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Cornell University Library The original of tiiis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924004047605 ELEMENTS DESCRIPTIVE GEOMETRY WITH APPLICATIONS TO SPHERICAL, PERSPECTIVE, AND ISOMETRIC PROJECTIONS, AND TO SHADES AND SHADOWS BY ALBEET E. gHURCH, LL.D. LATE PROFESSOR OF MATHEMATICS IN THE UNITED STATES MILITARY ACADEMT NEW YORK •:• CINCINNATI •:• CHICAGO AMERICAN BOOK COMPANY 5 Copyright, 1864, by BAENE8 & BUEE. COPYEIGHT, 1892 AHD 1902, BY MAEGAEET A. BLUNT. CHURCH. DESOK. GEOM. TEXT. M. P. II PUBLISHEKS' NOTE This book was originally published in 1864. The preface to the first edition states : " Without any effort to enlarge or originate, the author has striven to give with a natural arrangement and in clear and concise language, the elemen- tary principles and propositions of this branch of science, of so much interest to the mathematical student, and so neces- sary to both the civil and military engineer." Professor Church succeeded so well in his efforts to pro- duce a practical and well-adapted treatise that now, nearly forty years after its publication, it is still in use as a text- book in the United States Military Academy and in many other academies, technical schools, and colleges. The continued demand for this favorite text-book has rendered necessary a renewal of the plates. The publishers have submitted the book to several eminent mathematicians, and by their advice have retained the text unchanged as not being capable of improvement. The new edition, how- ever, appears with larger and more legible type, displayed to greater advantage on the page. CONTENTS PART I ORTHOGRAPHIC PROJECTION FAOE Preliminary Definitions 9 Representation of Points 10 Representation of Planes 12 Representation of Right Lines 13 Revolution of Objects 15 Revolution of the Vertical Plane 16 Notation used in the Description of Drawings 20 Manner of delineating the Different Lines used .... 21 Elementary Problems relating to the Right Line and Plane . . 22 Construction and Classification of Lines 42 Projection of Curves 46 Tangents and Normals to Lines 47 Generation and Properties of the Helix 49 Generation and Classification of Surfaces 50 Generation and Properties of Cylindrical Surfaces .... 52 Generation and Properties of Conical Surfaces .... 54 Warped Surfaces with a Plane Directer 57 Generation and Properties of the Hyberbolic Paraboloid ... 59 Warped Surfaces with Three Linear Directrices .... 62 Generation and Properties of the Helicoid 63 Surfaces of Revolution 65 The Hyperboloid of Revolution of One Nappe .... 66 Tangent Planes and Surfaces ; Normal Lines and Planes . . 72 Problems relating to Tangent Planes to Single Curved Surfaces . 77 6 CONTENTS PAGE Problems relating to Tangent Planes to Warped Surfaces . . 86 Problems relating to Tangent Planes to Double Curved Surfaces . 90 Points in which Surfaces are pierced by Lines .... 98 Intersection of Surfaces by Planes. Development of Single Curved Surfaces 99 Intersection of Curved Surfaces 113 Development of an Oblique Cone 119 Practical Problems 122 PART II SPHERICAL PROJECTIONS Preliminary Definitions . . • 129 Orthographic Projections of the Sphere 132 Stereographic Projections of the Sphere 138 Globular Projections 147 Gnomonic Projection] . 148 Cylindrical Projection 149 Conic Projection 149 Construction of Maps 151 Lorgna's Map 152 Mercator's Chart 152 Flamstead's Method 154 The Polyconic Method 155 PART III SHADES AND SHADOWS Preliminary Definitions I57 Shadows of Points and Lines I59 Brilliant Points Ig2 Practical Problems Ig3 CONTENTS PART IV LINEAR PERSPECTIVE PAGE Preliminary Definitions and Principles ...... 176 Perspectives of Points and Right Lines. Vanishing Points of Right Lines 177 Perspectives of Curves 181 Line of Apparent Contour 182 Vanishing Points of Rays of Light and of Projections of Rays . 183 Perspectives of the Shadows of Points and Right Lines on Planes . 185 Practical Problems 186 PART V ISOMETRIC PROJECTIONS Preliminary Definitions and Principles ...... 207 Isometric Projections of Points and Lines 209 Practical Problems 210 PART I ORTHOGRAPHIC PROJECTIONS Preliminaky Definitions 1. Descriptive Q-eometry 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 magni- tudes 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 magnitude 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 plane, the projections are Orthographic. When 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. 10 ORTHOGRAPHIC PROJECTIONS 3. It is manifest that if a straight line be drawn through a given point and the point of sight, the point in which this line pierces the plane of projection will present to the eye the same appearance as the point itself, and will therefore be the projection of the point on this plane. The line thus drawn is the projecting line of the 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, mil 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 situ- ated, 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. The fourth is below the horizontal and in front of the vertical, as marked in the figure. Representation of Points 6. Let M, Fig. 1, be any point in space. Through it draw Mm perpendicular to the horizontal, and Mm' perpendicular to the vertical plane ; m will be the projection of M on the horizontal, and m' that on the vertical plane (Art. 4). ORTHOGRAPHIC PROJECTIONS 11 Hence, the horizontal projection of a point is the foot of a perpendicular through the point to the horizontal plane ; and the vertical projection of a point is the foot of a perpen- dicular 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 completely determined. For if at the horizontal projec- tion m 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 ground line at o. 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. 