Cornell Mniversity Library BOUGHT WITH THE INCOME FROM THE SAGE ENDOWMENT FUND THE GIFT OF Henry W. Saaqe 1891 A-274-793. | oe 7 fora. 9724 ell University Libr: The theory of engineering drawing ili iil T 35 iii iii THE THEORY ENGINEERING DRAWING BY ALPHONSE A. ADLER, B.S., M.E. Member American Society of Mechanical Engineers; Instructor in Mechanical Drawing and Designing, Polytechnic Institute, Brooklyn, N. Y. NEW YORK D. VAN NOSTRAND COMPANY 25 PARK PLACE 1912 EN. A171 43 Copyright, 1912 BY D. VAN NOSTRAND COMPANY THE SCIENTIFIC PRESS ROBERT DRUMMOND ANO COMPANY BROOKLYN, N. Y. PREFACE ALTHOUGH the subject matter of this volume is, in large measure, identical with that of many treatises on descriptive geometry, the author has called it ‘ Theory of Engineering Drawing,” believing that this title indicates better than could ny other, the ultimate purpose of the book. That texts on descriptive geometry appear with some degree of frequency, with but few, if any, additions to the theory, indicates that teachers are aware of certain weaknesses in existing methods of presenting the subject. It is precisely these weakness that the present work aims to correct. The author emphasizes the fact that the student is concerned with the representation on a plane of objects in space of three dimensions. The analysis, important as it is, has for its primary purpose the development of methods for such representation and the interpretation of the resulting drawings. It is nowhere regarded as an end in itself. The number of fundamental prin- ciples has been reduced to a minimum; indeed it will be found that the entire text is based on the problem of finding the piercing point of a given line on a given surface, and a few additional operations. The accepted method of presenting the subject, is to start with a set of definitions,to consider in detail the ortho- graphic projection of a point, and then, on the foundation thus laid, to build the theory of the projection of lines, surfaces, and solids. Logical and beautiful as this systematic developmeen may be, it nevertheless presents certain inherent difficulties, chief of which is that the student is confronted at the outset with that most abstract of all abstractions, the mathematical point. In this volume the order of presentation is reversed and the reader is asked to consider first some concrete object, a box, for instance, the study of which furnishes material of use in the later discussion of its bounding surfaces and lines. The “ Theory of Engineering Drawing ”’ is divided into four ili iv PREFACE parts. Part I treats of oblique projection, orthographic projec- tion, and a special case of the latter, axonometric projection. The student is advised to give special attention to the classification at the end of this section, because it gives a complete outline of the entire subject. Part II contains a variety of problems of such nature as to be easily understood by those whose training has not extended to the more highly specialized branches of com- mercial or engineering practice. Part III considers convergent projective line drawing, more familiar under the name of perspec- tive. Part IV has to do with the pictorial effects of illumination, since a knowledge of shades and shadows is frequently required in the preparation of complicated drawings. No claim is made to originality of subject matter, but it is not possible to acknowledge indebtedness to individual writers, for the topics discussed have been widely studied, and an historical review is here out of place. The author wishes, however, to express his sense of obligation to Professor William J. Berry of the Depart- ment of Mathematics in the Polytechnic Institute of Brooklyn for his criticism of Chapters IX and X, and other assistance, and to Mr. Ernest J. Streubel, M.A., of the Department of English for his untiring efforts in preparing the manuscript for the press. Po.tyrecanic InstiruTe or BROOKLYN, October, 1912. ART, 101. 102. 103. 104. - 201. 202. 203. 204. 205. 206. 207. 208. 209. 210. 211. 212, 301. 302. CONTENTS PART I THE PRINCIPLES OF PARALLEL PROJECTING-LINE DRAWING CHAPTER I INTRODUCTORY PAGE Nature of drawittg':.¢ o sas we-0 iss we yeas See vie oe yea de ten cee ees 3 Science and art of drawing........... 0. cece eee eens 4 Magnitude of objects .... 0.0... ccc eee nee eens 4 Commercial application of drawing. .... 0.0... sce e eee eee e eee 4 CHAPTER II OBLIQUE PROJECTION Nature of oblique projection... .......... 0c ccc cc cee eee 6 Oblique projection of lines parallel to the plane of projection. ..... 7 Oblique projection considered as a shadow...................05 8 Oblique projection of lines perpendicular to the plane of projection. 9 Oblique projection of the combination of parallel and perpendicular lines to the plane of projection................ 0.0 c eee eee eee 10 Oblique projection of circles................ 00 eee eng Wied 11 Oblique projection of inclined lines and angles................... 12 Representation of visible and invisible lines..................... 13 Drawings to SCales ‘cic: aa saneta cana eew ested sen amie aap eg ae PAA 14 Examples of oblique projection. ............ 00.002 cece eee eens 14 Distortion of oblique projection... 0.0.20... 0... cece eee eee 20 Commercial application of oblique projection...............50 ee 21 CHAPTER IIT ORTHOGRAPHIC PROJECTION Nature of orthographic projection.............. A lskoie Sadat aloha Ye 25 Theory of orthographic projection... .. 0.0.0.0... .0 cece eee e eee 26 Revolution of the horizontal plane. ........... 0... cece e eee eee 27 303. vi ART. 304. 305. 306. 307. 308. 309. 310. 311. 312. 313. 314. 315. 316. 317. 318. 401. 402. 403. 404. 405. 406. 407. 408. 409. 410. 411. 412. 413. 501. 502. 503. 504. 505. CONTENTS POSItON OF THE CY Es sc jsud eae ees $9 ao Cae waree ad aula eee eine OBE RE Relation of size of object to size of projection .......... ctdoh eee Location of object with respect to the planes of projection........ Location of projections with respect to each other.............-- Dimensions on a projection... 2.2... 2.0.0 eee eee Comparison between oblique and orthographic projection......... Orthographic projection considered as a shadow................- Profile planes vgs seen bees Mnakegdeue rama aee ras SAeRaoS Locationvof ‘profilessn « o4se444 3 Mars eaaige nye shied ces Oi) bec Ale wae Se Section plane ix cow samba ys Coan b24 HOW EEE ta eda Be Oe welene TAKE Supplementary plane... cciseavecadev ves ees eras ereca dares des Anglesiof projection... .... a cedsassanseed cesar ena de pte a Oe en oes Location of observer in constructing projections................. Application of angles of projections to drawing.............-...- Commercial application of orthographic projection.............-. CHAPTER IV AXONOMETRIC PROJECTION Nature of isometric projections... 2... 0.0.00... cece eee eee eee Theory of isometric projection........... 0.0.00. cee cee eee eee Isometric projection and isometric drawing.................2..5. DiPeGhlOn Gh AXES 2.4.n10c.cckenss deta eterna am aie Sed das Dane ee BRS Isometric projection of circles... . 2.2.0.2... 0000 c cece eee eee Isometric projection of inclined lines and angles................. Isometric graduation of a circle... 2.0.0... ee eee eee Examples of isometric drawing............... 000s Dimetric projection and dimetric drawing....................-. Trimetric projection and trimetric drawing..................... Axonometric projection and axonometric drawing................ Commercial application of axonometric projection............... REPRESENTATION OF LINES AND POINTS Introd uctORy to wa0G.; teria eae SV ba dia 6a wna dena ewes Representation of the line... 0.0... cece eee eee Line fixed in space by its projections... 20.0... 0000... cee cece es Orthographic representation of a line............. 0000... cece eee Transfer of diagrams from orthographic to oblique projection,..... PAGE 27 28 28 29 29 29 30 30 31 33 34 36 36 37 38 Classification of projections. . 0... ccc cece eee eee e eee enn PART II GEOMETRICAL PROBLEMS IN ORTHOGRAPHIC PROJECTION CHAPTER V CONTENTS vil ART. PAGE 506. Piercing points of lines on the principal planes................-- 66 507. Nomenclature of projections... 0.00... eee ccc cece teen eens 68 508. Representation of points... 0.0.0... cece cee ccc teens 68 509. Points lying in the principal planes............... 00.0 eee e eee 69 510. Mechanical representation of the principal planes................ 69 511. Lines parallel to the planes of projection..................0000. 70 512. Lines lying in the planes of projection.....................0000 71 513. Lines perpendicular to the planes of projection .................. 72 514. Lines in all angles... 1... eect eee nnaee 72 515. Lines with coincident projections... ................. 0c cence 74 516. Points in all angles... 2... ccc cece eect eens 75 517. Points with coincident projections... .............00 cece eee eee 75 518, Lines in profile planes. ..,,.csccscecc cece cette eee eee e teenies 75 CHAPTER VI REPRESENTATION OF PLANES 601. Traces of planes parallel to the principal planes................. 80 602. Traces of planes parallel to the ground line..................... 80 603. Traces of planes perpendicular to one of the principal planes. ..... 2 604. Traces of planes perpendicular to both principal planes............ 83 605. Traces of planes inclined to both principal planes................ 83 606. Traces of planes intersecting the ground line.................... 84 607. Plane fixed in space by its traces... 2.0... eee eee eee eee 84 608. Transfer of diagrams from orthographic to oblique projection.... 84 609. Traces of planes in all angles.......00.. 0... eee 86 610, Projecting plane of lines............. ccc ccc este eee ete eee ne 86 CHAPTER VII ELEMENTARY CONSIDERATIONS OF LINES AND PLANES 701. Projection of lines parallel in space...............0..02.22.20-00055 89 702. Projection of lines intersecting in space................0 00000 eee 89 703. Projection of lines not intersecting in space..................0.. 90 704. Projection of lines in oblique planes.................0.00 eee eee 91 705. Projection of lines parallel to the principal planes and lying in an Oblique lane! esa sci 5 snes ies cies: dee w iatins How a aa Sok aE Ra nO aera 92 706. Projection of lines perpendicular to given planes................. 95 707. Revolution of a point about aline............ cece eee ees 96 CHAPTER VIII PROBLEMS INVOLVING THE POINT, THE LINE, AND THE PLANE SOi. AntroductOryns saeiage otedede tha eak Sapnnidsks oedaues gaa omaletie 99 802. Solution of problems. ..........00 006 e cece eee eet 99 803. Problem 1. To draw a line through a given point parallel to a given vill CONTENTS ART. PAGE 804. Problem 2. To draw a line intersecting a given line at a given point.. 100 805. Problem 3. To find where a given line pierces the principal PIT OD Sea casos a toc hued hh eed eee ab apoedc we pb NG RL yee ee Saale ey dete ants 101 806. Problem 4. To pass an oblique plane through a given oblique line.. 102 807. Special cases of the preceding problem ......................44- 102 808. Problem 5. To pass an oblique plane through a given point....... 103 809. Problem 6. To find the intersection of two planes, oblique to cach other and to the principal planes...............-..0--.2 00005 104 810. Special case of the preceding problem...................00002005 104 811. Problem 7. To find the corresponding projection of a given point lying in a given oblique plane, when one of its projections is PVT 628 encsod. ein Sikri Gear d ena Beeb deOe Rate ese aaa BREET ee dace ee 104 812. Special case of the preceding problem.................0-. 000005 105 813. Problem 8. To draw a plane which contains a given point and is parallel to a given plane............. Fa Ttecd Gyr nic atin AALS 106 814. Problem 9. To draw a line perpendicular to a given plane through GiPIVED POM be 2 c.5 x4 cae scp aed meas eee eS ae ate oeR AGERE 107 815. Special case of the preceding problem......................0005 108 816. Problem 10. To draw a plane through a given point perpendicular to Se LV OI MSs» guid a, Syadesley a we ah Etheies chase past NH ne Rese FAMeede AS S 108 817. Problem 11. To pass a plane through three given points not in the saine straight ling,c..03cs neste seeusenoewe sun iew da ven See ee 109 818. Problem 12. To revolve a given point, not in the principal planes, about a line lying in one of the principal planes. ............... 110 819. Problem 13. To find the true distance between two points in space as given by their projections. First method. Casel. ...... lil SOW CaCO ia seit a ea to Bu cereale oh aor oe batten a Ate De ead a Bee aes ae a aut hen 112 821. Problem 13. To find the true distance between two points in space as given by their projections. Second method. Casel. .. 113 2 MOA cm wa zen Na Senate IN Ne ete A hla Sency Suialclty dake cells 2a cenauty utes Beageae 113 $23. Problem 14. To find where a given line pierces a given plane...... 114 824. Problem 15. To find the distance of a given point from a given DIAC tc a c epeets tad a arta acne e oioann Seemann enNlevng ray eR oka cambern en 115 825. . Problem 17. To find the angle between two given intersecting 827. 828. 829. 830. 831. Problem 16. To find the distance from a given point to a given line.... 115 NGS): euslenant oases Ai heaves HOMAGE eS ReRCRM eb e, eock iste ede 2 116 Problem 18. To find the angle between two given planes ..... . Le Problem 19. To find the angle between a given plane and one sat the principal planes 25 acess ag seca tcc sonn Ava saene awe ees 118 Problem 20. To draw a plane parallel to a given plane at a given Gistance: IrOm IG: 225 cE gcd amcanueeaewk Wee warms weoeh Sued on 119 Problem 21. To project a given line on a given plane............ 120 Problem 22. To find the angle between a given line and a given DIANE ss ao icenise eee Gt FOR Maou che yan i Sad Oe So ee ws 121 . Problem 23. To find the shortest distance between a pair of skew TIMES 1d ete elt die aia gi sw hg Mad ane ede ew eA each Sen mills an da 122 . Application to other problems. .........0. 000... 0 00 cece eee ee 125 CONTENTS ix ART, ‘ PAGE 834. Problem 24. Through a given point, draw a line of a given length, making given angles with the planesof projection... .......... 125 835. Problem 25. Through a given point, draw a plane, making given angles with the principal planes........0.. 6.000. c cece eee 127 836. Problem 26. Through a given line, in a given plane, draw another line, intersecting it at a given angle. ..............00 eevee ee 129 837. Problem 27. Through a given line, in a given plane, pass another plane making a given angle with the given plane.............. 130 838. Problem 28. To construct the projections of a circle lying in a given oblique plane, of a given diameter, its centre in the plane being TO WA: ta ar icdan ce Menane'ns Ea ar ade te MALO disk sheaves Baume: 131 CHAPTER IX CLALEL. FICATION OF LINES 901, Introdtietoty «ces eases sa cae Gait esa tiie ais el oemcd mune Rees Shunde 6 144 QO Sta SHC MNS cctesss a cosvers ikss, sya peed diol Se qeandua idea aieddotadskaed dace seabed bes 144 908. Singly curved line.... 0.0... eee ett ee ennnees 144 904. Representation of straight and singly curved lines............... 144 90D; CIRCLE oan Wciataemasiae Gah manene sob mcnsnen wns AqAlk A tied Pawns aremlounmerse 145 DOG. HUNGER. cc cums casacanscod sivaariaaagexamwkehaddwmnnn sawhitn 146 G07 Baral Ol aici asain eo naa Sere manrnineaisantirayreeieh ane amino aetigeee 147 908 Ely perb Ola i ccecc mae Hy yogi eegamaia nian 4 gorerdginge mats beeeraA See 147 9093 Cy cloidh s' G a E F aft 3/4 i Fig. 49. the line. For the line AG, the projecting perpendicular from the point A is AE, and E is then the horizontal projection of the point A in space.* As for the point G on the line, that already lies in the horizontal plane and is its own horizontal projection. Thus, two horizontal projections of two points on the line are established, and, hence, the horizontal projection of the line is determined by joining these two projections. This is true, because all the perpendiculars from the various points on the line lie in a plane, which is virtually the horizontal projecting plane of the line. It cuts the horizontal plane of projection in a line EG, which is the horizontal projection of the line AG in space. *It is to be noted that the projection of a point is found at the place where the projecting line pierces the plane of projection. 64 GEOMETRICAL PROBLEMS IN PROJECTION Putting this in another form, the projection EG gives the same mental impression to an abserver viewing the horizontal plane from above as does the line AG itself; in fact, EG is a drawing of the line AG in space. Directing attention for a moment to the vertical plane, it is found that the construction of the vertical projection consists in dropping a series of perpendiculars from the line AG to that plane. For point G, the perpendicular to the vertical plane is GH, and H is thus the vertical projection of G. The point A lies in the vertical plane and, hence, is its own vertical projection. A line joining A and H is the vertical projection of AG. Again, a plane passed through AG, perpendicular to the vertical plane, Fic. 49, will cut from it the line AH, which is the vertical projection as has been determined. Thus, AH is a drawing of AG because it conveys the same mental impression to an observer who views it in the way the projection was made. 503. Line fixed in space by its projections. The location of the principal planes is entirely arbitrary, as is also the line in question; but, when both are once assumed, the line is fixed in space by its projections on the principal planes. If a plane be passed through the horizontal projection EG (Fig. 49), perpen- dicular to the horizontal plane, it will. contain the line AG in space, since the method is just the reverse of that employed in finding the projection. Similarly, a plane through the vertical projection, perpendicular to the vertical plane, will also contain REPRESENTATION OF LINES AND POINTS 65 the linc AG. It naturally follows that the line AG is the inter- section of the horizontal and vertical projecting planes and, therefore, the line is absolutely fixed, with reference to the prin- cipal planes, by its projections on those planes. 504. Orthographic representation of aline. The line that has been considered so far is again represented in the left-hand view of Fig. 50, in oblique projection. The edges of the cube have been omitted here in order to concentrate attention to the line in question. AG is that line, as before, EG is its horizontal projection and AH is its vertical projection.* Suppose that the plane HH is revolved in the direction of the arrows, 90° from its present position, until it coincides with the Fic. 50. vertical plane VV. The view on the right of Fig. 50 shows the resultant state of affairs in orthographic projection. AH is the vertical projection and EG is the horizontal projection. AE in one view is the equivalent of AE in the other; HG and EH in one view are the equivalents of HG and EH in the other. Both views represent the same line AG in space. At first sight, it may appear that the oblique projection is sufficiently clear, and such is the case; but, in the solution of problems, the orthographic projection as shown on the right possesses many advantages. In due time, this mode of representation will be considered, alone, without the oblique projection. * The projections of a line may be considered as shadows on their respective planes. The light comes in parallel rays, perpendicularly directed to the planes of projection. See also Arts. 203 and 310 in this connection. 66 GEOMETRICAL PROBLEMS IN PROJECTION 505. Transfer of diagrams from orthographic to oblique projection. It is desirable to know how to transfer diagrams from one kind of projection to the other. If the orthographic projection appears confusing, the transfer to oblique projection may be of service. On the right of Fig. 50, is given the ortho- graphic projection of the line AG. To construct the oblique projection from this, draw the principal planes as shown in the left-hand diagram. From any point E on the ground line in the oblique projection, lay off EA, vertically, equal to EA in the orthographic projection. On the sloping line (ground line), lay off EH equal to EH in the orthographic projection. Then draw HG in the horizontal plane, equal to HG in the orthographic projection. A line joining A and G in the oblique projection Fia. 50. gives the actual line in space; hence, in the oblique projection. the actual line and both of its projections are shown. In the orthographic projection, only the projections are given; the line itself must be imagined. Compare the method of constructing the oblique projection with Art. 207 and note the similarity. 506. Piercing points of lines on the principal planes. If AG (Fig. 50) be considered as a limited portion of line indefi- nitely extended in both directions, then A and G are the piercing points on the vertical and horizontal planes respectively. On observation, it will be found that the vertical piercing point lies on the vertical projection of the line, and also on a perpendicular to the ground line XY from the point where the horizontal pro- jection intersects the ground line; hence, it is at their intersection. REPRESENTATION OF LINES AND POINTS 67 Likewise, the horizontal piercing point lies on the horizontal projection of the line and also on the perpendicular erected at the intersection of the vertical projection with the ground line; hence, again, it is at the intersection of these two lines. In Fig. 51 let ab be the horizontal projection of a line AB in space, and a’b’ be the corresponding vertical projection. The line, if extended, would pierce the horizontal plane at c and the vertical plane at d’. This line is shown as an oblique projection in Fig. 52. All the lines that are required in the mental process are shown in this latter view. The actual construction in ortho- graphic projection is given in Fig. 51. To locate the vertical piercing point prolong both projections, and at the point where the horizontal projection intersects the ground line, erect a per- \ a}--——-2 a ° \ 1 $+} 4 x = 7 Y 1 | % 1 | 1 || Vv ‘ c Fig. 51. Fia. 52. pendicular until it intersects the prolongation of the vertical projection. This locates the vertical piercing point. To find the horizontal piercing point, prolong the vertical projection until it intersects the ground line, and at this point, erect a per- pendicular to the ground line. Then, the point at which this perpendicular intersects the prolongation of the horizontal pro- pection is the horizontal piercing point. A convenient way of looking at cases of this kind is to assume that XY is the edge of the horizontal plane while viewing the verti- cal projection, therefore, AB must pierce the horizontal plane somewhere in a line perpendicular to the vertical plane at the point ce’. In viewing the horizontal plane, XY now represents the vertical plane on edge. Here, again, the line AB must pierce 68 GEOMETRICAL PROBLEMS IN PROJECTION the vertical plane somewhere in a line perpendicular to the hori- zontal plane at the point d. 507. Nomenclature of projections. In what follows, the actual object will be designated by the capital letters, as the line AB for instance. The horizontal projections will be indicated by the small letters as the line ab, and the vertical projections by the small prime letters as the line a’b’. It should always be remembered that in orthographic projection, the projections alone are given, the actual object is to be imagined. 508. Representation of points. A single point in space is located with respect to the principal planes as shown in Fig. 53. A is the actual point while a is its horizontal, and a’ its vertical projection. The distance of A above the horizontal plane is equal to the length of its projecting perpendicular Aa and this is equal to a’o because Aa and a’‘o are both perpendicular to the hori- Fig. 53. zontal plane HH. Also, Aa’ and ao are perpendicular to the verti- cal plane, therefore, the figure Aaoa’ is a rectangle, whose opposite sides are necessarily equal. Hence, also, the distance of A from the vertical plane is equal to the length of its projecting perpen- dicular Aa’ which also equals ao. Performing the usual revolu- tion of the horizontal plane, a will reach a” on a line a’a” which is perpendicular to the ground line XY. It has been shown that a’o is perpendicular to XY and it only remains to prove that oa” is a continuation of a’o. This must be so, because a revolves about the ground line as an axis, in a plane determined by the two intersecting lines a’o and ao. This plane cuts from the ver- REPRESENTATION OF LINES AND POINTS 69 tical plane the line a’a’’ which is perpendicular to XY because a portion (a’o) of it is perpendicular. As the point a revolves about XY as an axis, it describes a circle, whose radius is oa, and hence oa” must equal oa. In projection,* this is shown on the right-hand diagram of Fig. 53. Both figures are lettered to correspond as far as con- sistent. The actual point A is omitted, however, in the right- hand diagram because the very object of this scheme of repre- sentation is to locate the point from two arbitrary planes (prin- cipal planes), solely by their projections on those planes. This latter is an exceedingly important fact and should always be borne in mind. 509. Points lying in the principal planes. If a point lies in one of the principal planes, it is its own projection in that plane and its corresponding projection lies in the ground line. Fig. 54 shows such cases in projection. A is a point a’ -----2 r a b ce lying in the vertical plane, at a distance = 1 a’a above the horizontal plane; its ver- is tical projection is a’ and its correspond- Fic. 54. ing horizontal projection is a. B is another point, lying in the horizontal plane, at a distance bb’ from the vertical plane; b is its horizontal and b’ the correspond- ing vertical projection. If a point lies in both planes, the point coincides with both of its projections and must therefore be in the ground line. C is such a point, and its two projections are indicated by cc’, both letters being affixed to the one point. The use of these cases will appear-as the subject develops. 510. Mechanical representation of the principal planes. For the time being, the reader may find it desirable to construct two planes ¢ so that lines and points may be actually represented * Hereafter, orthographic projection will usually be designated simply as “in projection.” t For classroom work, a more serviceable device can be made of hinged screens, constructed of a fine mesh wire. Wires can be easily inserted to represent lines and the projections drawn with chalk. The revolution of the planes can be accomplished by properly hinging the planes so that they can be made to lie approximately flat. 70 GEOMETRICAL PROBLEMS IN PROJECTION with reference to them. If two cards be slit as shown in Fig. 58, they can be put together so as to represent two planes at right angles to each other. Lines may be represented by use of match-sticks and points by pin-heads, the pin being so inserted as to represent its projecting perpendicular to one plane. The idea is recommended until the student be- comes familiar with the involved opera- tions. As soon as this familiarity is obtained, the cards should be dispensed with, and the operations reasoned out fin space, as far as possible, without the use of any diagrams. Fia. 55. 511. Lines parallel to the planes of projection. Assume a line parallel to the horizontal plane. Evidently this line cannot pierce the horizontal plane on account of the parallelism. It will pierce the vertical plane, however, if it is inclined to that plane. Fia. 56. In Fig. 56 this line is shown both in oblique and orthographic projection. The piercing point on the vertical plane is found by erecting a perpendicular at the point where the horizontal projection intersects the ground line. It will also lie on the ver- tical projection of the line. As the line is parallel to the horizontal plane, its vertical projection is parallel to the ground line. The piercing point will therefore lie at the intersection of the two lines; that is, of the perpendicular from the ground line and the vertical projection of the line. In both views of the figure, AB is the given line and b’ is its piercing point. If, on the other hand, the line is parallel to the vertical plane REPRESENTATION OF LINES AND POINTS 71 (Fig. 57) and inclined to the horizontal plane, it will pierce the horizontal plane at some one point. Its horizontal projection is now parallel to the ground line XY. The horizontal piercing Fia. 57. ‘point is at b, as shown, and is found in much the same way as that in the illustration immediately preceding. A case where the line is parallel to both planes is shown in x Fia. 58. Fig. 58. This line cannot pierce either plane and, therefore, both its projections must be parallel to the ground line as depicted. 512. Lines lying in the planes of projection. If a line lies in the plane of projection, it is its own projection in that plane, and its corresponding projection lies in the ground line. When the line lies xe in both planes of projection it must therefore coincide with the ground line. Fig. 59 shows in projection the three cases possible. The first is a line lying in the horizontal plane, ab is its horizontal projection and oo o. ee ee! ff oe ° Qu Fia. 59. 72 GEOMETRICAL PROBLEMS IN PROJECTION ab’ is its corresponding vertical projection. The second is a d' line lying in the vertical plane, c’d’ is its vertical projection and cd is its | ; corresponding horizontal projection. xe Sy The third case shows a line in both ; planes and its horizontal and vertical ie projections coincide with the ground Fic. 59. line and also the line itself; the coincident projections are indicated as shown at ee’ and ff’, read ef and e’f’. 