro- jected down on Aa^ gives a^. The ray Cc — Co' pierces the same plane at r,V„ hence ^/^ is the indefinite shadow of AC^A'C on this plane, and ep — e'p', so much ot a^c^ as is within the limits of the top of step i, is its real portion. Then by (175, 6°) ea^ — e'al — e'^a^' is the remainder of the shadow on the front of step i. The point ee' can, however, be found directly by begin- ning with the projection e"E" of the ray which meets the front edge of step i. Then project E" over to E' on A'C, and draw the projection E'e' of the same ray, which gives e' , 4°. The remaining shadows, — cs projected down from c' , and parallel to CD, is the shadow of CD — C on the top of step 3. qr and vc are parallel to ep, and by reason of the equality of the steps, c, q and v are all in the same line, parallel to ^C—^' 6"' — A'\C",^\ncQ this line is parallel to the plane q"e" of the front edges of the steps. Likewise, pq—p'q' and rv — r'v^ are parallel to a^e — a^e'. 5°. Plan construction from the side elevation only, — Take, for example, the point r'V, the point of shadow on the bottom edge of step 3 (the same thing as the back edge of .step 2). Drawing the ray r"R!'y we find the point, R!\ SHADES AND SHADOWS. 1 97 whose shadow is r", project R!' at R!'" ; by counter-revo- lution about 6^ as a centre, bring R!'" to R"y whence pro- ject it by R!"R parallel to the ground line to R^ the hori- zontal projection of R'. Then the horizontal projection Rr of the same ray, RW'y intersects the given edge of step 2 at r, the horizontal projection of r". Similarly, a is found from ^", c from c" , etc. Examples. — 1°. Change the proportions of the steps, also enlarge the scale. Ex. 2°. Change the direction of the light (174). Ex. 3°. Let the piers of the steps converge towards the back of the steps — that is, let the plane JiB'C, continuing vertical, be oblique to V. 176. Summing the four last problems, we deduce the following principle and consequent rule for the solution of every problem of shadows upon the planes of projection^ or up07t any plane parallel or perpendicular to them. One projection of every such plane is a line, viz., that upon the plane of projection to which the given plane is perpen- dicular. Thus, ABy Fig. 4, is the horizontal projection of the face of the lence, A'B' ; /r/. Fig. 5, is the vertical pro- jection of the top of step i, etc. Hence, comparing the similar constructions of a and b\ Fig. 2 ; ^ and c' , Fig. 3 ; /', etc.. Fig. 4; a,c,a;, etc.. Fig. 5, we have for all like cases the Rule. — Note first the intersection of that projection of the given plane which is a line^ with the like projection of the ray employed (see c' or ^„ Fig. 5, etc.), and then project the point so found upon the other projection of the same ray. (See r, or ^/, Fig. 5, etc.) 177. All the lines casting shadows have thus far been straight; but the operations would have been precisely similar had those lines been curved. That is, the shadow of any curve upon any plane situated as in (176) is found by assuming any convenient number of points upon it, and finding their shadows, just as in the preceding problems, and then joining them. iqS shades and shadows. The shadow of any plane curve upon a plane parallel to it will, in general, be an equal curve. That of a circle, on any other than a parallel plane, is generally an ellipse. If, however, the plane of any curve be parallel to the direction of the light, it will be a plane of rays, and its trace, the shadow of the curve, will be straight. Examples. — i". In PI. XVH., Fig. 2, let the tablet be replaced by a hori- zontal circle at any height from the block. Ex. 2°. Also by any vertical semicircle having its diameter in the top of the block. Ex. 3°. Let the panel be cylindrical. Ex. 4°. In Fig. 3, let the arms bear semicircles on DF, AC, and A'B', as diameters. Ex. 5°. In Fig. 4, let the arms of the turnstile be curved. Ex. 6°. In Fig. 5, let an arc, tangent to jD"C' and B"A" at C and A", be the profile of the pier. 178. It is only important to add that in constructing the shadows on different vertical projections (elevations) of the same fixed object, the light is supposed to turn equally with the observer as he turns from facing one vertical plane to view another. Thus, if, in PI. XVII., Fig. 5, the side elevation bore any visible shadows, the three projec- tions of the light used in finding them would be in the direction of the three arrows r„ r/, r/\ which indicates a revolution, by the light, of 90° around a vertical axis, as the observer also turns 90° from viewing the vertical plane V, to viewing TUT, which is V^. PROBLEM V. — To find the shadows of chimneys on side and end plane roof surfaces. In Space. — Here most of the surfaces are not perpen- dicular to either plane of projection, and hence will not have any linear projection. Hence we find where a ray pierces any of them by (D. G., Prob. XL), that is, by pass- ing, not a ray alone, as in the four preceding problems, but SHADES AND SHADOWS. I99 a plane of rays alsoy through a point casting a shadow. Then, where the ray through that point pierces the trace upon the roof of the plane of rays through the same point will be the shadow of that point upon the roof. In Projection.— ?\. XVII., Fig. 6. ROMQNS is the hori- zontal, and M'N'S'QR' the vertical projection of the roof, and ABC—T'A'C and 'DF—o'D'P are the chimneys whose shadows upon the roof are to be found. 1°. The shadow of the left-hand chimney. — Taking the plane of the foot of the roof as H, and the edge ^-5 — A' per- pendicular to V, and Aa — A' a' for the direction of the light, A'h'h is the plane of rays containing AB — ^', and therefore perpendicular to V. It cuts the ridge of the roof at g'gy and its foot at h'h and h'h'\ giving hg — h'g' and h"g — Ji!g' as its traces on the front and back roof slopes. Then by (D. G., Prob. XL) the rays Aa—A'a' and Bb—A'b' meet the roof at their intersections with these traces, giving aa' and bb' as the shadows oi AA' and BA' ^ and thus agb — a'g'b' as the shadow oi AB — A' upon the roof. The edge BC being parallel to the side roofs, its shadow, be — b'c' ^ is equal and parallel to BC — A'C . Finally, as the edges at A and C are vertical, the planes of rays containing them are so also, and consequently their shadows are hori- zontally projected (D. G., 27, 8°) in the horizontal traces Aa and Cc of these planes. The vertical projections of these shadows are a'p' and c'q'C". 2°. The shadow of the right-hand chimney, — ODs and FEt being the traces of the front and back of this chimney upon the roof, project s and t at / and t' , and draw s'oo" and t'F" parallel to the ground line MR! ^ to obtain the vertical projections of these traces. Then Dn — o'n' ^ the shadow of a part of D — o'D' on the front roof, is parallel to Aa—p'a' (175, 4°), dd' is the shadow oiDU on the plane of the roof surface SQR—S'Q'R' (Prob. IV.), and is found as aa' was. That is, D'm'm is the plane of rays containing DE — D' ^ and m"l — m'l' (parallel to h"g — h'g'), and Im — I'm' , are its traces on the back and end roofs, in- tersected by the rays Dd—D'd' and Ee — D'e' at dd' and ee'^ 200 SHADES AND SHADOWS. the shadows of DD and EE'. Hence nv — n'v\ part of nd — n'd! y is the remainder of the shadow of D — o'D\ Finding//"' as bb' was found, Fjf—F"j'f' is the shadow of F—F'F; thence fk—fk', parallel to FE—FD\ is the shadow of a part of that line, and ke — k'e' is the remainder of its shadow. By drawing a ray inversely from kk' to meet FE — F D\ we shall find what part of this line has fk^' k! for its shadow. All the shadows on the back part of the roof, or behind the right-hand chimney, are invisible in vertical projection. Examples. — 1°. Vary the proportions of the roof in any manner. Ex. 2°. Change the direction of the li^ht. Ex. 3°. Let iV5and its parallels be oblique to V. Ex. 4". Add an abacus to each chimney [see Fig. 7, which indicates that the portions ED and AF, only, of the front and left-hand under edges of the abacus, cast shadows on external objects, the portions ^^ff, ,5(7 casting shadows on the front, and CD on the left side of the chimney]. Ex. 5°. A very interesting example is found by means of the following modifications of PI. XVH., Fig. 6. Let each chimney be cylindrical, and have a cylindrical abacus. It will be convenient to make the figure large enough to fill half the plate or more, and to use an auxiliary vertical projec- tion on a plane perpendicular to M'K! and revolved into H. 179. Fig. 6 shows clearly how to ascertain which of a group of intersecting plane surfaces in various positions are in light or shade. For example, had the ridge been raised to the height ^/iV/, and the direction of the light changed to Aa^ — ^'^', the traces of the plane of rays ^V^V^ on the new roof slopes would have been £^Ji and ^Jt'\ and the same ray Aa^ would have pierced both of these traces, at a^ and a^. In every such case^ the surface first pierced by the ray is in the lighty and the other is in the dark, and their intersection {g^N^) is an edge of shade or shade line (166). Likewise, had the ray Dd — D'd' pierced the front roof before piercing QSR, the latter would have been in the dark, and QS an edge of shade. On the other hand, with the figure as it is, the whole of SHADES AND SHADOWS. 201 the roof is in the light, except as shadowed by the chim- neys. The principles and problems now given will enable the student to solve every possible problem of shadows on plane surfaces, whether cast by straight or curved lines. B — Shades and Shadows on Developable Surfaces. 1 80. The elements of shade on a cylinder and cone, the only developable surfaces necessary to consider here, are the elements of contact of tangent planes of rays. When the luminous point is near, the problem therefore merely becomes an application of (D. G., Prob. XXX.). When it is indefinitely remote, and the rays consequently parallel (173), the problem is an application of (D. G., Probs. XXXI., XXXII.), the tangent plane in each case being then a plane of rays. Examples. — 1°. Given a cylinder oblique both to H and V, and the direc- tion of the light, to find the elements of shade on its convex surface. Ex. 2°. Find the elements of shade on an upright cone. Ex. 3°. Find the elements of shade upon an inverted cone. Ex. 4°. Find the elements of shade upon a cylinder in any horizontal posi- tion, perpendicular, parallel, or oblique to V. Ex. 5°. Find the elements of shade upon a cone in the same position. Ex. 6°. Find the elements of shade upon a cone whose axis is oblique both to H and V. PROBLEM VI. — To find the shadow of a rectangular abacus upon a cylindrical column and the shade of the column. In Space. — A cylindrical surface being the only one, be- sides a plane which has parallel elements, it is the only curved surface capable of being perpendicular to a plane. When thus situated, its projection upon the plane to which it is perpendicular is a curved line, viz., its right section. The shadows of any points on a cylinder so situated may 202 SHADES AND SHADOWS. therefore be found in the same manner as upon planes simi- larly placed. In Projectioji.— Let EAD^-A'D', PI. XVIII., Fig. lo, be the square abacus of a cylindrical column, half shown at FbG — FGG' y Cc — Cc' being the direction of the light ; Cc is also the horizontal trace of the vertical tangent plane of rays, whose element of contact c — d'c' is the visible element of shade of the column. The portions CA and AE of under edges of the abacus then cast shadows on the col- umn. Thus the ray Aa — A' a' pierces the column at a point whose horizontal projection is ^, and whose vertical projec- tion, a', is found by projecting a upon Aa\ giving aa' the shadow of AA^ upon the column. Any other points, as bb\ can be similarly found, a'f is the vertical projection of the intersection of the plane of rays containing AE — A\ and thence perpendicular to V, with the cylinder ; that is, aF — a'f is the shadow oi AE — A' on the cylinder. 1 8 1. From the foregoing solution, we have at once the rule for finding shadows on all cylinders which are perpen- dicular to a plane of projection. Rule. — Find first that projection of each point of shadow which is on the plane to which the cylinder is per- pendicular, by noting the intersection of the like projection of a ray with the curve which is the like projection of the cylinder ; then project this point upon the other projection of the same ray. Any variations in WiQ form of the abacus or column will change only the form of the shadow, not the manner of finding it. Examples. — 1°. Let the abacus be C)'lindrical, and then find the complete intersection of the whole cylinder of rays with the entire surface of the column. Ex. 2°. Let it be hexagonal ; octagonal. Ex. 3°. Let the column be clustered. [For convenience, the following, having the same solution, are added to complete the series.] Ex. 4°. Let the abacus and column both be square. Ex. 5°. Let the abacus be square or circular, and the column hexagonal. SHADES AND SHADOWS. 203 Ex. 6°. Let the column in each of the preceding cases be perpendicular to V. Ex. 7°. Let it be parallel to the ground line. PROBLEM VIL— r^ pid the shadow of the upper base of a vertical hollow half cylinder upon the visible interior surface. In Space. — The solution is the same as in the last problem. In Projection.— V\. XVIII., Fig. 12. ADH—A'H' is the half cylinder. The shadow is cast by a portion of the edge of shade A—G'A', and by the portion, AD—A'U, of the upper circle of the interior surface, which is also an edge of shade, since it separates the illuminated annular surface AEHF from the darkened portion of the interior. First, A — G'G" casts the shadow Aa on the base of the cylinder; then A — G"A' casts the shadow a^-g'a\ parallel to A — G"A'; and lastly, the shadow a'c'D' is found by as- suming points, as CC on AD — A'D' , and drawing rays, as Cc — Cc' y giving points, as cc\ where (181) ^ is first found. Examples. — 1°. Let the cylinder be perpendicular to V. Ex. 2°. Find the complete intersection of the entire cylinder of rays whose base is the whole upper circle, with the whole surface of the given cylinder produced upward. This will be an ellipse, of which D'c'd is an arc. Ex. 3°. Let the cylinder be parallel to the ground line, with the plane EF parallel to V. Ex. 4°. In the last example, let the plane EF be parallel to H. PROBLEM VIII. — To find the shadow of a vertical cylindri- cal turret upon a concave cylindrical roof surface. In Space. — The complete line of shade upon a material cylinder, limited by bases, consists, besides the elements of shade of its convex surface, of that portion of the circum- ference of each base which separates those parts of the base and convex surface, one of which is in the light and the other in the dark. 204 SHADES AND SHADOWS. The intersections of the planes of rays tangent to the cylinder casting the shadow, with the one receiving the shadow, will be the shadows of the elements of shade (i8o) of the former. Their construction is an application of the general problem : " To find the intersection of a plane and cylinder." (D. G., Prob. XXXIII.) The rays from all points of that portion of either base which is a line of shade form a cylinder of rays. The in- tersection of this cylinder with the cylinder receiving the shadow, is the shadow of that base, and is found as in (59). In Projection.— Y\. XVIIL, Fig. 8. Let QEF—QE'F' be the turret, and let ABCD—A'B'CD' be the projections of a concave roof surface, whose elements are parallel to the ground line CD^ and whose profile in the plane 010' is the arc 01 — //', shown after revolution about 10' to the right, at dl'. Let Oo — O'o' be the direction of the light. Then — 1°. The shadow of an element of shade of the turret, — The vertical line at F is one of these elements, and Ff is the horizontal trace of the plane of rays containing it, and is therefore (Prob. V., 1°) the horizontal projection of its shadow on the roof. To find the vertical projection of the same shadow, assume any element, as be of the roof surface, revolve b to b" ^ whence it is projected to ^''', giving b"'o" for the vertical projection of the same element. Then project c upon o"b"' at c' ^ the vertical projection of c. Other points being similarly found, and /, on the lower edge of the roof, being projected at /', the vertical projection I'c'h' of the shadow of an element of shade can be sketched. The shadow of the opposite element at E — omitted to avoid confusing the figure — can be similarly found. 2°. The shadow of the upper base of the turret. — The ray Oo — O'o' , through the centre of this base, pierces H at ^, the centre of the equal geometrical shadow on H, the upper base. The problem then becomes : To find the intersection of the cylinder of r2iysfnd — O with the roof cylinder. The light being taken at 45° (174), and crossing the second angle, both traces of the plane of rays through the axis SHADES AND SHADOWS. 205 Oo — O'o' and parallel to the axis of the roof-surface (59) co- incide in 00" parallel to CD, and o"a {o"al' = 45°) is the re- volved trace of this plane on the profile plane 010' , Then assuming any trace, as nn" , parallel to CD, of a plane A'n"p"'y parallel to 00" a, we ^nA p"'q' , and by counter-revolution (1°) pq, the two projections of the element of the roof contained in this plane, and nN—n'F' that of the cylinder of rays con- tained in the same plane. These elements intersect at qq' the shadow of the point NF upon the roof. Other points, as hh\ being similarly found and joined, give the shadow of the upper base of the turret upon the roof. Examples. — 1°. Bring the horizontal projection forward so as to bring o in front of the ground line. Ex. 2°. Change the direction of the light so that the traces as 00" of the planes of rays shall not coincide. Ex. 3°. Let the elements of the roof surface be oblique to V. Ex. 4°. Substitute an inverted frustum for .a cylindrical turret. PROBLEM IX. — To find the shadow of a square abacus upon a conical column, and the shade of the column^ In Space.