'2 c~J~ U/C~'2~~ JlZ;LG. 6 /O I~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~..... ll,,~,r ~~""~~~,3'Xf ELEMENTS OF DESCRIPTIVE GEOIIETRY, WITH ITS APPLICATIONS TO'TIE~RICAL PROJECTIONS, SHADES AND SHADOWS, PERSPECTIVE AND ISOMETRIC PROJECTIONS. BY ALBERT E. CHURCH, LL. D., 08OFESSOR OF MATHEMATICS IN THE U. S. MILITARY ACADE.MY, AUTHOR OF ELEMENTS OF THE DIFFERENTIALT AND INTEGRAL CALCULUS, AND OF ELEMENTS OF ANALYTICAL (IEOMETRV. A. S. BARNES & COMPANY, NEW YORK AND CHICAGO. Vulable Works by Leadi Authors IN THE HIGHER MATHEM ATICS. A. E. CPHUTRCII~ Ll.D. Prof. Jfalhenatics in the U. S..31rilitayo Acadeny, Wt0s Voint. CI'EIURCI-H'S NA IALYTICAL X GEOMIETRIY. CXIfU- CIH'S C AJLCTULUS. CI-IUJIUCI-R'S:DESCPRIIPTIVE GEEOM:ETETY EI~WAIW D - 1i. COUITEfNAY, LL.D., 4ate.rof. Mtathematics in the [fniyersily of F'i;.inaa. C OURTENAY S CALTCULUS. CIHAS. W. HACIi.LEY, S. T. D., Zate 2-'of. of.athemat/is teand qshi'onomr h iC (ou/uwtbia College. HACKLEY'S TRIG-ON O Lxii'l ETR. W. II. H. C. BAWITLETT, I L.D., Prof. of.Jat. di Exp. Philos. in the U. 8. Militaiy.4cad.,'s~t Zoint BARTLETT- S SYNTT-HEIETIC MMIECI-ANITICS. BARTTLETTUL'S ANALYTICAL M1ECHA1NIICS. BAIRTLEUT'T'S ACOUSTICS AND OiPrICLS BART LETrT'S ASTRONOMY. DA'VIES c& PECII, toepartment of XZathenatics, Columbnia Coet7e. M:ATII E MATIT C A.L DICTIONARY. CHAIRLES gDAVIES, LL.D., late of the United gtates #,itilitaty le1cademy and of Cotimnbia Cotlegp A COMPLETE COUIRSE TN 1M ATH9EMLATI]CS See A. S. BAi-NES & Co.'s Descriptive Catalogue. Entered, according to Act of Congress, in the year 1864, by BARNES & BURR, In the Clerk's Office of the District Court of the United States for the Southera District New York. C. D. G. PREFACE..0.6 * -- THESE pages have been written and are published with lhe single object of presenting, in proper form to be used as a text-book, the course of Descriptive Geometry, as taught at the U. S. Military Academy. Without any effort to enlarge or originlate, the author has striven to give, wit~h a natural arrangement and in clear and concise language, the elementary principles and propositions of this branch of science, of so much interest to the mathematical student, and so necessary to both the civil and military engineer. Though indebted for many of the ideas to the early instructions of L.is predecessor and friend, Professor Davies, whose text-books on this subject were among the first in the English language, the author has been much aided by a frelquent reference to the French works of Leroy and Oiivier, and to the elaborate American work of Professor Warren. It is intended to include, in an edition to be issued at an early day, the application of the subject to shades and shadows, and perspective. U. S. MILITARY ACADEMY, j October, 1864 CONTENTS. PART I. ORTHOGRAPHIC PROJECTION. PA69 Preliminary definitions........................................ 1 Representation of points........................................ 2 Representation of planes....................................... 4 Representation of right lines................................... 5 Revolution of objects.......................................... 7 Revolution of the vertical plane................................ 8 Notation used in the description of drawings..................... 12 Manner of delineating the different lines used.................... 12 Elementary problems relating to the right line and plane......... 13 Construction and classification of lines........................... 33 Projection of curves.......................................... 36 Tangents and normals to lines................................. 37 Generation and properties of the helix......................... 39 Generation and classification of surfaces........................ 40 Generation and properties of cylindrical surfaces................. 42 Generation and properties of conical surfaces............. 44 Warped surfaces with a plane directer......................... 47 Generation and properties of the hyberbolic paraboloid............ 49 Warped surfaces with three linear directrices.................... 51 Generation and properties of the helicoid....................... 53 Surfaces of revolution....................................... 54 The hyperboloid of revolution of one nappe...................... 56 Tangent planes and surfaces; normal lines and planes........ 61 Problems relating to tangent planes to single curved surfaces...... 66 Problems relating to tangent planes to warped surfaces.......... 74 Problems relating to tangent planes to double curved surfaces... 78 Points in which surfaces are pierced by lines................... 85 Intersection of surfaces by planes. Development of single curved surfaces.................................................. 86 V] CONTENTS. Intersection of curved surfaces.....................9............ 9 Development of an oblique cone................................ 104 Practical problems......................................... 107 PART II. SPHERICAL PROJECTIONS. Preliminary definitions........................................ 113 Orthographic projections of the sphere........................ 116 Stereographic projections of the sphere........................ 121 G(lobular projections....................................... 130 Gnomonic projection..............................131 Cylindrical projection....................1.............. 132 Conic proj ection.......................................... 132 Construction of maps...................1................. 134 Lorgna's map............................................ 135 Mercator's chart.............................................. 135 Flamstead's method........................................... 137 The Polyconic method...................................... 138 PART III. SHADES AND SHADOWS. Preliminary definitions........................................ 140 Shadows of points and lines.................142 Brilliant points..................................... 145 Practical problems........................................... 146 PART IV. LINEAR PERSPECTIVE. Preliminary definitions and principles.......................... 157 Perspectives of points and right lines. Vanishing points ofrightlines 158 Perspectives of curves......................................... 162 Line of apparent contour....................................... 163 CONTENTS. Vil PAG.E Vanishing points of rays of light and of projections of rays........ 164 Perspectives of the shadows of points and right lines on planes.... 165 Practical problems............................................. 167 PART V. ISOMETRIC PROJECTIONS. Preliminary definitions and principles.......................... 186 Isometric projections of points and lines......................... 188 Practical problems.......................................... 189 PART I. ORTHOGRAPHIC PROJECTIONS. PRELIMINARY DEFINITIONS. 1. DESCRIPTIVE GEOMETRY is that branch of Mathematics which has for its object the explanation of the methods of representing by drawings: First. All geometrical magnitudes. Second. The solution of problems relating to these magnitudes in space. These drawings are so made as to present to the eye, situated at a particular point, the same appearance as the magnitude or object itself, were it placed in the proper position. The representations thus made are the projections of the maqnitude or object. The planes upon which these projections are usually made are the planes of projection. The point, at which the eye is situated, is the point of sight. 2. When the point of sight is in a perpendicular, drawn to the plane of projection, through any point of the drawing, and at an infinite distance from this plaue, the projections are Ortho. graphic. 2 D ESCRIPTIVE GEOIETR,''. WThen the point of sight is within a finite distance of the drawing, the projections are Scenographic, commonly called the Perspective of the magnitude or object 3. It is manifest that, if a straight line be drawn through,: given point and the point of sight, the point, in which this line pierces the plane of projection, will present to the eye the sarne appearance as the point itself, and therefore be the projection of the point on this plane. The line thus drawn is the projecting line of tLie point. 4. In the Orthographic Projection, since the point of sight is at an infinite distance, the projecting lines drawn from any points of an object, of finite magnitude, to this point, will be parallel to each other and perpendicular to the plane of projection. In this projection two planes are used, at right angles to each other, the one horizontal and the other vertical, called respectively the horizontal and vertical plane of projection. 5. In Fig. 1, let the planes represented by ABF' and BAD be the two planes of projection, the first the horizontal and the second the vertical. Their line of intersection AB is the ground line. These planes form by their intersection four diedral angles. The first angle, in which the point of sight is always situated, is above the horizontal and in front of the vertical plane. The second is above the horizontal and behind the vertical. The third is below the horizontal and behind the vertical. I'he Ofuurth, below the horizontal and in front of the vertical, a. rmarked in the figure. REPRESENTATION OF POINTS. 6. Let M. Fig. 1, be any point in space. Through it draw DESC RIPTIVE GEOMETRY. 3 NIMm perpendicular to the horizontal, and Mm' perpendiculaI to the vertical plane; m will be the projection of M on the hcrizontal, and in' that on the vertical plane, Art. (4). Hence, th.e horizontal projection of a point is the foot of aperpendicular throutgh the point, to the horizontal plane; and the vertical pro-;ection of a point is thefoot of a perpendicular through it to the vertical plane. The lines Mm and Mm' are the horizontal and vertical projecting lines of the point. 7. Through the lines Mm and Mm' pass a plane. It will be perpendicular to both planes of projection, since it contains a right line perpendicular to each, and therefore perpendicular to the ground line AB. It intersects these planes in the lines mo and m'o, both perpendicular to AB at the same point, forming the rectangle Mo. By an inspection of the figure it is seen that Mm = m'o, and Mm' = mo; that is, the distance of the point M, from the horizontal plane, is equal to the distance of its vertical projection from the ground line; and the distance of the point from the vertical plane is equal to that of its horizontal projection from the ground line. 8. If the two projections of a point are given, the point is complletely determined; for if at the horizontal projection In, a perpendicular be erected to the horizontal plane, it will contain the point M. A perpendicular to the vertical plane at m', will also contain M; hence the point M is determined by the intersection of these two perpendiculars. If M be in the horizontal plane, Mm = 0. and the point is its own horizontal projection. The vertical projection will be in the around line at o. 4 DESCRIPTIVE GEOMEtRY. If M be in the vertical plane, it will be its own vertical projection, and its horizontal projection will be in the ground line at o. If M be in the ground line, it will be its own horizontal, and also its own vertical projection. REPRESENTATION OF PLANES. 9. Let tTt', Fig. 2, be a plane, oblique to the ground line, intersecting the planes of projection in the lines tT and t'T respectively. It will be completely determined in position by its two lines tT and t'T. Its intersection with the horizontal plane is the horizontal trace of the plane, and its intersection with the vertical plane is the vertical trace. Hence, a plane is given by its traces. Neither trace of this plane can be parallel to the ground line; for if it should be, the plane would be parallel to the ground line, which is contrary to the hypothesis. The two traces must intersect the ground line at the samen point, for if they intersect it at different points, the plane would intersect it in two points, which is impossible. If the plane be parallel to the ground line, as in the same figure, its traces must be parallel to the ground line; for if they are not parallel they must intersect it; in which case the plane would have at least one point in common with the ground line, which is contrary to the hypothesis. If the plane be parallel to either plane of projection, it will have but one trace, which will be on the other plane and parallel to the ground line. 10. If the given plane be perpendicular to the horizontal plane, its vertical trace will be perpendicular to the ground line, as t'T, in Fig. 2; for the vertical plane is also perpendicular to the horizontal plane; hence the intersection of the two planes, whihll is the vertical trace, must be perpendicular to the hor' DESCRIPTIVE GEOME'ITRY. # zontal plane, and therefore to the ground line which passes through its foot. Likewise, if a plane be perpendicular to the vertical plane, its horizontal trace will be perpendicular to the ground line. If the plane simply pass through the ground line, its position is not determined. If two planes are parallel, their traces on the same plane of projection are parallel, for these traces are the intersections or the parallel planes by a third plane. REPRESENTATION OF RIGHT LINES. 11. Let MN, Fig. 3, be any right line in space. Through it pass a plane Mmn perpendicular to the horizontal plane; mnn will be its horizontal, and pip', perpendicular to AB, Art. (10), its vertical trace. Also through MN pass a plane Mm'n', perpendicular to the vertical plane: m'n' will be its vertical, and o'o its horizontal trace. The traces mn and nm'n' are the projections of the line. Hence, the horizontal projection of a right line is the horizontal trace of a plane passed through the line perpendicular to the horizontal plane; and the vertical projection of a right line is the vertical trace of a plane, through the line perpendicular to the vertical plane. The planes Mmn and Mm'n' are respectively the horizontal and vertical projecting planes of the line. 12. The two projections of the line being given, the line will, in general, be completely determined; for if through the horizontal projection we pass a plane perpendicular to the horizontal plane, it will contain the line; and if through the vertical projection we pass a plane perpendicular to the vertical plane, it will also contain the line. The intersection of these planes must, therefore, be the line. Hence we say, a right line is given by its projections. 0j L)D SGRIPTIVE GEOMElIRY. 1 3. The projections mn and im'n' are also manifestly made up of the projections of all the points of the line MN. Hence, if a right line pass through a point in space, its projections will vuass through the projections of the point. Likewise, if any two points in space be joined by a right line, the projections of this line will be the right lines joining the projections of the points on the same plane. 14. If the right line be perpendicular to either plane of projection, its projection on that plane will be a point, and its projection on the other plane will be perpendicular to the ground line. Thus in Fig. 4, Mm is perpendicular to the horizontal plane: m is its horizontal, and m'o its vertical projection. If the line be parallel to either plane of projection, its projection on that plane will evidently be parallel to the line itself, and its projection on the other plane will be parallel to the ground line. For the plane which projects it on the second plane must be parallel to the first; its trace must therefore be parallel to the ground line, Art. (9); but this trace is the projection of the line. Thus MN is parallel to the horizontal plane, and m'n' is parallel to AB. Also, the definite portion MN, of the line, is equal to its projection mn, since they are opposite sides of the rectangle Mn. If the line is parallel to both planes of projection, or to the griound line, both projections will be parallel to the ground line. If the line lie in either plane of projection, its projection on that plane will be the line itself, and its projection on the other plane will be the ground line. Thus in Fig. 5, MN, in the vertical plane, is its own vertical projection, and mn or AB is its horizontal projection. 15. If the two projections of a right line are perpendicular to the ground line, the line is undetermined, as the two pro DESCRIPTIVE GEOMIETRY. 7 jecting planes coincide, forlmirg only one plane, and do not by their intersection determine the line as in Art. (12). All lines in this plane will leave the same projections. Thus in Fig. 5, tmn and m'n' are both perpendicular to AB; and any line in the plane MNo will have these for its projections, If, however, the projections of two points of the linle are given, the line will then be determined; that is, if mam' and ant' are given, the two points AI anld N will be determined, and, of coulsc, the right line which joins them. All lines and points, situated in a plane perpendicular to either plane of projection, will be projected on this plane in the corresponding tra:ce of the plane. 16. If two right lines are parallel, their projections on the same plane will be parcoallel. For their projecting planes are parallel, since they contain parallel lines and are perpendicular to the same plane; hence their traces will be parallel, Art. (10); but these traces are the projections REVOLUTION OF OBJECTS. 17. Any geometrical magnitude or object is said to be revolved about a right line as an axis, when it is so moved that each of its points describes the circumference of a circle whose plane is perpendicular to the axis, and whose centre is in the axis. By this revolution, it is evident that the relative position of the points of the object is not changed, each point remaining at the same distance from any of the other points. Thus, if the point M, Fig. 6, be revolved about an axis DD', in the horizonal plane, it will describe the circumference of a circle whose centre is at o and whose radius is Mo; and since the point must remain in the plane perpendicular to DD', when it reaches the horizontal plane it will be at p or p', in the perpendicular mop, at a distance from o equal to Mo; that is, it J MDESCRIPTIVE GEOMETRY. will be found in a straight line passing through its horzontali projection perpendicular to the axis, and at a distance from the axis equal to the hypothenuse of a right-angled triangle of which the base (mo) is the distance from the horizontal projection to the axis, and the altitude (Mm) equal to the distance of the point from the horizontal plane, or equal to the distance (m'r) of its vertical projection from the ground line. Likewise, if a point be revolved about an axis in the vertical plane until it reaches the vertical plane, its revolved position will be found by the same rule, changing the word horizontal into vertical, and the reverse. If the axis pass through the horizontal projection of the point, in the first case, the base of the triangle will be 0; the hypothenuse becomes equal to the altitude, and the distance to be laid off will be simply the distance from the vertical projection of the point to the ground line. In like manner, its revolved position will be found when, in the second case, the axis passes through the vertical projection of the point. REVOLUTION OF THE VERTICAL PLANE. 18. In order to represent both projections of an object on the same sheet of paper or plane, after the projections are made as in the preceding articles, the vertical plane is revolved about the ground line as an axis until it coincides with the horizontal plane, that portion of it which is above the ground line falling beyond it, in the position ABC'D', Fig. 1, and that part which is below coming up in front, in the position ABF'E'. In this new position of the planes it will be observed, that the planes being regarded as indefinite in extent, all that part of the plane of the paper which is in front of the ground line will represent not only that part of the horizontal plane which is in fiont of the ground line, but also that part of the vertical plane which is below the horizontal plane; while the part be DIt SCRIPTIVE GEOMAETRY. 9 yond the ground line represents that part of the horizontal plane which is behind the vertical plane, and also that part of the vertical plane which is above the horizontal plane. 19. After the vertical plane is revolved as in the preceding ar title, the point mn', in Fig. 1, will take the position m" in the line mo produced, and the two projections m and m' will then be in the same straight line, perpendicular to AB. Hence, in every drawing thus made, the two projections of the sam.e point?must be in the same strtai/ht line, perpendicular to the ground line. If; thenr, AB, Fig. 7, be the ground line, and it be required to represent or assaeme a point in space, we first assume mn for its horizontal projection; through m erect a perpendicular to AB, and assume any point, as mn' on this perpendicular, for the vertical projection. The point will then be fully determined, Art. (8). The point thus assumed is in the first angle, above the hlorizontal plane at a distance equal to?n'o, and in fiont of the vertical plane at a distance equal to oo, Art. (7). If the point be in the second agle, its horizontal projection m must be on that part of the horizontal plane beyond the ground line, and its vertical projection m' on that part of the vertical plane above the ground line, Art. (5). When the latter plane is revolved to its proper position, m' will fall into that part of the horizontal plane beyond the ground line, and the two projections will be as in (2), Fig. 7, mo representing the distance of the point behind the vertical plane, and m'o its distance above the horizontal plane. If the point be in the thir'd angle, its horizontal projection will be beyond the ground line, and its vertical projection on the part of the vertical plane below the horizontal plane. The vertical plane being revolved to its proper position, m' comes in front of AB, and the two projections will be as in (3). If the point be in the fburth, angle, the two projections will be as in. (4). 10 DESCRIPTIVE GEOMETRY. 20. To represent, or assume a plane in space, we draw at pleas ure any straight line, as tT, Fig. 8, to represent its horizontal trace; then through T, draw any other straight line, as Tt', to represent its vertical trace. It is absolutely necessary that these traces intersect AB at the same point, if either intersects it, Art. (9). The plane and traces being indefinite in extent, the portion included in the first angle is represented by tTt'. The portion in the second angle by t'Tt". That in the third by t"'l't"'. That in the fourth by t'"Tt. If the plane be parallel to the ground line and not parallel to either plane of projection, both traces must be assumed parallel to AB, as in Fig. 9; Tt being the horizontal, and Tt' the vertical trace, Art. (9). If the plane be parallel to either plane of projection, the trace on the other plane is alone assumed, and that parallel to AB. If the plane be perpendicular to either plane of projection, its trace on this plane may be assumed at pleasure, while its trace on the other plane must be perpendicular to AB, as in (2), Fig. 9; Tt is the horizontal, and Tt' the vertical trace of a plane perpendicular to the horizontal plane. Also in (3), Tt is the horizontal and Tt' the vertical trace of a plane perpendicular to the vertical plane, Art. (10). 21. To represent or assume a straight line, both projections liay be drawn at pleasure, as in (1), Fig. 10, mn is the horizontal, and mn'n thle vertical projection of a portion of a straight line in the first aigyle. In (2), mnn is the horizontal and m'n' the vertical projection of a portion of a straight line in the second an gle. In (3), the line is represented in t}he third arngle, and in (4), in theovurfh angle. In Fig. 11 are, (1), tile projections of a right line parallel to the horizontal plane; (2), those of a right line parallel to the vertical plane, (3), tiose of a right line perpendicular to the hori DESCRIPTIVE GEOMETRY. 1 1 zontal plane; and (4), those of a right line perpendicular to the vertical plane, Art. (14.) 22. To assume a point upon a given right line; since the pro ections of the point must be on tlie projections of the line, Art (13), and in the same perpendicular to the ground line, Art. (19); we assume the horizontal projection as m, Fig. 17, on mn, and at this point erect mrm' perpendicular to AB: m', where it intersects rT'n', will be the vertical projection of the point. 23. If two lines intersect, their projections will intersect; for the point of intersection being on each of the lines, its horizontal projection must be on the horizontal projection of each of the lines, Art. (13), and hence at their intersection. For the same reason, the vertical projection of the point must be at the intersection of their vertical projections. These two points being the projection of the same point, must be in the same straight line perpendicular to the ground line, Art. (19). Hence if any two lines intersect in space, the right line joining the points in wh.ich their projections intersect, must be perpendicular to the ground line. Therefore, to assume two right lines ewhich intersect, we draw at pleasure both projections of the first line, and the horizontal projection of the second line intersecting that of the first; through this point of intersection erect a perpendicular to the ground line until it intersects the vertical projection of this line; through this point draw at pleasure the vertical projection of the second line. Thus, in Fig. 16, assume mn and rn'n', also too; through m erect mm' perpendicular to AB, and through nz' draw m'o' at pleasure. Two parallel right lines are assumed by drawing their pro jections respectively parallel, Art. (16). 12 DESCRIPTIVE GEOMP'I'RY. NOTATION TO BE USED IN TIlE DESCRIPTION OF DRAWINGS. 24. Points representeuo as in Fig. 7, will be described a, tlhe point (mm'), tihe letter at the horizontal projection being always written alld read first, or simply as the point M. Planes given by their traces, as in Fig. 8, will be described as the plane tTt', the middle letter being tile one at the intersection of the two traces, and the other letter of the horizontal trace being the first in. order. If the traces are parallel to the ground line, or do not intersect it within the limits ol the drawing, tile same notation will be used, the middle letter being placed on both traces; in the first case, at the lefthand extremity, and in the second case, at the extremity nearest the ground line. Lines given by their projections, as in Fig. 10, will be described as the line (mn, m'n'), the letters on the horizontal projection being first in order; or simply the line MN. The planes of projection will often be described by the capitals EI and V; H denoting the horizontal, and V the vertical plane. The ground line will be in general denoted by AB. MANNER OF DELINEATING THE DIFFERENT LINES USED IN THE REPRESENTATION OF MAGNITUDES, OR IN THE CONSTRUCTION OF PROBLEMS. 25. The projections of the same point will be connected by a dotted line, thus........ Traces of planes which are given or required, when they can be seen from the point of sight,-that is, when the view is not obstructed, either by the planes of projection or by some intervening opaque object,-are draawn full. When not seen, o lwhen they are the traces of auxiliary planes, not the project DESCRIPTIVE GEOMETRY. 1 5 ing planes of right lines, they will be drawn broken and dotted, thus: Lines, or portions of lines, either given or required, when seen, will have their projections full; When not seen, or auxiliary, these projections will be broken, thus: In the construction of problems, planes or surfaces which are required will be regarded as transparent, not concealing other parts previously drawn. All lines or surfaces are regarded as indefinite in extent, unless limited by their form, or a definite portion is considered for a special purpose. Thus the ground line and projections of lines in Fig. 10 are supposed to be produced indefinitely, the lines delineated simply indicating the directions. CONSTRUTCTTON OF ELEMENTARY PROBLEMS RELATING TO THE Rltlt'T LINE AND PLANE. 26. Hlaving explained the manner of representing with accuracy, points, planes, and right lines, we are now prepared to represent the solution of a number of important problems, reating to these magnitudes in space. In every problem certain points and magnitudes are given, fioin which certain other points or magnitudes are to be constructed. Let a iight line be first drawn on the paper or slate to rep resent the ground line; then assume, as in Art. (19), &c., the 14 DESCRIPTIVE GEOMETRY. representatives of the given objects. The proper solution of the problem will now consist of two distinct parts. The first is a clear statement of the principles and reasoning to be employed in the construction of the drawing. This is the anallysis of the problem. The second is the construction, in proper order, of the different lines which are used and required in the problem. This is the construction of the problem. 27. PROBLEM 1. To find the points in which a given right line pierces t/he planes of projection. Let AB, Fig. 12, be the ground line, and (izn, mn'n'), or simply MIN, the given line. First. To find the point in which this line pierces the horizontal plane. Analysis. Since the required point is in the horizontal plane, its vertical projection is in the ground line, Art. (8); and since the point is in the given line, its vertical projection will be in the vertical projection of this line, Art. (13); hence it imist be at the intersection of this vertical projection with tljhe ground line. The horizontal projection of the required point must be in a straight line drawn through its vertical projection, perpendlicular to the ground line, Art. (19), and also in the horizontal projection of the given line; hence it will be at the intersection of these two lines. But the point being in the horizontal plane, is the same as its horizontal projection, Art. (8); hence the rule:.Produce the vertical projection of the line until it intersects the ground line; at the point of intersection erect a perpendicular to the ground line, and produce it until it intersects the horizontal orojection of the line; this point of intersection is the required uDoint. Construction. Produce zm'n' to in'; at m' erect the perpendicular )n'm, and produce it to m. This is the required point. econecd. In the above analysis, by changing the word "verti cal" illto " horizontal," and the reverse, we have the analysis and Di:SCRIPTIVE GEOMETRY. 15 rule for finding the point in which the given line pIierces the vertical plane. Constfirction. Produce mn to o; at o erect the perpendicular o0', and produce it to o'. This is the required point. 28. PROBLEM 2. TO find the lengyth of a?right line joining two given points in space. Let AB, Fig. 13, be the ground line, andi (mnm') and (n,') the two given points. Alnalysis. Since the required line contains the two points, its projection must contain the projections of the points, Art. (13). Hence, if we join the horizontal projections of the points by a right line, it will be the horizontal projection of the line; and if we join the vertical projections of the points, we shall have its vertical projection. If we now revolve the horizontal projecting plane of the line about its horizontal trace until it coincides with the horizontal plane, and find the revolved position of the points, and join them by a right line, it will be the required distance, since the points do not change their relative position during the revolution, Art. (17). Construction. Draw mn and m'n'. MN will be the required line. Now revolve its horizontal projecting plane about mn until it coincides with H; the points M and N will fall at mn" and rn", at distances from m and n equal to rm' and S'n' respectively, Art. (17); join m" and n"; m"n" will be the required distance. Since the point o, in which the line produced pierces H, is in ihe axis, it remains fixed. The line m"n" produced must then pass through o, and the accuracy of the drawing may thus be verified. 29. Second qmethod for the same problem. Analysis. If we revolve the horizontal projecting plane of the 16 DESCRIPTIVE GEOMETRY. line about the projecting perpendicular of either of its points until it becomes parallel to the vertical plane, the line will, in its revolved position, be projected on this plane in its true length, Art. (14). If we then construct this vertical projection, it will be tile required distance. (7onstructio?. Revolve the projecting plane about the peripendicular at m. The point n describes the arc nl, until it comes into the line ml parallel to AB; I will be the horizontal projection oi N in its revolved position. Its vertical projection must be in 11' perpendicular to AB; and since, during the revolution, the point N remains at tile same distance above H, its vertical projection must also be in the line n'l' parallel to AB, Art. (7), therefore it will be at 1'. The point M being in the axis remains fixed, and its vertical projection remains at in'; An'lT is then the vertical projection of MN in its revolved position, and the true distance.;By examining the drawing, it will be seen that the true distance is the hypothenuse of a right-angled triangle whose base is the horizontal projection of the line, and altitude the difference between the distances of its two extremities from the horizontal plane. Also, that the angle at the base is equal to the angle made by the line with its projection, or the angle made by the line with the horizontal plane. Also, that the length of the line is always greater than that of its projection, unless it is parallel to the plane of projection. 30. Every right line of a plane must pierce any other plane, to which it is not parallel, in the common intersection of the'wo; hence every right line of a plane, not parallel to the horiontal plane of projection, will pierce it in the horizontal trace of the plane and if not parallel to the vertical plane, will pierce it in the vertical trace: hence, to assume a straight line in a giveu plane. take a point in each trace, and join the two by a right line; or otherwise, draw the horizontal projection at pleasure; at tlh. DESCRIPTIVE GE()METRY. 17 points where it intersects the ground line and the horizontal trace erect perpendiculars to the ground line; join the point where the first intersects the vertical trace with the point where the second ntersects tile ground line: this will be the vertical projection of he line. Thus, in Fig. 14, draw m2nu, also mm' and In': join m'n'; it will be the required vertical projection. 31. PROBLEM 3. To pass a plane through three given points. Let Mi, N, and P, Fig. 15, be the three points. Analysis. If we join either two of the points by a right line, it will lie in the required plane, and pierce the planes of projection in the traces of this plane, Art. (30). If we join one of these points with the third point, we shall have a second line of the plane. If we find the points in which these lines pierce the planes of projection, we shall have two points of each trace. The traces, and therefore the plane, will be fully determined. Construction. Join m and n by the straight line ran; also mn' and n' by m'n'. MN will be the line joining the first two points. This pierces H at h, and V at v, as in Problem 1. Draw also np, and n'l)'; NP will be tile second line. It pierces H at t and V at t'. Join h and t, by the straight line ht; it is the required horizontal trace. Join v and t'; t'v is the vertical trace: Or pro,duce ht until it meets AB, and join this point with either v or t' for the vertical trace, Art. (9). If either MN or NP should be parallel to AB, the plane, and consequently its traces, will be parallel to AB, Art. (9), and it will be necessary to find only one point in each trace. 32. If it be required to pass a plane through two right lines which either intersect or are parallel, we have simply to find the points in which these lines pierce the planes of projection, as in the preceding problem. If the lines do not pierce the planes of 2 18 DI SCRIPTIVE GEOMETRY. projection, within the limits of the drawing, then any two points of the lines may be joined by a right line, and a point in each trace be determined, by finding the points in which this line pierces the planes of projection. 33. A plane may be passed through a point and right line, by joining the point with any point of the line by a right line, and then passing a plane through these lines; or by drawing through the point a line parallel to the given line, and then passing a plane through the parallels, as above. 34. PROBLEM 4. To find the angle between two right lines which intersect. Let MN and MO, Fig. 16, be the two right lines, assumed as in Art. (23). Analysis. Since the lines intersect, pass a plane through them, and revolve this plane about its horizontal trace until it coincides with the horizontal plane, and find the revolved position of the two lines. Since they do not change their relative position, their angle, in this new position, will be the required angle. Construction. The line MN pierces H at n, and the line MO at o, Art. (27); no is then the horizontal trace of the plane containing the two lines, Art. (32). Revolve this plane about no until it coincides with H. The point M falls at p, Art. (17). The points n and o, being in the axis, remain fixed; np will then be the revolved position of MN, and po of NO, and the angle npo will be the required angle. 35. Second method for the same problem. Analysis. Revolve the plane of the two lines about its horizontal trace until it becomes perpendicular to the horizontal plane; then revolve it about its new vertical trace until it co DESCRIPTI VE GEOME I. TRY, 19 incides with the vertical plane; the angle will then be in the vertical plane in its true size. Constructionz. First revolve the plane about no until it becomes perpendicular to IH. Tt' will be the new vertical trace, the point M will be horizontally projected at s, and vertically at I', s'r being equal to sp. Now revolve the plane about Tt' until it coincides with V; o will revolke to o'", s to s", and n to n", while the point M, or (ss'), will be found at w; n"w and o"w will be the revolved positions of the two lines, and n"wo" the iequired angle. By examining the drawing, it will be seen that if the angle is oblique, it is less than its projection, unless both lines are parallel to the plane of projection, in which case the angle is equal to its projection. Let the problem be constructed with one of the lines parallel to the ground line. 36. If two right lines be perpendicular to each other mz space, and one of them parallel to the uplane of projection, their oprojections will be perpendicular. For the projecting plane of the line which is not parallel to the plane of projection is perpendicular to the second line, and also to its projection, since this projection is parallel to the line itself, Art. (14);.and since this projection is perpendicular to this projecting plane, it is perpendicular to its trace, which is the projection of the first line. 37. PROBLEM 5. To find the position of a line bisecting the angle formed by two right lines, one of which is perpendicular to either plane of projection. Let MN and OP, Fig. 17, be the two lines, the latter being perpendicular to the vertical plane. Analysis. If the plane of the two lines be revolved about the second, until it becomes parallel to the horizontal plane, the angle 20 D)E CRIPTIVE G i:OMEITRIY. will be projected on this plane in its true size, and max' be bisected by a right line. If the plane be then revolved to its primiltive position, and the true position of one point of the bisecting line be detelmined, and joined wih the vertex of the given angle, we shall have the required line. Cotstrzuction. Let the plane of the two lines be revolved about OP, until it becomes parallel to H. Any point of MN, as M, will describe the arc of a circle parallel to V, and be horizontally projected at mn", and om" will be the projection of MN, and mrz"op will be the true size of the angle. Bisect it by oq, which will be the horizontal projection of the bisecting line in its revolved posl tion. Join mn" with any point of op, as p; this will be the horizontal projection of a line of the given plane, in it.s revolved position, which intersects the bisecting line in a point thorlzontallY projected at q. When the plane resumes its prim1it1ve posi]tion, this line will be horizontally projected in mp. anei the point, of which q is the horizontal projection, will be horizontally projected at r, and vertically at r'; hence or will be the honrzontal, anti o'm' the vertical projection of the required line. Or the plane of the two lines may be revolved about its vertical trace, and the true position determined as indicated in the figurle. 38. PROBLEM 6. To finll the intersection of two planes Let tTt' and sSs', Fig. 18, be the two planes. _Analysis. Since the line of intersection is a right line, contained in each plane, it must pierce the horizontal plane in the horizontal trace of each plane, Art. (30); that is, at the intersection of the two trauces. F r the same reason, it must pierce the vertical plane at the intersection of the vertical traces. If these two points be joined by a right line, it will be the required intersection. Constriuctiozn. The required line pierces H at o and V at p" o is its own horizontal prl jection, and p' is horizontally proiected at p; hence po is the horizontal projection of the lrt )DESCRIPTIVE GEOMETRY. 2 1 quired line; o is vertically projected at o'; p' is its own vertical projection; and o'p' is the vertical projection of the required line. 39. Second:7ethodi for the same problem. When either the horizonta. or vertical traces do not intersect within the limits of the drawing. Let tTt' and sS"', Fig 19, be tile planes; tTl and sS not intersecting within the limits of the drawing. An1alysis. If we pass any plane parallel to the vertical plane, it will intersect each of the given planes in a line parallel to its vertical trace, and these two lines will intersect in a point of the required intersection. A second point may be determinedl in the same way, and the right line joining these two points will be the required line. Conlstructioin. Draw pq parallel to AB; it will be the trace of an auxiliary plane. It intersects the two given planes in lines, which pierce H at p and q, and are vertically projected in p'o' and q'o'; o' is the vertical, and o the horizontal projectiotl of their intersection. Draw mnn also parallel to AB, and thus determine L. OL is the required line. Let the problem be constructed when both planes are parallel to the ground line. 40. PROBLEM 7. To find the point in which a given right liter pierces a givenplane. Let MN, Fig. 20, be the given line, and tTt' the given plane. Analysis. If through the line any plane be passed, it will intersect the given plane in a right line, whichl must contain the required point, Art. (30) This point must also be on the given line; hence it will be at the intersection of the two lilies. Construction. Let the auxiliary plane be the horizontal projecting plane of the line; np is its horizontal and pt' its vertical trace, Art. (10). It intersects tTt' in a right line, which pierces II at o and V at t', of which o't' is the vertical proqje' 2' -' DESCRIPTIV'E GEOMET RY. tion, Art. (38). The point in', in which o't' intersects m'n', is the vertical projection of the required point; and mb is its hlori. zontal projection. The accuracy of the drawing may be verified by using the vertical projecting plane of MN, as an auxiliary Ilane, and determining m directly as represented in the figure. Let the problem be constructed when the given line is paralle to the ground line. 41. Second Method for the same problem. When the plane is given by any two of its right lines, find the points in which these two lines pierce either projecting plane of the given line, and join these points by a straight line; this will intersect the given line in the required point. Con.struction. Let MN and OP, Fig. 21, be the lines of the given plane, intersecting at L, and QR be the given line. The line MN pierces the horizontal projecting plane of QR at a point of which m is the horizontal, and m' the vertical projection. 