513. Lines perpendicular to the planes of projection. If a line is perpendicular to the horizontal plane, its projection on that plane is a point, because both projecting perpendiculars from the extremities must coincide with the perpendicular line. The vertical projection, however, shows the line in its true length, perpendicular to the ground line, as such a line is parallel to the Vv 3 tae B | “e oof te = a 1 | | o 2 ‘ + s—Y a3 734 ab d He Fia. 60. vertical plane. Fig. 60 shows this in projection as AB, and, in addition a profile plane is added which indicates the fact more clearly. CD isa line perpendicular to the vertical plane with one extremity of the line in that plane. In the profile plane, the lines AB and CD appear to intersect, but this is not necessarily the case. The construction of the projections of the line CD is identical with that immediately preceding. 514. Lines in all angles. So far, the discussion of lines and points in space has been limited entirely to the first angle. If a line is indefinite in extent, it may pass through several of the four angles. A case in each angle will be taken and the salient features of its projections will be pointed out. Fig. 61 shows a line passing through the first angle. It con- tinues beyond into the second and fourth angles. In projection, REPRESENTATION OF LINES AND POINTS 73 the condition is shown on the right. The projections ab and a’b’ show those of the limited position that traverses the first angle. Fig. 61. The dotted extensions show the projection of the indefinite line, and are continued, at pleasure, to any extent. A line in the second angle is shown in Fig. 62. The horizontal projection is ab and the ver- tical projection is a’b’. If the horizontal plane be re- volved into coincidence with the vertical plane, the view in projection will show that, in this particular case, the projections of the lines cross Fic. 62. each other. Only the limited portion in the second angle is shown, although the line may be indefinitely extended in both directions. A third angle line is illustrated in Fig. 63. It may be observed, Fig. 63. in comparison with a line in the first angle, that the horizontal and vertical projections are interchanged for the limited portion 74 GEOMETRICAL PROBLEMS IN PROJECTION of the line shown. In other words the horizontal projection of the line is above the ground line and the vertical projection is below it. The case may be contrasted with Fig. 61. The last case is shown in the fourth angle and Fig. 64 depicts / -------4a t 2 — x H 3 oef—-—Vot & Fic. 64. this condition. Both projections now cross each other below the ground line. Contrast this with Fig. 62. The similarity of the foregoing and Arts. 315, 316 and 317 may be noted. 515. Lines with coincident projections. Lines, both of whose projections lie in the ground line have been previously considered (511). If a line passes through the ground line from o coe Fig. 65. the second to the fourth angle, so that any point on the line is equidistant from both planes, then the two projections of the line will be coincident. Fig. 65 illustrates such a condition. The line is not indeterminate by having coincident projections, however, because the oblique projections can be ¢éonstructed REPRESENTATION OF LINES AND POINTS 79 from the orthographic representation. The projection on the profile plane shows that this line bisects the second and fourth angles. If the line passes through the ground line, from the second to the fourth angles, but is not equidistant from the principal planes, then both projections will pass through the same point on the ground line. 516. Points in all angles. Fig. 66 shows four points A, B, C, and D, lying, one in each angle. The necessary construction lines are shown. On the right, the condition is depicted in projection, the number close to the projection indicating the angle in which the point is located. Observation will show that the first and third angle projec- Vv a’ ec ; ' b BY a . re I, be- a ! { | 1 i ge aba | | | Cc av. y | éc! od oa éa' Fic. 66. tions are similar in general appearance, but with projections interchanged. The same is true of the second and fourth angles; although, in the latter cases, both projections fall to one or the other side of the ground line. 517. Points with coincident projections. It may be further observed in Fig. 66 that in the second and fourth angles, the projection b, b’ and d, d’ may be coincident. This simply means that the points are equidistant from the planes of projection. The case of the point in the ground line has been noted (508), such points lie in no particular angle, unless a new set of principal planes be introduced. It can then be considered under any case at pleasure, depending upon the location of the prin- cipal planes. 518. Lines in profile planes. A line may be located so that both of its projections are perpendicular to the ground line. 76 GEOMETRICAL PROBLEMS IN PROJECTION The projections must therefore be coincident,* since they pass through the same point on the ground line. Fig. 67 shows an example of this kind. Although the actual line in space can be determined from its horizontal projection ab and its vertical Fia. 67. projection a’b’, still this is only true because a limited portion of the line was chosen for the projections. The profile shown on the extreme right clearly indicates the condition. The location of the profile with respect to the projection should also be noted. The view, is that obtained by looking from right to left, and is therefore located on the right side of the projection. The number- ing on the angles should also help the interpretation. - e D B a d H i A G! : f 241 H 2 3 x 7 ee? 4 K F G Fia. 6°. Fig. 68 shows a profile (on the left-hand diagram) of one line in each angle. The diagram on the right shows the lines in projection. The numbers indicate the angle in which the line is located. * Compare the coincident projections in this case with those of Art. 515. 1. oR CO DO Ono 11. 12. 13. 14, 15. 16. 17. 18. 19. 20. 21. 22. REPRESENTATION OF LINES AND POINTS 77 QUESTIONS ON CHAPTER V Discuss the point, the line, and the surface, and show how the material object is made up of them. . What are the mathematical elements of a material object? . Why are the mathematical elements considered as concepts? . How is the outline of a material object determined? . What is meant by the graphical representation of mathematical con- cepts? . What are the principal planes of projection? . What is the ground line? . How many dihedral angles are formed by the principal planes and how are they numbered? Make a diagram. . How is a line orthographically projected on the principal planes? . How many projections are required to fix the line with reference to the principal planes? Why? Show how one of the principal planes is revolved so as to represent a line in orthographic projection. Assume a line in the first angle in orthographic projection and show how the transfer is made to an oblique projection. Under what conditions will a line pierce a plane of projection? Assume a line that is inclined to both planes of projection and show how the piercing points are determined orthographically. Give the reasoning of the operation. Do the orthographic projections of a line represent the actual line in space? Why? Show a point in oblique projection and also the projecting lines to the principal planes. Draw the corresponding diagram in orthographic projection showing clearly how one of the principal planes is re- volved. Draw, in projection, a point lying in the horizontal plane; a point lying in the vertical plane; a point lying in both planes. Observe nomenclature in indicating the points. Indicate in what angle the points shown in Fig. 5-A are located. Draw a line parallel to the horizontal plane but inclined to the vertical plane in orthographic projection. Transfer ara diagram to oblique projection. ! , i Draw a line parallel to the vertical plane bp but inclined to the horizontal plane in orthographic projection. Transfer dia- gram to oblique projection. Draw a line parallel to both principal planes in orthographic pro- jection. Transfer diagram to oblique projection. When a line is parallel to both principal planes is it parallel to the ground line? Why? Fic. 5-A. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. GEOMETRICAL PROBLEMS IN PROJECTION . In Fig. 5-B, make the oblique projection of the line represented. . Draw the projections of a line lying in the horizontal plane. . Draw the projections of a line lying in the vertical plane. . Draw the projections of a line lying in the ground line, observing the nomenclature in the representation. cha ae pore a—, Fig. 5-B. Fie. 5-C. In Fig. 5-C give the location of the lines represented by the ortho- graphic projections. Construct the corresponding oblique projec- tions. Show two lines, one perpendicular to each of the planes and also draw a profile plane for each indicating the advantageous use in such cases. When a line is perpendicular to the horizontal plane, how is it pro- jected on that plane? When a line is perpendicular to the vertical plane, why is the horizon- tal projection equal to it in length? Draw a line in the second angle, in orthographic projection. Transfer the diagram to oblique projection. Draw a line in the third angle, in orthographic projection. Transfer the diagram to oblique projection. Draw a line in the fourth angle, in orthographic projection. Transfer the diagram to oblique projection. In Fig. 5-D, specify in which angles each of the lines are situated. , € 1 ' es La 1 h os if x + “> of vid | oy cog’ e! Fic. 5-D. Draw lines similar to 5-D in orthographic projection and transfer the diagrams to oblique projection. Make the orthographic projection of a line with coincident projections and show by a profile plane what it means. Take the case where the line passes through one point on the ground line and is per- pendicular to it, and the other case where it is inclined to the ground line through one point. 37. 38. 39. 40. 41, 42. REPRESENTATION OF LINES AND POINTS 79 Locate a point in each angle and observe the method of indicating them. Make an oblique projection of the points given in Question 37. Make an orthographic projection of points with coincident pro- jections and show under what conditions they become coincident. In what angles are coincident projections of points possible? Show by profile. Make the oblique projections of the lines shown in Fig. 68 of the text. Use arrows to indicate the lines. Why is it advantageous to use a profile plane, when the lines are indefinite in extent, and lie in the profile plane? CHAPTER VI REPRESENTATION OF PLANES 601. Traces of planes parallel to the principal planes. Let Fig. 69 represent the two principal planes by HH and VV intersecting in the ground line XY. Let, also, RR be another plane passing through the first and fourth angles and parallel to VV. The plane RR intersects the horizontal plane HH in a line tr, which is called the trace of the plane RR. As RR is parallel Fic. 69. to VV, the case is that of two parallel planes cut by a third plane, and, from solid geometry, it is known that their intersections are parallel. If the horizontal plane is revolved, by the usual method, into coincidence with the vertical plane, the resulting diagram as shown on the right will be the orthographic represen- tation of the trace of a plane which is parallel to the vertical plane. If, as in Fig. 70, the plane is parallel to the horizontal plane, the condition of two parallel planes cut by a third plane again presents itself, the vertical plane of projection now being the cutting plane. The trace t’r’ is then above the ground line, and, as before, parallel to it. 602. Traces of planes parallel to the ground line. When a plane is parallel to the ground line, and inclined to both planes 80 REPRESENTATION OF PLANES 81 of projection, it must intersect the principal planes. The line of intersection on each principal plane will be parallel to the ground ty ae \ Fie. 70. line because the given plane is parallel to the ground line and, hence, cannot intersect it. Fig. 71 shows this condition in oblique Fia. 71. and orthographic projection, in which tr is the horizontal trace and t’r’ is the vertical trace. A special case of this occurs if the plane passes through the ground line. Both traces then coincide with the ground line and the orthographic .repre- sentation becomes indeterminate unless the profile plane is attached. H Fig. 72 is a profile and shows several 7 planes passing through the ground line, each R- of which is now determined. It may be possible, however, to introduce a new hori- Fra. 72. zontal plane H’H’, parallel to the principal horizontal plane. Such an artifice will bring the case into that immediately preceding. It may then be shown as an ordinary 82 GEOMETRICAL PROBLEMS IN PROJECTION orthographic projection, just as though the original horizontal plane were absent. The same would still be true if a new ver- tical plane were added instead of the horizontal plane, or, if an entirely new set of principal planes were chosen so that the new principal planes would be parallel to the old ones. 603. Traces of planes perpendicular to one of the prin= cipal planes. Fig. 73 shows a plane that is perpendicular to Fia. 73. the horizontal plane but inclined to the vertical plane. It may be imagined as a door in a wall of a room. The angle with the vertical plane (or wall in this case) can be changed at will by swinging it about the hinges, yet its plane always remains per- Fia. 74, pendicular to the horizontal plane (floor of the room). The inter- section with the vertical plane is perpendicular to the horizontal plane because it is the intersection of two planes (the given plane and the vertical plane), each of which is perpendicular to the horizontal. As a consequence the vertical trace is perpen- dicular to the ground line, because, when a line is perpendicular REPRESENTATION OF PLANES 83 to a plane it is perpendicular to every line through its foot (from geometry). The orthographic representation of three distinct planes is shown on the right; the cases selected all show Tt’ perpendicular to the ground line. What is true regarding planes perpendicular to the horizontal plane is equally true for planes similarly related with respect to the vertical plane. Fig. 74 gives an illustration of such a case. Here, the horizontal trace is perpendicular to the ground line but the vertical trace may make any angle, at will, with the ground line. 604. Traces of planes perpendicular to both principal planes. When a plane is perpendicular to both principal planes, Fia. 75. its two traces are perpendicular to the ground line. Such a plane is a profile plane and is shown in Fig. 75. 605. Traces of planes inclined to both principal planes. It may be inferred from the preceding, that, if the plane is Fic. 76. inclined to both principal planes, neither trace can be perpen- dicular to the ground line. Fig. 76 shows such a case, the 84 GEOMETRICAL PROBLEMS IN PROJECTION orthographic representation of which in projection is shown on the right. Fig. 77 shows two cards, each slit half way, as indicated. These can be fitted together to represent the principal planes. If at any point, a and a’, on XY 8 slits tt and t’t’ be made, it will k= Y be found on assembling the ‘ Pa cards that a third card can be x——_“__y inserted. a In the upper card, another t slit s’s’ may be made, through x we S|" the point a’, with its direction t parallel tott. As in the previous instance, a card can again be inserted. This case is of in- terest, however, because both traces become coincident on the revolution of the horizontal plane, and fall as one straight line tTt’, as shown in projection in Fig. 77. ay & Fia. 77. 606. Traces of planes intersecting the ground line. It must have been observed in the cases where the given plane is inclined to the ground line that both traces pass through the same point on the ground line. This becomes further evident when it is considered that the ground line can intersect a plane at but one point, if at all. This one point lies in the ground line, and, hence, it has coincident projections (509). 607. Plane fixed in space by its traces. Two intersect- ing lines determine a plane (from solid geometry). Hence, the two traces of a plane fix a plane with reference to the principal planes because the traces meet at the same point on the ground line. If the diagram is such that the traces do not intersect in the ground line within the limits of the drawing, it is assumed that they will do so if sufficiently produced. The limiting posi- tion of a plane whose traces cannot be made to intersect the ground line is evidently a plane parallel to it. This plane is then parallel to the ground line and its traces must also be parallel to it (602). 608. Transfer of diagrams from orthographic to oblique projection. Let the right-hand diagram of Fig. REPRESENTATION OF PLANES 85 78 represent a plane in orthographic projection. Through any point S, on the ground line, pass a profile plane sSs’, inter- secting the two traces tT and Tt’ at points o and p. To transfer the orthographic to the oblique projection, lay off the principal planes HH and VV as shown, intersecting in the ground line Fie. 78. XY. Lay off any point S on the ground line in the oblique pro- jection and then make So and Sp of the orthographic equal to the similarly lettered lines in the oblique projection. The profile plane sSs’ is therefore determined in the oblique projection. Make TS of one diagram equal to the TS of the other, and com- Fia. 79. plete by drawing the traces through T, 0 and T, p. To increase the clarity of the diagram, a rectangular plane may be shown as though it passes through the principal planes. Fig. 78 has this plane added. _ A case where the two traces are coincident is shown in Fig. 79. Two profile planes are required, but only one trace of each 86 GEOMETRICAL PROBLEMS IN PROJECTION is needed. All the necessary construction lines are shown in this diagram. 609. Traces of planes in all angles. As planes are indef- inite in extent, so are their traces; and, therefore, the traces are not limited to any one angle. In the discussion of most problems, it may be possible to choose the principal planes so as to limit the discussion to one angle—usually the first. The advantage to be gained thereby is the greater clarity of the diagram, as then the number of construction lines is reduced to a minimum. Third angle projection may also be used, but the transfer to oblique projection is undesirable. Second and fourth angles are avoided because the projections overlap (317, 514, 516). Fig. 80 shows the complete traces of a given plane T. The ey \t Ny x Nr Ns i 7 oe 1 \ ny “A Ne : a ‘ me Ne Fig. 80. Fia. 81. full lines indicate the traces in the first angle; the dotted lines show the continuation in the remaining three angles. Fig. 81 represents each quadrant separately for the same plane that is shown in Fig. 80. The appended numbers indicate the angle to which the given plane is limited. 610. Projecting plane of lines. It is now evident that the finding of the projection of a line is nothing more nor less than the finding of the trace of its projecting plane. The two perpendiculars from a line to the plane of projection are necessarily parallel and therefore determine a plane. This plane is the projecting plane of the line and its intersection (or trace) with the plane of projection contains the projection of the line. Manifestly, any number of lines contained in this projecting plane would have the same projection on any one of the principal planes, and that projection therefore does not fix the line in space. REPRESENTATION OF PLANES 87 If the projection on the corresponding plane of projection be taken into consideration, the projecting planes will intersect, and this Ze oe 7 ow” ge iL {4 Fia. 82. intersection will be the given line in space. Fig. 82 illustrates the point in question. QUESTIONS ON CHAPTER VI 1. What is the trace of a plane? , 2. Draw a plane parallel to the vertical plane passing through the first and fourth angles, and show the resulting trace. Make diagram in oblique and orthographic projection. 3. Take the same plane of Question 2 and show it passing through the second and third angles. 4. Show how a plane is represented when it is parallel to the horizontal plane and passes through the first and second angles. Make diagram in oblique and orthographic projection. 5. Take the same plane of Question 4 and show the trace of the plane when it passes through the third and fourth angles. 6. Show how a plane is represented when it is parallel to the ground line and passes through the first angle. 7. How is a plane represented when it is parallel to the ground line and passes through the second angle? Third angle? _ Fourth Angle?. 8. Show how a plane passing through the ground line is indeterminate in orthographic projection. 9. When a plane passes through the ground line show how the profile plane might be used to advantage in representing the plane. 7 10. When a plane passes through the ground line, show how an auxiliary principal plane may be used to obtain determinate traces. 11. When a plane is perpendicular to the horizontal plane and inclined to the vertical plane, show how this is represented orthographically. Make, also, the oblique projection of it. 88 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. GEOMETRICAL PROBLEMS IN PROJECTION When a plane is perpendicular to the vertical plane and inclined to the horizontal plane show how this is represented orthographically. Make, also, the oblique projection of it. When a plane is perpendicular to both principal planes draw the orthographic traces of it. Why is the ground line perpendicular to both traces in Question 13? Is this plane a profile plane? When a plane is inclined to the ground line show how the traces are represented. Why do both traces of a plane intersect the ground line at a point when the plane is inclined to it? Why do the traces fix the plane with reference to the principal planes? \e’ af yes / Bp t Q_R Ss cl x ( RK 7 Y \r \ \gy Y Fia. 6-A. When a plane is parallel to the ground line, why are the traces of the plane parallel to it? Show an oblique plane in all four angles of projection (use only the limited portion in one angle and use the same plane as in the illustration). In what angles are the planes whose traces are shown in Fig, 6-A? ty te wf A Y v7 ‘ ~e Fic. 6-B. Fic. 6-C. Show how a line in space and its projecting perpendiculars determine a plane which is the projecting plane of the line. Is the projection of the line the trace of the projecting plane of the line? Is it possible to have two separate lines whose projections are coin- cident on one plane? How are such lines determined? Given the traces of a plane in orthographic projection as shown in Fig. 6-B construct the oblique projection of it. Construct the oblique projection of a plane having coincident projec- tions (Fig. 6-C). CHAPTER VII ELEMENTARY CONSIDERATIONS OF LINES AND PLANES 701. Projection of lines parallel in space. When two lines in space are parallel, their projecting planes are parallel, and their intersection with any third plane will result in parallel lines. If this third plane be a plane of projection, then the traces of the two projecting planes will result in parallel projections. Fig. 83 shows two lines, AB and CD in space. The lines are Fig. 83. shown by their horizontal projections ab and cd, which are parallel to each other, and by their vertical projections a’b’ and c’d’, which are also parallel to each other. A perfectly general case is represented pictorially on a single plane of projection in Fig. 84. 702. Projection of lines intersecting in space. If two lines in space intersect, their projections e , intersect, because the two lines in space must KG meet in a point. Further, the projection of = a! id this point must be common to the projections xH—_j—-y of the lines. Xu tty In Fig. 85 two such lines, AB and CB are | shown, represented, as usual, by their pro- ce jections. O is the intersection in space, Fie. 85, indicated by its horizontal and vertical projections o and 0’, 89 90 GEOMETRICAL PROBLEMS IN PROJECTION respectively. EF and GH (Fig. 86) are two other lines, chosen to show how in the horizontal plane of projection, the two pro- jections may coincide, because the plane of the two lines happens also to be the horizontal projecting plane. The case is not inde- terminate, however, as the vertical projection locates the point M in space. The reverse of this is also true, that is, the vertical Sess =e ea, > — y ak----- 4 @q----— Fig. 86. Fig. 87. Fie. 88. instead of the horizontal projections may be coincident. General cases of the above are represented pictorially in Figs. 87 and 88. Should the horizontal and the vertical projections be coin- cident, the lines do not intersect but are themselves coincident in space and thus form only one line. 703. Projection of lines not intersecting in space.* ‘There are two possible cases of lines that do not intersect in space. N | he o:. & --\--70 - oe. Sh ° Ry . te = OFF ~ Gna [eae ° Fie. 89. Fig. 90. The case in which the lines are parallel to each other has previously been discussed (701). If the two lines cannot be made to lie in the same plane, they will pass each other without intersecting. Hence, if in one plane of projection, the projections intersect, they cannot do so in the corresponding projection. This fact is depicted in Fig. 89. AB and CD are the two * Called skew lines. CONSIDERATIONS OF LINES AND PLANES 91 lines in space. Two distinct points E and F, on the lines are shown in the horizontal projection as e and f; their vertical projections are, however, coincident. Similarly, G and H are also two distinct points on the lines, shown as g’ and h’ in the vertical projection, and coincident, as g and h in the horizontal projection. The pictorial representation is given in Fig. 90. 704. Projection of lines in oblique planes. When a third plane is inclined to the principal planes, it cuts them in lines of intersection, known as traces (601). Any line, when inclined to the principal planes will pierce them ina point. Hence, if a plane is to contain a given line, the piercing points of the line must lie in the traces of the plane. Viewing this in another way, a plane may be passed through two parallel or two inter- secting lines. On the resulting plane, any number of lines may be drawn, intersecting the given pair. Hence, an inclined line must pass through the trace, if it is contained in the plane.* Fig. 91 shows a plane tTt’ indicated by its horizontal trace tT and its vertical trace Tt’. It is required to draw a line AB in this plane. Suppose the hori- zontal piercing point is assumed at b, its corresponding vertical projection will lie in the ground line (509). Also, if its vertical piercing point is assumed at a’, its corresponding horizontal pro- jection will be a. With two horizontal projections of given Fre. 91. points on a line and two vertical projections, the direction of the line is determined, for, if the horizontal and vertical projecting planes be erected, their inter- section determines the given line (610). As a check on the correctness of the above, another line CD may be assumed. If the two lines lie in the same plane and are not parallel, they must intersect. This point of intersection * When the plane is parallel to the ground line, a line in this plane parallel to the principal planes cannot pierce in the traces of the plane. See Art. 602. +It must be remembered that the principal planes must be at right angles to each other to determine this intersection. In the revolved position, the planes would not intersect in the required line. 92 GEOMETRICAL PROBLEMS IN PROJECTION is M shown horizontally projected at m and vertically, at m’. The line joining the two is perpendicular to the ground line (508). This is illustrated in oblique projection in Fig. 92. It should be here noted that if the line is to be contained by the plane, only the direction of one projection can be assumed, and that the corresponding projection must be found by the principles so far developed. A converse of this problem is to draw a plane so that it shall contain a given line. As an unlimited number of planes can be passed through any given line, the direction of the traces is not fixed. Suppose AB (Fig. 91) is the given line, then through the horizontal piercing point b draw any trace tT, and from T where this line intersects the ground line, draw Tt’, through the vertical Vv Fig. 92. piercing point a’. The point T may be located anywhere along XY. All of these planes will contain the line AB, if their traces pass through the piercing points of the line. Still another feature of Fig. 91 is the fact that from it can be proved that two intersecting lines determine a plane. If AB and CD be the two lines intersecting at M, their horizontal piercing points are b and c and their vertical piercing points are a’ and d’ respectively. Two points fix the direction of a line, and, hence, the direction of the traces is fixed; tT is drawn through cb and Tt’ is drawn through a’d’. The check lies in the fact that both traces intersect at one point T on the ground line (606). 