12 ORTMOGBAPHIG PBOJECTIONS 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 respectively. It will be completely determined in position by its two lines ^T and ^'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 same point. For if they should 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 he 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 he perpendicular to the horizontal plane, its vertical trace will he perpendicular to the ground line, as t'T, in Fig. 2. For the vertical plane is also per- pendicular to the horizontal plane; hence the intersection of the two planes, which is the vertical trace, must be per- ORTHOGUAPHIC PROJECTIONS 13 pendicular to the horizontal 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 of 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; mn will be its horizontal, and pp', 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 m'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 14 OBTHOQRAPHIC PROJECTIONS vertical plane, it will also contain the line. The intersec- tion of these planes must therefore be the line. Hence we say a right line is given hy its projections. 13. The projections mn and m'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 pass 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 he 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 M.n. If the line is parallel to both planes of projection or to the ground 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 ORTHOGMAPHIC PROJECTIONS 15 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 perpendicu- lar to the ground line, the line is undetermined, as the two projecting planes coincide, forming only one plane, and do not by their intersection determine the line as in Art. 12. All lines in this plane will have the same projections. Thus, in Fig. 5, mn and m'n' are both perpendicular to AB ; and any line in the plane MNo will have these for its pro- jections. If, however, the projections of two points of the line are given, the line will then be determined ; that is, if mn' and nn' are given, the two points, M and N, will be determined, and, of course, 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 trace of the plane. 16. If two right lines are parallel, their projections on the same plane will be parallel. For their projecting planes are parallel, since they contain parallel lines and are perpendic- ular to the same plane ; hence their traces will be parallel (Art. 10) ; but these traces are the projections. Rbvoltjtion 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 center 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 remain- ing at the same distance from any of the other points. Thus, 16 ORTHOOEAPHIC PROJECTIONS if the point M, Fig. 6, be revolved about an axis DD', in the horizontal plane, it will describe the circumference of a circle whose center 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 per- pendicular mop, at a distance from o equal to Mo ; that is, it will be found in a straight line passing through its horizontal projection perpendicular to the axis, and at a distance from the axis equal to the hypotenuse 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 th^ 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 verti- cal 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 ; the hypotenuse 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 projec- tion of the point. Revolution op 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 ORTBOGSAPHIC PROJECTIONS 17 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 of 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 front of the ground line, but also that part of the vertical plane which is below the horizontal plane ; while the part beyond 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 preced- ing article, the point to', in Fig. 1, will take the position to" in the line mo produced, and the two projections m and m' mil then be in the same straight line, perpendicular to AB. Hence, in every drawing thus made, the two projections of the same point must he in the same straight line, perpendicular to the ground line. If, then, AB, Fig. 7, be the ground line, and it be required to represent or assume a point in space, we first assume m for its horizontal projection ; through to erect a perpendicu- lar to AB, and assume any point, as to', on this perpendicular, for the vertical projection. The point will then be fully de- termined (Art. 8). The point thus assumed is in the first angle, above the horizontal plane at a distance equal to m'o, and in front of the vertical plane at a distance equal to too (Art. 7). If the point be in the second angle, its horizontal projec- tion TO must be on that part of the horizontal plane beyond the ground line, and its vertical projection to' on that part DESCR. 6EOM. 2 18 OBTEOGRAPHIC PROJECTIONS 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 tvi^o projections vcill be as in (2), Fig. 7, mo repre- senting the distance of the point behind the vertical plane, and m'o its distance above the horizontal plane. If the point be in the third 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 fourth angle, the two projections will be as in (4). 20. To represent, or assume a plane in space, we draw at pleasure any straight line, as tT, Fig. 8, to represent its hori- zontal trace ; then through T draw any other straight line, as Tt', to represent its vertical trace. It is absolutely neces- sary 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 por- tion in the second angle by t'Tt"; that in the third by t"Tt"'; 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. ORTHOGRAPHIC PROJECTIONS 19 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 ; T< is the horizontal, and Tt' the vertical trace of a plane perpendicular to the horizontal plane. Also in (3), Ti 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 projec- tions may be drawn at pleasure, as in (1), Fig. 10, mn is the horizontal, and m'n' the vertical projection of a portion of a straight line in the first angle. In (2), mn is the horizontal and m'n' the vertical projec- tion of a portion of a straight line in the second angle. In (3), the line is represented in the third angle, and in (4), in the fourth angle. In Fig. 11 are, (1) the projections of a right line parallel to the horizontal plane ; (2) those of a right line parallel to the vertical plane, (3) those of a right line perpendicular to the horizontal plane ; and (4) those of a right line perpen- dicular to the vertical plane (Art. 14). 