0' pierces the same plane at P, and p'm' is the vertical projection of the line joining these two points. This intersects q'r' at r', which is the vertical projection of the required point, and ir its horizontal projection. 42. If either projection of a point of an oblique plane be given, the other projection may at once be determined by a simple application of the principles of the preceding problem. Thus let mn, Fig. 22, be the horizontal projection of a point of the plane ITt'. If at im a perpendicular be erected to HI, it will pierce tTt' in the only point of the plane which can be horizontally projected at m; m is the horizontal, and mn"mn' the vertical projection of this perpendicular. Through it pass any plane, as that whose horizontal trace is no. Since this plane is perpendicular to H, nn' will be its vertical trace. It intersects tTt' in a right line, of which o'n' is the vertical projection; hence n' is the required vertical projection, Art. (40.) DESCRIPTIVE GEOMETRY. 2 The auxiliary plane may be passed parallel to tT; mp will be its horizontal, and pp' its vertical trace. It intersects tTt' in a line parallel to tT, which pierces V at p', of which mp is the horizontal, and nm'' the vertical projection, and m' will be the required vertical projection. In a similar way, if the vertical projection be given, the hori zontal can be found. 43. Irf a right line is perpendicular to a plane, its projectionts will be respectively perpendicular to the traces of the plane. For the horizontal projecting plane of the line is perpendicular to the given plane, since it contains a line perpendicular to it. This projecting plane is also perpendicular to the horizontal plane, Art. (11). It is therefore perpendicular to the intersection of these two planes, which is the horizontal trace of the given plane. Hence the horizontal projection of the line, which is a line of this projecting plane, must be perpendicular to the horizontal trace. In the same way it may be proved that the vertical projection, of the line will be perpendicular to the vertical trace. Conversely, if the projections of a right line are respectively ier pendicular to the traces of a plane, the line will be perpendicular to the plane. For, if through the horizontal projection of the line, its hori. zontal projecting plane be passed, it will be perpendicular to the horizontal trace of the given plane, and therefore perpendicular to the plane. In the same way it may be proved that the vertical projecting plane of the line is perpendicular to the given plane; therefore the intersection of these two planes, which is the given line, is perpendicular to the given plane. Hence, to assume a right line perpedicular to a plane, we draw its projections perpendicular to the traces of the plane respectively. Also, to assume a plane perpendicular to a. right line, we draw '24 DESCRIPTIVE GEOM2FETRY. the two traces from any point in the ground line, perpen(liculat to the projections of the line. 44. PROBLEM 8. To draw through a given point a right lise pvependicular to a given plane, and to find the distance Qf t]h voint from the plane. Let M, Fig. 23, be the given point, and ITt' the plane. Analysis. Silce the required perpendicular is to pass through the given point, its projections must pass through the projections of the point, Art. (13); and since it is to be perpendicular to the plane, these projections must be respectively perpendicular to the traces of the plane, Art. (43). Hence, if through the horizontal projection of the point, a right line be dirawn perpendicuiar to the horizontal trace, and through the vertical projection, a right line perpendicular to the vertical trace, they will be respectively the horizontal and vertical projections of the required line. If the point in which this perpendicular pierces the plane be found, the distance between this point and the given point will be the required distance or length of the perpendicular. Construction. Through m draw mn perpendicular to tT, and through m', m'r', perpendicular to t'T. MN will be the required perpendicular. N is the point in which MN pierces the plane, Art. (40), and m"n" the length of the perpendicular, Art. (28). Let the problem be constructed when the plane is parallel to the ground line; also when it is perpendicular to it. 45. PROBLEM 9. To project a given right line on any oblique -lane, and to show the true position of this projection. Let MN, Fig. 24, be the given line, and tTt' the given plane. Analysis. If through any two points of the line perpendicaiars be drawn to the plane, and the points in which they l ESCRIP'TIVE GEOMETl'rY. 2) pierce the plane be found, these will be two )points of the required projection, and the right line joining( them will be the required line. If l!ow the plane be revolved about its horizontal trace to coincide with the horizontal plane, or aboutt its vertica1 trace until it coincides with the vertical p)lane, and t!e revolved position of these two points be fbunol and joined by right line, this will show the true position of the line in the oblique plane. ConstruCtion. Assume the two points M and P, Art. (22), and draw the perpendiculars AIR and PS, Art. (44). The first pierces the plane at Ix, and the second at S, Art. (40); and rs will hte the horizontal, and r's' the vertical projection of the requiredl projection. The point N, in which the given line pierces the plane, will also be one point of the required projection. Now revolve the plane about tT until it coincides with II. 11 is found at r', Art. (17), and S at s", and r"e" is the true position of RS in its own plane: r"s" produced must pass through thle point in which the projection pierces H. If the given line be parallel to the plane, it will only be necessary to determine the projection of one point on the plane, and througlh this to draw a line parallel to the given line, Art. (14). 46. PROBLEM 10. Through a given point, to pass a plane perpendicular to a yiven r'ight line. Let M, Fig. 25, be the given point, and NO the given line. Anctlysis. Since the plane is to be perpendicular to the line, its traces must be respectively perpendicular to the projections of the line, Art. (43). We thus know the direction of the traces. Through the point, draw a line parallel to the horizontal trace; it will be a line of the required plane, and will pierce the vertical plane in a point of the vertical trace. Thlrough this point draw a right line perpendicular to the vertical projection of the line; it will be the vertical trace of the required plane. Through the point in which this trace intersects the ground line, draw a rigaht 2(3 DESCRIPT'IVE GEOMETRY. line perpendicular to the horizontal projection of the line; it will be the horizontal trace. Construction. Through m, draw mp, perpendicular to no; it will be the horizontal projection of a line through AI, parallel to the horizontal trace; and since this line is parallel to II, its veI'tieal projection will be m'p', parallel to AB. This line pierces V at po, Art. (27). Draw p'T perpendicular to n'o', and Tt perpendicular to no; tTp' Iwill be tle required plane. Or, through AM draw MS parallel to the vertical trace. It pierces H at s, which must be a point of the horizontal trace, and the accuracy of the drawing may thus be tested. 47. PROBLEM 11. To pa.s's a platne through a given point, parallel to two given right lines. Let AM, Fig. 26, be the point, and NO, and PQ, the two given lines. Anaclyis. Through the given point, draw a line parallel to each of the given lines. The plane of these two lines will be the required plane, since it contains a line parallel to each of the given.lines. Costfruction. Through rn draw m.s, parallel to no, and throughl m', nm's' parallel to n'o'. The line MS will be parallel to NO, Art (i6). In the same way, construct MR parallel to QP. These lines pierce II at s and t respectively, and MR pierces V at "'; hencIe tT,' is the required plane, Art. (32). Let the problemn he constructed when one of the given lines is parallel to the ground line. Let the problem, to pass a plane through a given point parallel to a given plane, also be constructed. 48. PROBLEM 12. To ptas.s a plarne through a given right line, parallel to another'ight t line. Let MN, Fig. 27, be the line through which the plane is to be Dassed, and PQ the other given line. DESCRIPTIVE GEOMETRY. 27 Analysis. Through any point of the first line, drain a line parallel to the second. Through this auxiliary line and the first, pass a plane. It will be the required plane. Construction. Through R, on the first line, draw RO parallel to PQ It pierces H at o, and V at t'. MN pierces II at m%, ald V at n'; hence oTt' is the required plane. Let the problem be constructed when either line is parallel te the ground line. 49. PROBLEM 13. To find the shortest distance from a given point to a given rigqht line. Let M, Fig. 28, be the given point, and NO the given straight line. Alnclysis. The required distance is the length of a perpendi, ular from the point to the line. If through the given point art,the line we pass a plane, and revolve this plane about eithe trace until it coincides with the corresponding plane of projection. the line and point will not change their relative positions; hence, if through the revolved position of the point we draw a perpendicular to the revolved position of the line, it will be the required distance. Con.struction. Through M draw MP parallel to NO. It pierces H at p. NO pierces H at o. po is then the horizontal trace of the plane through M and NO, Art. (32). Revolve this plane about op until it coincides with H. M falls at m", Art. (17). Since p remains fixed, pm" is the revolved position of MP NO being parallel to MP before revolution, will be parallel after; and as o is in the axis, oq" parallel to pm" will be the revolved position of NO. Draw i"q" perpendicular to oq"; it will be the, required distance. When the plane is revolved back to its primitive position, mi" is horizontally projected at %n, and q" at q hence MQ is the perpendicular in its true position. 2 8 DDKSCRTPTIVE GEOMETRY.,50. Secon)Cd method for the same problem. Ainalysis. If through the given point a plane be paosed per. pendicular to the given line, Art. (46), and the point in whicli the given line pierces the plane be found, Art. (40), and joined with the given point, we shall have the required distance, tht true length of which can be found as in Art. (28). Let the problem be constructed in accordance with this analysis. Let the problem also be constructed when the given line is parallel to the horizontal plane. 51. PROBLEM 14. To find the angle which a given right line mnakes with a given plane. Let MN, Fig. 29, be the given line, and tTt' the given plane. Analysis. The angle made by the line with the plane, is the same as that made by the line with its projection on the plane. Hence, if through any point of the line a perpendicular bie drawn to the plane, the foot of this perpendicular will be one poi.nt of the projection. If this point be joined with the point in which the given line pierces the plane, we shall have the projection of the line on the plane, Art. (45). This projection, the perpendicular, and a portion of the given line, form a right-angled triangle; of which the projection is the base, and the angle at the base is the required angle. IBut the angle at the vertex, that is, the anygle between the perpendiculur and given line, is the coilnplement of the required angle; hence, if we find the latter angle, and subtract it from a right angle, we shall have the required angle. Conlstruction. Through MI draw the perpendicular MP to tTt' Art. (44). It pierces HI in p. The given line pierces H in o, and op is the horizontal trace of the plane of the two lines, Art. (32). Revolve this plane about op, and determnine their angle pin'o, as in Art. (34). Its complement, pren', is equal to the re uired anule. DESCRIPTIVE GEOMETRY. 29 Let the problem be constructed when the plane is parallel tc the ground line. 52. PtROBLE:M 15. To find the angle between two given planes. Let sSs', Fig. 30, and tTt', be the two planes, intersecting in tllhe line ON, Art. (38). Acnalysis. If we pass a plane perpendicular to the intersection of the two planes, it will be perpendicular to both; and cut from each a right line perpendicular to this intersection at a common point. The angle between these lines Xwill be the measure of the required angle. Coinstruction. Draw pq perpendicular to on; it will be the horizontal trace of a plane perpendicular to ON, Art. (43). This plane intersects the given planes in right lines, onl of which lpierces HI at p, and the other at q. If right lines be drawn from these points to the point in which the auxiliary plane inter sects ON, they will be the lines cut fiom the planes, and the angle between them will be the required angle. The vertical trace of the auxiliary lplane may be dllawn as in Art. (43), and the vertex of the angle found as in Art. (40), and then the angle as in Art. (34). Or otherwise, thus: Suppose a right line to be drawn from r to the vertex of the angle, it will be perpendicular to ON, since it is contained in a plane perpendicular to. it; it wili also be perpendicular to Iq, since it is in the horizontal projecting plane of ON, which is perpendicular to pq, Art. (43). If this projecting plane be revolved about no until it coincides with H, r' will fall at n"; and since o is fixed, on" will be the revolved position of ON, and rm", perpendicular to on", will be the revolved position of the line joining r with the vertex. If now the plane of the two lines be revolved about pq until it coincides with H, mn" will be at v, rv being equal tc rtn", and pvq will be the required angle, Art. (34). The point in", from its true position, is tlorizontally projectea at in, and vertically at m', and pmq is the horizontal, and p'm'q' the vertical projection of the angle. 50 1)ESCRIPTIVE GEOMETRY Let the problem be constructed when both planes are parallel to the ground line. 53. If the angle between a given plane and either plane of projection, as the horizontal, be required, we simply pass a plane perpendicular to the horizontal trace, as in Fig. 31. This plane cuts on firom H, and( ON firom tTt', and the angle norn", found by revolving the auxiliary plane about on, Art. (34), will be the required angle. In the same way the angle p'q'p", between the given plane and vertical plane, may be found. 54. PKOBLEM 16. Either trace of a plane being given, ah, the anclle which the plane msakes with the correesponding plane o; pr)ojection, to construct the other trace. Let tT, Fig. 31, be the horizontal trace of the plane, and del the angle which the plane makes with the horizontal plane. Analysis. If a right line be drawn through any point of the given trace, perpendicular to it, it will be the horizontal trace of a plane perpendicular to the given trace, and if at the same point a line be drawn, making with this line an angie equal to the given angle, this will be the revolved position of a line cut from the required plane by this perpendicular plane, Art. (53). If this line be revolved to its true position, and the poignt in which it pierces the vertical plane be found, this will be a point of the required vertical trace. If this point be joined with the point where the horizontal trace intersects the ground line, we shall have the vertical trace. C(Yonstreuction. Through o draw no perpendicular to tT; also on", making the angle non" = def; on" will be the revolved position of a line of the required plane. When this line is revolved to its true position. it pierces V at n'. and n'T is the re. quired trace. DESCRIPTIVE GE' IMETRY. 31 If the given trace does not intersect the ground line within the limits of the drawing, the same construction may be made at a second point of the trace, and thus another point of the vertical trace be determined. 55. PROBLEM 17. To find th.e shortest lile,which can be drawn, termrinating in two 9right lines, not in the sam.'ae plan.e. Let AN, Fig. 32, and OP, be the two right lines. Anzalysis. The required line is manifestly a right line, perpendicular to both of the given lines. If through one of the lines we pass a plane parallel to the other, and then project this second line on this plane, this projection will be parallel to the line itself, Art. (14), and therefore not larallel to the first lille. It will then intersect the first line'in a point. If at this point we erect a perpendicular to the plane, it will be contained in the projecting plane of the first line, be perpendicular to both lines, and intersect them both. That portion included between them is the required line. Construction. Through MN pass a plane parallel to OP, Art. (48): Ymr is its horizontal, and k'n' its vertical Irace. Through any point of OP, as Q, draw QU perpendicular to this plane, Art. (49). It pierces the plane at U, Art. (40); and this is one point of the projection of OP on the l)arallel plane. Through U draw UX parallel to OP; it will be the projection of OP on the plane. It intersects MIN in X, which is the point through which the required line is to be drawn; and XY, perpendicular to the plane, is the required line, the true length of which is x"y", Art. (28). Let the problem be constructed with one of the lines parallel to the ground line. Also with one of the lines perpendicular to either plane of projectionm 342 DESCRIPTIVE GEOMETIRl'. 56. Secondc construction of the same problem. Let MN and OP, Fig. 33, be the right lines. Through MN pass a plane parallel to OP, Art. (48): mr is its horizontal trace. Through p conceive a perpendicular to be drawn to this plane. The point in which it pierces thle lplane will be one point of the projection of OP on the plane. To and this point, through the perpendicular pass a plane perpendicular to OP; pq will be its horizontal trace, Art. (43). This plane will intersect the parallel plane in a right line, which pierces H at q. It intersects the horizontal projecting plane of OP in a right line perpendicular to OP at p. To determine this line, revolve the projecting plane of OP about op until it coincides with H. Any point of OP, as L, falls at 1", and pl" is the revolved position of OP. This projecting plane intersects the parallel plane in a right litie, which pierces i1 at k, and is parallel to OP; ku, parallel to pl", is the revolved position of this parallel line; pu, perpendicular to pi", is the revolved posi. tion of the intersection of the projecting plane and perpendicular plane; and u is the revolved position of a point of the line of intersection of the perpendicular and parallel plane. Now revolve the plane perpendicular to OP about pq as an axis, until it coincides with H. The point, of which u is the revolved position, falls at u", and u"q is the revolved position of the line of intersection of the perpendicular and parallel plane; pp" is the revolved position of the line through p perpendicular to the paralel plane, and is equal to the distance required; and p"' is the revolved position of the projection of p on the parallel plane. In the counter-revolution, the point p" will be horizontally projected, somewhere in the perpendicular to the axis pyq; p" x" is the horizontal projection of the projection of Ol otl the parallel plane, and xy, perpendicular to mr, is the horizontal, and x'y' the vertical projection of the required line, DESCRIPTIVE GEIOMETRY. 33 GONSTRUCTION AND CLASSIFICATION OF LINES. 57. Every/ line may be generuted by the continaed motion of a point. if the generating point be taken in any position on th line, and then be moved to its next position, these two points may be regarded as frming an infinitely small fright line, or elementary line. The two points are consecutive points, or points havng no distance between them, and may practically be considered as one point. The line mlay thus be regarded as made up of an infinite number of infinitely small elements, each element indicating the direction of the motion of the point while generating that part ryf the line. 58. The law which directs the motion of the generating point, determines the nature and class of the line. If the point moves always in the same direction, that is, so that the elements of the line are all in the same direction, the line generated is a right line. If the point moves so as continually to change its direction from point to point, the line generated is a curved line or curve. If all the elements of a curve are in the same plane, the curve is of single curvature. If no three consecutive elements, that is, if no four consecutive points are in the same plane, the curve is of double curvature. We thus have three general classes of lines. I. RTsHT LINES: all of whose points lie in the same direction. II. CURVES OF SINGLE CULRVATURE: all of whose points lie in the same plane. III. CURVES OF DOUBLE CURVATURE: no four consecutive points of which lie in the same plane. i9. The simplest curves of single curvature are: 3 34 DESCRIPTIVE GEOMETRY. I. The circurnference of a circle, which may be generated by a point moving in the same plane, so as to remain at the same distance from a given point. II. A parabola, which may be generated by a point moving in the same plane, so that its distance from a given point shall be constantly equal to its distance from a given right line. The given point is the focus, the given right line the directri.r. If through the focus a right line be drawn perpendicular to the directrix, it is the axis of the parabola; and the point in which the axis intersects the curve is the vertex. From the definition, the curve may readily be constructed by points, thus: Let F, Fig. 34, be the focus, and CD the directrix. Through F draw FC perpendicular to CD. It will be the axis. The point V, midway between F and C, is a point of the curve, and is the vertex. Take any point on the axis, as P, and erect the perpendicular PM to the axis. With F as a centre, and CP as a radius, describe an arc cutting PM in the two points M and:M'. These will be points of the curve, since FM = CP = DM, also FM' = CP = D'M'. In the same way all the points may be constructed. III. An ellipse, which may be generated by a point moving in the same plane, so that the sum of its distances from two fixed points shall be constantly equal to a given right line. The two fixed points are thefoci. The curve may be constructed by points, thus: Let F and F', Fig. 35, be the two foci, and VV' the given right line, so placed that VF = V'F'. Take any point as P between F and F'. With F as a centre, andV'P as a radius, describe an arc. With F' as a centre, and VP as a radius, describe a second arc, intersecting the first in the points M and M'. These will be points of the required curve, since DESCRIPTIVE GEOMETRY 35 ME + MF' = VP + V'P = VV'; also M F + MF' = VV'. In the same way all the points may be constructed. V and V' are evidently points of the curve, since VF + VF' = VF' + VF' = VV'; also V'F + V'F - VV'. The point C, midway between the foci, is the centre of the curve. The line VV', passing through the foci, and terminating in the curve, is the transverse axis of the curve. The points V and V' are the vertices of the curve. ID' perpendicular to VV', at the centre, is the conjugate axis of the curve. If the two axes are given, the foci may be constructed thus: With D the extremity of the conjugate axis as a centre, and CV the semi-transverse axis as a radius, describe an arc cutting VV' in F and F'. These points will be the foci, for DF + DF' = 2CV - VV'. IV. The hyperbola, which may be generated by moving a point tin the same plane, so that the difference of its distances from two fixed points shall be equal to a given line. The two fixed points are the foci. The curve may be constructed by points, thus: Let F and F', Fig. 36, be the two foci, and VV' the given line, so placed that FV = F!V'. With F' as a centre, and any radius greater than F'V, as F'O, describe an arc. With F as a centre, and a radius FM, equal to F'O - VV', describe a second are, intersecting the first in the points M and M. These will be points of the required curve, since F'M - FM =-F'O - FM =VV'; also F'M' - FM' = VV In the same way any number of points may be determined. DESCRIPTIVE GEOMETIY. it is manifest, also, that if the greater radius be used with F as a centre, another branch, N'V'N', exactly equal to AlVM', will be described. V and V' are evidently points of the curve, since F'V - FV = VV'= F= V'- F'V', and are the vertices of the hyperbola. The point C, midway between the foci, is the centre, and VV' Is the transverse axis. A perpendicular, DD', to the transverse axis at the centre, is the indefinite conjugate axis. It evidently does not intersect the curve. PROJECTION OF CURVEs. 60. If all the points of a curve be projected upon the horizontal plane, and these projections be joined by a line, this line is the horizontal p,'qjection of the curve. Likewise, if the vertical projections of all the points of a curve. be joined by a line, it will be the vertical projection of the curve. 61, The two projections of a curve being given, the curve will, zn general, be completely determined. For in the same perpendicular to the ground line, two points, one on each projection, may be assumed, and the corresponding point of the curve determined, as in Art. (8). Thus nm and m', Fig. 37, being assumed in a perpendicular to AB, M will be a point of the curve, and in the same way every point of the curve may in general be determined. 62. If the plane of a curve of single curvature is perpendicular to either plane of projection, the projection of the curve on that plane will be a right line, and all of its points will be projected into the trace of the plane on this plane of projection. DESCRIPTIVE GEOMETRY. 37 If the plane of the curve be perpendicular to the ground line, both projections wiil be right lines, perpendicular to the ground line, and the curve will be undetermined, as in Art. (15). If the plane of the curve be parallel to either plane of projection, its projection on that plane will be equal to itself, since each clement of the curve will be projected into an equal element Art. (14). Its plrojection on the other plane will be a riglht line, parallel to the ground line. The projection of a curve of double curvature can in no case be a right line. 63. The points in which a curve pierces either plane of projection can be found by the same rule as in Art. (27). Thus o, Fig. 37, is the point in which the curve MN pierces IH, and p' the point in which it pierces V. TANGENTS AND NORMALS TO LINES. 64. If a right line be drawn through any point of a curve, as I, Fig. 38, intersecting it in another point, as M', and then the second point be moved along the curve towards M, until it coincides with it, the line, during the motion containing both points, will become tangent to the curve at M, which is the point of contact. As when the point M' becomes consecutive with M, the line thus containing the element of the curve at M, Art. (57), may, for all prao(tical purposes, be regarded as the tangent, we say that a right line is tangent to another line, when it contains two consecutive 2points of that line. If a right line continually approaches a curve, and becomes tangent to it, at an infinite distance, it is called an asymptote q/ the curve. Two curves are tangent to each other, when they contain two consecutive points, or have, at a common point, a commonZ tan(ge7nt. If a right line is tangent to a curve of single curvature, it will 38 DESCRIPTIVE GEOMETRY. be contained in the plane of the curve, for it passes through twe points in that plane, viz., the two consecutive points of the curve. Also, if a right line is tangent to another right line, it will coincide with it, as the two lines have two points in common. Tie expression, "a tangent to a curve," or "a tangent," will hereafter be understood to mean a rectilineal tangent, unless otherwise mentioned. 65. If two lines are tangent in space, their projections on the same plane will be tangent to each other. Fol the projections of the two consecutive points, common to the two lines, will also be consecutive points, common to the projections of both lines, Art. (60). The converse of this is not necessarily true. But if both the horizontal and vertical projections are tangent at points, which are the projections of a common point of the two lines, Art. (23), the lines wiil be tangent in space; for the projecting perpendiculars, at the common consecutive points, will intersect in two consecuFive points common to the two lines. 66. If a right line be drawn perpendicular to a tangent at its point of contact, as MO, Fig. 38, it is a normal to the curve. As an infinite number of perpendiculars can be thus drawn, all in a plane perpendicular to MT at M, there will be an infinite number of normals at the same point. If the curve be a plane curve, that is, a curve of single curvature, the term "normal" will be understood to mean that normal which is in the plane of the curve, unless otherwise mentioned. i7. If we conceive a curve to be rolled on its tangent at any point, until each of its elements in succession comes into this tangent, the curve is said to be rectified; that is a right line, equal to it in length, has been found. DESCRIPTIVE GEOMETRY. 39 Since the tangent to a curve at a point contains the element of the curve, the angle which the cuive, at this point, makes with any line or plane will be the same as that made by the tangent. THE HELIX. 68. If a point be moved uniformly around a right line, remaining always at the same distance from it, and having at the same time a uniform motion in the direction of the line, it will generate a curve of double curvature, called a helix. The right line is the axis of the curve. Since all the points of the curve are equally distant from the axis, the projection of the curve on a plane perpendicular to this axis will be the circumference of a circle. Thus let ir, Fig. 39, be the horizontal, and a'n' the vertical projection of the axis, and P the generating point, and suppose that while the point moves once around the axis, it moves through the vertical distance m'n'; prqs will be the horizontal projection of the curve. To determine the vertical projection, divide prqs into any number of equal parts, as 16, and also the line m'n' into the same number, as in the figure. Through these points of division draw lines parallel to AB. Since the motion of the point is uniform, while it moves one-eighth of the way round the axis it will ascend one-eighth of the distance m'n', and be horizontally projected at x, and vertically at x'. When the point is horizontally projected at ir, it will be vertically projected at r'; and in the same way the points y', q', &c., may be determined, and p'r'q's' will be the required vertical projection. 69. It is evident from the nature of the motion of the genera. ting point, that in generating any two equal portions of the curve, it ascends the same vertical distance; thtat is, any two eletnentaly 40 DESCRIPTIVE GEOMETRY. arcs of the curve will make equal angles with the horizontal plane. Thus, if CD]) (a), Fig. 39, be any element of the curve, tlle angle which it makes with the horizontal plane will be DCe, cr the angle at the base of a right-angled triangle of which Ce = cd, Fig. 39, is the base and De the altitude. But fiom the nature of the motion, Ce is to De as any arc px is to the corresponding ascent xx'2. fhence, if we rectify the arc.7p, Art. (67), and with this as a base construct a right-angled triangle, having x'x" for its altitude, the angle at the base will be the angle which the arc, or its tangent at any point, makes with the horizontal plane. Therefore, to draw a tangent at any point as X, we draw xz tangent to the circle pxi' at x; it will be the horizontal projection of the required tangent. On this, from x, lay off the rectified arc xp to z; z will be the point where the tangent pierces It, and z'x' will be its vertical projection. Since the angle which a tangent to the helix makes with the horizontal plane is constant, and since each element of the curve is equal to the hypothenuse of a right-angled triangle of which the base is its horizontal projection, the angle at the base, the constant angle, and the altitude, the ascent of the point while gen-erating the element; it follows, that when the helix is rolled out on its tangent, the sum of the elements, or length of any portion of the curve, will be equal to the hypothenuse of a right-angled triangle, of which the base is its horizontal projeceion rectified, and altitude, the ascent of the generating point while generating the portion considered. Thus the length of the are pX is equal tu the length of the portion of the tangent ZX. GENERATION AND,CLASSIFICATION OF SURFACES. 70. A surface may be generated by the continued Emotion (!f a lifne. The moving line is the (/eneratrix of the surfa:ce; and the different positions of the generatrix are the elements. If the generatrix be taken in any position, and then be moved DESCRIPTIVE GEOMETRY. 41 to its next position on the surface, these two positiol s are consecutive positions of the generatrix, or consecutive elements of the surface, and may practically be regarded as one element. 71. The form of the generatrix, and the law which directs its motion, determine the nature and class of the surface. Surfaces may be divided into two general classes. First. Those which can be generated by right lines; or which have rectilinear elements. Second. Those which can only be generated by curves, and which can' have no rectilinear elements. These are DOUBLE CURVED SURFACES. Those which can be generated by right lines are: First. PLANES, which may be generated by a right line moving so as to touch another right line, having' alltits positions parallel to its first position. Second. SINGLE CURVED SURFACES, which may be generated by a right line, moving so that any two of its consecutive positions shall be in the same plane. Third. WA11PED SUIRFACES, which may be generated by a right line moving so that no two of its consecutive positions shall be in the same plane. 72. Single curved sulfaces are of three kinds. I.'Those in which all the positions of the rectilinear generatrix are parallel. II. Those in which all the positions of the rectilinear generatrix intersect in a common point. 1.II. Those in which the consecutive positions of the rectilinear generatrix intersect two and two, no three positions intersecting in a common point. i2 DESCRIPTIVE GEOMETRY. CYLINDRICAL SURFACES, OR CYLINDERS. 73. Single curved surfaces of the first kind are Cylindrical sur "aces, or Cylinders. Every cylinder may be generated by movitn a right line so as to touch a curve, and have all its positions parallel. The moving line is the rectilinear generatrix. The culve is the directrix. The different positions of the generatrix are the rectalinear elements of the surface. Thus, Fig. 40, if the right line MN be moved along the curve rolo, haling all its positions parallel to its first position, it will generate a cylinder. If tthe cylinder be intersected by any plane not parallel to the rectilinear elements, the curve of intersection may be taken as a directrix, and any rectilinear element as the generatrix, and the surface be re-generated. This curve of intersection may also be the base of the cylinder. The intersection of the cylinder by the horizontal plane is asually taken as the base. If this base have a centre, the right line through it, parallel to the rectilinear elements, is the axis of the cylinder. A definite portion of the surface included by two parallel planes is sometimes considered; in which case the lower curve of intersection is the lower btoe, and the other the upper base. Cylinders are distingui-hed by the name of their bases; as a cylinder with a circular base; a cylinder with an elliptical base. If the rectilinear elements are perpendicular to the plane of the base, the cylinder is a right cylinder, and the base a right section. A cylinder may also be generated by moving the curvilinear directrix, as a generatrix, along any one of the rectilinear elements, as a dirlectrix, the curve remaining always parallel to its first position. DESCRIPTIVE GEOMETRY. 43 If the curvilirear directrix be changed to a right line, the cylinder becomes a plane. It is manifest that if a plane parallel to the rectilinear elements intersects the cylinder, the lines of intersection will be rectilinear (:lements, which will intersect the base. 74. It will be seen that the projecting lines of the different points of a curve, Art. (60), form a right cylinder, the base of which, in the plane of projection, is the projection of the curve. These cylinders are respectively the horizontal and vertical proiecting cylinders of the curve, and by their intersection determine the curve. 75. A cylinder is represented by projecting one or more of the curves of its surface, and its principal rectilinear elements. When these elements are not parallel to the horizontal plane, it is usually represented thus: Draw the base, as molo, Fig. 40, in the horizontal plane. Tangent to this, draw right lines lx and kr, parallel to the horizontal projection of the generatrix; these will be the horizontal projections of the extreme rectilinear elements, as seen froln the point of sight, thus forming the horizontal projection of the cylinder. Draw tangents to the base, perpendicular to the ground line, as mm', oo'; through the points 9m' and o', draw lines m'n/' and o's', parallel to the vertical projection of the generatrix, thus forming the vertical projection of the cylinder; m'o' being the vertical projection of the base. 76. To assume a point of the surface, we first assume one of itv projections, as the horizontal. Through this point, erect a per pendicular to the horizontal plane. It will pierce the surface in the only points which can be horizontally projected at the point t.aken. Through this perpendicular, }pas. a plane parallel to the 14 DESCRIPTIVE GE;OMElTRY. rectilinear elements; it will intersect the cylinder, in eletnents Art. (73), which will be intersected by the perpendicular in the required points. Co2struction. Let p, Fig. 40, be the horizontal projection as sulmed. Through p, draw pq parallel to lx; it will be the horizontal trace of the auxiliary plane. This plane intersects the cylinder in two elements; one of which pierces H at q, and the o~ther at u; and q'y', and U'z", will be the vertical projections of these elements, v'p' the vertical projection of the perpendicular, and p' and p" the vertical projections of the two points of the surface, horizontally projected at p. To assume a rectilinear element, we have simply to draw a line parallel to the rectilinear generatrix, through any point of the base, or of the surface. CONICAL SURFACES, OR CONES. 77. Single curved surfaces of the second kind, are Conical sur. faces, or Cones. Every cone may be generated by moving a right line so as continually to touch a given curve, and pass through a given point not in the plane of the curve. The moving line is the rectilinear generatrix; the curve, the directrix; the given point, the vertex of the cone; and the different positions of the generatrix, the rectilinear elements. The generatrix being indefinite in length, will generate two parts of the surface, on different sides of the vertex, which arc called nappes; one, the upper, the other, the lower naoppe. Thus, if the right line MS, Fig. 41, move along the curve rlo and continually pass through S, it will generate a cone. If the cone be intersected by any plane not passing through the vertex, the curve of intersection may be taken as a directrix, and any rectilinear element as a genera trix, and the cone be regenerated. This curve of intersection may also be the base of the DESCRIPTIVE GEOMETRY. 4., cone. The intersection of the cone by the horizontal plane, is usually taken as the base. If a definite portion of the cone included by two parallel plaw.es is considered, it is called a frustum of a cone; one of the limlit iog curves being the lower, and the other the upper base of the i;rustuln. Cones are distinguished by the names of their bases; as a cone with a circular base; a cone with a parabolic base, &c. If the rectilinear elements all make the same angle with a right line passing through the vertex, the cone is a right cone, the right line being its axis. A cone may also be generated by moving a curve so as continually to touch a right line, and change its size according to.a proper law. If the curvilinear directrix of a cone be changed to a right line, or if the vertex be taken in the plane of the curve, the cone will become a plane. If the vertex be removed to an infinite distance, the cone will evidently become a cylinder. If a cone be intersected by a plane through the vertex, the lines of' intersection will be rectilinear elements, intersecting thle base. 78 A cone is represented by projecting the vertex, one of the curves onl its surface, and its principal rectilinear elements. Thlls, let S, Fig. 41, be the vertex. Draw the base, mlo, in the horizontal plane, and tangents to this base through s, as sl and sk; thus forming the horizontal projection of the cone. Draw tangents to the base, perpendicular to the ground line, as mm', oo'; and through m' and o', draw the right lines sm's' and o's', thus foilling the vertical projection of the cone. 79. To assume a point of the surface, we first assume one of itl projections, as the horizontal. Through this erect a perpen 46 DESCRIPTIVE GEOMETRY. dicular to the horizontal plane; it will pierce the surface in the only points which can be horizontally projected at the point taken. Through this perpendicular and the vertex pass a plane. It will intersect the cone in elements which will be intersected by he perpendicular in the required points. Conzstruction. Let p be the horizontal projection. Draw ps. It will be the horizontal trace of the auxiliary plane, which interects the cone in two elements; one of which pierces H at q, and the other at r, and q's' and?r's' are the vertical projections of these elements, and p' and p" are the vertical projections of the two points of the surface. To assume a rectilinear element, we have simply to draw through any point of the base, or of the surface, a right line to the vertex. 80. Single curved surfaces of the third kind, may be generated by drawing a system of tangents to any curve of double curvature. These tangents will evidently be rectilinear elements of a single curved surface. For if we conceive a series of c(lnsecttive points of a curve of double curvature, as a, b. c, d, &c., the tan-!~ent which contains a, and b, Art. (64), is intersected by the one which contains b and c, at b; that which contains b and c, by the one which contains c and d, at c; and so on, each tangent inter secting the preceding consecutive one, but not the others, since no two elements of the curve, not consecutive, are, in general, in the same plane, Art. (58). 81. If the curve to which the tangents are drawn, is a helix, Art. (68), the surface may be represented thus: Let pxy, Fig. 42, be the horizontal, and p'x'y' the vertical projection of the helical directrix. Since the rectilinear elements are all tangent to this directrix, any one may be assumed, as in Art. (69); hence XZ, YLU, &c., are elements of the surface. DESCRIP'IIVE G(}oMETRY. 47 A point of the surface may be assumed by taking a point or, any assumed element. These elements pierce H in the points z, u, v, &c., and zuvw is the curve in which the surface is intersected by the horizontal p ane, and may be regarded as the base of the surface; and it is evident that if the surface be intersected by any plane parallel to this base, the curve of intersection will be equal to the base. WARPED SURFACES WITH A PLANE DIRECTER. 