705. Projection of lines parallel to the principal planes and lying in an oblique plane. If a line is horizontal, it is parallel to the horizontal plane, and its vertical projection must be CONSIDERATIONS OF LINES AND PLANES 93 parallel to the ground line (511). If this line lies in an oblique plane, it can have only one piercing point and that with the vertical plane, since it is parallel to the horizontal. Thus, in Fig. 93, let tTt’ be the given oblique plane, represented, of course, by its traces. The given line is AB and a’b’ is its vertical projection, the piercing point being at a’. The corresponding horizontal projection of a’ is a. When a line is horizontal, as in this case, it may be considered as being cut from the plane tTt by a hori- zontal plane, and, as such, must be parallel to the principal horizontal plane. These two parallel horizontal planes are cut by the given oblique plane tTt’ and, from geometry, their lines of intersection are parallel. Hence, the horizontal projection Fie. 93. of the horizontal line must be parallel to the horizontal trace, of the given plane because parallel lines in space have parallel projections. Accordingly, from a, draw ab, parallel to Tt, and the horizontal line AB is thus shown by its projections. A line, parallel to the vertical plane, drawn in an oblique plane follows the same analysis, and differs only in an interchange of the operation. That is to say, the horizontal projection is then parallel to the ground line and pierces in a point on the horizontal trace of the given plane; its vertical projection must be parallel to the vertical trace of the given plane. In Fig. 93, CD is a line parallel to the vertical plane and cd is the horizontal projection, piercing the horizontal plane at ¢ vertically projected at c’; c’d’ is, therefore, the required vertical projection. 94 : GEOMETRICAL PROBLEMS IN PROJECTION A check on the problem lies in the fact that these two lines must intersect because they lie in the same plane by hypothesis. The point of intersection M is shown as m in the horizontal a' ¥ mm, BY S. D 4 ii a a ct a oe ~ < aaa C. Ck SS a c ESS a mm = Vv Bie: a Fia. 94., projection, and as m’ in the vertical projection. The line joining these points is perpendicular to the ground line. The oblique projection of this is shown in Fig. 94. Fig. 95 is a still further step in this problem. Three lines , ch - ~f 1 o we ad m! ce mi | xy | eI a’ i ET call aS F € i = : 3 { I b i | wf ° m Wa. n ~ ad y L Fia. 95. AB, CD, and EF are shown, all being in the plane tTt’. AB is parallel to the vertical plane, CD is parallel to the horizontal plane and EF is any other line. It will be observed that the three lines intersect so as to form a triangle MNO shown by having the area shaded in both its projections. CONSIDERATIONS OF LINES AND PLANES 95 706. Projections of lines perpendicular to given planes. If a line is perpendicular to a given plane, the projections of the line are perpendicular to the corresponding traces of the plane. Let, in Fig. 96, LL be any plane and MM any other plane intersecting it in the trace tr. Also, let AB be any line, perpen- dicular to MM, and ab be the projection of AB on LL. It is desired to show that the projection ab is perpendicular to the trace tr. Any plane through AB is perpendicular to the plane MM, because it contains a line perpendicular to the plane by hypothesis. Also, any plane through Bb, a perpendicular to the plane LL, is perpendicular to LL. Hence, any plane contain- B Fia. 96. ing both lines (it can do so because they intersect at B) will be perpendicular both to LL and MM. As the plane through ABb is perpendicular to the two planes LL and MM, it is also perpendicular to a line common to the two planes, such as tr. Thus, ab is perpendicular to tr. In fact, any line perpendicular to a plane will have its projection on any other plane perpendicular to the trace of the plane, because, instead of LL being assumed as the plane of projection and BA a perpendicular to another plane MM, the conditions may be reversed and MM be assumed as the plane of projection and Bb the perpendicular. The converse of this is also true. If Ao be assumed the pro- jection of Bb on MM, a plane passed through the line Ao per- pendicular to the trace tr will contain the lines AB and Bb, where 96 GEOMETRICAL PROBLEMS IN PROJECTION Bb is the given line and AB then the projecting perpendicular to the plane MM. 707. Revolution of a point about a line. Frequently it is desirable to know the relation of a point with respect to a line, for, if the line and point are given by their projections, the true relation may not be apparent. To do this, revolve the point about the line so that a plane through the point and the line will either coincide or be parallel to the plane of projection, then they will be projected in their true relation to each other. The actual distance between the point and the line is then shown as the perpendicular distance from the point to the line. Consider the diagram in Fig. 97* and assume that the point A is to be revolved about the point B. The projection of A on y a|------—— Fia. 97. the line Oa” is a and, to an observer looking down from above the line Oa”, the apparent distance between A and B is aB. Oa’ is the apparent distance of A from B to an observer, looking orthographically, from the right at a plane perpendicular to the plane of the paper through Oa’. Neither projection gives the true relation between A and B from a single projection. If the point A is revolved about B as an axis, with BA as a radius, until it coincides with the line Oa’, A will either be found at a” or at a’, depending upon the direction of rotation. During the revolution, A always remains in the plane of the paper and de- scribes a circle, the plane of which is perpendicular to the axis, through B, the centre of the circle. * This diagram is a profile plane of the given point and of the principal planes. CONSIDERATIONS OF LINES AND PLANES 97 A more general case is shown in Fig. 98 where BB is an axis lying in a plane, and A is any point in space not in the plane containing BB. If A-be revolved about BB as an axis, it will describe a circle, the plane of which will be perpendicular to the axis. In other words, A will fall somewhere on a line Ba’”’ perpendicular to BB. The line Ba” must’ be perpendicular to BB because it is the trace of a perpendicular plane (706). This point is a’’ and Ba” is equal to the radius BA. Contrast this B Fig. 98. with a, the orthographic projection of A on the plane containing BB. Evidently, then, a is at a lesser distance from BB than a”. Indeed, BA equals Ba’ and is equal to the hypothenuse of a right triangle, whose base is the perpendicular distance of the projection of the point from the axis, and whose altitude is the distance of the point above the plane containing the line. The angle AaB is a right angle, because a is the orthographic projection of A. QUESTIONS ON CHAPTER VII 1. If two lines in space are parallel, prove that their projections on any plane are parallel. 2. When two lines in space intersect, prove that their projections, on any plane, intersect. 3. Draw two lines in space that are not parallel and still do not intersect. 4, Show a case of two non-intersecting lines whose horizontal projections are parallel to each other and whose vertical projections intersect. ° Show also, that the horizontal projecting planes of these lines are parallel. 98 a ot 10. 11, 12. 13. 14. 15. GEOMETRICAL PROBLEMS IN PROJECTION . Make an oblique projection of the lines considered in Question 4. . Prove that when a line lies in a plane it must pass through the traces of the plane. . Given one projection of a line in a plane, find the corresponding pro- jection. . Given a plane, draw intersecting lines in the plane and show by the construction that the point of intersection satisfies the ortho- graphic representation of a point. . Show how two intersecting lines determine a plane by aid of an oblique projection. In a given oblique plane, draw a fine parallel to the horizontal plane and show by the construction that this line pierces the vertical plane only. Give reasons for the construction. In Question 9, draw another line parallel to the vertical plane and show that this second line intersects the first in a point. Given an oblique plane, draw three lines; one parallel to the hori- zontal plane, one parallel to the vertical plane and the last inclined to both planes. Show, by the construction, that the three lines form a triangle (or meet in a point in an exceptional case). Prove that when a line is perpendicular to a plane the projection of this line on any other plane is perpendicular to the trace of the plane. Show the general case and also an example in orthographic projection. A line lies in a given plane and a point is situated outside of the plane. Show how the point is revolved about the line until it is contained in the plane. Prove that the point, while revolving about the line, describes a circle the plane of which is perpendicular to the axis (the line about which it revolves). CHAPTER VIII PROBLEMS INVOLVING THE POINT, THE LINE, AND THE PLANE 801. Introductory. A thorough knowledge of the preceding three chapters is necessary in order to apply the principles, there developed, in the solution of certain problems. The commercial application of these problems frequently calls for extended knowl- edge in special fields of engineering, and for this reason, the application, in general, has been avoided. Countless problems of a commercial nature may be used as illustrations. All of these indicate, in various ways, the im- portance of the subject. In general, the commercial problem may always be reduced to one containing the mathematical essentials (reduced to points, lines and planes). The solution, then, may be accomplished by the methods to be shown sub- sequently. 802. Solution of problems. In the solution of the following problems, three distinct steps may be noted: the statement, the analysis, and the construction. The statement of the problem gives a clear account of what is to be done and includes the necessary data. The analysis entails a review of the principles involved, and proceeds, logically, from the given data to the required conclusion. On completion of the analysis, the problem is solved to all intents and purposes. The construction is the graphical presentation of the analysis. It is by means of a drawing and its description that the given data is associated with its solution. It may be emphasized again that the drawings are made orthographically, and that the actual points, lines and planes are to be imagined. By a slight change in the assumed data, the resultant con- struction may appear widely different from the diagrams in the book. Here, then, is an opportunity to make several con- structions for the several assumptions, and to prove that all 99 100 GEOMETRICAL PROBLEMS IN PROJECTION follow the general analysis. The simpler constructions might be taken and transformed from orthographic to oblique projec- tion; this will show the projections as well as the actual points, lines and planes in space. By performing this transformation (from orthographic to oblique projection), the student will soon be able to picture the entire problem in space, without recourse to any diagrams. To bring forcibly to the student’s attention the difference between the analysis and the construction, it may be well to note that the analysis gives the reasoning in its most general terms, while the construction is specific, in so far as it takes the assumed data and gives the solution for that particular case only. 803. Problem 1. To draw a line through a given point, parallel to a given line. Analysis. If two lines in space are parallel, their projecting planes are parallel and their intersection with the principal planes are parallel (701). Hence, through the projections of the given point, draw lines parallel to the projections of the given lines. Construction. Let AB, Fig. 99, be the given line in space, eat represented by its horizontal pro- u e jection ab and its vertical projection | a a’b’. Further, let G be the given x ¥ point, similarly represented by its | horizontal projection g and its ver- ! Va we tical projection g’. Through the { 9 horizontal projection g, draw cd rj parallel to ab and through g’, draw Fre. 99. c’d’ parallel to a’b’. As the length of the line is not specified, any line that satisfies the condition of parallelism is permissible. Therefore, CD is the line in space that is parallel to AB through the point G. A pictorial representation of this is shown in Fig. 84. 804. Problem 2. To draw a line intersecting a given line at a given point. , Analysis. If two lines in space intersect, they intersect in a point that is common to the two lines. Therefore, their pro- jecting planes will intersect in a line which is the projecting line of the given point (702). Hence, through the projections of the THE POINT, THE LINE, AND THE PLANE 101 given point, draw any lines, intersecting the projections of the given lines. Construction. Let AB, Fig. 100, be the given line, shown horizontally projected as ab and ver- tically projected as a’b’. Let, also, ‘G be the given point, situated on the line AB. As no direction is specified for the intersecting line, draw any line cd through the horizontal projec- tion g and this will be the horizontal projection of the required line. Similarly, any other line c’d’ through g’ will be the vertical projection of Fic. 100. the required line. Hence, CD and AB are two lines in space, intersecting at the point G. The pictorial representation of this case is depicted in Figs. 87 and 88. ra 805. Problem 3. To find where a given line pierces the principal planes. Analysis. If a line is oblique to the principal planes, it will pierce each of these in a point, the corresponding projection of which will be in the ground line. Hence, a piercing point in any principal plane must be on the projection of the line in that plane. It must also be on a perpendicular erected at the point where the corresponding projection crosses the ground line. Therefore, the required piercing point is at their intersection. Construction. Let AB, Fig. 101, be a limited portion of an indefinite line, shown by its horizontal a’ projection ab and its vertical projection be ! a’b’. The portion chosen AB will not _& | | pierce the principal planes, but its continua- = oe tion, in both directions, will. Prolong the 6 vertical projection a’b’ to c’ and at c’, erect | a perpendicular to XY, the ground line, and i continue it, until it intersects the prolonga- woe tion of ab atc. This will be the horizontal piercing point. In the same way, prolong ab to d, at d, erect a perpendicular to the ground line as dd’, the intersection of which with the prolonga- tion of a’b’, at d’ will give the vertical piercing point. Hence, Fie. 101. 102 GEOMETRICAL PROBLEMS IN PROJECTION if CD be considered as the line, it will pierce the horizontal plane of projection at c and the vertical plane of projection at d’. A pic- torial representation in oblique projection is shown in Fig. 52. 806. Problem 4. To pass an oblique plane, through a given oblique line. Analysis. Ifa plane is oblique to the principal planes, it must intersect the ground line at a point (606); and if it is to contain a line, the piercing points of the line must lie in the traces of the plane (704). Therefore, to draw an oblique plane containing a given oblique line, join the piercing points of the line with any point of the ground line and the resulting lines will be the traces of the required plane. Construction. Let AB, Fig. 102, be the given line. This ie se Sy Ss r H Xi_4 io ST #s - gS Zz H i (jer z aT a le ee Fic. 102. Fig. 103. line pierces the horizontal plane at a and the vertical plane at b’. Assume any point T on the ground line XY, and join T with a and also with b’.. Ta and Tb’ are the required traces, indicated, as usual, by tTt’. Thus, the plane T contains the line AB. The construction in oblique projection is given in Fig. 103. 807. Special cases of the preceding problem. If the line is parallel to both planes of projection, the traces of the plane will be parallel to the ground line (602), and the profile plane may be advantageously used in the drawing. If the line is parallel to only one of the principal planes, join the one piercing point with any point on the ground line, which results in one trace of the required plane. Through the point on the ground line, draw the corresponding trace, parallel to the THE POINT, THE LINE, AND THE PLANE 103 projection of the line in the plane containing this trace. The case is evidently that considered in Art. 705. 808. Problem 5. To pass an oblique plane, through a given point. Analysis. If the oblique plane is to contain a given point, it will also contain a line through the point. Hence, through the given point, draw an oblique line and find the piercing points of this line on the principal planes. Join these piercing points with the ground line and the result will be the required traces of the plane. Construction. Let C, Fig. 104, be the required point, and let AB be a line drawn through this point. AB pierces the hori- \ oho Fiag. 104. Fia. 105. zontal plane at a and the vertical plane at b’. Join a and b’ with any assumed point T and tTt’ will be the required traces. The plane T, contains the point C, because it contains a line AB through the point C. Fig. 105 shows this pictorially. Nore. An infinite number of planes may be passed through the point, hence the point T was assumed. It might have Leen assumed on the opposite side of the point and would still have contained the given point, or the auxiliary line through the point. Also, it is possible to draw a line through the given point ‘parallel to the ground line. A profile construction will then be of service (602). » Again, a perpendicular line may be drawn through the given point and a plane be passed so that it is perpendicular to one or both planes of projection (603, 604). 104 GEOMETRICAL PROBLEMS IN PROJECTION 809. Problem 6. To find the intersection of two planes, oblique to each other and to the principal planes. Analysis. If two planes are oblique to each other, they intersect in a line. Any line in a plane must pass through the traces of the plane. As the line of intersection is common to the two planes, it must pass through the traces of both planes and hence it passes through the intersection of these traces. Construction. Let T and S, Fig. 106, be the given planes. The horizontal piercing point of their line of intersection is at b, Fig. 106. Fria. 107. vertically projected at b’; the vertical piercing point is at a’, horizontally projected at a. Join ab and a’b’, as they are the projections of the line AB, which is the intersection of the planes TandS&S._ In oblique projection, this appears as shown in Fig. 107. 810. Special case of the preced= ing problem. If the two planes are chosen so that the traces in one plane do not intersect within the limits of the drawing, then draw an auxiliary plane R (Fig. 108) and find the inter- section, AB as shown. From c’ draw c’d’, parallel to a’b’ and from c, draw cd, parallel to ab. The line of inter- section of the given planes is thus deter- mined. Fie. 108. 811. Problem 7. To find the corresponding projection of a given point lying in a given oblique plane, when one of its pro- jections is given. Analysis. If a line lies in a given plane and also contains THE POINT, THE LINE, AND THE PLANE 105 a given point, the projections of this line will also contain the projections of the given point. Hence, through the projection of the given point, draw the projection of a line lying in the given plane. Then find the corresponding projection of the line. The required projection of the given point will lie on the inter- section of a perpendicular to the ground line, through the given projection of the point with the corresponding projection of the line. Construction. Let tTt’, Fig. 109, be the traces of the given plane and c the horizontal projection of the given point. Draw ab, the horizontal projection of a line in the plane, through c the horizontal projection of the given point. The horizontal Fic, 109. Fig. 110. piercing point of the line AB is in the trace Tt at a, and its cor- responding projection lies in the ground line at a’. Further, the vertical piercing point lies on a perpendicular to the ground line from the point b and also in the trace Tt’, hence, it is at b’ and a’b’ is thus the corresponding projection of the line ab. The required projection of the given point lies on a line through c, perpendicular to the ground line, and also on a’b’; hence, it is at their intersection c’. The point C, in space, is con- tained in the plane T, and c and c’, are corresponding pro- jections. The oblique projection of this problem is given in Fig. 110. 812. Special case of the preceding problem. The point in the above problem was purposely chosen in the first angle, in order to obtain a simple case. It may be located anywhere, 106 GEOMETRICAL PROBLEMS IN PROJECTION however, because the planes are indefinite in extent. For instance, in Fig. 111 the vertical projection is selected below the ground line. However, a single projection does not locate a point in space. It may be assumed as lying either in the third or fourth angles (515). Subsequent operations are dependent upon the angle in which the point is chosen. 4 Assume, for instance, that the point % ae x is in the fourth angle; the traces of j ea ; the given plane must then also be | | 1 Not” X— rl assumed as being in the fourth angle. Thus, with the given plane T, Tt is were “Xb the horizontal trace, and Tt””’ is the é vertical trace (609). In completing Fic. 111. the construction by the usual method, let c’ be the assumed vertical pro- jection, and through it, draw a’b’ as the vertical projection of the assumed line’ through the given point and lying in the given plane. This line pierces the horizontal plane at b and the vertical plane at a’. Hence, ab and a’b’ are the corresponding projections of the line AB in space which is situated in the fourth angle. The corresponding horizontal projection of c’ is c, and thus the point C in space is determined. Had the point been assumed in the third angle, then the traces Tt” and Tt’” would have been the ones to use, Tt” being the horizontal, and Tt’” the vertical trace. The construction would then, in general, be the same as the previous. 813. Problem 8. To draw a plane which contains a given point and is parallel to a given plane. Analysis. The traces of the required plane must be parallel to the traces of the given plane. A line may be drawn through the given point parallel to an assumed line in the given plane. This line will then pierce the principal planes in the traces of the required plane. Hence, in the given plane, draw any line. Through the given point, draw a line parallel to it, and find the piercing points of this line on the principal planes. Through these piercing points draw the traces of the required plane parallel to the corresponding traces of the given plane. The plane so drawn is parallel to the given plane. Construction. Let T, Fig. 112 be the given plane, and G THE POINT, THE LINE, AND THE PLANE 107 the given point. In the plane T, draw any line CD as shown by cd and c’d’ its horizontal and vertical projections respectively. Through G, draw AB, parallel to CD and ab and a’b’ will be the projections of this line. The piercing points are a and b’ on the horizontal and vertical planes respectively. Draw b’S parallel to t’'T and aS parallel to tT, then sSs’ will be the traces 4 Ne NS / d as a Toiiy | \ d ‘ rh \ - af WV PY bn ty a J Fig. 113. of the required plane, parallel to the plane T and containing a given point G. A check on the accuracy of the construction is furnished by having the two traces meet at S. Also, only one plane will satisfy these conditions because the point S cannot be selected at random. This construction is represented picto- tially in Fig. 113. 814. Problem 9. To draw a line perpendicular to a given plane through a given point. ae ; Analysis. If a line is perpendicular Se o , to a given plane the projections of the NS uti | line are perpendicular to the corresponding ae traces of the plane. Hence, draw through # > | ¥ ! | | each projection of the given point, a ge é | | line perpendicular to the corresponding trace 4 ~ (706). 3 Construction. Let T, Fig. 114, be Fic. 114. the given plane, and C, the given point. Through c, the horizontal projection of the given point, draw ab, perpendicular to Tt; and through c’, the vertical pro- jection of the given point, draw a’b’, perpendicular to Tt’. 108 GEOMETRICAL PROBLEMS IN PROJECTION Thus, AB is perpendicular to the plane T. An oblique projection of this problem is given in Fig. 115. 815. Special case of the preceding problem. If the point is chosen in the third angle, then it must be observed that Tt” is the horizontal trace (Fig. 116) and Tt’” is the vertical trace. Again, the line AB is drawn perpendicular to T, by making a’b’, through c’, perpendicular to Tt’”’, and ab, through c, perpendicular to Tt”’. 816. Problem 10. To draw a plane through a given point perpendicular to a given line. Analysis. The traces of the required plane must be perpen- ‘ ‘ IN al Ny \ x { “A NCAP Y Fig. 115. Fic. 116. dicular to the corresponding projections of the given line (706). Through one projection of the given point, draw an auxiliary line parallel to the trace; the corresponding projection of this line will be parallel to the ground line, because it is a line parallel to that plane in which its projection is parallel to the trace (705). Find the piercing point of this auxiliary line, and through this point, draw a line perpendicular to the corre- sponding projection of the given line. Where this intersects the ground line, draw another line, perpendicular to the corresponding projection of the given line. The traces are thus determined. Construction. Let, in Fig. 117, AB be the given line, and C the given point. For convenience, assume a horizontal line in the plane as the auxiliary line. Then, through c, draw cd, THE POINT, THE LINE, AND THE PLANE 109 perpendicular to ab, and through c’, draw c’d’, parallel to the ground line. The piercing point of this auxiliary line is d’, and only one point on either trace is required. Hence, through d’, draw Tt’, perpendicular to a’b’, and from T, draw Tt, per- pendicular to ab. T is, therefore, the required plane, and Tt must be parallel to cd because both must be perpendicular to ab. Fig. 118 shows a pictorial representation of the same problem. Norse. Instead of having assumed a line parallel to the horizontal plane, a line parallel to the vertical plane might have been assumed. In the latter case, the vertical projection would have been perpendicular to the vertical projection of the line, ., if i Ge ie q a VY ff ye w ea! Xie s ' A 4 4 Fig. 117. Fie, 118. and the horizontal projection would therefore have been parallel to the ground line. Also, a point in the horizontal plane would have fixed the traces, instead of a point in the vertical plane as shown in the problem. 817. Problem 11. To pass a plane through three given points not in the same straight line. : Analysis. If two of the points be joined by a line, a plane may be passed through this line and revolved so that it contains the third point. In this position, the plane will contain a line joining the third point with any point on the first line. Hence, join two points by a line, and from any point on this line, draw another line, through the remaining point. Find the piercing points of these two lines, and thus establish the traces of the Tequired plane. 110 GEOMETRICAL PROBLEMS IN PROJECTION Construction. Let AB and C, in Fig. 119, be the three given points. Join A and B and find where this line pierces the prin- cipal planes at d and e’.. Assume any point H, on the line AB, and join H and C; this line pierces the principal planes at g and f’, Join dg and f’e’ and the traces obtained are those of the required plane. A check on the accuracy of the work is furnished by the fact that both traces must meet at one point on the ground line, as shown at T. Hence, the plane T contains the points A, B and C. Fig. 120 is an oblique projection of this problem. Note. In the construction of this and other problems, it may be desirable to work the problem backwards in order to Fia. 119. Fig. 120. obtain a simpler drawing. It is quite difficult to select three points of a plane, at random, so that the traces of the plane shall meet the ground line within the limits of the drawing. In working the problem backwards, the traces are first assumed, then, any two distinct lines are drawn in the plane, and, finally, three points are selected on the two assumed lines of the plane. It is good practice, however, to assume three random points and proceed with the problem in the regular way. Under these conditions, the piercing points are liable to be in any angle, and, as such, furnish practice in angles other than the first. 818. Problem 12. To revolve a given point, not in the principal planes, about a line lying in one of the principal planes. Analysis. If a point revolves about a line, it describes a circle, the plane of which is perpendicular to the axis of revolu- THE POINT, THE LINE, AND THE PLANE 111 tion. As the given line is the axis, the point will fall somewhere in the trace of a plane, through the point, perpendicular to the axis. The radius of the circle is the perpendicular distance from the point to the line; and is equal to the hypothenuse of a tri- angle, whose base is the distance from the projection of the point to line in the plane, and whose altitude is the distance of the corresponding projection of the point from the plane containing the line (707). Construction. Assume that the given line AB, Fig. 121, lies in the horizontal plane and therefore is its own projection, ab, in that plane; its corresponding projection is a’b’ and lies in the ground line. Also, let C be the , given point, shown by its projections c *h and c’. Through c, draw cp, perpendic- yy ular to the line ab; cp.is then the trace a’ a , somewhere along this line. The radius of i, % ce the circle is found by making an auxiliary \e" view, in which c’o is the distance of the Fic. 121. point above the horizontal plane, and oq=cp is the distance of the horizontal projection of the given point from the axis. Hence, c’q is the radius of the circle, and, therefore, lay off pe’’=c’g. The revolved position of the point C in space is, therefore, c’’. The distance pce” might have been revolved in the opposite direction and thus would have fallen on the opposite side of- the axis. This is immaterial, however, as the point is always revolved so as to make a clear diagram. Nore. In the construction of this problem, the line was assumed as lying in the horizontal plane. It might have been assumed as lying in the vertical plane, and, in this case, the operations would have been identical, the only difference being that the trace of the plane containing the path of the point would then lie in the vertical plane. The auxiliary diagram for deter- mining the radius of the circle may be constructed below the ground line, or, if desirable, in an entirely separate diagram. 