—SincG the lines casting shadows are here straight, the problem becomes an application of the general one: " To find the intersection of a plane and a cone." In Projection,— Fl XVIII.,, Fig. 9. FGM—G'E is the abacus, ABODE— ACD'E' the column, and B'R—V'R a ray of light. I °. To find the shade of the coluimt, — To condense the figure, let DHE—DEV be a cone of the same axis and taper as the column. The horizontal trace of the ray B'R— V'R of the ray through its vertex is R, then Rt, tangent to DHE, is the horizontal trace of a plane of rays, tangent to this cone, and tB'—t"V'{t"V'TLoX. drawn) is one of the two elements ot shade of the cone. TQ, parallel to tSR, is the horizontal trace of the parallel plane of rays tangent to the column, and Tt— Tt' is an element of shade of the column. QF, parallel to SV, is the vertical trace of this tangent plane of rays. 206 SHADES AND SHADOWS. 2°. The point of shadow in the meridian plane of rays FB' R. This plane cuts from the column the element ab—a'b\ and from the edge GF— G' of the abacus, the point i% G' ; the ray Ff—Gf from which pierces ab—a'b' at ff\ the required point of shadow, that of Fy G' . 3°. The shadow of the edge GL—G', — The vertical projec- tion of this shadow falls on the vertical trace G'g' of the plane of rays through the given edge. Any point of its horizontal projection is found essentially as Tc'h' in Fig. 8 was; thus, assuming any point, as m' or f on this shadow, project it, as at m or f upon the horizontal projection, as AD or ab, of the element, as AD' or a'b'y on which it is found. 4°. The highest pointy kk\ of the shadow of GM— G'E' , — This point is the intersection of the plane of rays through this edge with the element BH~B'H\ Taking now BB'V as an auxiliary vertical plane of projection, G"H'\ parallel to BB\ and at a height HH" ^—B'H' , is the new vertical pro- jection of the under surface of the abacus ; G" is that of GG' ; g'\ at a height equal to g'yy that of gg' ; and hence BH" is that of the element BH, and G"g" that of the ray Gg—G'g'. Then V^ intersection of G"g" and BH" , is the auxiliary projection of the point required, and k and k' — making B'k' = kk" — are its required projections. 5°. The point of shadow on the element of shade Tt^ Tt' . — The plane P'QT cvX^ the vertical plane through GM—G'E' in the line MI-MT parallel to F Q, and thus (D. G., Prob. XL) it cuts the edge GM—G'E' at T'L Then the ray through // meets Tt—T't' at ii' , the required point of shadow. Any other points can be found by this method. For any secant from R to DHE—DE is the horizontal trace of a plane of ra3^s containing elements of the cone DHE— V'DE, But as this cone is similar to and concentric with the column, and with its base equal to the upper base of the column, B'xX, for example, gives the element xX of the column contained in a secant plane whose trace Rx, regarded as on the plane D'E' ^ cuts the edge G'E' in the point whose shadow falls on Xx. 6°. Indirect cofistruction of points, — This is here done by SHADES AND SHADOWS. 20/ the method of one auxiliary shadow ^ an application of (175, 3°), and stated thus. If any two surfaces, M and N, intersect in any line C, the shadow on ^ of a given line being the one most easily found, then where the shadow /, of L on iV, meets the intersection C, is a point of the shadow of L upon M, In applying this often very useful method to the present problem, M is the conical surface of the column ; N is any horizontal plane as U'W; 6' is the circle U'W'—UrW c\x\. from the column by this plane, and L is the edge GM—G'E' of the abacus. The ray Gg—G'g\ for example, pierces the plane U' W at g'g, giving go, parallel to GM, for the shadow (/) of GM on this plane. Then nji' and 00' , where this aux- iliary shadow intersects the circle UrW—U'Wy are two points of the shadow of GM—G'E' upon the column. By this method, therefore, points of shadow are found before knowing the points which cast them. The latter are, how- ever, found by drawing rays inversely from nn' and oo\ which will meet GM—G'E' in the points casting these shadows. In the figure, the ray from 7in' will only meet MG produced, showing nn' to be an imaginary point. Examples. — 1°. Let the abacus be C)'lindrical. Ex. 2°. Construct points of shadow on the column by means of secant planes of rays cutting it in elements. Ex. 3°. Apply the method of (6°) to Prob. V. and to Prob. VHI. PROBLEM X. — To find the shadow of a vertical cylinder upon an upright cone. In Space. — This problem, as a practical one, might be : to find the shadow of a cylindrical chimney or turret upon a conical tower roof. As in Prob. VII., the line of shade of the cylinder consists of the elements of shade and of half the circumference of the upper base. The shadows of the elements of shade are the intersec- tions of the planes of rays containing them, with the cone. That of the edge of shade of the upper base is the intersec- tion of the cylinder of rays containing it with the cone (58). 208 SHADES AND SHADOWS. In Projection.— ?\. XVIIL, Fig. ii. CED-P'E' is the cylinder; VAB—VA'B' the cone, and Oo—0'o'\% a ray of light. The circle oq, equal to CED^ and whose centre is the horizontal trace, o, of the ray Oo—O'o' through the centre 00' of the upper base of the given cylinder, is the base of the cylinder of rays through CED^D'E', Proceeding as in (58), R is the horizontal trace of the ray through the vertex of the cone. Hence any secant from R to the bases of the given cone and the cylinder of rays y is the horizontal trace of a plane of rays cutting elements from both, whose intersec- tions are points of shadow ; provided that the elements of the cylinder of rays contain points of the semicircle CED and that those of the cone are the ones first met by the rays. Thus, the plane of rays whose horizontal trace is RA cuts from the cone two elements, one of which is A V—A'V\ and from the cylinder of rays two elements, one of which, aE—a'E'y contains a point of CED, Then ee', intersection of the ray Ea—E'a! with the element A V—A'V of the cone, is a point of shadow on the cone. Other points, as cc' and dd'y are similarly found. The horizontal projections of the shadows of the element of shade at Dy are Ds on H, and sdon the cone. The latter is a hyperbolic arc (36, 3°) whose vertical projection can be immediately found either by elements or by horizontal cir- cles of the cone which shall intersect it between ss^ and dd\ (Prob. VIIL, r^'. Prob. IX., ;;^< //•'.) Examples.— 1°. Change the direction of the light. Ex. 2°. Change the relative position of the bodies, keeping both upright. Ex. 3°. Make the cylinder horizontal. Ex. 4°. Make the axis of the cone horizontal. [In this example, and in 3, an auxiliary vertical plane perpendicular to the horizontal axis will be useful.] Ex. 5'. Apply the method of Prob. IX., 6°. O— Shades and Shadows on Warped Surfaces. 182. We have under this head, general methods, based on general principles ; and special ones, based on the prop- erties of each surface. The former are here stated. SHADES AND SHADOWS. 20g Shades, — Every plane containing an element of a warped surface is tangent to the surface at some point of that ele- ment (139). If then it be also a plane of raj^s, its point of contact will be a point of shade. Otherwise: we have shown (159) how to pass a plane through a given line and tangent to a warped surface. If the line be a ray, the auxiliary parallel tangent cylinder there described will be a cylinder of rays, and hence its curve of contact, the curve of shade. 183. Shadows on Warped Surfaces.— Th^ general method for finding these would be to pass planes of rays, each of which should contain an element of the surface, and should cut the line casting the shadow in some point (D. G., Prob. XL, or 54, 55). Then the ray through this point would meet the element in the shadow of the point. This article and the last will be sufficient for the few cases in which these shades and shadows need be found. Shades and shadows on the one warped surface of revo- lution (65) are found by methods which will be explained as applied to double-curved surfaces. Other problems are sufficiently represented by the fol- lowing problem of the screw. 184. Description of the screw. — PL XVIII., Fig. 13. Let the isosceles right-angled triangle G'A'C'y whose base G'C is vertical, uniformly revolve about the vertical axis o — 00' in the same plane with it, and at the same time vertically and uniformly ascend a distance equal to G'C, so as to occupy, after one such ascending revolution, the position CA"C". Every point of the triangle will thus generate a helix (100). The whole triangle will generate a solid called the thread of a triangular one-threaded screw. The upper and under surfaces of this thread are generated by A'C and A'G' respectively, and are zones of oblique helicoids (127, 3°), the former comprised between the inner and outer helices gen- erated respectively by CC and by AA'; the latter, between the helices generated by CG' and AA'. 210 SHADES AND SHADOWS. The circle of radius oA is the horizontal projection of the outer helix, and the one of radius oC is that of the inner helix. The cylinder, of radius ^6" and axis o—oo', to which the thread is attached, is the core of the screw. 185. Any element is thus assumed. Produce ^^T — A'C till it meets the axis at 0', Then assuming STo for example, project 5 at S' on the outer helix, and at o'\ as far above o' as S' is vertically above A' y giving S'T'o" for the vertical projection of STo ; when, as a check upon the accuracy of the figure, T^ projected from T on the inner helix, will also fall on So". PROBLEM XI. — To find the shades and shadows on a tri- angular threaded screw whose axis is vertical. On account of the length of some of the topics in this comparatively complex problem, we will give the solution, " in space," and ** in projection," for each separately. 1°, A point of shade on the inner helix and underside of the thread by means of the declivity cone. In Space. — The tangent plane to a helicoid at any point of a given helix is determined by the tangent to the helix, and the element, at that point (138). The inclination of each of these lines to H — here perpendicular to the axis — is con- stant for all points of the same helix ; hence the like is true for all tangent planes at points of the same helix. If, then, the line of declivity (Prob. VII.) of any one of these planes, cut from it by a meridian plane (33) of the screw, be revolved about the axis — 00' of the screw, it will generate ^ cone all the tangent planes to which will be parallel to correspond- ing tangent planes on the given helix. Hence a tangent plane of rays to the helicoid (182) will be parallel to the tangent plane of rays, easily found, to this declivity cone. In Projection. — PI. XVIII., Fig. 13. Making Dd equal to the arc Dj\ of the inner helix (102) ^and b, horizontal traces SHADES AND SHADOWS. 211 of the tangent, and of the element BD — B'D\ give dbc the horizontal trace of the tangent plane to the under surface of the thread at DD', Drawing oc perpendicular to db^ pro- jecting it at c' ^ and producing B'D' to v' on o'o produced, gives oc — v'c' the generatrix of the declivity cone, whose horizontal trace is the circle oc^ and to which the plane db is tangent. All other like tangent planes to the under surface of a thread, at points on the inner helix, are parallel to tan- gent planes to this cone. Now cob is the constant angle between the line of de- clivity and element of contact of these tangent planes; hence, drawing rh tangent to circle oc from r, the horizontal trace of the ray or — v'r\ through the vertex ov' of the de- clivity cone, it is the horizontal trace of the plane of rays tangent to this cone ; hence, making moh equal cob and om and ob in like directions from oh and oc^ we have om for the element of contact of a plane of rays parallel to the plane r/i and tangent to the inner helix at the point 7nm' of the required curve of shade. The other tangent from r to circle ocy not shown, will afford another point of shade. Points of shade on the outer, or on any intermediate helix can be similarly found. 2°. A point of shade on the outer helix and under surface by the method of helical transposition of a tangent plane. In Space. — Imagine a tangent plane to the helicoid at any point of the given helix. If it be also a plane, of rays (182), the horizontal trace of a ray through its intersection with the axis will be in the horizontal trace of the plane. Other- wise this plane will intersect the cone generated by this ray in two elements, which may be considered as revolved positions of this ray. If then the tangent plane receive the same helical motion as the generatrix of the helicoid, it will continue tangent at successive points of the given helix, and when it has so revolved until either of the revolved rays, by an equal angular motion, again coincides with a ray, it will become a tangent plane of rays^ when (182) its contact will be a point of shade. In Projection. — Making ^^ = arc AJ, we have au' the 212 SHADES AND SHADOWS. horizontal trace of the tangent plane at A A' to the under surface of the thread. This plane cuts the axis o — oo' at oe\ where the element A'u' in it meets the axis. The ray o/—e'/\ through oe\ pierces H at /, not in the horizontal trace au' of the tangent plane ; which is thus not a plane of rays, but one that cuts the cone of vertex 0/ and horizontal trace M/N, generated by of—e'f, in the elements oM and oN, Now moving the plane au' helically upward till oN is again horizontally projected in of, it will then be a tangent plane of rays, and its point of contact, AA' , having an equal angular motion, Aon = No/, will be found at nn\ another point of shade. By this method two points on any helix can be found. For by helically revolving the plane till oM coincides with of, we shall find a point of shade on the outer helix and on the back of the screw. By reason of the uniformity of the screw, m can also be projected at m'\ etc., and ;/ at n'\ etc. Examples. — 1°. Find do^/i of the points derivable from the plane dd as mm' was ; and from the plane att as nn' was. Ex. 2°. Find points on an outer helix by (1°). Ex. 3". Find points on an inner helix by (2°). Ex. 4°. Taking the light more nearly horizontal, find a point of shade on the u^per surface of the thread, by each of the methods given in the problem. 3°. T/ie shadow of the curve of shade on the upper surface of a thread. In Space, — Points of this shadow are found by the method called that of two auxiliary shadows, which, by refer- ence to Fig. n, can easily be stated as a general principle thus : If the shadows,/ and q, of two lines /*and Q in space, upon a plane M, intersect as at a, the ray through a must intersect both of the lines P and Q (it being the element common to the two intersecting cylinders of rays whose bases are / and q, and whose elements are parallel), and A, where this ray intersects Q, will be the shadow on Q, of ^„ where the same ray intersects P. In Projection. — Taking the horizontal plane s't' {z=: M, Fig. n), in order to condense the figure, the shadow of SHADES AND SHADOWS. 213 mn — m"n" { — P, Fig", n) upon this plane is pg; that of any assumed element (185) ST—S'T' {= Q, Fig. n)j on Which it is supposed that some point of the shadow of inn—m"n" will fall, is St, Then drawing the X2iy yUU" , y'U'U'" ixom yy' , where pq and st intersect, the point UU' , where it crosses ST—S'T, is the shadow of the point U'U" where it meets the curve of shade. 4°. The shadow of the outer helix on the thread below. In Space. — This shadow can be found as was the preced- ing, but for variety, it shall be found by direct construction by the general principle of (183). In Projection. — Assume any element FE — F'E' (185) on which a point of this shadow may fall. Its horizontal trace is k, and that of the ray Be" — Ee'" intersecting it is e" y hence ke" is the like trace of a plane of rays through it. The intersection of this plane with the outer helix could be found by (55), but by reason of the uniform helical mo- tion of every point of the screw, a like helical revolution of the plane ke" is better. Then revolve it thus till perpen- dicular to V, when its vertical trace will coincide with E^o"" , the like helically revolved position of Eo — E'o'" (2°), found by making EE^ — Ifi, where /, is on ol perpendicular to ke'\ and therefore the line of declivity of that plane. The plane ke" y thus made perpendicular to V, cuts the outer helix at i"', thence projected at i". Returning to the primitive position, oB returns to ol^; oE^ to oE ; i" to /, by making i"i — Efi — Bl^^ and i is then projected at i'. Thence the ray ig — i'g meets the element EF — E'F' , in whose plane of rays it is by construction, at gg' a point of the required shadow. Applying the method of (3°) to the helical arcs S'B' and i"'B"y we could find x where the shadow leaves the screw. Or it could be found indirectly by joining^' with a point similarly found on an element, as o""E/y produced below E,E/. The shadows on different threads are equal and similarly placed. Those not falling on the screw fall on the planes of projection, and are found as in Prob. I. 214 SHADES AND SHADOWS. Examples. — 1°. Find the shadow of the curve of shade — or the outer helix — upon the outer helix below by the method of (3°). Ex. 2°. Find the shadow of a hexagonal nut upon the screw. D— Shades and Shadovrs on Double-Curved Surfaces. 186. The line of shade on a double-curved surface is wholly curved. Any point of it is imagined and con- structed by direct methods^ either as the point of contact of a tangent plane of rays, or of a ray tangent to a section of the surface made by a secant plane of rays. Points of shade are often best found bv indirect methods, which will be explained. 