22. To assume a point upon a given right line, since the projections of the point must be on the 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 mm' perpendicular to AB : m', where it intersects m'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 inter- section. For the same reason, the vertical projection of the point must be at the intersection of their vertical projec- 20 ORTHOGRAPHIC PROJECTIONS tions. 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 inter- sect in space, the right line joining the points in which their projections intersect must be perpendicular to the ground line. Therefore, to assume two right lines which 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 perpen- dicular 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 m'n\ also mo ; through m erect mm' per- pendicular to AB, and through m' draw m'o' at pleasure. Two parallel right lines are assumed by drawing their projections respectively parallel (Art. 16). Notation to be used in the Description op Drawings 24. Points represented as in Fig. 7 will be described as the point (mm'), the letter at the horizontal projection being always written and 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 the 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 of the drawing, the same notation will be used, the middle letter being placed on both traces; in the first case, at the left-hand extremity, and in the second case at the extremity nearest the ground line. OBTHOGliAPBIC PROJECTIONS 21 Lines given by their projections, as in Fig. 10, will be described as the line' (mn, m'n''), the letters on the horizon- tal projection being first in order, or simply the line MN. The planes of projection will often be described by the capitals H and V; H denoting the horizontal, and V the vertical, plane. The ground line will be in general denoted by AB. Manner op Delineating the Different Lines used IN THE Representation of Magnitudes, or in the Construction op 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 drawn full. When not seen, or when they are the traces of auxiliary planes, not the projecting 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 con- 22 ORTHOGRAPHIC PROJECTIONS sidered 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. Construction of Elementary Problems relating to THE Right Line and Plane 26. Having explained the manner of representing with accuracy, points, planes, and right lines, we are now pre- pared to represent the solution of a number of important problems relating to these magnitudes in space. In every problem certain points and magnitudes are given, from which certain other points or magnitudes are to be constructed. Let a right line be first drawn on the paper or slate to represent the ground line ; then assume, as in Art. 19, etc., the 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 draw- ing. This is the analysis 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 the planes of projection. Let AB, Fig. 12, be the ground line, and (mn, m'n'^, or simply MN, 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) ; ORTHOGRAPHIC PROJECTIONS 23 and since the point is in the given line, its vertical pro- jection will be in the vertical projection of this line (Art. 13) ; hence it must be at the intersection of this vertical projection with the ground line. The horizontal projection of the required point must be in a straight line drawn through its vertical projection, perpendicular 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 in- tersects the ground line; at the point of intersection erect a perpendicular to the ground line, and produce it until it intersects the horizontal projection of the line; this point of intersection is the required point. Construction. Produce m'n' to w! ; at rnJ erect the per- pendicular m'm, and produce it to m. This is the required point. Second. In the above analysis, by changing the word "vertical" into "horizontal," and the reverse, we have the analysis and rule for finding the point in which the given line pierces the vertical plane. Construction. Produce mn to o ; at o erect the perpen- dicular oo', and produce it to o' . This is the required point. 28. Problem 2. To find the length of a right line joining two given points in space. Let AB, Fig. 13, be the ground line, and (mm') and (ww') the two given points. Analysis. 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 24 ORTHOGRAPHIC PROJECTIONS 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 dis- tance, since the points do not change their relative position during the revolution (Art. 17). Construction. Draw mn and m'n'. MN will be the re- quired line. Now revolve its horizontal projecting plane about mn until it coincides with H ; the points M and N will fall at m" and n", at distances from m and n equal to rm' and sn' respec- tively (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 the 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 method for the same problem. Analysis. If we revolve the horizontal projecting plane of the 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 the required distance. Construction. Revolve the projecting plane about the perpendicular 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 of N in its revolved position. Its vertical projection must be in IV perpendicular to AB; and OBTkOGBAPHIC PROJECTIONS 25 since during the revolution the point N remains at the same distance above H, its vertical projection must also be in the line n'V parallel to AB (Art. 7), therefore it will be at I'. The point M being in the axis remains fixed, and its vertical projection remains at m' ; m'V 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 hypotenuse 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 intersec- tion of the two ; hence every right line of a plane, not parallel to the horizontal 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 given plane, take a point in each trace and join the two by a right line, or other- wise, draw the horizontal projection at pleasure ; at the 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 intersects the ground line — this will be the vertical projection of the line. 26 ORTHOGRAPHIC PROJECTIONS Thus, in Fig. 14, draw mn, also mm' and nn'; join m'n'; it will be the required vertical projection. 