82. There is a great variety of warped surfaces, differing froin each other in their mode of generation and properties. The most simple are those which may be generated by a right line generatrix, moving so as to touch two other lines as directrices, and parallel to a given plane, called a plane directer. Such surfaces are warped surfaces, with two lineur directrices and a plane directer. They, as all other warped surfaces, may be represented by proJecting one or more curves of the surface, and the principal rec-'ilinear elements. 83. The directrices and plane directer being given, a rectslinear element, passing through any point of either directrix, may be determined, by passing a plane through this point, parallel to the plane directer, and finding the point in which this plane cuts the other directrix, and joining this with the given point. Construction. Let MN and PQ, Fig. 43, be any two linear directrices, tTt' the plane directer, and O any point of the first direc.tix. Assume any line of the plane directer, as CD, Art. (30), and through the different points of this line, draw right lines SE, SF, &c., to any point, as s of tT. Through O draw a system of lines OR, OY, &c., parallel respectively to SE, SF, &c. These will 48 DESCRIPTIVE GEOMETRY. form a plane through 0, parallel to tTt', and pierce the horizonta projecting cylinder of PQ, in the points R, Y, W, &c. These points being joined, will form the curve RW, which intersects PQ in X, and this is the point in which the auxiliary plane cuts the directrix PQ. OX will then be the required element. If the curve r'w' should intersect p'q' in more than one point, two or more elements passing through O would thus be determined. 84. If an element be required parallel to a given fright line, this line being in the plane directer, or parallel to it, we draw through the different points of either directrix, lines parallel to the given line. These form the rectilinear elements of a cylinder, parallel to the given line. If the points in which the second directrix pierces this cylinder be found, and lines be drawn through themi parallel to the given line, each will touch both directrices, and be an element required. Cogst'uction. Let the surface be given as in the preceding Article, and let FS, Fig. 44, be the given line. Through the points 0, K, L, &c., draw OX, KY, LZ, &c., parallel to FS. These lines pierce the horizontal projecting cylinder of the directrix PQ. in the points X, Y, Z, &c., which, being joined, form the curve XY, intersecting PQ in W. Through W draw WR, parallel to FS; it is a required element, and there may be two or -nore as in the preceding Article. 85. A warped surface, with a plane directer, having one directrix a right line and the other a curve, is called a Conoid. Thle elements of the conoid pass through all the points of the rectiinear directrix, instead of a single point, as in the cone. If the rectilinear directrix is perpendicular to the plane directe:/ it is a right conoid, and this directrix is the line of striction. DESCRIPTIVE GEOMETRY. 49 THE HYPERBOLIC PARABOLOID. 86. A warped surface with a plane directer and two rectilineac iirectrices. is a. Hyperbolic Paraboloid, since its intersection by a plane may be proved to be either an hyperbola or parabola. Its rectili'iear elements may be constructed by the principles in Arts. (83 a,,. 84). In the first case, two right lines through the given point will determine the auxiliary plane, and the point in which it is pierced by the second directrix may be determined at once, as in Art. (41). In the second case, the cylinder becomes a plane, and the point in which it is pierced by the second directrix is also determined as in Art. (41). 87. The rectilinear elements qf a hyperbolic paraboloid divide the directrices proportionally; for these elements are in a system of planes parallel to the plane directer and to each other, and these planes divide the directrices into proportional parts at the points where they are intersected by the elements. (Davies' Legendre, Book vi., Prop. xv.) Conversely. If two right lines be divided into any snumber of proportional parts, the right lines joining the corresponding points of division will lie in a system of parallel planes, and be elements of an hyperbolic paraboloid, the plane directer of which is parallel to any two of these dividing lines. Thus, let MN and OP, Fig. 45, be any two rectilinear directrices. Take any distance, as ml, and lay it off on mn any number of times, as ml, 1k, kg, &c. Take also any distance, as vr, and lay it off on po any number of times, as pr, rs, su, qo, &c. Join the corresponding points of division by right lines, lr, ks, gu, &c.; these will be the horizontal projections of rectilinear elements of the surface. Through m, 1, k, &c., and p, r, s, &c., erect perpendiculars to AB, to in', I', k', &c., and p', r', s', &c., and join 4 50 1)ESCRLPTIVE GEOMETRY. the corresponding points by the right lines i'r', k's', g'it &cA these will be the vertical projections of the elements. 88. To assume any point on this s.rface, and in general, on any warped surface, we first assume its horizontal projection, and at this, erect a perpendicular to the horizontal plane. Through this perpendicular pass a plane; it will intersect the rectilinear elements in points which, joined, will give a line of the surface, and the points in which this line intersects the perpendicular will be the required points on the surface. Let the construction be made upon either of the figures, 43, 44, or 45. 89. If any two rectilinear elements of an hyperbolic paraboloid be taken as directrices, with a plane directer parallel to the first directrices, and a surface be thus generated, it will be identical with the first surface. To prove this, we have only to prove that any point of an element of the second generation is also a point of the first. Thus, let MN and OP, Fig. 46, be the directrices of the first generation, and NO and MP any two rectilinear elements. Through M draw MW parallel to NO. The plane WMP will be parallel to the plane directer of the first generation, and may be taken for it. Let NO and MP be taken as the new directrices, and let ST be an element of the second generation, the plane directer being parallel to MN and OP. Through U,any point of ST, pass a plane parallel to WMP, cutting the directrices MN and OP in Q and R. Join QR, it will be an element of the first generation Art. (Od). Draw Nn and Qq parallel to ST, piercing the plane WMP in n and q. Also draw Oo and Rr parallel to ST, piercing WMP in o and r; and draw no, intersecting MP in T, since Nn, ST, and Oo are in the same plane. M, q, and n will be in the same right line, as also P, r, and o; and since MN and Nn, inter DIKSCRIPIIVE GE()METRY 51 secting at N, are parallel to the plane directer of the second gen. oration, their plane will be parallel to it, as also the plane of PO and Oo; hence, these planes being parallel, their intersections, Mnt and Po, with the plane WMP, will be parallel. Draw qr. Since Qq and Nn are parallel, we have AMQ: QN M: qn. Also, PR: RO:: Pr ro. But Art. (87), MQ: QN:: PR: RO; hence, Mg: qn:: Pr: ro; and the right line. qr, must pass through T, anld the plane of the three parallels, Qg, ST, and Rr, contains the element QR, which must therefore intersect ST at U. H-lence, ally point of a rectilinear element of the second generation is also a point of an element of the first generation, and the two surfaces are identical. It follows from this, that through any point of an hyperbolic parabo. loid two rectilinear elements can always be drawn. WARPED SURFAOES WITH THREE LINEAR DIRECTRICES. 90. A second class of warped surfaces consists of those which may be generated by moving a right line so as to touch three lines as directrices; or wa)ped sztufaces with three linear directrices. A rectilinear element of this cldss of surfaces passing through a given point on one of the directrices, may be found by drawing through this point a system of right lines intersecting either of the other directrices; these form the surface of a cone which will 5* T:I)OSCRIPTIVE GEOMETRY. be pierced by the third directrix in one or more points, throuigh which and the given point right lines being drawn will touch the three directrices, and be required elements. Conzstruction. Let MN, OP, and QR, Fig. 47, be any three linear directrices, and M a given point on the first. Through M draw the lines MO, MS, MP, &c., intersecting OP in 0, S, P, &c. They pierce the horizontal projecting cylinder of QR in X, Y, Z, &c., forming a curve, XYZ, which intersects QR in U, the poinit in which QR pierces tile surface of the auxiliary cone. MU is then a required element, two or more of which would be deterrmined if XZ should intersect QR in more than one point. 91. The simplest of the above class of surfaces is a surjace which may be venerated by moving a right lilne so as to touch three rectilinear directrices. It is an Hyperboloid of one nappTe, as many of its intersections by planes are hyperbolas, while the surface itself is unbroken. The construction of the preceding Article is much simplified for this surface, as the auxiliary cone becomes a plane; and the point in which this plane is pierced by the third directrix is found as in Art. (40), or (41). THE HELICOID. 92. If a right line be moved uniformly along another right line, as a directrix, always making the same angle with it, and at the same time having a uniform angular motion aroun(l it, a warped surface will be generated, called a Helicoid. The rectilinear directrix is the axis of the surface. Thus, Fig, 48, let the horizontal plane he taken perpendicular to the axis, o being its horizontal, and o'n' its vertical projection, and let OP be the generatrix, parallel to the vertical plane. DESCRIPTIVE GEOMETRY. 53q It is evident froim thile nature of the motion of the generatrix, that each of its points will generate a helix, Art. (68). That generated by the point P, constructed as in Art. (68), will be horizontally projected in prq, and vertically in p'r'q'. To ass'ume a rectilinear element of the surwface, we first assume its horizontal projection, as ox. Through x erect the perpendicular xx', and fiom o' lay off the distance o'o", equal to x'x"; o"x' will be the required vertical projection. In the same way, the element (oy, o"'y') may be assumed. To assutme any point on the sec:fiice, we first assume a rectilinear element, and then take any point in this element, as M. The elements pierce the horizontal plane in the points p, u, w, &c., and the curve paw is its intersection by the surface. The surface is manifestly a warped surface, since, from the tnature of its generation, no two consecutive positions of the generatrix can be parallel or intersect. This surfaice is an important one, as it forms the curved surface of the thread of the ordinary screw. If the generatrix is perpendicular to the axis, the helicoid beconmes a particular case of the right conoid, Art. (85), the horizontal plane being the plane directer, and any helix the curvi. linear directrix. 93. Other varieties of warped surfaces may be generated by moving a right line so as to touch two lines, hlaving its differe1nt positions, in succession, parallel to the different rcctilinear el. ements of a cone: By moving it parallel to a given plane, so as to touch two surfaces, or one surface and a line; or so as to touch three surfaces-two surfaces and a line-one surface and two lines; and in general, so as to fulfil any three reasonable conditions. It should be remarked, that in all these cases the directrices should be so chosen that the surface generated will be neither a plane nor single curved surface. Also, that these surfaces, as all 54 DESCRIPTIVE GEOMETRY. others, may be generated by curves moved in accordance with a law peculiar to each variety. 94. If a curve be moved in any way so as not to generate a surface of either of the above classes, it will generate a double curved surface, the simplest variety of which is the surface of a sphere. 95. An examination of the modes of generating surfaces, above described, will indicate the following test for ascertaining the general class to which any given surface belongs. 1. If a straight ruler can be made to touch the surface in every direction, it must be a plane. 2. If the ruler can be made to touch a curved surface in any one direction, and then be moved so as to touch in a position very near to the first, and the two right lines thus indicated are parallel or intersect, the surface must be single curved. 3. If the lines thus indicated, neither intersect nor are parallel, the surface must be warped. 4. If the ruler cannot be made to touch the surface in any direction, the surface must be double curved. SURFACES OF REVOLUTION. 96. A surface of revolution, is a surface which may be generated by revolving a line about a right line os an axis, Art. (17). From the nature of this generation, it is evident that any intersection of such a surface by a plane perpendicular to the axis, is the circumference of a circle. Hence, the surface may also bec generated by moving the circumference of a circle with its plane perpendicular to the axis, and centre in the axis, and whose radius changes in accordance with a prescribed law. DESiCRIPTIVE GEOMETRY. 55 If the surface be intersected by a plane passing through the RniS, the line of intersection is a meridian line, and the pldne a meridian plane; and it is also evident that all meridian lines of the same surface are equal, and that the surface may be generated by revolving any one of these meridian lines about the axis 97. If two surfaces of revolution, having a common axis intersect, the line of intersection must be the circumference of a a', whose plane is perpendicular to the axis, and centre in the axis, For if a plane be passed through any point of the intersection and the common axis, it will cut friom each surface a meridiar line, Art. (96), and these meridian lines will have the point in common. If these lines be revolved about the common axis, each will generate the surface to which it belongs, while the common point will generate the circumference of a circle common to the two surfaces, and therefore their intersection. Should the meridian lines intersect in more than one point, the surfaces will intersect in two or more circumferences. 98. The simplest curved surface of revolution is that which may be generated by a right line revolving about another right line to which it is parallel. This is evidently a cylindrical sur — face, Art. (73), and if the plane of the base be perpendicular to the axis, it is a right cylinder with a circular base. If a right line be revolved about another right line which it intersects, it will generate a conical surface, Art. (77), which is evidently a right cone, the axis being the line with which the rectilinear elements make equal angles. These are the only two single curved surfaces of revolution. 56 DESCRIPTIVE GEOMETRY. THE HYPERBOLOID OF REVOLUTION OF ONE NAPPE. 99. If a right line be revolved about another right line, niot in the same plane with it, it will generate a warped surface, which is an fylqperboloid of' revolution of onle n(tqpe, Art. (91). To prove this, let us take the horizontal plane perpenrdicular to the axis, aind the vertical plane parallel to the generatrix in its firlt position, an(d let c, Fig. 49, be the horizontal, and c'nzm' the vertical projection of the axis, and MP the generatrix: cm will be the horizontal, and m' the vertical projection of the shortest distance between these two lines, Art. (55). As MP revolves about the axis, CAM will remain perpendicular to it, and M will describe a circumference which is horizontally projected in mxqy, and vertically in y'x'; and as CM is horizontal, its horizontal projection will be perpendicular to the horizointal projection of MP, in all of its positions, Art. (36), and remain of the same length; hence the horizontal projection of Ml, in any position, will be tangent to the circle mxy. No two consecutive positions can therefore be parallel. Neither can they intersect; for from the nature of the motion, any two must be separated at any point by tile elementary arc of the circle described by that point. The surface is therefore a waipoed surface. The point P, in which MP pierces HI, generates the circle pwq, which may be regarded as the base of the surface. The circle generated by M, is the smallest circle of the surface, and is the circle of the gorge. 100. To assume a rectilinear element, we take any point in the base, pwq, and through it draw a tangent to xmy, as wz; this wvll oe the horizontal projection of an element. Through z erect the perpendicular zz'; z' will be the vertical projection of the point DESCRIPTIVE GEOMETRY. 57 in which the element crosses the circle of the gorge, andl aw'z' will be the vertical projection of the element. To assume a point of the surface, we first assume a rectilinear element, as above, and then take any point of this element:. Art. (22). 101. If through the point M, a second right line, as MQ, be' drawn parallel to the vertical plane, and making. with the horizontal plane the same angle as MP, and this. line be revolved about tile same axis, it will generate the samle surface. For it any plane be passed perpendicular to the axis; as the plane whose vertical trace is e'g', it will cut MP and MQ in two points, E and G, equally distant fiom the axis, and these points will, in the revolution of M P arid MQ, generate the same circurmference; hence the two surfaces must be identical. The surface having two generations by different right lines, it follows that through any poit, of the su7face two rectilinear elements can be drawn. 102. Since the points E and G generate the same circumfer ence, it follows that as MP revolves, MQ remaining fixed, the point E will, at some tilne of its motion, coincide with G, and the generatrix, MP, intersect MQ in G. In the same way any other point, as F, will come into the point K of MQ, giving another element of the first generation, intersecting MQ at K; and so for each of the points of MPI in succession. In this case, kl will be the horizontal projection of the element of the first generation. Hence, if the generatrix of either generation remain fixed, it will in.tersect all the elements of the other generation. If, then, any three elements of either generation be taken as lirectrices, and an element of the other generation be moved so as to touch them, it will generate the surface. It is therefore ad hyperboloid (f one )dappe, Art. (91) An hyperboloid of revolution of one nappe may thus be gen 58 DESCRIPTIVE GEOMETRY. crated, by moving a right line so as to touch three other right lines, equally distant from a fourth (the axis), the distances being measured in the same plane perpendicular to the fourth, and the dlirectrices making equal angles with this plane. 103. If the vertical plane is not parallel to the generatrix in its first position, the circle of the gorge may be constructed thus: Let WZ be the first position of the generatrix, and through c draw cz perpendicular to wz; it will be the horizontal projection of the radius of the required circle. The point of which z is the horizontal projection, is vertically projected at z', and mzx will be. the horizontal, and y'x' the vertical projection of the circle of the gorge. With c as a centre, and cw as a radius, describe the base pwq, and the surface will be fully represented. 104. To construct a meridian curve of this surface, we pass a plane through the axis, parallel to the vertical plane. It will inteyfect the horizontal circles, generated by the different points of thegeneratrix, in points of the required curve, which will be vertical!y projected into a curve equal to itself, Art. (62). Thus the horizontal plane, whose vertical trace is e'g', intersects the generatrix in E, and eo is the horizontal projection of the circle generated by this point, and O is the point in which this circle lierces thle meridian plane. In the same way, the points whose vertical projections are n', x', n", o"f, &c., are determined. The plane whose vertical trace is e''o", at the same distance fr'om y'x' as e'o', evidently determines a point o" at the same distance from y'x', as o'; hence the chord o'o" is bisected by y'x', and the curve o'x'o" is symmetrical with the line y'x'. The distance e'o' equal to ce - me, is the difference between the hypothenuse ce and base me of a right-angled triangle, having the altitude mc. As the point o' is further removed from x', the DESCRIPTIVE GEOMETRY. 51) altitude of the corresponding triangle remains the same, while the hypothenuse and base both increase. If we denote the altitude by a, and the base and hypothenuse by b and h respesctively, we ca.ve a5 h2 - b2 = a, and h -b = it + b from which it is evident that the difference, e'o', continually di minishes as the point o/ recedes from x'; that is, the curve x'n'o continually approaches the lines p'm' and q'm', and will touch them at an infinite distance. These lines are t}hen asymptotes to the curve, Art. (64). If, now, any element of the first generation, as the one passing through I, be drawn, it will intersect the element of the second generation, MQ, in R, and the corresponding element on the op posite side of the circle of the gorge in U. Since this element' has but one point in common with the meridian curve, and no point of it, or of the surface, can be vertically projected on the right of this curve, the vertical projection'l'r/ must be tangent to the curve at i'. But since ri is equal to iu, r'i' must be equal to i'u'; or the part of the tangent included between the asymptotes is bisected at the point of contact. This is a property peculiar to the hyperbola (Analyt. Geo., Art. 164), and the meridian curve is therefore an hyperbola; YX being its transverse axis, and the axis of the surface its conjugate, Art (59). If this hyperbola be revolved about its conjugate axis, it will generate the barface, Art. (96), and hence its name. This is the only warped surface of revolution. 105. The most simple double curved surfaces of revolution are: I. A spherical surface or sphere, which may be generated by ievolving a circumference about its diameter. II. An ellipsoid of r'evolution, or spheroid, which may be gener ted by revolving an ellipse about either axis. When the t30 DESCRIPTIVE GEOMETRY. ellipse is revolved about the transverse axis, the surfalce is a pr&o tl(le spheroid; when about the conjugate axis, an oblate spheroid. III. A paraboloid of revolution, which may be generated by revolving a parabola about its axis. IV. An hyperboloid of revolution of two nappes, which may be generated by revolving an hyperbola about its transverse axis. 106. These surfaces of revolution are usually represented by taking the horizontal plane perpendicular to the axis, and then drawing the intersection of the surface with the horizontal plane; or the horizontal projection of the greatest horizontal circle of the surface for the horizontal projection, and then projecting on the vertical plane the meridian line which is parallel to that plane, for the vertical projection. 107. To assume a point on a surface thus represented, we first assume either projection, as the horizontal, and erect a perpendicular to the horizontal plane, as in Art. (76). Through this perpendicular pass a meridian plane; it will cut fiom the surface a meridian line, which will intersect the perpendicular in the required point or points. Construction. Let the surface be a prolate spheroid, Fig. 50, and let c be the horizontal, and c'd' the vertical projection of the axis, won the horizontal projection of its largest circle, and d'm'c'n' the vertical projection of the meridian curve parallel to the vertical plane, and let p be the assumed horizontal projection; p will be the horizontal, and s'p' the vertical projection of the perpendiculaI to HII: cp will be the horizontal trace of the auxiliary meridian plane. If this plane be now revolved about the axis until it becomes parallel to the vertical plane, p will describe the arc pr, and r will be the horizontal, and r'r" the vertical projection of the revolved position of the perpendicular. -The meridian curve, in its revolved position. will be vertically projected into its DEISCRIPTIVE GEOMETRY. 61 equal, d'mn c'n', Art. (61), and r' and r" will be the vertical projections of the two points of intersection in their revolved position. When the meridian plane is revolved to its primitive position, these points will describe the arcs of horizontal circles, projected on H in 9p, and on V in rtp' and r"p", and p' and p" will be the vertical projections of the required points of the surface. TANGENT PLANES AND TANGENT SURFACES. NORMAL LINES AND NORMAL PLANES. 108. A plane is tangent to a surface when it has at least one point in common with the su2:face, through which, if a1y irnter. sectiny plane be passed, the right line cut from the plane will be tangent to the line cut from the surface at the point. This point is the point of contact. It follows from this definition, that the tangent plane is the locus of, or place inl which are to be found, all rikcht lines tangent to lines of the sntlface at the point of contact; and since any two of these right lines are sufficient to determine the tangent plane, we have the folliowing general rule for passing a plane tangent to any surface at a given point: Draw anty two lines of the surfcace intersectiuy at the point. Tangent to each of these, at the same point, 1,-'ow a Tight line. The plane of these two tangents will be the required plane. 109. A right line is normal to a surface, at a point, when it is verpendicular to the tangent plane at that point. A plane is normal to a suiface, when it is perpendicular to the tangent plane at the point of contact. Since any plane passed through a normal line will be perpendicular to the tangent plane there may be an infinite number of normal planes to a surface at a point, while there can be but one normal line. 62 DESCRIPTIVE GEOMETRY. 110. The tangent plane at any point of a surface with recti linear elements, must contain the rectilinear elements that piss throzugh the point of contact. For the tangent to each rectilinear element is the element itself, Art. (64), and this tangent must lit in the tangent plane, Art. (108). 111. A plane which contains a rectilinear element andits conseclttive one, of a single curved surface, will be tangent to the surface at every point of this element, or all along the element. For if through any point of the element any intersecting plane be passed, it will intersect the consecutive element in a point consecutive with the first point. The right line joining these two points will lie in the given plane, and be tangent to the line cut from the surface, Art. (64). Hence the given plane will be tangent at the assumed point. This element is the element of contact. Conversely, if a plane be tangent to a single curved surface, it must, in general, contain two consecutive rectilinear elenmelt.s. For if through any point of the element contained in the tangent plane, Art. (110), we pass an intersecting plane, it will cut from the surface a line, and from the plane a right line, which will have two consecutive points in comnion, Art. (108). Thlrough the point consecutive with the assumed point, draw the consecutive element to the first element; it must lie in the plane of the second point and first element, Art. (70), that is, in the tangent plane. 112. It follows from these principles, that if a plane be tangen to a single curved surface, and the element of contact be interected by any other plane, the right line cut from the tangent plane will be tangent to the line cut from the surface. Itence, if tile base of the surface be in the horizontal plane, the horizontal truace of the tan..gent plane m-us't be tlanent to this base, at the point D)ESCRIPTIVE GEOMETRY. 6, in which the element of contact pierces the horizontal plane; and the same principle is true if the base lie in the vertical plane. 113. A plane tangent to a warped surface, although it contains the rectilinear element passing through the point of contact, cannot contain its consecutive element, and therefore can, in general, be tangent at no other point of the element. 114. If a plane. contain a rectilinear element of a warped surface, and be not parallel to the other elemenlts, it will be tanyent to the sur7face at some point of this element. For this plane will intersect each of the other rectilinear elements in a point; and these points being joined, will form a line which will intersect the given element. If at the point of intersection a tangent be drawn to this line, it will lie in the tangent plane, Art. (108). The given element being its own tangent, Art. (64), also lies in the tangent plane. The plane of these two tangents, that is, the intersectilng plane, is therefore tangent to the surface at this point, Art. (108). Thus, Fig. 51, if a is an element, b, c, &c., b', c', &c., the consecutive elements on each side of a, the plane through a will cut these elements in the curve cbab'c', which crosses the element a at the point a. The tangent mn at this point, and the element a, determine the tangent plane. It is thus seen, that, in general, a tangent plane to a warped surface is also an intersecting plane. If the intersecting plane be parallel to the rectilinear elements, there will be no curve of intersection formed as above, and the plane will not be tangent. 115. A plane tangent to a surface of revolution, is perpendicular to the meridian plane passing through the point of contact. For this tangent plane contains the tangent to the circle of b ~64 DESCRIPTIVE GEOME'TRY. the surface at this point, Art. (108), and this tangent is perpen. dicular to the radius of this circle, and also to a line drawn through tile point parallel to the axis, since it lies in a plane perpendicular to the axis; and therefore it is perpendicular to the plane of these two lines, which is the meridian plane. 116. Two curved surfaces are tangent to each other when they have at least one point in common, through which if any intersecting plane be passed, the lines cut from the surfaces will b(e tangent to each other at this point. This will evidently be the ease when they have a common tangent plane at this point. 117. If two single curved surfaces are tangent to each other, at a point of a common element, they will be tangent all along this element. For the common tangent plane will contain this element, and be tangent to each surface at every point of the element, Art. (111). This principle is not true with warped surfaces. 118. But if two warped surfaces, having two directrices, have a common plane directer, a common rectilinear eleLment, and two comnmon tangent planes, the points of contact being on Ut]e common element, they will be tangent all along this element. For if through each of the points of contact any intersecting plane be passed, it will intersect the surfaces in two lines, which wiill have, besides the given point of contact, a second consecutive point in common, Art. (108). If, now, the common element be moved pon the lines cut from either surface, as dilectrices, and parallel to the common plane directer, into its consecutive position, containing these second consecutive points, it will evidently lie in both surfaces, and the two surfaces will thus contain two consecutive rectilinear elements. If, now, any plane be passed, intersect DESCRIPTIVE GEOMETRY. 65 ing these elements, two lines will be cut fiom the surfaces, having two consecutive points in common, and therefore tangent to each other; hence the surfaces will be tangent all along the common element, Art. (116). 119. Also, if two warped surfaces, having three directrices, have o common e!emzent and three commron tangent planes, the points o/ contact hTei,,o!,n this element, they will be tangent all along this element. For if through each point of contact any intersecting plane be passed, it will intersect the surfaces in two lines, which, besides the given point of contact, will have a second consecutive point in common. If the common element be moved upon the three lines cut from either surface, as directrices, to its consecutive position, so as to contain the second consecutive points, it will evidently lie in both surfaces; hence the two surfaces contain two consecutive rectilinear elements, and will be tangent all along the common element. 120. It follows from the principle in Art. (114), that the projecting plane of every rectilinear element of a warped surface is tangent to the surface at a point. If through these points of tangency, projecting lines be drawn, they will form a cylinder tangent to the warped surface, which may be regarded as the projecting cylinder of the surface; and the traces of these planes, or the projections of the elements, will all be tangent to the base of this cylinder. This is seen in the horizontal projection of the hyperboloid of revolution of one nappe, Fig. 49; also in both projections of the hyperbolic paraboloid, Fig. 45. 121. If two surfaces of revolution, having a common axis, are tangent to each other, they will be tangent at every point of a circumference of a circle, perpendicular to the axis. For if 5 66 DESCRIPTIVE GEOMETRY. ti-rough the point of contact a meridian plane be passed, it will cut fiom the surfaces meridian lines, which will be tangent at this point, Art. (108). If these lines be revolved about the common axis, each will generate the surface to which it belongs, while the point of tatgency will generate a circumference common to the two surfaces, which is their line of contact. SOLUTION OF PROBLEMS RELATING TO TANGENT PLANES TO SINGLE CURVED SURFACES. 122. The solution of all these problems depends mainly upon the principles, that a plane tangent to a single curved surface is tangent all along a rectilinear element, Art. (111); and that if such surface and tangent plane be intersected by any plane, the lines of intersection will be tangent to each other. 123. PROBLEM 18. To pass a plane tangent to a cylinder at a given point on the surface. Let the cylinder be given as in Art. (75), Fig. 40, and let P be the point, assumed as in Art. (76). Analysis. Since the required plane must contain the rectilinear element through the given point, and its horizontal trace must be tangent to the base at the point where this element pierces the horizontal plane, Art. (112), we draw the element, and at the point where it intersects the base, a tangent; this will be the horizontal trace. The vertical trace must contain the point where this element pierces the vertical plane, and also the point where the horizontal trace intersects the ground line. A right line joining these two points will be the vertical trace. When this element does not pierce the vertical plane, within the limits of the drawing, we draw through any one of its points a line parallel to the horizontal trace; it will be a line of the required plane, and pierce the vertical plane in a point of the vertical trace. Constructions. Draw the element PQ; it pierces II at q. Al DESCRIPTIVE GEOMETRY. 67 this point draw qT, tangent to rmlu; it is the required horizontal trace. Through P draw PZ, parallel to qT; it pierces V at z', and z'T is the vertical trace. mmi'n' is the plane tangent to the surface along the element MN. 124. PROBLEM 19. To pass a plane therough a given point, without the sujh.ce, tangent to a cylinder. Let the cylinder be given as in the preceding problem. Analysis. Since the plane must contain a rectilinear element; if we draw a line through the given point, parallel to the rectilinear elements of the cylinder, it must lie in the tangent plane, and the point in which it pierces the horizontal plane, will be one point of the horizontal trace. If through this point we draw a tangent to the base, it will be the required horizontal trace. A line through the point of contact, parallel to the rectilinear elements, will be the element of contact; and the vertical trace may be constrlict(ed as in the preceding problem: Or a point of this trace may be obtained by finding the point in which the auxiliary line pierces the vertical plane. Two or more tangent planes may be passed, if two or more tangents can be drawn to the base, from the point in which the auxiliary line pierces the horizontal plane. Let the construction be made. in accordance with the above analysis. 125. PROBLEM 20. To pass a plane, tangent to a cylinder, and parallel to a given right line. Let the cylinder be given as in Fig. 52, and let RS be the given line..Analysis. Since the required plane must be parallel to the rectilinear elements of the cylinder, as well as to the given line; if a plane be passed through this line, parallel to a rectilinear element, it will be parallel to the required plane, and its traces parallel to the required traces, Art. (10). Hence, a tangent to 68 R1)1 SRIPT1V GE(< MV'IRY. the base, parallel to the horizontal trace of this auxiliary plane, will be the required horizontal trace. The elemlent of contact and vertical trace may be found as in the preceding problem. Construction. Through RS pass the plane sT'r', p)arallel to NMN, as in Art. (48). Tangent to mlo and parallel to sT', draw qTr; it is the horizontal trace of the required plane, and Tt' parallel to T'r' is the vertical trace. QP is the element of contact. The point z', determined as in the preceding problem, will aid in verifying the accuracy of the drawing. When more than one tangent can be drawn parallel to sT', there will be more than one solution to the problem. 126. PROBLEM 21. To pass a plane tacngent to a r'qcht cylinder with a circular base, having its axis parallel to the groundzc line at a qiven point on the suif(ace. Let cd, Fig 53, be the horizontal, and ef' the vertical projection of the circular base, and GK the axis; then lcdi will be the horizontal, and u'e'f'b' the vertical projection of the cylinder. Let p be the horizontal projection of the point. To determline its vertical projection, at p erect a perpendicular to H, Art. (76). Through this pass the plane tTt' perpendicular to AB. It intersects the cylinder in the circumference of a circle equal to the base. of which M is the centre. This circumference will intersect the perpendicular in two points of the surface. Revolve this plane about tT until it coincides with H; M falls at mn" Art. (17). With mn" as a centre and cy as a radius, describe the circle qsr; it will be the revolved position of the circle cut from the cylinder; q and r will be the revolved positions of the required points, and p' and p" their vertical projections. Let the plane be passed tangent at P. Analysis. Since the plane must contain a rectilinear elemllet, it will be parallel to the ground line, and its traces, therefore parallel to the ground line, Art. (9). If through the given point a plane be passed perpendicular to the axis, and a tangent be DESCRIPTIVE GEOMETRY. 69 drawn to its intersection with the cylinder, at the point, it will be a line of the required plane. If through the points in which this line pierces tile planes of projection, lines be drawn parallel to the ground line, they will be the required traces. Constr?.ction. Through P pass the plane tTt', and revolve it as above. At q draw qx tangent to qas; it is the revolved position of thle tangent. When the plane is revolved to its primitive position, this tangent pierces HI at x, and V at y', Ty' being equal to Ts'; and xz and y'w' are, the required traces. 127. PROBLE:M 22. To pass a plane tangen t to a right cylinder with a circular base, having its axis parallel to the ground line, through a given point without the surface. Let the cylinder be given as in Fig. 53, and let N be the given point. Analysis. Since the required plane must contain a rectilinear element, it must be parallel to the ground line. If through the given point a plane be passed perpendicular to the axis, it will cut from the cylinder a circumference equal to the base; and ii through the point a tangent be drawn to this circumference, it will be a line of the required plane, and the traces may be deter mined as in the preceding problemn. Since two tangents can be drawn, there may be two tangent planes. Let the construction be made in accordance with the anal)ysis. 128. PROBLEM 23. To pass a plane pctrallel to a given right line, and tangent to a right cylinder, with a circular buse, with its axis parallel to the ground line. Let the cylinder be given as in Fig. 54, and let MN be tl] given line. A4Ealysis. Since the plane must be parallel to the axis, ii through the given line we pass a plane parallel to the axis, it will be parallel to the required plane, and to the g.lollnnl line. 70 DESCRIPTIVE GEOMETRY. A pllane perpendicular to the ground line will cult fr'om tile cylinder a circumference; from the required tangent plane a tangent to this circumference, Art. (108); and from the parallel plane a right line parallel to the tangent. If, then, we construct the circumference, and draw to it a tangent parallel to the intersection of the parallel plane, this tangent will be a line of the requiled plane, from which the traces may be found as in the preceding problems. Construction. Through m and n' draw the lines mp and n'(', parallel to AB. They will be the traces of the parallel plane, Art. (48). Let tTt' be the plane perpendicular to AB. It cuts from the cylinder a circle whose centre is 0, and fronl the parallel plane a right line which pierces H at p and V at q', Art. (38). Revolve this plane about tT, until it coincides with H; xyz will be the revolved position of the circle, and sp that of the line cut from the parallel plane; zu tangent to xyz, and parallel to sp, will be the revolved position of a line of the required plane, which, in its true position, pierces H at u, and V at w', and uv and w'v' are the traces of the required plane. Since another parallel tangent can be drawn, there will be two solutions. 129. PROBLEM 24. To pass a plane tangent to a cone, througn a given point on the sufrace. Let the cone be given as in Fig. 41, and let P, assumed as in Art. (79), be the given point. Anaclysis. The required plane must contain the rectilinear element, passing through the giveir point, Art. (110). If, thenl, we draw this element, Art. (79), and at the point where it pierces the horizontal plane, draw a tangent to the base, it will be the horizental trace of the required plane, and points of the vertical trace may be found as in the similar case for the cylinder. Let the construction be made in accordance with the analvsis. DESCRIPTIVE GEO)METRY. 7 1 130. PROBLEM 25. Through a point without the surjfce of (A cone, to pass a plane tangent to the cone. Let the cone be given as in Fig. 55, and let P be the given p)Oillt. Analysis. Since the required plane must contain a rectilinear element, it will pass through the vertex; hence, if we join the given point with the vertex by a right line, it will be a line of the required plane, and pierce the horizontal plane in a point of the required horizontal trace, which may then be drawn tangent to the base. If the point of contact with the base be joined to the vertex by a right line, it will be the element of contact. The vertical trace may be found as in the preceding problems. If more than one tangent can be drawn to the base, there will be more than one solution. Construction. Join P with S, by the line PS; it pierces II at u. Draw uq tangent to mlo; it is the required horizontal trace. SQ is the element of contact which pierces V at w', and z'T is the vertical trace. ux will be the horizontal trace of a second tangent plane, through P. 131. PROBLEM 26. To pass a plane tangent to a cone, and parallel to a given right line. Let the cone be given as in Fig. 55, and let NR be the given right line. Analysis. If through the vertex we draw a line parallel to the given line, it must lie in the required plane, and pierce the horizontal plane in a point of the horizontal trace. Through this point draw a tangent to the base, it will be the required horizontal trace; and the element of contact and vertical trace may be found as in the preceding ploblem. WThen more than one tangent can be drawn to the base, there will be more than one solution. If the parallel line tllrough the vertex pierces the horizontal, 72 DECRIPTrVE GEOMETRY. plane within the base, no tangent can be drawn, and the problem is impossible. Let the construction be made in accordance with the analvsis. 