187. Points of shadow on a double-curved surface of revolution are in general easily found directly by construct- ing the intersection of rays through points casting shadows with the sections of the surface made by planes of rays through the latter points ; also by the indirect methods of (Prob. IX., 6°) and (Prob. XL, 3°). PROBLEM XII. — To find the curve of shade upon a sphere. In Space. — This curve will be the great circle of contact of a cylinder of rays, which will also be one of revolution, since on a sphere the plane of the circle of shade, called the plane of shade, is perpendicular to the light. By these principles, the shade as seen on each projection can be found separately. In Projection. — PL XIX., Fig. 14. Let the circle of radius OB be the vertical projection of a sphere, V being the plane of the paper, and supposed to contain the centre of the sphere, and let OV be the vertical projection of a ray of light, taken at 45° (174). Then L'C is the vertical trace of a plane of rays perpendicular to V through the centre of the sphere (thence called the perpendicular plane of rays)^ and AB, per- pendicular to L'C, is the like trace both of the plane of SHADES AND SHADOWS. 21$ shade, and of that plane of rays which makes with V the same angle that the light does (thence called the oblique plane of rays). The two latter planes are perpendicular to each other and to the plane LC, If now the plane L'ChQ revolved about its trace L'Cto coincide with V, the revolved traces of the other two planes upon it will appear as perpendiculars to each other at O. Make OL' = BF, considered simply as the hypothenuse of the isosceles right-angled triangle FOB, and L'L perpen- dicular to OL and equal to OB, and LO will be the revolved position of the ray through Oy taken in the given direction (Prob. XII.), that is, of the trace of the oblique plane of rays. Then OD, perpendicular to LO, is the revolved trace of the plane of shade, and the projection of the curve of shade upon the plane L'C{27, 8°). Hence, drawing ab, cd, etc., planes parallel to FC, the circles of radius Oa', Oc', etc., cut by them from the sphere, give, by their intersections with OD, the auxiliary projec- tion of the curve of shade, the points ff, ee' , DD' of the curve of shade, as may be more evident by considering the tangent rays parallel to LO at/, e, D, The vertical projection of the curve of shade, being an ellipse whose axes are AB and 2OD', its three other quarters may be found from the quarter Be'D'. The darkened portion of the figure represents the visible portion of the shade of the sphere. Examples, — 1°. Find the horizontal projection of the curve of shade in the same manner. Ex. 1". Change the direction of the light. [Both projections of a ray must then be given.} PROBLEM XIII. — To find the curve of shade on a torus. Repeating the arrangement of topics employed in Prob- lem XI., and for a like reason, we have — Direct Methods. — 1°. Points of shade on the apparent contours. 2l6 SHADES AND SHADOWS. In Space. — PI. XIX., Fig. 15. The torus here meant is of the kind used in architectural mouldings, and is gener- ated by a rectangle 0'0"C'D' , to which a semicircle is tan- gent as at CA'D'y\\i\s compound figure revolving about the side O'O", The contours are then those forming the projections, viz., the greatest parallel (Prob. XLIX.) A TB — A'B' y and the meridian in or parallel to V. The points of shade on these contours are the points of contact of tangent planes of rays, perpendicular respec- tively to H and V. In Projection.— RO — R'O' being the projections of a ray, the tangent at 7", parallel to RO, and at a' and b' ^ parallel to R!0' y are the traces of the planes just mentioned, and give the required points TT , and a' a and b'b (186). 2°. The highest and lowest points. In Space. — These points, the axis of the torus being ver- tical, are the points of contact of rays of light with that meridian which is in a meridian plane of rays. In Projection. — RO is the horizontal trace of the meridian plane of rays. Revolving it about O — O'O" into V, its meridian coincides with that of the vertical projection. The ray RO — R'O' appears after the same revolution at R!'0—R!"0'. Then radii, as o'r'", at 0' and ;/, centres of the semicircular ends of the meridian, and perpendicular to R"'0' y give the revolved points of contact as r"'r" of rays. By counter-revolution, r"'r" returns to rr' , the lowest point. The highest point is similarly found. 3°. Intermediate points by tangent planes of rays. In Space. — Any plane containing a ray is a plane of rays. If, then, a plane be passed through a ray, and perpendicular to any meridian plane of the torus, a tangent to the meri- dian in that plane, parallel to the trace on the latter of the perpendicular plane, will be the trace of a tangent plane of rays perpendicular to the meridian plane. Hence by (80) the contact of this tangent and meridian will be the point of contact of a tangent plane of rays with the torus, and hence a point of shade. In Projection. — Assume any meridian plane Oc. 00' is its SHADES AND SHADOWS. 21/ own projection upon this plane. RR' is there projected by the line RR^ — R'R^. Revolving the meridian plane about 0—0' O" into V, the projected ray appears at OR,"—0'R,"', when o'c''' perpendicular to O'Rl" gives c"c" the revolved, and thence, as in (i°) cc\ the primitive position of the point of contact of a tangent plane of rays perpendicular to the meridian plane Oc, Other points can be similarly found. Indirect Methods. — 1°. By auxiliary circumscribed cones. Ill Space. — This method stated as a principle applicable of all double curved, and the warped (Theor. V.) surface of revolution, is as follows : If two surfaces, M principal, and N auxiliary, are tangent to each other on a circle of contact (as a cone and sphere may be), then the intersection of the curve or plane of shade of N (supposed to be easily found) with the circle of contact of J/ and N, is a point of shade common to both, and thus a point of the required curve of shade of M. In Projection. — PL XIX., Fig. i6. Representing the torus of Fig. 15 again, to avoid confusion of lines, assume M'N' — J/iV anywhere between the highest and lowest points of shade, as the circle of contact of a cone whose axis is O — O'V , that of the torus, and whose generatrix is M'V\ tangent at M' to the meridian A'M'B' of the torus. Then n is the trace of the ray On — V'n' upon the plane MN\ and consequently nt is the trace, on this plane, of the tangent plane of rays to the cone, and Ot — V't' (not necessarily drawn) is the element of shade of this cone. Then tt' , inter- section of this element of shade of the cone, with the circle of contact JJ/iV — M'N' of the cone and torus, is a point of shade on the torus. 2°. Auxiliary tangent spheres. In Space. — This solution is the same in principle as the last, only instead of actually finding the circle of shade of the sphere, we only find where its plane cuts the circle of contact of the sphere and cone. In Projection. — Assume, as before, K L' — KqL as the circle of contact of a sphere with the torus. The centre 2l8 SHADES AND SHADOWS. 00' of this sphere will then be where K'o', through the centre of the semicircular portion M'A'D' of the meridian meets the axis 0—0' V of the torus; and the circle O'K' is the vertical projection of the sphere. Now O'Q', perpendicular to V'n', is the vertical trace of the plane of shade of the sphere (D.G., Theor. II.). Project- ing Q at Q, the line Qq, perpendicular to On, is the trace of the same plane on the plane K' L^, Hence qq' , intersection of Qq — Q'q' with KqL — K' L , is the intersection of the plane of shade, ancj hence of the curve of shade of the sphere with the circle of contact of the sphere and torus, and hence is a point of shade on the torus. If the axis of the torus O — O'O" were not in V, the meridian plane parallel to V would have been used just as Y itself has been here. Examples. — 1°. Let O — 0' 0" be in space, and find the complete curve of shade. Ex. 2°. Given both projections of a sphere, find its curve of shade as in the torus. [Here the given and auxiliary spheres will coincide.] Ex. 3°. Find the curve of shade on the inner or concavo-convex portion of an annular torus (73), when its axis is vertical. See Fig. 18. Ex. 4°. Find the curve of shade on the inner or concavo-convex portion of an annular torus (73), when the axis is perpendicular to V. Ex. 5°. Find the curve of shade on an ellipsoid of revolution, whose axis of revolution is vertical, by the Various methods of the last problem. Ex. 6°. Find the curve of shade on an ellipsoid of revolution, whose axis of revolution is vertical, by means of an auxiliary vertical plane of projection, Vi, parallel to the light [on which, since the curve of shade is plane,it will be projected in a straight line joining the points of contact of tangents parallel to the projections of rays on Vi]. PROBLEM XIV.— To find the shadow of a niche on itself. In Space.— K niche, as usually formed, is a hollow in a wall, composed of a vertical semi-cylinder of revolution capped by a quarter sphere, generated by the quarter revolution of the upper base of the cylindrical portion about its diameter. The total shadow is in four parts : that of the vertical SHADES AND SHADOWS. 2ig edge of the cylindrical part upon its lower base ; that of the same upon the cylindrical surface ; that of the vertical semicircle of the face upon the cylindrical part; and, finally, that of this semicircle upon the spherical part. The three former are easily found as in previous similar problems. The latter is the intersection of the cylinder of rays through the first semicircle, with the spherical sur- face and would be an equal semicircle, in case of a full hemisphere. For if parallel rays, AD, CB, etc., PI. XIX., Fig {n), are passed through all points of any great circle, as AB, of a sphere, but oblique to its plane, all but the two which, like mn-, are tangents, will be secants containing chords, as AD, ad, etc., of the sphere. But the plane PQ, perpendicular to these chords and through the centre of the sphere, will bisect them all. Hence AB and CD^ being thus symmetrical with PQ, are alike great circles ; which intersect in a common diameter at O. The method of (Prob. IX., 6°) is also readily employed. In Projection.— ¥\, XIX., Fig. 17. AbB-A'B'CD' is the cylindrical, and AOBb — A'E'B'O' the spherical part of the niche, the face of which is parallel to V ; and RL — R!0' is the direction of the light. 1°. By direct construction. — Taking the ray RL — R' 0\ its trace on the upper base of the cylinder is L,0' , Hence making O'L" = OL, and perpendicular to R!0' , we have R'L' the revolved position of the ray RL — RO' about the trace RO — R'O' of a plane of rays through it and per- pendicular to the face of the niche. The like revolved posi- tion of the semicircle of the sphere in the plane R'O' is R!T'B', hence a", its intersection with the revolved ray RL" , is the revolved position of the point of shallow, which, by counter-revolution, is found at a' . Again assume m'p' , parallel to R!0', as the trace, upon the face of the niche, of a plane of rays, perpendicular to that face, and hence to V. This plane cuts from the spherical part the semicircle whose vertical projection is m'p' , and whose position, after revolution about the trace m'p' into the face, is m'l"p\ Then m'l" parallel to RL" is the revolved ray at m' , and f 220 SHADES AND SHADOWS. is therefore the revolved shadow of m\ which, by counter- revolution, returns to I' . The horizontal projection /, of /', is found by making 77, on ll\ equal to I'l" . Other points may be similarly found. At TT , contact of a tangent plane of rays perpendicular to V, the points casting and receiving shadow unite, then Tl'a' is the ver- tical projection of a half of the semi-ellipse of shadow, all below A'B' being imaginary. 2°. The indirect method. — Assume any plane FK parallel to the face of the niche. It cuts from the spherical part the semicircle FK—Fn'K\ and from the cylinder of rays through the front semicircle, the semicircle equal to A' EB\ and whose centre is MM^ where the plane FK meets the ray OM — 0'M\ axis of this cylinder. Then drawing an arc n'y' with centre M and radius 0'A\ its intersection ti'n with FK—Fn'K\ is (Prob. IX., 6°) a point of the shadow on the spherical surface. 3°. The point of shadow on AbB — A'B'. — This is found with practical accuracy by indirect means ; also by an easy, direct construction. Thus, suppose the spherical part produced below the plane A'B\ as in finding a\ the intersection s's of the quar- ter ellipse Ta' with A'B' gives the point desired. Also, suppose the cylindrical part to be produced upward, and find the shadow, from hh' upward, of the semicircle AB — A'E'B' (i8i) upon this surface. The intersection of this shadow with A'B' — AbB will again give the same point s's. But again: 6^T', perpendicular to R'O' (Theor. II.), is the trace, on the face of the niche, of the plane of the circle of shadow; hence I'k' — Ik, where I'k' is parallel to O'T, is (27, 5°) a line in this plane. Therefore, k'k is a point of the trace of this plane on the upper base of the cylinder, and OO' is another. Then Ok is that trace, and ss' , its intersec- tion with AbB — A'B', is the point of shadow on that circle. Examples. — 1°. Substitute for the niche a hollow hemisphere with visible interior and its great circle parallel to V. Ex. 2^. In (i) let the great circle be parallel to H. SHADES AND SHADOWS. 221 Ex. 3". Let AbB be a semi-ellipse, and substitute for thfe quarter sphere a quarter ellipsoid of revolution having AB — A'B' for its longer axis. [Use the indirect method, and an auxiliary plane of projection perpendicular to the ground line.} PROBLEM XV.— To find the brilliant point of a double^ curved surface. In Space. — The absolute brilliant point of a surface is the one which receives the most light — that is, the one at which a ray is normal to the surface. The apparent brilliant point, relative to any given ob- server, is the one at which the most light is reflected to his eye in accordance with the principle that the incident and reflected rays make equal angles with the reflecting surface. Vertical projections being the ones generally shaded in a finished drawing, since they seem more like the reality, the reflected rays mean those which are perpendicular to the vertical plane of projection. The brilliant point is then the one at which a ray of light and a perpendicular to V make equal angles with the reflect- ing surface — that is, with a tangent plane to it, and hence also with a normal to it. In Projection. — 1°. On the sphere. — PL XIX., Fig. 14. The vertical projections of the incident and reflected rays are respectively L'O and O. Their revolved positions about the trace FC, into V, are LO and OB ; hence /, where the bisecting line Op of the angle LOB meets FBC the revolved great circle FOC, is the revolved position of the brilliant point. By counter-revolution, f is its vertical projection. 2°. On a concave tower roof. — PI. XIX., Fig. 18. A'CB' — A'O'B' is the half roof, V being taken through its axis ; OR — OR' is a ray of light, and OC — O the direction of reflected rays. Revolve the plane R'OC about OC into H, when OR— OR' will appear at OR!'— OR!", and N"0, bisecting the angle R'OC, is the revolved position of the bisector of the angle between the incident and reflected rays at O, 222 SHADES AND SHADOWS. By counter-Vevolution, ON"— ON'" appears in ROC at ON^ON, Once more, revolving ON— ON' about the ver- tical axis 00' and into Y, it appears at ON^ — ON^ '. Now Q' being the centre of the roof arc A'O'^ draw Q'q'" parallel to ON^' , and project q'" at q"y and q"q"' will be the revolved position of the brilliant point. By equal counter-revolu- tions of ON^—ON^ and q"q'", the latter returns to qq' , its real position. Hence, in making a finished shading of the vertical pro- jection, q' would be the lightest point of the shading, from which the latter, very light at first, would be made gradually darker until it reached the line of shade. 1 88. To find the brilliant point on an upright screw ^ find the point of contact of a tangent plane which is perpendicu- lar to the line bisecting the angle between a ray of light and a perpendicular to V. The brilliant /d?/;// is replaced on surfaces with which a tangent plane has an element of contact, by a brilliant line or element. But in most cases, as in that of an upright cylinder or cone, there will be no tangent plane perpendicu- lar to the bisecting line just described (see NO — N'O^ Fig. 1 8), and it will be practically sufficient, in the case sup- posed, to take the brilliant element in that vertical plane whose horizontal trace bisects the angle between the hori- zontal projections of an incident and reflected ray through a point on the axis of the body.* Examples.— 1°. Find the brilliant point on an ellipsoid. Ex. 2°. On a torus. * For numerous other examples, including those with divergent rays of. light, see my volume on Shades and Shadows. LINEAR PERSPECTIVE. First Principles. 189. Hitherto the eye has been supposed to be at an infinite distance (5) from each of the planes of projection, and the projecting lines (5) to be perpendicular to them. We shall now consider the case in which the eye is at a finite distance from the plane of projection, so that the pro- jecting lines will radiate from the eye to the determining points of the object. This case corresponds to the natural conditions under which objects are actually seen. The projecting lines are then the rays remitted from points of the object to the eye, and are thence called visual raysy and the system of projec- tion is that described in (6, II.). 190. Perspective, then, descriptively de fitted, is the repre- sentation of any object, upon some surface, usually plane, by a figure that shall appear to the eye situated at a certain point, just as the object itself would when seen from the same point. 