%vtv&\\ UttioetBitg 2Itbtart|- ..P.ttJb..l.i.sii.er.s.. Cornell University Library arV19534 Descriptive geometi 3 1924 031 281 094 ■ -i.anx The original of tliis book is in tlie Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031281094 DESCRIPTIVE GEOMETRY BY ADAM V. MILLAR Assistant Professor of Drawing University of Wisconsin EDWARD S. MACLIN Professor of Industrial Education ' University of Tennessee 1 and LORRAINE J. MARKWARDT Engineer in Forest Products U. S. Department oE Agriculture TBACY & KIIiGOBB, PRINTERS MADISON, WISCONSIN. 1919 Copyright, 1919 BY A. V. MILLAR E. S. MACLIN AND L. J. MARKWARDT PREFACE In preparing this text in descriptive geometry, the authors have endeavored to present the subject in a simple, logical man- ner and in close conformity to commercial drafting practice, as well as to develop in the student the faculty of visualizing mag- nitudes in space. One of the principal factors in accomplishing these desiderata is the omission of the ground line. When the views of several points are given without the ground line being shown, the dis- tances of the points from the horizontal or vertical planes of projection are not determined. The front view, however, does show the relative heights of the points and the top view shows the relative distances of the points from the vertical plane. It is the relative distances of points of an object from a plane with which we are concerned, since the distance of the whole object from the plane of projection does not change the ortho- graphic projection of the object on that plane. When it becomes necessary to locate points which are at given distances from the planes of projection the ground line must of course be used. In all cases, however, even when the ground line is not shown on the drawing, it is understood to be at right angles to the line joining the top and front views of any point. By the omission of the ground line, and therefore the traces of a plane, the student's attention is concentrated on the object or magnitude in space, and not on the planes of projection. This teaches the student to visualize the object rather than mem- orize the views of the object and leads to more original work. The subject is thus made easier because memorizing construc- tions and centering attention on the views of the object rather than on the object itself are the greatest obstacles which the student encounters in mastering the subject. Since the ground line is omitted in commercial work, the subject as presented in this text conforms to common practice.' [iiil While the first quadrant or angle is used in nearly all current text books on descriptive geometry, it is worthy of note that the third quadrant is used almost exclusively in the drafting oflSces of this country and at the various universities as well in the mechanical drawing courses which are a pre-requisite to descrip- tive geometry. It seems logical and advisable therefore to pre- sent the subject in the third quadrant as is done in this text. Since the elimination of the ground line dispenses with the necessity of using a particular horizontal or vertical plane, it is possible to represent objects such as cones in their natural posi- tions rather than inverting them to bring their bases in a hori- zontal plane of projection. This removes the chief objection to the use of the third quadrant. Another salient feature of this text is the extensive use of auxiliary views, that is views other than the top, front, or end views. These are taken from any direction which will aid in construction or give a clearer idea of the form of the object. These views train the visualizing powers of the student. In beginning the study of descriptive geometry, students usually experience considerable difficulty in assuming, on their own initiative, satisfactory layouts for the given exercises and problems, and much valuable time is consumed in this preliminary work. The system of miniature layouts for black board work and the co-ordinate system of presenting problems for drafting room work are features of this text which go far toward reliev- ing these difficulties and allowing the student to concentrate his attention on the solution of the problem. Although the method used here of •presenting the principles of descriptive geometry is new in American texts on the subject, it is used to some extent by French authors such as Javary, Pillet, "F. J.", and others. It has been used and developed at the University of Wisconsin for the past six years, with results which have been most satisfactory and gratifying. The authors have consulted many descriptive geometries in preparing the following text, but are particularly indebted for suggestions and problems to the following: Javary, Pillet, "F. J.", MacCord, Smith, Church, Higbee, and Ferris. [Iv] CONTENTS PAGE Introduction ix FIRST PRINCIPLES Representing an object 5 Projection 6 CHAPTER I THE ELEMENTARY PRINCIPLES OF POINT. STRAIGHT LINE, AND PLANE Views or projections of an object 7 Planes op projection 8 Relative position of views 9 Points Views of points ]2 Picturing magnitudes from the drawing 13 Letter notation 14 Straight Lines Views of a line 16 Point on line 17 Co-ordinate system 19 Revolution and counter revolution 20 Revolution about an axis parallel to H 21 Revolution about an axis parallel to V 22 The true length of a line By means of an auxiliarj' view 24 By means of rotation 27 The angle which a line makes with H and V By means of an auxiliary view 24 By means of rotation 27 Planes Representation of planes , 30 Points and lines on planes 31 Plane containing point and parallel to two lines 33 The angle which a plane makes with H 34 To represent a plane which makes a given angle with H 35 The angle which a plane makes with V 3() To represent a plane which makes a given angle wjth V 37 M Plane figures To find the true size of an angle 39 To bisect an angle 41 Through a given point to draw a line which makes a given angle with a given line 43 To construct a square on an oblique plane 44 Intersections of lines with planes To find where a line pierces a plane 46 To find the line of intersection of two planes 50 Perpendicular Relations Between Lines and Planes A line perpendicular to a plane 52 To find the distance from a point to a plane 53 To represent a plane which is perpendicular to a line 55 The projection of a line on a plane 57 The angle which a line makes with a plane 59 The angle between two planes 61 The common perpendicular to two lines 62 Special Problems A line making given angles with both H and V 66 A plane making given angles with both H and V 67 CHAPTER II APPLICATIONS OF THE ELEMENTARY PRINCIPLES OF THE POINT. LINE, AND PLANE Shades and shadows 75 Plane sections and developments of prisms and pyramids . . 82 Intersection of prisms and pyramids 88 CHAPTER III CURVED LINES AND SURFACES Generation and classification of lines 91 Tangents and normals to lines 93 Corves ob' single curvature 94 Curves op double curvature 100 Generation and classification of surfaces 102 Surfaces of revolution 103 Tangent planes to surfaces 105 Single curved surfaces 108 Cylinders 108 Cones Ill Convolutes 114 [vi] "Warped surfaces 115 Hyperboloids of revolution of one sheet 115 Helicoids 118 Hyperbolic Paraboloids 120 Conoids 123 Cylindroids 124 Double curved surfaces 125 CHAPTER IV PLANE SECTIONS AND DEVELOPMENTS OF CURVED SURFACES HlGHT CYLINDERS .' . . . 133 Oblique cylinders 136 Oblique cones 138 Warped surfaces 140 Double curved surfaces , 140 Approximate development of warped surfaces 142 Approximate development of a sphere 144 CHAPTER V INTERSECTIONS OF CURVED SURFACES Intersection of any two curved surfaces 148 Intersection of two right cylinders 148 Intersection of right cone and right cylinder 150 Intersection of two oblique cylinders 151 Intersection of cone and cylinder 164 Intersection of two cones 156 Intersection of sphere with cone or cylinder 156 Intersection of torus and cylinder 157 General method for finding intersections of surfaces 157 [vli] INTRODUCTION It may help some instructors who contemplate using the fol- lowing text-book to know how the authors have successfully used the book in their classes. With this in view, the follow- ing general method for conducting the course is suggested. Each instructor will, no doubt, need to alter the method to some extent to suit the conditions under which he works. At the University of Wisconsin, descriptive geometry is given as a three credit course for one semester of eighteen weeks. Bach week's work consists of one general lecture for all students in the course, one recitation and two two-hour drafting periods for each section. One of the two-hour draft- ing periods is sometimes turned into a one-hour recitation period. At the lecture, the general principles involved in the next lesson are explained , general announcements made, and prob- lems assigned for a home plate which is to be handed in at the beginning of the recitation period. At the recitation, the stu- dents are drilled in the analyses of the problems and then sent to the black board with some particular problem to solve. Black board problems are assigned from the miniature layouts which accompany many of the lists of problems in the text. This in- sures a satisfactory figure and requires less time than making layouts from co-ordinates. In the drafting room, other and usually more difficult prob- lems are assigned. The number solved in each drafting period varies from one to four in accordance with the difficulty of the problem. The co-ordinate system of presenting layouts for drafting room problems will be found very convenient. The student solves the problems and if he has time letters the statements of the problems. The work is done on a 11" x 15" sheet and is left in pencil. Neatness, clearness, and, accuracy [ix] are demanded. With the exception of a few plates, the work is completed and handed in at the close of each two-hour drafting period. The plates are then corrected, graded, and returned to the student at the next drafting period. This method of giving a plate to be completed each time the student comes to the draft- ing room has the following advantages: the student comes bet- ter prepared for his work, he wastes no time in the drafting room, and the attendance is improved. Unannounced written quizzes are given in the drafting room about every three weeks.- Neatness and clearness of the con- struction counts for 15% of the grade. M DESCRIPTIVE GEOMETRY FIRST PRINCIPLES 1. Representing" an object. Some objects may be described by a written or oral statement in such a way that a clear idea of the size and form of the object may be obtained. For example, the statement "a cube having an edge 3" long" gives a clear idea of the form and size of the object. But when the abject becomes more complicated and it is necessary to show not only its form and size, but also its position in space or its position with reference to some other object, the description by means of an oral or written statement becomes very difficult. A picture or drawing of the object will usually give the desired informa- tion regarding it more quickly and acurately than any other form of expression. For this reason the draftsman makes use of working drawings to convey his ideas to the manufacturer. It is the purpose of descriptive geometry to explain the meth- ods employed in making such a drawing. Usually these princi- ples are explained by reference to geometrical magnitudes, such as points, lines, and planes, but their application to material things can easily be made. The study of descriptive geometry should enable a person to form a mental picture of an object of three dimensions from a drawing made on a plane sheet. This process is called visualizing the object. Visualizing not only enables a person to read a com- pleted drawing, but it also helps the designer to form a mental picture of one part of an object and its position with respect to other parts even before a drawing of the object is made. There are different methods of representing an object by means of a drawing, but all of the methods use the principle of projection as explained in the following article. b DESCRIPTIVE GEOMETRY 2. Projection. If from a point S, Fig. 1, straight lines are drawnlthrough a series of points A, B, C, , the points a, h, c, in which these lines pierce the plane T, are the projections of the points A, B, C on this plane. S is the point at -which the eye is supposed to be located, and is called the point of sight. A, B, C, . . . .are points such as corners of an object in space. Sa, Sb, Se are lines of sight or projecting lines of the points A, B, C -rS / / / /^\ / ^' / , is! I ao T !k Pig. 1.— Perspective. Fig. 2.— Oblique Fig. Z.— Orthographic projection. projection. T is the picture plane or the plane of projection. Leaving color out of consideration, the projections a, 6, c, present the same appearance to the eye, situated at the point of sight, as the points A, B, C, in space. In perspective, Fig. 1, the point of sight is at a finite dis- tance from the plane of projection. The projecting lines diverge. Such a view is a true picture of the object, similar to the view obtained with a camera. In oblique and orthographic projection, Figs. 2 and 3, the point of sight is at an infinite distance from the plane of pro- jection. The projecting lines are parallel. The projection is Oblique when the projecting lines are par- allel to each other and oblique to the plane of projection. In per- spective and oblique projections, the picture from one position is considered sufficient to represent the object. The projection is orthographic when the projecting lines are perpendicular to the plane of projection. Since this form of projection is more commonly used than any other in engineering work, the greater part of this book will be devoted to the prin- ciples involved in orthographic projection. CHAPTER I THE ELEMENTARY PRINCIPLES OF THE POINT, STRAIGHT LINE, AND PLANE 3. A point in space is not completely determined by its ortho- graphic projection on one plane, for the distance of the point from the plane is not shown by its projection. All points A, B, C, Fig. 4, which lie in a vertical straight line have the same projection on a horizontal plane. There are two methods of representing definitely a point in space. One method is to give its projection on a plane and also its y^* distance from that plane. This I rj method is used in making contour j maps; a contour line being a line C* joining the projections of all points of an object which are a given dis- aib \^ tance above or below a base plane. ^ The other method requires the use „ , , . , of two (or more) different planes Fig. 4.— a, 6, c, projectoms of . points A, B, C, on plane T. «* proDection. When two projec- tions are used, one is usually con- sidered the principal projection. The other is supplementary, showing the distance of the point from the plane upon which the principal projection is made. The method in which two projec- tions of an object are used is the more common. 4. Views or projections of an object. The top view of an object is the view obtained by looking directly down fi-om the top. The point of sight is imagined to be an infinite distance directly above the object. The projecting lines for different points of the object are vertical and the plane of projection upon which this view or drawing is made is horizontal. This plane is called the horizontal plane of projection or H. The top view is also called the horizontal projection of the object and in architectural work it is called the plan. 8 DESCRIPTIVE GEOMETRY The front view of an object is the view obtained by looking directly from the front. The projecting lines for'this view are horizontal and are therefore at right angles to the projecting lines used in obtaining the top view. The plane upon which the front view is made is in a vertical position, and is called the vertical plane of projection or V. The front view is also called the vertical projection of the object, and in architectural work it is called the front elevation. The horizontal and vertical planes of projection are at right angles to each other. Pig. 5. Their line of intersection is called the gfround line or G. L. In most drawings, the object represented is considered to be below H and back of V. An object in this position is P said to be in the third quadrant or third angrle. In this case the planes of projection are between the points of sight and the ob- ject. In some drawings the object is considered to be above H and in front of V. In such a position the object is said to be in the first quadrant or first angle. Here the object is between the points of sight and the planes of projection. The end or side view of an object is the view obtained by looking directly from the left or right. The projecting lines for this view are parallel to the ground line. The plane upon which this view is made is perpendicular to the ground line and is called an end or profile plane or P. The end view is also called the profile projection and in architectural work the end or side elevation. 5. In this text, the term auxiliary view will be used for any other than the top, front, or end view. If the top and front views of an object are given, an auxiliary view taken by look- FiG. 5.—Trincipal planes of projection. PROBLEMS IN POINT, STRAIGHT LINE, AND PLANE 9 ing parallel to H or V can be derived from these views. It must be remembered, however, that the top and front views do not g^ive information enoug'h to construct easily an auxiliary view taken by looking- obliquely to both H and V. In Fig. 6, the right square pyramid is given by its top and front views. Three auxiliary views are shown, in each case the pro- FiG. 6.— Views of Pyramid. Orthographic Projection. jecting lines being parallel to H. In the front view, h'h' repre- sents a horizontal plane upon which the pyramid stands. This plane is represented in the different auxiliary views by hi hi, h^ hi, and h^ h^. In each view, it is placed at right angles to the line of sight, but its distance from the top view is arbitrary. The altitude of the pyramid shows true length in all except the top view where it appears as a point. Notice that in different views, diiferent edges are hidden and that none of the edges show true length in any of the views. In the first auxiliary view, the planesof two of the faces of the pyramid each appear as straight lines, one as the line a^ Vi and the other as 61 i)i. 10 DESCRIPTIVE GEOMETRY In the following pages, the terms top view, front view, end or side view, and auxiliary view will be used. It must not be for- gotten that these terms are synonymous, respectively, to hori- zontal projection, vertical projection, profile projection, and auxiliary projection. 6. Relative position of views. In Fig. 7, the H, V, and P planes are shown in their proper relative positions. H and V are at right angles to each other, and P is at right angles to both of them, and therefore, at right angles to their line of in- tersection G. L. The point A is shown as being below H and back of V, that is, in the third quadrant, a is the top, a' the front, andtti the right end view of the point A. If all these views Fig. 7.— Views of point A. Fig.— 8. are shown on one sheet or drawing the different planes of pro- jection H, V, P, etc., must all be considered as taken from their natural positions with respect to the object, and placed with the side which is turned away from the point of sight against the drawing surface. Notice that this was done in Fig. 6. The planes are usually placed with respect to each other as shown in Fig. 8. The ground line, G. L., is placed parallel to the edge of the T-square and the line joining a and a' at right angles to G. L. The end views are usually placed opposite the front view, PROBLEMS IN POINT, STRAIGHT LINE, AND PLANE 11 the right end view to the right of the front view as shown in Pig. 8 andthe left end view, when shown, to the left of the front view. Sometimes it is more convenient to place the end views to the right and left of the top view. If the object is in the first quadrant, that is above H and in front of V, the front view is placed above the top view on the drawing. If the object represented is large or is drawn to a large scale, it frequently happens that each view is made on a separate sheet. This is usually the case in architectural work, a separate sheet being used for each floor plan, and still other sheets for front elevation, side elevation, etc. 12 DESCRIPTIVE GEOMETRY POINTS 7. Views or projections of points. In Fig. 7, a is the top, a' the front, and ai the right end view of the point A. The point is below H and back of V, that is, in the third quadrant. When all these views are made on one sheet, they are usually placed with respect to each other as shown in Fig. 8. The distance from the top view of the point to the ground line is the same as the distance from the point in space to V. This dis- tance also shows in the end view. The distance from the front view of the point to the ground line is the same as the distance from the point in space to H. When only one point in space is under consideration, the ground line must be shown on the draw- ing in order to fix definitely the position of the point; otherwise there wuold be nothing from which distances could be measured. Fig. 9 represents two points A and B, and their positions with respect to H, V, and P. Fig. 10, shows the top, front, and end A- -la. ,1 dr H A. a'^ A. ■^ M b. a Fig. 9.— Points A and B and their Views Fig. 10.