132. PROBLEM 27. Topass a plane tangent to a single curved surface with a helical directrix, at a given point. Let the surface be given as in Fig. 42, and let 1I be tile given point, Art. (81). An.alysis. The tangent plane must, in general, be tangent all along the rectilinear element, through thle point of contact. Its horizontal trace must therefore be tangent to the base, Art. (112). If tlliough the point where this element pierces the horizontal plane, a tangent be drawn to the base, it will be the required horizontal trace, and the vertical trace may be determined as ill the case of the cylinder or cone. Constrauction. The element RX pierces H at Z. At this point draw the tangent tT; it is the horizontal trace. Tt' is the vertical trace. To pass a plane through a given point without this sut:fce, tangent to it, we pass a plane through the point parallel to the base, and draw a tangent to the curve of intersection, Art. (81), through the point. This tangent, with the element of the surface through its point of contact, will determine the tangent plane. 133. PROBLEM 28. To pass a plane tangent to a single curvewd surface with a helical directrix, and parallel to a given rig/ht line. Let the surface be given as in Fig. 42, and let MN be the given line. Analysis. If with any point of the right line, as a vertex, we c(nstruct a cone, whose elements make the same angle with the Ihorizonltal plane as the elements of the surface; and pass a plane through the line tangent to this cone, it will be parallel to the required plane. The traces of the required plane may then be constructed as in Art. (125). DESCRIPTIVE GEOMETRY. 73 Construction?. Take n' as the vertex of thle auxiliary cone, and draw n'o", making with AB an angle equal to o's'T; o"'cs" will be the base of the cone in II. Through sm draw mc tangent to o"cs". It will be the horizontal trace of the parallel plane, and t"T', parallel to it, and tangent to u2vw, is the required horizontal trace. Let the pupil construct the vertical trace. 134. It is a remarkable property of this surface, with a helical directrix, that any plane passing through a rectilirear elemrent, is tangen-t to the suft'ace, at the point where the element intersects the directrix. For this plane will intersect the other elements, thus forming a curve, Art. (114). This curve will intersect the given element at the point where the element touches the directrix. The tangent to the curve at this point, and tile element, both lie in the given plane; it is therefore tangent to the surface at the point, Art. (114). This plane does not, in general, contain the consecutive element, and is therefore not necessarily tangent all along the element. The projecting planes of all the elements are tangent to the surface, and the cylinder formed by the projecting lines of the points of contact is therefore tangent to the surface, Art. (120). Its base is the circle plxq. It should be observed, also, that at any point on the helical directrix, an infinite number of planes can be passed tangent to the surface, as at the vertex of a cone. Only one of these planes will contain two consecutive rectilinear elements. 135. By an examination of the preceding problems, it will be seen that, with two remarkable exceptions, only one tangent plane can be drawn to a single clurved surface at a given point. That the number which can be drawn through a given point without the surface, and tangent along an element, will be limited. That the number which can be drawn parallel to a given right line, and tangent along an element, is also limited. 74 DESCRIPTIVE GEOMblTRY. That, in general, a plane cannot be passed through a given right line and tangent to a single curved surface. If, however, the given line lies on the convex side of the surface, and is parallel to the rectilinear elements of a cylinder, or passes through the vertex of a cone, or is tangent to a line of the surface, the problem is possible. PROBLEMS RELATING TO TANGENT PLANES TO WARPED SURFACES. 136. Since a tangent plane to a warped surface must contain the rectilinear element passing through the point of contact, Art. (110), we can at once determine one line of the tangent plane. A second line may then be determined, in accordance with the rule in Art. (108). When the surface has two different generations by right lines, the plane of the two rectilinear elements passing through the given point will be the required plane. 137. PROBLEM 29. To pass a plane tangent to a hyperbolic paraboloid, at a given point of the surface. Let MN an(] PQ, Fig..56, be the directrices, and MP and NQ any two elements of the surface, and let 0, assumed as in Art. (88), be the given point. Anrtlysis. Siulce through the given point a rectilinear element of each generation can be drawn, Art. (89), we have simply to construct these two elements, and pass a plane through them, Art. (136). Cornstruction,. Throiugh O draw OE and OF, parallel respectlvely to NQ and MP. These will determine a plane parallel to the plane directer of the first generation, Art. (86). This plane cuts the directrix PQ in the point U, Art. (41). Join U with O and we have an element of the first generation, Art. (83). Let NQ and Ml' be taken as directrices of the second generation DESCRIPTIVIE GEOMETRY. 75 Through O draw OC and OD parallel respectively to MN and PQ. They will determine a plane parallel to the plane directei of tile second generation, Art. (86). This plane cuts MP in \7, and OW will be an element of the second generation, and the tangent plane is determined, as in Art. (32). 138. PROBLEM 30. To pass a plane tangent to an hlyperboloid of revolutiov of one nappe, at a given point of the suofiace. Let the surface be given as in Fig. 57, and let O be the given point. Analysis. Same as in the preceding problem. Constrtectioln. OZ is the element of the first generation passing through O, Art. (100). It pierces H at r. Draw OS, Art. (101). It is an element of the second generation passing through 0. This element pierces 1H at w, and wr is the horizontal trace of the required plane, and Tt' is its vertical trace. Since the meridian plane through O must be perpendicular to the tangent plane, Art. (115), its trace cx must be perpendicular to wr. 139. PROBLEM 31. To pass a plane tangent to a helicoid. at a point of the surface. Let the surface be given as in Fig. 48, and let M be the given point. Ancalysis. The tangent plane must contain the rectilinear el ement passing through the given point, and also the tangent to the helix at this point, Art. (108). The plane of these two lines will then be the required plane. Construction. MX is the rectilinear element through M. I. pierces H at u; cmd is the horizontal projection of the helix through AM. Draw the tangert to this helix at M, as in Art, (69). mnz will be its horizontal pr(jection; Z the point in which it pierces the holiz,,ntal plane thlrough C, and mn'z' its vertical 76. nDESCRIPTIVE GEO( METR Y. projection. This tangent pierces H at k; hence kt is the horn zontal trace of the required plane, and t'T is the vertical trace. Since a tangent plane to the surface contains a rectilinear eleinent, it is evident that it cannot make a less angle swith the lhorizontal plane than the elements; nor a greater angle than 90~ the angle made by the projecting plane of any rectilincar elemeInt, which is tangent at the point where the element intersects the axis. 140. PROBLEM 32. To pass a plane tangent to a helicoid, and perpendicular to a given right line. -Anaclysis. If, with any point of the axis as a vertex, we construct a cone whose rectilinear elements shall make with the horizontal plane the same angle as that made by thie rectilinear elements of the given surface, and through the vertex of this cone pass a plane perpendicular to the given line, Art. (46), it will, if the problem be possible, cut from the cone two elements, each of which will be parallel to a rectilinear element of the helicoid, and have the same horizontal projection. If through either of these elements a plane be passed parallel to the auxiliary plane, it will be tatngent to the surface, Art. (114), and perpendicular to the given line. Let the problem be constructed in accordance with the analysis. 141. An auxiliary surface may sometimes be used to advantage in passing a plane tangent to a warped surface at a given point. Thus, let MN and PQ, Fig. 58, be the two directrices of a warped surface having V for its plane directer, and let 0 be the given point. At the points X and Y, in which the rectilinear element through 0 intersects the directrices, (lraw a tangent-to each directrix, as XZ and YU. On these tangents as directrices, move XY parallel to V. It will generate an hyperbolic paraboloid, Art. (86), having, with the given surface, the com1mon) element XY, I)ESCRIPTIVE GEOMETRY. 77 and a common tangent plane at each of the points X and Y, since the plane of the two lines XY and XZ, and also that of XY atnd YTJ, is tangent to both surfaces, Art. (108). The two surfaces are therefore tangent all along the common element, XY, Art. (] 18) If, then, at O, we pass a plane tangent to the hyperbolic paraboloid, it will be also tangent to the given surface at the same point. If the given surface have three curvilinear directrices, a tangent may be drawn to each, at the point in which the rectilinear element through the given point intersects it; and then this element may be moved on these three tangents as directrices, generating a hyperboloid of one nappe, Art. (91), which will bot tangent to the given surface all along a common element, Art. (119). A plane tangent to this auxiliary surface at the given point, will also be tangent to the given surface. 142. An infinite number of planes may, in general, be passed through a point without a warped surface, and tangent to it. For if, through the point, a system of planes be passed intersecting the surface, tangents may be drawn fiom the point to the curves of intersection, anld these will form the surface of a cone tangent to, the warped surface. Any plane tangent to this cone will be tangent to the warped surface, and pass through the point. Also, an infinite number of planes may. in general, be passed tangent to a warped surface, and parallel to a right line. For i, tile surface be intersected by a system of planes pzarallel to the line, and tangents parallel to the line be drawn to the sections, they will form the surface of a cylinder, tangent to the warped surface. Any plane tangent to this cylinder will be tangent to the warped surface, and parallel to the given line. 143. To pass a plane through a given right line, and tangent to a woar2ped su'face, it is only necessary to produce the line until 78 LD)SCRIPTIVE GEOMETRY. it pierces the surface, and through the point thus determined draw the rectilinear element of the surface. This, with the given line, will determine a plane tangent to the surface at some point of the element, Art. (114). If tllere be two rectilinear elements passing through this point, each will give a tangent plane; and the number of tangent planes will depend upon the number of points in which the line pierces tihe surface. If the given line be parallel to a rectilinear element, it pierces the surface at an infinite distance, and the tangent plane will be determined by the two parallel lines. PROBLEMS RELATING TO TANGENT PLANES TO DOUBLE CURVED SURFACES. 144. These problems are, in general, solved either by a direct application of the rule in Art. (108), taking care to intersect the surface by planes, so as to obtain the two simplest curves of the surface intersecting at the point of contact, or by means of Inore simple auxiliary surfaces tangent to the given surface. 145. PROBLEM 33. To pass a plane tangent to a sphere, at a Oiven point. Let C, Fig. 59, be the centre of the sphere; prq its horizontal, and p'.'q' its vertical projection. Let M be the point assumed, as in Art. (107). Analysis. Since the radius of the sphere, drawn to the point of contact, is perpendicular to the tangent to any great circle a this point, and since these tangents all lie in the tangent plane, Art. (108), this?(radius must be perpendicularc to the tangent plune. WVe have then simply to pass a plane perpendicular to this radius at the given point, and it will be the required plane. (Jonstr'uction. This may be made directly, as in Art. (46), o DESCRIPTIVE GEOMETRY. 7T9 otherwise, thus: Through M and C pass a plane perpendicular to II; cm will be its horizontal trace. Revolve this plane about cmn as an axis, until it co)incides with H; c"m" will be the revolved position of the radius, Art. (28). At nm" draw sm"t perpendicular to c"m". It is the revolved position of a line of the required plane. It pierces IH at I, and since the horizontal trace must be perpendicular to cm, Art. (43). tT is the required horizontal trace. Through M draw MN parallel to tT. It pierces V at n'; and Tn', perpendicular to c'rm', is the vertical tiace. 146. PROBLEM 34. To pass a plane tangent to an ellipsoid oJ revolution, at a given point. Let the surface be given as in Art. (107), Fig. 50, arid let P be the point. Analysis. If, at the given point, we draw a tangent to the meridian curve of the surface, and a second tangent to the circle of the surface at this point, the plane of these two lines will be the required plane, Art. (108). Construction. Through P pass a meridian plane; cp will be its horizontal trace. Revolve this plane about the axis of the surface until it is parallel to V. R will be the revolved position of the point of contact. At r' draw r'x' tangent to c'n'd'ni'. It will be the vertical projection of the revolved position of a tangent to the meridian curve at R. When the plane is revolved to its primitive position, the point, of which y' is the vertical projection, remains fixed, and yIp' will be the vertical projection of the tan gent, and cp its horizontal projection. It pierces H at x, one point of the hori ontal trace of the required plane. pr is the horizontal, and p'r' the vertical projection of an arc of the circle of the surface containing P. At p draw pu perpendicular to cqp. It will be the horizontal projection of the tangent to the circle at P, and p'u' is its vertical projection This pierces V at u', a point of the vertical trace. Through x draw xT, parallel to pu and through T draw Tu'; xTu' will be the required plane. 80 DESCRIPTIVE GEOMETRY. 147. Second method for the same problem. Analysis. If the tangent to the meridian curve at P revolve about the axis of the surface, it will generate a right cone tangent to the surface in a circumference containing -the given point, Art. (121). If, at this point, a tangent plane be drawn to the coiie, it will be tangent also to the surface. Construction. Draw the tangent at P, as in the preceding article. It pierces H at x, and this point, during the revolution, describes the base of the cone whose vertex is at (cy'). The element of the cone through P pierces H at x, and xTt' is the required plane. 148. Bv the same methods a tangent plane may be passed to any surface of revolution at a given point. Since the tangent plane and horizontal plane are both perpendicular to the meridian plane through the point of contact, their intersection, which is the horizontal trace, will be perpendicular to the meridian plane and to its horizontal trace. While -at a -given point on a -double curved surface only, one tangent plane can be passed, it may be proved as in Art. (142), that from a point without the surface, an infinite number of such planes can be passed. 149. PROBLEM 35. To pass a plane' through a girz e right line,7id tangent to a s]phere. Let the ground line pass through the centre of the sphere, and let C, Fig. 60, be the centre, and def the circle, cut fiom the sphere by the horizontal plane, and let MN be the given line. Analysis. If-we take any point of the given line as the veiter of a right cone, tangent to the sphere, and pass a plane thronti the line tangent to this cone, it will be tangent also to the spheru Construction. Take mn as the vertex of the auxiliary cone. and draw the tangents md and me, and the right line mC. The tan gents will be the two rectilinear elements of th}e cone in the hori DESCRIPTIVE GEOMETRY. 81 zontal plane, and mC will be its axis. Since the line of contact of the two surfaces is a circunmference, whose plane is perpendicnlar to mC, Art. (121), this plane will be perpendicular to H, and de will be its horizontal trace. This circumference may be regarded as the base of the cone; and, if we find the point in which its plane is pierced by MN, and draw fiom this point a tangent to the base, it will be a line of the required plane, Art. (112). O is this point, Art. (15). If the plane of the base be revolved abollt de as a-l axis, until it coincides with H, 0C) will fall at o", o"o'being equal to o'q, Art. (17). The circle of contact will take the position dp"e, de being its diameter. Draw o"ip"; it will be the revolved position of the tangent. It pierces H at t, one point of the horizontal trace; MN, pierces H at m; hence, m is the required horizontal trace. MN pierces V at n', a point of the vertical trace. A second point, t', may be determined as in Art. (123), and n't' is the vertical trace. When the plane of the circle of contact is in its true position, p" is horizontally projected at p, and vertically at p', p'r being equal to pp"; hence, CP will be the radius passing through P. Since the tangent plane must'be perpendicular to this radius, Cp and Cp' must be respectively perpendicular to tm and t'n'. Since a second tangent can be drawn from O to the base of the cone, another tangent plane may be constructed. 150. Second method for the same problem. Let the sphere and the right line MN be given as in Fig. 61..Analysis. If any two points of the given line be taken, each as the vertex of a cone tangent to the sphere, each cone will be tangent in the circumference of a circle, and the planes of these circles will intersect in a right line, which will pierce the surface of the sphere at the intersection of the circumferences, anal these points will be common to the three surfaces. If, through either of these points and the given line, we pass a plane, it will be tangent to both cones and to the sphere. 6 82 DESCRIPTIVE GEOME'TIRY. C'onstruction. Take the.points m and n' as the vertices of the two auxiliary cones; de is the horizontal projection of the circle of contact of the first cone with the sphere, and fg is the vertical projection of the circle of contact of the second cone and sphere, Art. (15). The planes of these two circles intersect in a right line, of which de is the horizontal, and fg the vertical projection, Art. (15); and this line pierces H at o. Now revolve the circle of which de is the diameter, about de as an axis, until it coincides with H. It takes the position dsex. Any point of the line (de, fg), as R, falls at r", and or" will be the revolved position or the line of intersection of.the two planes, and p" and q" will be the revolved positions of the two points in which it pierces the surface of the sphere. After the counter-revolution, these points are horizontally, projected at p and q, and vertically at p and q'. CI' is the radius of the sphere at P. The traces tT, and t'n', of the plane tangent at P, may now be drawn as in the preceding article; or by drawing rmet perpendicular to C(p, and n't' perpendicular to Cp', Art. (43). A second tangent plane at Q may be determined in the same way. 151. Third method for the same problem. Let C, Fig. 62, be the centre, and def the horizontal, and d'hy' the vertical projection of the sphere, and let MN be the given line. Analysis. Conceive the sphere to be circumscribed by a cylinder of revolution, whose axis is parallel to the given line. The line of contact will be the circumference of a great circle perpendicular to the axis and given line, Art. (121). A plane through the right line tangent to this cylinder will be tangent also to the sphere. The plane of the circle of contact will intersect the given line in a point, and the required tangent plane in a right line drawn from this point tangent to the circle. The plane of.his tangent and the given line will be the required plane. Without constructing the cylinder, we have then simply to pass a DESCRIPTIVE GEOMETRY. 83 plane through the centre of the sphere perpendicuiar to the given line, and from the point in which it intersects the line, to draw a tangent to the circle cut from the sphere by the same plane, and pass a plane through this tangent and the given line. CoJzstructiont. Through C draw the two lines CP and CQ, as in Art. (46). The plane of these two lines is perpendicular to MN, and intersects it inll O, Art. (41). The horizontal trace of this plane may now be determined as in Art. (46), and the plane revolved about this trace until it coincides with H, and the revolved position of the point and circle be found, and the tangent drawn. Otherwise thus: Revolve this plane about CP, until it becomes parallel to H. The point 0, in its revolved position, will be horizontally projected at r, and the circle will be horizontally projected in def, Art. (62). From r draw the two tangents ik and'l. They will be the horizontal projections of the revolved positions of the two tangents to the great circle, cut from the sphere by the perplendicular plane, and k and I will be the horizontal projections of the revolved positions of the two points of contact of the required tangent plane.. After the counter-revolution, R will be horizontally projected at o; and since X remains fixed, oy will be the horizontal, and o'y' the vertical projection of the first. tangent, and Y its point of contact. Since the second tangent does not intersect PC within the limits of the drawing, draw the auxiliary line g.p, and find the true horizontal projection of the point which in revolved position is horizontally projected in g, as in Art. (37). It will be at g", and og" will be the horizontal projection of the second tangent, and z the horizontal projection of the second point of contact. S is the point in which this tangent intersects CQ, o's' is its vertical projection, and Z is the second point of contact. A plane through MN, and each of these points will be tangent to the sphere. 8I- W)'SCRIPTIVE GEOMETRY. 152. PROBL,EM 36. To pass a plane through a gsven right line and tangent to any surface of revolution. Let the horizontal plane be taken perpendicular to the axis, of which c, Fig. 63, is tile horizontal, and c'd' the vertical projection. Let poq be the intersection of the surface by the horizontal plane, p'd'q' the vertical projection of the meridian curve parallel to the vertical plane, and MN the given line. Analysis. If this line revolve about the axis of the surface, it xill generate a hyperboloid of revolution of one nappe, Art. (99), having the same axis as the given surface. If we now conceive the plane to be passed tangent to the surface, it will also be tangent to the hyperboloid at a point of the given right line, Art. (114); and since the meridian plane through the point of contact on each surface must be perpendicular to the common tangent plane, Art. (115), these meridian planes lmust forln one and the same plane. This plane will cut fiom the given surface a meridian curve, from the hyperboloid an hyperbo'a, Art. (104), and from the tangent plane a right line tangent to these curves at the required points of contact, Art. (108). The plane of this tangent and the given line will therefore be the required plane. Construction. Construct the hyperboloid as in Art. (99). yzx will be the horizontal, and x'y'x" the vertical projection of one branch of the mneridian curve parallel to V. If the meridian plane through the required points of contact be revolved about the commnon axis until it becomes parallel to V, lhe corresponding meridian curves will be projected, one into the curve p'd'q', and the other into the hyperbola x'y'x". Tangent to these curves draw x'r'; X will be the revolved position of the point of contact on the hyperbola, and R that on the meridian curve of the given surface. When the meridian plane is revolved to its true position, X will be horizontally projected in mn at s; sce will be the horizontal trace of the meridian plane, and u will be thec horizontal, and a' the vertical projection of the point of contact on the given surface. A plane through this point and MN will be the tangent plane. DESCRIPTIVE GE(-METRY. 8; 153. In general, through a given right line a limited number Df planes only can be passed tangent to a double curved surface For let the surface be intersected by a system of planes parallel to the given line, and tangents be drawn to the sections also parallel to the line. These will form a cylinder tangert to the sulrface. Any plane through the right line tangent to this cylin der will be tangent to the surface; and the number of tangent planes will be deterinined by the numnber of tangents which can be drawn fiom a point of the given line to a section made by a plane through this point. POINTS IN WHICH SURFACES ARE PIERCED BY LINES. 154. The points in which a right line pierces a surface are easily found by passing through the line any plane intersecting the surface. It will cut firom the surface a line, which will intersect the given line in the required points. This auxiliary plane shouldl be so chosen as to intersect the surface in the simplest line possible. Itf the given surface be a cylinder, a plane through the right line parallel to the rectilinear elements should be used. It will intersect the cylinder in one or more rectilinear elements. which will intersect the given line in the required points. If the given surface be a cone, the auxiliary plane should pass through the vertex. If a sphere, the auxiliary plane should pass through the centre, as' the line of intersection will be a circumference with a radius equal to that of the sphere. 155. If the given line be a curve of single curvature, its plane will intersect the surface, if at all, in a line which will contain the required points. If the plane does not intersect the surface, or if the line of intersection does not intersect, or is not tangent to the 86 DESCRIPTIVE GEOMETRY. given curve, there will, of course, be no points common to the curve and surface. If the given line be a curve of double curvature, a cylinder or cone may be passed through it intersecting the given surface. If tilhe line of intersection intersects the given line, the points thus determined will be the required points. These problems will be easily solved when the more general problem of finding the intersection of* surfaces has been discussed. INTERSECTION OF SURFACES BY PLANES. DEVELOPMENT OF SINGLE CURVED SURFACES. 156. The solution of the problem of the intersection of surfaces consists in finding two lines, one on each surface, which intersect. The points of intersection will be points in both surfaces, and therefore points of their line of intersection. 157. To find the intersection of a plane with a surJace, we intersect the plane and surface by a system qf auxiliary planes. Each plane will cut from the given plane a right line, and firom the surface a line, the inTtersection ofJ which will be points of the required line. The system of auxiliary planes should be so chosen as to cut from the surface the simplest lines; a rectilinear element, if possible, or the circumference of a circle, &c. The curve of intersection may be drawn with greater accuracy by determining at each of the points thus found a tangent to the curve, and then drawing the projections of the curve tangent to the projections of these tangents, Art. (65). This tangent must lie in the intersecting plane, that is, in the plane of the curve, Art. (64). It must also lie in the tangent plane to the surface at the given point, Art. (108). Hence, if we pass a plane tangent to the giver surface at the given point, and DESCRIPTIVE GEOMETRY. 87 determine its intersection with the intersecting plane; this will be the required tangent. 168. Since the tangent plane to a single curved surface in general contains two consecutive rectilinear elements, Art. (111) it will contain the elementary portion of the surface generated by the generatrix in moving from the first element to the second. Now if the surface be rolled over until the consecutive element following the second comes into the tangent plane, the portion of the plane limited by the first and third elements will equal the portion of the surface limited by the same elements. If we con tinue to roll the surface on the tangent plane until any following element comes into it, the portion of the plane included between this element and the first will be equal to the portion of the surface limited by the same elements. Therefore if a single curved surface be rolled over on any one of its tangent planes until each of its rectilinear elements has come into this plane, the portion of the plane thus touched by the suq.face, and limited by the extreme elements, will be a plane surface equal to the given surface, and is the development of the surface. As a tangent plane to a warped surface cannot contain two consecutive rectilinear elements, Art. (113), the elementary surface limited by these two elements cannot be brought into a plane without breaking the continuity of the surface..A warped surface, therefore, cannot be developed. Neither can a double curved surface be developed; as any elementary portion of the surface will be limited by two curves, and cannot be brought into a plane without breaking the continuity of the surface. 159. In order to determine the position of the different rectilinear elements of a single curved surface, as they come into the tangent plane, or plane of development, it will always be necessary D DFSCRIPTIVE GEOMETRIY. to find some curve upon the surface which will develop into a right line, or circle, or some simple known curve, upon which the rectified distances between these elements can be laid off. 160. PROBLEM 37. To find the intersection of a right cylinder with a circular base by a plane. Let nblo, Fig. (64). be the base of the cylinder, and c the horizontal, and c'd' the vertical projection of the axis; then rmlo will be the hor'izontal, and s'n"o"tu' the vertical projection of the cylinder, Art. (75). Let tTt' be the intersecting plane. Anvalysis. Intersect the cylinder and pa,ine by a system of auxiliary planes parallel to the axis, and also to the horizontal trace of the given plane. These will cut from the cylinder recti-. linear elements, and from the plane right lines parallel to the horiztmtal trace. The intersection of these lines will be points of the required curve, Art. (157). This curve, as also the intersection of any cylinder or cone with a circular or an elliptical base, by a plane cutting all the rectilinear elements, is an ellipse. Analyt. Geo., Arts. (82 and 200). Construction. Draw xy parallel to tT, it will be the horizontal trace of one of the auxiliary planes; yy' is its vertical trace. It intersects the cylinder in two elements, one of which pierces H at x, and tile other at z, and s"x' and z"z' are their vertical projections. It intersects the plane in the right line XY, Art. (38), and X and Z, the intersections of this line with the two elements, are points of the curve. In the same way any number of points may be fobund. If a point of the curve on any particular element be required, we have simply to pass an auxiliary plane through this element Thus, to construct the point on (o, o"u'), draw the trace ow, and construct as above the point 0. Since this point lies on the curve, and also on the extreme element, and since no point of the curve can be vertically pro DESCRIPTIVE GEOME'IRY. jected outside of o"u', the vertical projectivn of the cur e must be tangent to o"u' at o'. Or, the reason may be given thus, in many like cases: If a tangent be drawn to the curve at O, it will lie in the tangent plane to the surface at 0C, Art. (108); and since this tangent plane is perpendicular to the vertical plane, the tallgent will be vertically projected into its trace o"u', which must. therefore be tangent to mn'o'l' at o', Art. (65.). Also, qm'"s' must be tangent to m'o'l' at mn'. Ift a plane be passed through the axis perpendicular to the intersecting plane, it will evidently cut finom tile plaine a right line, which will bisect all the chords of the curve perpendicular to it, and this line will be the transverse axis of the ellipse. Ic is the horizontal trace of such a plane. It cuts the given plane in KG, Art. (38), and the cylinder in two elements hoiizontally projected at k and 1, and KL is the transvelse axis, and K and L the ver-< tices of the ellipse. K is the lowest, and L the highest point of the curve. Since the curve lies on the surface of the cylinder, its horizontal projection will be in the base mlo. To draw a tangent to the curve at any point, as X; draw xr tangent to xol at x; it is the horizontal trace of a tangent plane to the cylinder at X, Art. (123). This plane inltersects tTt' in a, right line, which pierces H at r; and since X is also a point oL. the intersection, RX will be the required tangent, Art. (157). II a tangent be drawn at each of the points determined as above, the projections of the curve can be drawn, with great accuracy, tangent to the projections of these tangents, at the projections of the points of tangency. The part of the curve MXO, between the points M and O, lies in front of the extreme elements (m, m"s') and (o, o"u'), and is seen, and therefore m'x'o' is drawn full, and m'l'o' broken. To represent the curve in its true dimensions let its plane be revolved about tT until it coincides with H. The revolved position of each point, as X, at x"', will be determined as in Art. (17) k"' and L"' will be the revolved position of the vertices, rx"' will 190 DESCRIPTIVE GEOMETRY. be the revolved position of the tangent, and l"'x"'k"' the ellipse in its true size. 161. PROBLEM 38. To develop a right cylinder with a circular base, and trace upon the development the curve of intersection of th cylinder by an oblique plane. Let the cylinder and curve be given as in the preceding probtem, and let the plane of development be the tangent plane at L. Analysis. Since the plane of the base is perpendicular to the. element of contact of the tangent plane, it is evident that as the cylinder is rolled out on this plane, this base will develop into a right line, on which we can lay off the rectified distances between the several elements, and then draw them each parallel to the element of contact. Points of the development of the curve are found by laying off on the development of each element, from the point where it meets the rectified base, a distance equal to the distance of the point from the base. Construction. The plane of development being coincident with the plane of the paper, let lL, Fig. 65, be the element of contact. Draw 11 perpendicular to IL, and lay off 11 equal to the rectified circumference Iw - - - - 1, Fig. 64. It will be the development of the base. Lay off lw equal to the arc Iw, and draw wW parallel to IL. It is the development of the element which pierces H at w. Likewise for each of the elements lay off wz, zm, &c., equal respectively to the rectified arcs we, zm, &c., in Fig. 64, and draw zZ, maM, &c. The portion of the plane included between lL and IL will be the development of the cylinder. On lL lay off IL = I"',L will be one point of the developed curve; on'wW lay off wW = w"w', W will be a second point; and thus each point may be determined, and L - - K - - L will be the developed curve. The line rx, the sub-tangent, will take the position rx on the developed base, and Xr will be the tangent at X in the plane of development. This must be tangent at X, DESCRIPTIVE GEOMETRY. 91 since the tangent -after development must contain the same two consecutive points which it contains in space, and therefore be tangent to the development of the curve, Art. (64). 162. PROBLEM 39. To find the intersection of. an oblique cylinder by a plane. Let the cylinder be given as in Fig. 66, and let tTt', perpendicular to the rectilinear elements, be the intersecting plane. Anazlysis. Intersect the cylinder and plane by a system of auxiliary planes parallel to the rectilinear elements, and perpendicular to the horizontal plane. These planes will each intersect the cylinder in two rectilinear elements, and the plane in a right line, the intersection of which will be points of the curve Construtction. Let eq, parallel to li, be the horizontal trace ol an auxiliary plane; qq' will be its vertical trace. It intersects the cylinder in two elements, one of which pierces H at r, the other at s, and these are vertically projected in r'g' and s'h', and horizontally in eq. It intersects tTt' in a right line, which pierces II at e, and is vertically projected in e'q', and horizontally in eq. These lines intersect in Z and Y, Art. (23), points of the curve. In the same way any number of points may be determined. The auxiliary planes, being parallel, must intersect tTt' in parallel lines, the vertical projections of which will be parallel to e'q'. By the plane, whose horizontal trace is mn, the points U and X are determined. The vertical projection of the curve must be tangent to m'n' at u'. The points in which the horizontal projection is tangent to ii and kf, are determined by using these lines as the traces of auxiliary planes. To draw a tangent to the curve at any point, as X, pass a plane tangent to the cylinder at X; av is its horizontal trace, and vx the horizontal, and v'x' the vertical projection of the tangent. A. sufficient number of points and tangents being thus determined, the projections of the curve can be drawn with accuracy 92 aoDESCRIP'rIIVE GEO()METRY. The part cyd is full, being the horizontal projection of that part of the curve which lies above the extreme elements LI and KF. For a similar reason, u'x'w' is fuIll. To show the curve in its true dimensions, revolve the plane about tT until it coincides with H. The revolved position of each point may be found as in Art. (17), and c"y"d"z" will be tile curve in its true size. The section thus made is a right section. 163. If it be required to develop the cylinder on a tangent plane along any element, as KF, we first make a right section as above. We know this will develop into a right line perpendicular to KF. On this we lay off the rectified arcs of the section included between the several elements, and then draw these elements parallel to KF. The developed base, or any carve on the suiface, may be traced on the plane of development by laying off on each element, from the developed position of the point where it intersects the right section, the distance from this point to the point where the element intersects the base or curve. A line through the extremities of these distances will be the required development. 164. PROBLEM 40. To find the intersection of a right cone with a circular base by a plane. Let the cone be given as in Fig. 67, and let tTt' be the given plane, the vertical plane being assumed perpendicular to it. Analysis. Intersect the cone by a system of planes through the vertex and perpendicular to the vertical plane. The elements cut from the cone by each plane, Art. (157), will intersect the right line cut from the given plane in points of the required curve. Construction.. Let lk be the horizontal trace of an auxiliar} plane, ks' will be its vertical trace. It intersects the cone in two DIESCRIPTIVE GEOMETRY. 93 elements, one of which pierces I in I, and the other in i, horizontally projected in Is and is respectively. It intersects the given plane in a right line perpendicular to V, vertically projected at x', and horizontally in xv; hence x and v are the horizonta_ projections of two points of the curve, both vertically projected at r'. In the same way any number of points can be found, as Y (wy'), &c. The plane whose horizontal trace is mo, perpendicular to tTt', intersects it in a right line vertically projected in Tt', which evidently bisects all chords of the curve perpendicular to it, and is therefore an axis of the curve. This plane cuts from the cone the elements SM and SO, which are intersected by the axis in the points Z and U, which are the vertices. To draw a tangent to the curlve at any point, as X, pass a plane tangent to the cone at X, ir is its horizontal trace. It intersects tTt' in (rx, Tx'), which is therefore the required tangent, Art. (157). The horizontal projection, uxzv, can now be drawn. u'd' is its vertical projection. To represent the curve in its true dimensions, we may revolve it about tT until it coincides with H, or about Tt' until it coincides with V, and determine it as in Art. (17). Otherwise, thus: Revolve it about UZ until its plane becomes parallel to V, it will then be vertically projected in its true dimensions, Art. (62). The points U and Z being in the axis, will be projected at tu' and z' respectlve:y. X will be vertically projected at x", x'x" being equal to px. Y at y", y'y" being equal to qy. (wy') at w", &c., and u'x"z'x'! will be the curve in its true size. 165. If a right cone, with a circular base, be intersected by a plane, as in Fig. 67, making a less angle with the plane of the base than the elements do, the curve of intersection is an ellipse, An(ilyt. Geo, Art. (82.) If it make the same angle, or is parallel to one of the elements, the curve is a parabola. If it make a P94 DESCRIPTIVE GlOMETRY. greater an.gle, the curve is an hyperbola. Hence these three curves are known by the general name, conic sections. 166. PROBLEM 41. To develop a right cone with a circular 6ase. Let the cone and its intersection by an oblique plane be given as in the preceding problem, Fig. 67, and let the plane of development be the tangent plane along the element MS; the half of the cone in front being rolled to the left, and the other half to tho right. analysis. Since the base of the cone is everywhere equally distant from the vertex, as the cone is rolled out, each point of this base will be in the circumference of a circle described with the vertex as a centre, and a radius equal to the distance from the vertex to' any point of the base. By laying off on this Circumference the rectified are of the base, contained between any two elements, and drawing rigrtt lines from the extremities to the vertex, we have, in the plane of development, the position of these elements. Laying off on the proper elements the distances fronr the vertex to the different points of the curve of intersection, and tracing a curve through the extremities, we have the development of the curve of intersection. Constmruction. With SM = s'm', Figs. 67 and 68, describe the arc OMO. It is the development of the base. Lay off MG - mzy, and draw SG; it is the developed position of the element SG. In the same way lay off GL = gl, LO = lo, and draw SL and SO. OSM is the development of one-half the cone. In like manner the other half may be developed. On SM lay off SZ _s'z'. Z will be the position of the point in the plane of development. To obtain the true distance fionm S to Y, revolve SY about the axis of the cone until it becomes parallel to V, as in Art. (28); s'e' will be its true length. On SG lay off SY -_ s'e'; on SL, SX = s'd'; on SM, SU = s't'. Z Y, X, nind U will be the position of these points on the plane of DESURIPTIVE GEOMETRY. 95 development, and UX —- U will be the development of the curve of intersection. Through L draw LR perpendicular to LS, and make it equal to ir. RX will be the developed tangent. 