191. Such representation embraces two branches : first, the construction of the correct form of the perspective 224 LINEAR PERSPECTIVE. figure ; second, the. addition of proper shading and color- ing. The former branch, only, concerns us now, and is called linear perspective. The latter, or the representation of natu- ral effects of shade and color by the rules of art, is aerial perspective. 192. Any plane passing through the point of sight is called a visual plane ^ all lines in it and containing that point being visual rays. A plane can always be passed through a given point and straight line, hence a visual plane can always be passed through any given straight line, and as it will con- tain the visual rays from all points of that line, its trace on the perspective plane will be the perspective of that line. If the line be parallel to the perspective plane, eVj perspec- tive will evidently be parallel to the line itself. 193. The visible boundary of an object is called its ap- parent contour^ and in the case of any curved surface, it is the curve of contact of a circumscribed tangent cone, whose vertex is the eye. 194. Linear Perspective, therefore, geometrically de- finedy consists in finding the intersection of the perspective plane with the visual cone (or pyramid) whose vertex is at the eye or point of sights whose elements (edges) are visual rays, and whose base is the apparent contour of the given object. Problems in perspective being thus, including the pre- liminary construction of apparent contours, either problems of tangency, or of the intersection of a plane and cone, the subject is obviously an application of descriptive geometry. 195. The angle, at the eye, subtended by any object, is called the visual angle, and its size determines the apparent size of the object. Unless otherwise indicated, this angle will be understood as the greatest horizontal angle subtended by the object. LINEAR PERSPECTIVE. 22$ A — Perspectives of Plane-sided Bodies — Various Methods. PROBLEM I. — To find the perspective of a monument com- posed of a square prism^ capped by a square pyramid. In Space. — If threads were stretched from the principal points of the object to a point representing the position of the eye, the figure formed by joining the points in which these threads would pierce the perspective plane would (189) be the perspective of the object. Models illustrating the definition (194) of perspective are actually made in this way. But perspective figures could hot generally be thus con- structed. We substitute for the actual object its two pro- jections which (13, second) are equivalent to it, likewise for the eye and the visual cone their equivalent projections, and for the perspective plane its traces. In Projection.— ?\. XX., Fig. i. Let ABDV—A'G'CV represent the given monument, EE' the position of the eye, and AA'Y^ perpendicular to the ground line B'A'O, the perspective plane having the same position. The perspective figure in ^^'F will appear by revolv- ing that plane about its vertical trace A'Y till it coincides with V ; but as such a revolution would confound together the perspective and the vertical projection, the perspective plane is first translated to the new and parallel position XOZ, at any sufficient distance from AA'Y, before revolu- tion. Again : remembering that the spectator at E faces the perspective plane ^^'F perpendicularly — that is, in the direction Ep — the points, for example, is at his right. It must so appear in the perspective figure, as at b^ in order not to reverse the figure, hence the new position XOZ of AA'Y must be revolved to the left about OZ to coincide with V. These preliminaries, alike for all cases under the present method, being explained, then, for example, VE — V'E is the 226 LINEAR PERSPECTIVE. visual ray from the vertex VV\ It pierces the perspective plane at //', perspective of VV\ which is thence translated along //j—/'//, parallel to the ground line, to/,// on XOZ. Thence, by the revolution of XOZy already described,/,// describes the horizontal quadrant /lA— A^, giving v the de- sired perspective of VV. Likewise b is the perspective of BB\ as may be traced by the full lettering of its construction, and all the other points may be similarly found. Having taken VE perpendicular to A A' Y^ to represent the observer as directly in front of the object, EE' is pro- jected on XOZ at /,^', which appears after revolution at E", If then the perspective figure be viewed by placing the eye on a perpendicular to the paper at E"y and at a distance from it equal to Ep, the figure will perfectly represent the given object, and will exactly do so at no other position of the eye. Examples. — 1°. Let the pyramid be octagonal. Ex. 2°. Replace it by a hexagonal prism. Remarks on the Method of Three Planes, 196. 1°. Its extreme simplicity. — It obviously conforms, in the construction of every point, to the most elementary defi- nition (194) of perspective, so that one point having been found, as v, nothing remains but to repeat the same process for every other point required. This method is therefore peculiarly appropriate to those who are making constant or frequent use of perpendicular projections, but who comparatively seldom need to make perspective constructions. 2°. The apparently large amotmt of paper covered. — But in practice, let UXYZ represent a sheet of drawing-paper to be occupied by the perspective figure only. Suppose this sheet to be stretched on a large drawing-board or table ; the auxiliary projections would then be pencilled, only, on cheap paper and temporarily fastened at the side of the LINEAR PERSPECTIVE. 22/ dra\ying-paper, as indicated by RQYD, Besides, on be- coming accustomed to the method, the operations can be greatly condensed by supposing the given object to be in the second angle (7) which will transfer all the construction lines now in front of the ground line B'O to the opposite side of B'O, where they might in practice be distinguished by colored lines. Moreover, the perspective plane AA'Y can then be revolved to the right about A'Y, into V, without translation. 3°. The convergence of lines, as ab and gc, which in space are parallel, but are seen obliquely. The correct relative direction of lines in a perspective figure being of prime im- portance to an undistorted figure, it will be useful to con- struct the points to which sucK lines converge. This is easily done as in the next problem. PROBLEM 1 1. — To construct the perspective of a pier, with the points of convergence of the perspectives of its parallel lines. In Space. — The dispositions in space, as indicated by the figure, being the same as in the last problem, if a visual ray be drawn parallel to any line of the given object, it will meet that line only at infinity. The intersection of this ray with the perspective plane will, as in any other case, be the perspective of the point from which the ray proceeds — that is, of a point at infinity. But a visual ray, R, which is parallel to one of several given parallels, Z, Z„ Z„ etc., will be parallel to them all, and will meet them all at one and the same infinitely distant point, /; hence the intersection of this ray with the perspective plane will be the perspec- tive i of that common infinitely distant point /. Hence the perspectives /, /„ /„ of the parallels Z, Z„ Z,, etc., will meet at i perspective of the point / common to all the parallels. /;/ Projectio7i.—?\. XX., Fig. 2. A CFM—A'CPM' repre- sents a stone pier, whose sides, ^(7 and FM,are parallel, and whose top is partly level, and at BCFK—B'C'G'K' partly in- clined to H. 228 LINEAR PERSPECTIVE. The perspective figure, as illustrated in full by a, per- spective of A A', may be wholly found as in the last problem. But now let ER—E'R! be that visual ray which is parallel to CB—CB'y and to all lines parallel to the latter line. This ray therefore meets CB—CB' and all parallels to it at infinity. It also meets the perspective plane CD'R[ at RR\ which is thence translated to the new position XOR( of that plane, at R.R^y and after revolution as before about OR^ into V, appears at V, Hence V is the perspective of the point at infinity where CB — CB' and all parallels to it meet each other, and cb and^>^, perspectives of CB—CB' and FK—G'K\ therefore meet at V. In like manner, ER — E'r' is the visual ray which is parallel to all horizontal lines which are parallel toBA — B'A\ This ray pierces the perspective plane at Rr\ which is trans- lated to R^r/j and revolved in the arc R^R^ — r/H to H, the perspective of the infinitely distant point at which ^^ — B'A' and all parallels to it meet. At N, therefore, the indefinite perspectives, as d/i and da of such parallels, will meet. Having V and Hy also c and a, the point b is the inter- section oi cV and Hay thus found without the visual ray from BB'. Also ^, for example, might have been found as the intersection of bH (supposing b already known) with either n^ay perpendicular, or ;/^, parallel to the ground line. 197. The points V and Hy and others which might be similarly found for other sets of parallels, are no others than those commonly called vanishing points. They are obvious- ly useful, and are found by the following. Rule. — To find the vanishing point of any systan of parallel lineSy find the intersection of the visual ray parallel to them with the perspective plane. This intersection will be the vanishing point in which the perspectives of all those paral- lels will meet. NotCy that in like manner the trace of any visual plane upon the perspective plane is the perspective of its infinitely distant intersection with any parallel planes, and is thence called the vanishing line of those planes. LINEAR PERSPECTIVE. 229 198. The perspectives of shadows can be found from their projections, just as the perspectives of the objects cast- ing them are. It is, however, necessary to remember that, in making the perspective, the observer is facing the per- spective plane, and that the light therefore should have the same position relative to the horizontal and the perspective planes that it usually has, relative to the two planes of pro- jection (178). PRO B LE M III. — To find the perspective of a square block and of its shadow on the horizontal plane. In Space. -^By visual rays. Having the shadow itself, the visual ray from any point of it will pierce the perspective plane in the perspective of that point. By vanishing points^ shadows parallel to the lines casting them will have the same vanishing points as those lines, and may be partly found on this principle. "hi Projection,— ?\, XX., Fig. 3. The block, ABCD—A'B'G' represents the block. Combining the methods by visual rays and vanishing points, make Of=^ FA ; and fa, equal and parallel \.o PA\ is the perspective of FA', this line be- ing in the perspective plane. Also make On^ = Fn, and Or^-= Fr, and the perspectives of the vertical edges at B and D will be in perpendiculars to the ground line at n^ and r^. But, now, ER—E'R\\hQ visual ray parallel to AB—A'B\ pierces the perspective plane at RR', which, by translation and revolution, as in Fig. 2, making XV=FR, gives F, the vanishing point (197) of AB — A'B' and of all parallels to it. Similarly, 5„ vanishing point of AD, etc., is found, XS, being made equal to FS, where 5 indicates the inter- section of ES, parallel to AD, with R'Fr. This done, fV and aV, fS^ and aS^ are the indefinite perspectives oi AB—F'G' and AB—A'B\ AD—F'H' and AD—A'D\ respectively. They are then limited at g and b, and h and d, by the perpendiculars njb and r^d. The Shadow, — AA^ — A' A,' is the proper direction of the 230 LINEAR PERSPECTIVE, light according to (174) and (178), giving A, A/', for example, as the shadow of AA\ Then, as partly shown by the ray A^Ey translating g to ^^ and revolving it to ^^, the shadow a^ of a — perspective of the real shadow A^A/ — is found as all the points of Fig. i were. Thence, as A^B^ is parallel to ABf its perspective a^^^ will have the same vanishing point Vy as ad, and will be limited at d^ by />^d^. Likewise the shadow of be proceeds from /^, towards 5,. 199. The line from a point to its shadow is a ray of light ; hence aa^ and dd^ are the perspectives of the rays^^j — A^A/ and BB^ — B'B/. These lines could have been found first, and then used in finding ^, and d^, by finding their vanishing points by the rule (197). Likewise /^i and^^„*perspectives of AA^ — FA/ and BB^ — , is the intersection of / V, with o' V'y or with rd, parallel to aa^. Finally r, perspective of Cy is the intersection of/V with n'V. LINEAR PERSPECTIVE. 239 The Shadow, — By (Theor. XIIL), and no particular direc- tion of the light being given, assume H' on the horizon as the vanishing point of horizontal projections of rays of light, and then R, making H'R perpendicular to GL^ as the vanish- ing point of the rays themselves. Then bR is the perspec- tive of the ray through the upper corner B of the prism, and hH' is the perspective of the horizontal projection of the same ray, hence b^y their intersection, is the perspec- tive of the shadow of B {b) on H, and hb^ is that of bh on H. Then ^,F/ will be the shadow, partly visible, of the upper edge BC. Observing how previous vanishing points have been found (197), project H' at //"then HE^ and RE' will be the projections of a visual ray parallel to the rays of light, and will thus show the direction of the light. PROBLEM VIII. — To Jind the perspective of a wall whose front face is sloping^ and whose horizontal edges are oblique to V ; also its shadow on H. In Space. — This problem will illustrate the method of diagonals and perpendiculars when H is revolved 180° (200), and the advantage of occasional recourse to other methods at certain points. In Projection. — PL XXL, Fig. 8. GL is the ground line, E'D the horizon, and EE' the point of sight, since, when the horizontal plane is revolved 180°, the horizontal projection of this point must be as far behind the ground line as the eye really is in front of the perspective plane. Also, as in (Prob. Yl.\E'D=Et. ABFI is the horizontal projection of the wall after the revolution of H. Then ACBJ, its sloping face, being the side towards E, is the real front of the wall. In fact, if, for example, A and B were as far behind the ground line, on the perpendiculars Ao and BL produced, as they now are in front of it, the line AB would then be in its real position. 240 LINEAR PERSPECTIVE. Finally, d'r' is the vertical trace of the horizontal plane of the top of the wall. So much done, the perspective of B^ for example, in the base of the wall, is b, the intersection of inDy perspective of the diagonal Bm, with LE\ perspective of the perpendicu- lar BL, The perspectives a and g oi A and F in the base are similarly found, as shown. The perspective, r, of CC in the upper base, is also simi- larly found, being careful to observe (203) that the diagonal Ce—C'e'y and perpendicular Cn — C from CC\ pierce the vertical or perspective plane at the height of the point CC whence they proceed — that is, at e' and C in the trace d'r' of the top of the wall. The points/ and y, perspectives of -F and y in the top of the wall, might be found in the same manner. But having g already, / may be found as the intersection of gf, per- spective of the vertical edge at F, with d'D, perspective of the diagonal FG^ or with the perspective (not shown) of the perpendicular at F. At / the perspectives of the diagonal and perpendicular at J would intersect quite acutely, and we use the vertical visual plane whose horizontal trace is JE, and whose vertical trace hj intersects the diagonal r'D, for example, perspective of Jr, at/. The Shadow. — Proceeding, as in the last problem, to bring all points within the limits of the figure, assume //", the vanishing point of horizontal projections of rays (205) and R that of rays, and find k, perspective of the point J in the base of the wall, by the intersection of the diagonal rD and trace hj. Then jR is the perspective of the ray from J (j) and kH\s> the perspective of its horizontal projection; hence j\, their intersection, is the perspective of the shadow ofy on H ; and bj\ is thence the shadow of bj. The shadows t^ of / and of/ (not shown) may be similarly found, which will determine the direction of the small visible portion of the shadow oi FI. Examples. — 1°. Let £' and Z> change places, and use the diagonals then appropriate. Ex. 2°. Let £' be above (fr\ and let the two projections of a ray of light be given. LINEAR PERSPECTIVE. 