31. Problem 3. To pass a plane through three given points. Let M, 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 mn; also m' 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'p'; NP will be the second line. It pierces H at ^ 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, produce 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 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 may be ORTHOGRAPHIC PROJECTIONS 27 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 (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 MO, 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 hori- zontal plane ; then revolve it about its new vertical trace until it coincides with the vertical plane ; the angle will then be in the vertical plane in its true size. 28 OBTHOGBAPHIC PROJECTIONS Construction. First revolve the plane about no until it becomes perpendicular to H. Tt' will be the nevsr vertical trace, the point M will be horizontally projected at s, and vertically at s', s'r being equal to sp. Now revolve the plane about T^' until it coincides with V; will revolve to o", s to s", and n to n", while the point M, or (^ss'y, will be found at w; n"w and o"w will be the re- volved positions of the two lines, and n"wo" the required angle. By examining the drawing, it will be seen that if the angle is oblique, it is less than its projection, unless both lines axe 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 he perpendicular to each other in space, and one of them parallel to the plane of projection, their projections will he perpendicular. For the project- ing 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 perpen- dicular 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 perpendic- ular 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 will be projected on this plane in its true size, and may be bisected by a right line. If the plane be then re- ORTHOGBAPHIC PROJECTIONS 29 volved to its primitive position, and the true position of one point of the bisecting line be determined, and joined with the vertex of the given angle, we shall have the required line. Construction. 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 m", and om" will be the projection of MN, and m"op will be the true size of the angle. Bisect it by oq, which will be the horizontal projection of the bisect- ing line in its revolved position. Join m" with any point of op, as p ; this will be the horizontal projection of a/'iue of the given plane, in its revolved position, which intersects the bisecting line in a point horizontally projected at q. When the plane resumes its primitive position, this line will be horizontally projected in mp, and the point, of which q is the horizontal projection, will be horizontally projected at r, and vertically at r'; hence or will be the horizontal, and 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 figure. 38. Problem 6. To find 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 traces. For 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. Construction. The required line pierces H at o and V at p' : is its own horizontal projection, and p' is horizontally 30 OBTHOGRAPHIC PBOJECTIONS projected at p ; hence po is the horizontal projection of the required 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 method for the same problem. When either the horizontal or vertical traces do not intersect within the limits of the drawing. Let tTt' and sSs', Fig. 19, be the planes; tT and sS not intersecting within the limits of the drawing. Analysis. If we pass any plane parallel to the vertical plane, iu .will intersect each of the given planes in a line parallel to \ts vertical trace, and these two lines will inter- sect in a point of the required intersection. A second point may be determined in the same way, and the right line join- ing these two points will be the required line. Construction. Draw^g' parallel to AB; it will be the trace of an auxiliary plane. It intersects the two given planes in lines which pierce H at ^ and q, and are vertically projected iup'o' and q'o'; o' is the vertical, and o the horizontal pro- jection of their intersection. Draw mn 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 line pierces a given plane. 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, which 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 lines. Construction. Let the auxiliary plane be the horizontal projecting plane of the line ; np is its horizontal and pt' its ORTHOGRAPHIC PROJECTIONS 31 vertical trace (Art. 10). It intersects tTt' in a right line, which pierces H at o and V at t', of which o^t' is the vertical projection (Art. 38). The point m', in which o't' intersects m'n, is the vertical projection of the required point; and m is its horizontal projection. The accuracy of the drawing may be verified by using the vertical projecting plane of MN, as an auxiliary plane, and determining m directly as repre- sented in the figure. Let the problem be constructed when the given line is parallel 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. Oonstruction. 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. OP 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, r being 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 m, Fig. 22, be the horizontal projection of a point of the plane tTt'. If at m a perpendicular be erected to H, it will pierce tTt' in the only point of the plane which can be horizontally projected at m; m is the horizontal, and m"m' the vertical projection of this perpendicular. Through it pass any plane, as that whose horizontal trace is no. Since 32 OBTHOGBAPHIC PBOJECTIONS 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 ver- tical projection; hence ml is the required vertical projection (Art. 40). The auxiliary plane may be passed parallel to f£ ; mp will be its horizontal, and pp' its vertical, trace. It intersects tit' in a line parallel to ^T, which pierces V at p', of which mp is the horizontal, and m'p' the vertical, projection ; and m' will be the required vertical projection. In a similar way, if the vertical projection be given, the horizontal can be found. 