1 | 1 | xe—7 of the plane of the revolving point and Hele jee the revolved position of the point will fall LA | | 819. Problem 13: To find the true distance between two points in space as given by their projections. First method. Analysis. The true distance is equal to the length of the 112 GEOMETRICAL PROBLEMS IN PROJECTION line joining the two points. If, then, one projecting plane of the line be revolved until it is parallel to the corresponding plane of projection, the line will be shown in its true length on the plane to which it is parallel. Construction. Case 1. When both points are above the plane of projection—Let AB, in Fig. 122, be the given line. For convenience, revolve the horizontal projecting plane about the projecting perpendicular from the point A on the line. The point B will describe a circle, the plane of which is perpendicular to the axis about which it revolves. As the plane of the circle (or arc) is parallel to the horizontal plane, it is projected as the arc be. The corresponding projection is b’c’, because its J Fig. 122. Fia. 128. plane is perpendicular to the vertical plane. In the position ac, the projecting plane of AB is parallel to the vertical plane, and c’ is the vertical projection of the point b, when so revolved. Hence, a’c’ is the true length of the line AB in space. 820. Case 2. When the points are situated on opposite sides of the principal plane.—Let AB, in Fig. 123, be the given line. Revolve the horizontal projecting plane of the line about the horizontal projecting perpendicular from the point A on the line. The point B will describe an are which is horizontally pro- jected as be and vertically projected as b’c’. The ultimate position of b’ after revolution is at c’, whereas a’ remains fixed. Hence, a’c’ is the true length of the line AB. THE POINT, THE LINE, AND THE PLANE 113 821. Problem 13. To find the true distance between two points in space as given by their projections. Second method. Analysis. The true distance is equal to the length of a line joining the two points. If a plane be passed through the line and revolved about the trace into one of the principal-planes, the distance between the points will remain unchanged, and in its revolved position, the line will be shown in its true length. Construction. Case 1. When both points are above the plane of projection.—Let A and B Fig. 124 be the two points in question. For convenience, use the hori- zontal projecting plane of the line as the revolving plane; its trace on the horizontal —— plane is ab, which is also the projection of the line. The points A and B, while yo Ss? revolving about the line ab, will describe circles, the planes of which are perpendic- ular to the axis, and, hence, in their re- a volved position, will lie along lines aa” and bb’. In this case, the distance of the \ projections of the points from the axis is A zero, because the projecting plane of the ae. 124. line joining the points was used. The altitudes of the triangles are the distances of the points above the horizontal plane, and, hence, they are also the hypothenuses of the triangles which are the radii of the circles.* Therefore, lay off aa’ =a’o and bb’=b’p along lines aa” and bb”, which are perpendicular to ab. Hence, ab’ is the true distance between the points A and B in space. 822. Case 2. When the points are situated on opposite sides of the principal plane——Let AB, Fig. 125, be the two points. From the projections it will be seen that A is in the first angle and B is in the fourth angle. If, as in the previous case, the horizontal projecting plane is used as the revolving plane then the horizontal Fic. 125. projection ab is the trace of the revolv- ing plane as before. The point A falls to a’ where aa”=a’o. Similarly, B falls to b’’ where bb’ =b’p, * See the somewhat similar case in Problem 12. , ” 5 114 GEOMETRICAL PROBLEMS IN PROJECTION but, as must be noticed, this point falls on that side of ab, oppo- site to the point a”. A little reflection will show that such must be the case, but it may be brought out by what follows: The line AB pierces the horizontal plane at m and this point must remain’ fixed during the revolution. That it does, is shown by the fact that the revolved position of the line a’’b” passes through this point. Norr. In both cases, the vertical projecting plane might have been used as the revolving plane and after the operation is performed, the true length of the line is again obtained. It must of necessity be equal to that given by the method here indicated. 823. Problem 14. To find where a given line pierces a given plane. Analysis. If an auxiliary plane be passed through the Fig. 126. Fig. 127. given line, so that it intersects the given plane, it will cut from it a line that contains the given point. The given point must also lie on the given line, hence it lies on their intersection. Construction. Let T, Fig. 126, be the given plane and AB the given line. For convenience, use the horizontal projecting plane of the line as the auxiliary plane; the horizontal trace is cb and the vertical trace is cc’, The auxiliary plane cuts from the plane T the line CD. The vertical projection is shown as c’d’ and the horizontal projection cd is contained in the trace of the horizontal projecting plane, because that plane was pur- THE POINT, THE LINE, AND THE PLANE 115 posely taken as the cutting plane through the line. It is only in the vertical projection that the intersection m’ is determined; its horizontal projection m is indeterminate in that plane (except from the fact that it is a corresponding projection) since the given line and the line of intersection have the same horizontal projecting plane. An oblique projection of this problem is given in Fig. 127. 824. Problem 15. To find the distance of a given point from a given plane. Analysis. The perpendicular distance from the given point to the given plane is the required distance. Hence, draw a per- pendicular from the given point to the given plane, and find where this perpendicular pierces the given plane. If the line joining the given point and the piercing point be revolved into one of the planes of projection, the line will be shown in its true length. Construction. Let T, Fig. 128, be the given plane, and A, the given point. From A, draw AB, perpendicular to the plane T; the pro- jections of AB are therefore perpendicu- lar to the traces of the plane. If the horizontal projecting plane of the line AB be used as the auxiliary plane, it cuts the given plane in the line CD, and pierces it at the point B. Revolve the projecting plane of AB about its horizontal trace ab into the horizontal plane of projection. A will fall to a”, where aa’’=a’o, and B will fall to b”, where bb’’=b’p. Therefore, a’’b” is the a distance from the point A to the plane T. Fie. 128. 825. Problem 16. To find the distance from a given point to a given line. Analysis. Through the given point pass a plane perpen- dicular to the given line. The distance between the piercing point of the given line on this plane and the given point is the required distance. Join these two points by a line and revolve this line into one of the planes of projection; the line will then be seen in its true length. 116 GEOMETRICAL PROBLEMS IN PROJECTION Construction. Let AB, Fig. 129, be the given line, and G, the given point. Through G, draw a plane perpendicular to AB (816), by drawing ge perpendicular to ab, and g’c’, parallel to the ground line; the piercing point ye of this line on the vertical plane , ’ is c’.. Hence, the traces Tt’ and ¢ Tt, perpendicular to a’b’ and ab, respectively, through the point c’, will be the traces of the required io] 2 y plane. AB pierces this plane at : B, found by using the horizontal yy, 9g projecting plane of AB as a cutting d S plane, this cutting plane inter- ¢ Se / secting in a lme DE. The point B must be on both AB and DE. 9 The projected distance, then, is Fic. 129. the distance between the points B and G; the true distance is found by revolving BG into the horizontal plane. The latter operation is accomplished by using the horizontal projecting plane of BG and revolving it about its trace bg; G falls to g”, where gg” = g’p, and B falls to b’, where bb’’=b’o. Thus, g’b” is the true distance between the point G and the line AB. 826. Problem 17. To find the angle between two given intersecting lines. Analysis. If a plane be passed through these lines and revolved into one of the planes of projection, the angle will be shown in its true size. Hence, find the piercing points of the given lines on one of the planes of projection; the line joining these piercing points will be the trace of the plane containing the lines. Revolve into that plane and the revolved position of the two lines shows the true angle. Construction. Let AB and AC, Fig. 130, be the two given lines intersecting at A. These lines pierce the horizontal plane of projection at b and c and be is the trace of a plane containing the two lines. If A be considered as a point revolving about the line be, it then describes a circle, the plane of which is per- pendicular to be and the point A will coincide with the horizontal plane somewhere along the line 0a”. The radius of the circle THE POINT, THE LINE, AND THE PLANE 117 described by A is equal to the hypothenuse of a right triangle, where the distance ao, the projec- tion of a from the axis, is the base, and a’p, the distance of the point above the plane, is the altitude. This is shown in the triangle a’pq, where pq is equal to ao and therefore a’q is the required radius. Hence, make oa’’=a’q, and a” is the re- volved position of the point A in space. The piercing points b and c of the given lines do not change their relative positions. Thus a’’b and a”’c are the revolved position of the given lines, and the angle ba’’c is the true angle. Fria. 130. 827. Problem 18. To find the angle between two given planes. Analysis. If a plane be passed perpendicular to the line of intersection of the two given planes, it will cut a line from each plane, the included angle of which will be the true angle. Re- volve this plane, containing the lines, about its trace on the principal plane, until it coincides with that plane, and the angle will be shown in its true size. Construction. Let T and S, Fig. 131, be the two given planes, intersecting, as shown, in the line AB. Construct a sup- plementary plane to the right of the main diagram. H’H’ is the new horizontal plane, shown as a line parallel to ab. The line AB pierces the vertical plane at a distance a’a above the horizontal plane. Accordingly, a’a is laid off on V’V’, perpendicu- lar to H’H’; it also pierces the horizontal plane at b shown in both views. The supplementary view shows ba’ in its true relation to the horizontal plane, and is nothing more or less than a side view of the horizontal projecting plane of the line AB. If, in this supplementary view, a perpendicular plane cd be drawn, it will intersect the line AB in c, and the horizontal plane in the trace dfe shown on end. The lettering in both views is such that similar letters indicate similar points. Hence, dfe is the trace of the plane as shown in the main diagram, and ec and dc are the two lines cut from the planes T and S by the plane 118 GEOMETRICAL PROBLEMS IN PROJECTION cde. When the plane cde is revolved into coincidence with the horizontal plane, c falls to c’’ in the supplementary view and is projected back to the main diagram as ce’. Therefore, ec’’d is the true angle between the planes, because e and d remain fixed in the revolution. 828. Problem 19. To find the angle between a given plane and one of the principal planes. : Analysis. If an auxiliary plane be passed through the given plane and the principal plane so that the auxiliary plane is per- f - S\al 4! Fig. 131. pendicular to the intersection of the given plane and the prin- cipal plane, it will cut from each a line, the included angle of which will be the true angle. If, then, this auxiliary plane be revolved into the principal plane, the angle will be shown in its true size. Construction. Let T, Fig. 132, be the given plane. The angle that this plane makes with the horizontal plane is to be determined. Draw the auxiliary plane R, so that its horizontal trace is perpendicular to the horizontal trace of the given plane; the vertical trace of the auxiliary plane must as a consequence be perpendicular to the ground line as Rr’. A triangle rRr’ is THE POINT, THE LINE, AND THE PLANE 119 cut by the auxiliary plane from the given plane and the two principal planes. If this triangle be revolved into the horizontal plane, about rR as an axis, the point r’ will fall to r’ with Rr’ as aradius. Also, the angle rRr’” must be a right angle, because it is cut from the principal planes, which are at right angles to each other. Hence, Rrr” is the angle which the plane T makes with the horizontal plane of projection. The construction for obtaining the angle with the vertical plane is identical, and is shown on the right-hand side, with plane S as the given plane. All construction lines are added and no comment should be necessary. Nots. The similarity of Probs. 18 and 19 should be noted. In Prob. 19 the horizontal trace is the intersection of the given a ae” , Fre. 132. plane and the horizontal plane; hence, the auxiliary plane is passed perpendicular to the trace. A similar reasoning applies to the vertical trace. 829. Problem 20. To draw a plane parallel to a given plane, at a given distance from it. Analysis. The required distance between the two planes is the perpendicular distance, and the resulting traces must be parallel to the traces of the given plane. If a plane be passed perpendicular to either trace, it will cut from the principal planes and the given plane a right angled triangle, the hypothenuse of which will be the line cut from the given plane. If, further, the triangle be revolved into the plane containing the trace and the required distance between the planes be laid off perpendicular to the hypothenuse cut from the given plane, it will establish a point on the hypothenuse of the required plane. A line parallel 120 GEOMETRICAL PROBLEMS IN PROJECTION to the hypothenuse through the established point will give the revolved position of a triangle cut from the required plane. On the counter revolution, this triangle will determine a point in each plane, through which the required traces must pass. Hence, if lines be drawn through the points so found, parallel to the traces of the given plane, the traces of the required plane are established. Construction. Let T, Fig. 133, be the given plane, and rg the required distance between the parallel planes. Pass a plane rOr’, perpendicular to the hori- zontal trace Tt; its vertical trace Or is, therefore, perpendicular to the ground line. The re- volved position of the triangle cut from the two principal planes and the given plane is rOr”. Lay off r’’g, perpendicular to rr’, and equal to the required distance between the planes. Draw uu” parallel to rr” and Fic. 133. the triangle cut from the prin- cipal planes and the required plane is obtained. On counter revolution, u’’ becomes u’ and u remains fixed; Su’ and Su, parallel respectively to Tr’ and Tr, are the required traces. Therefore, the distance between the planes T and S is equal to r’’g. 830. Problem 21. To project a given line on a given plane.. Analysis. If perpendiculars be dropped from the given line upon the given plane, the points, so found, are the projec- tions of the corresponding points on the line. Hence, a line joining the projections on the given plane is the required pro- jection of the line on that plane. Construction. Let T, Fig. 134, be the given plane, and AB the given line. From A, draw a perpendicular to the plane T; its horizontal projection is ac and its vertical projection is a’c’. To find where AC pierces the given plane, use the hori- zontal projecting plane of AC as the cutting plane; FE is the line so cut, and C is the resultant piercing point. Thus, C is the projection of A on the plane T. A construction, similar in THE POINT, THE LINE, AND THE PLANE 121 detail, will show that D is the projection of B on the plane T. Hence, CD is the projection of AB on the plane T. 831. Problem 22. To find the angle between a given line and a given plane. Analysis. The angle made by a given line and a given plane is the same as the angle made by the given line and its projection \ Fig. 134. on that plane. If from any point on the given line, another line be drawn parallel to the projection on the given plane, it will also be the required angle. The projection of the given line on the. given plane is perpendicular to a projecting perpendicular from the given line to the given plane. Hence, any line, parallel to the projection and lying in the projecting plane of the given line to the given plane is also perpendicular to this projecting perpendicular. Therefore, pass a plane through the given line and the projecting perpendicular from the given line to the given plane. Revolve this plane into one of the planes of pro- 122 GEOMETRICAL PROBLEMS IN PROJECTION jection, and, from any point on the line, draw a perpendicular to the projecting perpendicular. The angle between this line and the given line is the required angle. Construction. Let T, Fig. 135, be the given plane, and AB the given line. From B, draw a perpendicular to the plane T, by making the projections respectively perpendicular to the traces of the given plane. Find the piercing points e and f, on the horizontal plane, of the given line and this perpendicular. Revolve the plane containing the lines BE and BF; B falls to b”, on a line b’”p, perpendicular to ef. The distance b’p is equal to the hypothenuse of a right triangle, where bp is the base and J o ~ fs a, ' ~. SS ~. -7 v4 7 “ ~ S. | Qs--— Fig, 135. b’o is the altitude; b’oq is such a triangle, where og=bp. Hence, b’’p is laid off equal to b’q. If from any point d, a line de be drawn, perpendicualr to b’f, then b’dc is the required angle, as de is parallel to the projection of AB on the plane T in its revolved position. 832. Problem 23. To find the shortest distance between a pair of skew * lines. Analysis. The required line is the perpendicular distance between the two lines. If through one of the given lines, another line be drawn parallel to the other given line, the intersecting lines will establish a plane which is parallel to one of the given lines. *Skew lines are lines which are not parallel and which do not intersect. THE POINT, THE LINE, AND THE PLANE 123 ‘The length of a perpendicular from any point on the one given line to the plane containing the other given line, is the required distance. i Construction. Let AB and CD, Fig. 136, be the two given O€j- anes eb ee SO Fia. 136. lines. Through any point O, on AB, draw FE, parallel to CD, and determine the piercing points of the lines AB and FE; a, e and f’,b’ are these piercing points, and, as such, determine the plane T. CD is then parallel to the plane T. From any point G, on CD, draw GH, perpendicular to the plane T, and 124 GEOMETRICAL PROBLEMS IN PROJECTION find its piercing point on that plane. This point is H, found by drawing gh perpendicular to Tt and g’h’ perpendicular to Tt’; the horizontal projecting plane cuts from the plane T, a line MN, on which is found H, the piercing point. GH is there- fore the required distance, but to find its true length, revolve \" Fic. 136. the horizontal projecting plane of GH into the horizontal plane. On revolution, H falls to h’’, where hh” is equal to the distance h’ above the ground line; and, similarly, G falls to g’”. There- fore, h’’g” is the true distance between the lines AB and CD. THE POINT, THE LINE, AND THE PLANE 125 Note. In order to find the point on each of the lines at which this perpendicular may be drawn, project the one given line on the plane containing the other. Where the projections cross, the point will be found. ADDITIONAL CONSTRUCTIONS 833. Application to other problems. The foregoing prob- lems may be combined so as to form additional ones. In such cases, the analysis is apt to be rather long, and in the remaining few problems it has been omitted. The construction of the Bde \ \ 4 \ \ @ a 7 ge $ “ee i SS 7 es / a td Fic. 137. problem might be followed by the student and then an analysis worked up for the particular problem afterward. 834. Problem 24. Through a given point, draw a line of a given length, making given angles with the planes of projection. Construction. Consider the problem solved, and let AB, Fig. 137, be the required line, through the point A. The con- struction will first be shown and then the final position of the line AB will be analytically considered. From a’, draw a’c’ of a length 1, making the angle « with the ground line, and draw ac, parallel to the ground line. The horizontal projecting plane of AB has been revolved about the horizontal projecting perpen- dicular, until it is parallel to the vertical plane, and, therefore 126 GEOMETRICAL PROBLEMS IN PROJECTION a’c’ is shown in its true length and inclination to the horizontal plane. On the counter-revolution of the projecting plane of AB, it will be observed that the point A remains fixed, because it. lies in the axis; B describes a circle, however, whose plane is parallel to the horizontal plane, antl is, therefore, projected as the arc cb, while its vertical projection (or trace, if it be consid- ered as a plane instead of the moving point) is b’c’, a line parallel to the ground line. The angle that the line AB makes with the horizontal plane is now fixed, but the point B is not finally located as the remaining condition of making the angle @ with the hori- zontal plane is yet conditional. In order to lay off the angle that the line makes with the Um x qi x Fia. 137. vertical plane, draw a line ad, through a, making the angle 6 with the ground line, and of a length 1. From a’, draw a’d’, so that d’ is located from its corresponding projection d. Draw bd, through d, parallel to the ground line, and where its inter- section with the arc cb, locates b, the final position of the hori- zontal projection of the actual line. The corresponding pro- jection b’, may be located by drawing the arc d’b’, and finding where it intersects the line b’c’, through c’, parallel to the ground line. The points b and b’ should be corresponding projections, if the construction has been carried out accurately. In reviewing the latter process, it is found that the vertical projecting plane THE POINT, THE LINE, AND THE PLANE 127 of the line has been revolved about the vertical projecting per- pendicular through A, until it was parallel to the horizontal plane. The horizontal projection is then ad, and this is shown in its true length and inclination to the vertical plane. It may also be noted that the process of finding the ultimate position of the line is simply to note the projections of the path of the moving point, when the projecting planes of the line are revolved. That is, when the horizontal projecting plane of the line is revolved, the line makes a constant angle with the hori- zontal plane; and the path of the moving point is indicated by its projections. Similarly, when the vertical projecting plane of the line is revolved, the line makes a constant angle with the vertical plane; and the path of the moving point is again given 7 B b we ee a 4 NK, | I 8 1 I ! i 5 Ls a Y . | I ' x | | / Z Fig. 138. by its projections. Where the paths intersect, on the proper planes, it is evidently the condition that satisfies the problem. There are four possible solutions for any single point in space, and they are shown in Fig. 138. ach is of the required length, and makes the required angles with the planes of projection. The student may try to work out the construction in each case and show that it is true. 835. Problem 25. Through a given point, draw a plane, making given angles with the principal planes. Construction. Prior to the solution of this problem, it is desirable to investigate the property of a line from any point. on the ground line, perpendicular to the plane. If the angles that this line makes with the principal planes can be determined, the actual construction of it, in projection, is then similar to the preceding problem. To draw the required plane, hence, resolves itself simply into passing a plane through a given point, per- pendicular to a given line (816). 128 GEOMETRICAL PROBLEMS IN PROJECTION Let, in Fig. 189, T be the required plane, making the required angles a with the horizontal plane, and @ with the vertical plane. To find the angle made with the horizontal plane, pass a plane perpendicular to the horizontal trace as AOB, through any as- sumed point O on the ground line; OA is thus perpendicular to Tt and OB is perpendicular to the ground line. This plane cuts from the plane T, a line AB, and the angle BAO is the required angle « Similarly, pass a plane COD, through O perpen- dicular to the vertical trace, then CO is perpendicular to Tt and OD is perpendicular to the ground line; DCO is the required _ angle @, which the plane T makes with the vertical plane. The planes AOB and COD intersect inaline OP. OP is perpendicular Fie. 139. to the plane T because each plane AOB and COD is perpendicular to the plane T (since they are each perpendicular to a trace, which is a line in the plane), and, hence, the line common to the two planes (OP) must be perpendicular to the plane. The angles OPC and OPA are, therefore, right angles, and, as a result, angle POA=90°—«, and angle POC=90°—®@. Hence, to draw a perpendicular to the required plane, draw a line making angles with the principal planes equal to the complements (90° minus the angle) of the corresponding angles. It is evident that any line will do, as all such lines, when measured in the same way, will be parallel, hence, it is not necessary, although convenient, that this line should pass through the ground line. To complete the problem, let AB, Fig. 140, be a line making THE POINT, THE LINE, AND THE PLANE 129 angles 90°—a with the horizontal plane, and 90°—8 with the vertical plane. Through G, the given point, draw a perpendicular plane to AB, and the resultant plane T is the required plane , , * d £ "1 meee NM $ a ‘ g a \r_y a oe x Sif Sh 6 Fic. 140. making an angle «, with the horizontal plane, and an angle 8, with the vertical plane. Nors. As there are four solutions to the problem of drawing a line making given angles with the principal planes, there are also four solutions to this problem. The student may show these cases and check the accuracy by finding the angle between a given plane and the principal planes (828). 836. Problem 26. Through a given line, ina given plane, draw another line, intersecting it at a given point, and ata given angle. Construction. Let AB, Fig. 141, be the given line, T the given plane and G the given point. Revolve the limited portion of the plane tTt’ into coincidence with the horizontal plane. Tt remains fixed but Tt’ revolves to Tt”. To find the direction of Tt’’, consider any point b’ on the original position of the trace Tt’. The distance Tb’ must equal Tb” as 130 GEOMETRICAL PROBLEMS IN PROJECTION this length does not change on revolution; since Tt is the axis of revolution, the point B describes a circle, the plane of which is perpendicular to the axis, and, therefore, b’’ must also lie on a line bb”, from b, perpendicular to the trace Tt. In the revolved position of the plane, the point a remains unchanged, while B goes to b’. Hence, ab” is the revolved position of the given line AB. The given point G moves to g”, on a line gg”, perpendicular to Tt and must also be on the line ab’. Through g”, draw the line cd”, making the required angle a with it. On counter-revolution, c remains fixed, and d” moves to d’.. Thus CD is the required line, making an angle a with another line AB, and lying in the given plane T. SNe 3 /o-t Fie. 142. 837. Problem 27. Through a given line in a given plane, pass another plane, making a given angle with the given plane. Construction. Let T, Fig. 142, be the given plane, AB the given line in that plane, and « the required angle between the planes. Construct a supplementary view of the line AB; H’'H’ is the new horizontal plane, the inclination of ab’ is shown by the similar letters on both diagrams. The distance of b’ above the horizontal plane must also equal the distance b’ above H’H’ and so on. Through h, in the supplementary view, draw a plane hf perpendicular to ab’. Revolve h about f to g and locate g as shown on the horizontal projection ab, of the main THE POINT, THE LINE, AND THE PLANE 131 diagram. The line eg is cut from the plane T, by the auxiliary plane egf, hence, lay off the angle # as shown. This gives the direction gf of the line cut from the required plane. The point f lies on gf and also on ef which is perpendicular to ab. Hence, join af and produce to S; join S and b’ and thus estab- ! nm t uw L Ne gilt wm Fia. 143. lish the plane S. The plane S passes through the given line AB and makes an angle a with the given plane T (827). 