241 PROBLEM IX. — To construct the perspective of an inte- rior containing various objects^ by the method called that of scales. In Space, — This method, which most fully avoids the ap- pearance of the projections of the given objects, depends mostly upon a principle, the simplest application of which is illustrated by a diagonal and a perpendicular. Thus a diagonal, as DE — D'E , Fig. 6, cuts off, as we have seen, equal distances Et = E'D' from a perpendicular, as Et — E' ^ and from a horizontal line E'D' in the perspective plane. If, then, Fig. 6, we knew that the point C,A' was at the height F'A' above H, and at the distance CA back of V, we could, for example, at once lay off, by a scale, ^7/, equal to the known distance A C (numerically given), and draw the perspective diagonal h' D' and perpendicular y^'^S"', without drawing any planj and hence zvithout necessity for translating the perspective plane. The present problem illustrates this highly useful method, together with that of reduced vanishing points. In Projection. — PI. XXL, Fig. 9. ABEF represents, on a scale of three feet to an inch, the section in the perspec- tive plane of a room nine feet long and nine feet high. CD is the horizon at the height of the eye (192), five feet from the floor. C is the centre of the picture (207), five feet to the right of AE. Supposing the room to be seven and ahalfic^t deep, and that F, one of the vanishing points of diagonals (not shown), is twelve feet to the right of C on the horizon. Then the diagonal q, V from g„ yi feet to the left of B, will cut off from BC, the perspective of the perpendicular at B, the per- spective Bb of 7i feet of depth into the room from the ground line AB, But Y being inaccessible, make CD = iCV, and corre- spondingly Bq = iBq„ and qD will give the same point b as before. For the triangles CXV (where X represents the eye itself 12 feet in front of C) and the triangle Bq^B,, whose perspective is Bq^b, are similar, and with parallel 242 LINEAR PERSPECTIVE. bases CV and B^^; hence (by Fig. 7, and as will at once appear on drawing the plan ABB^A^ of the room), dividing these bases proportionally, CD = iCVy and Bq = \Bq^y DXy in space, will be parallel to the line qB^y of which qb is the perspective ; hence D is the vanishing point of this line (197). Having ^, the further side of the room is the square bfeay whose sides are limited by the perpendiculars FC, ECy and A C. The screen. — Let this be 3 feet to the right of A ; i\ feet into the room ; i^ feet wide, and 2\ feet high. Then draw the perpendiculars, dC and 4^6", and limit them at H and Gy by the parallel ^/^ to AB^ from g, found by the line vD^ drawn from a point \ foot from B, Make AI^ = 2J feet, and limit the perpendicular I^C at k on the vertical M, and draw the horizontal kKy meeting the verticals at //"and Gy which completes the screen. T/ie mirror and rod. — Let there be a mirror 4 feet high and li feet wide on the left wall* of the room, 3 feet from the floor, and 4^ feet back of AE. Then on AE make A^ and ^7 at the heights of 3 feet and 7 feet, and the mirror will be included between 36'and yC. The construction can be completed as before. The rod ro parallel to AB is assumed. Its true length would be from m to the intersection of Co produced with a parallel to AB from m. The drop light and shadows. — Let the light be 3 J feet to the left of BFy 3 feet back of the perspective plane ABEFy and i\ feet Ipng. Its intersection with the ceiling will then be the intersection / of the perpendicular 3IC from the point 3J on Z^^, with the line 4iA or with tyy where / is the intersection of \D with FCy and Ft is therefore the per- spective of 3 feet from F on EC. Then Z, perspectively 3J feet below /, is the intersection of the vertical IL with the horizontal Z^Z, where L^ is the intersection of the perpen- dicular l\C with the vertical tL^ on the right-hand wall. brawing now the vertical 7/,, intersecting L^L produced at /„ and the parallel to ^^ from Z,, intersecting IL pro- LINEAR PERSPECTIVE. 243 duced at /,, we have the projections, /, on the left-hand wall, and /j on the floor, of the light at L. This done, the shadow k of K, upon the floor, is the intersection of the perspective ray LK with its horizontal projection Ifik; and /, like shadow of /, is similarly found. Also s, shadow of o upon the left-hand wall, is the inter- section of the perspective ray Lo with its projection l^rs, 209. Ill extendmg this method of scales to the case of ob- jects oblique to V, as in Figs. 7 and 8, we may proceed in either of two ways. First, We can ascertain, by scale, from temporary plans on waste paper, the necessary dimensions and distances estimated on parallels and perpendiculars to the perspective plane. Thus in Fig. 8, Lo is the length of AB estimated in the direction of the ground line, and LB and Ao are the perpendicular distances of B and A from the perspective plane, or their depth into the picture. Second. See Fig. 7, drawing the arc DD^ from A as a. centre, the chord D^Dy and any parallels to it, will cut off" equal distances from Ay both on AD and on the ground line. This is the general principle of operation, of which the cut- ting off" of equal distances, on the perpendicular and on the ground line, by a diagonal (as at Ah =AC, Fig. 6), is a par- ticular case. Example. — Let a square be subdivided into any desired number of small squares, any of which at pleasure shall be, or shall circumscribe the bases of various geometrical solids, of various heights, and find the perspective of the whole. [See the next problem for cylinders and cones.] PROBLEM X. — To find the perspectives of two circles lying in H, the centre of one of them beings witjp the eye, in a profile plane. In Space, — In this problem, we shall use the method of diagonals and perpendiculars, aided by the principle that the perspectives of parallels to the ground line have, them- 244 LINEAR PERSPECTIVE. selves, a like direction, and illustrating the convenience of circumscribing squares in finding the perspectives of circles. The perspective plane is here translated forward. In Projection,— Y\, XXL, Fig. lo. Let PQRS and P^QJR.^S^ be the circumscribing squares of two equal cir- cles, tangent to the ground line, which contains the sides RS and R^S^ of the squares, sr^ is the ground line of the translated position of V, the perspective plane, E'D the horizon, EE' the point ,of sight, and D (E'D =EK) the vanishing point of diagonals. The translated position s of the trace 5 of the perpen- dicular PS and diagonal QS, is one point of their perspec- tives sE' and sDy respectively. Then q, the intersection of sD with rE\ the perspective of QRy is the perspective of 0, and enables us to draw gp^ parallel to sr^ for the perspective of QP, Likewise, j„ r„ ^„ p^ are found, as shown. . Next, in both figures alike, as shown by like letters dis- tinguished from each other by subscript numbers, draw the perpendiculars BMN, CK, FJH, whose perspectives nE\ kE' y hE' , meet the diagonal sD at m, o, /, perspectives of M, (9, F. These points enable us to draw mj\ aoi, fb, parallel to sr^, giving J, perspective of y, on hE'; a and /, perspectives of A and /, on sE' and rE\ and ^, perspective of B, on nE' . Through the eight points thus found by means of only one diagonal, the ellipse ab/k^ perspective of the circle ABFK, can be drawn. Special points, — We see that co is less than ok. That is, the perspective o of the centre O is not the centre v of the perspective ; v being the middle point of ck. Draw the perspective diagonal Dvg-y translate g- to G, and the diagonal from G will give V, the point whose perspec- tive is Vy and fVU the chord whose perspective is the trans- verse axis tvu of the perspective ellipse. T and U can also be found as the points of contact of the horizontal traces, -£'7' and EUj of vertical visual planes. In either case, t and u can be found in advance, and thence the perispective ellipse constructed on its axes ck and ///. LINEAR PERSPECTIVE. 245 The axes of the other ellipse are less easily found.* By precisely similar constructions, the perspectives of circles parallel to H and above it, or in planes perpendicular to the ground line, could be found. In fact, to realize the former case, it is only necessary to invert the figure, which will then, however, represent H as revolved 180°. Examples.— 1°. Construct Fig. 10 by the method of Fig. 8. Ex. 2°. In Fig. 10 find the perspective of a circle in a horizontal plane whose vertical trace is anywhere above the horizon. [If equal to ABIJC, and vertically over it, points of its perspective would be in vertical lines at abcf »?.] Ex. 3°. Find the perspectives of two circles, one in a profile plane whose traces are j^and SP, the other in the profile plane r^RiQi. Ex. 4°. Find the perspective of a cylinder. [See Ex. 2 (note) and draw vertical common tangents to the upper and lower bases.] Ex. 5°. Find the perspective of a cylinder whose axis is parallel to the ground line. Ex. 6°. Find the perspective of a horizontal cylinder whose axis is oblique toV. Ex. 7°. In Exs. 4, 5, 6, substitute a cone for the cylinder.f B— Pierspectives of Developable Surfaces. 210. The apparent contour of the convex surface con- sists of the element of contact of tangent visual planes (D. G., Prob. XXX., etc.). These planes are tangent to the visual cone whose base is a base of the surface, hence the perspectives of the ele- ments of apparent contour are tangent to the perspective of either base. Problems of this kind relating to single bodies are so easily solved by any of the methods already explained,:}: that we will take here but one, of a little more complication. * See my Higher Perspective, Prob. LXV. f See my Elementary Linear Perspective. X See either my Elementary or Higher Perspective. 246 LINEAR PERSPECTIVE. PROBLEM XL — To find the perspective of an elbow arch and of its shadows. In Space. — The elbow arch differs from the complete groined arch in that the two equal intersecting arched passages which form it, terminate at their meeting, instead of completely penetrating or crossing each other. Thus, if PI. XXI L, Fig. 12, were the plan of a groined arch, composed of the semi-cylinders, one of which had the vertical semicircles projected on JG and IKiox its bases, and the other those projected on JK and 6^/ for its bases, the horizontal projection of the real portions of the former cylinder would be JoG and Kol ; while the horizontal pro- jection of the real portions of the other cylinder would be y^isTand Gol ; and the two cylinders would intersect in the two equal vertical semi-elliptical groins Jol and KoG. But, taking the same figure as the plan of an elbow arch formed by the same cylinders, one groin as KoG would dis- appear, and the horizontal projection of their real portions would be JGI and JKI, where 6^/ and// would be elements of one cylinder, and KI and Oo would be those of the other. All the operations required in case of the groined arch will, therefore, be exhibited with less repetition, and on larger and less broken surfaces. The problem will further illustrate the use of reduced vanishing points, lines in the direction (9/, Fig. 12, being substituted for diagonals, as y/, and hence giving a vanish- ing point half as far as that of the diagonals from the centre of the picture ; also the construction of shadows on cylinders parallel and perpendicular to V, and on profile planes. In Projection. — 1°. The outlines. — Let ^^/T^ be the inter- section of the perspective plane with the perpendicular passage at the beginning y, Fig. 12, of its intersection with the parallel one. This section is then seen in its real size on the scale of the drawing. vS is the centre of the picture and SD^ the horizon. Z>, is the vanishing point of lines parallel to (9/, Fig. 12. ^5 and BS being then the perspectives of the LINEAR PERSPECTIVE. 247 perpendicular edges JG and KI, Fig. 12, of the floor of the arch, they are limited at D by 7X>„ perspective of (9/, and at C by DC, parallel to AB. The verticals at ^'and Z>, equal in space to those at A and B, are therefore limited at G and / by the perspective per- pendiculars ES and FS (not shown). Supposing BN to be the thickness of the exterior walls, make NU=iNA and UD, and NS will intersect at the exterior corner A^^, whence the parallel N^C to DC czti be drawn. The side semicircle, EhG, can be found independently by lines parallel to OG, Fig. 12 (not drawn), but is here found from the groin. 2°. The groin. — Divide Ojf, Figs. 11, 12, into four equal parts, when parallels to 01 from Z, Q, Py will meet the groin at /, 0, p. Now Qo is in the level of O, the crown of the arch, but Z/and Pj\ are in the level of g^. Fig. 11, found by drop- ping the vertical QQ^ to intersect the semicircle EOF, and drawing the horizontal Q^jf^. This done, 0, summit of the groin, is the intersection of the perpendicular (9S and gA? perspective of Qo, Fig. 12. Also, / is the intersection of Q^S and LD^, and / that of Q^S and PD,, Greatest apparent height of groin. — This point is on the element of apparent contour of the parallel cylinder (210). To find it, suppose the structure to be revolved 90° to the left, for example, about a vertical axis through 0. Then, it being square in plan, the vertical projections of the equal cylinders will replace each other, and ^y will represent the perspective plane, and a point S^ (not shown) at the distance of 2SD^ to the left of S^, will represent the point of sight. The tangent S^W to EOF \y\S\. then be the revolved trace of the visual plane tangent to the parallel cylinder, and V a point of its trace on the perspective plane, which, by counter- revolution, appears at Vw and is met by the perspective perpendicular WS at w, the required apparent highest point. The lateral semicircle. — This is here thus found from the 248 LINEAR PERSPECTIVE. groin, j and.y, are the intersections of the perpendicular y,5 with the horizontal parallels Ij and pj\, the latter con- founded in the figure with Vw ; and hy the apparent highest point, is the intersection of the perpendicular W^S with Vw, 3°. The shadows, — That of EOF on the perpendicular cylin- der and side wall is found by planes of rays perpendicular to Y, each of which will cut a point from the semicircle EOF, and an element receiving its shadow, from this cylinder. Thus, SR being the vertical projection of the ray of light through the point of sight, assume R as its vertical trace and consequently the vanishing point of rays, whence H (Theor. XIII.) will be the vanishing point of horizontal pro- jections of rays. Then XY, parallel to SR, is the trace on the perspective plane of a plane of rays, containing the ele- ment YS of the cylinder, and the point X, whose shadow x is the intersection of YS with the perspective ray XR, Zy where the shadow begins, is the contact of the tan- gent, parallel to XY, Points of shadow of EOF upon the side wall, as x^, shadow of JT,, may be similarly found, as shown. Also thus: R^TSO, being a plane perpendicular to the ground line, SR and SR^ are the projections of a ray of light through 5, and of its projection on the plane R^ TO, Hence RR,^ is the trace, on the perspective plane, of the plane of rays containing these lines, and consequently R^ is the vanishing point of all lines parallel to the projection SR^ of rays upon planes perpendicular to the ground line. Then XJz is the trace on the perspective plane of a plane of rays through Jf„ and perpendicular to R^ TO, hence the ray X^R meets its projection kR^ on the plane BK, at x^, per- spective of the shadow of X^ on BK, And again thus: Xj, being an ordinate of EOF, the shadow of t on the side wall Bk is /„ intersection of tR and FR^y by the last method. Then t^x^, parallel to tX^, is the shadow of tX^y limited at x^ by its intersection either with X,Ry kR,, or Y,S, The graphical use of various methods is the advantage of each in cases where the adoption of others would give LINEAR PERSPECTIVE. 249 inaccurately acute intersection, as at the intersection of /,a', with kR^y or occasion confusion of new lines with previous ones. 4°. The shadows on the parallel cylinder ^ rear wally and floor. — These are cast by both the front and side circles, and the vertical AE. Drawing AH^ the portion AA^ is the shadow of AE on the floor. At A^ it rises, parallel to AE, on the rear wall, and is limited by the ray ER 2it e, shadow of .fi"; also found as follows. The projection of O^, centre of EOF, upon the rear wall is o^, on GI. Then 0^ intersection of the ray O^R with its projection 0^0^, on the rear wall,, and parallel to SR, is the shadow of O^ on that wall. Thence o^e, parallel to O^E and limited by the ray ER, is the like shadow of O^E, and radius of the circular shadow of EO, limited by DI, A tangent from R^ to the side circle is the perspective of the trace upon the plane -^y of that circle, of a plane of rays tangent to the parallel cylinder, and hence gives j (accidentally on If) as the point (analogous to Z on EOF), where the shadow on the parallel cylinder begins. Secants from R^, as RySr and R^m, are traces of similar but secant planes of rays, cutting points from the side circle, whose shadows fall on the horizontal elements as su and nq, cut from the cylinder or the rear wall by these planes. Thus //, shadow of r, is the intersection of su with the ray rR, and q, shadow of m, is the intersection of nq with mR, The shadows vqe and x^e are tangent to each other and to A^e at e, as the lines casting them are at E. Examples. — 1°. Replace the elbow arch by a complete groined arch, and make the figure fill the plate. Ex. 2°. Find the perspective of two intersecting solid cylinders ; one of them vertical, the other parallel to the ground line, with their shadows. Perspectives of Warped Surfaces. 211. There is seldom occasion to find these. That of the warped hyperboloid of revolution is best found in the way 250 LINEAR PERSPECTIVE. to be explained for a similar double-curved surface. Screws and other practical examples of warped surfaces are always limited by simple and definite outlines, whose perspectives, easily found by methods already explained, will be, or will determine the perspectives of the given surfaces. Perspectives of Double-Curved Surfaces. 212. The method of three planes is in many cases prefer- able for problems under this head. Also, in nearly all cases, an important part of the problem is the pre.liminary construction of the apparent contour, whose perspective is that of the given body. 213. The principle of (190) has important applications to the perspectives of double-curved surfaces. Thus, a sphere itself can never appear otherwise than round, or of circular outline, however viewed. Accordingly, if the perspective of a sphere be actually viewed from the real positions of the eye as given by its projections, or vertical projection and points of distance (207), this perspective will appear circular, though it may really be any one of the conic sections, and will so appear when viewed from other points than the actual point of sight. For the visual cone, whose vertex is the eye, and base its circle of contact with the sphere, will be a cone of revolution. Now the perspective plane may be taken so as to cut this cone in any one of the conic sections. But the section of the visual cone by the perspective plane is the perspective of the given sphere, which may thus be any one of the conic sections. In fact, the perspective of a sphere will only be as well as appear circular, when the visual ray to its centre is per- pendicular to the perspective plane. Examples. — 1°. Find the perspectives of four spheres, the centre of one of them being in the perpendicular from the eye to the perspective plane, that of another being at any other point in the plane of the horizon, that of the third LINEAR PERSPECTIVE. 