43. If a right line is perpendicular to a plane, its pro- jections will he 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 perpen- dicular 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 hori- zontal 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 pro- jection of the line will be perpendicular to the vertical trace. Conversely — If the projections of a right line are respec- tively perpendicular to the traces of a plane, the line will he perpendicular to tJie plane. For if through the hori- zontal projection of the line its horizontal 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 project- ing plane of the line is perpendicular to the given plane; OETBOGBAPHIC PROJECTIONS 33 therefore the intersection of these two planes, which is the given line, is perpendicular to the given plane. Hence, to assume a right line perpendicular 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 the two traces from any point in the ground line perpendicular to the projections of the line. 41. Problem 8. To draw through a given point a right line perpendicular to a given plane, and to find the distance of the point from the plane. Let M, Fig. 23, be the given point, and tTt' the plane. Analysis. Since 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 respec- tively perpendicular to the traces of the plane (Art. 43). Hence, if through the horizontal projection of the point a right line be drawn perpendicular to the horizontal trace, and through the vertical projection a right line perpendicu- lar to the vertical trace, they will be respectively the hori- zontal 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 ^T, and through m', m'n', 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 perpendicu- lar (Art. 28). Let the problem be constructed when the plane is parallel to the ground line ; also when it is perpendicular to it. DESCR. GEOM. 3 34 ORTHOGRAPHIC PROJECTIONS 45. Problem 9. To project a given right line on any oblique plane, and to show the true position of this projection. Let MN, Fig. 24, be the giveu line, and f£t' the given plane. Analysis. If through any two points of the line perpen- diculars be drawn to the plane, and the points in which they pierce the plane be foun^ these will be two points of the required projection, and the right line joining them will be the required line. If now the plane be revolved about its horizontal trace to coincide with the horizontal plane, or about its vertical trace until it coincides with the vertical plane, and the revolved position of these two points be found and joined by a 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 MR and PS (Art. 44). The first pierces the plane at R, and the second at S (Art. 40); and rs will be the horizontal, and r' s' the vertical projection of the required 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 H. R is found at r" (Art. 17), and S at s", and r"s" is the true position of RS in its own plane ; r"s" produced must pass through the 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 through this to draw a line parallel to the given line (Art. 14). 46. Pkoblem 10. Through a given point to pass a plane perpendicular to a given right line. Let M, Fig. 25, be the given point, and NO the given line. ORTHOGRAPHIC PROJECTIONS 35 Analysis. Since the plane is to be perpendicular to the line, its traces must be respectively perpendicular to the pro- jections 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. Through 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 th& point in which this trace intersects the ground line draw a right line perpen- dicular 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 hqrizontal projection of a line through M, parallel to the hprizjontal trace ^ and since this line is parallel to H, its vertical projection will be m'p' parallel to AB. This line pierces V at jo' (Art. 27). Draw ^'T perpendicular to n'o', and Tt perpendicular to no ; tTp' will be the required plane. Or through M 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 pass a ]»lane through a given point parallel to two given right lines. Let M, Fig. 26, be the point, and NO and PQ the two given lines. Analysis. 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. Construction. Through m draw ms parallel to no, and through m', m's' parallel to n'o'. The line MS will be parallel to NO (Art. 16). In the same way construct MR parallel 36 ORTHOGRAPHIC PROJECTIONS to QP. These lines pierce H at s and t respectively, and MR pierces V at r' ; hence tTr' is the required plane (Art. 32). Let the problem be 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 pass a plane through a given right line parallel to another right line. Let MN, Fig. 27, be the line through which the plane is to be passed, and PQ the other given line. Analysis. Through any point of the first line draw 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 H at m, and V at n' ; hence oTt' is the required plane. Let the problem be constructed when either line is paral- lel to the ground line. 49. Problem 13. To find the shortest distance from a given point to a given right line. Let M, Fig. 28, be the given point, and NO the given straight line. Analysis. The required distance is the length of a per- pendicular from the point to the line. If through the given point and the line we pass a plane, and revolve this plane about either 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 posi- tion of the line, it will be the required distance. ORTHOGBAPHIC PROJECTIONS 37 Construction. Through M draw MP parallel to NO. It pierces H at p. NO pierces H at o. po is then the hori- zontal 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 'm"q" perpendicular to oq"; it will be the required distance. When the plane is revolved back to its primitive position, m" is horizontally projected at m, and q" at q, hence MQ is the perpendicular in its true position. 