— Views of A and B views of the points A and B with the ground line omitted. The top view shows that B is the distance d farther back than A, and the front view shows that B is the distance d^ higher than A. The end view shows both distances d and ^2- The top and front views both show that B is the distance d^ to the right of A. If POINTS 13 two views of the points A and B are given, the third can be de- rived from these two. When the ground line is omitted, the drawing does not show how far the points, such as A and B, Fig. 10, are from H, V. and P, but it does give their true relative position. Thus the drawing. Fig. 10, shows that B is higher, farther back, and farther to the right than A. It is the relative position of points of an object which deter- mines the forms of its views or projections. As an illustration, suppose a three inch cube placed in a certain position and the top, front, and end views drawn. If each corner of the cube be lowered two inches, the top view of the cube will remain un- changed. The front and end views will remain the same in form, but may be placed lower on the sheet. But if one corner of the cube be placed one inch, another two inches, and a third three inches lower than they were originally, all views of the cube will be changed in form. The view of an object, therefore, does not depend upon the distances of its points from the pic- ture plane, but rather upon the relative distances of its points from that plane. Hereafter the ground line will rarely be shown on the drawing, but will be referred to at times to indicate direction. On the drawing, the ground line is at right angles to the line joining the top and front views of a point, and is usually parallel to the edge of the T-square. In space, it must be remembered as the line of intersection of H and V. 8. Picturing magnitudes in space from the drawing. In order to solve problems intelligently, the student should learn to picture to himself points, lines, and objects in their proper po- sitions with reference to the drawing. When the drawing is in a horizontal position, it is best to consider the plane of the drawing as H, or the plane upon which the top view is made. The points of the magnitude should then be pictured as being directly above or below their top views. For example, in Fig. 10, the point B can be pictured as being on the plane of the drawing at b. Then A is the distance .<^2 directly below its top view a. This gives a definite picture of the points A and B in space. 14 DESCRIPTIVE GEOMETRY It is more difficult to picture the maghitude from its front view. To do this, hold the plane of the drawing, Fig. 10, in a ver- tical position. The point A can be pictured as coinciding with a! and B as located the distance d directly behind &'. This again gives a definite picture of the points A and B in space. While the plane of the drawing is in a vertical position, picture the point B directly behind 6' the distance d. Now if the plane of the drawing be placed in its natural horizontal position, the point will be directly below 6' the distance d. Therefore to pic- ture magnitudes from their front views while the plane of the drawing is in a horizontal position, distance back must be thought of as distance down, while distance toward the front is distance up or above the front view. One of the greatest benefits derived from the study of descrip- tive geometry is the ability to picture an object from the draw- ing. This is called "reading the drawing." 9. In the solution of a problem in descriptive geometry there are two parts, the analysis and the construction. The analysis of the problem states the general method of solution and refers usually to magnitudes in space and not to their views. The analysis for any problem is the same for all different positions of the magnitudes in space. The construction of the problem deals with the making of the drawing and explains how the lines are drawn to secure the required result. The construction changes with different positions of the magnitudes in space. The student should endeavor to keep these two parts of the problem as distinct as possible. 10. Letter notation. In descriptive geometry it is custom- ary when referring to an object to designate points in space, such as corners, by capital letters, as A, B, C, etc. The top views or horizontal projections of these points are designated by lower case letters, as a, b, c, etc., while the front views are designated by lower case letters primed, as a', b' , c' , etc. If other than the top and front views of the object are shown, its corners are designated by lower case letters with subscripts, as Oi, 6i, Ci, etc., a2, 63, C2, etc. POINTS 15 The lines which join two views of the same point are shown in the cuts of this book as line dotted lines, but it is suggested that on pencil drawings these be made light full lines. 11. Problems. Solve in orthographic projection the follow- ing problems. 1. Given the top view of two points, A and B. Draw the front view of the points when A is V' higher than B. 2. Given the top and front views of two points. Draw the left end view of the points. 3. Given the front and right end views of two points. Draw the top view of the points. 4. Given the top and front views of three points. Draw the right end view of the points. 5. Given the front and left end views of three points. Draw the top view of the points. 6. Given the front view of two points A and B. Draw the top view of the points when A is 1" farther back than B. 7. Draw the top, front, and right end views of two points A and B. B is 1" higher, 2" farther back, and 1^" to the right of A. 16 DESCRIPTIVE GEOMETRY STRAIGHT LINES 12. Views of a line. The word "line" will be taken to mean "straight line" unless statement to the contrary is made. A line appears as a line in all views with the exception of the view taken by looking in the direction of the line itself; in this case it is a point. At least two views of a line must be shown before the direc- tion and length of the line are determined. The picture planes upon which these views are taken must not be parallel to each other. It is not suflBcient to show just the top view or just the front view, but both top and front views must be shown, or one of these with some other such as the end view. If two views of a line are shown, other views can be derived from them. Fig. 11 shows "the top, front, and right end views of a line AB. The top view of the line is obtained by looking directly down from the top. This view shows that the line slopes back- Pie. 11.— Three Views of line A B. Fig. 12.— Point C on line A B. ward to the right. The front view is obtained by looking directly from the front, and shows that the line slopes downward to the right. The right end view is obtained by looking from the right, in the direction of the ground line, and shows that the line slopes backward and downward, d is the distance which B is farther back than A. This distance shows in the top view and also in the end view. STRAIGHT LINE 17 If the top and front views of a line are given, the right end view can be found as follows: Since the front and end views both show height, draw lines through a! and 6' parallel to G. L. At any distance to the right of the front view and on the line through a' , select a point a^ . Through a\ draw a line perpen- dicular to G. L. Take the distance d which 6 is back of a in the top view and set it off along the horizontal line through b' to the right of the vertical line through ai. This locates &i. Then tti 6 1 is the right end view of the line AB. Quite often in practical work, distances from front to back in the top view and from left to right in the end view are given from the center line of the object. 13. Views of lines whicli are in various positions with reference to H, V, and P. If a line is oblique to H , V, and P, its top and front views are oblique to G. L. In a manner similar to the above, describe the directions of the top and front views of lines in the following positions: (a) Parallel to G. L. (b) Parallel to H and oblique to V. (c) Parallel to V and oblique to H. (d) Perpendicular to H. (e) Perpendicular to V. (f) A line of profile. (A plane which is at right angles to the G. L. and cuts it at any point is a profile plane. A line of profile is any line which lies in a profile plane) . 14. Point on line. If a point is on a line in space, the top view of the point is on the top view of the line, the front view of the point is on the front view of the line, etc-. Fig. 13. If the top and front views of a line AB are given, and also the top view of a point C on the line AB, the front view of C is found by drawing a perpendicular to G. L. through c and ex- tending it to cut a! V at c'. If the front view of C had been given the top view could have been located in a similar manner. If the top and front views of a line are oblique to G. L., as AB, Fig. 13, the direction of the line is definitely determined 18 DESCRIPTIVE GEOMETRY without having particular points of the line lettered or desig- nated. If, however, the top and front views of a line are per- piendicular to G. L., as they are with a line of profile AB, Fig. 12, both views of two points of the line must be lettered before the direction of the line is fixed. For example, suppose the top and front views of a line of profile are given as in Fig. 12, but having the front view unlettered. If the upper end of the front view is lettered a and the lower end 6', a definite line is repre- sented. If, however, the lower end is lettered a' and the upper end 6', a line of different direction is represented. This can be done since lines joining the top and front views of all points on a line of profile coincide. If the front view c ' of a point on a given line of profile AB, Fig, 12, is represented, the top view of C cannot be found by the usual method of a perpendicular to G. L. The end view Ci of the point is on the end view a-i b^, of the line. From the end view of the point, its top view c is located. If the top view of a point on the line had been given, the front view of the point could have been determined in like manner from the end view. Fig. 13.— Point C on line A B. Fig. 14. — Inter- secting lines Fig. \o.— Lines non-intersecting. 5' Fig. 1^.— Parallel lines. Intersecting^ lines. If two lines intersect in space. Fig. 14, the line joining the point of intersection d of their top views with the point of intersection d' of their front views must be perpendicular to G. L. Non-intersecting^ lines. Two non-intersecting lines, Fig. 15, may be in such a position that their top views and also their STRAIGHT LINE 19 front views are intersecting. The line joining the point of in- tersection d of their top views with the point of intersection g' of their front views is not perpendicular to G. L. Parallel lines. If two lines are parallel in space, Pig. 16, a view of the lines from any position, other than from a point on the plane of the lines, will show them as parallel lines. Hence any view of a parallelogram, from a point outside of the plane of the figure, is a parallelogram. 15. Coordinate system. In order to locate the problem in a convenient position on the sheet, the top and front views of the given points are determined by coordinates. Unless other- wise stated, the origin of coordinates is to be taken at the center of the space allotted to the problem. The base line is a horizon- tal line through the origin. The first coordinate gives the dis- tance of the point to the right or left of the origin measured in the direction of the base line, to the right when positive, to the left when negative. The second coordinate locates the top view of the point, above the base line when positive, below when nega- tive. The third coordinate locates the front view of the point, above the base line when positive, below when negative. The problems for which coordinates are given in this text can be conveniently solved in a 5" x 7" rectangle unless statement to the contrary is made. The base line should be drawn through the center and parallel to the longer side of the rectangle. Dis- tances are given in inches and should be used full scale unless otherwise stated. For blackboard work, multiply the coordinate by five. The cuts, accompanying lists of problems, are num- bered to correspond with the problems to which they belong. It is suggested that they be used as layouts for blackboard work in the place of using the coordinates of the stated problem. 16. Problems. 1. Draw the left end view of the line joining the points A (J, 1}, —1^) andB(2, i, -i). 2. A ( — U, 1, —J), B ( —J, 0,'— 1|), and C (IJ, f , — 2J) are three cor- ners oi a parallelogram. Draw its top and front views. 20 DESCRIPTIVE GEOMETRY 3. A ( -i, li, -li), B (i, i, -J), and C ( -2, 0, — li) are three cor- ners of a parallelogram. Draw its right end view. 4. Given two lines A ( — J, 0, — 2i), B (IJ, If, —J), and C ( — U, If, — IJ), E (1, f, — 2i). Do the lines intersect? 5. If P(0, li, ?) lies on the line A(0, If, — |) B(0, i, -2), locate its front view. 6. Locate the top and front views of a point E so that the line C ( —2, li, —i) E (?, ?, ?) will be parallel to the line A (— i, IJi —1) B (— i, i, -2i). a / ■< ' '1^ 3 a 1 .«"! C ^^ b >n \ ! I? a' \i£ -ic' i" ^ a' c^ 4 ^h 5 ^ (5 a \a/' > C C i b 3" tfv X"-~ -'e' i,' i &' REVOLUTION AND COUNTER-REVOLUTION 17. An object is said to revolve about a straight line as an axis when each of its points moves in the circumfer- ence of a circle whose center is in the axis and whose plane is perpendicular to the axis. When an object is revolved about a straight line as an axis, the relative position of its points is not changed. The object can thus be brought into a simpler position with reference to the planes of projection. The views of the object in this position are easily found and from these the views of the object in its original position are located by the counter-revolution of its points. REVOLUTION AND COUNTER-REVOLUTION 21 If a point revolves about an axis, a view of its path taken by looking in the direction of the axis is a circle. A view of its path taking by looking in a direction perpendicular to the axis is a straight line at right angles to the axis and equal in length to the diameter of the circle. If an axis of revolution is perpendicular to H, describe the top, front, and end views of the path of a point revolving about it. In a similar manner, describe the path of a point revolving about an axis which is perpendicular to V; one perpendicular to P. 18. To revolve a point about a horizontal axis. Let P, Fig. 17, be the given point and SS the given axis.. An auxiliary view taken by looking in the direction of the axis SS, shows the path of the point P as a circle or the arc of a circle. The top view of this path is a straight line passing, through p at right angles to ss. The line Ai^i in the auxiliary view represents a horizontal plane at the level of the axis SS, h^hi is drawn at right angles to ssandat any convenient dis- tance from the top view. The point Si on this line is the auxiliary view of the axis. In the front view, this horizontal plane is represented by the line h'h' . Since the auxiliary view is taken by looking in the direction of SS, the aux- iliary view of P will be loca- ted on a line through p par- allel to SS. The distance d is taken from the front view and used to locate Pi in the auxiliary view, since height shows in both of these views. Withsi as center and SiPi as radius, the circle representing the path of the point in the auxiliary view is drawn. If the point is stopped at any position on the circle in the auxiliary view, its top and front views are found as follows: Through the auxiliary view of the Fig. 17. — P revolved about axis SS. 22 DESCRIPTIVE GEOMETRY point, draw a line parallel to ss until it intersects the top view of the circle, which is a line through p at right angles to ss. This intersection is the top view of the point. The front view of the point is directly below the top view, and the same distance above or below h'h' that its auxiliary view is above or below hihi. ps and p' i are the top and front views of the point when it is at the level of the axis 88. If the point makes a complete revolution about the axis, the auxiliary view of the path is a circle, the top view a straight line, and the front view an ellipse. 19. To revolve a point about an axis which is parallel to V. Let P, Fig. 18, be the given point and FF the given axis. An auxiliary view taken by looking in the direction of the axis FF, shows the path of the point P as a circle or the arc of a circle. The front view of this path is a straight line pass- ing through p' at right angles to // . The top view of the path is an ellipse or the arc of an ellipse. The line ViVi in the auxiliary view represents a vertical plane containing the axis FF. ViVi is drawn at right angles to // and at any convenient distance from the front view. The point A on this line is the auxiliary view of the axis. In the top view, the vertical plane is represented by the line vv. Since the auxiliary view is taken by look- ing in the direction of FF, the aux- iliary view of P will be located on a line through jt>' parallel to/ jf'. The distance d is taken from the top view and used to locate pi in the auxiliary view, since distance from front to back is shown in both of these views, With fi as center and /iPi as radius, the circle representing the path of the point in the auxiliary view is drawn. If the point is stopped at any position on the circle in the aux- FiG. 18.- -P revolved about axis FF. REVOLUTION AND COUNTER-REVOLUTION 23 iliary view, its top and front views are found as follows: Through the auxiliary view of the point, draw a line parallel to ff' until it intersects the front view of the circle which is a line through ip' at right angles io ff . This intersection is the front view of the point. The top view of the point is directly above the front view and the same distance in front or back of vv that the auxiliary view of the point is tvovixViVx. p' s and pg are the front and top views of the point when it is the same distance from V as the axis FF. If the axis of revolution is oblique to H and V, both the top and front views of the path of the point are ellipses. In this case the point cannot be revolved to a definite position in its path without the use of an auxiliary axis or two successive auxiliary views. 20. Problems. 1. Rotate the point P ( — |, IJ, —J) about S ( — 2f, If, — IJ) S (J, J, — -11) as an axis until it reaches the level of SS. Letter the topand front views of the point in this position. 2. In problem 1, rotate P until it is J" lower than SS and letter its top and front views in this position. „ 3. Rotate the point P (IJ, |, —1) through an angle of 60° about F { — J, 1, — I) F (2 1,1, — 2 J) as an axis. Letter the top and front views of the point in its final position. 4. S {—J, i, — IJ) S {2|, 1|, — 1 J ) is the center line of a shaft which car- ries a 11" wheel at its center. If the wheel has eight spokes, letter the top and front views of the end of each spoke. (Use center lines only). 5. F ( — 2f, 1, —25) F (J, 1, — }) is the center line of a shaft which carries a li" wheel at its mid paint. It the wheel has eight spokes, let- ter the top a!nd front views of the end of each spoke. (Use center lines only). s 1 1 1 f 3 f s^ 4 1 1 1 j f 5 ^ 1 1 1 jf s' A'i '•t' <^ ^/' l-l'l- — 1 1 .. Ir' ?^ ^ 24 DESCRIPTIVE GEOMETRY THE TRUE LENGTH OF A LINE 21. In orthographic projection, two views of a line will de- termine its direction in space, providing the plane upon which these views are taken are not parallel to each other or the line is not perpendicular to the line of intersection of the picture planes. If two points of the line are designated, th^ line will have a definite length between these points. Any view in which the line of sight is perpendicular to the given line will show the given line in its true length. In order to conveniently make an auxiliary view from two given views, the line of sight must be parallel to the plane upon which one of the given views is taken . Therefore to find the length of a line when two of its views are given, take an auxiliary view with the line of sight perpendicu- lar to the given line and parallel to the plane of one of the given views. 22. Given the top and front views of a line. Find the true length of the line and the angle which it makes with H. Let AB, Fig. 19, be the given line. Analysis. The angle which a line makes with H is the angle between the line and its top view. A view taken by looking at right angles to a plane containing the line and the top view will show the true length of the line and the angle which the line makes with H. In this case the line of sight will be parallel to H and perpendicular to the given line. Construction. In Pig. 19, h' h' is the front view of a hori- zontal plane which passes through A. Picture the line AB in space from its top view ah, A coinciding with a and B the distance d di- rectly below 6. Take an auxiliary view by looking at right angles to the plane a6B. To do this draw lines at right angles to ab at the points a and 6. At any point in the line through a, select the point tti, arid draw hihi through this point parallel to a6. Ai^i is the auxiliary view of the horizontal plane which has h'Ji' for a THE TRUE LENGTH OF A LINE 25 Fig. 19.— ^h angle AB makes with R. front view. 61 is located on the line through 6 at right angles to db and the distance d below AjAi. Then a^bi is the true length of AB and ^h the angle which AB makes with H. If AB had been parallel to V, the plane ah B would have been parallel to V, and the front view would have shown the true length of the line and its angle with H. If AB had been a line of profile, the end view would have shown its true length and its angle with H and V. 23. Given the top view ab of a line and the angle ^h which AB makes with H, to find the front view of AB. There are any number of lines which make 30" with H and have ab for a top view. In Fig. 19, let at be the given top view. At a and 6, draw lines perpendicular to ab. At any point in the line through a, select the point a^ and draw h^h^ through this point parallel to db. Through a-i, draw the line a^bi making the angle 6a with hih^. This gives the distance d which B is lower than A. Draw perpendiculars to G. L. through a and b. At any point on the perpendicular to G. L. through a, select the point a' and draw h'h' through this point parallel to G. L. Take the distance d from the auxiliary view and set it below h' h' locating the point b' on the perpendicular to G. L. through b. Then a'b' is the required front view of AB. If the top view and true length of AB had been given, the front view in this case could also have been found by the use of the auxiliary view. The point 61 would have been located from the length of a^b-i rather than from the angle ^h. 24. Given the top and front views of a line. Find the true leng-th of the line and the angle which it makes with V. Let AB, Fig. 20, be the given line. 26 DESCRIPTIVE GEOMETRY front view a'&', A directly back of b ' . The angle which a line makes with V is the angle between the line and its front view. A view taken by looking at right angles to a plane containing the line and its front view will show the true length of the line and the angle which it makes with V. In this case the line of sight will be parallel to V and perpendicular to the given line. Construction. In Fig. 20, vv is the top view of a vertical plane which passes through A. Picture the line AB from its coinciding with a' and B the distance d If the drawing is in a horizontal position, distance back is pictured as distance below the front view. Take an aux- iliary view by looking at right angles to the plane a' &'B. To do this draw lines at right angles to a'b' at the points a' and &'. At any point in the line through a' , select the point Oi and draw tiiDi through this point parallel to a'b' . v^Vi is the auxiliary view of the vertical plane which has vv for a top view. 61 is located on the line through 6' at right angles to a'b' and the distanced below i^i^i. Then aj&i is the true length of AB and ^v the angle which AB makes with V. If AB had been parallel to H, the plane a'&'B would have been parallel to H, and the top view would have shown the true length of the line and its angle with V. If the front view of the line is given and the angle which the line makes with V, the top view can be drawn as follows: From the front view and the angle which the line makes with V, draw the auxiliary view. Fig. 20. This view shows the distance d which one end of the line is further back than the other. With this distance, the top view can be determined. There are any number of lines which make a given angle with V, and have a given front view. Fig. ^0.