838. Problem 28. To construct the projections of a circle lying in a given oblique plane, of a given diameter, its centre in the given plane being known. Construction. Let T, Fig. 143, be the given plane, and G 132 GEOMETRICAL PROBLEMS IN PROJECTION the given point lying in the plane T. When g’, is assumed, for instance, g is found by drawing a horizontal line g’a’ and then ga is its corresponding projection; g can therefore be determined as shown. When the plane of a circle is inclined to a plane of mw t v7 nm l ™m NE fyi mn Fig. 143. projection, it is projected as an ellipse. An ellipse is determined, and can be constructed, when its major and minor axes are given.* To construct the horizontal projection, revolve the plane T about Tt, until it coincides with the horizontal plane as Tt”. The direction of Tt’ is found by drawing aa’’ perpendicular to Tt and laying off Ta’=Ta” (836). The centre of the circle in * For methods of constructing the ellipse see Art. 906. THE POINT, THE LINE, AND THE PLANE 133 its revolved position is found by drawing a”g’’ parallel to Tt and gg’’ perpendicular to Tt; g” is this revolved position. With the given radius, draw c’’d’e’’f”, the circle of the given diameter. Join e’’d’’ and f’’c’’; prolong these lines to o and q, and also draw g’’p parallel to these. Thus, three parallel lines in the revolved position of the plane T, are established and on counter revolution, they will remain parallel. The direction gp, of one of them is known, hence, make fq and eo parallel to gp. The points cdef are the corresponding positions of c’d’e’’f” and determine the horizontal projection of the circle. The line ec remains equal to ec” but fd is shorter than fd’, hence, the major and minor axes of an ellipse (projection of the circle) are . determined. The ellipse may now be drawn by any convenient method and the horizontal projection of, the circle in the plane T will be complete. By revolving the plane T into coincidence with the vertical plane, Tt’’”’ is found to be the revolved position of the trace Tt and g’”, the revolved position of the centre. The construction is identical with the construction of the horizontal projection and will become apparent, on inspection, as the necessary lines are shown indicating the mode of procedure. As a result, k’I’m‘n’ determine the major and minor axes of the ellipse, which is the vertical projection of the circle in the plane T. Nors. As a check on the accuracy of the work, tangents may be drawn in one projection and the corresponding projection must be tangent at the corresponding point of tangency. QUESTIONS ON CHAPTER VIII om . Mention the three distinct steps into ors the solution of a problem may be divided. What is the statement of a problem? What is the analysis of a problem? What is the construction of a problem? What type of projection is generally used in the construction of a problem? Sue Co NS Note. In the following problems, the construction, except in a few isolated cases, is to be entirely limited to the first angle of pro- jection. 6. Draw a line through a given point, parallel to a given line. Cive analysis and construction. 134 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27, 28. 29. 30. 31. 32. GECMETRICAL PROBLEMS IN PROJECTION . Draw a line intersecting a given line at a given point. Give analysis and construction. . Find where a given oblique line pierces the planes of projection. Give analysis and construction. . Transfer the diagram of Question 8 to oblique projection. . Pass an oblique plane through a given oblique line. Give analysis and construction. Make an oblique projection of the diagram in Question 10. Pass a plane obliquely through the principal planes and through a line parallel to the ground line. Give analysis and construction. Hint: Use a profile plane in the construction. Transfer the diagram of Question 12 to an oblique projection. Pass an oblique plane through a line which is parallel to the hori- zontal plane but inclined to the vertical plane. Give analysis and construction. Transfer the diagram of Question 14 to an oblique projection. Pass an oblique plane through a line which is parallel to the vertical plane but inclined to the horizontal plane. Give analysis and construction. Transfer the diagram of Question 16 to an oblique projection. Pass an oblique plane through a given point. Give analysis and construction. Make an oblique projection of the diagram in Question 18. Find the intersection of two planes, oblique to each other and to the principal planes. Give analysis and construction. Transfer the diagram of Question 20 to an oblique projection. Find the intersection of two planes, oblique to each other and to the principal planes. Take the case where the traces do not intersect on one of the principal planes. Give analysis and construction. Find the corresponding projection of a given point lying in a given oblique plane, when one projection is given. Give analysis and construction. Transfer the diagram of Question 23 to an oblique projection. Draw a plane which contains a given point and is parallel to a given plane. Give analysis and construction. Transfer the diagram of Question 25 to an oblique projection. Draw a line through a given point, perpendicular to a given plane. Give analysis and construction. Transfer the diagram of Question 27 to an oblique projection. Draw a plane through a given point, perpendicular to a given line. Give analysis and construction. Transfer the diagram of Question 29 to an oblique projection. Pass a plane through three given points, not in the same straight line. Give analysis and construction. Transfer the diagram of Question 31 to an oblique projection. - Revolve a given point, not in the principal planes, about a line lying in one of the principal planes. Give analysis and construc- tion. THE POINT, THE LINE, AND THE PLANE 135 34, Transfer the diagram of Question 33 to an oblique projection and show also the plane of the revolving point. 35. Find the true distance between two points in space, when both points are in the first angle of projection. Use the method of revolving the projecting plane of the line until it is parallel to one plane of projection. Give analysis and construction. 36. Make an oblique projection of the diagram in Question 35 and show the projecting plane of the line. ~ 387. Find the true distance between two points in space, when one point is in the first angle and the other is in the fourth angle. Use the method of revolving the projecting plane of the line until it is parallel to one plane of projection. Give analysis and construction. 88. Make an oblique projection of the diagram in Question 37 and show the projecting plane of the line. 39. Find the true distance between two points in space, when both points are in the first angle of projection. Use the method of revolving the projecting plane of the line into one of the planes of projection. Give analysis and construction. 40. Transfer the diagram of Question 39 to an oblique projection and show the projecting plane of the line and the path described by the revolving points. 41. Find the true distance between two points in space, when one point is in the first angle and the other is in the fourth angle of pro- jection. Use the method of revolving the projecting plane of the line into one of the planes of projection. Give analysis and construction. 42. Transfer the diagram of Question 39 to an oblique projection and show the projecting plane of the line and the path described by _ the revolving points. 43. Find where a given line pierces a given plane. Give analysis and construction. 44, Transfer the diagram of Question 43 to an oblique projection. 45. Find the distance of a given point from a given plane. Give analysis and construction. 46. Transfer the diagram of Question 45 to an oblique projection. Omit the portion of the construction requiring the revolution of the points. 47. Find the distance from a given point to a given line. Give analysis and construction. 48. Transfer the diagram of Question 47 to an oblique projection. Omit the portion of the construction requiring the revolution of the points. 49. Find the angle between two given intersecting lines. Give analysis and construction. 50. Make an oblique projection of the diagram in Question 49. 51. Find the angle between two given planes. Give analysis and con- struction. 52. Make an oblique projection of the diagram in Question 51. 136 GEOMETRICAL PROBLEMS IN PROJECTION 53. Find the angle between a given plane and the horizontal plane of projection. Give analysis and construction. 54. Transfer the diagram of Question 53 to an oblique projection. 55. Find the angle between a given plane and the vertical plane of pro- jection. Give analysis and construction. 56. Transfer the diagram of Question 55 to an oblique projection. 57. Draw a plane parallel to a given plane at a given distance from it. Give analysis and construction. 58. Project a given line ona given plane. Give analysis and construction. 59. Make an oblique projection of the diagram in Question 58. 60. Find the angle between a given line and a given plane. Give analysis and construction. 61. Make an oblique projection of the diagram in Question 60. 62. Find the shortest distance between a pair of skew lines. Give analysis and construction. 63. Make an oblique projection of the diagram in Question 62. 64. Through a given point, draw a line of a given length, making given angles with the planes of projection. 65. Show the construction for the three remaining cases of the problem in Question 64. 66. Through a given point, draw a plane, making given angles with the. principal planes. 67. Prove that the construction in Question 66 is correct by finding the angles that the given plane makes with the principal planes. (Note that there are four possible cases of this problem.) 68. Show the construction for the three remaining cases of the problem in Question 66. 69. Prove that the constructions in Question 68 are correct by finding the angles that the given plane makes with the principal planes. 70. Through a given line, in a given plane, draw another line intersecting it at a given point, and at a given angle. 71. Transfer the diagram of Question 70 to an oblique projection. 72. Through a given line in a given plane, pass another plane, making given angles with the given plane. 73. Make an oblique projection of the diagram in Question 72. 74. Construct the projections of a circle lying in a given oblique plane, the diameter and its centre in the given plane being known. Note. The following exercises embrace operations in all four angles. 75. Given a line the first angle and a point in the second angle, draw a line through the given point, parallel to the given line. 76. Given a line in the first angle and a point in the third angle, draw a line through the given point, parallel to the given line. Given a line in the first angle and a point in the fourth angle, draw a line through the given point, parallel to the given line. 78. Given a line in the second angle and a point in the third angle, draw a line through the given point, parallel to the given line. 7 ee 79. 80. 81. 82. 83. 84. 85. 86. 87. 88. 89. 90. 91. 92. 93. 94. 95. 96. 97. 98. 99. THE POINT, THE LINE, AND THE PLANE 137 Given a line in the third angle and a point in the fourth angle, draw a line through the given point, parallel to the given line. Given a line in the fourth angle and a point in the second angle, draw a line through the given point, parallel to the given line. Draw two intersecting lines in the second angle . Draw two intersecting lines in the third angle. Draw two intersecting lines in the fourth angle. Find where a given line pierces the principal planes when the limited portion of the line is in the second angle. Find where a given line pierces the principal planes when the limited portion of the line is in the third angle. Find where a given line pierces the principal planes when the limited portion of the line is in the fourth angle. Show the second angle traces of a plane passed through a second angle oblique line. Show the third angle traces of a plane passed through a third angle oblique line. ; Show the fourth angle traces of a plane passed through a fourth angle oblique line. Given a line in the second angle and parallel to the ground line, pass an oblique plane through it. Use a profile plane as part of the construction. Given a line in the third angle and parallel to the ground line, pass an oblique plane through it. Use a profile plane as a part of the construction. Given a line in the fourth angle and parallel to the ground ling, pass an oblique plane through it. Use a profile plane as a part of the construction. Given a point in the second angle, pass an oblique plane through it. Show only second angle traces. Given a point in the third angle, pass an oblique plane through it. Show only third angle traces. Given a point in the fourth angle, pass an oblique plane through it. Show only fourth angle traces. Given the second angle traces of two oblique planes, find their intersection in the second angle. Given the third angle traces of two oblique planes, find their inter- section in the third angle. Given the fourth angle traces of two oblique planes, find their intersection in the fourth angle. Given the first angle traces of an oblique plane and one projection of a second angle point, find the corresponding projection. 100. Given the first angle traces of an oblique plane, and one projection of a third angle point, find the corresponding projection. 101. Given the first angle traces of an oblique plane, and one projection of a fourth angle point, find the corresponding projection. 102. Given the second angle traces of an oblique plane, and one projec- tion of a third angle point, find the corresponding projection. 138 103. 104. 105. 106. 107. 108. 109. 110. 111. 112. 113. 114. 115. 116. 117. 118. 119. 120. GEOMETRICAL PROBLEMS IN PROJECTION Given the third angle traces of an oblique plane, and one pro- jection of a fourth angle point, find the corresponding pro- jection. Given the first angle traces of an oblique plane, and a point in the second angle, pass a plane through the given point and parallel to the given plane. Given the first angle traces of an oblique plane, and a point in the third angle, pass a plane through the given point and parallel to the given plane. Given the first angle traces of an oblique plane, and a point in the fourth angle, pass a plane through the given point and parallel to the given plane. Given the second angle traces of an oblique plane, and a point in the third angle, pass a plane through the given point and parallel to the given plane. Given the third angle traces of an oblique plane, and a point in the fourth angle, pass a plane through the given point and parallel to the given plane. Given the first angle traces of an oblique plane and a point in the second angle, draw a line through the given point perpendicular to the given plane. Given the first angle traces of an oblique plane and a point in the third angle, draw a line through the given point perpendicular to the given plane. Given the first angle traces of an oblique plane and a point in the fourth angle, draw a line through the given point perpendicular .to the given plane. Given the second angle traces of an oblique plane and a point in the fourth angle, draw a line through the given point perpendicular to the given plane. Given the first angle traces of a plane perpendicular to the hori- zontal plane but inclined to the vertical plane, and a point in the third angle, draw a line through the given point perpendicular to the given plane. Given the third angle traces of a plane perpendicular to the vertical plane but inclined to the horizontal plane, and a point in the fourth angle, draw a line through the given point, perpendicular to the given plane. Given the first angle traces of a plane parallel to the ground line and a point in the third angle, draw a line through the given point perpendicular to the given plane. Draw a profile plane of the diagram in Question 115. Make an oblique projection of the diagram in Question 115. Given the fourth angle traces of a plane parallel to the ground line and a point in the second angle, draw a line through the given point perpendicular to the given plane. Draw a profile plane of the diagram in Question 118. Make an oblique projection of the diagram in Question 118. 121. 122. 123. 124, 125. 126. 127. 128. 129. 130. 131, 132. 133. 134. 135. 136. 137, 138. 139. 140. 141, 142, THE POINT, THE LINE, AND THE PLANE 139 Given a line in the first angle and a point in the second angle, pass plane through the given point perpendicular to the given ne. Given a line in the first angle and a point in the third angle, pass a plane through the given point perpendicular to the given line. Given a line in the first angle and a point in the fourth angle, pass a plane through the given point perpendicular to the given line. Given a line in the second angle and a point in the third angle, oe a plane through the given point perpendicular to the given ine. Given a line in the third angle and a point in the fourth angle, pass a plane through the given point perpendicular to the given line. Given three points in the second angle, pass a plane through them. Given three points in the third angle, pass a plane through them. Given three points in the fourth angle, pass a plane through them. Given two points in the first angle and one point in the second angle, pass a plane through them. Given a point in the first angle, one in the second angle, and one in the third angle, pass a plane through them. Given a point in the second angle, one in the third angle, and one in the fourth angle, pass a plane through them. Given a line in the horizontal plane and a point in the second angle, revolve the point about the line until it coincides with the hori- zontal plane. Given a line in the horizontal plane and a point in the third angle, revolve the point about the line until it coincides with the hori- zontal plane. Given a line in the horizontal plane and a point in the fourth angle revolve the point about the line until it coincides with the hori- zontal plane. Given a line in the vertical plane and a point in the second angle, revolve the point about the line until it coincides with the vertical plane. Given a line in the vertical plane and a point in the fourth angle, revolve the point about the line until it coincides with the vertical plane. Given two points in the second angle, find the true distance between them. Given two points in the third angle, find the true distance between them. Given two points in the fourth angle, find the true distance between them. Given one point in the first angle and one point in the second angle, find the true distance between them. Given one point in the first angle and one point in the third angle, find the true distance between them. Given one point in the second angle and one point in the third angle, find the true distance between them. 140 GEOMETRICAL PROBLEMS IN PROJECTION 143. Given one point in the second angle and one point in the fourth angle, find the true distance between them. 144, Given the first angle traces of a plane and a line in the second angle, find where the given line pierces the given plane. 145. Given the first angle traces of a plane and a line in the third angle, find where the given line pierces the given plane. 146. Given the first angle traces of a plane and a line in the fourth angle, find where the given line pierces the given plane. 147. Given the second angle traces of a plane and a line in the third angle, find where the given line pierces the given plane. 148. Given the second angle traces of a plane and a line in the fourth angle, find where the given line pierces the given plane. 149. Given the third angle traces of a plane and a line in the fourth angle, find where the given line pierces the given plane. 150. Given the first angle traces of a plane and a point in the second angle, find the distance from the point to the plane. 151. Given the first angle traces of a plane and a point in the third angle, find the distance from the point to the plane. 152. Given the first angle traces of a plane and a point in the fourth angle, find the distance from the point to the plane. 153. Given the second angle traces of a plane and a point in the third angle, find the distance from the point to the plane. 154. Given the second angle traces of a plane and a point in the fourth + angle, find the distance form the point to the plane. 155. Given the third angle traces of a plane and a point in the fourth angle, find the distance from the point to the plane. 156. Given a line in the first angle and a point in the second angle, find the distance from the point to the line. 157. Given a line in the first angle and a point in the third angle, find the distance from the point to the line. 158. Given a line in the first angle and a point in the fourth angle, find the distance from the point to the line. | 159. Given a line in the second angle and a point in the third angle, find the distance from the point to the line. 160. Given a line in the second angle and a point in the fourth angle, find the distance from the point to the line. 161. Given a line in the third angle and a point in the fourth angle, find the distance from the point to the line. 162. Given two intersecting lines in the second angle, find the angle between them. 163. Given two intersecting lines in the third angle, find the angle between them. 164. Given two intersecting lines in the fourth angle, find the angle between them, 165. Given two intersecting lines, one in the first and one in the second angle, find the angle between them. 166. Given two intersecting lines, one in the first and one in the third angle, find the angle between them. 167. 168. 169. 170. 171. 172. 173, 174, 175. 176. 177. 178, 179. 180. 181. 182. 183. 184, 185. 186. 187. 188. 189. 190. THE POINT, THE LINE, AND THE PLANE 141 Given two intersecting lines, one in the first and one in the fourth angle, find the angle between them. Given two intersecting lines, one in the second and one in the third angle, find the angle between them. Given two intersecting lines, one in the second and one in the fourth angle, find the angle between them. Given two intersecting lines, one in the third and one in the fourth angle, find the angle between them. Given the second angle traces of two intersecting planes, find the angle between them. Given the third angle traces of two intersecting planes, find the angle between them. Given the fourth angle traces of two intersecting planes, find the angle between them. Given the first angle traces of one plane and the second angle traces of another plane, find the angle between them. Given the first angle traces of one plane and the third angle traces of another plane, find the angle between them. Given the first angle traces of one plane and the fourth angle traces of another plane, find the angle between them. Given the second angle traces of one plane and the third angle traces of another plane, find the angle between them. Given the second angle traces of one plane and the fourth angle traces of another plane, find the angle between them. Given the third angle traces of one plane and the fourth angle traces of another plane, find the angle between them. Given a plane in the second angle, find the angle between the given plane and the horizontal plane. Given a plane in the second angle, find the angle between the given plane and the vertical plane. Given a plane in the third angle, find the angle between the given plane and the horizontal plane. Given a plane in the third angle, find the angle between the given plane and the vertical plane. Given a plane in the fourth angle, find the angle between the given plane and the horizontal plane. Given a plane in the fourth angle, find the angle between the given « plane and the vertical plane. Given the second angle traces of a plane, draw another parallel plane at a given distance from it. Given the third angle traces of a plane, draw another parallel plane at a given distance from it. Given the fourth angle traces of a plane, draw another parallel plane at a given distance from it. Given the first angle traces of a plane and a line in the second angle, project the given line on the given plane. Given the first angle traces of a plane and a line in the third angle, project the given line on the given plane. 142 191. 192. 193. 194. 195. 196. 197. 198. 199. 200. 201. 202. 203. 204. 205. 206. 207. 208. 209. 210. 211. 212, 213. 214. GEOMETRICAL PROBLEMS IN PROJECTION Given the first angle traces of a plane and a line in the fourth angle, project the given line on the given plane. Given the second angle traces of a plane and a line in the third angle, project the given line on the given plane. Given the second angle traces of a plane and a line in the fourth angle, project the given line on the given plane. Given the third angle traces of a plane and a line in the fourth angle, project the given line on the given plane. Given the fourth angle traces of a plane and a line in the second angle, project the given line on the given plane. Given the first angle traces of a plane and a line in the second angle, find the angle between the given line and the given plane. Given the first angle traces of a plane and a line in the third angle, find the angle between the given line and the given plane. Given the first angle traces of a plane and a line in the fourth angle, find the angle between the given line and the given plane. Given the second angle traces of a plane and a line in the third angle, find the angle between the given line and the given plane. Given the second angle traces of a plane and a line in the fourth angle, find the angle between the given line and the given plane: Given the third angle traces of a plane and a line in the fourth angle, find the angle between the given line and the given plane. Given the fourth angle traces of a plane and a line in the second angle, find the angle between the given line and the given plane. Given the fourth angle traces of a plane and a line in the third angle, find the angle between the given line and the given plane. Given a line in the first angle and another line in the second angle (skew lines). find the shortest distance between them. Given a line in the first angle and another line in the third angle (skew lines), find the shortest distance between them. Given a line in the first angle and another line in the fourth angle (skew lines), find the shortest distance between them. Given a line in the second angle and another line in the third angle (skew lines), find the shortest ditstance between them. Given a line in the second angle and another line in the fourth angle (skew lines); find the shortest distance between them. Given a line in the third angle and another line in the fourth angle (skew lines), find the shortest distance between them. Through a given point in the second angle, draw a line of a given length, making given angles with the planes of projection. Through a given point in the third angle, draw a line of a given length, making given angles with the planes of projection. Through a given point in the fourth angle, draw a line of a given length, making given angles with the planes of projection. Through a given point in the second angle, draw a plane making given angles with the principal planes. Through a given point in the third angle, draw a plane making given angles with the principal planes. THE POINT, THE LINE, AND THE PLANE 143 215. Through a given point in the fourth angle, draw a plane making given angles with the principal planes. 216. Given the second angle traces of a plane, a line, and a point in the plane, draw another line through the given point, making a given angle with the given line. 217. Given the third angle traces of a plane, a line. and a point in the plane, draw another line through the given. point, making a given angle with the given line. 218. Given the fourth angle traces of a plane, a line and a point in the plane, draw another line through the given point, making a given angle with the given line. 219. Given the second angle traces of a plane, and a line in the plane, draw through the line another plane making a given angle with the given plane. 220. Given the third angle traces of a plane, and a line in the plane, draw through the line another plane making a given angle with the given plane. 221. Given the fourth angle traces of a plane, and a line in the plane, draw through the line another plane making a given angle with the given plane. 222. Given the second angle traces of a plane, the diameter and the centre of a circle, construct the projections of the circle. 223. Given the third angle traces of a plane, the diameter and the centre of a circle, construct the projections of the circle. 224. Given the fourth angie traces. of a plane, the diameter, and the centre of a circle, construct the projections of the circle, CHAPTER IX CLASSIFICATION OF LINES 901. Introductory. Lines are of an infinite variety of forms. The frequent occurrence in engineering of certain varieties makes it desirable to know their properties as well as their method of construction. It must be remembered that lines and points are mathematical concepts and that they have no material existence. That is to say, a line may have so many feet of length but as it has no width or thickness, its volume, therefore, is zero. Hence, it cannot exist except in the imagination. Like- wise, a point is a still further reduction and has position only; it has no dimensions at all. Of course, the material representa- tion of lines and points requires finite Cimensions, but when speaking of them, or representing them, it is the associated idea, rather than the representation, which is desired. 