167. PROBLEM 42. To find the intersection of any cone by a vlane. Let the cone and plane tTt' be given as in Fig. 69. Analysis. Intersect the cone by a system of planes through the vertex and perpendicular to the horizontal plane. Each of these planes will intersect the cone in one or more rectilinear elements, and the given plane in a right line, the intersection of which will be points of the curve. Since these auxiliary planes are perpendicular to the horizontal plane, they will intersect in a right line through the vertex, perpendicular to the horizontal plane, and the point in which this line pierces the cutting plane will be a point common to all the right lines cut from this plane. Construction. Find the point in which the perpendicular to I, through S, pierces tTt', as in Art. (42). v' is its vertical projection. The vertical projections of the lines cut from tTt' all pass through this point. Let sp be the horizontal trace of an auxiliary plane. It intersects the cone in the elements SE and SD, and the cutting plane in the right line (ps, p'v'). This line intersects the elements in R and Y, which are points of the required curve. In the same way any number of points may be found. To find the point of the curve on any particular element, as SM, we pass an auxiliary plane through this element. sm is its horizontal trace, and Z and X the two points on this element. [he vertical projection of the curve is tangent to s'm' at z', and to s'o' at u'. The points q and w, in which the horizontal projection is tangent to sl and sn, are found by using as auxiliary planes the two planes whose traces are sl and sn. To draw a tangent to the curve at any point, as X, pass a plane 96 DESCRIPTIVE GEOMETR'Y. tangent to the cone at X. ic is its horizontal trace, Art. (129). It intersects tTt' in cX, which is therefore the required tangent, Art. (157). The part of the curve which lies above the two extreme elemnents SL and SN, is seen, and therefore its projection, wyzq, is full. For a similar reason the projection, z'q'x'u', of that part of the curve which lies in front of the two extreme elements, SM and SO, is full. To show the curve in its true dimensions, revolve the plane albout tT until it coincides with H, and determine each point, as Q at q", as in Art. (17). Or the position of (sv') may be found at v". Then if the points c, a, b, p, &c., be each joined with this point by right lines, we have the revolved positions of the lines cut from the given plane by the auxiliary planes, and the points y", z", r", X//, in which these lines are intersected by the perpendiculars to the axis, yy", zz", &c., are points of the revolved position of the curve. x"c is the revolved position of the tangent. 168. The intersection of the single curved surface, with a helical directrix, by a plane, may be found by intersecting by a system of auxiliary planes tangent to the projecting cylinder of the helical directrix. These intersect the surface in rectilinear elements, and the plane in right lines, the intersection of which will be points of the required curve. 169. PROBLEM 43. To find the intersection of any surface oj revolution by a plane. Let the surface be a hyperboloid of revolution of one nappe given as in Fig. 70, and let tTt' be the cutting plane. Analysis. If a meridian plane be passed perpendicular to the cutting plane, it will intersect it in a right line, which will divide the curve symmetrically, and be an axis. If the points in which this line pierces the surface be found, these will be the vertices of DESCRIPTIVE GEOMETRY. 9'7 the curve. Now intersect by a systelll of planes perpendicular to the axis of the surface; each plane will cut from tle surface a circumlterence and from the given plane a right line, the intersection of which will be points of the required curve. Constrctiop Draw cn perpendicular to tT, it will be the hori zontal trace of the auxiliary meridian plane. This plane intersects tTt' in the right line NC, Art. (38), and the surf ce in a meridian curve which is intersected by NC ill the two vertices of the requilreda cullrve; or in its highest and lowest points. To find these points, revolve the meridian plane about the axis until it becomes parallel to V. The meridian curve will be vertically projected into tile hyperbola, which limits the ver:ical pl'ojection of the surface, Art. (62), and the line NC into d'c'. The points s' and r' will be the vertical projections of the revolved positions of the vertices. After the counter revolution, these points are horizontally projected at k and 1, and vertically at k' and 1'. Let u'z' be the vertical trace of an auxiliary plane perpendicular to the axis. It cuts the surface in a circle horizontally projected in wuz, and the plane in a right line, whlich piercing V at e' is horizontally projected in ez; hence U and Z are points of the curve. In the same way any number of points may be determined. The points o' andl v', in which the vertical projection of the curve is tangent to the hyperbola which limits the projection of the surface, are the vertical projections of the points in which the line, cut out of the given planle by the meridian plane parallel to the vertical plane, intersects the meridian curve cut out by the same plane. The points upon any particular circle may be determined by Ising the plane of this circle as an auxiliary plane. If the curve crosses the circle of the gorge, the points in which it crosses are determined by using the plane of this circle, and, in this case, the horizontal projection of the curve must be tangent to the horizonltal projection of the circle of the gorge, at the points x and y. The method given above for determining the vertices of the 7 98 I)ESCRIPTIVE GEOMETRY. curve is applicable to any surface of revolution. In this particular surface it may be modified thus: Let the line NL revolve about the axis; it will generate a cone of revolution whose base is dni, and vertex C. This cone intersects the hyperboloid in two circumferences, Art. (97), and the points in which NL intersects these circumferences will be the points required. To construct them; through any rectilinear element of the hyperboloid, as MQ, and the vertex of the cone, pass a plane; qJ will be its horizontal trace, Art. (33). It intersects the cone in two elements horizontally projected in ci and cg, and the points a and b are points in the horizontal projections of the circles in which the hyperboloid and cone intersect, and K and L are the required vertices. To draw a tangent to the curve at Z, pass a plane tangent to the surface at Z, as in Art. (1 38); its intersection with tTt' will be the tangent, Art. (157). The curve may be represented in its true dimensions as in Art. (162). Let the intersection of an oblique plane witlt? a sphere, an ellipsoid of revolution and paraboloid of revolution, be constructed in accordance with the principles of the preceding problem. 170. To find the intersection of any warped surface with a plane directer, by an oblique plane, intersect by planes parallel to the plane directer. Each will cut from the surface one or more rectilinear eleinents, and from the plane a right line, the intersection of which will be points of the required curve. 171. To find the intersection of a helicoid by a plane, the surface being given as in Art. (92); intersect by a system of auxiliary planes through the axis. These will cut from the surface rectilinear elements, and from the plane right lines, the intersection of which will be points.of the required curve. DES(CRt[I'TIVE GEOM()EI tl'I. 99 Let the construction be made, and the curve and its tangent represented in true dimensions. INTERSECTION OF CURVED SURFACES. 1. 72. To find the it tersection of any two curved surfaces, we intersect them by a systemn of auxiliary surfaces. Each auxiliary surface will cut from the given surfaces lines the intersection of which will be points of the required line. Art. (156). The system of auxiliary surfaces' should be so chosen as to cut from the given surfaces the simplest lines, rectilinear elements if possible, or the circurnfelences of circles, &c. To draw a tanrlent to the curve of intersection at any point; pass a plane tangent to each surface at this point. The intersection of these two plnes will be the required tacgyent, since it must lie in each of the tangelit planes. Art. (108). In cozistructing this curve of intersection, great care should be taken to determine those points in which its jrojectionms are taingent to the limiting lines of the projections of the surfaces; and also those points in which the curve itself is tangent to other lines of either surface, as these points aid much in drawing the curve with accuracy. 173. PROBLEM 44. TO find the intersection of a cylinder and cone. Let the surfaces be given as in Fig. 71; the base of the cylinder bai being in the horizontal plane, and its rectilinear elements parallel to the vertical plane; the base of the cone m'l'o', in the. vertical plane, an(l S its vertex. A,nalysis. Intersect the two surfaces by a system of auxiliary planes passing through the vertex of the cone and parallel to the rectilinear elements of the cylinder. Each plane will intersect cachi of the surfaces in two rectilinear elements, the intersection of which 100 I)DESCRIPTIVE GE( M El' liY. will be points of the required curve. These llane. will iiteseci in a right- line passing through the vertex of the cone and parallel'o the rectilinear elements of the cylinder, and the point, in which this line pierces the horizontal plane, will be a point common to lthe horizontal traces of all the auxiliary planes. Constructiotn. Through S draw ST parallel to BE. It pierces 11 in t. Through t draw any right line, as th; it may be taken as the horizontal trace of an auxiliary plane, the vertical trace of which is hg' parallel to b'e'. This plane intersects the cylinder in two elements which pierce II at g and i, and are horizontally projected in gg" and ii". The same plane intersects the cone in two elements which pierce V in m' and In', and are horizontally pro. jected in ms and rns. These elenments intersect in the points X, Y Z and U, which are points of the required curve. In the same way any number of points may be deterllined. The horizontal projection of the required cuirve is tangent to ins at the points x and y; since ms is the horizontal projection of one of the extreme elements of the cone. The points of' tangency on si may be determined by using an auxiliary plane, which shalll coNtain the element SL. The points of tangency on diw are obtained in the same way, by using the plane whose horizontal trace is td. The points in whlichi the vertical projection of the curve is tan gent to the vertical projecliois of the extreme elements of both cone and cylinder will be determined, by using as auxiliary planes those which contain these elemnents. A tangent to the curve at any of the points, thus determined, may be constructed by finding the intersection of two planes, one tangent to the cylinder and the other to the conle, at this point, Art. (172). The plane, of which tc is the horizontal trace, is tangent to the cylinder along the element CQ, and intersects the cone in the two elements SP and SQ. If at P a plane be passed tangent to the 3one, it will be tangent along SP, which is also its intersection Aith the tangent plane to the cylinder. SP is then tangent to DESCRIPTIVE GEOMETRY. 1( 1 cht curve at P, and for a similar reason SQ is tangent at Q. Helnce the two projections sp and sq are tangent to x-py —qu at p and q respectively; and the vertical projections of the same elemeults will be tangent at p' and q'. The plane of which tc' is the horizontal trace, is tangent to the cone along the element SK, and intersects the cylinder in two elements, the horizontal pro(itions of which are tangent to.p —q?) at k and v. The projections of the curve can now be drawn with great accuracy. The horizontal projection of that part which lie- above the two elements SM and SL, and also above the element D W, is drawn full. Likewise the vertical projection of that part which lies in front of the element IB-E, and also in front of the two ex treme elementa of the cone is full. If two auxiliary planes be passed tangent to the cylinder, and both intersect the cone, it is evident that the cylinder will penetrate the cone, so as to forn two distinct curves of intersection. It' one intersects the cone and the other does not, a portion only of the cylinder enters the cone, and there will be a continuous curve of intersection, as in the figure. If neither of these planes intersects the cone, and the cone lies between them, the cone will penetrate the cylinder, making two distinct curlves. If both planes are tangent to the cone, all the rectilinear eldo -1uents of both surfaces will be cut. 174. The intersection of two cylinders may be found, by passing a plane through a rectilinear element of one cylindler parallel to the rectilinear elements of the other, and then intersectingo tile cylinders by a system of planes parallel to this plane. The h,,ri zontal traces of these planes will be parallel, and the conlstruction will be in all respects similar to that of the precedinlg problem. The intersection of two cones may also be found by using a system of planes, through the vertices of both cones. The right line, 102 DESCRIPTIVE GEOMETRY. which joins these vertices, will lie in all of these planes, and pierce the horizontal plane in a point common to all the horizontal traces, WTe may ascertain whether the cylinders or cones intersect in two distinct curves, or only one, in the same manner as in Art. (1I 73), by passing auxiliary planes tangent to either cylinder or cot e. 175. PROBLEM 45. To find the intersection of a cone and helzcoid. Let mnl, Fig. 72, be the base of the cone, S its vertex; and let prq be the horizontal, and p'o"q' the vertical projection of the helical directrix, (s, o's') being the axis, and the rectilinear genera trix being parallel to the horizontal plane, Art. (92). Analysis. Intersect the surfaces by a system of auxiliary planes through the axis of the helicoid. These planes cut from both surfaces rectilinear elements, which intersect in points of the required curve. Constr, action. Draw sk as the horizontal trace of an auxiliary plane. This plane intersects the cone in the element SK, and the helicoid in an element of which sg is the horizontal and h'g' the vertical projection. These intersect in X, a point of the required curve. In the same way, Y, Z, and other points, are determined. At the points m' andl z', thle vertical projection of the curve is tangent to.r'm' and s'l', Art. (160). To draw a tangent to the carve at X; pass a plane tangent tc each of the surfaces at X, Arts. (129) and (139): kht is the horizontal trace of the plane tangent to the cone, and tw, parallel to sg, that of the plane tangent to the helicoid, and their intersection UX, is the required tangent line. 176. PROBLEM 46. To find the intersection of a cylinder anld hemisphere. Let amnn Fig. 73, be the base of the cylinder and MX a rectili DESCRIPTIVE GEOM'ETRY. 103 near element. Let ced be the horizontal and c'f'd' the vertical projection of the hemisphere. Analysis. Intersect the surfaces by a system of auxiliary plaines parallel to the rectilinear elements of the cylinder and perpendicular to the horizontal plane. Each plane will cut fiom the cylinder Two rectilinear elements, and from the sphere a semicircumfuernci the intersection of which will be points of the required curve. Construction. Take pn as the trace of an auxiliary plane. It cuts from the cylinder the two elements PY and NZ, anld from the hemisphere a semicircumference of which i is the centre and i9 the radius. Revolve this plane about pn until it coincides with H. pi" will be the revolved position of PY, Art. (28), and nzm" parallel to pi" the revolved position of NZ; ycz" the revolved position of the semicircumference; y" and z" the revolved position of the required points, and Y aznd Z the points in their true position. In the same way any number of points may be determined. The points of tangency x and tw are found by using mx and luw as the traces of auxiliary planes, and r' and u', by using or and su. A tangent at ny point may be constructed by finding the intersection of two planes tangent to the cylinder and sphere as in Arts. (123) and (145). The tangents at Z and Y are evidently parallel to H. 177. PROBLEM 47. To find the intersection of a cone and hem,sphere. Let mlo, Fig. 74, be the base of the cone, and S its vertex, at the centre of the sphere; abc being the horizontal and a'd'c' the vertical projection of the hemisphere. Analysis. Intersect the surfaces by a system of planes passing through the vertex and perpendicular to the horizontal plane. Each plane will cut from the cone two rectilinear elements, and from the hemisphere a semicircumference, the intersection of which will be points of the required curve. Construction. Take sp as the horizontal trace of one of the auxil. 104 D ESO IPTIVE GEOM ETRY. iary'planes. It intersects the cone in the two elements SP and SQ, and the hlemisphere in a semicircle whose centre is at S. Revolve this plane about the horizontal projecting line of S, until it becomes parallel to V. The element SP will be vertically projected in s'p"', SQ in s'q", the semicircle in ('d'c', and z" and y" will be the vertical projections of the revolved positions of the points of intersection. In the counter revolution these points describe the arcs of horizontal circles and in their tIrue position will be verti tally projected at r' and y', and horizontally at x and y. In the same way any number of points may be tound. The points of tangency u' and z' are obuntd by using auxiliary planes which cut out the extreme elements SM and SO. The points in which the vertical projection of the curve is tangent to the semiciircle t'd'c', aie found by using the auxiliary plane whose trace is st. To draw a tangent to the curve at any point, we pass a plane tangent to the sphere at this point as in Art. (145) and also one tangent to the cone at the same point as in Art (129), and determine their intersection. 178. PROBLEM 48. To develop an oblique cone with any base. Let the cone be given as in the preceding problem, Fig. 74, and let it be developed on the plane tangent along the element S'. Anulysis. If thle cone be intersected by a sphere having its centre at the vertex, all the points of the curve of intersec ion will he at a distance froin the vertex equal to the radius of the sphere; hence, when the cone is developed, this curve will (levelop into the are of a circle having its centre at the position of the vertex, and i s radius equal to that of the sphere. On this we can lay off tlhe rectified arcs of the curve of intersection, included between th( several rectilinear elements, Art. (159), and then draw these clements to the position of the vertex. The developed base, or any curve on the suiface, mlay be traced on the plane of development, by laying off on each element, friom DESCRIPTIVE GkOM1ETRYV. 10 the vertex, the distance from the vertex to the point whlere the element intersects the base or curve. A line through the extremities of these distances will be the required development. Construtction. Find as in the preceding problem the cutrve XUY -—. With S, Fig. 75, as a centre, and sa as a radius, descrihbt the are XUR. It is the indefinite development of the intersection of the sphere and cone. Draw SX for the position of the element SX. To find the distance between any two points measured on the curve XUY- —, we first develop its horizontal projecting cylinder on a plane tangent to it at X, as in Art. (161). X'U'R' Fig. a, is the development of the curve. On XUR lay off XU=X'U', UR=U'R', &c, and draw SU, SR, &c. These will be the positions of the elements on the plane of development. On these lay off SP=s'p"', SM-s'm'', &c., and join the points PM, &c., and we have the development of the base of the cone..179. PROBLEM 49. To find the intersection of two suifaces of revolution, whose axes are in the same plane. First, let the axes intersect and let one of the surfaces be an ellipsoid of revolution and the other a paraboloid; and let the horizontal plane be taken perpendicular to the axis of the ellipsoid and the vertical plane parallel to the axes; (c,c'd'), Fig. 76, being the axis of the ellipsoid, and (cl, s'l') that of the paraboloid. Let the ellipsoid be represented as in Art. (107) and let z)f'r' be the vertical projection of the paraboloid. Analysis. Intersect the two surfaces by a system of auxiliary spheres having their centres at the point of intersection of the axes. Each sphere will intersect each surface in the circumference of a circle perpendicular to its axis, Art. (97). and the points of intersection of these circumferences will be points of the required curve. Construction. Witht s' as a centre and any radius, s'q', describe the circle q'p'r'; it will be the vertical projection of an auxiliary sphere. This sphere intersects the ellipsoid in a circunltelellce 106 DESCRIPTIVE GEOMETRY. vertically projected in p'q', and horizontally in pxq. It intersects the paraboloid in a circumference vertically projected in r'v'. These circumferences intersect in two points vertically projected at x', and horizontally at x and x". In the same way any number of points may be found. The points on the greatest circle of the ellipsoid are found by using s'n' as a radius. These points are horizontally projected at u and u", points of tangency of xzx" with norn. The points W and Z are the points in which thle meridian curves parallel to V intersect, and are points of the required curve. Each point of the curve z'x'w' is the vertical projection of two points of the curve of intersection, one in front and the other behind the plane of the axes. A tangent may be drawn to the curve at any point as X, as in Art. (172). Otherwise thus: If to each surface a normal line be drawn at X, the plane of these two normals will be perpendicular to the tangent plane to each surface at X, and therefore perpendicular to their intersection, which is the required tangent line. Art. (172). Hence if through X a right line be drawn perpendicular to this normal plane, it will be the required tangent. Since the meridian plane to a surface of revolution is normal to the surface, Art. (115), the normal to each surface at X must lie in the meridian plane of the surface and therefore intersect the axis. To construct the normal to the ellipsoid, revolve the meridian curve throulg X, about the axis of the surface, until it becomes parallel to V; it will be projected into c'n'd' and the point X will be vertically projected at q'. Perpendicular to the tangent at q' draw q'k'. It will be the vertical projection of the normal in its revolved position. After thle counter-revolution, K remaining fixed, k'x' will be the vertical, and cx the horizontal projection of the normal. In the same way the normal LX to the paraboloid may be constructed. The line k'l' is the vertical projection of the intersection of the DESCRIPTIVE GEOMETRY. 107 plane of these two normals with the plane of the axes, and is parallel to the vertical trace of the first plane; hence x't' perpendicular to this line is the vertical projection of the require(l tangent, xi is the horizontal projection of the intersection of the normal plane with the plane of the circle PXQ, and this is parallel to the horizontal trace of the normal plane; hence xt perpendicular to this is the horizontal projection of the required tangent. Second. If the axes of the two surfaces are parallel, the construction is more simple, as the auxiliary spheres become planes perpendicular to the axes, or parallel to the horizontal plane. Let the construction be made in this case. PRACTICAL PROBLEMS. 180. In the preceding articles we have all the elementary principles and rules, relating to the orthographic projection. The student who has thoroughly mastered them will have no difiiculty in their application. Let this application now be made to the solution of the following simple problems. 181. PROBLEM 50. Having given two of thefaces of a triedral angle, and the diedral angle opposite one of them, to construct the triedral angle. Let dsf and fse", Fig. 77, be the two given faces and A the given angle oppositefse", dsf being in the horizontal plane, and the vertical plane perpendicular to the edge sf. Construct, as in Art. (54), de' the vertical trace of a plane whose horizontal trace is sd and making with H the angle A; sde' will be the true position of the required face. Revolve fse" about sf until the point e" comes into de' at e'. This must be the point in which the third edge in true position pierces V Join ef; it will be the vertical trace o' the plane of the face f.se", in true position, and SE will be the third eldc. 1 08 DESCRIPTIVE GE(OM'rETRY'. Revolve sde' about sd until it coincides with II; e' falls at e"' Art. (17), and dse"' is the true size of the third or required face. e'fd is the diedral angle opposite the face dse'; and p?,q, deter mined as in Art. (52), is the diedral angle opposite dsf. 182. PROBLEM 51. Having given two diedral angles formed by the faces of a triedral angle, and the face opposite one of them, to construct the angle. Let A and B, Fig. 78, be the two diedral angles, and dse"' the face opposite B. Make e'dJ = A. Revolve dse"' about ds until e"' comes into de' at e'. This will be the point wheie the edge se"', in true position, pierces V, and SE will be this edge. Draw e'?n making e'me _- B, and ievolve e'm about e'e; it will generate a right cone whose rectilinear elements all make, with H, an angle equal to B. Through s pass the plane.fe' tangent to this cone, Art. (130). It, with the faces fsd and sde', will forln the required angle. fse" is the true size of the face opposite A, e" being the revolved position of e', determined as in Art. (17), and the third diedral angle, formed by ejfs and e'ds may be found as in Art (52). 183. PROBLEM 52. Given two faces of a triedral angle and their included diedral angle, to construct the angle. Let dse"' and dsf, Fig. 79, be the two given faces, and A the' given angle. t Make e'df equal A. de' will be the vertical trace of the plane of the face dse"', in its true position. Revolve dse"' about sd urtil e"' comes into de' at e', the point where the edge se"', in true position, pierces V. Draw elf. It is the vertical trace of the plane of the third or required face, andfse" is its true size. eon", Art. (53), is the diedral angle opposite e'ds, and the third diedral angle can be found as in Art. (52). DESCRIPTIVE GEOMETRY. 1019 184. PR,,I x M 53. Given one face and the two adjacent diedral,ngles of a triedral angle, to cdnstruct the angle. Let ds F',g. 79, be the given face and A and B the two adjacent diedral angles. Make e'df = A. Construct as in Art. (54), fe', the vertical trace of a plane sfe', making with H an angle = B. SE will be the third edge. dse"' and fse" the true size of the other faces and the third diedral angle is found as in Art. (52). 185. -PROBLEM 54. Given the threefaces of a triedral ad)gle, to construct the angle. Let dse"', Fig. 80, dsf and lse", be the three given faces, sd and sf being the two edges in the horizontal plane. Make se" = se"'. Revolve the facef'e" aboutf8;e" describes an are whose plane is perpendicular to H, Art. (17), of which e"e is the horizontal and ee' the vertical trace. Also revolve dse"' about ds; e'l' describes an arc in the vertical plane. These two arcs intersect at e', the point where the th;rd edge pierces V and SE is this edge. Join e'd and ef. These are the vertical traces of the planes ot the laces dse"' andfse", in true position. The diedral angles mlay now be found as in the preceding articles. 186. PROBLEM.55. Given the three diedrol angles formed by the faces of a toiedr7al angle. to constrtct the angle. Let A, B, and C, Fig. 81, be the diedral angles. Make e'df = A. Draw ds perpendicular to AB and take e'ds as the plane of one of the faces. If we now construct a plane which shall make, with H and e'ds, angles respectively equal to B and C, it, with these planes, will form the required triedral angle. To do this; with d as a centre and any radius as dm, describe a sphere; rnn'q will be its vertical projection. Tangent to mrn'q draw o'u making o'ud - B, and revolve it about o'd. It will 1 0 DESCRIU'TIVE GEOMETRY. generate a cone whose vertex is o', tangent to the sphere and all of twhose rectilinear elements- make with LI an angle equal to B. Also tangent to mn'q, draw p'r' making with de' an angle equal to C, and revolve it about p'd. It will generate a cone whose vertex is p', tangent to the sphere, and all of whose rectilinear elements make with the plane sde' an angle equal to C. If now dthrough a' and p' a plane be passed tangent to the sphere, it will be tangeenr to bohi cones, and be the plane of the required third face. p'o' is tLe vertical trace of this plane, andfs tangent to the base uxy is the horizcntal trace, SE is the third edge and dsf, dse"' and fse" the three )iices in true size. i 87. By a. reference'c, Spherical Trigonemetry, it will be seen that the preceding six problemls are simple cotnstructions of the required parts of a spherical triangle, when either three are given. Thus in problem 50, two sides a and c, arid an angle A opp)osite one, are given, and the others constructed. In problem 51, two angles A and B, and a side b, opyosite one, are given, &c. 188. PROBLEM 56. To construct a tlriang'lar pyraimid, having given the base and the three lateral edges. Let cde, Fig. 82, be the base in the horizontal plane, AB being taken perpendicular to cd; and let cS, dS, and eS be the three edges. With c as a centre and cS as a radius, describe a sphere, intersecting H in the circle mon. l'he required vertex must be in the surface of this sphere. With d as a centre and dS as a radius, describe a second sphere intersecting H in the circle qm:p. The re quired vertex must also be on this surface. These two spheres intersect in a circle of which mn is the horizontal, and tn's'n' the vertical projection, Art. (97). With e as a centre, and eS as a radius, describe a third sphere, intersecting the second in a circle of which qp is the horizontal projection. These two circles inter DESCRIPT'IVE G(EO)ME'KRY. 111 sect in a right line perpendicular to H at s, and vertically projected in r's'. This line intersects the first circle in S, which must be a point common to the three spheres, and therefore the vertex of the required pyramid. Join S with c, d and e, and we have the lateral edges, in true position. 189. PROBLEM 57. To circumscribe a sphere about a triangukr pyramid. Let mno, Fig. 83, be the base of the pyramid in the horizontal plane, and S its vertex. AAnalysis. Since each edge of the pyramid must be a chord of the required sphere, if either edge be bisected by a plane perpendicular to it, this plane will contain the centre of the sphere. Hence, it three such planes be constructed intersecting in a point, this must be the required centre, and the radius will be the right line joining the centre with the vertex of either triedral angle. Construction. Bisect mn and no, by the perpendiculars rc and pc. These will be thle horizontal traces of two bisecting and perpendicular planes.'They intersect in a right line perpendicular to H at c. Through U the middle point of SO pass a plane perpendicular to it, Art. (46). tTt' is this plane. It is pierced by the perpendicular (c, de') at C, Art. (42), which is the intersection of the three bisecting planes, and therefore the centre of the required sphere. CO is its radius, the true length of which is c'o", Art. (29). With c and c' as centres, and with'o" as a radius, describe circles. They will be, respectively, the horizontal and vertical prujections of the sphere. 190. PROBLEM 58. To inscribe a sphere in a given triangular pyramid. Let the pyramid S-mno, Fig. 84, be given as in the preceding problem. Analysis. The centre of the required sphere must be equally 112 DESCRIPTIVE GEOMETR' - distant from the foul faces of the pyramid, and therefore must be in a plane bisecting the diedral angle formed by either two of its faces.- Hence, if we bisect three of the diedral angles, by planes intersecting in a point, this point must be the centre of the required sphere, and the radius will be the distance from the centre to either face. Construction. Find the angle sps" made by the face Son with IH, Art. (53). Bisect this by the line pu, and revolve the plane sps" to its true position. The line pu, in its true position, and on will determine a plane bisecting the diedral angle sps". In the same way determine the planes bisecting the diedral angles srs"', sqs"". These planes, with the base mne, form a second pyramid, the vertex of which is the intersection of the three planes, and therefore the required centre. Intersect this pyramid by a plane parallel to HI, whose vertical trace is t'y'. This plane intersects the faces in lines parallel to rnn, no, and ow respectively. These lines form a triangle whose vertices are in the edges. To determine these lines, lay offpv= y'y", draw vz parallel to ps. z will be the revolved position of the point in which tile parallel plane intersects pu in its true position. z" is the horizontal projection of this point, and z'y of the line parallel to no. In the same way ty and tx are determined. Draw ox and ny; they will be the horizontal projections of two of the edges. These intersect in c the horizontal projection of the vertex. n'y' is the vertical projection of the edge which pierces H at n, c' the vertical projection of the centre, and c'd' the radius. With c and c' as centres describe circles with c'd' as a radius. They will be the horizontal and vertical projections of the required sphere. PART I1. SPHERICAL PROJECTIONS. PRELIMINARY DEFINITIONS. 191. One of the most interesting applications of the principles of Descriptive Geometry is to the representation, upon a single plane, of the different circles of the earth's surface, regarded as a perfect sphere. These representations are Spherical Projections. The plane of projection which is generally taken as that of one of the great circles of the sphere, is the primitive plane; and this great circle is the primitive circle. The axis of the earth is the right line about which the earth is known daily to revolve. The two points in which it pierces the surface are the poles, one being taken as the North and the other as the South Pole. The axis of a circle of the sphere is the right line through its centre perpendicular to its plane. and the points in which it pierces the surface are the poles of the circle. The polar distance of a point of the sphere is its distance froni either pole of the primitive circle. The polar distance of a circle of the sphere is the distance of any point of its circumference from either of its poles. 8 114 DESCRIPTIVE GEOMETRY. 192. rhe lines on the earth's surface, usually represented, are: 1. The Equator, the circumference of a great circle whose plane is perpendicular to the axis. 2. The Ecliptic, the circumference of a great circle making an angle of 23~0, nearly, with the equator. It intersects the equator in two points, called the Equinoctial Points. 3. The Meridians, the circumferences of great circles whose planes pass through the axis; and are therefore perpendicular to the plane of the equator. Of these meridians two are distinguished: the Equinoctial Colure, which passes through the equinoctial points; and the Solstitial Colure, whose plane is perpendicular to that of the equinoc tial colure. The solstitial colure intersects the ecliptic in two points, called the Solstitial Points. 4. The Parallels of Latitude, the circumferences of small circles parallel to the equator. Four of these are distinguished: The Arctic Circle, 23~~ from the north pole; The Antarctic Circle, 23~~ from the south pole The Tropic of Cancer, 232~ north of the equator; The Tropic of Capricorn, 23l~ south of the equator. The first two are also called Polar Circles. 193. The Latitude of a point on the earth's surface, is its distance from the equator, measured on a meridian passing through the point. The Horizon of a point or place, on the earth's surface, is the circumference of a great circle whose plane is perpendicular to the radius passing through the point. Let M, Fig. 85, be any point on the earth's surface. Through this point and the axis pass a plane, and let MNS be the circum ference cut from the sphere; N the north and S the south pole ECE' the intersection of the plane with the equator, and PCQ SPHERICAL PROJECtrIoNS. 115 perpendicular to CM. its intersection with the horizon of the given point. Then ME is the latitude of the point, and NQ the distance of the pole N from the horizon. NQ also measures the angle NCQ, the inclination of the axis to the horizon. But NQ = ME, since each is obtained by subtracting NM from a quadrant; that is, the distance from either pole of the earth to the horizon of a place, is equal to the latitude of that place. 194. Let NS, Fig. 85, and MR be the axes of two circles intersecting the plane NCM in the lines EE' and PQ respectively. ECP is then the angle made by the planes of these circles. But ECP = NCM, since each is obtained by subtracting MCE from a right angle; that is, the angle between any two circles of the sphere, is equal to the angle formed by their axes. 195. If a plane be passed through the axes of any circle of the sphere and of the primitive circle, its intersection with the primitive plane is the line of measures of the given circle. This auxiliary plane is perpendicular to the planes of both circles, and therefore to their intersection; hence the line of measures, a line of this plane, must be pelrpendicular to the intersection of the given with the primitive circle, and must also pass through the centre of the primitive circle. Thus, if EE', Fig. 85, is the intersection of a circle with the primitive plane NESE', NS is its line of measures. Also NS is the line of measures of any small circle whose inter section with the primitive plane is parallel to EE'. 116 DESCRIPTIVE GEOMETRY. ORTHOGRAPHIC PROJECTIONS OF THE SPHERE. 196. When the point of sight is taken in the axis of the prini. tive circle, and at an infinite distance from this circle, the projec. tions of the sphere are orthographic, Art. (2). If E, Fig. 85, be any point, e will be its orthographic projection on the plane of a circle whose axis is CM. But Ce = Ed; that is, the orthographic projection of any point of the surface of a sphere is at a distance.from the centre qf the primnitive circle equal to the sine of its polar distance. 197. The circumference of a circle, oblique to the primitive plane, is projected into an ellipse. For the projecting lines of its different points form the surface of a cylinder whose intersection with the primitive plane is its projection, Art. (74), and this intersection is an ellipse, Art. (160). If the plane of the circumference be perpendicular to the primitive plane, its projection is a right line, Art. (62). If the plane of the circumference be parallel to the primitive plane, its projection is an equal circumference, Art. (62). The projection of every diameter of the circle which is oblique to the primitive plane, will be a right line less than this diameter, Art. (29), while the projection of that one which is parallel to the primitive plane, will be equal to itself, Art. (14). This projection will then be longer than any other right line which can be drawn in the ellipse, and is therefore its transverse axis. Analyt. Geo., Art. (127). The projection of that diameter which is perpendicular to the one which is parallel to the primitive plane, will be perpenldicular SPHERICAL PROJECTIONS. 117 to this transverse axis, Art. (36), and pass through the centre, and therefore be the conjugate axis of the ellipse, Art. (59). This last diameter is perpendicular to the intersection of the plane of the given circle and the primitive plane, and therefore makes with the primitive plane the same angle as the circle; and one-half its projection, or the semni-conj.gyate (xi.s' of the ellipse, is evidently the cosine qf this inclination, computed to the radius of the given circle. Hence, to project any circle orthographically, we have simply to find the projection of that diameter which is parallel to the primitive plarne, and through its middle point draw a right line perpendicular to it, and make it equal to the cosine of the angle made by the circle'with the primitive plane. The first line is the transverse, and the second the semi-conjugate axis of the required ellipse, which may then be accurately constructed as in Art. (59). It should be remarked that the conjugate axis of the ellipse always lies on the line of measures of the circle to be projected, Art. (195). 198. The line of measures of a circle evidently contains the projections of both poles of the circle, Art. (195); and since the arc which measures the distance of either pole from the pole of the primitive circle, measures also the inclination of the two cil cles, Art. (194), it follows that either pole of a circle is ortho graphically projected in its line of measures, (it ( distance from the centre of the prim tive circle equal to tiLe sine of its inclination, Art. (196). 199. PROBLEM 59. To project the sphere u7po)l the plane of any one of its great circles. Let EpE'q, Fig. 86, be the primitive circle intersecting th equator in the points E and E', and making with it an angle denoted by A Let E and E' also be assnlled as the equilloctial 1] 18 DESCRIPTIVE GEOMETRY. points. The line EE' is then the intersection of the primitive plane by the equator, ecliptic, and equinoctial colure, and pq perpendicular to it is the line of measures of all these circles. Let us first project the hemisphere lying between the primitive plane and the north pole. Since EE' is that diameter of the equator which lies in the primitive plane, it is its own projection, and therefore the transverse axis of the ellipse into which the equator is projected. From q lay off qm' = A, and draw m'm perpendicular to pq. Cm = cos A, and is the semi-conjugate axis, Art. (197). On this and EE' describe the semi-ellipse EmE'; it is the projection of that part of the equator lying above the primitive plane. n is the projection of the north pole, En' being made equal to A, and Cn = n'x its sine, Art. (198). EE' is also the transverse axis of the projection of the ecliptic. If the portion of the ecliptic on the hemisphere under consideration, lies between the equator and the north pole, it will make an angle with the primitive plane greater than that of the equator by 231-~. It it lies between the equator and the south pole, it will make a less angle by 23_. Taking the former case, lay off m'o' = 231~, then go = A + 23-10, and Co = cos (A + 23-}~) the semi-conjugate axis, with which and EE' describe the semi. ellipse EoE', the projection of one half the ecliptic. The equinoctial colure, making with the equator a right angle, makes with the primitive plane an angle equal to 900 + A. EE' is the transverse axis of its projection, and Cn = cos qn' = cos (900 + A) = the semi-conjugate axis. And the semi-ellipse EnE' is the projection of that half above the primitive plane. The solstitial colure being perpendicular to the equator and equinoctial colure is perpendicular to EE', and therefore to the primitive plane; hence its projection is the right line qp, Art. (197). To project any meridian, as that which makes with the solstitial colure, an angle denoted by B; pass a plane tangent to the sphere at the north pole. It will intersect the planes of the given SPHERICAL PROJECTIONS. 