—0\ angle AB makes with V. « THE TRUE LENGTH OF A LINE 27 If the front view and the true length of the line are given, the auxiliary view and then the top view can be found. 25. Second method. Given the top and front views of a line, find the true length of the line and the ang^le which it makes with H. Let AB, Pig. 21, be the given line. Analysis. Rotate the given line about an axis which cuts it and is perpendicular to H until the line is parallel to V. In this position, the front view will show the true length of the line and the angle which the line makes with H. Construction. In Pig. 21, ac is the top and a'c' the front view of an axis perpendicular to H. When AB is rotated about ac Fig. 21— ^h Angle AB makes with H. Fig. 22— ^v Angle AB makes with V. AC as an axis, A remains fixed and B moves in the arc of a cir- cle to Bi, abi being drawn parallel to G. L. The top view of the path in which B moves is the arc 66 1 and the front view the line 6'6i' parallel to G. L. Since AB, is parallel to V, the front view a'b' 1 shows the true length of the line AB. Oh is the true size of the angle which AB makes with H. If the top view, rt6. Pig. 21, of a line AB and the angle ^h which the line makes with H are given, the front view of AB can be found as follows. With a as center and ab a,^ radius, draw the arc 66 1; abi being parallel to G. L. Draw perpendicu- 28 DESCRIPTIVE GEOMETEY * larsto G. L. through a and 61 . From any point as a' in the perpen- dicular through o, draw a line a'b' i making the angle ^h with Gr. L. and cutting the perpendicular through 61 at &'i. When Bi is revolved back to the position B, the front view b' j moves parallel to G. L. until it intersects at 6' a line perpendicular to G. L. through 6. Then a'b' is the required front view. 26. Second method. Given the top and front views of a line, find the true leng-th of the line and the angle which it makes with V. Let AB, Pig. 22, be the given line. Analysis. Rotate the given line about an axis which cuts' it and is perpendicular to V until the line is parallel to H. In this position, the top view will show the true length of the line and the angle which the line makes with V. Construction. In Fig. 22, a'c' is the front and ac the top view of an axis perpendicular to V. When AB is rotated about AC as an axis, A remains fixed and B moves in the arc of a cir- cle to Bj , a'6' 1 being drawn parallel to G. L. The front view of the path in which B moves is the arc ' 6 ' 1 , and the top view the line bbi parallel to G. L. Since ABi is parallel to H, the top view a6) shows the true length of line AB. Oy is the true size of the angle which AB makes with V. If the front view, a'b' , Fig. 22, of a line AB and the angle Oy which the line makes with V are given, the top view of AB can be found as follows. With a' as center and a'b' as radius, draw the are 6'i' 1 ; a'b' i being parallel to G. L. Draw perpen- diculars to G. L. through a' and 6' 1. From any point as a in the perpendicular through a' , draw a line aftj making the angle Ov with G. L. and cutting the perpendicular through b' 1 at ftj. When Bj is revolved back to the position B, the top view 61 moves parallel to G. L. until it intersects at 6 a line perpendicular to G. L. through b' . Then ab is the required top view. 27. Problems. 1. Find the true length of the line A( —J, J, — IJ) B(i, 1^, — 2^) and the angle which it makes with H. 2. Find the true length of the line A( — |, f, —2) B(^, 1|, — |) and the angle which it makes with V. THE TRUE LENGTH OP A LINE 29 3. Find the true length of the line C(0, 1|, ^i) E{0, J, — 2^) and the angle which it malces with H and V. 4. Draw the front view of the line E( —J, If, ?) G(l, i, ?) when its true length is 2^". 5. Draw the top view of the line A( —A, ?, —\) B(l|, ?, — IJ) when the line makes 30° with V. 6. The three guy wires of a derrick run from the upper end of the mast 0(0', 12',— 6') to points of support at A(— 22', 3',— 15'), B(U', 34', —23'), and C(24', 2', — 19'). What is the maximum length of a boom fastened 3' from the lower end of a 30' mast which will clear all the guy wires? How long is each guy wire? (Scale i" — 1' —0"- Use a 7"xlO" rectangle for this problem). 7. A( -6', 4', -32'), B( -6', 28', -32'), C( IS', 28', -32'), and E( 18', 4', — 32') are the corners of the plate of a hip roof which has its peak at the point 0( 6', 16', ?). If the hip rafters make 30° with the horizontal, find their true lengths. Draw the front elevation of the roof. (Scale J"= r — 0". Use a T'xlO" rectangle for this problem.) b 2 ^,* 3 c 4 e 5 -/ a ^ e \ cK ^ *' ct y' c' e' !^ \*' 30 DESCRIPTIVE GEOMETKY PLANES .^-' M 28. A plane must usually be considered as indefinite in ex- tent. Points or lines may be drawn on the plane, but do not bound it. In space, a plane is fixed or represented by three points, not in the same straight line, a point and a line, two in- tersecting lines, or two parallel lines. If the top and front views of these magnitudes are given, the plane which' they determine will have its inclination with H and V definitely fixed. The points A, B, and C, Fig. 23, represent a plane in space. It is evident that a line joining A and B, B and 0, or A and C will lie in this plane. If a plane is represented by a point and a line, other lines of a plane can be found by joining the given point with points on the given line. If the plane is represented by two lines, either parallel or intersecting, any number of other lines of the plane can be found by joining a point in one of the given lines with a point in the other. When the lines are inter- secting. Fig. 14, the line which joins the top and front views of their point of intersection must be at right angles to the ground line. If the lines are parallel, Fig. 16, all views of the lines will show them parallel. If the lines are neither parallel nor intersecting. Fig. 15, they do not represent a plane. If a plane is perpendicular to H, all lines of the plane which are of indefinite length, have the same top view. The front views of these lines will be either parallel or intersecting lines. If a plane is perpendicular to V, all lines of the plane which are of indefinite length have the same front view. The top views of these lines will be either parallel or intersecting lines. Fig. 23.— Pomts A, B, G represent a plane. PLANES 31 If a plane is perpendicular to both H and V, all lines of the plane which are of indefinite length have the same top view and also the same front view. If one line of a plane is parallel to G. L., the plane is parallel to G. L. A line is parallel to G. L. when both its top and front views are parallel to G. L. This line with any point outside the line will represent a plane parallel to G. L. 29. Lines on the plane. The two intersecting lines MN and OP, Pig. 24, represent a plane. The line AB, joining the point A in MN with the point B in OP, is another line of the plane. ah is its top and a'h' its front view. A line which lies in a given plane and is parallel to H, is called a horizontal line of that plane. In Fig. 25, SS is such a line which lies in the plane of the lines MN and OP. Its front view s' s' is drawn first and then its top view ss is found by locating the top views of the points X and Y where it crosses the given • lines MN and OP. A line which lies in a given plane and is par- allel to V is called a frontal line of that plane. FF, Fig. 26 is such a line which lies in the plane of MN and OP. Its top view ff is drawn first and then its front view /'/' is found. These lines which lie in a plane Fig. 24.-4B line in plane of and are parallel to H or V, play MN and OP. an important part in the solu- tion of many problems. 30. Points on planes. A point of a plane can be represented by placing the views of the point on the views of some line of the plane. If the top view of the point P which lies on the plane is given, the front view of P can be found as follows: Draw the top view of any line through the top view of P. Find the front view of the line by finding the front view of the points where it crosses the lines representing the plane. Then the front 32 DESCRIPTIVE GEOMETRY view of P is on the front view of this line. If the front view of P had been given, the top view could have been found in a similar manner. All points of the plane which are a given distance, say 1" higher or lower than any point P, lie on a horizontal line of the plane. The true distance which the point is higher or lower than P is shown in the front view. Likewise, all points of the plane Fig. 25.— SS horizontal line of plane. Fig. 26.— FF frontal line of plane. which are a giving distance farther back or farther toward the front than a point P lie on a frontal line of the plane. The true distance which the point is farther to the front or back than P is shown in the top view. 31. Problems. In the following problems the given planes may be represented by the tpp and front views of any two of their lines unless otherwise stated. 1. Having given the top view of a line which lies in a given plane, to find its front view. 2. Having given the front view of a line which lies in a given plane, to find its top view. 3. Having given the top view of a point which lies in a given plane, to find its front view. 4. Having given the front view of a point which lies in a given plane, to find its top view. 5. A point and two parallel lines are given by their top and front views. Is the point in the plane of the two parallel lines? PLANES 33 6. Having given a plane which is parallel to the ground lipe and the top view of a point of this plane, find the front view of the point. 7. Having given a plane which is parallel to the ground line, and the top view of a line of this plane, find the front view of the line. 8. Having given a plane which is parallel to G. L., and the front view of a line of this plane which is also parallel to G. L, find the top view of the line. 9. Two lines are in such a position that neither their top nor their front views intersect within the limits of the drawing. Determine graphically whether or not the lines intersect. 10. A plane is represented by two intersecting lines. The top view of a line which passes through their point of intersection and lies in the given plane is given. Find the front view of the line. 11. A plane is represented by two parallel lines. The top view of an- other line of this plane which is parallel to the first two lines is given. Find the front view of the line. 12. Determine graphically whether or not the lines A ( — IJ, IJ, — 2J) B (I, If, —2) and C (-f, {, — f) E (IJ, |, — |) intersect. Do not prolong the lines AB and CE. 13. Draw the top and front views of a horizontal line and a frontal line on the plane M ( — li, i, — IJ), N (1, 1|, — f), O (1, J, — 2J). 14. Find the top and front views of the loous of all points which lie on a given oblique plane ABC and are 1" lower than A. 15. Find the top and front views of the locus of all points which lie on a given oblique plane ABC and are H" farther back than B. 16. Find the top and front views of a point which lies on a given oblique plane ABC and is }" higher than A and 2" farther back than B. 32. A point and two lines are griven by their top and front views. Represent a plane which contains the s^iven point and is parallel to the given lines. Analysis. Through the given point draw a line parallel to each of the given lines. The plane of these two lines is the re- quired plane. Let the construction be made in accordance with the above analysis. 33. Problems. 1. Represent a plane which contains a given line and Is parallel to another given line. 2. Represent a plane which contains a given point and is parallel to G. L. 3. Two lines parallel to the G. L. are given by their top and front views. Represent a plane which contains a given point and is parallel to the given lines. 34 DESCRIPTIVE GEOMETRY 4. Two lines, one oblique to G. L. and the other a line of profile, are given by their views. Represent a plane which contains a given point and is parallel to the given lines. 5. Represent a plane which contains a given point and passes at equal distances from two other given points. (Two solutions). 6. Represent a plane which contains a given line and passes at equal distance from two given points. {Two solutions.) 7. Represent a plane which contains a given point and passes at equal distances from three other given points. (Four solutions). 34. To find the ang-le which a g-iven oblique plane makes with a horizontal plane. Fig. 27 represents a triangular prism resting on one of its lateral faces. •cyX>.- Fig. 27. -Angle plane makes with H. Fig. 28-— Angle plane ABC mahe-i with, H. Let the given oblique plane be the face ABCE of the prism, and let the face upon which the prism rests be the horizontal plane. Analysis. A vievv taken by looking in the direction of the line of intersection of the planes will show each plane as a straight line. The angle between these lines is the angle between the planes. Construction, ah is the top anda'&' the front view of the lateral edge of the prism, which is the line of intersection of the given planes. In the auxiliary view, hxhi which is at right angles to ai, represents the horizontal plane upon which the prism rests, h'h' is the front view of this plane. The distance d which the edge CE is above the plane of the opposite face is PLANES 35 taken from the front view and used to locate CiCi in the auxiliary view. aiCi represents the plane of the face of the prism in the auxiliary view and 6a is the angle which this face makes with the horizontal plane upon which the prism rests. 35. To find the ang'le which a given oblique plane makes with a horizontal plane when the oblique plane is repre- sented by two intersecting lines AB and AC. Fig. 28. In the front view the horizontal plane is represented by one line h' h' parallel to G. L. The top view of this horizontal plane is unlimited in extent, covering all of H, and therefore need not be represented by any straight lines. The intersection of this horizontal plane with the plane ABC is the horizontal line SS. In the auxiliary view taken by looking in the direction of the line SS, hihi, which is at right angles to ss, represents the hori- zontal plane. The distance d taken from the front view, locates ai in the auxiliary view. ai6i represents the oblique plane ABC in the auxiliary view and ^h the angle which it makes with the horizontal plane. 36. Conversely, to represent a plane which jnakes a given angle with H. There are any number of planes which make a given angle, say 30°, with H. Any horizontal line SS, Fig. 29, can be drawn and a plane represented which contains this line and makes 30° with H. In the auxiliary view taken by looking in the direction of SS, hihi at right angles to ss, represents Hand SiOi, at 30° with A 1^1, represents the required plane. The top view of the point A can be consid- ered to be any point on the line ab, parallel to ss. The front view of AB is located by taking the distance d from the auxiliary view and setting it above h' h'in the front view, a'b' being parallel to s's'. Fig. 29.— Plane 30° with H. The required plane is repre- 36 DBSCEIPTIVE GEOMETRY seated by the two parallel lines SS and AB or by SS and any point on the line AB. Other oblique lines of the plane can be drawn by joining points on AB with points on SS. There are two planes which contain SS and make 30° with H. The other one would be represented in the auxiliary view by a line passing through a^, sloping downward and making 30° with hihi. 37. To find the angle which a given oblique plane makes with a plane parallel with V. Let ABC, Fig. 30, represent the oblique plane and vv the top view of the plane which is parallel to V. Analysis. A view taken by looking in the direction of the line of intersection of the planes will show each plane as a straight line. The angle between these lines is the angle between the planes. Construction, ff is the top view and /'/' the front view of the line of intersection of the planes. In the auxiliary view, vi.V[ which is at right angles to /'/' , represents the plane which is parallel to V. The distance d which the point A is back of this plane is taken from the top view and used to locate ui in the auxiliary view, ai/i represents the oblique plane in the auxiliary view and ^v is the angle which this plane makes with the plane parallel with V. The oblique plane also makes this same angle ^v with V. In the auxiliary view, ai is placed nearer the front view than the line Dit>i when the point A is in front of the vertical plane as determined by the top view. If A is back of the vertical plane then ai is placed on the other side of ViVi. 38. Conversely, to represent a plane which makes a given angle with V. Fig. 30. — ^y angle plane makes with V. PLANES 37 There are any number of planes which make a given angle, say 45°, with V. Any frontal line FF, Pig. 31, can be drawn and a plane represented which contains this line and makes 45° with V. In the auxiliary view taken by looking in the direction of FF, viV[ at right angles to /'/' . represents V and /ifli , 45° with ViVi, represents the required plane. Each point of the line /lOi represents a line which is parallel to FF and lies in the required plane- (iibi represents such a line, a'b' parallel toff' is its front view. The top view of AB is located by taking the dis- tance d from the auxiliary view and placing it back of vv in the top view, ab being parallel to ff. The required plane is represented by the two parallel lines FF and AB, or by FF and any point on the line AB. Other oblique lines of the plane can be drawn by joining points on AB with points on FF. There are two planes which contain FF and make 45° with V. The other one slopes forward making 90° with the plane of FF and AB. Fig. si— Plane i6° with V. 39. Problems. B(-i, n 1. Find the angle which the plane A( — 2J, |, - C( i, i, — If) makes with H. 2. A(-2f , If, -4) B( -If, 3i, -4) C( -h 2, -4) E( -li, i, -4) is the base of a regular square pyramid whose altitude is 2J". If V is the ver- tex, find the angle which the face VBC makes with the plane of the base. (Use a 7"xlO" rectangle for this problem.) 3. Vertical bore holes are driven to a certain vein of ore at three points A( —550, 400, ?) B( —250, 700, ?) and C( 50, 200, ?). The bore holes strike the vein at 650', 1050' and 750' above sea level, respectively. If the direction of the G. L. is considered to be East and West, find the strike and dip of the vein. Scale 1" = 200'. (Use a 7" x 10" rectangle for this problem.) 38 DESCRIPTIVE GEOxMETEY Note. By strike is meant a horizontal line on the plane of one wall of the ore body. The direction of the strike is the same as the direction of outcrop of the vein along a horizontal surface. By dip is meant the angle of inclination of the wall with a horizontal surface. 4. S( — i, i, — IJ) S{ 2f , 1?, — IJ) is a horizontal line of a plane which makes 30° with H. Represent the plane in the top and front views. 5. The strike of a certain ore vein is N 45° E from the point S( 0, i, — 1^). Represent one wall of the vein when the dip is 45° to the NW. 6. Find the angle which the plane of A( 1, 3J, —2) B( 1, 1^, — 3^) and C( 0, 2i, — 2J) E( 2|, \i, — 3,|) makes with H and V. (Use a 7" x 10." rect- angle for this problem.) 7. Findthe angle which the plane A(—2|, f,— 2i)B { — 1^, li,— i) C(i, i, — If) makes with V. 8. Represent a plane which contains the line F( — 2|, 1, — 21) F( |, 1, — I) and makes 45° with V. / ,b a \li Cr '■a' c'i c s' PLANE FIGUBES 39 PLANE FIGURES 40. If two views of the magnitudes which represent a plane are given, the position of the plane is determined. These mag- nitudes may form some geometrical figure as an angle, a tri- angle, a square, etCi, or such a geometrical figure may be on the plane in addition to the magnitudes which represent the plane. A view of such a plane figure taken with the line of sight per- pendicular to the plane will show the figure in its true size and and shape. Consequently, if the plane of the figure is parallel to H, V, or P, the top, front, or end view respectively will show the true size of the figure. If the plane of the figure is oblique to both H and V, the true size of the figure can be found by ro- tating the plane, and the^ figure also, until ; it is parallel to, H when the top view will show the true size, or else parallel toV, when the front view will show its true size. . In order to rotate the plane of a^gure until it is parallel, to some other plane, the axis of rotation must be parallel to the line of intersection of the two planes. i*-,^. ,^ %^ Another method, for finding the true size of a figure whiSn tie plane of the figure is oblique to H aad V, is to tate tw© aux- • iliary views of it. The line of sight for the first auxiliaty view must be parallel to the picture plane of 'one of^the given views and also from such a direction that the plan^fj^ figure ap- pears as a line. The second auxiliary view is derived fgpm the first and has the line of sight perpendicular to the^kme of the figure, thus showing the true size of the figure. The second method is very convenient at times, but students frequently find it more difilcult than the method of revolution. It is therefore not recommended to students at this stage of the subject. • "" 41. Two interseeting lines are represented by their tap, and front views. Find the true size of the angl§,lietwe#tt them. ^ Let AO and BO, Fig. 32, be the given lines. 40 DESCRIPTIVE GEOMETEY Analysis. Draw a horizontal line in the plane of the given lines. Revolve the given lines about the horizontal line as an axis until the plane of the lines is parallel to H. The lines do not change their relative position during the revolution. The top view of the angle in the revolved position shows the true size of the angle between the lines. Construction, s' s' is the front and ss the top view of a hori- zontal line on the plane of AO and BO. In the auxiliary view taken by looking in the direction of the axis, Aj^i represents a syb Fig. 32.