50. Second method for the same problem. Analysis. If through the given point a plane be passed perpendicular to the given line (Art. 46), and the point in which the given line pierces the plane be found (Art. 40), and joined with the given point, we shall have the required dis- tance, the 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 makes 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 per- pendicular be drawn to the plane, the foot of this perpen- dicular will be one point of the projection. If this point be joined with the point in which the given line pierces 38 ORTHOGRAPHIC PROJECTIONS 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. But the angle at the vertex, that is, the angle between the perpendicular and given line, is the complement of the required angle ; hence, if we find the latter angle and subtract it from a right angle, we shall have the required angle. Construction. Through M draw the perpendicular MP to tTt' (Art. 44). It pierces H 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 determine their angle pmJ'o, as in Art. 34. Its com- plement, prmJ, is equal to the required angle. Let the problem be constructed when the plane is parallel to the ground line. 52. Problem 15. To find the angle between two given planes. Let sSs' (Fig. 30) and tTt' be the two planes intersecting in the line ON (Art. 38). Analysis. If we pass a plane perpendicular to the inter- section of the two planes, it will be perpendicular to both, and cut from each a right line perpendicular to this inter- section at a common point. The angle between these lines Avill be the measure of the required angle. Construction. 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, one of which pierces H at p, and the other at q. If right lines be drawn from these points to the point in which the auxiliary plane intersects ON, they will be the lines cut from the planes, and the angle between them will be the requii-ed angle. OETHOGBAFHIC PROJECTIONS 39 The vertical trace of the auxiliary plane may be drawn 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 ver- tex of the angle, it will be perpendicular to ON, since it is contained in a plane perpendicular to it ; it will also be per- pendicular to pq, 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, n' 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, m" will be at v, rv being equal to rm", and pvq will be the required angle (Art. 34). The point m", from its true position, is horizontally pro- jected at m, and vertically at m', and pmq is the horizontal, and p'm'q' the vertical projection of the angle. 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 from H, and ON from tTt', and the angle non", 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. Problem 16. Either trace of a plane being given, and the angle which the plane makes with the corresponding plane of projection, to construct the other trace. Let tT, Fig. 31, be the horizontal trace of the plane, and 40 ORTHOGSAPHIC PROJECTIONS def 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 angle equal to the given angle, this will be the revolved position of a line cut from the required plane by this perpen- dicular plane (Art. 53). If this line be revolved to its true position, and the point 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. Construction. Through o draw no perpendicular to tT ; also on'', making the angle nan" = 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 w', and m'T is the required trace. 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 the shortest line which can be drawn, terminating in two right lines, not in the same plane. Let MN, Fig. 32, and OP be the two right lines. Analysis. 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 parallel to the first line. It will then intersect the first line ORTHOGRAPHIC PROJECTIONS 41 ill a point. If, at this point, we erect a perpendicular to tlie 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) ; mr is its horizontal, and k'n' its vertical trace. 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 parallel plane. Through U draw UX parallel to OP ; ' it will be the projection of OP on the plane. It intersects MN 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 projection. 56. Second 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 the plane will be one point of the projection of OP on the plane. To find 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 hori- zontal 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 I", and pi" is the revolved position of 42 ORTHOGRAPHIC PROJECTIONS OP. This projecting plane intersects the parallel plane in a right line, which pierces H at k, and is parallel to OP ; ku, parallel to pi", is the revolved position of this parallel line ; pu, perpendicular to pi", is the revolved position 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 posi- tion of the line of intersection of the perpendicular and parallel plane; pp" is the revolved position of the line through p perpendicular to the parallel 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, some- where in the perpendicular to the axis pq; p"x" is the horizontal projection of the projection of OP on the parallel plane, and xy, perpendicular to mr, is the horizontal, and x'y' the vertical projection of the required line. Construction and Classification of Lines 57. Every line may he generated hy the continued motion of a point. If the generating point be taken in any position on the line, and then be moved to its next position, these two points may be regarded as forming an infinitely small right line, or elementary line. The two points are consecutive points, or points having no distance between them, and may practi-^ cally be considered as one point. The line may thus be regarded as made up of an infinite number of infinitely small elements, each element indicating ORTHOGRAPHIC PROJECTIONS 43 the direction of the motion of the point while generating that part of 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 in 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 con- secutive points, are in the same plane, the curve is of double curvature. We thus have three general classes of lines. I. Right lines : all of whose points lie in the same direction. II. Curves of single curvature : 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. 59. The simplest curves of single curvature are : I. The circumference 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 directrix. 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. 