902. Straight line. A straight line may be defined as the shortest distance between two points.* It may also be described as the locus (or path) of a generating point which moves in the same direction. Hence, a straight line is fixed in space by two points, or, by a point and a direction. 903. Singly curved line. A singly curved line is the locus (plural :-loci) of a generating point which moves in a varying direction but remains in a single plane. Sometimes the singly curved line is called a plane curve because all points on the curve must lie in the same plane. 904. Representation of straight and singly curved lines. Straight and singly curved lines are represented by their pro- jections. When singly curved lines are parallel to the plane * Frequently, this is called a right line. There seems no reason, however, why this new nomenclature should be used; hence, it is here avoided. 144 CLASSIFICATION OF LINES 145 of projection, they are projected in their true form and require only one plane of projection for their complete representation. If the plane of the curve is perpendicular to the principal planes, a profile will suffice. If the plane of the curve is inclined to the planes of projection, both horizontal and vertical projections may be necessary, unless a supplementary plane be used, which is parallel to the plane of the curve.* This latter condition is then similar to that obtained when the plane of the curve is parallel to one of the principal planes. 905. Circle. The circle is a plane curve, every point of which is equidistant from a fixed point called the centre. The path described by the moving point (or locus) is frequently designated as the ‘circumference of the circle.”’+ To define such curves consistently, it is necessary to limit the definition to the nature of the line forming the curve. Therefore, what is usually known as the circumference of the circle should simply be known as the circle and then subsequent definitions of other curves become consistently alike. In Fig. 144 a circle is shown; abedc is the curve or circle proper. The fixed point o is called the centre and every point on the circle is equidistant from it. This fixed distance is called the radius; oa, oc, and ob are all radii of the circle; be is a diameter and is a straight line through the centre and equal to two radii in length. The straight line, de, limited by the circle is a chord Fic. 144, and when it passes through the centre it becomes a diameter; when the same line is extended like gh it is a secant. A limited portion of the circle like ac is an arc; when equal to one-quarter of the whole circle it is a quadrant; when equal to one-half of the whole circle it is a semicircle as cab or bfc. The area included between two radii and the circle is a sector as aoc; that between any line like de and the circle is *In all these cases, the plane of the curve may be made to coincide with the plane of projection. This, of course, is a special case. + The introduction of this term makes it necessary to define the circle as the enclosed area. This condition is unfortunate, as in the case of the parabola, hyperbola, and numerous other curves, the curves are open dnd do not enclose an area. : 146 GEOMETRICAL FROBLEMS IN PROJECTION asegment. If any other circle is drawn with the same centre of the circles are concentric, otherwise eccentric; the distance between the centres in the latter case is the eccentricity. 906. Ellipse. The ellipse is a plane curve, in which the sum of the distances of any point on the curve from two fixed points is constant. Fig. 145 shows this curve as adbc; The major axis is ab, and its length is the constant distance of the definition; cd is the minor axis and is perpendicular to ab through its middle point o. The fixed points or foci (either one is a focus) are e and f and are located on the major axis. If ab be assumed as the constant distance, and e and f be assumed as the foci, the minor axis is determined by drawing ares from e and f with oa, equal to one-half of the major axis, Fie. 145. Fic. 146. as a radius. Thus ec+cf=ab and also ed+df=ab. To find any other point on the curve, assume any distance as bg, and with bg as a radius and f as a centre, draw an are fh. With the balance of the major axis ag as a radius, draw an arc eh from the focus e as a centre. These intersecting ares locate h, a point on the curve. Thus eh+hf=ab and therefore satisfies the definition of the curve. With the same radii just used, the three other points i, j and k are located. In general, four points are determined for any assumption except for the points a, c, b or d. It can be shown that a and b are points on the curve, because oa=ob and oe=of. Hence ae=fb, therefore, fa+ea= ab and also eb+fb=ab. In the construction of this, or any other curve, the student should avoid trying to save time by locating only a few points. This ‘is a mistaken idea, as, within reasonable limits, time is saved by drawing numerous points on the curve, particularly CLASSIFICATION OF LINES 147 where the curve changes its direction rapidly. The direction of the curve at any point should be known with a reasonable degree of accuracy.” Another method of drawing an ellipse is shown in Fig. 146. It is known as the trammel method. Take any straight ruler and make oc=a and od=b. By locating c on the minor axis and d on the major axis, a point is located at 0, as shown. This method is a very rapid one, and is the one generally used when a true ellipse is to be plotted, on account of the very few lines required in the construction. Both methods mentioned are theoretically accurate, but the latter method is perhaps used oftener than the former. In practice, ellipses are usually approximated by employing four circular arcs, of two different radii as indicated in Art. 405. The major and minor axes are laid off and a smooth looking curve drawn between these limits. Of course, the circular arcs do not produce a true ellipse, but as a rule, this method is a rapid one and answers the purpose in conveying the idea. 907. Parabola. The parabola is a plane curve, which is the locus of a point, moving so that the distance from a fixed point is a.ways equal to the distance from a fixed line. The fixed point is the focus and the fixed line is the directrix. ° A line through the focus and perpen- “ dicular to the directrix as cg, Fig. 147, is the axis with respect to which the curve 7% : ma is symmetrical. The intersection of the axis and the curve is the vertex shown at f. The point a on the curve is so situated that ac=ad; also b is so situated that Ficq. 147. be=be. Points beyond b become more and more remote, from both the axis and directrix. Hence, it is an open curve, extending to infinity. Discussions are usually limited to some finite portion of the curve. 908. Hyperbola. The hyperbola is a plane curve, traced by a point, which moves so that the difference of its distances from two fixed points is constant. The fixed points a and b, 148 GEOMETRICAL PROBLEMS IN PROJECTION Fig. 148, are the foci. The line ab, passing through them, is the transverse * axis; the point at which either curve crosses the axis as e or f is the vertex (plural:—vertices), and the line kl, perpendicular to ab at its middle point, is the conjugate axis. To draw the curve, lay off the foci a and b; also lay off ef, the constant distance, so that eo=of and o is the middle of ab. It must be observed that ef is always smaller than ab otherwise the curve cannot be constructed. Take any radius bc, greater than bf, and draw an indefinite are; from a, draw ac, so that ac —bc=ef, hence, ac=bce-+ef. The point c is thus on the curve. Similarly, draw an arc ad, from a, and another Fra. 148. arc bd=ad-+ef. This locates d, which, in this case, is on another curve. In spection will show that there are two branches of this curve. The point c has been selected so as to be on one branch, and d, on the other. To construct the curve, accurately, many more points must be located than shown. Both branches are open and symmetrical with respect to both axes, A tangent to the curve through o is called an asymptote when it touches the hyperbola in two points, each at an infinite distance from 0. As will be observed, there are two asymptotes. 909. Cycloid. The cycloid is a plane curve, traced by a point on a circle which rolls over a straight line. The straight line over which the circle rolls is the directrix; the point on the circle may be considered as the generating point. The curve is shown as abc...i in Fig. 149. To construct it, lay out the directrix ai, and to one side, draw an auxiliary circle equal in diameter to the rolling circle as shown at 1’ 2’, etc. On a line through the centre of the auxiliary circle, draw the line 3’-9 parallel to the directrix ai. Lay off the distance * Sometimes known as the ‘principal axis.” CLASSIFICATION OF LINES 149 1-9 on this line, equal to the length of the circle (3.14 diameter). Assume that when the centre of the rolling circle is at 1, a is the position of the generating point. After one-eighth of a revolution the centre has moved to 2, where the distance 1-2 is one-eighth of the distance 1-9. The corresponding position of the generating point is shown at 2’ in the auxiliary circle at the left. Hence, for one-eighth of a revolution the centre of the rolling circle has moved from 1 to 2 and the generating point has moved a distance vertically upward equal to the distance of 2’ above the line ai. Therefore, draw a line through 2’, parallel to ai; and then from 2 as a centre, draw an arc with the radius of the rolling circle intersecting this line in the point b, a point on the required curve. After a quarter of a revolution, the gen- erating point is above the directrix, at a height equal to the Fig. 149. distance of 3’ above ai. It is also on an arc from 8, with a radius equal again to the radius of the rolling circle; hence, c is the point. This process is continued until the generating point reaches its maximum height e after one-half of a revolution, when it begins to descend, to i, as shown, after completing one revolution. Further rolling of the circle causes the generating point to dupli- cate its former steps, as it continues along the directrix to infinity if desired. i The cycloid is the same curve that is produced by a mark on the rim of a car-wheel, while rolling along the track, only, here, no slipping is permitted. The cycloid is, thus, continuous, each arch being an exact duplicate of the preceding, 910. Epicycloid. The epicyloid is a plane curve, which is generated by a point on a circle which rolls on the outside of another circle. The directrix is here an arc of a circle. instead of a straight line. 150 GEOMETRICAL PROBLEMS IN PROJECTION The construction is indicated in Fig. 150. The centre of the rolling circle assumes successive positions as 1, 2, 3, etc. The length of the arc ai is equal to the length of the rolling circle. The points are located much the same as in the cycloid, and, as the necessary construction lines are shown, the student should have no difficulty in following the construction. 911. Hypocycloid. The hypocycloid is a plane curve. generated by a point on a circle which rolls on the inside o Fig. 151. another circle. The directrix of the epicycloid and the hypocy- cloid may be the same, the epicycloid is described on the outside of the arc while the hypocycloid is described on the inside. All the necessary construction lines required to draw the hypocycloid have been added in Fig. 151 and the method is CLASSIFICATION OF LINES 151 perhaps evident. No confusion should arise even though the curves are drawn in positions other than those shown. For instance, the cycloids (this includes the cycloids, epicycloid and hypocycloid) could be shown upside-down; the curves are gener- ated in much the same way and their properties, therefore, do not change. 912. Spiral. The spiral is a plane curve, generated by a point moving along a given line while the given line is revolving about some point on the line. An infinite number of spirals may exist, because the point may have a variable velocity along the line, while again the line may have a variable angular velocity about the point. The one having uniform motion of the point along the line, and uniform angular velocity about the point is called the Spiral of Archimedes. ‘Fig. 152 shows the Archimedian spiral which is perhaps the simplest type of spiral. If ox be assumed as the primitive position of a line revolving at o, and the point also starts at o, then o is the starting point of the curve. Suppose that after one-eighth of a revo- lution the generating point has moved a distance oa, then, after one-quarter of a revolution, the point will be at b, where ob=2Xoa, and soon. The spiral becomes Fic. 152. larger and larger as the revolution con- tinues. Had the line revolved in the other direction, the curve would have been the reverse of the one shown. 913. Doubly curved line. A doubly curved line is one whose direction is continually changing and whose points do not lie in one plane. A piece of wire may be twisted so as to furnish a good example of a doubly curved line. : 914. Representation of doubly curved lines. As no single plane can contain a doubly curved line, it becomes neces- sary to use two planes and project, orthographically, the line on them. Familiarity is had with the representation of points in space on the principal planes. The line may be conceived as being made up of an infinite number of points and each point can be located in space by its projections. 152° GEOMETRICAL PROBLEMS IN PROJECTION Fig. 153 represents a curve ABCD, shown by its horizontal projection abcd and its vertical projection a’b’c’d’. Any point on the curve, such as B, is found by erecting perpendiculars from b and b’ and extending them to their intersection; this will be the point sought. The principal planes must be at right angles to each other if it is desired to locate a point by Fra. 153. erecting these perpendiculars; other- wise, the curve must be imagined from its projections when the planes are revolved into coinci- dence, as is customary in orthographic projection. 915. Helix. The helix is a doubly curved line described by a point having motion around a line called the axis, and in addition, a motion along it. Unless it is noted otherwise, the helix will be. considered as having a uniform circular motion around the axis and also a uniform motion along it. The curve finds its most extensive application on that type of screw known as a machine screw. This is then a uniform cylindrical helix. Wood screws furnish examples of conical helices.* The helix is also frequently used in making springs of a type known as helical springs. Fig. 154 shows the construction of the helix. Assume that the drawing is made in the third angle of projection; the plan is therefore on top. The ground line is omitted because the dis- tance of the points from the principal planes is not required, but only their relative location to each other. To return, 1’, 2’, . . . 8’is the plan (horizontal projection) of the helix, showing the circular motion of the point around the axis ab. In the elevation (vertical projection), the starting point of the curve is shown at 1. If the point revolves in the direction of the arrow, it will, on making one-eighth of a revolution, be at 2’ in plan *The distinction might be made between cylindrical and conical helices by considering the curve as being drawn on the surface of a cylinder or a cone as the case may be. } Helical springs are frequently although incorrectly called spiral springs. The spiral has been previously defined; and a spring of that shape is a spiral spring. CLASSIFICATION OF LINES 153 and at 2 in elevation. After a quarter of a revolution, the point is at 3’ in plan, and at 3in elevation. The position 3 is its extreme movement to the right, for at 4, the point has moved to the left, although continually upward as shown in the elevation. On completion of one revolution, the point is at 9, ready to proceed with an identical curve beyond it. The distance between any point and the position of the point after one complete revolution is known as the pitch. The dis- tance p is that pitch, and may be given from any point on the curve to the succeeding position of that point after one 6, . a SK . end ile sw Ss. ~, ny 7S ss. ~S. SA —< Sis, < , ’ a Fig. 154. Fia. 155. revolution. The distance 01’, 02’, etc. is the radius of the helix. Fig. 155 is a double helix and consists simply of two distinct helices, generated so that the starting point of one helix is just one-half of a revolution ahead or behind the other helix. The portion of the helix that is in front of the axis is shown in full; that behind the axis is shown dotted to give the effect of a helix drawn on a cylinder. The pitch again is measured on any one curve and requires a complete revolution. It must not be measured from one curve to a similar position on the next curve as the two helices are entirely distinct. Fig. 155 will show this correctly. Double, and even triple and quadruple helices are used on screws; they are simply interwoven so as to be equal distances apart. 154 GEOMETRICAL PROBLEMS IN PROJECTION 916. Classification of lines. Straight Lines: Unidirectional. circle ellipse parabola hyperbola cycloid epicycloid hypocycloid . Curved Lines: spiral etc. Doubly Curved Lines: Di- , rection continually chang- | helix ing, but not in any one | etc. plane. Singly Curved Lines: Di- rection continually chang- ing, but always in a single plane. Lines 917. Tangent. A tangent is the limiting position of a secant as the points of secancy approach and ultimately reach coincidence. Suppose, in Fig. 156, pq is a secant to the curve and that it revolves about p as a tentre. At some time, the secant pq will assume some such position as pr; the point q has then moved tor and if it continues, it will pass through p and reach s. The secant then intersects on the opposite side of p. If the rotation be such that Fic. 156. q passes through r and ultimately coincides with p, then this limiting position of the secant, shown as pt, is the tangent to the curve and p is the point of tangency. It will be observed that the tangent is fixed by the point p, and the direction of the limiting position of the secant. The condition of tangency is a mutual relation. That is, the curve is tangent to the line or the line is tangent to the curve. Also, two or more curves may be tangent to each other because the tangent line may be considered (at the point of tangency) as the direction of the curve (see Art. 920). 918. Construction of a tangent. If a tangent is to be drawn to a curve from an outside point, the drafting room method is to use a ruler of some sort and place it a slight distance away from the point and then revolve it until it nearly touches the CLASSIFICATION OF LINES 155 curve; a straight line then drawn through the point and touching the curve will be the required tangent. If the problem is to draw a tangent at a given point on a curve, however, the quick method would be to estimate the direction so that the tangent appears to coincide with as much of the curve on one side as it does on the other. A more accurate method of constructing the tangent at a given point on a curve is shown in Fig. 157. Let a be the desired point of tangency. Draw through a, the secants ab, ac, ad, the number depending upon the degree of accuracy, but always more than here shown. With a as a centre, draw any indefinite arc eh, cutting the prolongations of the secants. Lay off the chord ab=ei, ac=fj, and ad=hk, but this etter chord is laid Fig. 157. off to the left of the arc, because its secant cuts the curve to the left of the desired point of tangency. A smooth curve may now be drawn through i, j, and k, and where this curve inter- sects the indefinite arc at g, draw ga, the required tangent. The proof of this is quite simple. If ab, ac and ad be considered as displacements of the points b, c and d from the positions that they should occupy when the secant becomes a tangent, then, when these points approach a, so as ultimately to coincide with it, the displacement is zero. The secant then becomes a tangent. The curve ijk is a curve of displacements from efh, the indefinite arc. Hence, the desired tangent must pass through g, as it lies on the curve efh and its displacement from that curve is therefore zero. 919. To find the point of tangency. On the other hand, if the tangent is drawn, and it is desired to find the point of 156 GEOMETRICAL PROBLEMS IN PROJECTION tangency, the problem becomes slightly different. Suppose TT, Fig. 158, is the given tangent, and atb is the given curve. It is desired to find the point of tangency t. Draw ab, cd and ef, < any chords parallel to TT. Lay off q hg, through a, perpendicular to TT, y and br, through b, perpendicular : also to TT. Make gh=qr=ab. In the other cases, make ij=op=cd, @ 6 and kl=mn=ef. Now draw a le a 7 smooth curve through hjlnpr. T Gtk t ™ : : x Where this curve intersects TT at 2 t, the desired point of tangency is n found. Again, the proof is quite D simple. The lengths of the vertical r lines above and below TT are equal Fic. 138. to the lengths of the corresponding chords. As the chords approach the tangent, they diminish in length and ultimately become zero. Where the curve crosses the tangent, the chord length is zero, and, hence, must be the point of tangency. 920. Direction of a curve. A curve continually changes its direction, but at any given point its direction is along the tangent to the curve by definition. It is proved in Mechanics, that if forces act on a particle so as to give it a curved motion, the particle will fly off along a tangent when the impressed forces cease to act.* 921. Angle between curves. The angle between two inter- secting curves is the same as the angles, made by the tangents at the point of intersection, because the tangents determine the direction of the curve at that point. Hence, to draw smooth curves, it is necessary that the tangent at the end of one curve should coincide with the tangent at the beginning of the next curve. 922. Intersection of lines. Two lines intersect when they have a point in common. When the term intersection is used in connection with a pair of straight lines, it necessarily implies that the lines make an angle with each other greater than zero; *See Newton’s laws of motion in a text-book on Physics. CLASSIFICATION OF LINES 157 since, otherwise, if there is a zero angular relation, the lines become coincident and have all points in common. When a line meets a curve, the angle between the line and the curve at the point of intersection is the same as the angle between the line and the tangent to the curve at that point. When this angle of intersection becomes zero, however, by having the line coincide with the tangent, then it is the special case of intersection known as tangency. Similarly when two curves intersect their angle of intersection is the same as the angle between their tangents at the point of intersection. Also, when this angle becomes zero then the curves are tangent to each other. Thus, in order to differentiate between the two types of intersection, angular intersection means intersection at an angle greater than zero; and, as a consequence, the intersection at a zero angle is known as tangential intersection. When inter- section is unmodified, then angular intersection is implied. 923. Order of contact of tangents. A tangent has been previously defined as the limiting position of a secant, as the points of secancy approach and ultimately coincide with each other. This contact, for simple tangency, is of the first order. Two curves may also be tangent to each other so as to have con- tact of the first order, as for example, two circles, internally or externally tangent. Suppose two curves acb and abe, Fig. 159, have first order contact at a, and cut each other at the point b. If the curve abe be made to revolve about the point a as a centre, so as to maintain simple tangency * and also to have the point b approach a, then at some stage of the revolution the point b can be made to coincide with a. Under this condition there is tangency Fic. 159, of a higher order because three points were made to ultimately coincide with each other; and it is called second order of contact. It is possible to have third order of contact with four coincident points; and so on. The *In order to make this a rigid demonstration, the centre of curvature of the two curves, at the point a must not be the same. See Art. 924. 158 GEOMETRICAL PROBLEMS IN PROJECTION order of contact is always one less than the number of points that approach coincidence. 924. Osculating circle. Centre of curvature. Let abc, Fig. 160, be a curve and b, a point through which a circle gdbe passes, cutting the curve abc in three points d,b and e. If the diameter of the circle is properly chosen, it may be revolved about b as a centre so that the points d and e will both approach and ultimately coincide with b at the same instant. This position of the circle is shown as bf. Hence, the circle bf is tangent to the curve abc, and is of the second order xd of contact. This circle is the oceulating b circle. As this osculating circle must Fig. 160. more nearly approach the curvature of the curve abe than any other circle, its radius at the point b is the radius of curvature. At every other point on the curve, there is a new osculating circle, of a new centre, and of a new radius. Thus, the osculating circle is the second order tangent circle at the point; and the radius of curvature may be defined as the radius of the osculating circle through the point, the centre of curvature being the centre of the osculating circle. 9 ay 925. Osculating plane. Ifa tangent be drawn to any doubly curved line, an infinite number of planes may be passed so as to contain the tangent. If some one position of the plane be selected so as to contain the tangent and a piercing point of the doubly curved line on it, then by proper revolution of the plane the piercing point can be made to approach and ultimately coincide with the point of tangency; in this position, the plane is an osculating plane. To put the matter differently, suppose it is desired to find the osculating plane at some point on a doubly curved line. In this case, draw a pair of secants to the doubly curved line which intersect at the given point. A plane passed through these secants will cut the doubly curved line in three points. As the secants approach tangency, the plane will approach osculation, and this osculating plane is identical with that of the former discussion for the same point on the curve. If the curve under CLASSIFICATION OF LINES 159 consideration be a plane curve, then the secants will lie in the plane of the curve, and, hence, the osculating plane of the curve will be the plane of the curve. 926. Point of inflexion. Inflexional tangent. Assume a curve dae, Fig. 161, and through some one point a draw the secant be. If the secant be revolved about the point a so that the points of intersection b and ¢ approach a, at some stage of the revolution they will coincide with it at the same instant if the point a is properly chosen. The point a must be such that three points on the curve ultimately coincide at the same instant. Further than this, the radius of curva- Fig. 161. ture (centre of the osculating circle) must make an abrupt change from one side of the curve to the other at this point. The point of inflexion therefore is a point at which the radius of curvature changes from one side of the curve to the other. The inflexional tangent is the tangent at the point of inflexion. It may also be noted that the inflexional tangent has a second order of contact (three coincident points) and therefore is the osculating line to the curve at the point of inflexion. T. é 927. Normal. A normal to a curve is a perpendicular to the tangent at the point of tangency. The principal normal lies in the osculating plane. As an infinite number of normals may be drawn to the tangent at the point of tangency, the nor- mal may revolve about the tangent so as to generate a plane which will be perpendicular to the tangent and thus establishes a normal plane. In the case of a circle, the radius at the point of tangency is normal to the tangent; in such cases, the tangent is easily drawn, as it must be perpendicular to the radius at the point of contact. 928. Rectification. When a curve is made to roll on a straight line, so that no slip occurs between the curve and the line, the distance measured on the line is equal to the corresponding length of the curve. This process of finding the length of the curve is called rectification. Commercially applied, the curve is measured by taking a divider and stepping off very small distances; the 160 GEOMETRICAL PROBLEMS IN PROJECTION number of steps multiplied by the distance between points will give approximately the length of the curve. It must be noticed that the divider measures the chord distance, instead of the arc distance and is therefore always less than the actual length of the curve, but when the distance is taken small enough, the accuracy of the final result is proportional to the care taken in making the measurement. 