— XO3 Y true size of angle AOB. horizontal plane at the level of the axis. The distance d taken from the front view locates the point Oj in the auxiliary view. The line o^Si is the auxiliary view of the plane AOB and ^h the angle through which it must be revolved to bring it into a hori- Lzontal position. When the plane is in a horizontal position, the auxiliary view of is at 02, its top view at 03, and its front view PLANE FIGURES 41 at o' 3. The points X and Y where the lines OA and OB inter- sect the axis do not move during the revolution. Therefore zosy is the true size of the angle AOB. The problem can be solved by revolving the plane of the angle until it is parallel to V about a frontal line as an axis. Then the front view shows the true size of the angle. 42. To bisect an angfle. When the angle is shown in its true size, the bisector can be drawn. The bisector can then be revolved back until the plane of the angle is in its original po- sition and the top and front views of the bisector found. In general, the views of the bisector will not bisect the views of the angle. The view of an angle can be larger than,, equal to, or smaller than the angle itself. The view of a right angfle is a rig-ht angle when one side of the angle is parallel to the plane upon which the view is taken. 43. By the method of the above article, the true size of any plane figure can be found if the figure is represented by two of its views. If, for example, the top and front views of a triangle or quadrilateral are given, the true size of the figure can be- found by revolving its corners about any horizontal axis which lies in the plane of the figure until each corner reaches the level of the axis. Joining the corners thus located gives the true size of the figure in the top view. 44. Problems. 1. Find the true size of the angle A( — 2^, i, — 2i) B( —1, IJ, -1) C( ^, i, — 2) by using a frontal line as an axis. 2. Draw the top and front views of the bisector of the angle B( — J, 0, -I) A( -h n -2i) C( -2}, 1, -i). 3. Draw the top and front views of the bisectors of all the angles of the triangle A( -2J, 0, —J) B{ -li. If, -2^) C( i, f, -If). 4. Draw the top and front views of the quadrilateral A( J, If, — 2|), B( 0, i, —4) C( — li, 0, — 3f) E( —21, f, ?) and find its true size and shape. (Use a T'xlO" rectangle for this problem.) 42 DESCRIPTIVE GEOMETRY 5. Draw the top and front views of the pyramid A( 1|, i, —4^) B( 3^, 2i, —4s) C( O; If, — 4J) E(i, 3, —2). By using an auxiliary view, find the angle between the planes AEB and ABC, the true lengths of the lines AE and BE, the slant height of the pyrami^, and the true size of the faee angle AEB. (Use a 7"xlO" rectangle for this problem.) 6. Find the true size of the angle between two guy wires which are anchored at A{ —65', 32.5', —72.5'), and B( 10', —10', —90') and run to tha top of a 50' vertical stack with base at C( —15', 57.5', —82.5'). .What is the length of the wires? (Use a 7"xlO" rectangle for this problem. Scale 1" = 20'.) 7. In Art. 27, Problem 7, find the true size of one side oE the roof and the angle between the hip rafters. (Use. a 7"xlO" rectangle for this problem.) / ^^^ 2 a 3 A < ■ 4 -^" a 1 1 c'k,_^ ^ eK. 1 1 1 1 cr / ^c ^\ et \i ?^'' c'<^ ^' PLANE FiaUKES 43 45. A point and a line are represented by their top and front views. Draw the top and front views of a line which passes through the given point and makes a given angle with the given line. Let P be the given point, 60° the given angle, and AB the given line, Fig. 33. Pig. 33.— P^ 60° with AB. Analysis. Revolve the plane of the point and line about a horizontal axis until it is parallel to H. From the revolved po- sition of the point, draw a line which makes the angle 60° with the revolved position of the given line. When this line is re- volved back with the plane to tlie original position of the plane, it will be the required line. Construction, s' s' is the front view and ss the top view of a horizontal line which lies on the plane of P and AB and cuts AB at the point C. When the plane is revolved about S8 as an axis, B moves to B3 while P and G remain fixed, being on the axis. 63C is the top view of the revolved position of AB. Through p draw peg, making the required angle 60° with ftgC. e and e' are the top and front views of the point B after the counter revo- lution of the plane. Then pe. is the top and p'e' the front view of the required line. 46. If a plane is represented by two views of the magnitudes which determine the plane, say the top and front views of two of its lines, the views of any figure lying on this plane can be found by the method of the last article. 44 DESCRIPTIVE GEOMETRY For example, let it be required to construct a 2" square lying on the plane ABC, ¥ig. 34, having- one corner of the square at A and another corner 4" lower than A. A hori- zontal line SS is first drawn, every point of which is |" lower than A; s' s' being found first and then ss. The plane ABC is re- 7e \c,' 0' Fig. 34. — Square on plane ABC. volved into a horizontal position about SS as an axis, fflg being the topviewof A after revolution. Withag asacenteranda2" radius, strike an arc cutting ss at 0. Construct a square with agO as one side. Revolve this square into the plane ABC about SS as an axis and locate its top and front views. Then AOE3G3 is thfe re- quired square. The square could take several other position^on the plane and still satisfy the conditions of the problem. PLANE FIGURES 45 .47. Problems. 1. Draw the top aad front views of a line which represents the short- est distance from the point P(— i, 0, —1) to the line A(— ^, IJ, —2^) B (-2i, 1, -I). 2. Draw the top and front views of a line which passes thru the point P{ —2, i, — 2J) and makes 45° with the line A(— 1, 1^, — |) B( —1, J, —If). 3. Find the distance between the two parallel lines A( — |, 1|, — 2|) B(|, f, -f) and C(4, 1|, -2) E( 2, 1, -J). 4. Find a point which is 2|" from A( — 2^, i, — 2^) and lies on the line B( -1, n, -f) C( i,^, ^11). 5. Draw the top and front views of a square with center at C( 1}, 24, — 2i) and side along the line A ( — 1 , If, — 2J) B( 3, If, —4). (Use a 7" x 10" rectangle for this problem.) 6. Draw the top and front views of an equilateral triangle with cen- ter at C( — li, 11, —3) and side along the line A( J, 1, —3) B( —3, IJ, — ^^41^). (Use a 7" X 10" rectangle for this problem.) 7. A circle with center C( IJ, 2i, — 2i) is tangent to the line A( —1, 1|, — 2i) B( 3, If, —4). Draw its top and front views. (Use a 7"xlO" rectangle for this problem. Note. The major axis of the top view is parallel to a horizontal line of the plane ABC, and that of the front view parallel to a frontal line of tlie plane. 8. A drain runs N 45°. E on a falling 20^ grade from the point P( — If, 0, — 3). Draw the top and front views of the shortest connection to this drain from the point 0( — J, 2, — 2^). "What is the grade of the connection and how much pipe is required? (Scale 1" = 20'. Use a 7"xlO" rectangle for this problem.) 9. What is the size of the opening which must be left in a root whose pitch is 46° to accommodate a 16" x 24" chimney? 46 DESCEIPTTVE GEOMETRY INTERSECTIONS OF LINES WITH PLANES 48. A plane is represented by two of its lines. Find the point in which a given oblique line pierce this plane. This problem should be thoroughly mastered. It is used in finding the line of intersection of two planes, the plane sections of all ruled surfaces, the intersections of surfaces with plane faces, and indirectly in finding the intersections of such curved surfaces as cylinders and cones. Let AB and AC, Fig. 35, be the lines -which represent the plane and MN the given oblique line. Fig. 35.— MN pierces plane of AB and AC at P. Analysis. Find the points where the lines which represent the given plane pierce the plane which contains the given line and is perpendicular to H. The line joining these two points is the line of intersection of the given plane with the plane which is parpendicular to H. This line of intersection cuts the given line in the required point. INTERSECTIONS OF LINES WITH PLANES 47 Construction. AB pierces the plane which contains MN and is perpendicular: to H at T and AC pierces it at X. x'y' , the front view of the line joining these two points, intersects m'n' at p', the front view of the required point; p is its top view. The point in which the given line pierces the plane ABC can be found also by means of a plane which eoutains MN and is perpendicular to V. The line of intersection of this plane with the plane ABC cuts MN at the piercing point. Fig. 36.— ^G pierces plane at P. 49. Second method. Let the points A, B, and C, Fig. 36, represent the plane and EG the given oblique line. Analysis. An auxiliary view taken by looking parallel to the plane will show the plane as one line and the oblique line as an- other line. Where these lines cross will be the auxiliary view of the point where the oblique line pierces the given plane. The top and front views of this piercing point can be found from the auxiliary view. Construction. Draw the front view s's' and then the top view ss of a horizontal line on the plane ABC. In the auxiliary view taken by looking in the direction of SS, hihi, at right 48 DESCRIPTIVE GEOMETRY angles to ss represents a horizontal plane at the level SS. Using this for a reference line, the lines ftiSj and e^gi which represent respectively the plane ABC and the line EG, are determined. Pi, the intersection of &iSi, and eigfj, is the auxiliary view of the point where EG pierces the plane ABO. Since P is on the line EG, its top view is p and its front view p' . The same piercing point could have been found by taking an auxiliary view looking in the direction of a frontal line of the plane ABC. The first method is the better when it is required to find where one or two lines pierce a plane. The second method is prefer- able when it is necessary to find where a great many lines pierce one plane, as is the case with the plane section of a cone or cylinder. 50. Problems. 1. Find the point where the line E( — 1^ i, —1) G{li, 1|, — 2i) pierces the plane A( IJ, 1, -I) B{ -l^ 1}, -1|) C(i, i, -2). 2. Find the point in which the line A( — 1| , IJ, — |) B{ 1^, J, — 2J) pierces the plane C( -1, f, -IJ) E( 1, i, -J) G( -1, 1|, -2|) K( 1, 11, -IJ). 3. Find the point in which the line C( —1, IJ, — 2J) E( IJ, 1|, — |) pierces the plane A( |, H, —If) B( — li, U, -If) G( IJ, i, -J). 4. Find the point in which the line A( —1, 1|, — |) B( 1, J, — 2i) pierces the plane C( —J, 1^, —2) E( — 2J, IJ, —2) G( 0, i, —J). 5. Find the point in which the line A( —1, 2, — 2J) B( —1, >-, —i) pierces the plane C{ —J, IJ, —J) E{ —2^, IJ, —2) G{ 0, 0, —2). 6. Find thelengbhof the part of the line A( -l|,3f, — 4^) B{ 2, |,— If) which is included between the planes C{ i, 3, — 3) E{ — 2^, 3, —3) G,( — i. If,— 4i)andM{-l, 2,-2) N( 2J, 2, -2) O (1, 1, -3). (Use a 7" x 10" rect- angle for this problem. ) 7. Draw the top and front views of a line which passes through P{ IJ li, — U) and touches the lines A( — li, i, — |) B( -{, IJ, — 2)andC( — |,' 1|, -i,) E( I, I, -2). 8. Draw the top and front views of a line which passes through F ( --2i, h -1), is parallel to the plane A( 2i, 1», — ?) B{ i. J, — f ) C( 2f, i, — 1|), and touches the line E( — H, IJ, — 2J) G{ 0, J, — ^). 9. A bore hole is driven at P( —250', 100', —100') N45°E dipping 60°- Pind the length of the bore hole to the point where it strikes the wall of a vein of ore whose strike is S( —250', 300', —300') S( 50', 150', —300') INTERSECTIONS OF LINES WITH PLANES 49 and whose dip is 45° to the South West. What is the difference in ele- vation between P and the point where the bore hole strikes the vein? (Scale 1" = 100'. Use a 7" x 10" retangle for this problem). 10. Fin4the points in which the line A{ —2, 3|, — 3|) B( 1, |, — 2f ) pierces the tetrahedron whose vertex is V( ^, 2J, — 1^) and whose base is C(^lh 2i, —i) E( -U, i, —4) G( i, i, -4). (Use a 7" xlO" rectangle for this problem). . yi* J, c P\ f' ^^M: 50 DESCRIPTIVE GEOMETRY 51. Two planes are each represented by two of their lines. Find the line of intersection of the planes. Let AB and BC, Fig, 37, be the lines of one plane and MN and OP the lines of the other plane. Fig. 37. — KG line, of intersection of two planes. \ Analysis. Find the- points where two lines of one plane pierce the other' plane. The line joining these piercing points is the required line of intersection of the planes. Construction. MN pierces the plane of AB and BC at the point K. (Art. 48). OP pierces the plane of AB fl,nd BC at the point G.. Then kg and fc'gr' are top and front views of the required line of intersection of the planes. Some times the lines which represent one plane do not cross the lines which represent the other plane within the limits of the drawing, or they cross in such a manner that the above method cannot be used to find the intersections of the planes. In this case points of the line of intersection can be found by drawing additional lines of one plane which do cross the lines of the other plane within the limits of the drawing. The piercing points of these additional lines determine the line of intersec- tion of the planes. INTERSECTIONS OF LINES WITH PLANES 51 52. Problems. Use a T xlO" rectangle, for each of the following problems: 1. Draw the top and front views of the line of intersection of the planes A( -2, 1, —3) B(i, 3|, — 1|) C{ —1, i, -^) E{li, 2|, -2|) and M (-1, 2|, -1) N (li, If, -3f) 0( -If, 2i, -IJ) P(i, 1, -H). 2. DraWilJafij top -and front views of the line of intersection of the planes A( -2i, i, -3) B(0, 3, -|) C(l|, If, -44) and M( -2J, 2J, -2^) N (2, 2i, -2i) 0{ -2^, 1|, -H) P(2i, li, -15). 3. Draw the top and front views of the line of intersection of the planes B^, —4, — 4) A( -2, 1^, -2|)C(2, i, -21)andE(l, 2i,— 1|) A( —2, li, -2|)G(H, l,-3i). 4. Draw the top and front views ot the line of intersection of the planes A( —2, 2J, —11) B(2,2i, -If) C{ -2, |, 4) E(2, |, -4) andM( — 1|, 2, -3i) N(ll, 2, -31) 0( -1|, li, -2f) P(lf, li, -2 f). 5. Draw the top and front views of the lines of intersection of the planes A{ -2i, li, -2J) B{1, 3i, -3|) C{1, If, -1%) and M( -2, 2J, -2) N(li H, -3i) 0{ -1, 3i, -If) P(2J, 2h -2|). *■ * 6. Draw the top and front views of a line which passes through the point P( —2, IJ, — 2|) and is parallel to both planes A(2, 1, — IJ) B(— 1, 1, -4J) CCli, 34, -41) and M{-1, 3i, -3f) N(2, i, -3|) 0( -2, J, -1|). •4 ai |iy i: "Wc a'l ' \ti 1 rtf' pif i/n: "", 1 c'i — — !«?' m ^"n a' 52 DESCRIPTIVE GEOMETRY PERPENDICULAR RELATIONS BETWEEN LINES AND PLANES 53. A line perpendicular to a plane. A line which is perpendicular to a plane is perpendicular to every line of the plane. Two lines which are perpendicular to each other will appear perpendicular in any view where one oi the lines is shown in its true length. Since all hori- zontal lines of a plane show true length in the top view, the top view of a perpendicular to a plane will, therefore, appear at right angles to the top view of all these horizontal lines. Like- wise, the front view of a perpendicular to a plane will appear at right angles to the front view of all frontal lines on the plane. For example, in Fig. 38, let it be re- quired to represent a perpendicular to the plane ABC from the point P. Draw any horizontal line SS and any frontal line FF on the plane. Draw po perpendicular to SS and p' o' perjjendicular to /'/' . Then the line PO is perpendicular to the plane ABC. In all the problems dealing with lines and planes at right angles to each other, the planes may be represented by horizon- tal and frontal lines unless the problem states definitely that they are to be repre- sented otherwise. If the plane were not represented by its horizontal and frontal lines, these lines could be easily drawn, but the additional lines would complicate the drawing. Another advantage of repre- senting a plane by its horizontal and frontal lines is that its po- sition in space can be more easily visualized. 54- Given the top and front views of a point and of two lines which represent a plane, find the distance from the point to the plane. Fig. 38.— Line OP perpendicular to plane ABC. PERPENDICULAR RELATIONS BETWEEN LINES AND PLANES 53 Let SS and FF, Fig. 39, be the lines which represent the plane and P the given point. Analysis. Draw a perpendicular from the given point to the given plane (Art. 53), and find where it pierces the plane (Art. 48). The true length of the perpendicular from the given point to the piercing point is the required distance. Gonstruction. pq,- Sit right angles to ss, and p'q' , at right angles to//, are the top and front views, respectively, of the perpendicular to the plane from the point P. In the auxiliary view taken by looking in the direction of SS, Si/ represents the Fig. 39.— P§ pierces plane at 0. oblique plane and pi the given point. PiQi at right angles to Si/i, is the auxiliary view of the perpendicular to the plane and Oj is the auxiliary view of the point where it pierces the plane. and o' are the top and front views respectively of the point where the perpendicular pierces the plane. PiOi is the true dis- tance from the point P to the given plane. 55. Conversely, to find the top and front views of a point which is a g-iven distance from a given plane. Let SS and FF,Fig, 40, be the lines which represent the given plane. Analysis. There are an infinite number of points which are a given distance, say l", from a given plane. To represent one of 54 DESCRIPTIVE GEOMETKY these points, select any point on the plane and erect a perpen- dicular to the plane at this point. On the true length of the perpendicular measure the given distance from the point where the perpendicular pierces the plane. This will locate the required point. Construction. Take an auxiliary view of the plane from such a position that the plane appears as a straight line Si/j. At any point of the line Si/i as iCi, erect a perpendicular to the line and measure the given distance along the perpendicularf rom the point Fig. 40.— P is 1" from plane. Xi , thus locating Pi , the auxiliary view of the required point. The top view of the point X is at any point along the line through Xi parallel to ss as x. A perpendicular to ss from x, and a parallel to ss from Pi intersect at p, the top view of the required point 1" from the plane, p' is its front view, p'x' should check per- pendicular to p'f. This construction can be used to locate the vertex of a right cone or pyramid or the upper . corners of a rectangular object when the base of the object is in an oblique plane. 56. Given the top and front views of a point and of an oblique line. Represent a plane which contains the point and is perpendicular to the line. PERPENDICULAR RELATIONS BETWEEN LINES AND PLANES 55 Fig. il,— Plane per pendicular to AB. Let P, Fig. 41, be the given point and AB the given line. Analysis. Since the plane is to be per- pendicnlar to the line, a horizontal line of the plane will have its top view perpendicu- lar to the top view of the line. Likewise, a frontal line of the plane will have its front view perpendicular to the front view of the line. Therefore to represent the plane, draw the top and front views of a horizontal line and of a frontal line which pass through the given point and are per- pendicular to the given line. These two lines will represent the required plane. Construction. Through p, draw ss per- pendicular to ah and through p' draw s' s' parallel to G. L. Through p' , draw /'/' perpendicular to «'&' and through p draw ff parallel to G. L. Then the plane of SS and FF is perpendicular to the line AB. Although the plane of SS and FF is perpendicular to the line AB, the line AB does not intersect either of the lines SS or FF. 57. Problems. 1. Find the distance from the point P(J, If, — f ) to the plane A( — 2%, \, -I) B( -i, H, -2i) C(li, 0, -1|.) 2. Find the distance from the point 0( — \, 1|, — J) to the plane A( — 2, 1}, -2) B(i, 11. -2) C(-2, }, -1) EU, \, -1.) 3. 0( — 1|, I, —2) is one corner of a cube which has its base in the plane C( —2%, H, — i,) A( —J, If, — 2i) B{ — |, J, -J). Find the size of the cube. 4. B is one corner ot a IJ" cube which has its base in the plane A( — 2^, ], — 2J) B( —1, 1|, —I) C( \, J, — 1|). Locate an adjacent corner which does not lie in the plane of the base. 5. Represent a plane which is parallel to and is f " A( -^, h -I) B( -li, ii, -2j) c{i, i, -li.) 6. Locate a point which isl" from the plane A( — f, -f) C{h 1|, -3) E(2, 1, -i.) 7. Represent a plane which contains the line M(J, 1, - and is perpendicular to the plane A( — 1|, 2, —3) B( - 2|, — 4). (Use a 7"xlO" rectangle for this problem.) from the plane If, --2i) B{1, I, -f)N(l|, 2,-2i) J, 3i, -2) C{H, 56 DESCRIPTIVE GEOMETRY 8. Find the locus of all poiats which are equidistant from the points A( —h H, —I) B(l, 3, —2) and C(2^, 2, —4). (Usea 7"xlO" rectanglefor this problem.) 9. A( —1, If, — 2i) B(0, \, — I) is one edge of a cube. Represent by- two lines the plane of the base of the cube. 10. Find the center of the circle in which P ( — l,^, — 1) moves when it rotates about A(0, 1|, — 2J) B(l, \, — |) as an axis. 11. A( — 2, 2J, — 3) B( —1, I, — 11) is the base of an isosceles triangle which has its vertex in the line C(^, ^, — 3) E(2, 3|, — 2J). Represent the triangle in the top and front views. (Usea7"xlO" rectangle for this problem.) b ^P a. io *(?' c'h >' a"- PEEPENDICULAB RELATIONS BETWEEN LINES AND PLANES 57 58. Given the top and front views of two lines of a plane and the top and front views of a third line which does not lie on this plane. Find the projection of the third line on the given plane. Let SS, and FF, Fig. 42, represent the given plane and AB any line which does not lie on this plane. Analysis. From any two points of the given line, erect perpendiculars to the given plane. A line joining the points in which these perpendiculars pierce the plane will be the required projection. The point in which the given line pierces the given plane is also a point on the projec- tion of the line on the plane. Construction. The top view of the per- pendicular to the plane SSFF from the point B is the line through 6 perpendicular to ss. The front view of this perpendicu- lar is the line through b' perpendicular to f'f . This perpendicular pierces the given plane at B. The perpendicular to the plane from A pierces the given plane at the point C. Then ce is the top and c'e' the front view of the projection of AB on the plane SSFF. Fig. 42.— C£ pro- jection of AB on plane. 