119 meridian and solstitial colure in lines perpendicular to the axis and making with each other an angle equal to B; and these lines will pierce the primitive plane in points of the intersections of the planes of these meridians with the primitive plane, Art. (30). To determine this tangent plane; revolve the solstitial colure about pq, as an axis, until it comes into the primitive plane. It will then coincide with Enz'pq, and the north pole will fall at n'. Draw n't tangent to En'p; it is the revolved position of the intersection of the required tangent plane by the plane of the solstitial colure. It pierces the primitive plane at t, and st perpendicular to Cn, Art. (145), is the trace of the tangent plane. Revolve this plane about ts until it coincides with the primitive plane. The north pole falls at n"; ti" being equal to tn', Art. (17). Through n", draw n"s, making with n"t an angle equal to B. This will be the revolved position of the intersection of the tangent plane by the plane of the given meridian. It pierces the primitive plane at s, and sC is the intersection of the meridian plane with the primitive, and yz is the transverse axis of the required projection, Art. (197). To find the semi-conjugate; through the north pole pass a plane perpendicular to yz; nv is its trace. Revolve this plane about nv until it coincides with the primitive plane. N falls at n"', and n'"t'n is the angle made by the meridian with the primitive plane, Art. (53). Lay off vk = CE, and draw ku parallel to yz. Cu is the required semi-conjugate axis, and the semi-ellipse yuz is the projection of that half of the meridian which lies above the primitive plane. Since the plane of any parallel of latitude, as the arctic circle, is parallel to the equator, it will be intersected by the plane of the equinoctial colure in a diameter parallel to EE', and to the primitive plane, and the projection of this diameter will be the transverse axis of the projection. To determine it; revolve the equinoctial colure about EE' as an axis until it coincides with EpE'q; N falls at p. From p lay off pa' = 23~~, the polar dis tance of the parallel, and draw a'b'. Tt will be the revolved posi. 1 2t) DESCRIPTIVE GEOMETRY. tion of that diameter of the arctic circle which is parallel to the primitive plane. When the colure is revolved to its true position a'b' will be projected into ab, the required transverse axis. From d its middle point lay off di = cos A, computed to the radius dca; it will be the semi-conjugate axis, and the ellipse aibh is the required projection. In the same way the tropic of Cancer or any other parallel may be projected. If the polar distance of the parallel is greater than 90~ - A, the inclination of the axis, the parallel will pass below the primitive plane and a part of its projection be drawn broken. T'he projection of the tropic of Cancer is tangent to EoE' at o, Art. (65). 200. Each point in the primitive circle is evidently the projection of two!Points of the surface of the sphere, one above and the other below the primitive plane. To represent these points distinctly and prevent the confusion of the drawing, we first project the upper hemisphere, as above, and then revolve the lower 1800, about a tangent to the primitive circle at E. It will then be above the primitive plane and may be projected in the sanme way as the first. Emn"E" is the projection of the other half of the equator; s of the south pole; Eo"E" of the other half of the ecliptic; EsE" of the equinoctial colure; y'sz' of the meridian, &c.'201. If the projection be made on the equator, the precediing problem is much simplified. Thus, let EpE'q, Fig. 87, he the equator. n is the projection of the north pole; EoE' of one half of the ecliptic, qo' being equal to 23~~, and no its cosine. Since the meridians are all perpendicular to the equator, EE' is the projection of the equinoctial, and pq of the solstitial colure; yz of the meridian making an angle of' 303 with the solstitial colure. Since the parallels of latitude are parallel to the primitive plane, SPHERICAL PROJEC'I'IINS. 12 alibi is the projection of the arctic circle, and -(qkl that of the tropic of Cancer, na being equal to the sine of'23-~, an1 nisin 661O, Art. (196). ogkl is tangent to EoE' at o.'202. If the projection be made on the equinoctial colure; let ENE'S, Fig. 88, be the primitive circle, and E and E' the equlinoctiai points. Since the equator is perpendicular to the primitive plane, EE' will be its projection. N is the north and S the south pole. EoE' is the projection of one half the ecliptic; Co being equal to cos 66~~. NS is the projection of the solstitial colure. Since the parallels are perpendicular to the primitive plane, a6 and a'b' are the pr,,jections of the polar circles; Na and Sa' being each equal to 23~~; and ly and l'g' the projections of the tropics. Ng and Sg' being each equal to 66~. NyS is the projection of one half the meridian, making an angle of 30~ with the solstitial colure, or 60~ with the primitive plane.. Cv being equal to cos 60~ = ~ CE'. 203. The projections may be made upon the ecliptic, and! horizon of a place, in the same way as in problem 59. In the former case, the angle A will equal 23~; and in the latter, since the angle included between the axis and horizon is equal to the latitude of the place, Art. (193), the angle A between the equator and horizon will be 90~ + the latitude. ST'rREOGRAPHIC PROJECTIONS OF THE SPHERE. 204. The natural appearance and beauty of a scenographic drawing will depend, very much, upon the position chosen by the draughtsman, or artist, for the point of sight. This should be sc selected that a person taking the drawing into his hand for examl 1.22 DESCRIPTIVE GEOMETRY. ination, will naturally place his eye at this point. From anmy other position of the eye, the drawing will appear to some extent distorted. Hence it is, that an orthographic drawing never appears perfectly natural, as it is impossible to place the eye of the ob. server at an infinite distance from it. In spherical projections, if the point of sight be taken at either pole of the primitive circle, the projections are Stereographic, and. in general, present the best appearance to the eye of an ordinary observer, as, in this case, the projections of all circles of the sphere, as will be seen in Art. (207), are circles. 205. The projection of each point on the surface of the sphere, will be that point in which a right line, through it and the point of sight, pierces the primitive plane, Art. (3). Let M, Fig. 89, be any point on the surface of the sphere. Through it and the axis of the primitive circle pass a plane. It will intersect the sphere in a great circle EME'S, and the primitive plane in a right line EE'; N and S being the poles of the.primitive circle and S the point of sight. NM is the polar distance and m the stereographic projection of M. Cm is the tangent of the arc Co, computed to the radius CS =CE, and Co is one half of NM. That is, the stereographic projection of any point of the surface of the sphere is at a distance from the centre of the primitive circle equal to the tangent of one half its polar distance. In this projection, it should be observed that the polar distance of a point is always its distance from the pole opposite the point of sight, andl often exceeds 900. 206. If a plane be passed through the vertex of an oblique cone with a circular base, perpendicular to this base and through its centre, such plane is a plincipal plane, and evidently bisects all chords of the cone drawn perpendicular to it. SPHERICAL PROJECTIONS. 123 Let SAB, Fig. 90, be such a plane intersecting the cone in the elements SA and SB, and the base in the diameter AB. If this cone be now intersected by a plane tTt', perpendicular to the principal plane, and making with one of the principal elements, as SA, an angle Sba equal to the angle SBA, which the otller makes with the plane of the base, the section is a su.b contrary sections and will be the circumference of a circle. For, through o, any point of ba, which is the orthographic projection of the curve of intersection on the principal plane, pass a plant parallel to the base. It cuts fi'om the cone the circumference of a circle, and intersects the plane of the sub contrary section in a right line perpendicular to SAB at o, and the two curves have, at this point, a common ordinate. The similar triangles aod and cob give the proportion ao: oc:: od: ob; or, ao X ob = oc X od. But oc X od is equal to the square of the common ordinate, since the parallel curve is a circle; hence ao X ob is. equal to the square of the ordinate of the sub contrary section, which must, therefore, be a circle. 207. To project any circle of the sphere; through its axis and the axis of the primitive circle pass a plane, and let ENE'S, Fig. 89, be the circle cut fiom the sphere by this plane; S the point of sight; RM the orthographic projection of the given circle on the cutting plane, Art. (62); CN the axis of the primitive circle orthographically projected in EE', and C1' the axis of the given circle. The projecting lines, drawn from points of the circumference to S, form a cone whose intersection by the primitive plane will evidently be the stereographic projection of the circumference, SRlM1 is the principal plane of this cone, and SR and SM the principal elements. The primitive p,lane is perpendicular to thifs 12 4 DESCRIPTIVE GEOMETRY. plane, and intersects the cone in a curve of which rmr is the or. thographic projection. But the angle Srm = SMR. Since each is measured by SR, ES being equal to E'S, hence this section is a sub contrary section, and therefore a circle whose diameter is mnr. That is, the stereographic projection (~f every circle on the surface of a sphere, whose plane does not pass through the point of sight, is a circle. mnr is also the line of measures of the given circle, Art. (195) and evidently contains the centre of its projection. The distance Cr = tan Co' = tan 2 (rI + PN), and Cm = tan Co = tan I (PR - PN). Hence, the extremities of a diameter of the projection of any circle, on the surface of the sphere, are in its line of measures, one at a distance from the centre of the primitive circle, equal to the tangent of ozne-half the sum of the polar distance and inclination of the circle, and the other at a distance equal to the tangent of onehaljf the diierence of these two arcs. When the polar distance is greater than the inclination, these extremities will evidently be on different sides of the centre of the primitive circle. When less, they will be on the same side. If the polar distance is equal to the inclination, the projection of the given circle will pass through the centre of the primitive circle. The polar distance and inclination of any circle being known, a diameter of its projection can thus be constructed, and thence the projection. 208. If the circle be parallel to the primitive plane, the Fub SPH ERICAL PROJECTIONS. 125 contrary and parallel sections coincide, and the projecti n is a circle whose centre is at the centre of the primitive circle, and radiius the distance of the projection of any point of the circumference from the centre of the primitive circle; that is, the tangent of haly the circle's polar distance, Art. (205). If the plane of the circle pass through the point of sight, the projecting cone becomes a plane, and the projection is a right line. 209. If a right line be tangent to a circle of the sphere, its projection will be tangent to the projectiosn of the circle. For, the projecting lines of the circumference form a cone, and those of the tangent, a plane tangent to this cone, along the projecting line of the point of contact; hence the intersections of the cone and plane by the primitive plane, are tangent to each other at the projection of the point of contact, Art. (112). But the first is the projection of the circle, and the second that of the tangent. 210. Let MR and MT, Fig. 91, be the tangents to two circles of the sphere at a cnmmon point, M. Let these tangents be projected on thie primitive plane by the planes RMS and TMS respectively, in the lines mr and mt, and let MaS and MbS be the circles cut from the sphere by these planes, and let SR and ST be the lines cut troil the tangent plane to the sphere at S. Since this tangent plane is parallel to the primitive plane, the lines SR and ST will be parallel respectively to mr and mat, and the angle RST = rmt. Join RT. Since RS and RM are each tangent to the circle MaS, they are equal, and for the same reason TM = TS; hence the two triangles, RMT and RST, are equal, and the angle RMT = RST - rmt, that is, the angle between any two tangents to circles of the sphere, at a common point, is equal to the angle of their projections. The angle between the circles is the same as that between their DESCRIP'IVE GEOMETRY. tangents, and since the projections of the tangents are tangent to the projections of the circles, the angle between the projections of the circles is the same as that between the projections of the tan gents; hence the angle between acny two circumnferences or arcs is equal to the angle between their projections. 211. If from the centres of the projections of two circles radis be drawn to the intersection of these projections, they will make the same angle as the circles in space. For, these radii being perpendicular to the tangents to the projections, at their common point, make the same angle as these tangents, and therefore as the projections of the arcs, or as the arcs themselves. 212. If the circle to be projected be a great circle, it will intersect the primitive circle in a diameter perpendicular to its line of measures, Art. (195). Let O, Fig. 92, be the centre of the projection of such a circle intersecting the primitive circle EPE'R in the diameter PR, CE being its line of measures, and P and R evidently points of the projection. Draw the radius OR. The primitive circle is its own projection; therefore the angle CRO is equal to the angle between the given and primitive circles, Ait. (211). CO is the tangent of this angle, and OR its secant. IHence the centre of the projection of a great circle, is in its line of measures, Art. (207), at a distance from the centre of the primitive circle equal to the tangent of its inclination, and'tile radius of' the pi ojection is the secant of this angle. 213. Let O0 Fig. 93, be the centre of the projection of a snmall circle perpendicular to the primitive plane, and intersecting it in PR. OC is its line of measures, and P and R points of the projection. Join CR and OR. OR is perpendicular to CR, since EPE'R is the projection of thle primitive circle, and PgPR that of SPHERICAL, PROJFCTIONS. 127 the given circle, at right angles with it, Art. (211). OR is therefore the tangent of:the arc ER, the polar distance of the given circle, and CO is its secant. Hence the centre of the projection of a small circle, peipendicular to the primitive plane, is in its line of measures, at a distance from the centre of the primitive circle equal to the secant of the pola(r distance, and the radius of the projection is the tangent of the polar distance. 214. Let P and R, Fig. 94, be the poles of any circle of the sphere; EPE'S being the circle cut from the sphere by the plane of the axes of the given and primitive circles, and MQ, the intersection of this plane with that of the given circle, and EE', its intersection with the primitive circle, the line of measures of the given circle. p is the projection of P and r of R. Cp is the tangent of Co, equal to one-half of NP, which measures the inclination of the given circle to the primitive, Art. (]94). Cr is the tangent of Co', the complement of Co, or is the co-tanlgent of Co. Hence the poles of any circle of the sphere are projected into the line of measures, the one furthest from the point of sight at a distance from the centre of the primitive circle equal to the tanyent of half the inclination, and the other at a distance equal to the co-tangent of half the inclination of the given to the primitive circle. 215. PROBLEM 60. To project the sphere upon the plane oJ any of its great circles, as the ecliptic. Let EpE'q, Fig. 95, be the primitive circle intersecting the equator in EE'. In this case, E and E' will be the equinoctial Dpoints, and in any other case may be taken as such. EE' will also be the intersection of the plane of the equinoctial colure with the primitive plane, and pq the line of measures of both these circles.'We will first project the hemisphere above the primitive plane, the point of sight being at the pole underneath. 128 DESCRIPTIVE GEOMETRY. Since the Equator makes an angle of 23~~ with the primitive plane, we draw E'o, making the angle CE'c = 23-~. Co is the tangent of this angle and E'o the secant; hence with o as a centre and E'o as a radius, describe the arc EmE'; it is the projection of the part of the equator above the primitive plane, Art. (212). From E lay off Ek = 23~ and draw kE'; Cn is the tangent of half the inclination of the equator, and n is the projection of the zorth pole, Art. (214). The Equinoctial colure makes with the primitive plane an angle = 90~ + 231o. Through E' draw E'x perpendicular to E'o. The angle CE'x = 900 + 23-~, and Cr = tan CE'r is its tangent and E'r its secant. With r as a centre, and E'r as a radius describe E'mnE. It is the projection of the thalf of the equinoctial colure above the primitive plane. It must pass tllrough n. The Sol-.titial Colure, being perpendicular to EE', passes through the point of sight and is projected into the right line pq. To project any other meridian'?, as that which maks an angle of 300 with the equinoctial colure; produce the arc E'nE until it intersects Cr produced. The point of intersection, which we denote by., will be the projection of the south pole, and since all the meridians pass through the poles, their projections will pass through n and s, and ns will be a chord common to the projections of all the meridians. If at its middle point r, the perpendicular r1 be erected, this will contain the centres of all these projections. If through n, nt be drawn making rnl - 30U, it will be the radius of the projection of that meridian which makes an angle of 30~ with the equinoctial colure, since rn is the radiius of the projection of this colure Art. (211); and I is the centre of the projection of the required meridian, and ynz the projection. Toproject a parallel of latitude, as the Arctic Circle; lay off Ei = 47~, the sum of the inclination, 23~0, and the polar distal.ct 23-,o Art. (207). Draw E'i; Co = tan ~Ei, and o is one cx tremity of a diameter of the projection. Since the inclination is equal to the polar distance, the other extremity is at C, Art. (207) and the circle on Co, as a diameter, is the required projection. SPH ERIOAL PROJECrI'O.NS. 129 For the Tropic: of cancer; lay off Ep = 23~1 + 66-~; l)p is the tangent of half Ep, and p one extremity of a diameter of the projection. From k lay off kh = 66~-~ the polar distance of the parallel. Then Eh = 43~ equal the difference between the polar distance and the inclinatiol, and Cv is the tangent of its half, and v the other extremity of the diameter, and the circle on pv, the required projection. Tl'he projection of this tropic is tangent to the ecliptic at p, Art. (65). 216. Since each point on the hemisphere, below the primitive plane, has a greater polar distance than 90~ and will therefore be projected without the primitive circle, Art. (205), and those circles near the point of sight will thus be projected into very large circles, we make a more natural representation of this hemisphere by revolving it 180~, as in the orthographic projection, about a tangent at E, the point of sight being moved to the pole of the primitive circle in its new position. The hemisphere is then above the primitive plane and is projected as in the preceding article; s being the projection of the south pole; Em"E" of the other half of the equator; EsE" of the equinoctial colure, &c. 217. If the projection be made on any other great circle than the ecliptic, as on that making with the equator an angle denoted by A, the construction will be the same, the angle A being used instead of 23-~. If on the horizon of a place, A must equal 90~ minus the latiitude. Art. (193). 218. If the projectior, be made on the equator, the preceding problem is much simplified. Thus let EpE'q, Fig. 96, be the equator, E and E' the equinoctial points. EE' is the intersection of the plane of the ecliptic with that of the equator and pq is its 130 DESCRIPTIVE GEOMET1RY. line of measures and EoE' its projection, m being the centre and mn - tan 231~. Since the meridians pass through the point of sight they are projected into right lines. EE' is the projection of the equinoctial and pq of the solstitial colure; and ynz of the meridian which makes an angle of 30~ with the solstitial colure, n being the pro jection of the north pole. The parallels of latitude, being parallel to the equator, are projected as in Art. (208); the Arctic Circle into arb; and the Tropic of Cancer into dof; na being equal to tan ~(23~~'); and nd = tan 1(661). 219. If the projection be on the solstitial colure; let ENE'S, Fig. 97, be the primitive circle; EE' its intersection with the plane of the equator. The Equator, being perpendicular to the primitive plane, passes through the point of sight and is projected into EE'. The Ecliptic, for the same reason, is projected into oo', oCE being equal to 23~~. NyS is the projection of the meridian, making with the primitive plane an angle of 30~, m being its centre and Cm — = tan 30~. adb and a'hb' are the projections of the polar circles, Cx and C;e' being each equal to the secant of 23~-~, and xb and x'b' each equal to the tangent of 23O~, Art. (213). ogf and o'g'd' are the projections of the tropics, each described with a radius equal to the tan 66~. GLOBULAR PROJECTIONS. 220. By an examination of an orthographic or stereographic projection, it will be observed that the projections of equal arcs, of great circles which pass through the pole of the primitive circle, are very unequal in length. In the orthographic, as the arc is re SPHERICAL PRO()JECTIONS. 131.moved from the pole its projection is diminished, and when neai the primitive circle becomes very small, while the reverse is the case in the stereographic. To avoid this inequality,.as far as possible, the point of sight is taken in the axis of the primitive circle, without the surface, and at a distanicefrorn it equll to the sine of 450 -R. Spherical projections, with this position of the point of sight are called Globular. Thus let the quadrant Ep, Fig. 98, be bisected at M, and S the point of sight. M is projected at m, and Cm will be equal to mE For we have the proportion: oS: oM:: CS: Cm; whence, Cm= — _S Om (1). Since oM = oC = qS = R/lR, we have oS= R + 2R/ and CS= R + R/ Substituting these in (1), and reducing, we have Cm = R - mE; and it will be found that the projections of any other two equal arcs of this quadrant are very nearly equal. This is the only advantage of this mode of projection, as the projections of the circles of the sphere, being the intersections of their projecting cones with the primitive plane will, in general, be ellipses. GNOMONIC PROJECTION. 221. If the sphere be projected on a tangent plane at any point, 132 DESCRIPTIVE GEOMETRY. the point of sight being at its centre, the projection is called Gnomonic. In this case the projections of all meridians are right lines, since their planes pass through the point of sight. If the point of contact be on the equator the projections of the parallels of latitude will be arcs of hyperbolas, Art. (165). If the point of contact be at either pole of the earth, these projections will be circles. By this mode of projection, the portions of the sphere distant from the point of contact will be very much exaggerated. CYLINDRICAL PROJECTION. 222. If a cylinder be passed tangent to a sphere along the equator, and the point of sight be taken at the centre of the sphere, and the circles of the sphere be projected on the cylinder, and the cylinder be then developed, we have a developed projection called the cylindrical pr(jectioi,. In this case the meridians will be projected into right lines, elements of the cylinder, which are developed into parallel lines perpendicular to the developed equator, Art. (161); and the parallels into circles which are developed into right lines perpendicular to the developed meridians, and at distances from the equator each equal to the tangent of the latitude of the parallel. CONIC PROJECTION. 223. If a cone be passed tangent to a sphere along one of its parallels of latitude, and the circles of the sphere be projected oi it, the point of sight being at the centre, and the cone be then developed, we have a developed projection called the conic projection. In this case the meridians will be projected into right lines. SPHERICAL PROJECTIONS. 133 elements of the cone, which are developed into riglht lines passing thlrough the vertex; and the parallels into circles whose develop,ments will be arcs described fr'om the vertex. each with a radius equal to the distance of the projection from the vertex, Art. (1(6.) Thus, in Fig. 99, let EPE' be a section of the sphere an(l V the vertex of the cone tangent along the parallel of which ab is the orthographic projection. Tlhe equator will be projected into a circle whose diameter is ee' and the. parallel MN into one whose diameter is mnn, the first being developed into an arc of which Ve is the radius, and the second into one of which Vm is the radius. The radius Va of the development of the circle of contact is evidently the tangent of its polar distance. The drawing in this, as in the cylindrical projection, for those portions of the sphere distant from the circle of contact, will evidently greatly exaggerate the parts projected. 224. This exaggeration may be lessened by making the proJection on a none passing through two circumferences equally distant, ore from the equator and the other from the pole. Thus, let ab and cd, Fig. 100, represent two circles equally distant from EE' and P, Ea being one fourth of the quadrant; in this case while tht parts Ea and cP will be exaggerated in projection, the part ac will be lessened. 225. When a small portion of the surface, between two given parallels, is to be represented, the conic method nay be well used, taking the cone tangent to the sphere along a parallel midlway between the given parallels, if the first method be used; or passing through two parallels, each distant from a limiting parallel one fourth the are of the given portion, if the second method b used. 134 DESCRIPTIVE GEUMETRY. CONSTRUCTION OF MAPS. 226. If it be desired to represent the entire surface of the earth by a map, either of the preceding methods may be used. In this case it is usual to divide the quadrant, from the pole to the equator, into nine equal parts and project the parallels of latitude through each of these points, as well as the polar circles and tropics; and also to divide the semi-equator into twelve equal parts and to project the meridians passing through these points. These meridians will be 15~ apart and are called hour-circles. The projections of the different points to be represented are then made and the map filled up in detail. 227. The Stereographic projection gives the most natural representation and, in general, is of the easiest construction. In the Globular, when the equator is taken as the primitive circle, the projections of the meridians are right lines, and of the parallels of latitude, circles; and this projection has the advantage that these parallels, which are equally distant in space, have their projections also very nearly equally distant. 228. A very simple construction, when the primitive circle is a meridian, is sometimes made thus: Divide the arcs EN, E'N, ES, and E'S, Fig. 101, each into nine equal parts, and the radii CN and CS also each into nine equal parts, then describe arcs of circles through each of the three corresponding points of division, for the representatives of the parallels. In the same way divide CE and CE', each into six equal parts, and, through the points of division and the poles N and S, describe arcs of circles for the representatives of the meridians. This representation, though called the equidistant projection, is SPHIERICAL PROJECTIONS. 13 t not strictly a projection. It differs little, however, from the globular projection. LORGNA'S MAP, 229. A map of the globe is sometimes made by describing a primitive circle witb a radius equal to R v/2, R being the radius of the sphere, and regarding this as the representative of the equator. Through its centre draw right lines making angles with each other of 150, for the meridians. To represent the parallels: With the centre of the primitive circle as a centre and with a radius equal to v/2Rh, h being the altitude of the zone, included between the pole and the parallel, describe a circle to represent each parallel in succession. The area of the primitive circle is evidently equal to the area of the hemisphere, and the area of each other circle to that of the corresponding zone. Hence, the area between any two circunmferences will be equal to that of the zone included between the corresponding parallels. Also the area of any quadrilateral formed by the arcs of two meridians and two parallels will be equal to its representative on the primitive plane. MERCATOR S CHART. 230. This chart, which is much used by navigators, is a modifi. cation of the cylindrical projection. It has the great advantage, that the course of a ship on the surface of the sphere which makes a constant angle with the meridians intersected by it, will be repre.sented on the chart by a right line. To secure this, as the length of the representative of a degree of longitude, as compared with the arc itself, is manifestly in creased as the distance from the equator increases, it is necessary 1%6 DESCRIPTIVE GEOSIETRY. that the representatives of the degrees of latitude should increase in the same ratio. But the length of a degree of longitude, at ally latitude, is known to be equal to the length of a degree at the equator rmultiplied by the cosine of the latitude, and since the representative of this degree at all latitudes is constant on the chart, being the dis tance between two parallel lines the representatives of two consecutive meridians, it follows that. as we depart from the equator, this representative, as compared with the arc itself; increases as the cosine of the latitude decreases, or increases as the secant increases, and hence the representative of the degree of latitude must increase in the same ratio; that is, this representative, at any given latitude, must equal its length at the equator multiplied by the natural secant of this latitude. By adding the representatives of the several degrees, or, still more accurately, of the several minutes of a quadrant, the distance of the representative of each parallel fiom the equator may be found, and the chart may then be thus constructed. l)raw a right line to represent the equator, then a system of equidistant parallel lines tbr the meridians; on either one of these lay off the proper distances computed as above for. each parallel to be represented, and through the extremities of these distances draw right lines perpendicular to the system first drawn. They represent the parallels. 231. All the maps constructed as in the preceding articles, though giving a general representation of the relative position of objects on the earth's surface, are defective in this, that there is no definitee relation between actual distances of points and the repreI - sentation of these distances on the maps, so that there cal}n be no scale on the map by which these actual distances can be deter. mined. As In detailed representations of smaller portions of the surface, this scale is absolutely necessary, other modes of constructing SPHFERICAL PROJECTToNS. 1 37 these maps have been devised, by which a near approximatioi: to an accurate scale is made. FLAMSTEAD S METHOD. 232. In this method, modified and improved, and now in very general use, a right line, AB, Fig. 102, is drawn, which represents the rectified arc of the meridian passing through the middle of the portion to be mapped. A point, c, is then assumed to represent the point in which the parallel midway between the extreme parallels intersects this meridian, and from this point, in both directions, equal distances, cx, cy, &c., are laid off, each representing the true length of one degree of the mneridian. Then with a radius equal to the tangent of the polar distance of this central parallel, the are dee is described to represent the parallel. Thlis are is the development of the line of contact of a cone tangent to the sphere. Also, with the same centre, arcs are described through each of the points of division x, y --- r, o, to represent the other parallels one degree distant from each other. Then, on each of these arcs, from AB, lay off both ways the arcs ca, ca', yv,, yi', 0o, os', &c., each equal to the length of a degree of longitude at the points c, o, &c., viz, the length of a degree at the equator multiplied by the natural cosine of the latitude of the point. Through the points a, v, s, &c., and a', v', s', &c., draw the lines avs, and a'v's'. They will represent the two meridians making an angle of one degree each, with the central meridian. In the same way, the representatives of the two meridians next to these may be constructed, by laying off on the same arcs from a, v, s, &c., distances each equal to ca, yv, as, &c., and drawing lines through the points thus determined, and so on, until the representatives of the extreme meridians of the portion to be mapped are drawn. In this map the scale along the central meridian and parallels will be acculrate. In other directions when the mlap does not re(i 138 DESCRIPTIVE GEOMETRY. resent a very extended portion of the sphere, a very near approximation to accuracy is made. THE POLYCONIC METHOD. 233. In this method, by which the elegant and accurate maps of the U. S. Coast Survey are constructed, the central meridian and parallel are represented as in the preceding article. The representatives of the other parallels are each described, through the proper point of division, with a radius equal to the tangent of its polar distance, thus giving the development of the lines of contact of so many tangent cones to the sphere. The length of each degree of longitude is then laid off as in the former method and the representatives of the meridians drawn. In this method, also, the scale, along the central meridian and parallels, is accurate and in other directions very nearly so. This has the advantage that the representatives of the meridians and parallels are perpendicular to each other as in space, which is not. the case in Flamstead's method. 234. When very small areas are mapped this method is thus modified in the coast survey office. The same process is used as above to construct the representatives of the meridians with accuracy. Then commencing with the central parallel, the distance cx, Fig. 102, between it and the consecutive one, as measured on AB, is laid off on each meridian in both directions and through the extremities lines drawn to represent the consecutive parallels. Then from these the same distances are laid off for the next and so on until all are constructed. The first set of parallels, described,as in the preceding article, which should be in pencil, are then erased. By this method equal meridian distances are everywhere in-luded between the parallels, and the scale accurate only in the SP H;RICAL PROJECTI()NS. 139 direction of the meridians, and central parallel. This is called the equidistant polyconic method. 235. When the polar distance of the parallel is much greater t'an 45~, the practical construction of its representative becomes difficult, as its centre will be so far distant. In such case, for both the polyconic method and that of Flamstead, tables are carefully computed giving the rectilineal co-ordinates of the points of the representatives of the parallels, for each minute of latitude and longitude, and these representatives can then be accurately constructed by points. The tables thus computed in the U. S. Coast Survey office are very much extended and of great value. PART III. SHADES AND SHADOWS. PRELIMINARY DEFINITIONS. 236. To represent a body with accuracy, it is not only necessary that the drawing should give the representations of the details of its form, but also of its colors, whether natural or artificial, Jr the effect of light and shade. Different portions of the same body will appear lighter or darker according as the light falls directly upon it or is excluded from it, by itself or some other body. A simple application of the elementary principles previously deduced will enable us to limit and represent these portions, and constitutes that part of the subject called Shades and Shadows. 237. It is a principle of Optics, that the effect of light, when in the same medium and unobstructed, is transmitted from each point of a luminous body in eyery direction, along right lines. These right lines are called rays of light. Any right line, therefore, drawn from a point of a luminous body will be regarded as a ray of light. The sun is the luminous body which is the principal source of light. It is at so great a distance, that rays drawn from any of its points to an object on the earth's surface, may, vithlout mate SHADL:S AND SHADOWS. 141 rial error, be regarded as parallel. In the construction of problems, in this part of our subject, they will be so taken. 238. Let SR, Fig. 103, indicate the direction of the parallel rays of light, coming from the source and falling on the opaque body B; and let n-mlop-q be a cylinder, whose rectilinear elements are rays of light parallel to SR, enveloping and touching this body. It is evident that all light, coming directly firom the source, will be excluded from that part of the interior of this cylinder which is behind B. This part of space from which the light is thus excluded by the opaque body, is the indefinite shadow of the body; and any object within this portion of the cylinder is in the shadow, or has a shadow cast upon it. The line of contact mnlop separates the surface of the opaque body into two parts. That part which is towards the source oi light, and on which the rays fall directly, is the illuminated part; and that part opposite the source of light from which the rays are excluded by the body itself, is the shade of the body. This line of contact of the tangent cylinder of rays, thus separating the illuminated part from the shade, is the line of shade;. Any plane tangent to this cylinder of rays will also be tangent to the opaque body at some pointof this line of contact, Art. (116); and, conversely, every plane tangent to the opaque body at a point of the line of shade will be tangent to this cylinder, and contain a rectilinear element, Art. (110), or ray of light, and thus be a plane of rays. Hence points of the line of shade on any opaque body may be obtained by passing planes of rays tangent to the body, and finding their points of contact. 239. If the cylinder n-mlop-q be intersected by any surface, as the plane ABCD, that part m'l'o'p' of this plane, which is in the shadow, is the shadow of B on this plane. That is, the shadow of an opaque body on a surface is that part of the surface froml 142 DESCRIPTIVE GEOMETRY. which the rays of light are excluded by the interposition of this body between it and the source of light. The bounding line of this shadow, or the line-which separates the shadow on a surface from the illuminated part of that surface, is the line of shadow. It is also the line of intersection of the cylinder of rays which envelops the opaque body, and the surface on which the shadow is cast. 240. When the opaque body is bounded by planes, the cylinder of rays, Kj which its shadow is determined, will be changed into several planes of rays, which will include the indefinite shadow; and the line of shade will be made up of righlt lines which, though not lines of mathematical tangency, are the outer lines in which these planes touch the opaque body, and still separte the illuminated part from the shade. The line of shadow, in this case, is also made up of the lines of intersection of these planes with the surface on which the shadow is cast, and still separates the illuminated part of this surface from the shadow. SHADOW OF POINTS AND LINES. 241. The indefinite shadow of a point may be regarddl as that part of a ray of light, drawn through it, which lies in the direction opposite to that of the source of light, and the point in which this ray pierces any surface is the shadow of the point on that surface. 242. The shadow of a right line will then be determined by drawing through its different points rays of light. These form a plane of rays, and the intersection of this plane with any surface if the shadow of the line on that surface. SHADES AND SHADOWS. 143 To determine whether a given plane is a plane of lays, it is only necessary to ascertain whether it contains a ray of light. This may be done by drawing through any one of its points a ray. If this pierces the planes of projection in the traces of the given plane, it is a plane of rays, Art. (30). 243. The shadow of a right line on a plane may be constructed by finding the shadows cast by ally two of its points on the plane, Art. (241), and joining these by a right line; or by joining the shadow of any one of its points with the point in which the line pierces the plane. Art. (30). If the right line be parallel to the plane, its shadow on the plane will be parallel to the line itself, and the shadow of a definite portion of it will be equal to this portion, Art. (14). If the line is a ray of light, its shadow on any surface will be a point. Thus let MN, Fig. 104, be a limited right line, and tTt' the plane on which its shadow is required, MR indicating the direction of the rays of light. Throltgh the extremities M and N, draw the rays MR and NS. They pierce tTt' in R and S, Art. (40), which are points of the shadow, Art. (241). Join these points by RS, it will be the requiredl slhldow. If the shadow be cast on the horizontal plane, the rays through the extrelmities M and N, Fig. (a), pierce H at m" and n", and m"n "" is the shadow. If the shadow be cast on the vertical plane, these rays pierce V at min" and n", Fig. (b), and m"n" is the shadow. If MN is parallel to either plane, then RS in Fig. 104, and m, "z" in Figs. (a) and (b) will be equal to MN. If right lines are parallel, their shadows on a plane are parallel, since they rec the intersections of parallel planes of rays by this plane. [44, DESCRIPTIVE GEOMETRY. 244. The shadow of a curve will be determined by passing through it a cylinder of rays. The intersection of this cylinder with any surface will be the shadow of the curve on that surface. The line of shadow on a surface, Art. (239), will always be the shadow of the line of shade; and if through any point of the line of shadow a ray be drawn to the source of light, it will intersect the line of shade in a point which casts the assumed point of shadow. From this it follows that the point of shadow cast by any line in slpace upon any other line, may be found by constructing the shadows of both lines on a plane; and drawing through the point of intersection of these shadows a ray, it will intersect the second line in the point of shadow cast upon it, by the point in which it intersects the first. 245. If two lines are tangent in space, their shadows, on any surface, will be tangent at the point of shadow cast ly the given point of contact. For the cylinders of rays through the Hilies will be tangent along the ray passing throulgh the point of contact, Art. (117); hence their intersections by any surface will be tangent at the point where this ray pierces the surface. These intersections are the shadows of the lines. 246. The shadow of a curve of single curvature, on a plane to which it is parallel, will be an equal curve, since each of its elements will cast a parallel and equal element of the shadow, Art. (243). If the plane of the curve be a plane of rays, its shadow on a plane will evidently be a right line. The shadow of the circumference of a circle, on a plane to which it is parallel, will be an equal circumference, whose centre is the shadow cast by the given centre. SHAI)ES AND S-IADOWz. 145 Also, when the plane on whiclh the shadow is cast makes a subcontrary section in the cylinder of rays through the circumferezlce. Art. (206). In all other positions, when its plane is not a plane of rays, its shadow will be an ellipse, Art. (160), and any two diameters of the circle per7pendicular to each other will cast conjugate diameters of the ellipse of shadow. For if at the extremities of either diameter tangents be drawn, they will be parallel to the other diameter, and their shadow parallel to the shadow of this diameter, Art. (243). But the shadows of the two tangents are tangent to the shadows of the circle, at the extremities of the shadow of the first diameter, Art. (245); hence the shadow of the second diameter is parallel to the tangents at the extremities of the shadow of the first. These shadows are therefore conjugate diameters of the ellipse, and with them the ellipse may be constructed. Ana. Geo., Art. (150). If the shadows of two perpendicular diameters be found, also perpendicular to each other, they will be the axes of the eilipse, which may be constructed as in Art. (59). BRILLIANT POINTS. 