44 OBTHOGBAPBIC PROJECTIONS From the definition, the curve may readilj' 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 center, 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 the foci. The curve may be con- structed by points, thus : Let F and F', Fig. 35, be the two foci, and VV the given right line, so placed that VF = VF'. Take any point as P between F and F'. With F as a center and VP as a radius, describe an arc. With F' as a center, and VP as a radius, describe a second arc, inter- secting the first in the points M and M'. These will be points of the required curve, since MF + MF' = VP + VP = VV ; also MF + M'F' = VV. In the same way all the points may be constructed. V and V are evidently points of the curve, since VF + VF' = VF' -f VF' = VV ; also VF' + VF = VV. The point C, midway between the foci, is the center of the curve. The line VV, passing through the foci and ter- minating in the curve, is the transverse axis of the curve. OEmOGtlAPmc PROJECTIONS 45 The points V and V are the vertices of the curve. DD' perpendicular to W at the center 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 center, 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 in 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 center, and any radius greater than F'V, as FO, describe an arc. With F as a center, and a radius FM, equal to F'O — VV, describe a second arc, 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. It is manifest also that if the greater radius be used with F as a center, another branch,, NVN', exactly equal to MVM', will be described. V and V are evidently points of the curve, since F'V - FV = VV = FV - F'V, and are the vertices of the hyperbola. 46 OETHOGBAPHIC PROJECTIONS The point C, midway between the foci, is the center, and VV is the transverse axis. A perpendicular, DD', to the transverse axis at the center is the indefinite conjugate axis. It evidently does not intersect the curve. Projection of Cueves 60. If all the points of a curve be projected upon the hori- zontal plane, and these projections be joined by a line, this line is the horizontal projection 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 in 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 m 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 maj-^ in general be determined. 62. If the plane of a curve of single curvature is perpen- dicular 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. If the plane of the curve be perpendicular to the ground line, both projections will 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 element of the curve will be projected into an ORTHOGRAPHIC PROJECTIONS 47 .equal element (Art. 14). Its projection on the other plane will be a right 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 0, Fig. 37, is the point in which the curve MN pierces H, 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 M, 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 con- taining 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 practical purposes, be regarded as the tangent, we say that a right line is tangent to another line when it con- tains two consecutive points 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 of the curve. Two curves are tangent to each other when they contain two consecutive points, or have, at a common point, a common tangent. If a right line is tangent to a curve of single curvature, it will be contained in the plane of the curve. For it passes through two points in that plane, viz. the two consecutive points of the curve. 48 ORTHOGRAPHIC PROJECTIONS Also, if a right line is tangent to another right line, it will coincide ivith it, as the two lines have two points in common. The expression "a tangent to a curve" or "a tangent" will hereafter be understood to mean a rectilineal tangent unless otherwise mentioned. 65. // two lines are tangent in space, their projections on the same plane will be tangent to each other. For the projections of the two consecutive points common to the two lines will also be consecutive points common to the projec- tions 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 will be tangent in space ; for the pro- jecting perpendiculars at the common consecutive points will intersect in two consecutive 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 other- wise mentioned. 67. If we conceive a curve to be rolled on its tangent at any point until each of its elements in succession conies into this tangent, the curve is said to be rectified ; that is, a right line equal to it in length has been found. Since the tangent to a curve at a point contains the element of the curve, the angle which the curve at this point makes with any line or plane will be the same as that made by the tangent. outhographic projections 49 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 m, Fig. 39, be the horizontal, and m'n' the verti- cal 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 r, it will be vertically projected at r'; and in the same way the points «/', q', etc., 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 generating point, that in generating any two equal portions of the curve, it ascends the same vertical distance ; that is, any two elementary arcs of the curve will make equal angles with the horizontal plane. Thus, if CD (a). Fig. 39, be any element of the curve, the angle which it makes with tht' DESCB. GEOM. — 4 50 ORTHOGRAPHIC PROJECTIONS horizontal plane will be DCe, or the angle at the base of a right-angled triangle of which Ce = cd, Fig. 39, is the base, and De the altitude. But from the nature of the motion, Ce is to De as any arc px is to the corresponding ascent x"x'. Hence, if we rectify the arc xp (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 xs tangent to the circle pxr at a;; it will be the hori- zontal projection of the required tangent. On this, from x, lay off the rectified arc xp to 0; z will be the point where the tangent pierces H, 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 hypotenuse of a right-angled tri- angle 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 generating 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 hypotenuse of a right-angled triangle, of which the base is its horizontal projection rectified, and altitude, the ascent of the generating point while generating the portion considered. Thus the length of the arc joX is equal to the length of the portion of the tangent ZX. Generation and Classification op Suepaces 70. A surface may he generated hy the continued motion of a line. The moving line is the generatrix of the surface ; and the diflferent positions of