59. Problems. 1. Find the projection of the line S{1, 0, — 2i) S(1J, J, — 2i) on the plane A(1J, 2i-, -2) B( -li, J, -2) C(li, |, -§). 2. Find the projection of the line P(i, h — 2i) F(1J, J, —If) on the plane A{14, 2J, -1*) B( -IJ, i, — li) om, i, -J). 3. Find the projection of the line 0(i, SJ, —1) P(]|, 2|, — 2J) on the plane A( -If, U, -H) B( -i, 3, -3i) C(li, h -21). (Use a T'xlO" rec- tangle for this problem). 4 Find the projection of the triangle X( —1, 2f , —3) Y( —J, 2i,— 4J) Z(f , 2f, — 3i) on the plane A{1J, 3, -1) B( —Ih h -1) C(1J, i —3). (Use a 7"xlO" rectangle for this problem). 58 DESCRIPTIVE GEOMETRY 5. Find the projection of the parallelogram M(0, 3, — 3|) N{1, 3|, — 3i) 0(1|, 21, -3) P{f , 2, — 3J) on the plane A( -1|, 1^, -3) B{0, 2^, -1) cm, h —2)- (Use a T'xlO" rectangle for this problem). h. ^^ '/•. 1 ^5 PERPENDICULAR RELATIONS BETWEEN LINES AND PLANES 59 60. Given the top and front views of two lines of a plane and the top and front views of a third line which does not lie on this plane, find the ang'le which the third line makes with the g^iven plane. Let SS and FF, Fig. 43, represent the given plane and AB any line which does not lie on this plane. Analysis. The angle which a line makes with a given plane is understood to be the angle which the line makes with its pro- jection on that plane. If a perpendicular be dropped to the plane from any point in the given line, the angle between this perpendicular and the given line is the complement of the angle which the line makes with the plane. Therefore, find the angle between the given line and a perpendicular to the plane from any point of the line construct its complement. This comple- ment is the required angle. Construction. Draw oc and a'c' perpendicular re- spectively to a horizon- tal line and a frontal line of of the plane SSFF. BC is a horizontal line on the plane BAG. When A is rotated about BC it reaches the level of the axis at 03. ba^c is the true size of the angle between AB and the perpendicular to the plane from A. At any point a; of the line ca^, draw xy perpendicular to cag. Then xya^ is the true size of the angle which the line AB makes with the plane SSFF. 61. Problems. 1. Find the angle which the line A( —2, 1^, — 2i) B((), IJ, — 2^) makes with the plane M( —11, {. -U) N( -1, IJ, — i) 0( -1, J, -2i). Fig. 43.-6 angle AB makes with plane. 60 DESCRIPTIVE GEOMETRY 2. Find the angle which the line A( — IJ, I, — IJ) B( — i, IJ, — i makes with the plane S(2i, If, —2) S(i, i, —2) T(2J, i, —1). 3. Find the angle which the line 0( —i, 3J, — 4J,) P{1, IJ, —3) makes with the plane A( — 3i,li,— 3) B{ — li,2^, — 1) C(0, i, —2). (Use a 7" x 10" rectangle for this problem). PERPENDICULAR RELATIONS BETWEEN LINES AND PLANES 61 62. Two planes are each represented by the top and front views of two of their lines, find the' ang-le between the planes. Let ABC and MNO, Fig. 44, be the given planes. Analysis. Prom any point in space drop a perpendicular to each plane. The angle between these perpendiculars is the same size as the angle between the planes. Construction. From any point P, draw PE, perpendicular to the plane ABC and PG perpendicular to the plane MNO. SS is a horizontal line in the plane of the perpendiculars. By rota- ting the perpendiculars about SS as an axis, the true size of the angle between them is found to be ep^g. There- fore ep3.9 is the true size of the angle between the planes ABC and MNO. If an auxiliary view of the two planes be taken by looking in the direction of their line of intersection, each plane appears as a line and the angle between the lines is the true size of the angle between the planes. If one of the pianos is parallel to H or V, the auxiliary view method (Arts. 34 and 37) is the best to use. If the line of intersection of the planes is oblique to both H and V. the auxiliary view taken by looking in the direction of this line cannot be constructed directly from the top and front views. In this case an auxiliary view can be taken by looking paj'allel to H and at right angles to the line of intersection of the planes. Then from this view a second auxiliary view can be constructed by looking in the direction of the line of intersec- tion. The second auxiliary view shows the true size of the angle between the planes. The construction for this method is more 6 angle between planes. 62 DESCRIPTIVE GEOMETRY difficult than for the method using the perpendiculars as ex- plained above. 63. Problems. 1. Find the angle betweea the planes A( — IJ, If, — 2) B(5, If, — 2) 0{-h h -h) and C( -If, H. -I) E( i, U, -f) P( -1, i, -21). 2. Find the angle between the planes A(— 3, If, — 2i) B( —1^, i, — 2J) C( -3, i, -H) and M(3, 1|, -2) N(li, |, -2) 0(3, f, -f) . 3. Find the angle between the planes G( — 3i, 2, — 3i) E( —If, 3 J, -2.J) P( — i, 2i, -4i) and A(0, 2|, -4}) B(l|, 4, -2}) C(3i, 2, -3i). (Use a 7"xlO" rectangle for this problem). 4. A hip roof with peakatO(lJ, 2, — 2f), and plate corners at A( — J, i, —4) B(— J, 3i, —4) C(2J, 3i —4) and ^E{^, h —4) is covered with tile. What should be the angle between the faces of the tile directly over the hip rafter? (Use a T'xlO" rectangle for this problem) . 64. Two oblique lines which do not lie in the same plane are given by their top and front views, find the top and front views and the true length of their common perpen- dicular. General Analysis. An auxiliary view taken from such a position that one of the given lines appears as a point, shows the common perpendicular in its true length. If one of the given lines is parallel to either H or V, but one auxiliary view is neces- sary to show this line as a point, Case 1. If both of the given lines are oblique to H and V, two successive auxiliary views are necessary to show one of the lines as a point, Case 2. Case 1. When one of the lines is parallel to either H orV. Let AB and SS, Fig. 45, be the given lines. Analysis. Take an auxiliary view of the lines by looking in the direction of the line which is parallel to H. In the auxiliary PEBPENDICULAE RELATIONS BETWEEN LINES AND PLANES 63 view, this line appears as a point and the other one as a line. A perpendicular from the point to this line is the auxiliary view of the common perpendicular to the two given lines, and is the true length of the perpendicular. From the auxiliary view of the perpendicular, its top and front views can be found. Fig. 45. — XY common perpendicular to AB and SS. Construction. Si and Uib^ are the auxiliary views of the given lines taken by looking in the direction of SS; AjAj rep- resents the horizontal plane at the level of SS. Xit/i, perpen- dicular to ai6i, is the auxiliary view of the common perpendicu- lar and is its true length, x is the top view of the point X. xy, perpendicular to ss, is the top and x'y' the front view of the re- quired perpendicular to the lines AB and SS. When one of the given lines is parallel to V instead of H as above, the auxiliary view should be taken by looking in the di- rection of this line. 65. Case 2. When both of the lines are oblique to H and V. Let AB and CE, Pig. 46, be the given lines. Analysis. Take a view of both lines by looking parallel to H and perpendicular to one of the lines. Using this auxiliary view in connection with the top view, a second auxiliary view can be drawn. The second auxiliary view is taken by looking in the direction of the line which is shown in its true length in the first auxiliary view. In the second auxiliary view, one line appears 64 DESCEIPTIVE GEOMETRY as a point and the other as a line. Prom the point draw a per- pendicular to the line. This is the second auxiliary view of the required perpendicular and shows its true length. From this view the first auxiliary, the top, and the front views of the per- pendicular can be drawn. Construction. The first auxiliary view is taken by looking parallel to H and perpendicular to the line AB. aifij and Cie^ are the auxiliary views of the given lines. Consider this' aux- iliary view as a new front view and disregard for the time being the original front view. The second auxiliary view is taken from the fir^t by looking in the direction-^of AB. In this view AB appears as the point a2&2 and CE appears as the line Ca^a. Prom the point 0262, draw the line x^y^ per- pendicular to the line 02^3. a;2 2/s is the second auxiliary view of the common perpen- dicular to the two lines and is its true length. From the point 2/2, draw a line parallel to a lb I intersecting c-^ei aty^. Draw yiXi perpendicular to fli&i. Xiyi is the first aux- iliary view of the common perpendicular . From this view the top view xy and then the front view x'y^ of the common perpendicular can be drawn. The distances, asd, used in locating points in the first auxiliary view are measured in the direction of the line joining the top and front views of the point. In a similar manner, the distances, as di, used in locating points in the second auxiliary view are measured in the direction of the line joining the top and first auxiliary views of a point. In general, distances used in locating points in any view are measured in the direction of the line join- FiG. 46. — XY common perpendicular to AB and CE. PERPENDICULAR RELATIONS BETWEEN LINES AND PLANES 65 ing the views of a point in the two next preceding views. This process can be continued indefinitely. 66. Problems: 1. Draw the top and front views of the common perpendicular to the lines S( — 1^, li, —1) S(0, i, —1) and A( — IJ, J, —21) B(0, li, — IJ). 2. Draw the top and front views of the common perpendicular to the lines F(— IJ, i, -2) F(0, h -li) and 0( — IJ, If, — IJ) P(0, |, -21). 3. Find the top and front views of the common perpendicular to the lines A( -1, IJ, -11,) B(l, |, -1) and C(l, f, -^) E(li, 2, -1). 4. Find the top and f roat views of the common perpendicular to the lines A(l, li, -IJ) B(l, i, -21) and C(-J, H, -2) E(0, h -li). 5. Find the length and location of the shortest connection which can be made between two pipes with center lines A( — IT, 11', — 29') B( — 6', 0', —34') and C( —14', 8', —20') E(0', 10', —36'). (Use center lines only- Scale i"=l" —0"- Use a 7"xlO" rectangle for this problem). H h a' 66 DESCRIPTIVE GEOMETRY SPECIAL PROBLEMS 67. To show the top and front views of a line AB which makes 30° with H and 45° with V. If the line is first taken parallel with V and 30° with H, it will haveafei for a top and a'b' i for a front view, a'b'i is the true length of the part of the line under consideration. If a line of the same length is placed parallel with H and 45° with V, it would have aftj for a top and a'6'2 for a front view. If the line ABi be rotated about an axis through A per- pendicular to H, its top view would re- main a constant length. 61 would move in the arc of a circle bib, and b\ would move along the straight line 6' 16' par- allel to G. L. The angle which the line ABi makes with H would remain 30°, but the angle which it makes with V would increase as Bi moves away from V. If the line ABj be rotated about an axis through A perpendicular to V, its front view would remain a constant length. b\ would move in the arc of a circle 6' 26' and 62 would move along the straight line 636 parallel to G. L. The angle which the line AB2 makes with V would remain 45°, but the angle which it makes with H would increase B2 moves away from H. The circles in which Bi and B2 move intersect at the point B, the line 66' being perpendicular to G. L. These circles would not have intersected if the lines ABi and AB2 had been of different lengths. Then ab is the top and a'b' the front view of a line which makes 30° with H and 45° with V. The stationary point A of the line can be selected at random. There arg four lines which contain this point A and make the given angles with H and V. The sum of the angles which a line AB makes with H and V can vary from 0° to 90°. When the sum is 0° the line is par- FlG. il.—AB 30° with Hand 45° withV. SPECIAL PROBLEMS 67 allel to G. L.; when it is 90° the line is perpendicular to Gr. L., that is, a line of profile. 68. To represent a plane which makes 45° with H and 60° with V. The angles which a plane makes with H and V are the com- plements respectively of the angles which a line perpendicular to the plane makes with H and V. Therefore to represent a plane which makes 45° with H and 60° with V, first represent a line AB which makes 45° with H and 30° with V (Art. 67). Then represent a plane which is perpendicular to the line AB (Art. 56). This is the required plane. The required plane can pass through any point in space and make the given angles with H and V. The sum of the angles which the plane makes with H and V can vary from 90° to 180°. 68 DESCRIPTIVE GEOMETRY GENERAL PROBLEMS IN POINT. LINE, AND PLANE In the following list of problems, those marked (a) can be conveniently solved in a 7"xl0" rectangle, those marked (6) in a 10"xl4" rectangle. In each case the base line should be drawn through the center of the rectangle and parallel to its short side. 1. {a) Th9 top view of a parallelogram ABCE is a 2" square. Draw the front view of the parallelogram when its center is 1" above, and the corner B i" below the corner A. Find the true size of the parallelogram. 2. (a) Draw the top and front views of a line which passes through the point P{0, 2, —2), is parallel to the plane A( —}, i, —2) Bt2, 3, —2) C(2, i, — 3J)| and has its top and front views parallel to each other. 3. (a) The top views of two points are 1" apart and their front views are 2" apart. What is the greatest and what is the least distance apart that the points can be placed? i. (a) Draw the top and front views of the locus of a point which is 3" from A{ — 3|, J, —2) and 2" from B(J, IJ, —4). 5. (a) Find a point P on the line M(1J, IJ, —J) N(-i, 2i, —2^) which is equidistant from the planes A( — IJ, 1, —1) B(1J, 3, —1) C{1|, 1, —1) and A( -li, 1, -1) B(H, 3, -1) E(J, 1, -3). 6. (a) Find the top and front views of a point O which is equidistant from A{ — IJ, 1, —1) 3(4, 1, —3) 0(1^, 3, —1) and E( — i. 3, —2). 7. (a) Rspresent a plane which is equidistant from a point A(li, 2, —1) and a line B( —2, 3, —1) C{J, 0, —3) and is parallel to the line E(2, 3, -1) F( -1, 0, -3). 8. (a) Represent a plane which is equidistant from the points A(l^, 2, —1), B( —2, 3, —1), and C(i, 0, —3) and is parallel to the line E(2, 3, —1) F( —1, 0, —3). (Three solutions). 9. (a) Throughthe points A(l, 1, —2), B{ —1, 2, —1), and C(0, 3, —3), draw three planes which are parallel and equidistant. 10. (a) Through the points A(l,l, —2), B{— 1,2, —1), and C(0, 3, —3), draw three planes which are parallel and equidistant, and such that one of the planes passes through P( — 2, 1, — 2). 11. (a) Through the points A(0, 2, 0), B(2, 1, —3), C( —1, 0. —3), and E( — 1|^, 2|, —2), draw four planes which are parallel and equidistant. 12. (a) Draw, respectively, through the points A(l, 1, —2) andB( —1, 2, — 1) and the line C(0, 3, — 3) E(^, 0, — 1), three planes which are par- allel and equidistant. 13. (a) Draw the top and front views of a line which touches the three lines: A(— 1, 1, —1) B (1, 3, —3), C( —2, 2, 0)E(0, 1, —3), andF(0, 3, 0) G(2. 1, -2). 14. (a) Draw the top and front views of a line which passes through PROBLEMS 69 P(2i, 3, 0) and touches the lines C( —2, 2, 0) E(0, 1, —3) and P(0, 3, 0) G(2, 1, -2). 15. (a) Through the lines A( —1,1, —1) B(l, 3, — 3),C(— 2, 2, 0) E(0, 1 — 3), and F(0, 3, 0) G(2, 1, — 2), draw three planes which have a common line of intersection. 16. (a) Draw the top and front views of a plane which contains A( — J, 0, —1) B( —2, li, —2) and passes 1" from P(0, 2, —2). 17. (a) A(— I, 0, —1) B(l, 1, ?) C(?, ?, —3) is a 3" equilateral triangle. Draw its top and front views. 18. (a) Draw the top and front views of a triangle A( — i, 1, —2) B(14, 2, —4) C(?, ?, ?). AC=3", BC=2", and the point C lies in the plane E{0, 0, —2) F(3, 0, —4) G(3, 2, -2). 19. (o) Draw the top and front views of the line MN which intersects the lines A(l, 2, —3) B(3, 1, — 3J) and A(l, 2, —3) C(0, 1, —4) and is par- allel to and 1" from the plane E(0, 0, —2) F(3, 0, —4) G(3, 2, —2). 20. (a) Draw the top and front views of the triangle A(14, 0, — 2^) B{ —1, ?, —2) C(?, ?, 0), when AB=2|", BC=3", and AC=3r- 21. (a) Locate the center and find the diameter of a sphere which is tangent to planes through the origin parallel to H and V, and has its center on the line A{ —2, 2, 0) B(2, 0, —3). 22. (a) At a point of outcrop 0(0, 1, — 1) the strike of a body of ore is north 45° east. The "dip" (the inclination of the face of the ore body to the horizontal) is 60° S. E. A tunnel is driven at P( — 1, 2, — 3) south 60° east on a rising 10^ grade. How far must the tunnel be driven to reach the ore body? Note. The "strike" means a horizontal line in the plane of the face of the ore body. 23. (a) Two ore veins have strike lines due N. E. and both veins dip 30° S. E. The strike lines are 200' apart. What is the shortest distance between the veins? 24. (a) From the point of outcrop A(0, 3, 0), run south 30° west 100' to a drillhole B. The drillhole C is 100' from both A and B. The drill strikes ore at B at a depth of 25' and at C at a depth of 50'. Determine the strike and dip of the vein. 25. (a) The strike of a vein passes through the origin and bears north 75° east. The vein dips 60° toward the north and west. A drillhole which passes through the origin, bears north 30° east and dips 75°. If the core from the drillhole shows 6' of vein matter, what is the true thickness of the vein? Scale ^"=1' — 0" 26. (a) The strike of a vein passes through the origin and bears north 45° east. A drillhole through the origin bears north 15° east and dips 75° 70 DESCRIPTIVE GEOMETRY If the core from the drillhole shows 6' — 0" of vein matter and the vein is linown to be 3' — 0" thick, what must be the dip of the vein? Scale i"=l'-0". 27. (6) Given a plane M( —2, 2, 0) N(l,5, 0) 0(1, 2, —3) and a line A(2f, i, 0) B{|, If, 0). A square which has AB for one side lies in a hori- zontal plane. Revolve this square about MN as an axis until it comes into the plane MNO and then draw the top and front views of a cube which has this square for a base. 28. (6) Given a plane MNO as in problem 27. A 2J" cube rests on this plane in such a position that two edges AB and EF of the base make angles of 15° with the horizontal line MN. The corner B is at the point ( — 1, ?, — 1). Draw the top and front views of the cube. 29. (6) Keeping the edge BA as in 28, draw the top and front views of the same cube with one corner in H. The entire base of the cube does not rest on the plane MNO. 30. (6) Keeping the edge BA as in 28, draw the top and front views of the cube when one of its corners is in a plane parallel to and If" from V. 31. (6) Keeping the corner B as in 28, draw the top and front views of the cube with one corner in H and another in V. 32. (b) One side of an equilateral triangle lies along the line A( — J, 3|, — 2|) B(3|, IJ, — 5|) and the vertex opposite this side is at the point 0(3, 5, — 3). Draw the top and front views of a regular pyramid having this triangle for a base and an altitude of 4J". 33. (6) A regular triangular pyramid whose altitude is 5" stands on a plane which contains M( — 2, 4, — 4J)N(2|, IJ — 4^) and slopes downward to the back making 45° with H. Draw the top and front views of the pyramid when one corner of the base lies in the line A( — 2J, 3, — 2J) B( — |, IJ, — 5J) and one side of the base lies along the line of intersection of the plane of the base and a plane perpendicular to AB at a point 1" from B. 34. (b) Draw the top and front views of a regular triangular pyramid standingon a plane through A( — IJ, 2J, — 3^) parallel to the lines B( — 3J, — 1> — 2i) C(2J, 3J, — 4)andE(— li, 3|, — 2^) F{lf, 0, — 5J). One side of the base lies in the line of intersection of the plane of the base and a plane through A perpendicular to the line BO. The vertex is at the point 0(2J, — I) — i)- Find the points in which BC and EF pierce the pyramid. The above problem can be varied by making the base a square, regu- lar pentagon, regular hexagon, or changing the pyramid to a cone with the base tangent to the line of intersection of the two planes. 35. (b) C( — li, 2{, —2) E(li, 4, — 4|) is a diagonal of a cube. One cor- ner of the cube is in a plane ,which makes 60° with H and contains a horizontal line passing through P( — 3i, ^, —J) and making 45° with V. Draw the top and front views of the cube. PROBLEMS 71 36. (6) C( — 3i, IJ, —1^) E{li, IJ, — li) is a diagonal of a cube which has one corner in a horizontal plane through the origin. Draw the top and front views of the cube. 37. (6) One edge of a 2i" cube lies along the line A{ — 2i, SJ, —4) B{1J, 0, — 2|) and an adjacent edge passes through the point P (J, IJ, ?) in such a direction that its top view makes 60° with the top view of AB. Draw the top and front views of the cube. 38. (6) The face of the cube in problem 37 which lies on the plane ABP is the base of a right square pyramid. Draw the top and front views of the pyramid when it has an altitude of 3J". 39. (6) The common perpendicular to the lines A( — 2, 4, — IJ) B{ — |, J, —43) and C( — 3|, 2i, — 2J) E( —J, 3|, — 4^) is the axis of a right rect- angular prism. The diagonals of the upper and lower bases coincide with AB and CE respectively. The shorter edge of the base is 1}" long. Draw the top and front views of the prism. 40. (6) Substitute A(2, i, — 2^) B(2, 2J, — 4^) and C( —34, 2|, —2) E(i, 5i, -4) in problem 39. 41. (6) Draw the top and front views of a line which passes through 42. (6) Find the top and front views of the shortest line which is par- allel to H and terminates in the lines A( — 1^, 4, — 2^) B( —J, li, — 3|) and C( -U, 1|, -2i) E( -2|, 3^, -3). 43. (6) A 1" square stick has its edges making 30° to V and oblique to H. Find the true size of the section of this stick by a plane parallel to V; also by a plane parallel to P. 44. (6) Draw the top and front views of a triangular pyramid O— ABC. The base is A( — f, 3i, —5) B(li, IJ, -3J) C(3i, 4^, —41). The altitude is 2|", while OA=34" and OB=3" are two of the lateral edges. 45. (b) Draw the top and front views of a tetrahedron two edges of which are 3", two 4", and two 5". The edges are to be oblique to H and V. 46. (6) Draw the top and front views of three spheres, each tangent to the other two and having the following radii: IJ", 2", 2\". Draw the views of a fourth sphere with radius 1" which will be tangent to the other three spheres. The plane of the centers of any three spheres is to be oblique to both H and V. 47. (a) Given an acute triangle on a horizontal plane. Find a point in space which joined with the vertices of the triangle will form a tri- rectangular tetrahedron. 