247. In looking upon the illuminated part of a curved surface, it will be observed that, in general, one or more points appear much more brilliant than the others. This is due to the fiAt that the ray of light, incident at each of these points, is reflected zmmediately to the point of sight. These points are called brilliant points; and since it is a princil)le of Optics that the incident and reflected rays, at any point of a surface, lie in the same normal plane on opposite sides of the normal at this point, and making equal angles with it, it follows that at a brilliant point, the ray of light and the right line drawn to the point of sight must fulfil these conditions. When the point of sight is at an infinite distance, as in the 1l 146 DEISClIPTIVE GEOMETRY. Orthographic projection, the brilliant point may be constructed thus: Through any point draw a ray of light and a right line to the point of sight. These will be parallel, respectively, to the corresponding lines at the required point, and make the same angle. Bisect the angle formed by these auxiliary lines as in Art. (37). The bisecting line will be parallel to the line which bisects the parallel and equal angle at the brilliant point. Hence, perpendicular to this bisecting line, and tangent to the stuface, pass a plane. Its point of contact will be the brilliant point; for a line through this point parallel to the bisecting line will be a normal to the surface, and will bisect the angle formed by a ray of light and right line drawn from this point to the point of sight. PRACTICAL PROBLEMS. 248. PROBLEM 61. To construct the shadow of a rectangutat pillar, on the planes of projection. Let vnnlo, Fig. 105, be the horizontal and mn'n'p'q' the vertical projection of the pillar, and let (7Xn", n'r) indicate the direction of the rays of light.. The upper base vertically projected in m',n', and the two vertical faces horizontally projected in mn and mo, are evidently illuminated, and together form the illuminated part of the pillar; and the edges (n, p'n'), (nl, n'), (lo, n'mn'), and (o, q'm') separate the illuminated part fromn the shade, and mlake up the line of shade, Art. (240). Since n is in the horizontal plane and n" the shadow of N, nn" is the shadow of (n, p'n') on H, Art. (243). The plane of rays through (nl, a') is perpendicular to V, and n"r, parallel to nl, is its horizontal and rl" its vertical trace; hence n"r is that part of the shadow of (nl, n') which is on H; andrl/", limited by the ray through (In'), that part on V. The ray (rx, rn') intersects (nl, a') in (xn'), and nx is the horizontal projection of that part which casts the equal shadow n"r, Art. (243), and xl of that part whi,-h casts rl". SHADES AND SH A nOWS. 147 l"o" is the shadow of (ol, nm'n') on V, parallel and equal to itself. The plane of rays through (o, q'm') is perpendicular to 11, and os is its horizontal and so" its vertical trace; and os the shadow of (o, q'rm') on H, and so" its shadow on V, q'y' is the vertical protection of the part which casts the shadow os and y'm' of that part which casts its equal so", Art. (243). The line of shadow is thus the broken line nn"r...o, and the portion o,! the planes within this line is the required shadow, and should be darkened as in the drawing. 249. PROBLEM 62. To construct the shadow of a rectangular abacus on the faces of a rectangular pillar. Let mnulo, Fig. 106,'be the horizontal, and m'n'l'o' the vertical projection of the abacus, and cdd'c' the horizontal, and c'd'e'f' the vertical projection of the pillar; MR indicating the direction of the rays. The lines of shade on the abacus, which cast the required shadows, are evidently the two edges (mno, m') and MN. The plane of rays through (mo, m') is perpendicular to V, and intersects the side face, horizontally projected in cc', in a right line perpendicular to V at s', which is the shadow on this face. m'r' is the vertical trace of this plane. The ray MR, through M, pierces the front face of the pillar in R, Art. (40), the shadow of M, and (cr, s'r') is the shadow of the part (opm, m') on this face. MN being parallel to this face, its shadow on it will be parallel to itself, and pass through. R; hence (rd, r'x') is the required shadow cast by its equal MQ. 250. PROBLEM 63. To construct the shadow of an upmzght cross upon the horizontal plane and upon itself. Let mnop, Fig. 107, be the horizontal, and c'n'r'#gf' the vertical projection of the cross. cc" is the shadow of the edge (c, c'd'), c" being the shadow of 148 DESCRIPTIVE GEOMETRY. (cd'); c"'n" is the shadow of its eqnal (cn, d'n'), Art. (243); n"h" of the edge (n, n'h'); h"o" of its equal (no, h'); o"y" of (oy, h'x'); (xy, x') of a part of (cd, r'), on the face of the cross vertically projected in lh'. y"r." is the shadow of the remaining part of (cd, r') on 11. (cx, I'x') is the shadow of (c, l'r') on the face vertically pro-ected in l'h', this being the intersection of a vertical plane of rays through (c, l'r') with this face. r"q" is the shadow of its equal (de, r'q'); q"k" of (e, k'q'); k"9p" of (pe, k'Vy'); p"m" of (p, m'g'); and m"t of a part of (pm, m'). All within the broken line thus determined on I is darkened, as also the part cxyd, the horizontal projection of the shadow cast by the upper part of the cross on the face vertically projected in l'h'. 251. PROBLEM 64. To construct the shade of a cylindrical column, and of its cylindrical abacus, and the shadow of the abacus on the column. Let mlo, Fig. 108, be the horizontal, and m'o'h'n' the vertical projection of the abacus, cde the horizontal, and cee'g' the vertical projection of the column. Pass two planes of rays tangent to the column. Since each contaitns a rectilinear element, they will be perpendicular to H; and Id aud kf, parallel to the horizontal projection of the ray of light, will be their horizontal traces, Art. (123); and (d, z'd!) and (f, f'r') the elements of contact, which are indefinite lines, or elements o0 shade, Art. (238.) In the same way the elements of shade (i, u'i'), &c., on the abacus, are determined. The line of shadow on the column will be cast by a portion of the lower circumference of the abacus towards the source of light. To determine any point of it, pass a vertical plane of rays, as that whose horizontal trace is yx. It intersects the column in a rectilinear element (x, x"x'), and the circumference in a point Y. Through this point draw the ray YX. It intersects the element in X, a point of the required shadow, Art. (241) ~ and in the same SHADES AND SHADOWS. 149 way any number of points may be found. C is the shadow of Q, and D of L, and c'x'd' is the vertical, and cxd the horizontal projection of that part of the curve of shadow which can be seen. That part of the shade on the abacus and column, and of the shadow on the column, which can be seen, is darkened in the drawing. 252. PROBLEM 65. To construct the shade of a right cylinder with a circular base, and its shadow on the horizontal plane and on its interior surface. Let mlo, Fig. 109, be the base of the cylinder, and (c, d'c') its axis. The two planes of rays, whose horizontal traces are 11" and kk", determine the two elements of shade (1, n'l') and (k, r'k'). These elements cast the shadows 11" and kk" on H, Art. (243). The semi-circumference of the upper base LUK casts its shadow on the interior of the cylinder, and the other semi-circumference on the horizontal plane, without the cylinder. The shadow of the latter is the equal semi-circumference k"o"l" whose centre is c"', Art. (246), and this, with the lines 11" and kk", limits the shadow of the cylinder on H. To determine the shadow on the interior surface, pass any vertical plane of rays, as that whose horizontal trace is yx. It intersects the cylinder in the element (x, x"x') and the semi-circumference in the point Y. The ray of light YX intersects (x, x"x') in X, a point of the required shadow; and in the same way any number of points may be determined. The shadow upon any rectilinear element, may be found by using an auxiliary plane which shall contain this element. Thus Z is the shadow of U on the element (z, z"z'), s" of S, and t" of T. This shadow evidently begins at L and K, the upper ends ol the elements of shade. The direction of the rays is so taken that a part of this semi. 150 DESCRIPTIVE GEOMETRY.'. circumference casts its shadow on H, within the'cylinder. This part is horizontally projected in st, and its shadow is the equal arc s"t" the continuation of k"o"l". Should this shadow not reach H, its lowest point will be obtained by using as an auxiliary plane that one which contains tlhe axis; and the point in which the vertical projection of the curve of shadow is tangent to op', by using the plane which contains the element (o, o'p'). That part of the shade between the elements (1, n'l') and (o, o'p') only, can be seen when looking on V, and is darkened in the drawing. 253. PROBLEM 66. To construct the shade of an oblique cone and its shadow on the horizontal plane. Let the cone be given as in Fig. 110, S being its vertex. If two planes of rays be passed tangent to the cone, the elements of contact will be the elements of shade, Art. (238). Since these planes must pass through S, Art. (130). they must contain the ray of. light SR. This pierces H at r, and rn and rp are the horizontal traces of these tangent planes, and SN and SP are the elements of contact, Art. (130). These elements cast the shadows nr and pr. which limit the shadow of the cone on H. In looking upon H, the shade between the elements SP and SQ only is seen; and in looking upon V, that between SN and SO. 254. PROBLEM 67. To construct the shade of an ellipsoid oj revolution and its shadow on the horizontal plane. Let the ellipsoid be given as in Fig. 111. The line of shade is the line of contact of a cylinder of rays tangent to the surface, and is an ellipse, Ana. Geo., Art. (233); and since the meridian plane of rays, whose horizontal trace is cs, is evidently a principal plane, Art. (206), of both the ellipsoid SHADES AND SHADOWS. 151 and cylinder, bisecting all chords of both surfaces perpendicular to it, its intersection with the plane of the ellipse will also bisect all chords of the ellipse perpendicular to it, and be an axis of the curve. This meridian plane cuts from the ellipsoid a meridian curve, and from the cylinder two rectilinear elements, rays of light, tangent to this curve, at the vertices of the axis, the hiyhest anda lowest pvoints of the curve of shade. To determine these points, revolve the meridian plane about (c, e'd'), until it becomes parallel to V. The meridian curve will then be vertically projected into e'm'd'mu'; CT will be the revolved position of CS, the axis of the cylinder, and the two tangents at r' and x', parallel to c't', will be the vertical projections, in revolved position, of the two elements cut from the cylinder, and R and X will be the revolved position of the two vertices. When the plane is revolved to its true position, these points will be at P and Y, Art. (107), and PY is the transverse axis of the required curve. The conjugate axis, being perpendicular to the meridian plane of rays, is parallel to H, and therefore horizontally projected into fg, perpendicular to yp, Art. (36), and vertically into f'q'. The projections of this curve are ellipses. Of the horizontal projection, fg is evidently the longest diameter, and therefore the transverse axis, and yp the conjugate, and the ellipse fvgy is the horizontal projection of the curve of shade. Since the tangents to the curve at the highest and lowest points P and Y are horizontal, their vertical projections imust be r'p' and x'y' tangent to the vertical projection of tihe curve at p' and y'. These tangents being parallel to f'g', p'y' and fig' must be conju-gate diameters of the ellipse which is the vertical projection of the curve; hence p'f'y'g', on these diameters, is the required vertical projection. Points of this curve may also be found otherwise, by intersecting the surface by any horizontal plane, as that of which i'k' is the vertical trace. It cuts from the surface a circumference, 15.D 2 DESCRIP TIVE GEOMETRY. and from the plane of the curve a line parallel to FG. These lines intersect in the required points. K and L are thus determined. The points of tangency, u' and w', are the vertical projectiolls of the points in which the curve of shade is intersected by the meridian plane parallel to V. The shadow cast by the curve of shade on H is an ellipse, the intersection of the tangent cylinder with H. The conjugate axis of this ellipse,f'g", is the shadow of FG parallel and equal to itself, and the semi-transverse axis, y"s, the shadow of CY, since these shadows are perpendicular to each other, and y"s bisects all chords perpendicular to it. 255. If the ellipsoid becomes a sphere, the construction of the curve of shade is simplified, as it then becomes the circumference of a great circle perpendicular to CS, Art. (121), and fy is the projection of that diameter which is parallel to H, and py the projection of the one which is perpendicular to it, Art. (197), and the vertical projection may be obtained in the same way. 256. If a plane of rays be passed tangent to any surface of revolution, the poin!t of contact will be a point of the curve of shade, Art. (238). If through this point of contact a meridian plane be passed, it will be perpendicular to the tangent plane, Art. (115), and cut from it a line tangent to the meridian curve at this point, Art. (108). The plane, which projects a ray of light on to this meridian plane, is also a plane of rays perpendicular to it, and therefore parallel to the tangent plane, and its intersection with the meridian plane will be parallel to the tangent cut from the tangent plane. But this intersection is the projection of the ray. Hence, if a tangent be drawn to any meridian ctarve of a surface of revolution parallel to the projection of a ray of light on the meridian plane, its point of contact will be a point of the line of SHADES AND SHAD(,WS. 1.53 shalde, and thus any number of points may be found on different meridians, and the line of shade drawn. The points U and W, in Fig. 111, may thus be determined by drawing tangents to the meridian curve which is parallel to V, pa'rallel to c's'. 257. PROBLEM 68. To find the brilliant point on any surface of revolution. Let the surface be an ellipsoid, given as in Fig. 111. Through any point of the axis as C, draw the ray of light CS and the line (co, c') to the point of sight in front of the vertical plane. Bisect their included angle as in Art. (37) by the line CQ. To determine a plane perpendicular to this line and tangent to the surface, Art. (247), through it pass the meridian plane of which eq is the horizontal trace. This plane cuts from the surface a meridian curve, and from the required tangent plane a right line tangent to this curve at the required point, and perpendicular to CQ. Revolve this plane about (c, c'd') until it becomes parallel to V. The meridian curve will be vertically projected in e'n'd'm', and CQ in ciq"', Art. (29). Tangent to e'rn'd' and perpendicular to c'q"' draw i'z"'; z"' is the vertical projection of the revolved position of the required point, horizontally projected at z", and when the plane is revolved to its true position at z. Z is then the required brilliant point.'258. PROBLEM 69. To construct the shade and shadow of an upright screw. Let ecf, Fig. 112, be the circular base of a right cylinder whose axis is oc', and let em'e' be an isosceles triangle in the plane of the element ee' and of the axis, its base coincident with the element. If this triangle be moved about the cylinder, its plane always containing the axis, with a uniform angular motion, and at the same time with a uniform motion in the direction of the axis, so 154 DESCRIPTIVE GEOMETRY. that after passing around once it will occupy the position e'm"e", -it will generate a volume called the thread of a screw. The cylinder is the cylinder of the screw. It is evident that the two sides m'e and m'e' will each generate a portion of a helicoid, Art. (92), the side m'e' generating the upper surface of the thread, and m'e the lower surface. The point m' generates the outer helix of the thread, and the point e the inner helix. The curve of shade on the lower surface of the thread may be constructed by passing planes of rays tangent to this surface, and joining the points of contact by a line, Art. (238). To find the point of this curve on the inner helix, pass a plane tangent to the lower helicoid at F; gd is the horizontal trace of this plane, the distance fd being equal to the rectified arc fee, Art. (139). It is a property of the helicoid that all tangent planes to the surface, at points of the same helix, make the same angle with the axis, or with the horizontal plane, when taken perpendicular to this axis. For each of these planes is determined by a tangent to the helix, and the rectilinear element through the point of contact; and these lines at all points of the helix make the same angle with the axis, Arts. (69) and (92), and with each other. Hence, if through the axis we pass a plane perpendicular to the tangent plane at F, it will cut from this plane a right line, intersecting oc' at o", horizontally projected in oi, and making with H the same angle as that made by a plane tangent at any point of the helix, and with the rectilinear element passing through the point of contact all angle, which is the same for all these planes; hence the angle iof will be the same for all these planes. If this line be now revolved about oc', it will generate a cone whose rectilinear elements make the same angle with H as the tangent plane. If a plane of rays be passed tangent to this cone, it will be parallel to the plane of rays tangent to the surface at a point on the inner helix. kl, tangent to the circle generated by oi, is the horizontal trace of this plane, Art. (131), and ol is the SHADES AND SHIADOWS. 155 horizontal projection of the line cut from the parallel tangent plane of rays by a perpendicular plane through oc'. If, then, we make the angle lox equal to iof ox will be the horizontal projection of the element containing the required point of contact, and X will be the point. In the same way, the point Y on the outer helix may be obtained, by first passing a plane tangent to the lower surface at M. eq is its horizontal trace, mq being equal to the rectified semi-circumference myn, and eop the constant angle. A tangent fr,)m t to the circle generated by op (so nearly coincident with ecf that it is not drawn) is the horizontal trace of the parallel plane of rays, and oy, making with the right line joining o and the point of contact of this tangent, an angle equal to eop, the horizontal projection of the element containing the point of contact, and Y the required point. Intermediate points of the curve qf shade may be determined by constructing, as above, the points of contact of tangent planes of rays on intermediate helices. Thus, pass a plane tangent to the lower helicoid at W, midway between F and N; wbz is the horizontal projection of the helix passinog through this point; fb" the, horizontal projection of the intersection of the tangent plane with the horizontal plane through f', wb" being equal to the rectified arc wb. Since this intersection is parallel to the horizontal trace of the tangent plane, ou will be the horizontal projection of the line cut from the tangent plane by the perpendicular plane through oc', and fou the constant angle. us is the horizontal projection of the circle cut from the auxiliary cone by the horizontal plane through f'm', and ts the horizontal projection of the intersection of the tangent plane of rays with the same plane; oz making with the right line joining o and s, an angle equal to fou, is then the horizontal projection of the element containing the point ot contact, o"'v' Art. (92) its vertical projection, and Z the required point. As the surface, of which e'n"n'If" is the vertical projection, is in all respects identical with that of which em'rL'f' is the projec 156 DESCRIPTIVE GEOMETRY. tion, y"x" will be the vertical projection of the curve of shade on the first surface, its horizontal projection being yx. To construct the shadow cast by this curve of shade on the surface of the thread below it, construct the shadow of (xy, x"y") on the horizontal plane through n'e', under the supposition that the rays are unobstructed. x-"'y"' is the horizontal projection of this shadow, Art. (244.) Then assume any element, as (ao, za'o,), on which it is supposed the shadow will fall, and construct its shadow on the same plane. h"r" is the horizontal projection of this shadow. Through the point in which these shadows intersect, horizontally projected at r", draw a ray of light, intersecting the element in R, which is the shadow of the curve on the element, Art. (244.) The curve evidently begins at (x, x") and is vertically projected in x"r'. The shadow cast by the helix (inn, m"n") on the surface of the thread may be constructed in the same way by first constructing its shadow on the same horizontal plane as above, and the shadow of an assumed element (od, o06') on the same plane, and from their point of intersection horizontally projected at d", drawing a ray; (d, d') will be a point of the required shadow. The curve of shadow on the surface, whose vertical projection is enf', will be equal to the curve whose vertical projection is x"r'6', and may be drawn as in the figure. 259. The brilliant point on the upper surface of the thread may be constructed by bisecting the angle formed by a ray of light and a line drawn to the point of sight, Art. (247), and then passing a plane perpendicular to the bisecting line and tangent to the surface, as in Art. (140), and finding its point of contact PART IV. LINEAR PERSPECTIVE. PRELIMINARY DEFINITIONS AND PRINCIPLES. 260. It has been observed, Art. (204), that the orthographic projections can never present to the eye of an observer a perfectly natural appearance, and hence this mode of representation is used only in drawings made for the development of the principles, and the solution of problems in Descriptive Geometry, and for the purposes oi mechanical or architectural constructions. Whenever an accurate picture of an object is desired, the scenographic method must be used, and the position of the point of sight chosen, as indicated in Art. (204). 261. That application of the principles of Descriptive Geometry which has for its object the accurate representation, upon a single vlane, of the details of the form and the principal lines of a body, is called Linear Perspective. The art by which a proper coloring is given to all parts of the representation, is called Aerial Perspective. This, properly, forms no part of a mathematical treatise, and is therefore left entirely tc the taste and skill of the artist. 158 DESCRIPTIVE GEOMETRY. 262. The plane upon which the representation of the body is made is called the plane of the picture; and a point is represented upon it, as in all other cases, Art. (3), by drawing through the point and the point of sight a right line. The point in which it pierces the plane of the picture is the perspective of the given point. These projecting lines are called visual rays, and when drawn to the points of any curve, form a visual cone, Art. (77). Any plane, passing through the point of sight, is made up oI visual rays, and is called a visual plane. The plane of the picture is usually taken between the object to be represented and the point of sight, in order that tile drawing may be of smaller dimensions than the object. It is also taken vertical, as in this position it will, in general, be parallel to many principal lines of the object. It may thus be used as the vertical plane of projection, and will be referred to as the plane V, Art. (24). The orthographic projection of the point of sight, on the plane of the picture, is called the principal point of the picture; and a horizontal line through this point and in the plane of the picture, is the horizon of the picture. PERSPECTIVES OF POINTS AND RIGHT LINES. VANISHING POINTS OF RIGHT LINES. 263. Let M, Fig. 113, be any point in space, AB the horizontal trace of the plane of the picture, or ground line, and S the point of sight. The visual ray SM, through M, pierces the plane of the picture V, in m, Art. (40), which is the perspective of the point M. 264. The indefinite perspective of a right line, as MN, Fig. 114, may be drawn by finding the perspectives of any two of its points LINEAR PERSPECTIVE. 159 as in the preceding article, and joining them by a right line; oi by passing through the line a visual plane, and constructing its trace on the plane of the picture. The point n', in which this line pierces V, is one point of this trace, Art. (30). If through S a line SP be drawn parallel to MN, it will lie in the visual plane, and pierce V at p', a second point of the trace, and p'n' will be tile perspective of the line. The point p' is called the vanishing point of the line MN. Hence, to construct the perspective of any right line, join its vanishzng point with the point where the line pierces the plane of the picture, by a right line. 265. The vanishing point of any right line is the point in which a line, parallel to it through the point of sight, pierces the plane of the picture. This point, as seen in the preceding article, is always one point of the perspective of the line; and since the auxiliary line intersects the given line only at an infinite distance, this point is the perspective of that point of the given line which is at an infinite distance. The principal point is evidently the vanishing point of all perpendiculars to the plane of the picture. 266. Any horizontal right line making an angle of 450 with the plane of the picture is a diagonal. A right line through S, Fig. 113, parallel to a diagonal, pierces V in the horizon, and at a distance from the principal point s', (q-al to the distance of S froln the plane of the picture. Since Iwo Fnich lines can be drawn, one on each side of the perpendicular Ss', there will be two vanishing points of diagonals, one for those which incline to the right, and another for those which incline to the left. These points are also called points of distance, since when as 160 DESCRIPTIVE GEOMETRY. sumed or fixed, the distance of the point of sight from the plane of the picture is determined. 267. If a plane be passed through the point of sight parallel to a given plane, it will contain all right lines drawn through this poillt parallel to lines of the plane. Its trace on the plane of the picture will therefore contain the vanishing points of all these lines, and of lines parallel to them. This trace is the vanishing lnoe of the plane. The horizon is evidently the vanishing line of all horizontal planes. A system of parallel right lines will have a common vanishing pcant'; since the line through the point of sight parallel to one is parallel to all; hence their perspectives all intersect at this point, Art. (265). If the lines are also parallel to the plane of the picture, the auxiliary line will be parallel to it, and their vanishing point at an infinite distance. Their perspectives will then be parallel, and the perspective of each line parallel to itself, and it will be necessary to determine only one point of its perspective. If a right line pass through the point of sight, it is a visual ray, and the point in which it pierces the plane of the picture is its perspective. 268. If a point be on a line, its perspective will be on the perspective of the line. If then through any point two right lines be drawn, and their perspectives found as in Art. (264), the irtersection of these perspectives will be the perspective of the given point. Hence, in practice, the perspective of a point is constructe(d by drawing the perspectives of a di(,gonal and perpendicular, which, in space, pass through the point, and finding the intersection 0o these perspectives. Thus let s', Fig. 113, be the principal point, s'd, the horizon d, one point of distance. aivd M any poirt in space. TIINEARI PERSPECTIVF. 161 MN is a diagonal throughl the ploint its vertical projection m'n being parallel to AB, Art. (14). It pierces V at n', vanishes at d, and it'd, is its perspective, Art. (264). The perpendicular through M pierces V at si', vanishes at s', and m's' is its perspective. These perspectives intersect at m, the p)erspective of M. If the point M should be in the horizontal plane, the diagonal and perpendicular would pierce V in AB. When the given point is near the horizontal plane through the point of sight, the perspectives of the diagonal and perpendicular through it are so nearly parallel that it is difficult to mark accurately their point of intersection. In this case, find the perspective of a vertical line through the given point; its intersection with.the perspective of the diagonal or perpendicular will be the required point. Thus in Fig. 113, find m,, the perspective of m the foot of a vertical line through M, and draw mrnn, intersecting mt s' in m, the perspective of M. 269. Since the object to be represented is usually behind the plane of the picture, Art. (262), and is given by its orthographic projections, these projections when made as in Art. (4), will occupy the same part of the drawing as the perspective, and cause confusion. To avoid this, in some degree, that portion of the horizontal plane which is occupied by the horizontal projection of the object, is revolved about AB 1800, until it comes in front of the plane of the picture. Thus, in Fig. 113, m comes to m", and the horizontal projection of the diagonal MN to the position m"n, its vertical projection being in its primitive position, -and the point n' being found as before, Art. (27). It will be observed that the horizontal projection of each diagonal, in this new position, lies in a direction contrary to that of its true position. The vanishing point will be determined by its true direction, Art. (266). 11 162 DESCRIPTIVE GEO)MSTRY. PERSPECTIVES OF CURVES. 270. The perspective of any curve may be found by joining its points with the point of sight by visual rays, thus forming a visual cone. The intersection of this cone by the plane of the picture will be the required perspective: Or the perspectives of its points may be found as in Art. (268), and joined by a line. If the curve be of single curvature, Art. (58), and its plane pass through the point of sight, its perspective will be a right line. 271. If two lines are tangent in space, their perspectives will be tangent. For the visual cones, by which their perspectives are determined, will be tangent along the common rectilinear element passing through the point of contact of the lines; hence their intersection by the plane of the picture will be tangent at the perspective of this point of contact. 272. If the circumference of a circle be parallel to the plane ot the picture, its perspective will be the circumference of a circle whose centre is the perspective of the given centre; also, if it be so situated that the plane of the picture makes in its visual cone a sub-contrary section, Art. (206). In all other cases, when behind the plane of the picture, and its plane does not pass through the point of sight, its perspective will be an ellipse. It is evident that if the condition be not imposed that the circle shall be behind the plane of the picture, this plane may be so taken as to intersect the visual cone in any of the conic sections. 273. Let mlo, Fig. 115, be a circle in the horizontal plane, revolved as in Art. (269), and let s' be the principal point, the LINEAR PERSPECTIVE. 163 point of sight being taken in a plane through the centre and perpendicular to AB, and let d, and d2 be the points of distance. o, is the perspective of o, and m1, of m, Art. (268), and o1m, of the diameter om. The perspectives of the two tangents at o and in will be tangent to the perspective of the circle at o, and ml, Art. (271); and since the tangents are parallel to AB, their perspectives will be parallel to AB, Art. (267), or perpendicular to oin,; hence o1ml is an uxis of the ellipse. Through its middle point el draw de,; it is the perspective of the diagonal fe, and e, is the perspective of e, and ucz, of the parallel chord uz, u's' and z,s' being the perspectives of the perpendiculars through u and z, and u,z, is the other axis of the ellipse. On these two axes the ellipse can be described as in Art. (59). The perspectives of the two perpendiculars 11' and kk' are tangent to the ellipse at 1, and k,. If the point of sight be taken in any other position, the perspective of the circle, Fig. 116, may be determined by points, as in Art. (270), and the. curve drawn through these points tangent to the perspectives of the tangents at o, m, l and k. The line om, will be a conjugate diameter to the line ulz1, since the tangents at o, and ml are parallel to AB and to u,z,. LINE OF APPARENT CONTOUR. 274. If a body, bounded by a curved surface, be enveloped by a tangent visual cone, the line of contact of this cone will be the outer line of the body, as seen from the point of sight, and is the line of apparent contour of the body. The perspective of this line, Art. (270), is evidently the bounding line of the picture or drawing, and is a principal line of the perspective. When the body is bounded by plane surfaces, the visual cone will be made up of visual planes; the line of apparent contour will not be a line of mathematical contact, but will still be the 164 DESCRIPTIVE GEOMETRY. outer line of the body as seen, and will be made up of right lines. When the body is irregular, or composed of broken surfaces. the line of contour may be composed partly of strict lines of contact, either straight or curved, and partly of lines not of contact, but still the outer lines of the body as seen. In all causes their perspectives will folm the boundary of the picture. To construct the perspective of any body, we must then determine the perspective of its apparent contour, and of such other principal lines as will aid in indicating its true form. 275. If a line of the surface intersect the line of contact of the visual cone, the perspective of this line will be tangent to the perspective of'the line of contact. For at the point of intersection draw a tangent to each of the two lines; these tangents lie in the tangent plane to the surface, and this plane is also a visual plane tangent to the cone. Its trace on the plane of the picture is the perspective of both tangents, and tangent to the perspectives of both lines at a common point. They are therefore tangent to each other. VANISHING POINTS OF RAYS OF LIGHT AND OF PROJECTIONS OF RAYS. 276. Since the rays of light are parallel they will have a common vanishing point, which may always be determined by drawing a visual ray of light, and finding the point in which it pierces V, Art. (265). In the construction of a drawing or picture, the direction of the light is usually chosen by the draughtsmnan or artist. This is done by assuming the vanishing point of rays at once, as the direction of the light is thus completely determined. Thus let r,, Fig. 117, be the vanishing point of rays, S being the:point of sight, sr will be the horizontal projection of the ray, and s'r, its vertical projection. LINEAR PERSPECTIVE. ] 6V 277. The horizontal projections of rays of light being parallel and horizontal lines, must vanish in the horizon, Art. ('267); and since a line through the point of sight parallel to the horizontal projection of a ray, must be in the same vertical plane with a rav through the same point, it must pierce the plane of the picture ir. tte vertical trace of this plane, that is, in the line r,r' at r'. Hence, having assumed the vanishing point of rays, through it draw a right line perpendicular to the horizon; it will intersect it in the vanishing point of. the horizontal projections of rays. 278. The orthographic projections of rays on all planes perpendicular to AB, as the plane tTt', are parallel. The line s'.s is the vanishing line of these planes, Art. (267), and must therefore contain the vanishing point of these projections. This point must also be in the trace of a plane of rays through S, perpendicular to these side planes. rm12,, parallel to AB, is this trace; hence r, is the vanishing point of the projections of rays on side planes. 279. The orthographic projections of rays on the plane of the picture, or on planes parallel to it, are parallel; and being in, or parallel to, the plane of the picture, will be parallel in perspective, Alt. (267), and all parallel to rfs', the projection of the ray through S on V. PERSPECTIVES OF THE SHADOWS OF POINTS AND RIGHT LINES ON PLANES. 280. Since the shadow of a point on a plane is the point in which a ray of light through the point pierces the plane, Art. (241), it must be the intersection of the ray with its orthographic projection on the plane. Hence, to construct the perspective of the shadow of a point on any plane, through the perspective of the point draw 166 DESCRIPTIVE GEOMETRY. the perspective of a ray. and through the perspective of the projection of the point on the plane, draw the perspective of the projectlon of the ray; the intersection of these two lines will be the perspective of the required shadow, Art. (268). Itf the shadow be on the horizontal plane, join the perspective of the point with the vanishing point of rays r1, Fig. 117, and the perspective of the horizontal projection of the point with the vanishing point of horizontal projections r'; the intersection of these two lines will be the perspective of the shadow of the given point. If the shadow be on any side plane, join the perspective of the projection of the point on this plane with the vanishing point of the projections of rays on side planes r2; the intersection of this line with the perspective of the ray, will be the perspective of the shadow. If the shadow be on any plane parallel to V, through the perspective of the projection of the point on this plane draw a line parallel to the projection of the ray on the plane of the picture s'r,; its intersection with the perspective of the ray will be the perspective of the shadow. If the shadow be on any vertical plane, draw through the perspective of the horizontal projection of the point, a right line to the vanishing point of horizontal projections. It will be the perspective of the horizontal trace of a vertical plane of rays through the point. At the point where it intersects the perspective of the horizontal trace of the given plane erect a vertical line; it will be the perspective of the intersection of the plane of rays with the given plane. The point where this intersects the perspective of the ray through the given point, will be the perspective of the shadow. 281. The perspective of the indefinite shadow cast by a right line on a plane, may be constructed by finding the perspective of the point in which the line pierces the plane, and joining it *with LINEAR PERSPECTIVE. 167 the perspective of the shadow cast by any other point of the right line; or by joining the perspectives of the shadows of any two points of the line by a right line. If the line be of definite length join the perspectives of the shadows of its two extremities by a right line. PRACTICAL PROBLEMS. 282. PROBLEM 70. To construct the perspective of an upright rectangular pillar, with its shade and shadow on the horizontal plane. Let Imno, Fig. 118, be the lower base of the pillar in the horizontal plane, revolved as in Art. (269), and p'q' the vertical projection of the upper base, s' the principal point, d, one of the points of distance, r, the vanishing point of rays, and r' of hori. zontal projections of rays. These important points will be thus represented in all the following problems. Since ml and no are perpendicular to V, they vanish at s', Art. (265), and m's' and n's' are their indefinite perspectives; id2 is the perspective of the diagonal ni, and n of the point n, Art. (268). nm,, parallel to AB, is the perspective of mn, Art. (267); o0 of the point o; n,o, of the edge no, Art. (264); o,l, of ol, and iml of ml. The edges of the upper base, which are horizontally projected in on and Im, pierce V at p' and q', and vanish at s'; hence p's' and q's' are their indefinite perspectives. The diagonal of the upper base, horizontally projected in ni, pierces V at i', i'd, is its perspective, and p, the perspective of the vertex of the upper base horizontally projected in n, and v, of that horizontally projected at 1. The vertical edges which pierce H in re, n, o, and i, are parallel to V, and their perspectives parallel to themselves; hence mnq,, n1rp,, o1u, l,v, are their perspectives terminating in the points i, p,, u,, v,, and qlpou,v, is the perspective of the upper base. The face of which n,o,u,p, is the perspective is in the shade, and therefore is darkened in the drawing. 168 DESCRIPTIVE GEOMETRY. The shadow on H, of the edge represented by n,p, is np2, Art (281). The shadow of the edge represented by plu, is parallel to the line itself, and therefore perpendicular to V, and vanishes at s'. It is limited at P2 and u2, Art. (281). uv,, parallel to AB, is the perspective of tile shadow of the edge represented by uav, and lv2, of that represented by l,1J,. That part of the drawing within the line np,.v,x,, is darkened, be ing the perspective of that part of the shadow on IH which is seer, 283. PROBLEM 71. To construct the perspective of a square pyramid with its pedestal, and the perspective of its shadow. Let mnol, Fig. 119, be the base of the pedestal revolved, as in, Art. (269), and mnn'mn' its front face in the plane of the picture, pqtu the horizontal, and p'u' the vertical projection of the base of the pyramid, and C its vertex. The face mna'?n' being in the plane of the picture is its own perspective. The four edges of the pedestal, which are perpendicular to V, pierce it in m, n, n' and m' and vanish at s'. The two diagonals rno and (mo, m'n') pierce V at -, and rn' and vanish at d,. The diagonals nl and (nl, n'm') pierce V at n and n' and vanish at d.2. Ilence o0 is the perspective of o, Art. (268); k1 of (on'); 1, of 1; and h, of (lm'); and the perspective of the pedestal is drawn as in Art. (282). The two perpendiculars (at, u') and (pq, p') pierce V at u' and p'; u's' and p's' are their perspectives intersecting rt'd, and n'd.2 in u, q,, p, and t, and u,p,qlt, is the perspective of the base of tile pyramid. The perpendicular through C pierces V at c', and a diagonal through the same point at h' and c's' and h'dl are their perspectives, and c, the perspective of C; and clul, cp,, cqz, andl c,tt are the perspectives of the edges of the pyramid. nn2 is the perspective of the shadow of nn' on H; n2k2 of the shadow of the edge represented by n'k, Art. (265). i, is the perspective of c, the horizontal tDrojection of the vertex' ir' of the LINEAR PERSPECTIVE. 