the generatrix are the elements. OETHOGRAPHIC PROJECTIONS 51 If the generatrix be taken in any position, and then be moved to its next position on tlie surface, these two positions are consecutive positions of the generatrix, or consecutive ele- ments 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 ivhich can he generated hy right lines, or which have rectilinear elements. Second. Those which can only he generated hy curves, and which can have no rectilinear elements. These are Double CUKVED SURFACES. Those which can be generated by right lines are: (1) Planes, which may be generated by a right line moving so as to touch another right line, having all its positions parallel to its first position. (2) Single curved surfaces, which may be generated by a right line, moving so that any two of its consecutive positions shall he in the same plane. (3) Warped surfaces, which may be generated by a right line moving so that no two of its conservative positions shall he in the same plane. 72. Single curved surfaces are of three kinds. I. Those in which all the positions of the rectilinear genera- trix are parallel. II. Those in which all the positions of the rectilinear gen- eratrix intersect in a common point. III. Those in which the consecutive positions of the rectilinear generatrix intersect two and two, no three posi- tions intersecting in a common point. 62 orthographic projections Cylindrical Suefaces, or Cylinders 73. Single curved surfaces of the first kind are cylindrioal surfaces, or cylinders. Every cylinder may be generated by moving a right line so as to touch a curve, and have all its positions parallel. Tlie moving line is the rectilinear generatrix. The curve is the directrix. The different positions of the generatrix are the rectilinear elements of the surface. Thus, Fig. 40, if the right line ^IN be moved along the curve mlo, having all its positions parallel to its first posi- tion, it will generate a cylinder. If the 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 gen- eratrix, and the surface be regenerated. This curve of inter- section may also be the base of the cylinder. The intersection of the cylinder by the horizontal plane is usually taken as the base. If this base have a center, 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 base, and the other the upper base. Cylinders are distinguished 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 curvi- ORTHOGRAPHIC PROJECTIONS 53 linear directrix, as a generatrix, along any one of the recti- linear elements, as a directrix, the curve remaining always parallel to its first position. If the curvilinear directrix be changed to a right line, the cylinder becomes a plane. It is manifest that if a plane parallel to the rectilinear ele- ments intersects the cylinder, the lines of intersection will be rectilinear elements which will intersect the base. 74. It will be seen that the projecting lines of the differ- ent 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 projecting 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 mlo, Fig. 40, in the horizontal plane. Tangent to this draw right lines Ix and Jcr parallel to the horizontal projection of the genei'atrix; these will be the horizontal projections of the extreme rectilinear elements as seen from 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 m' 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 its projections as the horizontal. Through this point 54 ORTHOGRAPHIC PROJECTIONS erect a perpendicular to the horizontal plane. It will pierce the surface in the only points which can be horizontally pro- jected at the point taken. Through this perpendicular pass a plane parallel to the rectilinear elements ; it will intersect the cylinder in elements (Art. 73) which will be intersected by the perpendicular in the required points. Construction. Let p. Fig. 40, be the horizontal projection assumed. Through p draw pq parallel to Ix ; it will be the horizontal trace of the auxiliary plane. This plane inter- sects the cylinder in two elements, one of which pierces H at q and the other at u; and q'y' and u'z" will be the ver- tical projections of these elements, v'p' the vertical projec- tion 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 surfaces., 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 ORTHOGRAPHIC PROJECTIONS 55 which are called nappes; one the upper, the other the lower, nappe. Thus, if the right line MS, Fig. 41, move along the curve mlo and continually pass through S, it Avill 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 generatrix, and the cone be regenerated. This curve of intersection may also be the base of the 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 planes is considered, it is called a, frustum of a cone; one of the limiting curves being the lower and the other the upper base of the frustum. Cones are distinguished by the names of their bases ; as a cone with a circular base, a cone with a parabolic base, etc^ 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 accord- ing 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 the base. 78. A cone is represented by projecting the vertex, one of the curves on its surface, and its principal rectilinear ele- 56 OBTHOGBAPHIC PBOJECTIONS ments. Thus let S, Fig. 41, be the vertex. Draw the base mlo in the horizontal plane, and tangents to this base through s, as d 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 m's' and o'g\ thus forming the vertical projection of the cone. 79. To assume a point of the surface, we first assume one of its projections as the horizontal. Through this erect a perpendicular to the horizontal plane ; it will pierce the sur- face 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 the perpendicular in the required points. Construction. Let p be the horizontal projection. Draw ps. It will be the horizontal trace of the auxiliary plane which intersects 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 rectiliiiear 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 gen- erated by drawing a system of tangents to any curve of double curvature. These tangents will evidently be recti- linear elements of a single curved surface. For if we con- ceive a series of consecutive points of a curve of double curvatiire, as a, J,