48. (b) Draw the top and front views of a 3" cube with one corner at 0(0, 3, — 2) and three edges along the lines joining O with the points A( -2, 2i, ?), B(l|, 5|, ?), and C(2i, i, ?). 72 DESCRIPTIVE GEOMETRY 49. {&) Draw the top and front views of an equilateral triangle with one corner at A( —J, 3J, -4|), another at 3(1}, 2J, — 2|), an4 a side along the line joining B with P(3|, 5^, ?). 50. (6) Draw the top and front views of a cube with one corner at A(0, 3, —2), another at B(2, IJ, —21), and a side along the line joining A with C{ —2, 2, ?). 51. (6) Given the points 0(J, |, —If), A(25, 2, —3), B( — li, 1|, — 3i), C( —2^, 3J, —3), and E{ —2, 5|, —5^). Draw the top and front views of a parallelopiped with edges along the lines OA, OB, and OC, and a 5" diagonal along OE. 52. (a) A ray of light passes through P(3, 3, —2) and is reflected from V at the point 0(1, 0, — f). Where will the ray strike H? 53. (a) Find the top and front views of a ray of light which emanates from A( —2, 3, —2) and after being reflected from H and Vpasses through B(2,2,-l). 54. (6) A ray of light from A( —3, 2^, — 3f) is reflected at B(0, 3f, — 2|) to C( —1, |, — 4J). Find the reflecting plane. 55. (6) Represent a plane which contains the line A( — If, 2, — 2}) B(l, 3", — 3|) and is in such a position that AB bisects the angle between a horizontal line and frontal line of the plane. Draw the horizontal and frontal lines through A. 56. (a) Draw the top, front, and left side views of the square with cen- ter at P(lf , 2i, —3) and side along the line A(2|, 1|, — 4i) B(2|, 4, — 2J). 57. (a) Draw three views of a squre with center at P(2, 2, — 2) and side along the line A(3i, 1, — 3J) B{3i, 3i, — li). 58. (a) A 2" X 2^" rectangle has its center on the line 0(i, i, — 4J) P{2, 3J, — 2) and lies in a plane which contains the line M( — J, l, — 3J) N(3, 2J, — 2|) and slopes downward to the back making 45° with H. Draw the top and front views of the rectangle when one side makes 30° with MN. 59. (6) Draw the top and front views of a 2J" cube with one edge along the line A(1J, 3|, —If) B(li, J, — 5J) and an adjacent edge pass- ing through the point P(0, 3i, — 3J). 60. (6) Draw the top and front views of a 2^" cube standing on the plane A(0, 3J, —2) B(0, 0, — 5^) P{ — IJ, 3, —Si). One side of the base is to make 30° with AB. 61. (b) S( — 1|,5, — 3i)S(i,2J, — 3J) is onelineof aplane whichmakes 30^ with H. Draw the top and front views of a 2| " cube which has its base in the plane. One side of the base is to make 30° with a frontal line of the plane. 62. (6) Given a plane A( —4, i, 0) B(l, J, —5) C(l, 5|, 0) and a point 0( — 3J, 3, — 2i). Draw the top and front views of a cube which has one PROBLEMS 73 corner at O, its base in the plane ABC, and one corner of the base in the line AC. 63. (&) Given aplane A( —4, —i, — |) B(l, —J, — 5f ) C(l, 4|, — f). Draw the top and front views of a 2J " cube which has its base in the given plane. One corner of the base is at M( — 1, ?, — 1^) and another corner of the base is 1" lower than the line AC. 64. (6) A plane which makes 30° with H and slopes downward to the right has S( — 4J, — i, — 4J) S(0, 4J, — 4|) for a horizontal. C(1J, 2, —^) is an upper corner of a cube which has its base in the given plane. One corner of the base is to be on the horizontal AB. Draw the top and front views of the cube. 65.(5) GivenaplaneA{— 4,— i,— 1)B(1,— i,— 5J)C(1,4|,— I). Draw the top and front views of a 2J" cube which has its base in the given plane. One corner of the base is to be on the line AC at the point 0( — 1, ?, ?) and one edge of the base is to make 30° with the line AC. 66. (&) S(— If, 5, — 3i) S(f, 2f, — 3J) is one line of a plane which makes 30° with H. Draw the top and front views of a 2J" cube which has its base in the plane. One side of the base is to make 30° with MN. 67. (&) Given a point P( — 1|, 3f, — 3|) and a line A( — 2J, 2, —3) B{ — J, 2|, — 4J). Draw the top and front views of a cube having AB for one edge and the plane ABP for the plane of the base. 68. (&) A( —1, 2, ?) B(J. 3, ?) is a line which makes 45° with H. This line is one edge of a cube. Draw the top and front views of the cube when none of the edges are parallel to either H or V. 69. (&) A( —1, 2, ?) B(i, 3, ?) is a line which is 2J" long. This line is one edge of a cube. Draw the top and front views of the cube when none of the edges are parallel to either H or V. 70. (6) M(— 3, 2i, ?) N( —2, 3^, ?) is a line which makes 30°, 45° or 60° with H. Drajr the top and front views of a cube having MN for one edge. One corner of the base is to be ^" lower than M. 71. (6) One edge of the base of a 2J" cube lies along the line A( — 2|, . If, — 2J) B(f , 2J, — 4i) and an adjacent edge passes through the point P( — }, 4, — 4J). Draw the top and front views of the cube. ■ 72. (b) A( —3, 2i, ?) B( — IJ, 3i, ?) is a line which makes 60° with H. Draw the top and front views of the right square pyramid having AB for an axis. One corner of the base is to be f " farther back than B, side of base 2i". 73. (6) A regular triangular pyramid whose altitude is 5" stands on a plane which contains S( — 2, 4, — 4^) S(2|, IJ, — 4J) and slopes downward to the back making 4.5° with H- Represent the pyramid by its top and front views when one corner of the base lies in the line A( — 2J, 3, — 2J) B( — f, li^, — 5J) and one side of the base lies along the line of intersec- 74 DESCKIPTIVE GEOMETRY tion ot the plane of the base and a plane perpendicular to AB at a point 1" from B. 74. (6) Oae ed^re of the base of a right square pyramid lies along the line A(li, 3J, — 2) B(1J, 0, — 5J) andan adjacent edge passes through the point P(0, 3, — 3J). Draw the top and front views of the pyramid with side of base 24" and altitude 3^". 75. (6) Given a plane A( —4, I, 0) B(l, J, —5) 0(1, 5i, 0) and a point 0( — 2, 4|, — 3). Draw the top and front views of a right square pyra- mid with vertex at O and base in the plane ABC. The side of the base is to be 2\", and one corner of the base to be in the line AC. 76. (b) 0( — 2, 4i, — 3) is the vertex of a right square pyramid which has its base on the plane A( —4, J, 0) B(l, i, —5) C(l, 5J, 0). One side of the base is to make 30° with the line AB. Draw the top and front views of the pyramid with side of base 2J". 77. (6) The base ot a right pyramid whose altitude is 3J" is an equi- lateral triangle with one side along the line A( — 4|, J, — 4J) B(^, IJ, — J) and one corner on the line M( — 4f , 2|, — 6|) N(i, 3|, —3). Draw the top and front views of the pyramid. 78. (6) Draw the top and front views of a right square pyramid with vertex at 0( — 2J, 2|, — 4J) and base in the plane A( — 4^, —1, — 1^) B(|, — 1, — 8J) C(J, 4, — IJ). One corner of the base is to be on the line AC, side of base 2| " . 79. (6) P( — 1, 1|, — 3|) is the center of an equilateral triangle hav- ing one side along the line A( — 4|, 0, — 5J) B(l^, 1|, — 1|). This tri- angle is the b ise of a ri^ht pyramid with an altitude of 31". Draw the top and front views of the pyramid. 80. (6) Draw the top and front views of a right square pyramid which has its base on the plane A{ —4, —1, —2) B(0, —1, —6) C(i), 3, —2). One corner of the base is on the line A,C, at the point 0( — 1, ?, ?) and one side of the base makes 30° with the line AC. Side of base of pyra- ' mid 2J", altitude 3"- 81. (6) a( —3, 2J. ?) b( — IJ, 3J, ?) is the top view of a line which is 3|" long. Draw the top and front views of the right square pyramid having AB for an axis. Oae side ot the base is to make 30° with a frontal on the plane of the base. Side of base 2|" 82. (6) P( — i, 3J, — 2^) is the vertex of a right square pyramid which hasitsbiseintheplaneA(— 3, 1, — 2J)B(5, 1, — 4J) C( —3, 3|,— 4|). One side of the base is to make 30° with AB. Draw the top and front views of the pyramid when the side of base is equal to the altitude. 83. (&) A( — IJ, IJ, —3) is one corner of the base of a right square prism which has its axis along the line B( — ^i, ii — 15) C(2J, 4, ?). The axis makes 30° with H, and is 4" long. Draw the top and front views ot the prism. CHAPTER II APPLICATIONS OF THE ELEMENTARY PRINCIPLES OF THE POINT, LINE, AND PLANE SHADES AND SHADOWS 69. In order that the views of an object shall more nearly represent the object as it appears, the shadows which the differ- ent parts of the object cast on other parts or on the planes of projection are sometimes shown. Shadows are used more in architecture than in any other branch of engineering but every engineer should know how to find the shadow of a simple ob- ject. Only the elementary principles of the subject will be out- lined here. 70. Definitions. If an opaque object is placed between a surface and the source of light, the object will cast a shadow upon the surface. This is due to the fact that the object inter- cepts the rays of light and prevents their reaching the surface upon which the shadow is cast. When the light strikes an opaque body, that portion of the surface of the body which is turned away from the source of light and which, therefore, does not receive any of the light rays is said to be in the shade. The source of light . is the sun and because of its distance from the earth the rays of light are assumed to be parallel. For the sake of uniformity, the direction of a ray of light is the direction of the diagonal of a cube which joins the upper, left hand, front corner with the lower, right hand, back corner when the cube has two faces parallel to V and two parallel to H. In other words a ray of light is considered as coming over the left shoulder of the observer so that both its top and front views make angles of 45° with the ground line. 71 . Shadows of points and lines. To find the shadow of a point on a surface, pass a ray of light through the point and find where this ray pierces the surface. This piercing point is the required shadow point. 76 DESCRIPTIVE GEOMETRY To find the shadow of a straight line on a plane, find the shadows of two points of the line on the plane and join them by a straight line. If the shadow of the straight line falls upon two or more planes, find the shadows of two points of the line on each plane and join the corresponding shadow points. The shadow of the straight line is a broken line which changes di- rection on the line of intersection of the planes upon which the shadow is cast. To find the shadow of a straight line on a curved surface, find the shadows of several points of the line and join these shadow points in the proper order. To find the shadow of a curved line on a surface, it is usually necessary to find the shadows of several points of the curve on the surface and then join the shadow points in the proper order. The shadow of a plane fig^ure on a plane which is parallel to the plane of the figure is the same size and form as the figure itself. 72. To find the shadow of a square prism on the plane of the base. Let the prism be given as shown in Pig. 48. Analysis. Since the shadow of the prism is to be cast upon the plane of the base, it is evident that the base of the prism coincides with its own shadow. The shadow of the top is found by passing rays of light through the points A, B, C, and D and finding where these rays pierce the plane of the base. The straight lines joining these piercing points, in the proper order, will form the outline of the shadow of the top. The shadows of the corners of the. upper base joined with the shadows of the corresponding corners of the lower base will be the shadows of the lateral edges (Art. 71). This completes the outline of the shadow of the prism. Construction, bb, is the top and b'b', is the front view of a ray of light which passes through B. The front view &'» of the point where this ray pierces the plane of the base is first found and then the top view h Ideated. Then B, is the shadow of the SHADES AND SHADOWS 77 point B upon the plane of the base of the prism. In like man- ner Cs, D,, and A,, are respectively the shadows cast by C, D, and A upon the plane of the base of the prism. Joining these m' n' p' o'Os t>s ^s ^s Fig. 48.— S/iado'io of nquare prism. points in order gives the shadow of the top on the plane of the base. Joining B, with N gives the shadow of the vertical edge BN. The shadows of the other vertical edges are found in the same manner. Then nhCad,p is the top view of the outline of the shadow of the prism on the plane of the base of the prism. The face COPD of the object is turned away from the light and is therefore in the shade. This face is visible in the front view, and the fact that it is in the shade is indicated as shown in the figure. 78 DESCRIPTIVE GEOMETRY From the figure, it is seen that a line and its shadow are par- allel when the line is parallel to the plane upon which the shadow is cast. 73. To find the shadow of an object on the plane of the base and on the object itself. Let the object be given as shown in Fig. 49. This problem illustrates the finding of the shadow which an opaque object will cast upon itself as well as the plane of the base. It also shows the method of finding the shadow on a vertical plane in addition to finding the shadow on a horizontal plane. Analysis. All of the edges shown are straight lines. If all of the shadow of any edge falls on one plane, it will be neces- sary to find the shadow of only two points of the edge and join them by a straight line. If the shadow of an edge falls on more than one plane, two shadow points should be found on each plane and joined by a straight line. The shadow of any point of an edge falls upon the plane which is first pierced by the ray of light passing through that point (Art. 71) . Construction. Bach point of the top view of the shadow which the object casts upon the plane of the base is found as illus- trated by points C and D, Pig. 48. For this shadow it is only necessary to find where the rays of light pierce a horizontal plane. The shadow which the upper part of the object casts upon the rectangular collar is found as illustrated by corner D, d's being first located then d,. The rectangular collar casts a shadow upon the vertical faces of the object. The horizontal edge BE of the collar casts part of its shadow, Kcs, on the plane of the base and the remainder, e'so'.e', on two of the vertical faces of the object, o, and e, are first located and then the front views o\ and e\. To determine the part CO of the edge BE which casts a shadow on the vertical face, draw the top views of rays of light through c, and Os, thus locating the points c and o. The front view shows the shadow which the collar casts upon the vertical faces of the object. SHADES AND SHADOWS 79 Fig. 49. — Shadow of object on plane of base and object itself. 80 DESCRIPTIVE GEOMETRY 74. To find the shadow of a cylindrical column and cap on a horizontal plane and also the shadow of the cap on the column. Let the column and the cap be given as shown in Fig 50. /sis. Since the upper surface of the cap is a horizontal Fig. 50. — Shadow of column and cap. circle, its shadow on a horizontal plane is a circle of the same radius as the cap. The shadow of the center of the circle is the center of the shadow of the circle. Finding the shadow of the center and drawing a circle with a radius equal to the radius of the cap, will give the shadow of the top of the cap on a horizontal plane. The shadow of the lower surface of the cap on a horizontal plane is found in the same way. Common tan- SHADES AND SHADOWS 81 gents to these circles drawn in the direction of the top view of a ray of light are the shadows of the extreme elements of the cap and complete the shadow of the cap on the horizontal plane. The shadow of the column on a horizontal plane is bounded by- two lines drawn in the direction of the top view of a ray of light and tangent to the base of the column. To find the shadow of the cap on the column, pass rays of light through points of the lower edge of the cap and find where they pierce the column. The top views h,, d,, etc., of these points are first determined and subsequently, the front views 6/, ds', etc- Aline joining these points is the boundary of the shadow of the cap on the column. The side of the column and cap turned away from the source of light is in the shade. GK is the portion of the column and EM the portion of the cap which are respectively in the shade and visible in the front view. 75. Problems. 1. One edge of a 2" cube lies on a horizontal plane and is oblique to V. The plane of one face of the cube makes 30° with H. Draw the top and front views of the cube and find its shadow on a horizontal plane. 2. Draw the top and front views of a 2" cube when one diagonal is perpendicular to H. Find the shadow of the cube on a horizontal plane which passes through the lower end of the diagonal . 3. A right square pyramid has its vertex in a horizontal plane. The side of the base is 2" The altitude is 3" long, makes 60° with H, and is oblique to V. Draw the top and front views of the pyramid and find its shadow on a horizontal plane. 4. Draw the top and front views of a sphere which has a radius of 1\". Find at least twelve points in its shadow on a horizontal plane. 82 DESCRIPTIVE GEOMETKY PLANE SECTIONS AND DEVELOPMENTS OF THE SURFACES OF PRISMS AND PYRAMIDS 76. By the development of a given surface, such as the sur- face of a prism or cone, is meant a plane area of such a form and size that it can be rolled or folded into the form of the given surface. To find the line of intersection of a rig-ht prism witli a given oblique plane and to develop the surface of the prism. Let the prism be given as in Fig. 51, and let SSF represent the oblique plane. Analysis. Take an auxiliary view of the prism and plane by looking in the direction of the horizontal line SS of the plane. In this view the plane is represented as one line and the points in which the lateral edges of the prism pierce the plane are apparent. Pig. 51. — Plane section and development of right prism. The development is made from the true size of the base of the prism and from the true lengths of its lateral edges. The true size of the base appears in top view while the true lengths of the lateral edges appear in both the front and auxiliary views. PLANE SECTIONS AND DEVELOPMENTS OF SURFACES 83 Construction. In the auxiliary view, ^i^i is drawn at right angles to ss and at any convenient distance from the top view. hihi represents a horizontal plane at the level of SS. The lower base of the prism is the distance dj below this plane. Sj is the auxiliary view of the horizontal line SS and /j represents the point P in this view. Sj/i represents the oblique plane and ^ii yi> Zi, its intersection with the edges of 'the prism. The top and front views of X, Y, and Z are found from their auxiliary views. It is best to place the development directly opposite the front view so that the true lengths of the elements can be pro- jected across from that view. Draw azffa and m2»W2 as exten- sions of a'c' and m'o' respectively. Lay off a2b2 = ab, 62''2 = bo, etc. Then ajWaWijaa is the development of the lateral sur- face of the prism. Projecting the points x' , y' , and z' across to the proper edges determines the points X2, y2, and 22. and these joined by lines represent the line of intersection on the development. This method also applies to the plane section and development of a right cylinder. Note. All the problems in plane sections of prisms and pyra- mids can be multiplied by changing the number of lateral faces of the surface. When desired, the bases of the surfaces can be shown true size on the development. In this text the bases are omitted. Fre- quently the bases are shown true size in either the top or front view. 77. To develop the lateral surface of an oblique prism. Let the prism be given with its base in a horizontal plane and its lateral edges oblique to the plane of the base as show in Fig. 52. Analysis. Take a section of the prism by a plane at right angles to the lateral edges. This section becomes a straight line on the development and the width of the faces from one lateral edge to the next can be measured along it. The lateral edges are drawn at right angles to this section line in the develop- 84 DESCRIPTIVE GEOMETRY meat, and the distances from the plane of the section to the upper and lower bases are laid off on the respective edges in the development. Straight lines joining in the proper order «the ends of the lateral edges complete the development of the prism. Construction. From the top and front views of the prism an auxiliary view is drawn. This view is taken from a direction Fig. 52. — Plane section and development of oblique prism. such that the lateral edges of the prism show true length, as CjOi, ttiWi, fiiJii. The plane which cuts the prism at right angles to the edges appears in the auxiliary view as aline at right angles to the edges and cutting them in the points Xi,yi, and Sj . This section can be taken anywhere between the bases, x, y, s is the top and a;', y' , z' the front view of this section. Any point, as Sj, in the auxiliary view of the cutting plane can be PLANE SECTIONS AND DEVELOPMENTS OF SURFACES 85 taken as the view of a horizontal axis SS. When the triangle XYZ is revolved into a horizontal position about SS as an axis, the top view^s, Vs, Ss shows its true size. On any straight line, as x^x^, lay off Xiyi=Xay3, 2/424=^323, etc. At the points ^i, Vi, S4, 3:4 erect perpendiculars to XtX^ and measure off on them the respective lengths of the edges obtained from the aux- iliary view. The distance d 2 locates the point O in the develop- ment at 02 and'di locates C at C2. The ends of the other edges are located in a similar manner and when these ends are joined in the proper order the development is completed. 