929. Involute and evolute. When a tangent rolls about a fixed curve, any point, on the tangent describes a second curve which is the involute of the first curve. Fig. 162 shows this in construction. Let aceg be the fixed curve, and ab, be the posi- tion of a taut string that is wound on the curve aceg. If a pencil point be attached to the string and unwound, the pencil point will describe the curve bdfh which is the involute of the curve aceg. The process is the same as though Fia. 162. the tangent revolved about the curve aecg and some point on the tangent acted as the generating point. Ata, the radius is ab; at c, the radius is cd, which is equal to ab plus the rectified arc ac (length of string). It may be observed that the curve aceg is the curve of centres for the curve bdfh. If the string be lengthened so that ai is the starting position, it will describe the curve ijkl which again is an involute. This second involute is parallel to the first, because the distance between the two curves measured along the rolling tangent (or radius of curvature) is constant between the two curves. The primitive curve aceg is the evolute. The tangent rolls on the evolute and any point on it describes an involute. As any point on the tangent will answer as the tracing point, it follows that every evolute has an infinite number of involutes, all of which are parallel curves. Reversing the process of the construction of the involute, the method of drawing the primitive curve or evolute is obtained. If normals are drawn to the involute, consecutive positions of CLASSIFICATION OF LINES 161 the normals will intersect. The locus of these successive inter- sections will regenerate the primitive curve from which they have been evolved.* Again, the length of the tangent to the evolute is the radius of curvature for the involute. As the radius of any circle is perpendicular to the tangent at that point, it follows that the involute is always normal, point for point, to the evolute, since the rolling tangent is the direction of the evolute at the point of contact and that again is normal to the involute. 930. Involute of the circle. The involute of the circle is a plane curve, described by a point on a tangent, while the tangent revolves about the circle. Let o, Fig. 163, be the centre of a circle whose ractius is oa. Let, also, a be the starting point of the involute. Divide the circle in any number of parts, always, however, more than are shown in the illustra- tion. Draw tangents to the various radii. On them, lay off the rectified are of the circle between the point of 7 @ tangency and the starting point. For instance, eb equals the rectified length of the arc ea; also fe equals the rectified semicircle fea; gd equals the rectified length of the arc gfea. The involute may be continued indefinitely for an infinite number of revolutions, but, discussion-is usually centred on some limited portion. The curve here shown is approximately the same as that described by the end of the thread, when a spool is unwound. Fic. 163. QUESTIONS ON CHAPTER IX . Why are lines and points considered as mathematical concepts? How is a straight line defined? By what two means may a straight line be fixed in space? What is a generating point? What is a locus? What is a singly curved line? . What is a plane curve? . How are plane curves represented? CONS oR oy * The evolute of a circle, therefore, is its centre. GEOMETRICAL PROBLEMS IN PROJECTION . Show the mode of representing curves when their planes are parallel to the plane of projection? When perpendicular? When inclined? . Is it desirable to use the plane of the curve as the plane of projection? . Define the circle. . What is the radius? Diameter? Sector? Segment? . When the chord passes through the centre of the circle, what does it become? . What is a secant? Quadrant? Semicircle? . Define concentric circles; eccentric circles; eccentricity. . What is an ellipse? . Define major axis of an ellipse; minor axis; foci. . Describe the accurate method of ¢rawing an ellipse by the intersection of circular ares. . Describe the trammel method of drawing an ellipse. . Draw an ellipse whose major axis is 3’ long and whose minor axis is 2’’ long. Use the accurate method of intersecting circular arcs. . Construct an ellipse whose major axis is 3’’ long and whose minor axis is 13’ long. Use the trammel method. . What is a parabola? . Define focus of a parabola; directrix; axis; vertex. . Is the parabola symmetrical about the axis? . Is the parabola an open or a closed curve? . Construct the parabola whose focus is 2’’ from the directrix. . What is a hyperbola? . Define foci of hyperbola; transverse axis; conjugate axis; vertex; asymptote. . How many branches has a hyperbola? Are they symmetrical about the transverse and conjugate axes? . How many asymptotes may be drawn to a hyperbola? . Is the hyperbola an open or a closed curve? . Construct a hyperbola whose distance between foci is 2’’ and whose constant difference is 3’. . What is a cycloid? . Define rolling circle of a cycloid; directrix. . Construct a cycloid whose diameter of rolling circle is 13’. Draw the curve for one revolution only. . What is an epicycloid? . What form of directrix has the epycycloid? . Construct the epicycloid, whose diameter of rolling circle is 13” and whose diameter of directrix is 8’. Draw the curve for one revolution only. . What is a hypocycloid? . What is the form of the directrix of the hypocycloid? . Construct a hypocycloid whose diameter of rolling circle is 2’”’ and whose diameter of directrix is 9’. Draw the curve for one revolu- tion only. . What is a spiral? CLASSIFICATION OF LINES 163 . Construct an Archimedian spiral which expands 13” in a complete revolution. Draw the spiral for two revolutions. . What is a doubly curved line? . Draw a doubly curved line with the principal planes in oblique projection. . Construct the orthographic projection from Question 45. . What is a helix? . Define uniform cylindrical helix; conical helix; diameter of helixy pitch. . Construct a helix whose diameter is 2’’, and whose pitch is 1”. Draw for two revolutions. . Construct a triple helix whose diameter is 2” and whose pitch is 3”. The helices are spaced equally and are to be drawn for 14 revolu- tions. . Make a classification of lines. . What is a tangent? . Is the tangent fixed in space by a point and a direction? . Show that the tangent is the limiting position of a secant. . How is a tangent drawn to a curve from a point outside? . Given a curve and a point on it, draw a tangent by the accurate method. Prove that the chord length is zero for the tangent position. . Given a curve and a tangent, determine the point of tangency. Prove. . What is the direction of a curve? . What is the angle between two intersecting curves? . When several curves are to be joined, show what must be done to make them smooth curves. . Define intersection of lines. . Show that the tangent intersects the curve at a zero angle. . Define order of contact of tangents. . If three points become coincident on tangency, what order of contact does the tangent have? . Define osculating circle. . Define centre of curvature. . Is the radius of the osculating circle to a curve the radius of curvature at that point? . Show how an osculating circle may have second order contact with a plane curve. . What is an osculating plane? . When is the osculating plane the plane of the curve? . Define point of inflexion. . What is an inflexional tangent? . Does the radius of curvature change from one side of the curve to the other at a point of inflexion? . Define principal normal. . Show that the centres of curvature at a point of inflexion lie on opposite sides of the normal to the curve through the point of inflexion. 164 GEOMETRICAL PROBLEMS IN PROJECTION 76. 77. 78. 79. 80. 81. 82. 83. When a normal is drawn to a curve is the one in the osculating plane the one generally understood? How many normals may be drawn to a doubly curved line at a given point? What is meant by rectification? Rectify a 2’ diameter circle. Compute the length of the rectified circle (=2x3.1416) and express the ratio of rectified to computed length as a percentage. Define involute and evolute. Show that all involutes to a curve are parallel curves. Show that the involute is always normal to the evolute at the point for which it corresponds. Show that the drawing of the evolute is the reverse process of drawing the involute. . Draw the involute of a circle. . Draw the involute to an ellipse. . Draw the evolute to an ellipse. . Draw the involute to a parabola. . Draw the evolute to a parabola. 5 . Draw the involute to one branch of a hyperbola. . Draw the evolute to one branch of a hyperbola. . Draw the involute to a cycloid. . Draw the evolute to a cycloid. . Draw the involute to an epicycloid. . Draw the evolute to an epicycloid. . Draw the involute to a hypocycloid. . Draw the evolute to a hypocycloid. . Draw the involute to an Archimedian spiral. . Draw the evolute to an Archimedian spiral. CHAPTER X CLASSIFICATION OF SURFACES 1001. Introductory. A surface may be generated by the successive positions of a line which moves so as to generate an area. As there are infinite varieties of lines and as their motion may again be in an infinite variety of ways, therefore, an infinite variety of possible surfaces result. In engineering it is usual to limit the choice of surfaces to such as may be easily reproduced and easily represented. Surfaces, like lines and points, are mathematical concepts because they have no material exist- ence. When curved surfaces, of a more or less complex nature, are to be represented, they may be shown to advantage, by the effects of light on them. Examples of this kind are treated in Chapters XIV and XV. 1002. Plane Surface. A plane surface is a surface such that when any two points in it are joined by a straight line, the line lies wholly within the surface. Thus, three points may be selected in a plane and two intersecting lines may be drawn through the three points; the intersecting lines lie in the plane and, therefore, may be used to determine it. Also, a line and an external point may determine a plane. The plane surface may also be conceived as being generated by a straight line, moving so as to touch another line, and con- tinually remaining parallel to its original position. Hence, also, two parallel lines determine a plane. In the latter case, the moving straight line may be consid- ered as a rectilinear generatrix, touching a rectilinear directrix, and occupying consecutive positions in its-motion. Any one position of the generatrix may be used as an element of the surface. 165 166 GEOMETRICAL PROBLEMS IN FROJECTION Fig. 164 shows a plane surface ABCD on which straight lines ab, cd, ef and gh are drawn, all of which must lie wholly within the plane, irrespective of the direction in which they are drawn. Any curve drawn on this surface is a plane curve. 1003. Conical surface. If a straight line passes through a given point in space and moves so as to touch a given fixed curve, the surface so generated is a conical surface. The straight line is the rectilinear generatrix, the fixed point is the vertex and the given fixed curve is the directrix, which need not be a closed curve. The generatrix in any one position is an element of the surface. Fig. 165 shows a conical surface, generated in the manner Fic. 164. Tic. 165. indicated. Either the upper or the lower curve may be consid- -ered as the directrix. In fact, any number of lines may be drawn on the resulting surface, whether the lines be singly or doubly curved, and any of which will fill the office of directrix. A plane curve is generally used as the directrix. With a generatrix line of indefinite extent, the conical sur- face generated is a single surface (not too surfaces as might appear); the vertex o is a point of union and not of separation. The portion of the surface from the vertex to either side is called a nappe;* hence, there are two nappes to a conical surface. 1004. Cone. The cone is a solid, bounded by a closed conical surface of one nappe and a plane cutting all the elements. The * Pronounced ‘‘nap.” CLASSIFICATION OF SURFACES 167 curve of intersection of the rectilinear elements and the plane cutting all the elements is the base. A circular cone has a circle for its base and the line joining the vertex with the centre of the base is the axis of the cone. If the axis is perpendicular to the plane of the base the cone is a right cone. When the base is a circle, and the axis is perpendicular to the plane of the base, the cone is a right circular cone or a cone of revolution.* A cone of revolution may be generated by revolving a right tri- angle about one of its legs as an axis. The hypothenuse is then the slant height of the cone. The perpendicular distance from the vertex to the plane of the base is the altitude of the cone. The foot of the perpendicular may fall outside of the centre of the base and in such a case, the cone is an oblique cone. The frustum (plural:—frusta) of a cone is the limited portion of the solid bounded by a closed conical surface and two parallel planes, each cutting all the elements, and giving rise to the upper base and the lower base of the frustum of a cone. The terms upper and lower base are relative; it is usual to consider the larger as the lower base and to represent the figure as resting on it. When the cutting planes are not parallel, then the solid is a truncated cone. 1005. Representation of the cone. A cone, like any other object, is represented by its projec- r tions on the principal planes. For con- venience in illustrating a cone, the plane of the base is assumed perpendicular to ; one of the principal planes, as then its 4 i . . . . c projection on that plane isa line. Fig. x y 166 shows a cone in orthographic projec- f 1 | tion. The vertex O is shown by its b projections o and 0’; d’c’ is the vertical projection of the base since the plane of the base is assumed perpendicular to the vertical plane. The extreme limiting ele- Fie. 166. ments o’c’ and o’d’ are also shown, thus completing the vertical projection. In the horizontal projection, any curve acbd is assumed as the projection of the base so * This distinction is made because a cone with an elliptical base may also be a right cone when the vertex is chosen so that it is on a perpendicular to the plane of the base, at the intersection of the major and minor axes. 168 GEOMETRICAL PROBLEMS IN PROJECTION that d and c are corresponding projections of d’ and c’.. From 0, the lines ob and 0a are drawn, tangent to acbd, thus completing the horizontal projection. It must here be emphasized, that acbd is not the actual base of the cone, but only its projection. It is impossible to assume two curves, one in each plane of projection, and call them corre- sponding projections of the same base. The corresponding points must be selected, so that they will lie in one plane, and that plane must be the plane of the base. If it be desired to show the base in both projections, when the plane of the base is in- clined to the principal planes, it is necessary to assume one pro- jection of the base. Lines are then drawn in that plane, through the projection of the base and the corresponding projections Fia. 167. of the lines are found. The points can then be determined as they must be situated on these lines. (Arts. 704 and 811.) 1006. To assume an element on the surface of a cone. To assume an element of a cone, assume the horizontal projec- tion oa in Fig. 167. There are two elements on the cone which have the same horizontal projection and they are shown as o’a’ and o’b’ in the vertical projection. If oa be considered visible, while viewing the horizontal plane, then o’a’ is its corre- sponding projection. If o’b’ be the one assumed projection then oa is on the far side and should in this case be drawn dotted. 1007. To assume a point on the surface of a cone. To assume a point on the surface of a cone, assume c’ in Fig. 168 as the vertical projection, somewhere within the projected area. CLASSIFICATION OF SURFACES 169 Draw the element o’b’ through c’ and find the corresponding projection of the element. If o’b’ is visible to the observer, then ob is the corresponding projection, and ¢ on it is the required projection. If o’b’ is on the far side, then d is the desired pro- jection. A slightly different case is shown in Fig. 169. If ¢ is assumed on the visible element oa then c’ is the corresponding projection on o’a’. Otherwise, if ob is dotted (invisible) then o’b’ is the corresponding element and d and d’ are corresponding projections. 1008. Cylindrical surface. When a straight line moves so that it remains continually parallel to itself and touches a given fixed curve, the surface generated is a cylindrical surface. Fic. 169. The straight line is the rectilinear generatrix; the fixed curve is the directrix and need not be a closed curve. The generatrix in any one position is an element of the surface. Fig. 170 shows a cylindrical surface, generated in the manner indicated. Any curve, drawn on the resultant surface, whether singly or doubly curved, may be considered as the directrix. The limiting curves that are shown in the figure may also be used as directrices. A plane curve is generally used as a directrix 1009. Cylinder. A cylinder is a solid bounded by a closed cylindrical surface and two parallel planes cutting all the elements. The planes cut curves from cylindrical surface which form the bases of the cylinder and may be termed upper and lower bases. if the cylinder is so situated that the nomenclature fits. When the planes of the bases are not parallel then it is called a truncated. cylinder. 170 GEOMETRICAL PROBLEMS IN PROJECTION Should the bases have a centre, a figure such as a circle for instance, then a line joining these centres is the axis of the cylinder. The axis must be parallel to the elements of the cylinder. Ifthe axis is inclined to the base, the cylinder is an oblique cylinder. On the other hand, of the axis is perpendicular to the plane of the base, it is a right cylinder and when the base is a circle, it is a right circular cylinder, or, a cylinder of revolution. The cylinder of revolution may be generated by revolving a rectangle about one of its sides as an axis. A right cylinder need not have a circular base, but the elements must be perpendicular to the plane of the base. 1010. Representation of the cylinder. A _ cylinder, represented orthographically is shown in Fig. 171. Suppose the base is assumed in the horizontal plane, then e’g’ may be taken as the vertical projection of the base. Also e’f’ and g’h’, parallel to each other, may be taken as the projections of the extreme limiting elements. Any curve, as aecg, may be drawn for the horizontal projection so long ase and g are corre- sponding projections of e’ and g’. In addi- tion, draw ab and cd parallel to each other and tangent to the curve aecg. The hori- zontal projection is thus completed. It is to be noted that aecg is the true base Fic. 171. because it lies in the horizontal plane. If the plane of the base does not coincide with the horizontal plane, then, as in Art. 1005, what applies to the selection of the projection of the base of the cone applies here. 1011. To assume an element on the surface of a cylinder. To assume an element on the surface of a cylinder, select any line, ab, Fig. 172, as the horizontal projection. As all parallel lines have parallel projections, then ab must be parallel to the extreme elements of the cylinder. If ab is assumed as a visible element, then a’b’ is its corresponding projection, and is shown dotted, because hidden from view on the vertical projection. ff c’d’ is a visible element, then cd should be dotted in the horizontal projection. CLASSIFICATION OF SURFACES 171 1012. To assume a point on the surface of a cylinder. To assume a point on the surface of a cylinder, select any point c, Fig. 178, in the horizontal projection, and draw the element ab through it. Find the corresponding projection a’b’ and on it, locate c’, the required projection. What has been said before (Art. 1005) about the two possible cases of an assumed projection, applies equally well here and should require no further mention. 1013. Convolute surface. A convolute surface is a surface generated by a line which moves so as to be continually tangent * to a line of double curvature. For purposes of illustration, the uniform cylindrical helix will be assumed as the line of double ’ d Fig. 172. Fig. 1738. Fig, 174. curvature, remembering, in all cases, that the helix may be variable in radius and in motion along the axis, so that its char- acteristics may be imparted to the resulting convolute. The manner in which this surface is generated may be gained from what follows. Let abcd, Fig. 174, be the horizontal pro- jection of a half portion octagonal prism on which is a piece of paper in the form of a right triangle is wound. The base of the triangle is therefore the perimeter of the prism and the hypoth- enuse will appear as a broken line on the sides of the prism. If the triangle be unwound from the prism .and the starting point of the curve described by the hypothenuse on the horizontal * Tt may be observed that the tangent to a line of double curvature must ile in the osculating plane (Art. 925). 172 GEOMETRICAL PROBLEMS IN PROJECTION plane be at o, then the portion of the triangle whose base is 0a will revolve about the edge a soas to describe the are o1. At the point 1 the triangle is free along the face ab and now swings about b as a centre and describes the are a’ 1-2. As the process goes on to the point 1A 2, the triangle is free on the face be and then swings about c as a centre and tacit describes the arc 2-3; and soon. In the 1 10 ; vertical projection, the successive positions \/ “a d of the hypothenuse are shown by a’l’, b b’2’, c’3’ etc. It will be noted that a’l’ intersects b’2’ at b’, and b’2’ intersects c’3’ at c’, etc.; but, a’L’ does not intersect c’3’ nor does b’2’ intersect d’4’.. Hence, the elements of the surface generated by the 4 hypothenuse intersect two and two. The Fig. 174. first element intersects the second; the second the third; the third the fourth, etc.; but the first does not intersect the third, or any beyond, nor does the second intersect the fourth or any elements beyond the fourth. When the prism approaches a cylinder as a limit by increas- ing the number of sides indefinitely, the hypothenuse wound Fig. 175. around the cylinder approaches a helix as a limit; the unwind- ing hypothenuse will become the generatrix, tangent to the helix, and will approach the desired convolute surface. The ultimate operation is a continuous one and may be seen in Fig. 175. The curve abcd described by the hypothenuse fd on the plane MM CLASSIFICATION OF SURFACES 173 is the involute of a circle, if the cylinder is a right circular cylinder. Fig. 176 may indicate the nature of the surface more clearly. Examples of this surface may be obtained in the machine shop on observing the spring-like chips, that issue on taking a heavy cut from steel or brass. The surfaces are perhaps not exact convolutes, but they resemble them enough to give the idea. It is not necessary to have the tangent stop abruptly at the helix, the tangent may be a line of indefinite extent, and, hence, the convolute surface extends both sides of the helical directrix. No portion of this surface intersects any other portion of the surface, Fic. 177. but all the convolutions are dis- tinct from each other. Fig. 177 will perhaps convey the final idea. 1014. Oblique helicoidal screw surface.* When the helical directrix of a convolute surface decreases in diameter, it will ultimately coincide with the axis and the helix will become a line. The oblique helicoidal screw sur- face, therefore, resolves itself into the surface generated by a recti- linear generatrix revolving about another line which it intersects, at a constant angle, the inter- section moving along the axis at a uniform rate. The application of this surface is shown in the construction of the V thread screw which in order to become the United States Standard screw, must make an angle of 60° at the V as shown in Fig. 178. *The helicoid proper is a warped surface (1016). If a straight line touches two concentric helices of different diameters and lies in a plane tangent to the inner helix’s cylinder, the line will generate a warped surface. When the diameter of the cylinder becomes zero, the helix becomes a line and the helicoidal surface is the same as that here given. t 174 GEOMETRICAL PROBLEMS IN PROJECTION 1015. Right helicoidal screw surface. If the diameter of the helical directrix still remains zero, and the rectilinear generatrix becomes perpendicular to the axisand revolves so that the in- tersection of the axis and the genera- trix moves along the axis at a constant rate,* a special case of the convolute is obtained. This special case of the Fie. 179, convolute is called a right helicoidal screw surface, and when applied gives the surface of a square threaded screw as shown in Fig. 179. In both cases of the oblique and right helicoidal screw sur- faces the helices at the outside and root (bottom) of the thread, are formed by the intersection of the screw surface and the outer and inner concentric cylinders. The pitch of both must be the same, as every point on the generatrix advances at a uniform rate. Hence, the angle of the tangent to the helix must vary on the inner and outer cylinders. For this reason, the helices have a different shape notwithstanding their equal pitch. 1016. Warped surface. A warped surface is a curved surface, generated by a rectillinear generatrix, moving so that no two successive elements lie in the same plane. Thus, the consecutive elements can neither be parallel nor intersect, hence, they are skew lines. An example of this surface may be obtained by taking a series of straight sticks and drilling a small hole through each end of every stick. If a string be passed through each end and secured so as to keep them together, the series of sticks may be laid on a flat surface and thus represent successive elements of a plane. It may also be curved so as to represent a cylinder. Lastly, it may be given a twist so that no single plane can be passed through the axis of successive sticks; this latter case would then represent a warped surface. Warped surfaces find comparatively little application in engineering because they are difficult to construct or to duplicate. At times, however, they are met with in the construction of *TIf the pitch becomes zero when the diameter of the helix becomes zero, it is the case of a line revolving about another line, through a fixed point; the surface is therefore « cone of revolution, if the generatrix is inclined to the axis. If the generatrix is normal to the axis, the surface is a plane. CLASSIFICATION OF SURFACES 175 “forms” for reinforced concrete work, where changes of shape occur as in tunnels, and similar constructions; in propeller screws for ships; in locomotive “ cow-catchers,” etc. 1017. Tangent plane. If any plane be passed through the vertex of a cone, it may cut the surface in two rectilinear elements under which condition it is a secant plane. If this secant plane be revolved about one of the rectilinear elements as an axis, the elements of secancy can be made to approach so as to coincide ultimately. This, then, is a limiting position of the secant plane, in which case it becomes a tangent plane, having contact with the cone all along one element. If through some point on the element of contact, two inter- secting curved lines be drawn on the surface of the cone, then, also, two secants may be drawn to these curved lines and inter- secting each other at the intersection of the curved lines. The limiting positions of these secants will be tangent lines to the cone, and as these tangents intersect, they determine a tangent plane. The tangent plane thus determined is identical with that obtained from the limiting position of a secant plane. Instead of drawing two intersecting curves on the surface of the cone, it is possible to select the element of contact and any curve on the cone intersecting it. The tangent plane in this case is determined by the element of contact and the limiting position of one secant to the curve through the intersection of the element and the curve. As another example, take a spherical surface and on it draw two intersecting lines (necessarily curved). Through the point of intersection, draw two secants, one to each curve and deter- mine the tangent positions. Again, the plane of the two inter- secting tangents is the tangent to the sphere. Hence, as a general definition, a tangent plane is the plane established by the limiting position of two intersecting secants as the points of secancy reach coincidence. 1018. Normal plane. The normal plane is any plane that is perpendicular to the tangent plane. If a normal plane is to be drawn to a sphere at a given point, for instance, then construct the tangent plane and draw through the point of tangency any plane perpendicular to the tangent plane. An infinite number of normal planes may be drawn, all passing through the given 176 GEOMETRICAL PROBLEMS IN PROJECTION point. The various normal planes will intersect in a common line, which is normal to the tangent plane at the point of tangency. 1019. Singly curved surface. A singly curved surface is a surface whose successive rectilinear elements may be made to coincide with a plane. Hence, a tangent plane must be in contact all along some one rectilinear element. As examples, the conical, cylindrical, convolute and the helicoidal screw surfaces may be mentioned. 1020. Doubly curved surface. A doubly curved surface is a surface whose tangent plane touches its surface at a point. Evidently, any surface which is not plane or singly curved must be doubly curved. The sphere is a familiar example of a doubly curved surface. 1021. Singly curved surface of revolution. A singly curved surface of revolution is a surface generated by a straight line revolving about another straight line in its own plane as an axis, so that every point on the revolving line describes a circle whose plane is perpendicular to the axis, and whose centre is in the axis. Thus, only two cases of singly curved surfaces can obtain, the conical and the cylindrical surfaces of revolution. 