251 being below, and that of the fourth above this plane. [By the method of either two or of three planes, all secant vertical visual planes (204) will cut the spheres in circles, great or small, tangent visual rays to which, easily shown on H or V in revolved position, will give points of apparent contour and their perspectives.] Ex. 2°. Find one or more points of the curve of shade on each sphere. [The most elementary method being to find first their projections.] Ex. 3°. Find the perspective of a hollow hemisphere with its visible interior and shadows, limited by a great circle parallel to V. Ex. 4°. Find the j)erspective of a hollow hemisphere with its visible interior and shadows, limited by a great circle parallel to H. PROBLEM XII. — To find the perspective of the concave dou- ble-curved surface of revolution^ called a piedouche or scotia. In Space, — This problem will illustrate the method of three planes, as applied to double-curved surfaces ; together with the preliminary construction for finding t\\Q projections of the contour and the shade. To aid in becoming familiar with the method, the eye is, for variety, placed to the left of the object, and the perspective plane translated about its vertical centre line instead of about its vertical trace. In Projection. — PI. XXII., Fig. 13. The given body, ADBC—A'B'C, is called a scotia, when of large diameter compared with its height, as when found in the base of a column, and a piedouche, or pedestal, when proportioned as in the figure. The meridian curve ATqj' may be com- pounded of unequal quadrants, or, as in the figure, may be a semi-ellipse on A'f as a diameter. D^A^ is the real, and d^a^ the translated position of the perspective plane, which is revolved about the vertical at O to the position d^e^, parallel to V. The place of the eye is EE' (at the left of Fig. 1 1), and GL is the ground line. 1°. The projections of the apparent contour. — This is the contour seen from EE'. Two points of it are UU and H {H' not lettered) the points of contact of the vertical visual planes represented by tangents from E, to tl^e smallest circle of the concave surface. Two others are the points of contact, near Ff and BB\ of visual planes perpendicular 252 LINEAR PERSPECTIVE. to V, and hence shown by tangents from E' to the meridian Aq'f and B'p'. Intermediate points are found by auxiliary tangent cones or spheres just as in (shadows, Prob. XI II.), only that the rays considered converge to EE' instead of being parallel. Thus, assuming any parallel p'q'—pq (33) of the surface, the tangent /'g' at//' meets the axis QC of the surface at the vertex Q' of the tangent cone having /y for its circle of contact. Then the visual ray Q'E' — CE meets the plane /y of that circle at R'Ry whence the tangents Re and Rc^ give the two points cc' and c^c^' of the required contour. The points //' and /, are similarly found. But note^ that only so much of this contour is real on an opaque solid of this form as is between the points as MM' on the lower part, and KK' on the upper part of the sur- face, where visual rays are tangent, not only to the surface of the body, but to the curve of contour itself; because, for example, tangents to the surface between M and T, on the right, are also secants, having first pierced the surface somewhere to the left of the chord MT, 2°. The eurve of shade. — Recalling the principle of. (178) let Rfi—R^Q be the projections of a ray of light. The curve of shade is then found by constructions precisely similar to those for obtaining the contour. Thus, for inter- mediate points, for example, taking the same cone p'q'Q! as before, the ray of light CR^ — Q'Rl pierces the plane of its base at R^R, so that the tangents from R to the base pq^ being the traces, on the plane of the base, of tangent planes of rays, give ss' and s^s^ two points of shade. 3°. The perspective, — This is now found essentially as in Prob. I. Thus, the visual ray DE — D'E' pierces the per- spective plane at DJD^, which is thence translated on D^d^ — D^d parallel to aj.^ to d^, and then revolved to d^ and projected at d, the perspective of DD'. Likewise m^ extremity of the visible contour of the con- cave surface, is the perspective of MM' , and all other LINEAR PERSPECTIVE. 253 points, though not fully shown, of the contour and shade are found in the same way. 4°. Method by perspectives of parallels. — Sometimes, as a substitute for the preliminary construction of the projec- tions of the contour y when this is not familiar, we may find by Prob. X. the perspectives of numerous horizontal circular sections of the given body, and then the perspective of the visible contour will be a curve which can be sketched tangent to the perspectives of all these circles. Examples. — 1°. Filling a plate with the perspective only of the piedouche [the projections being on adjacent temporary sheets], find the perspective of its curve of shade, and of the shadow of the upper cylinder [fillet] upon the concave surface. Ex. 2°. Find the perspective of an ellipsoid, a vase, an annular torus, or a bell, by the method (4°) just indicated.* Adhemar's General Method, 214. Viewing perspective as a direct application of de- scriptive geometry, the method of three planes (196) may be considered as its most general method. But in the present case, perspective is considered rather in relation to practical convenience of construction ^ com- bined with the best pictorial effects^ obtained without constraint as to the relative position of the object and the observer. The method of Adhemar is universal in fulfilling all of these conditions ; so that under the first no point of con- struction shall ever fall without the convenient limits of the paper.f * Referring the reader to various accessible works, among them both my Elementary and Higher Perspectives, for abundant perspective examples as usually treated, we will, by preference, next give what may not easily be found elsewhere, a fuller account of the practical method of scales by Adhemar. f "Before publishing -this method in the first edition of my work, I had wished to assure myself that it was always applicable, and to attain this object I imposed upon myself the condition of executing all the plates of my atlas [81 (folio) in number] without ever employing a single point without the plate.*' Adhemar, 3d Ed., i860. 254 LINEAR PERSPECTIVE. In relation to its principal determining lines, this method is called that of dimension and vanishing scales^ and has been partially illustrated in Prob. IX. PRO B LE M XIII. — To find the perspective of a point taken at pleasure in H, In Space. — PL XXII., Fig. 14, represents an auxiliary sheet containing a plan view, on a smaller scale, and used where it is desired to make a large perspective view. EP and EQ represent the extreme visual rays, chosen so as to include a suitable visual angle (195) generally not varying materially from 45°. Ec, the central visual ray, bisects the visual angle, and the perspective plane PQ is perpendicular to it. PS is the horizontal trace of a plane perpendicular to the trace and plane PQ, and ^ is a point in ^ at a per- pendicular distance AB from PS. Regarding EP and EQ2iS> horizontal traces of vertical visual planes, the two vertical edges of the picture, at Pand (2, are their traces on the perspective plane. In Projection. — Passing now to Fig. 15, PQ = 2iPQ of Fig. 14; PQRS is the perspective plane standing perpen- dicular to the paper at PQ, Fig. 14, VC is the horizon, and C the centre of the picture. Then take Cx = cP, Fig. 14, and drop xp, perpendicular to VC, and meeting PC, perspective of PS, Fig. 14, at /. Make/^ = ^^, Fig. 14, and draw Cq. Also make CV=ECf Fig. 14, and PB = PB, Fig. 14, and draw BV, and from d, its intersection with PC, draw l^a, parallel to PQ. Then a, intersection of da with Cq produced, is the perspective of A, Fig. 14, Cq being the perspective of the perpendicular from A. Demonstration. — Produce Cq to A on PQ, Fig. 15. Then PA : pq -.: PC '. pC '.: RC : xC{=Pc, Fig. 14) : : IPQ : iPQ (Fig. 14) :i PQ: PQ (Fig. 14), whence, finally, PA :pq :: PQ: PQ (Fig. 14). LINEAR PERSPECTIVE. 255 Thus, by laying off AB once on /^, we obtain the same result as if we had made PA = ^AB (Fig. 14), and PB and CV each f of their present size. It may be objected to this method that it determines longer lines, as ba, by shorter ones, as pq ; but in its applica- tions, where numerous lines are to be found, it will be seen that, by means of numerous mutual checks, points are very accurately found. PROBLEM XIV. — To find the perspective of any quadrila- teral in H. PI. XXII., Fig. 16, represents a temporary sheet con- taining the plan A BCD of the quadrilateral, the perspective plane at PQ and the visual angle PEQ {E, the intersection of kP and cQ) all on the same convenient scale. Ps and Qc^ are perpendiculars to PQ at P and Q. PI. XXI I., Fig. 17, represents the perspective plane PQVV„ where PQ= 3PQ, Fig. 16, the horizon FF„ centre of the picture P, vanishing scales PE and QE^ perspectives of Ps and Qc^ ; and pq^ scale of dimensions, all found as in Prob. XIII., Fig. 15. That is: EV,= EV,,= rE, Fig. 16, and simi- larly always ; and Ex = Pr, Fjg, 16. So much *done, the perspectives of the corners, AjB,C, Z>, of the given quadrangle, may be found by ordinates to Ps or Qc^y either from those corners, or from the intersections of the sides of the quadrangle with the visual rays Pk and Qc, or with PQ. The latter is more accurate, as determining required points between points of construction. Proceeding as in Prob. XIII., make PP ,= PF, Fig. 16, and draw F'V, and i^will be the perspective of F, Fig. 16. In like manner the intersections of all the sides with PE (Ps) or QE{Qc^ could be found, as shown at h. But produce the sides to give n, a, b, r, etc.. Fig. 16. Then, for example, make Qc'= Qc,, Fig. 16, and draw c'V,, and thence c,c parallel to PQ, and c is the perspective of Cy Fig. 16. Also find ^ and b similarly. 256 LINEAR PERSPECTIVE. Again, for example, make pn^ and pd respectively equal to Pn and Dd^ Fig. 16, and En^ and Ed produced, will re- spectively give n and verify D. Now draw aF and cn^ which give ^. iVand 5 are found by making Pe' and Ps' respec- tively equal to Pe and Ps, Fig. 16, drawing e'V and s'V to meet PE at e and s, perspectives of e and s, Fig. 16, and thence eN and ^5, parallel to PQ, and intersecting Er at TV and 5, perspectives of the same points on Fig. 16. Then Sb and nc give ^, etc. We have thus, first, accurately made a perspective figure much larger than the usual methods would produce from the given plan ; and, second, have done it without the need of any point of construction outside of the picture. PROBLEM XV. — To find the J>erspective of a stone bridge. In Space. — This concluding problem will illustrate the following points. 1°. The compact but accurate construction, by means of suitable checks, additional to those before given, of large perspective plans from small projection plans. 2°. The construction of perspective altitudes. 3°. The construction of perspective elevations by throw- ing them up from auxiliary perspective plans ; a method which is particularly useful in the perspective drawing of more complicated structures. 4°. The independence of the plan and perspective in re- spect to scales, and thus the entire generality of the method. In Projection. — PI. XXIII., Fig. 5, represents the required perspective figure, while Figs. (A), (B), (C), (D), are sup- posed to be, in practice, on temporary sheets fastened adja- cent to the drawing-paper. Fig. 18, though Fig. (C) or (C) and (D) may at pleasure be on the same sheet with Fig. 18, and Fig. (C) may be either above or below Fig. 18, as most convenient. In Fig. (A), the visual angle AEB includes two piers and a part of a third one. AB is the perspective plane, E the LINEAR PERSPECTIVE. 257 point of sight, v the vanishing point of lines parallel to the sides, as aby of the piers, and Ds' the trace of a vertical plane parallel to the visual plane EB ; also EA happens to coin- cide with the face si of a pier. The lines of construction relied upon are the lines, as sfy containing the points of the piers, the traces of the planes, as hgy containing the longer edges of the piers or the axes, as rr^y of the arches, and parallels, to AB, as ss' or yy„ from corners of piers, or from traces of the previous lines upo n the visual planes or upon Ds\ The points used in determining the perspectives of these lines are their traces, on the limiting visual planes, or upon ^^, (Fig. C), perpendicular to ABy or on Ds', or AB, Fig. (B) is a partial side and end elevation on the same scale as Fig. A. 1°. The perspective plany Fig. (C). This is found as fol- lows : Having made Cx = \ABy Fig. (A) (the letters in parentheses denoting figures), and drawn xp and /P, and taken V as in the two previous problems, and AB at pleas- ure; make Ay^ = AJ^ (A), and draw y,F, giving J^ on ACy perspective of ^i on AC (A). Then making qP — BD (A) or BD a fourth proportional to AB (A), ABy BD (A), we have DB^y the perspective of Ds' (A), since By^ on the hori- zon VCB^ is the vanishing point of all parallels to EB (A). Now draw J^Jy perspective of JyJ (A). Next make Ak^ — Ak (A), draw k^Vy giving k on ACy and kKy parallel to ABy giving K ; then KJ is the perspective of KJ (A). Finding Kk" and ii"y similarly, perspectives of Kb and ic (produced to meet Ds' produced), KK^y perspective of KK^ (A), is accurately found by the intersection of KJ with Kk" and ii". Also K^ can be directly found by an ordinate par- allel to Kk. All other points of the perspective plan, on AA^yACy BB^y or DB^y can be similarly found by means of the lines B.nd points of construction already stated. 2**. T/ie perspective elevationy Fig. 18. Project the points as Sy iy etc., oi AA^ upon DB^ by parallels to ABy as at /, i'y etc., and then erect the verticals //„ i"Uy PHy etc., parallel 258 LINEAR PERSPECTIVE. to BB^. Then, taking <:F, as the horizon, D^c becomes the new perspective of Ds' (A)'. At P', intersection of /Wwith D,c, make FU' = uu" (B). Then, on P'U' lay off from F , the heights of the several characteristic horizontal lines of the bridge above ss' (B) — that is, the heights of the horizontals through /, F \ CALlFu»,',:vL\ SPHERICAL PROJECTIONS. 215. Spherical projections are the projections of such and so many of the circles of the earth, considered as a perfect sphere, as will permit the representation of any given points of its surface, and of the zones which are marked by the limiting apparent positions of the sun. These projections are commonly made upon the plane of some great circle, but sometimes upon a tangent plane, and sometimes upon one or more circumscribing tangent cones, which, by development, will also afford plane representa- tions. 216. The necessary circles are: Jirsfj certain ^reat cir- cles. Of these are : T/ie equator y whose plane is perpendicular to the axis of the earth. The ecliptic^ the intersection of the earth's surface with a plane containing its centre and that of the sun, and making an angle of very nearly 23^^° with the plane of the equator. Meridians, which are all in planes containing the axis, and hence perpendicular to the equator. 217. There are also the ioViOVfrng smalt circles yidSi in planes parallel to that of the equator : Parallels^ taken at any convenient constant number of degrees apart. SPHERICAL PROJECTIONS. 261 The tropics ^ Cancer and Capricorn, tangent to the eclip- tiC; and therefore 23^°- di&tarft ffom the equator, and limit- ing the central or torrid zone, beyond which the sun's rays are never vertical. '♦'''.:/ The polar circles, a^-ctic .and antarctic, 23^° from the north and south poles, respectively," ahd whose distance, on a meridian, from the remotest point of the ecliptic, is 90°. These mark the frigid zones, outside of which the sun's greatest apparent altitude is never less than 23^°. ^iB. The point of sight may be at an infinite or at a jlnite distance. In the former case, the projection is always made upon a plane, is called orthographic , and can easily be constructed by any one familiar with orthographic or perpendicular projec- tions (5 — 6). The plane of projection is, in spherical projection, often called the primitive plane, and its intersection, or contact, with the sphere, the primitive circle, or point, respectively. 219. The orthographic projection Y^ usually made either upon a plane containing the axis, as that of a meridian, or perpendicular to it, as that of the equator. In the former case, the projection oi ^\ cry parallel would be a straight line perpendicular to the axis, and of every meridian J an ellipse, whose transverse axis would be the axis of the sphere and whose conjugate axis is the projection of the diameter perpendicular to the axis, upon the plane of projection. Thus, PL XXIII., Fig. 19. NESWis the meridian in the plane of tlfts paper, NS the axis, and the meridian perpen- dicular to the paper, and NdS the meridian at an angular distance equal to Nc'* to the west of O; recollecting that the circles NESW and EOW are equal and perpendicular to each other, and that NWS may be considered as NdS after revolution about NS into the plane iV^SW^, so that by (Prob. XXI., 2°) 0N(,= Oc') : Od :: nc' : nc. 262, SPHERICAL PROJECTIONS. 