78. To develop the lateral surface of an oblique pyramid and to sliow on the development the line of intersection of the surface with an oblique plane. Let the pyramid be given with its base in a horizontal plane Pig. 53, and let SSF be the oblique plane cutting the lateral surface. Analysis. Find the points where the lateral edges of the pyra- mid pierce the oblique plane and join these piercing points in proper order with straight lines. This gives the intersection of the surface with the oblique plane. Lay out the faces of the pyramid on the development, constructing each face as a triangle having given the three sides. The line of intersection with the oblique plane is laid off on the development by measuring along each edge the true distance from the vertex to the point in which this edge is cut by the oblique plane. Joining these points by straight lines, gives the intersection on the development. Construction. The auxiliary view is taken by. looking in the direction of the line SS. In this view, s^fy represents the oblique plane cutting the lateral edges in the points iCi , ^1 , and Si . xyz is the top and x'y'z' the front view of the intersection. The top view shows the base ABC in its true size. If each of the lateral edges of the pyramid be rotated about a vertical axis through the vertex until the edge comes parallel to V, the front view will show its true length. For example, in the top view c moves in a circle with center at v until it reaches C2, while in the front view a' moves in a line parallel to G. L. until it 86 DESCRIPTIVE GEOMETRY reaches c' 2. Thenv' c' 1 is the true length of the edge VC. In the front view, the point 2' , on the line d'c', moves parallel to G.L. until it reaches 22'- "W ' 3 ' 2 is the true length of the line VZ. In the development each triangle as v^a^c^, D3C363, etc., is con- structed by arcs, the length of three sides of the triangle being given. The true distances from the vertex to the points in which the oblique plane cuts the lateral edges are taken from the front view and laid off on the development, locating the line of intersection x^z^y^x^^. Fig. 53. — Plane section and development of oblique pyramid. In a regular pyramid, the development is a series of isosceles triangles, with the equal sides the length of the laterial edges of the pyramid and the bases the length of the sides of the base of the pyramid. PLANE SECTIONS AND DEVELOPMENTS OP SURFACES 87 79. Problems. Use a T'xlO" rectangle for each of the following problems. 1. Assume any four points which do not lie in a plane, as A( — If, 1, — 4),B(— li, 2i, — 2|),C(— J, li,-3i), andD(l, 2,— 3^), and draw through them four parallel lines in such a way that the plane section of the prism which has the four lines for edges will be a parallelogram. 2. Given a tetrahedron, 0(0, 3, 0) A( —2, 2J, —1) B{ —3,1, — 3)C(— 1, i, — 2-J) . Select a point on one of the edges and then draw the top and front views of the shortest path to be followed on the surface of the tetrahedron, beginning at this point, crossing three of the faces and re- turning to the same point. ' 3. Given two planes, A(|, I4, 0) B( —2^, 0, 0) C(— IJ, 0, — 2i) and A(J, IJ, 0) E{1, 0, 0) C( — IJ, 0, — 2J). Draw the top and front views of the shortest path on the planes from B to E. 4. Pass a plane so as to cut a regular hexagon from a 3" cube. 5. Cut a parallelogram having one side 2" long from the pyramid O-ABCE. 0(1J, 3J, 0) A( -2^, 4, -4) B(i, -J, -4) C(2J, H, -4) E(-lf, 3i, -4.) 6. Given a square prism with its axis perpendicular to H. Side of base 2J" Cut a parallelogram from the prism having one side 3" long and one angle 60°. 7. Given a square prism with its axis perpendicular to H. Side of base 21". Cut a parallelogram from the prism having one diagonal 6^" long and one angle 60°. Use a 10" X 14" rectangle for each of the following problems. The liase line should be drawn through the center of the rectangle and parallel to the longer side. 8. A triangle A{ -6f , 2/^, — 4|) B( - 5J, 0, — 4|) C( -4i, 2^, — 4|) is the base of an oblique prism. The top view of the edges make 45° with V. Thee dges are 45" long and make 60° with H. Develop the surface of the prism. 9. The base of a triangular pyramid is A( — 6, 2J, — 1) B( — 3f , 1^, — 1) C( — 4J, 4|, —1). The vertex is V( — 2|, 4, ?). The axis passes through the center of the circle ABC and makes 60° with H. A plane passes through P( — 2|, \\, — 2) and is perpendicular to the axis. Find the line of intersection of the pyramid and the plane and develop the pyramid, showing the line of intersection on the development. 88 DESCRIPTIVE GEOMETRY INTERSECTIONS OF THE SURFACES OF PRISMS AND PYRAMIDS 80. General method. Find where the edges of each solid pierces the faces of the other solid. A series of straight lines joining in order these piercing points is the required line of in- tersection. The construction for such a problem can be made by the method of finding where a line pierces a plane. If, however-, one of the surfaces is a prism, the best method for finding the intersection is to take an auxiliary view looking in the direction of the axis of the prism. This view will show the faces of the prism as lines and the intersections of these faces with the edges of the other surface are readily found. This method is particu- larly desirable when the axis of the prism is parallel to H or V. If the axis is oblique to both H and V, a second auxiliary view will be necessary to make the faces of the prism appear as lines. 81. To find the line of intersection of the surfaces of a prism and a pyramid by the auxiliary view method. Let the prism and pyramid be given as in Pig. 54. Analysis. The auxiliary view in the direction of the axis shows the prism as a square. From this view, the points in which the edges of the pyramid pierce the faces of the prism are found by inspection. To find the points in which an edge of the prism pierces the pyramid, use an auxiliary plane contain- ing this edge and parallel to the base of the pyramid. This plane cuts lines from the faces of the pyramid which intersect the edge of the prism in the required points. Construction. In the auxiliary view, the points in which the lateral edges of the pyramid pierce the faces of the prism are seen to be Xi, yi, Wi, etc. The top views a;, y, w, etc., of these points are found from the auxiliary view by projection and then the front views x' , y' , w' , etc., are found from the top views. hihi is the auxiliary view of a plane which is parallel to the INTERSECTIONS OP SURFACES 89 base of the pyramid and contains the edge PP of the prism. This plane cuts a square from the pyramid which has one corner at R and sides parallel to the sides of the base of the pyramid. The square cuts the edge PP of the prism in the required points Q and Z. Joining the points thus found in the proper order gives the required line of intersection. ^>c-^. ~-^- w ^ &- V < =-^^\ /v ?< ^^~^,^ J \^ — 'T' N y ^ ^X>^ \ z.^ i / Fig. 10.— The helicoid in a similar manner and then the curve p'q'r' is drawn tangent to the front views of these elements. In a similar manner the curve x'y'z' is determined. These curves show the contour of the helicoid in the front view. 120 DESCRIPTIVE GEOMETRY The portion of the helicoid usually represented is generated by a line of definite length as CE, and the top view of such a surface is the two concentric circles shown in Pig. 70. Some- times the generating line is extended indefinitely in one direc- tion from the point O, and the intersection of the surface with a horizontal plane, as the one through B, taken as the base. The base in this case is a spiral. Points of the spiral are found first in the front view, where the elements of the surface pierce the plane of the base, and are then projected to the top view where the spiral shows in its true size. The helicoid can be varied in form by changing the distance the generatrix CB is from the axis AB, the angle CB makes with the axis, or the pitch of the helix a' o'b' . In the surface of the ordinary screw thread, the generatrix cuts the axis. Both sur- faces of the screw thread may or may not be generated by one line or this line prolonged beyond the axis. 1 24. To represent a plane tangent to a helicoid at a point of the surface. Let the surface be represented as in Fig. 70, and let G be the given point. Since the surface is ruled, the straight line element FG will lie in the tangent plane. The helix traced by the point E passes through G, and therefore a tangent to the helix at G will lie in the tangent plane. These two lines determine the required plane. If the point G had not been on the helix traced by B, it would have been necessary to construct the helix through G and then draw a tangent to it. The helicoid being a warped surface, the plane which is tan- gent to it at G will not be tangent at any other point of the ele- ment FG. HYPERBOLIC PARABOLOIDS 125. A hyperbolic paraboloid is the path of a straight line moving along two straight lines not in the same plane, and re- maining parallel to a given plane. WARPED SURFACES 121 The moving line is called the generatrix, the two fixed straight lines the directrices, and the given plane the plane di- recter. Any position of the generatrix is called an element of the surface. This surface has a second rectilinear generation in which any two rectilinear elements of the first generation may be taken as directrices and a plane which is parallel to the first directrices as a plane directer. It follows that throuffli any point of a hyperbolic paraboloid, two rectilinear elements can always be drawn. Any intersection of the surface by a plane, which is not a straight line intersection, is a hyperbola or a parabola, hence the name of the surface. It is evident from the nature of the motion of the generating line that the surface is warped. The pilot or "cow catcher" of an American locomotive is usually of the form of the hyperbolic paraboloid. 126. To represent the surface. In Fig. 71, AB and CD are the rectilinear directrices and V is taken for the plane di- recter. The plane directer must either be indicated or represented. Rectilinear elements are drawn parallel to the plane directer and cutting the directrices. In this figure the top views of the ele- ments are first drawn and then their front views determined. A plane parallel with the plane directer will always cut the direct- rices in two points and a straight line joining these points will be an element of the surface, mn and m'n' are the top and front views respectively of an element of the surface. To represent a point of the surface, a rectilinear element, as MN, Fig. 71, is first drawn and then a point P of this element is taken. 127. To represent a plane which is tangent to the sur- face at a point of the surface. Let 0, Fig. 71, be the given point. Analysis. Through the given point pass planes parallel re- spectively to the elements of each generation . These planes will 122 DESCRIPTIVE GEOMETRY intersect the surface in the two rectilinear elements which deter- mine the tangent plane at the given point. Construction. The plane through O parallel to the elements of the first generation will cut the element VW from the sur- face. Through O draw OG parallel to CD, and OU parallel to AB. These lines will determine a plane parallel to the elements Fig. 71. — Hyperbolic paraboloid and tangent plane. of the second generation. This plane will intersect the element XY in Z. The line joining Z with O will be an element of the second generation. OW and OZ will therefore determine the required tangent plane to the surface at the point 0. WARPED SURFACES 123 CONOIDS 128. A conoid is the path of a straight line moving along two other lines, one straight and the other curved, and remaining parallel to a given plane. When the generatrix of a right helicoid touches the axis, the helieoid is also a conoid. In this form the coinoid has many practical applications. To peppesent the SUPfaee. In Fig. 72, MN is the straight lihe and A BCD the curved line directrix. V is the plane direc- ter. In this case, the top view of MN and the circle ABCD rep- resent the top view of the surface. The front view of the sur- face is the triangle m'a'c' . An element of the surface is represented by selecting some point of the curved directrix as E and drawing a straight line through this point parallel to the plane directer until it cuts the di- rectrix MN at M. Then em is the top and e' m' the front view of an element of the surface. To represent a point of the sur- face, an element is first represented and then the point placed on the element, o and o' are the views of a point represented in this man- ner. 129. To peppesent a plane which is tangent to the conoid at a g-iven point on the supface. Let 0. Fig. 72, be the given point. The tangent plane must contain the rectilinear element EM through the point of contact. It must also contain the line OP tangent to the elliptical section made by a horizontal plane through the point. EM and OP will therefore represent the required tangent plane. '^a' Pig. 12.— Conoid and tangent plane. 124 DESCRIPTIVE GEOMETRY CYLINDROIDS 130. A cylindroid is the path of a straight line moving along two curves, and remaining parallel to a given plane. The fender of some automobiles is a cylindroid or a surface similar in form to one. The surface is also used in architecture for joining the arched ceilings in two parallel corridors of dif- ferent levels. To represent the surface. In Fig. 73, ABC and DEG are the curved directrices and H is the plane directer. The surface is represented in the front view by the curved directrices and a series of horizontal lines. The top view of the surface consists of the top views of these straight line elements and the curved directrices. If enough straight line elements are drawn in the top view, the contour of the surface will show as a curved line touching all the top views of these elements. Fig. 73. — Cylindroid and tangent plane. An element of the surface is represented by selecting some point B of one directrix and drawing a straight line parallel to the plane directer H until it cuts the other directrix at E. DOUBLE CUKVED SURFACES 125 Then 6'e' is the front and be the top view of a rectilinear ele- ment of the surface. A point of the surface is represented by first representing an element of the surface as BE and then taking a point of this element, o and o' are the views of a point on the surface. 131. To represent a plane which is tang-ent to the sur- laee at a given point of the surface. Let P, Fig. 73, be the given point. The tangent plane must contain the rectilinear element CP through this point and the tangent to a plane section of the surface at P. NPM is a section of the surface by a plane perpendicular to V. pk is the top and p' k' the front view of the tangent to this curve at P. Then CP and PK represent the tan- gent plane to the surface at the point P. DOUBLE CURVED SURFACES 132. A double curved surface is a surface which can only be generated by the motion of a curved line. 8nch a surface has no straight line elements. The surface may be double convex as a sphere or ellipsoid, concave outward as the curved surface of a pulley, or concavo-convex as the annular torus or anchor ring. The more common form of a double curved surface is that of revolution ; that is a surface which has no straight line elements and can be formed by the rotation of some curve about an axis . The general form of the double curved surface, however, is not one of revolution. For example, the general form of the ellipsoid is a surface in which the three principal axes through the center are all of different length, while in the ellipsoid of revolution two of these axes are of equal length. Since the general form of the surface is not common in prac- tical work, only double curved surfaces of revolution will be considered in this text. 133. Some of the simple double curved surfaces of revolution are the following: 126 DESCRIPTIVE GEOMETRY A sphere is the path of a circle revolving about its diameter as an axis. An annular torus or anchor ring is the path of a circle re- volving about a straight line which lies in the plane of the circle but does not cut the circumference. An ellipsoid of revolution is the path of an ellipse revolv- ing about either axis. When the ellipse is revolved about its major axis, a prolate ellipsoid is generated; when about its minor axis, an oblate ellipsoid. These surfaces are sometimes called spheroids. A paraboloid of revolution is the path of a parabola re- volving about its axis. This surface is extensively used for light reflectors. A hyperboloid of revolution of two sheets is the path of a hyperbola revolving about the axis which passes through the foci. 1 34. To represent the surface. For convenience of repre- sentation, surfaces of revolution are usually taken with the axis of rotation perpendicular to one of the principal planes of pro- jection, although the surface may be placed in any other posi- tion. The view of the surface taken by looking in the direction of the axis of rotation is one or more circles. If the surface is limited the largest circle is shown, but if the surface is indefi- nite in extent its intersection with some plane at right angles to the axis is shown. The view of the surface taken by looking at right angles to the axis is the meridian line, the plane of which is parallel to the picture plane. For example, a prolate ellipsoid with its axis of rotation perpendicular to H, will have for its top view a circle with diameter equal to the minor axis of the gen- erating ellipse and for its front view an ellipse identical to the generating ellipse, and having its major axis perpendicular to G. L. Since there are no straight lines on a double curved surface of revolution, a circle of the surface with its plane perpendicu- lar to the axis may be considered the element. A view of the DOUBLE CURVED SURFACES 127 element looking in the direction of the axis is a circle, and the view looking at right angles to the axis is a straight line equal in length to the diameter of the circle, and having its extremi- ties in the contour of the surface . If one view of a point on a surface of revolution is given, the other view can be found as follows: Through the given view of the point draw the corresponding view of an element of the surface. Then find the other view of the element and place the required view of the point on it. It must not be overlooked that if the top view shows the true size of a circular element, the front will show it as a straight line parallel to G. L. and vice versa. 135. Problems. 1. Represent a point on the surface of a sphere. 2. Represent a point on the surface of a prolate ellipsoid. 3. Represent a point on the surface of an oblate ellipsoid. 4. Represent a point on the surface of a paraboloid of revolution. 5. Represent a point on the surface of a torus. 136. Tangrent planes to double curved surfaces of revolution. The tangent plane to a surface at a point on the surface is determined by two straight lines tangent at this point to two lines of the surface. The simplest curves passing through the point on a surface of revolution are usually the meridian line and the circle which is cut from the surface by a plane perpen- dicular to the axis of the surface. The tangent plane, therefore, is determined by two straight lines, one tangent to the circular section at the given point and the other tangent to the meridian section at that point. 137. To represent a plane which is tangent to an ellipsoid of revolution at a point of the surface. Let the surface be given as in Fig. 74 and let P be the given point represented by the method of Art. 134. Analysis- A tangent to the meridian curve at the given point and a tangent to the circle of the surface at this point will rep- resent the tangent plane (Art. 136). 128 DESCRIPTIVE GEOMETRY Construction, prx is the top and r'x' the front view of the circle of the surface through the given point P. pm is the top and p'm' the front view of the tangent to the circle at the point P. ap is the top view of the meridian section through P. If Fig. li.— Ellipsoid of revolution and tan- gent plane. the meridian curve through P is revolved about the axis of the surface until the plane of the curve is parallel to V, it will have the ellipse r'x'y' for its front view, P moving to X. At x' draw DOUBLE CURVED SURFACES 129 x' a' tangent to the ellipse icV 2/ ' . This cuts the front view of the axis of the surface at a'. Then x' a' is the front view of the revolved position of the tangent to the meridian section at P. In the counter revolution of the meridian plane, X moves to P and the point A, in the axis, remains fixed. Then ap is the top and a'p' the front view of the tangent to the meridian curve at P. PM and PA represent the required plane which is tangent to the surface at the point P. There are any number of planes tangent to a double curved surface of revolution from a point outside the surface. One or more of these planes can be determined by the general method employed in the previous paragraph. 138. Problems. 1. Represent a plane which is tangent to a sphere, (a) at a point of the surface; (b) and which contains a point outside the surface. 2. Represent a plane which is tangent to a prolate ellipsoid, (a) at a point of the surface; (b) and which contains a point outside the surface. 3. Represent a plane which is tangent to a paraboloid of revolution, (a) at a point of the surface; (b) and which contains a point outside the surface. 4. Represent a plane which is tangent to a torus, (a) at a point of the surface: (b) and which contains a point outside the surface. 139. To represent a plane which contains a g^iven straigrht line and is tang-ent to a sphere. Let AB, Fig. 75, be the given straight line and let C be the center of the given sphere. Analysis. Assume that the required plane is drawn through the given line and tangent to the sphere. Now if an auxiliary plane is passed through the center of the sphere and perpen- dicular to the line AB, it will cut a point from AB, a great circle from the sphere and a line from the tangent plane which will pass through the point on AB and be tangent to the great circle cut from the sphere. Therefore, to make the construction for the tangent plane, pass a plane through the center of the sphere and perpendicular to the line AB. Prom the point in which this plane cuts AB, draw a tangent to the great circle cut from the sphere. This tangent and the line AB determine the re- 130 DESCRIPTIVK GEOMETRY quired tangent plane. Construction. SS and FF represent a plane which contains the center of the sphere and is perpendicular to the line AB. P is the point in which the line AB pierces this plane. When the plane of SS and FF, which contains a great circle of the sphere and the point P, is revolved about SS as an axis until it is parallel to H, the point P moves to P^ and the great circle has the same top view as the sphere. 10202 is the top view of the revolved position of a tangent from P to the great circle. By counter revolution the views po and p'o' of the required tangent are found. ThenPO and AB represent the required tangent plane. Fig. 75. — Plane containing AB and tangent to sphere. There is another plane containing AB and tangent to the sphere. This plane is represented by AB and the other tangent from P to the great circle cut from the sphere by the plane of SS and FF. PROBLEMS 131 140; Problems. 1. Represent a plane which contains a line parallel to V and is tan- gent to a given sphere. 2. Represent a plane which contains a line parallel to H and is ta n gent to a given sphere. 3. Represent a plane which contains a line parallel to G. L. and is- tangent to a given sphere. 