169 noTizontal projection of a ray through C, and c,?r, the )perspective of the ray; hence c. is the perspective of the shadow ot' C on H, Art. (280). e, is the perspective of the projection of (3 on the upper base of the pedestal, er' of the projection of a ray on this Plane, and c, of the shadow cast by C on this plane; andl p,c, the perspective of the shadow cast by the edge CP on this plane, Art. (281). This shadow passes from the upper base at a point of which x, is the perspective; x,r, is the perspective of a ray through this point, intersecting ns' at x., the perspective of the shadow cast by one point of the edge CP on H, and xzc, is the perspc-:tive of the shadow. t,c, is the perspective of the shadow cast by the edge CT on the upper base of the pedestal; y, of the point at which it passes from this base; y, of the shallow of this point on H, and yc,. of the shadow of the edge CT. The face represented by cp q, is in the shade, as also that re_resented by nn'klo,, anrid both are darkened on the drawing. p,1.xkylt, bounds the darkened part on the perspective of thae upper base, and nn,.. cys.. l, that on H. 284. PROBLEM 72. To construct the perspective of a cylindrical: column with its square pedestal and abacus, and also the shade of the column and shadow of the abacus on the colrumn. Let inngl (Fig. 120) be the horizontal projection of both pedestal and abacus, Art. (269); mm'n'n the vertical projection of the pedestal, and e'l'tf' that of the abacus; upqt being the horizontal projection of the column, m'n' the vertical projection of its lower, and ef' of its upper base; the plane of the picture being coincident with that of the front faces of the pedestal and abacusLet the point of sight be taken as in Art. (273), in a plane through the centres of the upper and lower bases of the column and perpendicular to AB. Constrnct the perspective of the pedestal as in Art. (282), mnM'n' being its own perspective, and m'l, and n'g, the perspee 170 DESCRIPTIVE GEOMlE RY. tives of those edges of the upper face of the pedestal which pierce V at m' and n'. In the same way construct the perspective of the abacus: e'l'g~f' being its own perspective, and e's' andf's' the indefinite perspectives of the edges of the lower face which pierce V at e' and f'. u,t,qp,p is the perspective of the lower base of the column de termined as in Art. (273); u,q, being its semi-conjugate and tp, its semi-tranverse axis. In the same way the perspective of the upper base is determined, its conjugate axis being the perspective of that diameter which pierces V at u'", horizontally projected in uq, and its tranverse axis z v1 the perspective of the chord corresponding to tp. t,z, andp,v, tangent to these ellipses, Art. (275), and perpendicular to AB, are the perspectives of the elements of contour of the column, and complete its perspective, Art. (274). The elements of shade on the column are determined by two tangent planes of rays, Art. (251), and since these planes are vertical, their intersections with the plane of the upper face of the pedestal will be parallel to the horizontal projections of rays, and therefore vanish at r'. Since these intersections are also tangent to the lower circle of the column, their perspectives will be tangent to the ellipse t,u,p,. Hence, if through r' two tangents be drawn to t ulp,, their points of contact will be points of the perspectives of the elements of shade. a,k, is the perspective of the only one which is seen. The plane of rays by which this is determined intersects the lower edge elf' of the abacus in k,, through which draw k,r,. It intersects aLk.2 in ko, the perspective of the point of shadow cast by k,, and limiting the perspective of the element of shade. Draw r't,; it is the perspective of the intersection of a vertical plane of rays with the upper face of the pedestal. This plane intersects the column in an element represented by tiz,, and the edge represented by e's' in a point of which y, is the perspective. Through yl draw y,?, intersectinlg t,z, in 2,, the perspective of the shadow cast on the column by the point Y, and in the same way LINEAR PERSPECTIVE. 171 the perspective of the shadow cast by any point of the same edge may be determined. Through mn' draw m'r', and through e' draw e'r, intersecting the perspective of the element cut from the column in e2, the perspective of the shadow cast by e' on the column, and in the same way the perspective of the shadow cast by any other point of the edge etf' may be found, as w2 the perspective of the shadow cast by w". All of the lower face of the abacus is in the shade, and is darkened in the drawing. The line ye2-.. &2 is the perspective of the line of shadow on the column, and all above it is darkened, as all beyond aks2. ar' is the perspective of the shadow cast by the element of shade on the upper surface of the pedestal, and the part beyond it which is seen is therefore darkened. 285. PROBLEM 73. To construct the perspective of an upright cylindrical ring, with its shade and shadow on its interior surf.ce. Let mngl, Fig. 121, be the horizontal projection of the outer cylinder of the rillg, and ezh' its vertical projection; upqt the horizontal, and o,i,w, the vertical projection of the inner cylinder, the plane of the picture being- coincident with the plane of the front face of the ring. The two circles ezh' and oli,iw being in the plane of the picture are their own perspectives. To construct the perspectives of the back circles horizontally projected in lg and tq, draw the vertical radius (c, c'lh') and the two vertical tangents at Q and G. hc, a,q,, and bg, are the perspectives of these lines, Art.. (268), and c,, q,, and g, are the perspectives of -the points C, Q, and G. With c, as a centre, and cq, and c,g, as radii, describe the arcs tt" and v,g,, they will be the perspectives of arcs of the back circles. siy, and s,x, are the perspectives of the elements of contour tangent to the circles eyx, and v,g,, Art. (275). The element of shade on the outer cylinder is determined by a tangent plane of rays. This plane being perpendicular to V, it. 17 2 DESCRIPTIVE GEOMETRIY. veltical trace will be parallel to 9rs', Art. (279), and tangent tc yzlh' at zj, and zyvl will be the perspective of this element. Points of the shadow cast by the circle 0o,iwz on the interior cylinder will be found by passing planes of rays perpendicular to the plane of the picture. Each plane will intersect the circle in a point which casts a shadow on the element which the plane cuts from the cylinder, Art. (252). o,i, parallel to rs' is the vertical trace of such a plane, intersecting the circle in o0, and the cylinder in an element of which ils' is the perspective. ol!r is the perspective of a ray through o,, and o, is the perspective of the shadow. The perspective of the shadow evidently begins at k,. 286. PROBLEM 74. To construct tAe perspective of an inverted frustum of a cone with its shade and shadow. Let C, Fig. 122, be the vertex, kolm the horizontal, Art. (269), and k'l' the vertical projection of the upper base, ehgf the horizontal, and e'g' the vertical projection of the lower base, and k'e' the vertical projection of one of the extreme elements. The perspectives of the bases are determined as in Art. (273), l,o'ylm, of the upper, and e1h,a,f, that of the lower base. The perspective of the vertex is found as in Art. (268), CZ being the diagonal through it, piercing V at z'; and (co, c') the perpendicular piercing V at c', z'd2 and c's' are their perspectives intersecting at c,. Through c1 draw the two tangents c,w, and c,y,; they are the perspectives of the elements of contour, and with the curves klo'ylm, and e,h,,a,fl limit the perspective of the frustum. The elements of shade on the cone are determined by two tangent planes of rays intersecting in a ray of light passing through the vertex, Art. (131), r,c, is the perspective of this ray, and ur' the perspective of its horizontal projection, ul being the perspective of c; hence c2 is the perspective of the point in which this ray pierces H, and the tangents c.,i and c2p, are the perspectives of the horizontal tracts of the tangent planes, and ianl and LINEAR PERSPECTIVE. 173 p,q, of the elements of shade, Art. (253), the former only being seen. That part of the circumference of the upper base towards the source of light, and between the points of which n, and q, are the perspectives, casts its shadow on the interior of the cone. Points of this shadow may be determined by intersecting the cone by planes of rays through the vertex. Each plane intersects the circumference in a point, and that part of the cone opposite the source of light in a rectilinear element. A ray of light through the point intersects the element in a point of the required shadow. c2a, is the perspective of the horizontal trace of such a plane. The plane intersects the cone in two elements represented by aly, and b,c,, and b, is the perspective of the point in which the plane intersects the circumference, brl the perspective of a ray through this point, and b. the perspective of the required point of shadow. The curve of shadow evidently begins at the points of which n, and q, are the perspectives. The perspective of the lowest point of this shadow will be found i) using the line c,u,, as this is the perspective of the trace of that plane which cuts out the element furthest from the point casting the shadow. The p,.rspective of the point of shadow on any element is found by using the lile drawn through c2 and the lower extremity of the perspective of the element. Thus the point of tangency b,, Art. (275), is found by using c2a, as the perspective of the trace of the auxiliary plane. q,k,o' is the perspective of that part of the shadow on the interior which is seen, and is therefore darkened in the drawing as is the perspective of the'shade n,ila,y,. iln, and plq2 are the perspectives of the shadows cast by the elements of shade on II, the points ni and q, being determined by n,r, and qur,. The plane determined by c2a, intersects the circumference of the 1 74 DESCRIPTI'VE GEOMIETIRY. upper base in a point of which y, is the perspective, yg1r is the perspective of a ray through this point piercing H1 at a point of which y, is the perspective. This is a point of the perspective of the curve of shadow of the upper circumference on H; and in the same way any number of points may be determined. r?,v, drawn tangent to nly,m, will also be tangent to ny.2q, since this tangent is the perspective of the element of contour of the cylinder of rays through the upper circumference by which its shadow is determined, Art. (275). The curve n.y,gq2 is also tangent to i,n2 and pIq2 at n. and q2. 287. PROBLEM 75. TO construct the perspective of a niche with its shadow on its interior suiface. Let the niche be formed by a semi-cylinder, the lower base ot which is mlo, Fig. 123; and the upper base vertically projected in n'o', and the quarter sphere vertically projected in the semicircle?n't co', its lower semicircle being coincident with the upper base of the cylinder, and the plane of the picture so taken as to contain the elements ma-' and oo' and the fiont circle rn'k'o'. These elements and front circle being in the plane of the picture will be their own perspectives. An arc of an ellipse, mlo, is the perspective of the lower base of the cylindrical part, and m'no' that of the upper base, these being constructed as in Art. (273), 1, being the perspective of I and n, of the corresponding point of the upper base. These lines form the perspective of the niche. The lines which cast shadows on the interior of the niche are the element mm' and the arc m'k'e'. The shadow cast by mm' is determined by passing through it vertical plane of rays; tmr' is the perspective of its trace, Art. (277). This plane cuts from the cylinder an element represented by f2m2, the intersection of which by m'r, in mZ2 is the perspective -of the shadow cast by m' on the element; and f2Am is the perspective of the shadow cast by mnf', a part of mm', on the LINEAR PK;RSPECTIVE. 175 interior of the cylinder; mfA is the perspective of the shadow cast by rnf' on H. Points of the shadow cast by the front circumference, m'k'e', on the cylindrical part of the niche may be determined by intersecting the cylinder by vertical planes of rays. Each plane intersects the cylinder in a rectilinear element aind the arc in a point. A ray of light through this point intersects the element in a point of the curve of shadow. r'g is the perspective of the horizontal trace of such a plane. It intersects n'k'e' in g', and the cylinder in an element represented by il,2 and g2 is the perspective of the shadow cast by g' on the element. The shadow cast by a part of the front circumference on the spherical part of the niche is an equal arc of a great circle. For this shadow is determined by a cylinder of rays through the front circumference, and the part of each element of the cylinder between the point casting the shadow and its shadow is a chord of the sphere, and all these chords will be Iisected by a plane through c' perpendicular to them. These half chords may be regarded, one set as the ordinates of the arc casting the shadow, and the other as the ordinates of the shadow, and since the corresponding ordinates of the two arcs Ore equal, and in all respects like situated, the two arcs will be equal. Their planes evidently intersect in the radius c'e' perpendicular to the axis of the cylinder of rays, or to the ray of light, and also perpendicular to the projection s'r, of the ray of light on the plane of the picture, Art. (276). Points of this shadow may be constructed by intersecting the quarter sphere by planes parallel to the plane of the picture. Each plane will cut from the quarter sphere a semi-circumference. The firont semi-circumference will cast upon this plane a shadow parallel and equal to itself, Art. (246), and the intersection of this with the semi-circumference cut from the sphere will be a point of the required shadow, Art. (244). Draw y,z, parallel to m'o', it may be taken as the perspective of the intersection of an auxiliary plane with the upper tbase ol 17'6 DESCRIPTIVE GEOMETRY. the cylinder, and the semi-circumference y1x2z, is the perspective of the semi-circumference cut from the sphere. The centre of the front circle casts a shadow upon this plane whose perspective is c,, Art. (280), c" being the perspective of the projection of c' on this plane, and c"c, parallel to s'r: the perspective of the projection of the ray through c' on this plane. c2m3 parallel to c'm' is the perspective of the shadow cast by the radius c'm' on this plane, Art. (267), and m.X2 the perspective of the arc of shadow intersecting y,x2z, in x., a point of the perspective of the shadow. Otherwise thus: Draw k'k" parallel to s'r,, it may be taken as the vertical trace of a plane of rays perpendicular to V. This plane cuts from the quarter sphere a semi-circlumference. A ray of light through k' will intersect this semi-circunmference in a point of the curve of shadow. Revolve the plane about k'Ic" until it coincides with V. k'x"k" will be the revolved position of the semi-circumference. Revolve the ray of light through the point of sight about s'r, until it coincides with V; s"rl will be its revolved position, s's" being equal to s'd,, Art. (266). Throughl k' draw k'x" parallel to s"r,, it will be the revolved position of the ray through k', and x" will be the revolved position of a point of the shadow. Through x' draw x's'; it is the perspective of the perpendicular x'x" intersecting ktr, in xl, the perspective of the shadow cast by k'. The perspective of this curve of shadow evidently lbgins at a', the point of tangency of a line parallel to s'r,. The point at which the curve of shadow passes from the spheric,,l to the cylindrical part is evidently the point in which the intersection of the plane of the curve of shadow with the plane of the upper base meets the circumference of the upper base. To determine the perspective of this point, through x.2 draw x.2u, parallel to c'e'; it is the perspective of a line of the plane of the curve of shadow as also of the plane of the circle of which mx, is the perspective. u, is the perspective of the point where this line pierces the plane of the upper base, and c'u, the perspective of LINEAR PERSPEC'TIVE. 1ss the intersection of the plane of the circle of shladow with the plane of the upper base, and p, the perspective of the required )point. 288. PROBLEM 76. To construct the perspective of a sphere with its shade, and shadow on the horizontal plane. Let the plane of the picture be taken perpendicular to the right line, Joining tlhe point of sight with the centre of the sphere, and tangent to the sphere; then, Fig. 124, c will be the horizontal and s' the vertical projection of the centre. ca being equal to the radius and b'c"c"' the vertical projection of the sphere. The perspective of the sphere will be determined by a visual cone tangent to it, and the circle ill which this cone is intersected by the plane of the picture will be the outer line of the perspective, Art. (274). To construct this, pass a plane through S and C perpendicular to AB. It intersects the sphere in a great circle, and the cone in rectilinear elements tangent to this circle and piercing V in points of the required circumference. awn, is the vertical trace of this auxiliary plane. Revolve the plane about am, as an axis until it coincides with V; S, in revolved position, will be at d, and C at c", and qn's' described with the' centre c" and radius c"s' will be the revolved position of an arc of the circle cut from the sphere, and m'd, that of one of the tangents. In true position this tangent pierces V at m,, and the circle m,u,v, described with s' as a centre and s'm, as a radius will be the perspective of the sphere. The curve of shade on the spbhelre, the line of contact of a tangent cylinder of rays, is the circumt'erence of a great circle whose plane is perpendicular to the axis, Art. (121), or to the ray of light. The plane of rays through the point of sight whose vertical trace is s'r, is perpendicular to the plane of the circle of shade, and is evidently a principal plane, Art. (206), of the visual cone by which the perspective of this curve is determined, and since 12 178 DESCRIPTIVE GEOMETRY. the plane of the picture is perpendicular to this plane, its inter section with the cone will be an ellipse, all the chords of which perpendicular to s'r, are bisected by it. This line will then be an indefinite axis of the ellipse. Ana. Geo., Art. (85). This princi pal plane intersects the circle of shade in a diameter whose perspective will be this axis. To determine it, revolve the plane about s'r, until it coincides with V. S is found at s", s's" being equal to s'd,. Tlhe centre of the circle cut from the sphere will be at c"', n'l'p' will be the revolved position of the circle, and n't' perpendicular to s"r, will be the revolved position of the diameter. Two visual rays from its extremities pierce V at n, and p,, and n,p1 is its perspective and the required axis. 0o, the middle point of this axis, is the centre of the ellipse, and lik, perpendicular to njp,, the position of the other axis, and o' the revolved position of the point of which o, is the perspective. If through l,k, and S a plane be passed, it will intersect the circle of shade in a chord passing through the point of which o' is the revolved position, perpendicular to the plane of rays of which s'r, is the trace and equal to l'k'. The perspective of this chord will be the other axis. To determine it, revolve this plane about k,l, until it coincides with V. S will be found at s"', o,s"' being equal to o,s"; o' will be at o", o00o" being equal to oo', and k"l" perpendicular to s'ir, will be the revolved position of the chord, the perspective of which is k,l,, the other axis. The ellipse described upon these two axes is the perspective of the curve of shade. It is tangent to m,u,vt at u, and v,, since these are the perspectives of two points which are determined by two visual planes of rays tangent to the sphere, whose traces are r,u, and r,v,. These are the perspectives of the points in which the curve of shade crosses the apparent contour of the sphere, Art (275). The curve of shadow on the horizontal plane is cast by the curve of shade. c, is the perspective of the shadow cast by the centre, LINEAR PERSPE Cr'VE. 179 c, being the perspective of its horizontal projection, and c r' the perspective of the horizontal projection of a ray through the centre, Art. (280). The perspective of the shadow of each diameter of the curve of shade must pass through c2. If, then, we find the perspective tof the point in which a diameter pierces the horizontal plane and join it with c,, we shall have the indefinite perspective of the shadow of this diameter, and the perspectives of two rays through its extremities will intersect the perspective of this shadow in points of the perspective of the curve of shadow. All diameters of the circle of shade pierce the horizontal plane in the horizontal trace of the plane of this circle. This trace being a horizontal line must vanish in the horizon of the picture s'd, Art. (267), and being in the plane of the circle of shade must also vanish in the vanishing line of this plane; and hence the intersection of these two lines will be a point of the perspective of the horizontal trace. To construct the vanishing line of the plane of shade, through s" draw s"x' perpendicular to s"ir,; it is the revolved position of a line passing through S in the plane of rays whose vertical trace is rls' and parallel to the plane of shade. It pierces V at x', one point of the vertical trace of a plane through S parallel to the circle of shade. Through x' draw x'q, perpendicular to x'r,, Art. (43); it is the vertical trace, or required vanishing line. This intersects s'q, at q,. To determine another point of the perspective of the horizontal trace, find the perspective of the point in which that diameter of the circle of shade, which is parallel to the plane of the picture, pierces H. The horizontal projection of this diameter is parallel ti AB; hence, cil is its perspective. s's" perpendicular to r,s' is the perspective of the diameter itself, for these two lines are the orthographic projections of the diameter and ray of light through the centre, since the visual plane through each of them contains the projecting perpendicular Ss', Art. (36). i, is then the perspeo [t80 DiSCRaIP'rIl E GEOMETRY. tive of the point in which tb's diameter pierces H, and qc,i the perspective of the horizontal trace. Join i, with c,; i,c2 is the perspective of the shadow cast bly the diameter represented by glj. Join g, and f, with r, by right lines. These are the perspectivw:s of rays through the extremities of the diameter, and g2 a I f2 are the perspectives of points of the curve of shadow. Produce ule, to yl, y, is the perspective of the point in which the corresponding diameter pierces I, and y,c2 the perspective of its shadow, and e2 and U2 the perspectives of the shadows cast by its extremities. In the same way other points of the curve are constructed. The curve must be drawn tangent to ru, and r1v, at u2 and v2, since these are the perspectives of the points in which the curve of shadow crosses the elements of apparent contour of the cylinder of rays, by which the curve of shadow is determined. 289. PROBLEM'77. To construct the perspective of a?1roined arch with its shadows. To form the groined arch here considered, two equal right semicylinders, with semi-circular bases, are so placed that their axes shall be at right angles and bisect each other and the diameters of the semi-circular bases in the same horizontal plane. Thus, Fig. 125, let nolu be the horizontal, and ok'l the vertical projection of one semi-cylinder, and fghi the horizontal, and pp'q'q the vertical projection of the second cylinder, pp' and qq' being the vertical projections of the two semi-circular bases. These two cylinders intersect in two equal ellipses horizontally projected in the diagonals ab and ce. These ellipses are the groins. The arch is now formed by taking out fiom each cylinder that part of the other which lies within it. Thus, all that part of the cylinder whose elements are parallel to V, horizontally projected in akc and bke, is removed, as also that part of the other cylinder projected in ake and bkc. LINEAR PERSPECTIVE. 181 The arch thus formed is placed upon four pillars standing in *the four corners of a square, and whose horizontal projections are the four squares renai, uchv, etc. The elements ih and fy being coincident with the inner upper edges of the pillars which are horizontally projected in ia-ch and *fe-by, and.the elements no and ul coincident with the inner upper edges which are projected in na-eo and uc-bl. In this position of the arch the front circumference, or front base of the cylinder whose elements are perpendicular to V springs from the upper extremities,f the two edges of the front pillars which are horizontally projected at n and Au and is described on a diameter equal and parallel to nu. The back circumference, or other base of the same cylinder, springs from the upper extremities of the edges horizontally projected at o and 1. The side circumferences, or bases of the other cylinder, spring, the one from the upper ends of the two edges horizontally projected in i and f and the other from those projected in h and y, and their diameters are parallel and equal to if and hq/. The groins spring fiom the upper ends of the four inner edges of the pillars, the one from those projected in a and b, anal the other from those projected in c and e, and ab and ce are respectively equal and parallel to the transverse axes of' the groins, the common conjugate axis being equal and parallel to the radius k'k". The planes of the outer faces of the pillars are produced upwards, inclosing the mass of masonry supported by the arch. Lb construct the perspective of the arch thus placed let us take, Fig. 126, the plane of the picture coincident with the plane of the front faces of the front pillars, mm'n'n and uu'v'v being these front. faces and their own perspectives, and let tile principa;l point be at s' in a plane perpendicular to V and midway between the pillars. The perspectives of the front pillars are constructed precisely ab in Art. (282); a, and i, being the perspectives of the upper ends 182 DESCRIPTIVE GEOMETRY. of the two back edges of the first, and c, and h, those of the second pillar; md,, the perspective of the diagonal corresponding to mq in Fig. 125, intersects us', the perspective of the perperldicular corresponding to ul, in b,, which is the perspective of the lower end of the edge of the back pillar corresponding to b; 62b,, parallel to AB, is the perspective of the edge corresponding to by, and b,1, that corrresponding to bl. Through b,, g,, and 12 draw bMb,, g,9g, and 12ll until they meet u's' and v's' in b,, g,, and 1,, and draw b.q, and b,l,, thus completing the perspectives of the two faces of the back pillars which can be seen. In the- same way the perspective of the other back pillar may be constructed, e. being the perspective of the point corresponding to e. On n'u' as a diameter describe the semicircle n'k'u'; it is the front semicircle of the arch in the plane of the picture and its own perspective. The other base of the same cylinder being parallel to the plane of the picture will be in perspective a semicircle, and since it springs from the points of which o, and 1, are the perspectives, o,l will be the diameter and the semicircle on this its perspective. To obtain points of the perspectives of the groins and side circles, intersect the arch by horizontal planes. Each plane will cut from the cylinders rectilinear elements, the intersection of which will be points of the groins; and from the side faces of the arch, right lines which will intersect the elements parallel to the plane of the picture in points of the side circles. Let m"v" be the trace of such a plane. It intersects the cylinder whose axis is perpendicular to V in two elements which pierce V at w' and z', vanish at s', and are represented by w's' arMd z's'. The same plane intersects the side faces of the arch in lines represented by m"s' and v"s'. The two diagonals through m" and v", intersect the elements in points of the groins, see Fig. 125. m"d, and v"d2 are the perspectives of these diagonals intersecting w's' in w" and w"' and z's' in z" and z"', points of the perspectives ou the groins. w"z" and w"'z"' are the perspectives of the elements LINEAR PERSPECTIVR. 183 cut from the cylinder whose axis is parallel to V. These are intersected by m"s' and v"s' in points p, and q, of the perspectives of the side circles. The perspectives of the groins thus determined are drawn, one from a, to b,, the other from c, to e,, and the perspectives of the side circles, one from i, to f, and the other from h, to g,. Since all four of these curves cross the element of contour oI' the cylinder whose axis is parallel to the plane of the picture, their perspectives will be tangent to the perspective of this element. To determine it, through S pass a plane perpendicular to the axis of the cylinder, r2s' is its trace. It cuts from the cylinder a circle and from the visual plane tangent to the cylinder a right line tangent to the circle. Revolve this plane about r.s' until it coincides with V. The centre of the circle will be at m', and S at d,. The semicircle described with m' as a centre and k"n' as a radius is the revolved position of the circle cut from the cylinder, and the tangent to it from d, the revolved position of the line cut from the tangent plane. This pierces V at x', a point of the perspective of the element of contour. The line through x' parallel to AB, is the tangent to all the curves. nr' and i2r' are the perspectives of the indefinite shadows cast on H by the vertical edges of the left hand pillar, Art. (282). The point in which nr' intersects bg2 is the perspective of a point of the shadow of nn' on the front face of the back pillar. This line being parallel to this face, its shadow will be parallel to itself as also the perspective of the shadow, which is terminated at n2, the perspective of the shadow cast by n', Art. (280). From this point the shadow is cast on this face by the front circle,'and is the are of a circle as is its perspective, Art. (272). The shadow of the radius n'k" is parallel to itself, as also the perspective of this shadow passing through n2 and terminated at k2. The are described with.'.2 as a centre and k2n2 as a radius is then the perspective of the shadow of the front circle on the face. e2r' is the perspective of the shadow cast on H by the edge rep 184 DESCRIPTIVE GEOM'ETRY. resented by e,e;, a small part of which only lies in the limits o0 the drawing. Points of the shadow cast by the front circle oin the cylinlder, of which it is the base, are determined by intersecting the cylinder by planes of rays perpendicular to the plane of the picture. The traces of these planes are parallel to s'r,, Art. (279), and each plane cuts the front circumference in a point casting the shadow and the cylinder in a rectilinear element, which -is intersected ly the ray through ihe point in its shadow. Let y y" be the trace of such a plane. It intersects the cylinder in an element represented by y","', and the circumference in the point y', and y, is the perspective of the shadow. The perspective of the shadow begins at the point in which a trace parallel to y"y' is tangent to the circle. Points of the shadoto cast by the side circle on the left, on the cylinder of which it is a base, may be found by intersecting the cylinder by planes of rays perpendicular to the side faces of the arch. Each plane intersects the circumference in'a point casting the shadow and the cylinder in a rectilinear element which is intersected by the ray through the point in a point of the shadow. The intersections of these planes with the outer side face of the arch vanish at r-I, Art. (278). Let r,t,, be the perspective of one of these intersections. Tile plane intersects the side circumference in a point of which t, is the perspective and the cylinder in an element represented by p,q, and t, is the perspective of the point of shadow. The perspective of the curve evidently begins at the point of tangency of a line through r2 tangent to ipf. 290. Although in general the perspectives of objects are made as in the preceding articles on a plane, it is not unusual tr make them upon curved surfaces. Extended panoramic views are thus made upon a right cylinder with a circular base, the point of sight being in the axis, and the observer thus entirely surrounded by pictures of objects. LINEAR PERSPECTIVE. 185 Objects are also sometimes'represented on the interior of a spherical dome. ~ In all cases the perspectives of points are constructed by drawing through the points visual rays, atid finding the points in which these pierct the surface on which the representation is made.. The constructions involve only the principles contained in Arts (263), (264), and (268). PAiRT V. ISOMETRIC PROJECTIONS. PRELIMINARY DEFINITIONS AND PRINCIPLES. 291. Let three right lines be drawn intersecting in a colnllon point and perpendicular to each other, two of them being horizontal and the third vertical; as the three rectangular co-ordinate axes in space, ARna. Geo., Art. (42), or the three adjacent edges of a cube; then let a fourth right line be drawn through the same point, making equal angles with the first three, as the diagonal of a cube. If a plane be now passed perpendicular to this fourth line, and the right lines and other objects be orthographically projected upon it, the projections are called Isometric. The three right lines first drawn are the co-ordinrate axes; and the planes of these axes, taken two and two, are the co-ordinate planes. The common point is the origin. The fourth right line is the Isometric axis. If A designate the origin, the co-ordinate axes are designated, as in Ana. G'eo., as the axes AX, AY, and AZ, the latter being vertical; and the co-ordinates planes as the plane XY, XZ and YZ, the first being horizontal, and the other two vertical. 292. Since the co-ordinate axes make equal angles with each otller and with the plane of projection, it is evident that their pro ISOMETRIC PROJECTI(ONS. 1 87 jections will make equal angles with each other, two and two, that is, angles of 12'0. Hence, Fig. 128, if any three right lines, as Ax, Ay, and Az, be drawn through a point as A, making with each other angles of 120~, these may be taken as the projections of the co-ordinate axes, and are the directrices of the drawing. It is further evident, that if any equal distances be taken on the co-ordinate axes, or on lines parallel to either of them, their projections will be equal to each other, since each projection will be equal to the distance itself into the cosine of the angle of inclination of the axes with the plane of projection, Art. (I 97). The angle which the diagonal of a cube makes with either ad-'jacent edge is known to be 54~ 44'; therefore the angle which either edge, or either of the co-ordinate axes, makes with the plane of projection will be the complement of this angle, viz., 350 16'. 293. If a scale of equal parts, Fig. 127, be constructed, the unit of the scale being the projection of any definite part of either coordinate axis, as one inch, one.foot, etc., that is, one inch multiplied by the natural cosine qf 350 16'=.81647 of an inch; we may, from this scale, determine the true length of the isometric projection of any given portion of either of the co-ordinate axes, or of lines parallel to them, by taking from the scale the same number of units as the number of inches, or feet, etc., in the given distance. Conversely, the true length of any corresponding line in space may be found by applying its projection to the isometric scale, and taking the same number of inches, or feet, etc., as the number of parts covered on the scale. 294. Since in most of the frame work connected with machinery, and in the various kinds of buildings, the principal lines to be represented occupy a position similar to that of the co-orditlate axes, viz., perpendicular to each other, one system being vet 188 DESCRIPTIVE GEOMETRY. tical, and two others horizontal, the isometric projection is used to great advantage in their representation. A still greater advantage arises firom the fact that in a drawing thus made, all lines parallel to the directrices are constructed on the same scale, Art. (292). ISOMETRIC PROJECTIONS OF POINTS AND LINES. 295. If a point be given, as in Analytical Geomeiry, by its co-ordinates, or its three distances from the co-ordinate planes, Art. (40), its isometric projection may be easily constructed. Thus, Fig. 128, let Ax, Ay, and Az be tile directrvices, A beilg the projection of the origin. On Ax lay off Ap, equal to the same number of units, taken from the isometric scale, as there are units in the distance of the point from the co-ordinate plane YZ. Through p draw pm' parallel to Ay, and make it equal to the number of units in the distance of the point from the plane XZ. Through nm' draw m'rn parallel to Az, and make it equal to the third given distance, and m will be the required projection. 296. The projection of any right line parallel to either of the co-ordinate axes may.be constructed by finding, as above, the projection of one of its points, and drawing through this a line parallel to the proper directrix. If the line is parallel to neither of the axes, the projections of two of its points. may be found as above and joined by a right line. The projections of curves may be constructed by finding a sufficient number of the projections of their points. 297. If the circumference of a circle be in, or parallel to, either co-ordinate plane, its projection may be constructed thus: At the ISOMETRIC PROJECTIONS. 189 extremities of the two diameters, which are parallel to the co-orlinate axes, draw tangents, thus circumscribing the circle by a square. The projections of each set of tangents will be parallel to the projection of the parallel diameter and tangent to the projection of the circle, Art. (65); hence the projections of these two equal diameters will be equal conjugate diameters of the ellipse which is the projection of the circle, and since these projections are parallel to the directrices they will make with each other an angle of 1200. Upon these the ellipse may be described, taking care to make it tangent to the projections of the four tangents. PRACTICAL PROBLEMS, 298. PROBLEM 78. To construct the isometric projection of a rcube. Let the origin be taken at one of the upper corners of the cube, the base being horizontal, and let Ax, Ay, and Az, Fig. 129, be the directrices. From A, on the directrices, lay off the distances Ax, Ay, and Az, each equal to the same number of units, taken friom the isometric scale, as there are units of length in the edge of the cube. These will be the projections of the three edges of the cube which intersect at A. Through x, y, and z draw xe, xgy, ye, yc, zc, and zy parallel to the directrices, completing the three equal rhombuses, Axey, etc. These will be the projections of the three faces of the cube which are seen, and the representation will be complete. 299. The ellipses constructed upon the equal lines kl, qs; st uv, &c., Fig. 129, are evidently the projections of three circles inscribed in the squares which form the faces of the cube, and these lines are the equal conjugate diameters of the ellipses, Art. (297) Ik] being drawn through the middle point of Ax, and sq through 190 DESCRIPTIVE GEOMETRY. the middle point of Az, and parallel respectively to Az and Ax, mn being the projection of the centre of the circle. 300. PROBLEM 79. To construct the isometric projection of an upright rectangular beam with its shade and shadow on the hori. zontal plane. Let the origin A and the directrices, Fig. 130, be taken as in the preceding problem. On Ax lay off the distance Ax equal to the breadth of the beam in units taken from the isometric scale; on Ay, its thickness, and on Az, its length, and complete the three parallelograms Axcz, Aybz, and Axey, These will be the projections of the three faces 4f the beam which are seen. Let a' be assumed as the isometric projection of the point in which a ray of light through A pierces IH; then Aa' will be the isometric projection of the ray, and za' the projection of the shadow of the edge AZ, Art. (248). Through a' draw a'y' parallel and equal to Ay. It is the projection of the shadow of the edge AY. y'e' equal and parallel to ye is the projection of the shadow cast by YE, and de' that of the edge, DE. The face represented by Azby is in the shade. 301. PROBLEM 80. To construct the isometric projection of the framework of a simple horizontal platform resting on four rectan,gular supports. Let A, Fig. 131, be the upper corner of one of the horizontal beams, the directrices being as in the preceding problems. Lay off Ax, Ab, and Az equal respectively to the number of units in the length, breadth, and thickness of the horizontal beam, and complete the three parallelograms Aq, Ax', Ab', thus forming the representation of the beam. Lay off zc equal to the distance of the first support from the end of the beam, cd equal to its thickness, cc' parallel to Az equal to its height, and c'e' parallel to ISOMETRIC PROJECTIONS. 191 Ay its breadth, and complete the parallelograms de' and ce'. Lay off zf equal to the distance of the second support fi'om the end of the beam, and complete its projection in the same way. Lay off ba equal to the horizontal distance between the two Leams, and construct the projection of the second beam and its supports precisely as the first was constructed. From a to n lay off the distance from the end of the beam to the first cross-piece. Make nn' equal to its breadth, and np parallel to Az equal to its thickness, and draw lines through n, n', and p parallel to Ay. Lay off ar equal to the distance of the second cross-piece from the end of the beam, and construct its projection in the same way as the first. The faces of the frame-work in the shade and seen are darkened in the drawing. 302. PROBLEM 81. To construct the isometric projection of a house with a projecting roof. Let A, Fig. 132, be the upper end of the intersection of the front and side faces of the house, Ax, Ay, and Az the directrices. On Ax lay off the length, on Ay the breadth of the house, and on Az the height of the side walls, and complete the two parallelograms Ap and Ap', they are the projections of the side and front faces of the house. At b, the middle point of Ay, draw br parallel to Az, and make it equal to the height of the ridge of the roof above AY, and join rA and ry; these lines will be the projections of the intersections of the face ZY with the roof. From r lay off rr' equal to the distance that the roof projects beyond the walls of the house, and draw r's and r's' parallel respqctively to rA and ry. On Ay lay off Ad and ye each equal to rr', and draw dd' and ee' parallel to Az, intersecting Ar and ry produced in d' and e'. These will be points of the projections of the eaves. Draw r'r" parallel to Ax, and make it equal to the length of the ridge, and complete the 192 DESCRIPTIVE GEOMETRY. parallelograms:i:i and r't'. These will be the projections of the two inclined faces of the roof. From b lay off bc and bf, each equal to the distance of the side faces of the chimney'from the ridge, and draw cy and fg' parallel to Az, intersecting Ar and ry in g and g', and draw gh and g'h' parallel to?r'r". They will be the projections of the intersections of the planes of the side faces of the chimney with the roof. Make gh equal to the distance of the front face of the chimney from the front face of the house, and draw hi and ih' parallel to Ar and ry; they will be the projections of the intersection of the front face of the chimney with the roof. Make hk equal to the thickness of the chimney, and hl equal to its height, and complete its projection as in the Figure. From z lay off zu equal to the distance of the door from the edge Az, uu' equal to its width, and uw equal to its height, and complete the parallelogram u'w. Make zq equal to the distance of the windows above the base, qn and qo their distances from the edge, Az, nn', and oo' their width, and nm equal to their height, and complete the parallelograms. 303. A knowledge of the preceding simple principles and constructions will enable the draughtsman to make isometric drawings of the most complicated pieces of machinery, and most extended collections of buildings, walls, etc.; drawings which not only present to the eye of the observer a very good representation of the objects projected, but are of great use to the machinist and builder. <6x',K:A - 4 - )y/ 6/ j-C —-SA-C