220. In the orthographic projection of the sphere on the plane of the equator^ the parallels would be circles concentric with the equator, and the meridians, diameters of the circle representing the equator. The advantage oi \)ci\s xvi^ihod. to all acquainted with de- scriptive geometry, is its great theoretical simplicity. Its disadvantages are the practical difficulty of construct- ing the elliptical meridians and the more serious one of the great distortion, by crowding, the nearer we approach the great circle which is in the plane of projection. 221. The eye at a finite distance (2 1 8). All cases under this head are YG2\\y perspectives of the sphere (189) varying with the positions chosen for the eye, and the kind and position (215) of the surface on which the projection is made, all of which are quite numerous. 222. Stereographic projection, — In this case, the eye is at the pole of the primitive circle, as at the north or the south pole, if the projection be made on the plane of the equator. PL XXIIL, Fig. 20. By this projection, only the hemisphere beyond the primitive circle is projected from the same point of sight, and it is mechanically easy of execution, since it possesses the curious property that every circle of the sphere not in a plane through the eye is projected in a circle, since, as will be immediately evident by a figure, the section cut by the primitive plane from the visual cone (194) whose base is any circle of the sphere, is a sub-contrary section of that cone (96). The distortion by this method is, however, still considerable, and is greater by diminution of true size near the centre of the primitive circle, for E being the point of sight, NOS the primitive (218) or perspectivanplane, and NWS the hemisphere projected from E, the perspective ajb^ of ah is less in proportion to ab th^ c^d^, perspective of cd, is in proportion to cd. In other words, be being a semi-parallel, and hence e, in the perspective plane NS, being its own perspective, b^, the perspective of by is further below eb than d^ is below d. SPHERICAL PROJECTIONS. 263 223. Construction. — From the ease with which the stereo- graphic projection is made on the plane of a meridian, we give the construction. Let NS, Fig. 21, be the axis, EWikiQ equator, (9, shown after revolution 90° about NS at E, the point of sight, and NESW the given meridian. Then ab being the vertical projection of a parallel, a and b are their own perspectives as seen from O, and / is that of a^ as shown by the revolved construction aE, Then the centre doi the parallel afb is the intersection of the tangent at a with the axis SN. For call g the middle point of af. Then the triangles ^^/and EaJVare evidently similar, hence their doubles da/ and Ea/i are so also, W/i being made equal to Wa. But the angle aE/i is measured by Wa, which therefore measures the equal angle ad/. But Wa = i{SWa — Na) which last measure shows da to be a tangent. Again, to draw any meridian, as the one 30° distant from W, draw S/i making OS/i = 30°, then /i is the centre and /iS the radius of the required meridian NkS. This will be obvious on making a horizontal projection, when O will appear as at S, and the horizontal projections of meridians will be straight lines through the centre. 224. Globular projectio7i. — PI. XXIII., Fig. 22. In this, the eye is exterior to the sphere, and at a distance Sd from it, on the axis NS of the primitive circle E W equal to the sine of 45° = \f\y the radius of the sphere being unity. Result: Nb being 45°, ab = \f\, aO =: ^^ ; OS = i ; Sd= ^i but da : ab :: dO : Oc or 1+2 4/i : Vi :-• i+i/i : C , • , * = ,*M = 4 = ^'. That is, the globular projections Oc and Ec of the equal arcs Eb and Nb are equal, and the like is nearly true of other equal divisions of NbE. % 264 SPHERICAL PROJECTIONS. But, on the other hand, the projections of circles not parallel to the primitive plane will generally be ellipses. 225. Approximate globular^ called equidistant, projection. In this, which is often called globular projection, and sup- posing it made on the plane of a meridian, divide j5"^and NS into equal parts, also NbE and NW into the same number of equal parts as NO. This done, meridians will be represented by arcs, each of which is determined by three points, viz., the poles, and one of the points of equal division on the equator EW, The parallels will also each be determined by three points, viz., a point of division on ON, and the points similarly num- bered on NE and NW, reckoning from O, E and W, 226. The three foregoing projections are the only ones which are adapted for representing the entire surface of the earth. They can be sufficiently well compared by making maps of the northern half of the western hemisphere, on the same scale for each, on the given meridian NESIV in the plane of the paper. The comparative amount and place of distortion will then be apparent. 227. Remembering that a sphere is undevelopable, no map of the world can represent the surface of the earth on a uniform scale as in plotting a survey of a tract sufficiently small to be regarded as plane. The following methods, of more or less approximately uniform scale, are applicable to greater or less portions of the earth's surface. 228. Cylindrical projection, PI. XXIII., Fig. 23. Here the projection is made upon a circumscribed tangent cylin- # der WEMK, the eye being at the centre of the sphere. Any parallel as cd is thence projected by a visual cone whose vertex is O, in the circle CD, intersection of this cone SPHERICAL PROJECTIONS. 265 with the cylinder. The meridians are projected upon the cylinder by their own planes, in parallel elements of the cylinder. The cylinder is then supposed to be developed, when meridians and parallels will both appear as parallel straight lines, each crossing the other at right angles. It is obvious that this method would not apply within a certain distance of the poles, owing to the extreme distance of the projections of the parallels from the equator. Indeed, this projection is but very little used. 229. Mercators projection. — This is a modification of the pure cylindrical projection, and is almost exclusively used for mariners' charts, since it is so constructed that the path of a vessel sailing so as to cross the meridians at a constant angle will be projected in a straight line. It is very easily shown that the length of a parallel or of any part of it, ab or ad^. Fig. 19, is equal to the great circle WE^ or arc Wd, respectively, multiplied by the cosine pf the latitude Wa of that parallel. Hence, if, instead of making the meridians converge according to this law, they are drawn as equidistant parallel straight lines, the parallels, instead of being equidistant as on the sphere, must be drawn at distances apart which shall increase according to the same law. That is, arc Wd ~ t—^ = arc ad^ x sec. lat., or the COS. lar. distances between the parallels must increase as the secant of the latitude increases. ' In the accurate construction of such charts, tables giving the increments of the secant for each minute of latitude, usually up to 80°, are prepared. 230. An approximate graphical construction^ sufficient for use in gaining an idea of the appearance of a map on Mer- cator's projection, is as follows : Supposing the quadrant NE^ Fig. 23, to be on a large scale, OE = i foot, for ex- ample, divide OE into equal parts — the more, the greater the accuracy — for example, 30 — and number them from O, which will be marked zero, and draw the meridians parallel to OiV^ Z(i6 SPHERICAL PROJECTIONS. through the points of division. Divide the quadrant EN into the same number [30] of equal parts, therefore of 3° each, and number them from E as the zero point, and draw radii through the points of division. At I on OE^ the first point of division to the right of (9, draw a parallel to ON and note its intersections, a with radius Oi^ b with radius 02^ c with radius (93, etc. Then Oa, laid off from O on ON, will give a' , the point through which the parallel of 3° will be drawn, like all the parallels, parallel to OE. Likewise Ob, laid off from a' , will give b' on ON, a point of the par- allel of 6° ; Oc, laid off from b' ^ will give c' , a point of the parallel of 9°, etc. All the parallels and meridians thus located need not be actually drawn ; every fifth meridian will be enough. These being 15° apart, are called hour-circles, as they divide the circumterence into 24 equal parts. Parallels are usually 10° apart. To make them so, divide the quadrant into 18 or 36 equal parts for example, of 5° or 2|° each, and replace the line \abc by a parallel line ajb^c^, distant from ON by -j^ or -^-^ of OE, as the case may be. 231. G^iomonic projection, WlLlL\W.,Y\^.2i. In this, the eye is at the centre of the sphere, which is then projected upon a tangent plane ; most conveniently that at one of the poles, as BN, The meridians will then be projected in straight lines, making the same angles with each other at the point of contact -A^, of the plane, that the meridians themselves do. The parallels, as at b, will be projected in concentric circles around the same point, and whose radii, as NB, are equal to the tangents of the distance Nb of the parallel from the pole, called the polar distance, 232. A modification of this projection, Q,2ii\.^^ polar projection, consists in taking the plane of projection through the polar circle. The meridians will be projected as in the last case, but the parallels will be projected in smaller circles than before. SPHERICAL PROJECTIONS. 26/ Both of these projections are used to supplement Mer- cator's projection in representing the polar regions, to which that will not apply (228 — 9). 233. Conic projection, — In this projection, a cone tangent along some parallel is substituted for a tangent cylinder, the point of sight still remaining at the centre of the sphere. On developing the cone to form the map, the meridians will appear as elements of the cone, and the parallels as arcs having the development of the vertex for their centre. This method is qilite inaccurate for parallels far from the parallel of contact, and inconvenient for parallels of contact near the equator, on account of the remoteness of the ver- tex, in that case. To increase the average accuracy within certain limits, a secant cone, containing two parallels not far apart, is some- times substituted for a tangent cone. 234. Poly conic projection. — In this, a separate tangent cone is used for each of as many parallels as may be desired. Each of these parallels then appears in the development as an arc whose radius is the tangent of the polar distance of that parallel (231). The central meridian of the map is made straight, and equal to the true length of the arc of that meridian contained within the limits of the map. The scale is therefore exact and uniform on this meridian and on the developed paral- lels; which intersect this meridian at their true distance apart. The other meridians are curved, crossing the parallels at right angles, and are tangents at those points to the de- veloped elements of the respective cones whose bases are these parallels. See Figs, i and 2 in the text. In Fig. i, NS is the axis of the sphere, and ad, be, cf, three parallels, curves of con- tact of the three tangent cones whose vertices are F, Fj, F^, respectively. 268 SPHERICAL PROJECTIONS. i^ \ \ \ A "i \\ X ^ ''^^sxX — --.J^A X ~-^ \jV^ / d /\^'^\ \\a ^^R^\ ' ' , V V / v^ \^ «? \ TJ \\ \ ^ \\\ y v^ \^V\ \// ^ \^^ ^ mX- / r a i^*'"^ \ s SPHERldAt. PROJECTIONS. 6IS9 Thert in Fig. 2, OA the straight line iV5, make db^. be, equal to the arcs aby be. Fig. i. On the latter figure draw arcs with centre O and radii ad, be, ef, giving the distances 710, pVj sx on the respective parallels, corresponding to ab on a great circle, as the equator ; and lay them off, as indi-' cated by the lettering on Fig. 2, each way from a, b, c. The parallels will now be arcs through a, b, e, with radii Va, VJ>y V^e, equal to those marked by the same letters on Fig. I. The meridians will be curved through 0, r, x, and n, p, s, etc. Vo will be the developed meridian at for the cone V, and would be that meridian if the map were made on that cone only (233). V^r is likewise the developed meridian at r for cone V^, and V^x that at x for cone F,. The actual polyconic curved meridian, orx, is evidently tangent to these mono-conic meridians. For the accurate construction of maps by the polyconic and Mercator*s projection, extensive tables are used by the United States Government Bureau of Navigation. 235. A slight modification of the above, and called the equidistant polyeonie projeetion, transfers the accuracy of the scale from the parallels, except the centre one of the map, to the meridians. It consists in laying off ab, for example, each Way on the meridians, from /, r, etc., taken on bpr as the cen- tral parallel, in order to find points through which to draw the final parallels, those shown in Fig. 2 being in Such case temporary. This applies to small areas where the meridians are sensibly straight. 236. Flamstead's projeetion.— T\ii^ differs from the poly- conic, in that, in Fig. 2, for example, the parallels through a and c would have been drawn with the same centre F„ correctly used for the true development, br, of the central parallel. The distances on NS and on the parallels are made of their real size, but the meridians and parallels do not inter- sect at right angles, as they do in the polyconic projection. TRIHEDRALS. 237. The term trihedrals practically means here the graphical solution of spherical triangles, as will now be explained. A spherical triangle, ^^67, PL XXIII., Fig. 25, is bounded by three arcs of great circles of the same sphere, and hence having their centre at O^ the centre of the sphere. Such a triangle has six parts, its three sides, which are measured by the angles a, (i, y, subtended by them at the centre of the sphere ; and its three angles, which, considered as plane angles, are as at -^, for example, measured by the angles between the tangents at ^, to the arcs BA and CA, 238. But join O with -^4, ^, C, and we form a spherical pyra- mid, O — ABC, in which the angle at O is called a trihedral, it being formed by the intersections of three planes which include three diedral angles. Now the tangents at any ver- tex, as ^, to the sides, as BA and CA, are perpendicular to the radius OA, and therefore include between them the measure of the diedral angle between OCA and OB A . Also the plane angles, a, 0, y, of the trihedral measure the sides BC, AC, AB, of the triangle. Thus a spherical triangle is the base of a spherical pyra- mid whose vertex is a trihedral. The plane angles of this trihedral measure the sides, and its diedrals measure the angles of the spherical triangle. TRIHEDRALS. 2/1 If the three plane angles are each of 90°, the trihedral is said to be tri-rect angular, 239. Any three parts of a spherical triangle being given, its remaining parts can be found. In the graphical solution, all these are represented by the equivalent parts of the cor- responding trihedral. The solution '' in space" for the six cases thus arising, is summed up for all of them in the simple statement that the construction of the diedrals, equivalent to the angles of the triangle, is an application of (D. G., Prob. XVII.). "To find the angle between given planes," and that of the plane angles equivalent to the sides of the triangle, is an applica- tion of (D. G., Prob. XV.). '* To find the angle between given lines." . Also, in every case designate the sides of the triangle by a, (3y y, and its respectively opposite angles by A, B^ C, PROBLEM I. — Given the three sides of a spherical triangle^ that isy the three plane angles of a trihedral ; to find its angles^ that is^ the diedrals of the trihedral. PI. XXIII. , Fig. 26. Let one of the plane angles, «r, be taken in V, and with one of its sides perpendicular to H. Thus, O being the vertex, and OC one edge of the trihedral, let a, (3, y be the given plane angles ; O^' being equal to (9^, which gives a'^d'\ base of the face y. Describe arcs with C as a centre, and radius Ca'\ and with d' as a centre and radius a'^d'' ; and they will intersect at a^ giving Cda for the intersection of the trihedral with H, and hence aCd = the required angle C. Then, as is obvious on inspection, by (238) B and A are the two other required angles, Cp being perpendicular to Ca, as Cq is to Cd, and B^^Cc^ being the plane (237) of the angled/ where CB^^ is perpendicular to Od as CA''' is to Oa'\ Examples. — 1°. Let one edge of the trihedral be perpendicular to H. Ex. 2°. Let the angle /3 be in V. ^72 TRIHEDRALS. PROBLEM II. — Given two sides and the included angle of a spherical triangle, that is, two plane angles and the included diedraly of a trihedral; to find the remaining parts. PI. XXIIL, Fig. 27. Let a and (i and bCa, = C, be the given plane, and included diedral angles. Revolve COa'^ about CO till Ca" takes the position Ca, then find 0'^ by intersecting arcs with a and b as centres, and Oa" and Ob respectively as radii. Then aO"b = y. CAp, = Af and CBg, = £^ are then found as in the last problem. PROBLEM III. — Given two sides and an opposite angle of a spherical triangle, that ii, two plane angles and the diedral opposite one of them, in a trihedral ; to find the remaifiing parts, PI. XXIIL, Fig. 28. Let a and (i be the given plane angles and CBq the diedral opposite (i, and located as before by revolving its plane qCB", perpendicular to Ob, into H. Then bq determines the position of the base ab of the face y of the trihedral, which base is limited at a by the arc a"a, representing as before the revolution of the face Oca'^ about OCi to its primitive position OCa, This gives bCa =:. C, after which ^ = CAp, is found as before* a"'h = ab and Oa"'~ Oa" , giving a!" Ob = y. ~ ^40. The three remaining cases of the spherical triangle, or of the equivalent trihedral, may be solved as independent problems by constructions similar to the one just given ; or, by the principle of supplementary spherical triangles, they may be resolved into the three just given. By the former method it will be useful to remember that the planes of the diedral angles measured by A and B, Fig. 25, are both perpendicular to the face OAB ; hence, if passed through the same point on OCy they will intersect TRIHEDRALS. 