4. Rspresent a plane which contains a line of profile and is tangent to a given sphere. GENERAL PROBLEMS Solve the following problems in a 10." x 14" rectangle with the base line through the center and parallel to the shorter side, unless state- ment to the contrary is made. 1. P{\, 3, — 1) is the vertex of a cone whose base is a 2" circle in a hori- zontal plane with center at 0( — IJ, |, — 4^). Represent a plane which makes 60° with H, and is tangent to the cone. (Use a 7" x 10" rectangle). 2. A cone whose base is a 2" circle in V, has A( — J, |, — 3) B(1J, 4i — I) for an axis. Piad a plane which is tangent to the cone and makes 60° with V. (Use a 7" x 10" retangle). 3. The base of a cylinder is a 2J" circle in H. L(l^, J, — 4^) K( —J, 3, — 1^) is the axis. Represent a plane which is tangent to the cylinder and makes 60° with H. (Use a 7"xl0" rectangle). 4. A( — -IJ, 1, — I) B(|, 4J, — 3J) is the axis of a cylinder whose base is a 2J" circle in V. Find a plane which is tangent to the cylinder, and makes 60° with V. (Use a 7"x 10" rectangle). 5. Represent a plane which is tangent to a 3" sphere with center at 0( —h 4|, — 2i) and is parallel to the plane A( —3, 4J, — 5J) B(0,'l|, -54) C( -3i, IJ, -2i). 6. Inscribe a sphere in a given tetrahedron. 7. A( —3, 2, —51) B( —i, 3i, — 2|) C(l, 1, — 3|) is a plane which is tangent to a right circular cylinder whose axis is E( -J, 2J, — 5^) G(lf, 4, — 2|). Draw the top and front views of the element of contact. 8. Through a given point, pass a plane tangent to a cylinder of revo- lution, having given the axis and radius of the cylinder, without finding the views of the cylinder. 132 DESCRIPTIVE GEOMETRY 9. Having given the radius and the axis of a cylinder of revolution, pass a plane tangent to the cylinder and parallel to a given straight line, without finding the views of the cylinder. 10. Find a common normal to two cylinders of revolution. 11. To two given cylinders pass tangent planes which are parallel to each other. 12. To a given cone and cylinder, pass tangent planes which are par- allel to each other. 13. A( -3, IJ -5|) B( —h 4i, -2) and C(2^, 2}, -5f) D(3i, 5|, -2) are the axes of two cones whose bases are 2J" and 3" circles, respect- ively, which lie in a horizontal plane. To the given cones pass tangent planes which will be parallel to each other. 14. Through a given point, pass a plane tangent to two given spheres. 15. Pass a plane tangent to two given spheres and parallel to a given straight line. 16. Pass a plane tangent to three given spheres. n. Through a given point, pass a plane which is equidistant from three given spheres. 18. Through a given straight line pass a plane which is equidistant from two givea spheres. 19. Pass a plane which is equidistant from four given spheres. 20. Find a common tangent plane to a sphere and a cylinder of revolu- tion. 21. Find a common tangent plane to a sphere and a cone of revolu- tion. CHAPTER IV PLANE SECTIONS AND DEVELOPMENTS OF CDRVED SURFACES 141. Plane section. The line of intersection of an oblique plane with a curved surface which is ruled e^n be drawn by find- ing the points in which the rectilinear elements of the surface pierce the given oblique plane. The line joining in proper order these piercing points is the required line of intersection of the surface with the plane. These piercing points can be found by the usual method of finding where a straight line pierces a plane, or by the use of an auxiliary view. The auxiliary view should be taken from such a direction that the cutting plane will appear as a straight line. The rectilinear elements should be taken near enough together to give the line of intersection as accurately as desired. The intersection of an oblique plane with a double curved surface is found by using a system of auxiliary planes. Each of the auxiliary planes will cut a straight line from the oblique plane and a curved line from the double curved surface and these lines will intersect in points of the required line of inter- section . The auxiliary planes should be so taken that they will cut simple curves from the "double curved surface and these curves should be in simple positions with reference to the planes of projection. Development. By the development of a curved surface is meant a plane area of such form and size that it can be rolled or folded to again form the original surface. The development is obtained by rolling the surface upon a tangent plane until each part of the surface comes in contact with the plane. The part of the plane thus covered is the development of the surface. 142. To find the intepseetion of aright eireular cylinder with a given oblique plane and to develop the surface of the cylinder. 134 DESCRIPTIVE GEOMETRY Let the cylinder be given as in Fig. 76, and let SS and PF represent the oblique plane. Analysis. Find the points in which the rectilinear elements of the cylinder intersect the given oblique plane. A curved line joining these points in the proper order is the required line of intersection. Construction. A view taken by looking in the direction of the horizontal line SS, shows the cylinder as a rectangle and the oblique plane as the line Si/i. The distance at which the line c Fig. 76 -Plane section and development of right circular cylinder. ttiCi, representing the base of the cylinder, is placed from the top view is arbitrary. The distances d and di are taken from the front view and used to locate the line sj'i which represents the oblique plane in the auxiliary view. The front and auxiliary PLANE SECTIONS AND DEVELOPMENTS OF SURFACES 135 views of elements of the cylinder are drawn from points in the base as A, B, C, etc., as many elements being drawn as are - necessary to determine the curve of intersection to the desired accuracy. The points aJi, 2/1, Si, etc., where the auxiliary views of these elements intersect the line Sj/i are the auxiliary views of the points in which the elements pierce the cutting plane SS, FF. By taking the distances of the points x 1, yi, z^, etc., from the line ttiCi and setting them off on the proper elements from the base a'o', the front views x' , y' , z' , etc., of the points are lo- cated. Joining these points in the proper order gives the ellipse which is the front view of the intersection of the cylin- der with the plane. The top view of the intersection is the cir- cle a, 6, c. To develop the surface oJ the cylinder showing- the curve of intersection with the oblique plane. If the cylinder be rolled on a tangent plane until each element has come into the plane, the base will develop into the straight line a^Ci. The distance a2h2 = 3kva ab, &2«2=are be, etc. In the develop- ment, the elements will be perpendicular to a^Uz, since they are perpendicular to the plane of the base. To get the line of inter- section on the development, lay off aiX2=a'x', biVi = b' y' , etc. A smooth curve passing through the points X2, y^, 22. ^2, is the development of the line of intersection. 143. To draw a straight line tangent to the curve of intersection of a plane with a curved surface at any point on the curve. The curve is evidently a plane curve and therefore a tangent to it at any point must lie in the plane which cuts the curve from the surface. Since the curve lies on the sur- face, a line which is tangent to it at any point must lie in a tan- gent plane to the surface at that point. (Art'. 101). The required line is, therefore, the line of intersection of the cutting plane with the plane tangent to the surface at the point to which the tangent line is to be drawn. This method is general and applies to the plane section of all surfaces. The construction for the problems is not as a rule, 136 DESCRIPTIVE GEOMETRY difficult, since usually the tangent plane to the surface can easily be drawn. In this chapter of the text, the lines tangent to the plane sec- tions of the different surfaces will not be shown in the figures, but in every case they can, if desired, be drawn by the above method. 144. To find the intersection of an oblique cylinder with a g-iven oblique plane and to [develop the surface of the cylinder. Let the cylinder be given as in Fig. 77, and let SS and FP rep- resent the oblique plane. Analysis. Take a view of the cylinder and plane by looking in the direction of the horizontal line SS. This view shows the cutting plane as a line. The points in which this line crosses the auxiliary view of the elements of the cylinder are the aux- iliary views of points on the required curve of intersection. From the auxiliary view, the top and front views of the intersec- tion can be derived. Construction. In the auxiliarj' view, the base of the cylinder appears as the line ai&i at right angles to ss, and the center of the base as the point c^. The auxiliary view CiOj of the axis is determined and then the extreme elements of the cylinder are drawn parallel to CiOi through the points Ui and 6i. The upper end of the cylinder is shown as broken. Si/i is the auxiliary view of the oblique plane. The points where the auxiliary view of the elements cross the line sJi are the auxiliary views of the points where these elements pierce the cutting plane. The top views of these piercing points are found by first finding the top views of the elements and then locating the top views of the points on the top views of the corresponding elements. From the auxiliary and top views, the front views of the piercing points are found. A curved line joining these points in the proper order is the required line of intersection. The method for find- ing points on the curve of intersection would be the same, no PLANE SECTIONS AND DEVELOPMENTS OP SURFACES 137 matter what direction the oblique plane SS FF is taken. It was drawn at right angles to the axis in order to be used in the de- velopment of the cylinder. Fig. 11.— Plane section and development of oblique cylinder. To develop the surface of the cylinder. If the cylinder be rolled on a tangent plane, the section which is at right angles to the axis will become a straight line on the development and the elements of the cylinder will appear in their true lengths and lie at right angles to this line. The true size of this right sec tion is found by a second auxiliary view, taken by looking in the direction of the axis, as shown in Fig. 77, or by rotating the cutting plane about SS as an axis until it becomes parallel to H. This would give the true size of the figure in the top view. From the true size of the right section the distances are 138 DESCRIPTIVE GEOMETRY taken and laid off as follows along a straight line: arc 3/282 = y^Za, arc Z2U2=Zz'>*i^ etc. The true lengths of the elements of the cylinder are shown in the first auxiliary view. These lengths are taken and laid off on the development at right angles to y^Xs at the proper points along that line. In Fig. 77, the line yz^s in the development was taken in the continuation of XiPi in order that the true lengths of the elements could be projected across from the first auxiliary view to the development. In the development, the line y^Xg represents the right section and 636303 the curve of the base. Only half the cylinder is shown in the development. 145. To find the intereseetion of an oblique cone with a given oblique plane and to develop the surface of the cone. Let the cone be given as in Fig. 78, and let SS and FF repre- sent the oblique plane. Analysis. Take a view of the cone and plane by looking in the direction of the horizontal line SS. This view shows the cutting plane as a line. The points in which this line crosses the auxiliary view of the elements of the cone are the auxiliary views of the points on the required curve of intersection. From the auxiliary view, the top and front views of the intersection can be derived. Construction. ■Wiaiei is the auxiliary view of the cone and Si/x that of the cutting plane. It is best to divide each half of the circumference of the base of the cone into the same number of equal parts, beginning at A, the foot of the shortest ele- ment. Then each element in the auxiliary view represents two elements of the cone, one visible and one invisible. By draw- ing the auxiliary views of the elements, the auxiliary views Xi, 2/1, Si, etc., of the points of intersection are found. From the auxiliary views, the top and then the front views are de- rived. A curve joining in proper order the points thus found, is the required line of intersection. PLANE SECTIONS AND DEVELOPMENTS OF SDKPACES 139 To develop the surface of the cone. It is first necessary to find the true lengths of the elements VA, VB, VC, etc. In Fig. 78, this is done by rotating the elements until they are par- allel to the picture plane upon which the auxiliary view is taken. e a- FC Pig. 78. — Plane section and development of oblique cone. The true lengths are then found to be i'lai, Vibg, ViC^, etc. Through Vi draw a line in any direction and lay off on it viUi equal to «>i aj , the true length of the shortest element of the cone. With Vi as center and Vib^ as radius strike an indefinite arc. With a^ as center and radius equal to the rectified arc ah, strike an arc cutting the first arc at 64. Then Vib^ is the ele- ment VB laid off on the development. By using Vi as center and ViCb as radius and 64 as center and radius equal to the rectified arc be, the point C4 is located. By continuing the pro- cess the development as shown in Pig. 78 is found. The de- velopment for only half the cone is shown, the other half is similar. If the given cone had been right circular, the curve a^. . .€4 would have been the arc of a circle with center at Vi. 140 DESCRIPTIVE GEOMETRY To lay the curve of intersection of the cone with the plane SS FF off on the devolpment, the true lengths of the elements from the vertex to the cutting plane are found and laid off on the corresponding elements of the development. This is easily accomplished by drawing lines through the points Xi, t/i, Zi, etc., parallel to ajei, until they cut the true lengths of the cor- responding elements. The construction is shown for the point Z, V1Z3 being the true length from the vertex to the cutting «plane. The distances thus found laid off on the development gives the curve X4 . . . W^. 146. To find the intersection of a warped surface with a given oblique plane. Analysis. Since a warped surface is ruled, it is only neces- sary to find where the rectilinear elements of the surface pierce the plane. A line joining in proper order these points is the re- quired line of intersection. The most convenient method of finding these piercing points is by means of an auxiliary view which shows the cutting plane as a line. The construction is left as an exercise for the student. A warped surface cannot be developed. For an approximate development of such a surface, see Art. 148. 147. To find the intersection of any surface of revolu- tion with a given oblique plane. Let the surface be given as in Fig. 79 and let AB and AC rep- resent the oblique plane. Analysis. Cut the surface and plane by a system of auxiliary planes perpendicular to the axis of revolution of the surface. The auxiliary planes will cut circles from the surface which will intersect the lines cut from the given plane in points of the required curve of intersection. Oonstruction . Since the method is the same for all surfaces of revolution, let the construction be made for the torus as given in Fig. 79. The axis of the surface is perpendicular to H. The auxiliary cutting planes will therefore be parallel to H, and will be represented in the front view by lines parallel to G. L. h'h' PLANE SECTIONS AND DEVELOPMENTS OP SURPACES 141 represents one of these planes. It cuts the line SS from the plane ABC and two circles from the torus, one with a radius gV and the other with a radius q' u' ■ The top view shows these circles cutting ss at x, y, z, and w, the top views of four points on the required curve of intersection. The front views of these points are on the line h'h' . By taking other planes parallel to H, enough points can be found to determine the curve of the intersection to the desired accuracy. Fig. 79. — Plane section of a torus. When the surface is symmetrical, as is the case with the torus, it is convenient to begin the construction with the plane which contains the center of the front view of the surface. This plane contains the largest and smallest circles of the torus as shown in the top view. If the remaining planes be passed in pairs at 142 DESCRIPTIVE GEOMETRY equal distances above and below the center plane, each pair will cut circles having the same top view. This will avoid the neces- sity of drawing so many circles in the top view. The highest and lowest planes which touch the surface will contain but one circle each. A plane which contains the axis of the surface and is parallel to V, cuts a line from the plane ABO which intersects the small circles of the front view of the torus at the points where the curve of intersection touches these circles. A plane T which contains the axis of the surface and is perpendicular to a hori- zontal line SS of the plane ABC will locate the points E and G on the curve of intersection. The points E and G are determined by rotating the section made by the plane T until it is parallel to V, locating the points ei' and g/, and then rotating back to the original position. For most surfaces of revolution, this section will locate the highest and lowest points on the curve of intersection of the oblique plane with the surface. Instead of rotating the plane T until it is parallel to V. the points E and G could have been located by taking an auxiliary view of the surface and plane ABC by looking in the direction of the line SS. This, however, would have involved more construction than the method by rotation. 148. Approximate development of warped surfaces. Since two consecutive straight line elements of a warped surface do not lie in one plane, the surface cannot be developed. In practice, however, approximate developments of these surfaces are made in the following manner. Let the surface be a transitional piece joining a circle at one level with an ellipse at a diJBferent level, Fig. 80. The openings joined by the transitional piece need not be in parallel planes and they need not be of the form given in Fig. 80. If the sur- face is such that the straight line elements Al, B2, C3, etc., are neither parallel nor intersecting, it is warped and may be ap- proximately developed by the following method. PLANE SECTIONS AND DEVELOPMENTS OF SURFACES 143 Divide one quarter of the circle into any number of equal parts as six, Fig. 80. Divide the corresponding quarter of the ellipse into the same number of equal parts and join points on the ellipse with the points on the circle by the lines Al, B2, C3, etc. I ' t t \ \\ I I II 1 V \ Fig. 80. — Method for approximate development of warped surface. Draw the dashed diagonal lines Bl, C2, E3, etc. Find the true lengths of both of these sets of lines. A convenient method for getting these lengths is shown in the sketches to the right of the front view. The lengths of the top views of the lines are laid off on the base line from the foot of the perpendiculars from Oj and Pi . The development is obtained by drawing the line a-ili, in any direction and laying it off equal to the true length of Al. With i 2 as center and the true length of Bl as radius strike an arc . With Ha as center andafi as radius strike an arc cutting the first arc at 62. Then with 62 as center and B2 as radius and ^2 as center and 12 as radius strike ares intersecting at the point ^2- By continuing this process the development of the entire piece is found. Only one-quarter of the development is shown, but the remainder is found in a similar manner. 149. Approximate development of a sphere. Theoreti- cally, double curved surfaces can not be developed, but approxi- 144 DESCRIPTIVE GEOMETRY mate developments are made and used in practice . There are two methods used for developing a sphere. Method 1. The gore method. Let the sphere be given as shown in Fig- 81. Only half the surface is shown and only one- quarter of the sphere is developed, but this will sufilce to ex- plain the method. Divide the surface into, gores by the meridian planes A7, B7, 07, etc., sixteen is a good number a h. c, e, 1 1 1 1 1 1 _l 1 1 _1_ \ % % /\ % / 7 Fig. 81. — Qore method of developing a sphere. although the more parts taken the more accurate the develop- ment should be. Then divide the surface by a series of hori- zontal circles 1, 2, 3, etc. In the development the great circle 1 becomes the straight line ai, 6i .. .ej .. ., the distance «]&! = arc db, 6iei = arc 6c, etc. The line 17 in the development is at right angles to (fifii, and is equal in length to one-quai'ter of a great circle, the distance i2 = arc i'5', 55=arc 5'5', etc. Cen- ter lines for each gore are drawn from the proper points along the line a^ei and these center lines divided into parts similar to the first one by lines through points 2, 3, 4, etc., parallel to to a^ei. Step off from the center line of each gore along the lines 3, 3, 4, etc., distances obtained from the corresponding cir- cles in the top view, measured from the center line to the sides of the gore. Joining the points thus located gives the develop- ment shown in Fig. 81. The development of the complete sphere PLANE SECTIONS AND DEVELOPMENTS OF SURFACES 145 would consist of twice as many gores as shown, the half of each gore above the line ajei, being similar to the half shown below this line. Method II. The zone method. Let the sphere be given as in Fig. 82. Half the sphere is shown in the top and front views and the same half in the development. Divide the surface of the sphere into zones by the horizontal planes 1, 2, 3, etc. The greater the number of zones the greater should be the accuracy of the development. Each zojie is then developed as if it were the frustrum of a right cone. The radius Fig. 82. — Zone method for developing a sphere. of circle 1 in the development is the length of the line l'3' from the point l' to where l'3' cuts the the vertical axis of the sphere. Two circles pass through the point 2 in the develop- ment, one with a radius from 2' to the point where the line l'2' cuts the axis of the sphere, and the other with a radius from the point 2' to where 3'3' cuts this axis. By continuing in this man- ner, all the circles shown in the development are drawn. It is best to take a center line ab and have the centers of all these cir- cles on this line. Meridian planes are then passed through the axis of the sphere dividing the surface into any convenient num- 146 DESCRIPTIVK GEOMETRY ber of parts as shown in the top view. With dividers, step off the distances on circle 1 between these planes in the top view upon circle 1 of the development, beginning with the center line