2/3 each Other in a common perpendicular from that point to OAB, and hence to their traces on OAB. Now in Figs. 26, 27, 28, A/f and Bq are the revolved positions, in H, of just such traces, since CB'^ and CA''^ are both drawn through C. Hence the perpendiculars from C to Ap and Bq are the transposed positions of this perpendicular, and are therefore equal, or Ap and Bg are both tangent to an arc having C for its centre and either of these perpendiculars for its radius. 241. Bj/ the second method^ if, for example, the three diedral angles A^ B, C, are given, their suppkments will be ihQ plane angles «r, i3, y, of the supplementary trihedral, which returns to Problem I. Then finding the A^B, C oi this supplement- ary trihedral, their supplements will be the required plane angles a, /?, y, of the given trihedral. 242. Likewise for the other two cases ; having given, for example, A,B and the included plane angle 7, resolve this into Prob. IL, by taking the supplements of A and B and } , which will be the or, jij and C of the supplementary tri- hedral, in which find A, Bj and y, whose supplements will be the required a, /3, and C of the given trihedral. Example. — Solve the three cases above indicated. AXONOMETRIC AND OBLIQUE PROJECTIONS. 243. These special kinds of projection afford a class of figures having more or less perfectly the effect of true per- spectives, while they are much more easily made and are thus highly useful, especially in their application to small and mostly rectangular objects, or to objects drawn on a small scale, as in text-book illustrations of machines, appa- ratus, etc. 244. The principal lines of most buildings, machines, and structures are generally in three directions: one vertical, and the two others horizontal and at right angles to each other. Such objects are fully represented by their perpendicu- lar projections (6) on three planes, each parallel to two of the directions described, that is, upon a horizontal plane, and on two vertical planes at right angles to each other, together with sections formed by cutting the bodies by other planes, usually parallel to one or more of the prin- cipal planes ; the portions in front of such cutting planes being removed. 245. The figures thus produced are, however, entirely destitute of the pictorial or perspective effects which would AXONOMETRIC AND OBLIQUE FROJECTIONS. 275 render them much more intelligible to those who are un- familiar with descriptive geometry, that is, with the method of projections ; while it is comparatively difficult and tedi- ous to construct true perspectives. Hence the great utility, to those who are unfamiliar either with ordinary projections or perspective, of the special and rapid methods, now to be described, which give pictorial effects, more intelligible than plans, elevations, and sections, and are more easily made than true perspectives. 246. Axonometric projection is a perpendicular projectiony made upon a plane which is oblique to all three of the lines described in (244) — that is, to all the edges of a solid right angle ; otherwise called a tri-rectangular trihedral (238). Also the object is so placed that its vertical lines will appear as such in the figure. The pictorial effect is thus due to the position of the object, in being placed so that all its dimensions are oblique to the plane of projection, and therefore represented as in perspective, by lines not at right angles to each other. 247. Oblique projections are, on the contrary, those in which it is the projecting lines indicating the direction of vision (5 — 6) that are oblique to the plane of projection. Thus, PL XXIV., Fig. lo, if AB—A'B' be a perpendicu- lar to the vertical plane, as given by its common or perpen- dicular projections, and if then A'a\ equal to AB^ the true length of the given line, be called the projection of this given line, it shows that the lines which project the points of AB — A'B' upon V have changed, from being perpendicu- lar to V to being oblique to it, at an angle of 45°, as at Aa — A^a\ where the true size of the angle AaBy found by (Prob. XII.), will be seen to be 45°. The line A'a\ equal to ABy may extend in any direction from A\ Also A'a\ instead of being equal to AB, may be any simple part of it, as one half, or one third, etc. 248. Thus in two ways oblique projections yield pic- torial effects ; first, by the many directions in which the pro- 276 AXONOMETRIC AND OBLIQUE PROJECTIONS. jection of a given perpendicular to the plane of projection may lie ; second, by the various ratios between the length AB of the given line and its projection A' a'. The following problems will sufficiently illustrate both of the foregoing kinds of projections, or perspectives, as they are sometimes called from their pictorial effect. PROBLEM I. — Having given three axes^ the axonometric pro- jections of the three edges of a solid right angle ^ to find the scales of those edges — that is^ the ratio of distances on them to their true size, PL XXIV., Fig. I. Let O be the projection upon the paper, of the vertex of a tri-rectangular trihedral or solid right angle, and Oa^ Oby Oc, drawn at pleasure, and of inde- finite length, the projection^ of its edges. As O may be at any height from the paper, we may assume a as the trace of Oa upon the paper. Then as each edge is perpendicular to the plane of the other two (246), ab^ perpendicular to Ocy and ac, perpendicular to Ob^ are (Theor. IL) the traces of the planes ^6'<^and «(9^ respectively upon the paper. The trace bcj of bOcy will then be perpendicular to aO. The lines Oa and Ok in space form a right angle at O, which is inscribed in a semicircle having ak for its diame- ter, hence make XYy parallel to ak^ the ground line of a plane perpendicular to the paper, project ak upon it at a'b\ on which describe a semicircle, and project O upon it at O', when a'O'b' will show the angle mentioned, and 0'a\ the true length of Oa and its inclination O'a'o to the paper. Revolve Ob and Oc about a perpendicular to the paper at O, till they are, as at Ob^ and Oc,, parallel to XY, when their projections, O'b,' and 6>V/, will show likewise their true length and obliquity to the paper. Next, take any distance a'm,,= b,'n^ = c^'r, ; and their projections, a'm, b^n, c^r, on XY, will be the lengths of a'm, in the directions of the three axes Oa^ Ob, Oc, respectively. Or, to take a particular case, let the scale of the plans AXONOMETRIC AND OBLIQUE PROJECTIONS. 277 and elevations be one of two inches to a foot, then if a'ni,, etc., = 2 inches, a'rriy b^'tt, c^r will be scales of 2 inches to i foot on Oa, Ob, Oc, respectively ; as indicated on the same lines shown separately below. Fig. 2 repeats the same construction, varied by a differ- ence in the relative directions of the axes, and in the scales upon them. PROBLEM II. — To construct the axonometric projection of a monument embracing prismatic and pyramidal portio7is. Let the scales be those of Fig. i. PL XXIV., Fig. 3. Suppose the base to be a square each side equal to a'm^^ Fig. i. Then make An and Ar equal to the scales Ob,7i and Oc,r, AB, the height of the base, is the projection on XV of its real height a'B^ shown at OB and OB' on scales a'7n^ and Oa respectively. To find J, make b^'e^=c/f, Fig. i, the true perpendicular distance between the vertical planes ABCy ABD, and their parallels GFy and HFy, and lay off Be and Bf respectively equal to b^'e and ^//, Fig. i, the proper scale values of these distances. Also make Ey, the altitude of the sloping portion of the base, equal to a'y, the value on the scale Oay Fig. i, of the real altitude a'y^. Then draw yg, limited by making Cc = Be and drawing parallels r/, to eE, and Ig, to Ey. In like manner, h is found. The whole drawing is completed by operations similar to the foregoing, which have illustrated all the scales. All the altitudes Fy, Vs, etc., are from the scale Oa,= a'm^, Fig. i ; all the right-hand widths, as sty half width of base of the pyramid, from the scale ob = b/n^ Fig. i ; all the left-hand widths, y/iy sb, etc., from the scale Oc — c^r^ Fig. i, and all the real dimensions from scale a'm^y Fig. i. Thus, Vs = a'v projection of the real altitude a'v^y Fig. i, upon XY. 249. When two of the three axes are equally inclined to the plane of projection, the projection is called mono- 278 AXONOMETRIC AND OBLIQUE PROJECTIONS. dimetric. The third axis then makes equal angles, in pro- jection, with the other two. There are two general cases ; in one, both of the equally inclined axes are horizontal on the object, (244) in the other, one of them is vertical, and the other horizontal. Very particular cases are those in which one of the three axes is parallel to the paper. In these cases the plane of the other two is perpendicular to the paper, and they are thence projected in one line. In this case, none of the horizontal surfaces of the monument, for example, would be seen, and the pictorial effect would be less striking. 250. Isometric projection. — When all three of the axes are equally inclined to the plane of projection, the figure be- comes the familiar isometric projection. PL XXIV., Fig. 4. In this case, the axes are also equally inclined to edch other, therefore at angles of 120°. 251. Isometric scales and protractors. — Let aC and OC^ PI. XXIV., Fig. 4, be drawn to make angles of 45° with Oa. Then the circle of radius Ch will represent in its iruG form y and on the scale of the isometric figure, the circle inscribed in the upper base aObs of the prism. Then divide the quadrant V,^i into any desired number of equal parts, and produce the radii through the points of division, to meet Oa. Then o being the isometric projection of Cy these radii become transformed into their isometric radii ; Ca, Cn, etc., into ao, no^ etc. Then if each of the latter be equally divided into the same number of parts, it will form a scale for all lines parallel to it. Again : draw e^e perpendicular to Oa, and eH parallel to Ob. Then F and H on oa and os will be extremities of the semi-axes of the ellipise which is the isometric projection of the inscribed circle just mentioned. The intersections of this ellipse with ao, no^ ho, etc., will then form an angular protractor ; in the figure one measuring isometric angles of 15° each. AXONOMETRIC AND OBLIQUE PROJECTIONS. 279 252. These intersections, as /, q^ r, can also be found as follows. The semicircle on EF as a diameter shows in its true size, and parallel to the paper ^ as well as in its circular form, the semicircle whose projection is the semi-ellipse EHF. Then divide each quadrant of it, as Fniy into six equal parts ; as before, at ^,^„ so now at /^ ^,, r^, etc., and draw perpendiculars /,/, q^q^ r^r^ etc., to EF, These perpendicu- lars represent the arcs in which /„ ^„ r„ etc., revolve back to their isometric positions, and they will meet the isomet- ric ellipse EHF in the same points in which they meet the isometric radii hoq, op, or, etc. PROBLEM III. — To construct the axonotnetric projection of a vertical cylinder capped by a hemisphere of less diameter. PI. XXIV., Fig. 5. We will use the scales of Fig. 2, where as^, the scale of plans and elevations, may now repre- sent a scale of three feet to an inch, and scales Oa, Ob, Oc, the same properly reduced, as in Prob. I., to the axes similarly lettered. Then the radius OR, = or, parallel to Ob, Fig. 2, is 4': 4'' taken from the scale Ob ; OS, parallel to Oc, Fig. 2, is the same, from scale Oc, and the altitude Oo of the cylinder is 5': 2" from scale Oa. OG, the real radius parallel to be. Fig. 2, of the sphere, and equal to OF, is 3': 6", and is laid off on OR and OS, as taken from scales Ob and O respectively. OQ corresponds to Or, projected in O'c' , Fig. 2 ; hence laying off c'D, Fig. 2, equal to OD, its projection c' Q gives the semi-conjugate axis OQ, Fig. 5 ; and OP may be simi- larly found. We can now draw the base, GPE, of the hemisphere, and its circular contour, EFG, of radius OG. Any vertical semicircles of the hemisphere may be found as follows : First. The one on the diameter HK parallel to Ob, Fig. 2, and in a plane parallel to aOb, Fig. 2. Its vertical radius 28o AXONOMETRIC AND OBLIQUE PROJECTIONS. ON is the distance corresponding to the true size OF, taken from the scale Oa. Its semi-transverse axis is OF. Its semi- conjugate OT, perpendicular to OF, and hence parallel to TtiOc, Fig. 2, may be found from the projection of Oni (after revolution till parallel to the side plane on XY) as OQ was found from Or. Or, 6^7" may be thus found: With centre K and radius OF, describe an arc cutting TOS at s ; then sKt gives Kt = the desired semi-conjugate OT, by the prop- erty of the ellipse that if a straight line move so that two points of it at a fixed distance apart move on two fixed in- tersecting straight lines, any other point of the moving line will describe an ellipse. The required semi-ellipse HNK can now be drawn. Second. Any vertical semicircle, as that on the diameter LM. The bases of the cylinder being parallel to the plane 'bOc, Fig. 2, draw Oy, Fig. 2, parallel to LM ; then ay is in the plane of the paper, and hence of its real size ; hence, draw Oy parallel to ay, Fig. 2, and y, its intersection with EFG, will be an extremity of the semi-transverse axis of the required elliptical projection of the proposed semicircle on LM. Then Oz, its semi-conjugate axis, can be found as shown, ViS> OT was ; or it can be found, as OQ was, by draw- ing a perpendicular from O to ay. Fig. 2, and proceeding as described for OT, as found from Om, Fig. 2. The semi- ellipse LzNyM can therefore be readily sketched. Examples. — 1°. Let Oo lie in the direction Ob or Oc. Ex. 2°. Substitute for the given body a niche. Ex. 3°. Replace the cylinder, Fig. 5, by a cone with its axis successively in the directions Oa, Ob, Oc. PROBLEM IV. — To construct comparative views of one given object in elevation, axonometric, and oblique projectio7is. PL XXIV., Figs. 6 — 9. Let the object be a conical frus- tum, EFef mounted on a cylindrical axis. Fig. 6 represents this object so as to be considered either as a plan or an elevation, as both views would be alike if the AXONOMETRIC AND OBLIQUE PROJECTIONS. 28 1 axis AB were parallel to the ground line. This figure shows the lengths and diameters in their true size, but except by its title or an end elevation, could not, even with plan and elevation, be certainly distinguished from the frustum of a square pyramid mounted on a square prism having the same linear axis, AB. Fig. 7 is an axonometric projection, made from the real dimensions afforded by Fig. 6, by using the scales of Fig. 2. Here Cf, parallel to ac, Fig. 2, is the true size of the radius Cf, Fig. 6, shown on the scales Oa and Oc, Fig. 2, at C/i and Q-. Similar explanations apply to the remaining parallel circles. The parts oi AB are those of AB, Fig. 6, reduced to the scale 01?, Fig. 2. Fig. 8 is an oblique projection of the object when its axis is perpendicular to the paper at A, and the projecting lines make an angle of 45° with the paper, in a plane whose trace on the paper is AB, thus projecting the point B at B so that AB = AB, Fig. 6. The circles of the object thus being parallel to the paper, are projected as circles in their real size, and the figure is as easily made as Fig. 6. It has a somewhat lengthened appear- ance, because e/ is greater than its real size ef. Fig. 6 ; but this could be remedied by making the parts of AB shorter, in a uniform ratio, than the like parts of AB, Fig. 6. This modification supposes the projecting lines to make some an- gle greater than 45° with the paper. The circles would still remain circles of their present size. Fig. 9 represents the same object as before, but with the axis AB parallel, and the planes of the circles perpendicular to the paper. It illustrates the considerably increased com- plexity of construction with more distorted result than in Fig. 7 or 8. The diameters, as 2C/, are here parallel, and those represented by//,, are perpendicular to the paper, but as Cp = iC/, the projecting lines make a larger angle than 45° with the paper. Describing a circle on 2 Cf as a diameter, draw the ordi- nate CP, and representative Pp, of a projecting line. Thus the radius perpendicular to the paper at C, and revolved at 282 AXONOMETRIC AND OBLIQUE PROJECTIONS. CPy is transformed into the oblique projection Cp, and the line Pr into the line rp, since r in the paper is its own pro- jection, and/ is that of P. Reasoning thus, draw the diameter iq perpendicular to Pr ; the tangent at / to the circle CP, and parallel to Pr, will be transformed into a parallel to rp at v, to which the ellipti- cal projection of circle CP will be tangent, as at a. Finding q, m, and other points, and the opposite tangent at X, parallel to that through v, by similar constructions, or making Cp^, etc., = Cp, etc., the elliptical projection of circle CP can be sketched. . The remaining circles can be similarly found. The parallel tangents to, the small circles are found as that at v is, and are at more than their real distance apart. The contour of the conical part consists of tangents to its bases. Comparing Figs. 8 and 9, the advantage of making cir- cular outlines parallel to the plane of the paper is obvious.* Examples. — 1°. Construct Fig. 7 with AB either in the direction of axis Oa or Ob, Fig. 2. Ex. 2°. In Fig. 8 let the segments of AB be made two thirds, or one half their real size. Ex. 3°. Make an isometric projection of Fig. 6. * See my Elementary Projections and Shadows, or my Elementary Projec- tion Drawing. ^ ^ OF THE UNIVERSITY OF / £^LlFORHVb. SITY I I n^) yf^) vf^ "^'wj^g \ ^^ / , — 1 < ^ [iSMmi