1022. Doubly curved surface of revolution. A doubly curved surface of revolution is a surface generated by a plane curve revolving about a straight line in its own plane as an axis so that every point on the revolving curve describes a circle whose plane is perpendicular to the axis, and whose centre lies in the axis. Hence, there are infinite varieties of doubly curved sur- faces of revolution as the sphere, ellipsoid, hyperboloid, etc., generated by revolving the circle, ellipse, hyperbola, etc., about their axes. In the case of the parabola, the curve may revolve about the axis or the directrix in which cases two distinct types of surfaces are generated. Similarly with the hyperbola, the curve may generate the hyperboloid of one or two nappes depend- ing upon whether the conjugate or transverse axis is the axis of revolution, respectively. Sometimes, a distinction is made between the outside and inside surfaces of a doubly curved surface. For example, the outside surface of a sphere is called a doubly convex surface, whereas, the inside is an illustration of a doubly concave surface. CLASSIFICATION OF SURFACES 177 A circular ring made of round wire, and known as a torus, is an example of a doubly concavo-convex surface. 1023. Revolution of a skew line. An interesting surface is the one generated by a pair of skew i lines when one is made to revolve about the other as an axis. Fig. 180 gives such a case, and as no plane can be ; passed through successive elements, it is a warped surface. While revolving about the axis, the line generates the same type of surface as would be gen- a CO et AL = ANS Af] Xx surface of revolution, it will cut from A the surface a line which is the me- tidian line. The plane cutting the Fic. 180. meridian line is called the meridian plane. Any meridian line can be used as the generatrix of the sur- face of revolution, because all meridian lines are the same. The circle is the meridian line of a sphere, and for this particular surface, every section is a circle. In general, erated by a hyperbola when revolved every plane perpendicular to the axis will cut a about its conjugate axis. The surface is known as the hyperboloid of revo- circle from any surface of revolution, whether singly or doubly curved. ’ ot @ ere lution of one nappe, and, incidentally, is the only warped surface of revolu- tion. 1024. Meridian plane and me- ridian line. If a plane be passed through the axis of a doubly curved. 1025. Surfaces of revolution having a common axis. If two surfaces of revolution have a common axis and the surfaces intersect, tangentially or angularly, they do so all around Fra. 181. in a circle which is common to the two surfaces of revolution. Thus, the two surfaces shown in Fig. 181 intersect, angularly, in a circle having ab as a diameter, and intersect, tangentially, in a circle having cd as a diameter. 178 GEOMETRICAL PROBLEMS IN PROJECTION 1026. Representation of the doubly curved surface of revolution. Fig. 182 shows a doubly curved surface of revolution shown in two views. One view shows the same as that produced by a meridian plane cutting a meridian line and the other shows the same as concentric circles. When considered as a solid, no ground line is necessary as the distance from the principal planes is unimportant. Centre lines, ab, cd and ef should be shown, ab and ef being represented as two lines, be- cause both views are distinct from each other. The lines indi- cate that the object is symmetrical about the centre line as an axis. 1027. To assume a point on a doubly curved surface of revolution. Let c, Fig. 183, be assumed as one point on the surface.* With oc as a radius, draw the arc Ca, a is, there- fore, the revolved position of z c when the meridian plane tA through co has been revolved e to ao. Hence, a is at a’ or Fra. 183. a’ in the corresponding view. On counter-revolution, a’ de- scribes a circle, the plane of which is perpendicular to the axis, and the plane is shown by its trace a’c’; in the other view, a returns to c and, hence, c’ is the final position. It may also be at c’’ for the same reason, but then c is hidden in that view. If, on the other hand, d is chosen as one projection, its corre- a” cv * These views bear third angle relation to each other. CLASSIFICATION OF SURFACES 179 sponding projection is d’, if d is visible; or, it is d”, if d is hidden. 1028. Developable surface. When a curved surface can be rolled over a plane surface so that successive elements come in contact with the plane and that the area of the curved surface can be made to equal the plane surface by rectification, the surface is a developable one. Hence, any singly curved surface, like a cylinder, can be rolled out flat or developed. A sphere cannot be rolled out as a flat surface because it has point con- tact with a plane, and is, therefore, incapable of development. If a sphere is to be constructed from flat sheets, it may be ap- proximated by cutting it into the type of slices called lunes, resembling very much the slices made by passing meridian planes through the axis of a sphere. To approach more nearly the sphere, it would be necessary to take these lunes and hammer them so as to stretch the material to the proper curvature. Simi- larly, in the making of stove-pipe elbows, the elbows are made of limited portions of cylinders and cut to a wedge shape so as to approximate the doubly curved surface known as the torus. 1029. Ruled surface. Every surface on which a straight line may be drawn is called a ruled surface. A ruled surface may be plane, singly curved or doubly curved. Among the singly curved examples may be found the conical, cylindrical, convolute and helicoidal screw surfaces. The hyperboloid of revolution of one nappe furnishes a case of a doubly curved ruled surface (1023). 1030. Asymptotic surface. If a hyperbola and its asymp- totes move so that their plane continually remains parallel to itself, and any point on the curve or on the asymptotes touches a straight line as a directrix, the hyperbola will generate a hy- perbolic cylindrical surface and the asynaptotes will generate a pair of asymptotic planes. Also, if the hyperbola revolves about the transverse or conjugate axis, the hyperbola will generate a hyperboloid of revolution and the asymptote will generate a conical surface which is asymptotic to the hyperboloid. In all cases, the asymptotic surface is tangent at two lines, straight or curved, at an infinite distance apart and the surface passes within finite distance of the intersection of the axes of the curve. 180 GEOMETRICAL PROBLEMS IN 1031. Classification of surfaces. Ruled surfaces. Straight lines may be drawn on the resulting sur- face. Planes. Any two points when joined by a straight line lie wholly within the surface. Singly curved sur- faces. May be developed into a flat surface by rec- tification. Surfaces. { * Warped surfaces. No two consecutive elements lie in the same plane; hence, they are non-devel- opable. Doubly curved going Classification. QUESTIONS ON CHAPTER X 1. How are surfaces generated? 2. What is a plane surface? 3. What is a rectilinear generatrix? 4, What is a directrix? 5. What is an element of a surface? 6. What is a conical surface? PROJECTION Conical surfaces. Rectilinear ele- ments pass through a given point in space and touch a curved directrix. Cylindrical surfaces. Rectilinear ele- ments are parallel to each other and touch a curved di- rectrix. Convolute surfaces. Rectilinear ele- ments tangent to a line of double curva- ture. Consecutive elements intersect two and two; no three intersect in a common foint. Doubly curved surfaces of revolution. Generated by plane curves revolving about an axis in the plane of the curve. All All surfaces. Tan-| meridian lines equal and all sections per- gent plane) pendicular to axis are circles. touches surface 7 at a point. Unclassified doubly curved surfaces. others which do not fall within the fore- CLASSIFICATION OF SURFACES 181 . What is the directrix of a conical surface? . What is the vertex of a conical surface? . Is it necessary for the directrix of a conical surface to be closed? . What is a nappe of a cone? How many nappes are generated in a conical surface? . What is a cone? . What is the base of a cone? . Must all elements be cut by the plane of the base for a cone? . What is a circular cone? . What is the axis of a cone? . What is a right circular cone? Is it a cone of revolution? . What is the altitude of a cone? . What is the slant height of a cone? . What is the an oblique cone? . What is a frustum of a cone? . How are the two bases of a frustum of a cone usually designated? . What is a truncated cone? . In the representation of a cone, why is the plane of the base usually assumed perpendicular to the plane of projection? . Is it necessary that the base of a cone should be circular? . Draw a cone in orthographic projection and assume the plane of the base perpendicular to the vertical plane. . Draw a cone in projection and show how an element of the surface is assumed in both projections. State exactly where the element is chosen. . Draw a cone in projection and show how a point is assumed on its surface. Locate point in both projections. . What is a cylindrical surface? . Define generatrix of a cylindrical surface; directrix; element. . Is it necessary that the directrix of a cylindrical surface be a closed curve? . How is a cylinder differentiated from‘a cylindrical surface? . How many bases must a cylinder have? . What is the axis of a cylinder? . Must the axis of the cylinder be parallel to the elements? Why? . What is an oblique cylinder? . What is a right circular cylinder? Is this cylinder a cylinder of revolution? . Is it necessary that a right cylinder have a circle for the base? Why? . Draw an oblique cylinder whose base lies in the horizontal plane. . In Question 38, assume an element of the surface and state which element is chosen. . Which elements are the limiting elements in Question 38? . Draw a cylinder in projection and then assume a point on the surface of it. Show it in both projections. . What is a convolute surface? . What is an oblique helicoidal screw surface? Give a prominent example of it. 1&2 44, 45. 46. 47, 48. 49, 50. SL. 52. 53. 54, GEOMETRICAL PROBLEMS IN PROJECTION What is a right helicoidal screw surface? Give a prominent example of it. Show that the helicoidal screw surfaces are limiting helicoids as the inner helix becomes of zero diameter. What is a warped surface? Illustrate by sticks. What is a secant plane? Show how a tangent plane is the limiting position of a secant plane to a conical surface. Show how a tangent plane is the limiting position of a secant plane to a cylindrical surface. Show that a tangent line to the surface is the limiting position of a secant drawn to a curve on the surface. Show how two intersecting lines may be drawn on a curved surface and how the limiting positions of two secants drawn through this point of intersection determine a tangent plane to the surface. Show how one element of a conical surface and one limiting posi- tion of a secant determine the tangent plane to the conical surface. Show how one element of a cylindrical surface and one limiting position of a secant determine the tangent plane to the cylindrical surface. Show how two intersecting lines may be drawn on the surface of a sphere and how the limiting positions of two secants drawn through the intersection determine a tangent plane to the sphere. . Define tangent plane in terms of the two intersecting tangent lines at a point on a surface. . Define normal plane. . How many normal planes may be drawn through a given point on a surface? . What is a normal (line) to a surface? . Define singly curved surface. Give examples. . Define doubly curved surface. Give examples. . Define singly curved surface of revolution. Give examples. . Define doubly curved surface of revolution. Give examples. . What is a doubly convex surface? Give examples. . What is a doubly concave surface? Give examples. . What is a doubly concavo-convex surface. Give examples. . Describe the surface of a torus. Is this a concavo-convex surface? . What surface is obtained when a pair of skew lines are revolved about one of them as an axis? . Construct the surface of Question 67. . What is a meridian plane? . What is a meridian line? . Why can any meridian line be assumed as a generatrix for its partic- ular surface of revolution? . What curves are obtained by passing planes perpendicular to the axis of revolution? CLASSIFICATION OF SURFACES 183 . When two surfaces of revolution have the same axis, show that the intersection is a circle whether the surfaces intersect tangentially or angularly. . Show how a doubly curved surface of revolution may be represented without the ground line. Draw the proper centre lines. . Assume a point on the surface of a doubly curved surface of revolu- tion. . What is a developable surface? . Is development the rectification of a surface? . Are singly curved surfaces developable? . Are doubly curved surfaces developable? . Are warped surfaces developable? . What is a ruled surface? Give examples. . What is an asymptotic plane? Give an example. . If a hyperbola and its asymptotes revolve about the transverse axis, show why the asymptotic lines generate asymptotic cones to the resulting hyperboloids. . Show what changes occur in Question 83 when the conjugate axis is used. . Make a classification of surfaces. CHAPTER XI INTERSECTIONS OF SURFACES BY PLANES, AND THEIR DEVELOPMENT 1101. Introductory. When a line is inclined to a plane, it will if sufficiently produced, pierce the plane in a point. The general method involved has been shown (Art. 823), for straight lines, and consists of passing an auxiliary plane through the given line, so that it cuts the given plane in a line of intersection. The piercing point must be somewhere on this line of intersection and also on the given line; hence it is at their intersection. In the case of doubly curved lines, the passing of auxiliary planes through them is evidently impossible. Curved surfaces, instead of planes, are therefore used as the auxiliary surfaces. Let, for example, Fig. 184 show a doubly curved line, and let the object be to find the piercing Aa’ point of the doubly curved line NN on the principal planes. If a cylindrical surface be passed | ey a through the given line, the he N elements of which are perpen- SL NS - dicular to the horizontal plane, : it will have the curve ab as Fic. 184. its trace, which will also be the horizontal projection of the curve AB in space. Similarly, the vertical projecting cylindrical surface will cut the vertical plane in the line a’b’, and will be the vertical projection of the curve AB in space. If a perpendicular be erected at a, where ab crosses the ground line, it will intersect the vertical projection of the curve at a’, the vertical piercing point of the curve. The entire process in substance consists of this: the surface of the horizontal projecting cylinder cuts the vertical plane in the line aa’; the piercing point of the curve AB must 184 INTERSECTIONS OF SURFACES BY PLANES 185 lie on aa’ and also on AB, hence it is at their intersection a’. Likewise, b, the horizontal piercing point is found by a process identical with that immediately preceding. 1102. Lines of intersection of solids by planes. The extension of the foregoing is the entire scheme of finding the line of intersection of any surface with the cutting plane. Ele- ments of the surface pierce the cutting plane in points; the locus of the points so obtained, determine the line of intersection. A distinction must be made between a plane cutting a sur- face, and a plane cutting a solid. In the former case, the surface. alone, gives rise to the line of intersection, whether it be an open surface, or a closed surface; the cutting plane intersects the surface in a line which is the line of intersection. In the latter case, the area of the solid, exposed by the cutting plane, is a section of the solid. This is the scheme of using section planes for the elucidation of certain views in drawing (313). 1103. Development of surfaces. The development of a surface consists of the rolling out or rectification of the surface on a plane, so that the area on the plane is equal to the area of the surface before development. If this flat surface be rolled up, it will regenerate the original surface from which it has been evolved (1028). For instance, if a flat rectangular sheet of paper be rolled in a circular form, it will produce the surface of a cylinder of revolution. Similarly, a sector of a circle may be wound up so as to make a right circular cone. In both cases, the flat sur- face is the development of the surface of the cylinder or cone as the case may be. 1104. Developable surfaces. A prism may be rolled over a flat surface and each face successively comes in intimate con- tact with the flat surface; hence, its surface is developable. A cylinder may likewise be rolled over a flat surface and the con- secutive elements will successively coincide with the flat sur- face, this, again, is therefore a developable surface. The sur- faces of the pyramid and the cone are also developable for similar reasons. A sphere, when rolled over a flat surface, touches at only one point and not along any one element; hence, its surface is 186 GEOMETRICAL PROBLEMS IN PROJECTION not developable. In general, any surface of revolution which has a curvilinear * generatrix is non-developable. A warped surface is also a type of surface which is non-develop- able, because consecutive elements even though rectilinear, are so situated that no plane can contain them. Hence, to review, only singly curved surfaces, and such surfaces as are made up of intersecting planes, are developable. Doubly curved surfaces of any kind are only developable in an approximate way, by dividing the actual surface into a series of developable surfaces; the larger the series, the nearer the approximation. 1105. Problem 1. To find the line of intersection of the surfaces of a right octagonal prism with a plane inclined to its axis. ‘Construction. Let abcd, etc., Fig. 185, be the plan of the prism, resting, for convenience, on the horizontal plane. The elevation is shown by a’b’c’, etc. The plane T is perpendicular to the vertical plane and makes an angle « with the horizontal plane. In the supplementary view S, only half of the intersec- tion is shown, because the other half is symmetrical about the line a’’e’’. Draw from e’d’c’b’ and a’, lines perpendicular to Tt’ and draw a’’e’’”’, anywhere, but always parallel to Tt’. The axis of symmetry ea, shown in plan, is the projection of e’a’ in elevation, and e’a’ is equal to e’”’a’”’; the points e’” and a’ are therefore established on the axis. To find d’”’, draw dm, perpendicular to ea; dm is shown in its true length as it is parallel to the horizontal plane. Accordingly, set off m’’d’” = md, on a line from d’, perpendicular to Tt’, from e’’a’” as a base line. As nb=md, then n’’b’’=m’’d’’, and is set off from ea on a line from b’, perpendicular to Tt’. The final point c’” is located so that o’”’c’’’=oc and is set off from e’”’a’”, on a line from c’, perpendicular to Tt. The “half section” is *If a curvilinear generatrix moves so that its plane remains continually parallel to itself and touches a rectilinear directrix, the surface generated is a cylindrical surface. Hence, this surface cannot be included in this connection. t Maps are developments of the earth’s surface, made in various ways. This branch of the subject falls under Spherical Projections. The student who desires to pursue this branch is referred to the treatises on Topographical Drawing and Surveying. INTERSECTIONS OF SURFACES BY PLANES 187 then shown completely and is sectioned as is customary. The line of intersection is shown by e’’d’’c’’b’’’a’’’; the half section is the area included by the lines e’’d’"c'"b'’a’'n'0'"'m'"e’”’. 1106. Problem 2. To find the developed surfaces in the preceding problem.* : Construction. As the prism is a right prism, all the vertical edges are perpendicular to the base; the base will develop into a straight line and the vertical edges will be perpendicular to it, \ ” Fig. 185. spaced an equal distance apart, because all faces are equal. Hence, on the base line AA, Fig. 185, lay off the perimeter of the prism and divide into eight equal parts. The distance of a” above the base line AA is equal to the distance of a’ above the horizontal plane; b’” is ablove AA, a distance equal to b’ above the horizontal plane. In this way, all points are located. It will be observed that the development is symmetrical about the vertical through e’’. To prove the accuracy of the construction, the developed surface may be laid out on paper, creased along all the verticals * In the problems to follow the bases are not included in the development ag they are evident from the drawing. 188 GEOMETRICAL PROBLEMS IN PROJECTION and wound up in the form of the prism. A flat card will admit of being placed along the cut, proving that the section is a plane surface. The upper and lower portions are both developed; the dis- tance between them is arbitrary, only the cut on one must exactly match with the cut on the other. 1107. Problem 3. To find the line of intersection of the surface of a right circular cylinder with a plane inclined to its axis. ? Construction. Let Fig. 186 show the cylinder in plan and Fra. 186. elevation. The plane cutting it is an angle « with the horizontal plane and is perpendicular to the vertical plane. It is customary, in the application, to select the position of the plane and the object so that subsequent operations become most convenient, so long as the given conditions are satisfied. Pass a series of auxiliary planes through the axis as oe, od, etc., spaced, for con- venience, an equal angular distance apart. These planes cut rectilinear elements from the cylinder, shown by the verticals through e’d’c’, etc. Lay off an axis e’”’a’”, parallel to Tt’, so that e’”’ e’ and a’’’a’ are perpendicular to Tt’. To find d’”, it is INTERSECTIONS OF SURFACES BY PLANES 189 known that md is shown in its true length in the plan; hence, lay off m’’’d’’’ ==md from the axis e’’a’’’, on a line from d’, perpen- dicular to Tt’. Also, make 0’’c’” =oc and n/b’” =nb in the same way. Draw a smooth curve through e’’d’"0e'"b'"a’"b’"c'"" de” which will be found to be an ellipse. The ellipse may be awn by plotting points as shown, or, the major axis e’’a’”’ and the minor axis c’’”’c’”” may be laid off and the ellipse drawn by any method.* Both methods should produce identical results. The area of the ellipse is the section of the cylinder made by an oblique plane and the ellipse is the curve of intersection. As an example, a cylindrical glass of water may be tilted to the given angle, and the boundary of the surface of the water will be elliptical. 1108. Problem 4. To find the developed surface in the preceding problem. Construction. As the cylinder is a right cylinder, the ele- ments are perpendicular to the base, and the base will develop into a straight line, of a length, equal to the rectified length of the circular base. Divide the base AA Fig. 186, into the same number of parts as are cut by the auxiliary planes. Every element in the elevation is shown in its true size because it is perpendicular to the horizontal plane; hence, make a”’A equal to the distance of a’ above the horizontal plane; b’A equal to the distance of b’ above the horizontal plane, etc. Draw a smooth curve through the points so found and the development will appear as shown at D in Fig. 186. Both upper and lower portions are shown developed, either one may be wound up like the original cylinder and a flat card placed across the intersection, showing, that the sur- face is plane. 1109. Application of cylindrical sur- faces. Fig. 187 shows an elbow, approximat- ing a torus, made of sheet metal, by the use of short sections of cylinders. The ellipse is symmetrical about both axes, and, hence, the Fig. 187. upper portion of the cylinder may be added to the lower portion so as to give an offset. Indeed in this way elbows. * See Art. 906. 190 GEOMETRICAL PROBLEMS IN PROJECTION are made in practice. The torus is a doubly curved surface, and, hence, is not developable, except by the approximation. b b Le rr Fig. 188. Se + — shown. ‘To develop these individual sheets, pass a plane ab, perpendicular to the axis. The circle cut thereby will then develop into a straight. line. Add corresponding distances. above and below the base line so. established and the development may be completed. Fig. 188 shows. the sheets as they appear in the: development. A sufficient number of points must always be found on any curve, so that no doubt occurs as to its form. In the ‘llustration, the number of points is always less than actually required so as to avoid the confusion incident to a large number of construction lines. 1110. Problem 5. To find the line of intersection of the surfaces of a right octagonal pyramid with a plane inclined to. its axis. Construction. Let Fig. 189 represent the pyramid in plan and elevation, and let T be the cutting plane. The plane T cuts the extreme edge o’e’ at e’, horizontally projected at e, one point of the required intersection. It also cuts the edge o’d’ at d’, horizontally projected at d, thus locating two points on the required intersection. In the same way b and a are located. The point c cannot be found in just this way. If the pyramid be turned a quarter of a revolution, the point c’ will be at q’ and c’q’ will be the distance from the axis where the edge o’c’ pierces the plane T. Hence, lay off oc=c’q’ and complete the horizontal projection of the intersection. To find the true shape of the intersection, draw the supple- mentary view S. Lay off e’’’a’” as the axis, parallel to Tt’; the length of the axis is such that e’a’=e’”a’”. From the horizontal projection, dm is made equal to d’’m’” in the supple- mentary view, the latter being set off from the axis e’’’a’”’, and on a line from d’, perpendicular to Tt’. Also, co=c’”0’” and bn=b’’n’”. Thus, the true intersection is shown in the supple- mentary view S. INTERSECTIONS OF SURFACES BY PLANES 191 1111. Problem 6. To find the developed surfaces in the preceding problem. Construction. The extreme element o’s’, Fig. 189, is shown in its true length on the vertical plane because it is parallel to that plane. Accordingly, with any point 0” as a centre, draw an indefinite arc through AA, so that o’A=o’s’. Every edge of the pyramid meeting at the vertex has the same length and all are therefore equal to o’’A. The base of each triangular face is shown in its true length in the plan, and, hence, with any one of them as a length, set the distance off as a chord on AA Fig. 189. by the aid of a divider, so that the number of steps is equal to the number of faces. Draw these chords, indicate the edges, and it then remains to show where the cutting plane intersects the edges. The extreme edge o’e’ is shown in its true length, hence, lay off oe’ =o’e’. The edge o’d’ is not shown in its true length, but if the pyramid is revolved so that this edge reaches the position o’s’, then d’ will reach p’ and o’p’ is the desired true length as it is now parallel to the vertical plane; therefore, set off o’p’=0'd”. In this way, 0’q’=0'’c”, and o'r’ = ob’. The edge o’a’ is shown in its true length, therefore, o’a’ =o’a"; and a’A on one side must equal a”A on the other, 192 GEOMETRICAL PROBLEMS IN PROJECTION as, on rolling up, the edges must correspond, being the same initially. Join ab’, b’’c”, etc., to complete the development. The proof, as before, lies in the actual construction of the model and in showing that the cut is a plane surface. 1112. Problem 7. To find the line of intersection of the surface of a right circular cone with a plane inclined to its axis. Construction. Let Fig. 190 show plan and elevation of the cone, and let T be the cutting plane. Through the axis, pass a WY Fig. 190. series of planes as ao, bo, co, etc. These planes cut rectilinear elements from the cone, shown as 0’a’, o’b’, o’c’, etc., in the eleva- tion. At first, it is a good plan to draw the horizontal projection of the line of intersection, as many points are readily located thereon. For instance, e’ is the point where the extreme ele- ment o’e’ pierces the plane T, horizontally projected at e. Sim- ilarly, the actual element OD pierces the plane so that d and d’ are corresponding projections. The points b and b’ and the points a and a’ are found in an identical manner, but, as in the case of the pyramid, the point ¢ cannot be located in the same way. If the cone be revolved so that the element o’c’ occupies INTERSECTIONS OF SURFACES BY PLANES 193 the extreme position o’s’, then c’ will be found at q’ and c’q’ will be the radius of the circle which the point c’ describes; hence, make oc=c’q’ and the resultant curve, which is an ellipse, may be drawn. The true line of intersection is shown in the supplementary view S. As in previous instances, first the axis e’”’a’” is drawn parallel to Tt’, and equal in length to e’a’ so that e’’’e’ and a’’a’ are both perpendicular to Tt’. From the axis, on a line from d’, lay off d’’m’” =dm; also o’” c’’”’=oc and b’’n’”’” =bn. A smooth curve then results in an ellipse. : If accuracy is desired, it is better to lay off the major and minor axes of the ellipse and then draw the ellipse by other methods,* since great care must be used in this construction. The major axis is shown as e’’a’”’. To find the minor axis, bisect e’a’ at u’and draw u’v’, the trace of a plane perpendicular to the axis of the cone. This plane cuts the conical surface in a circle, of which w’v’ is the radius. The minor axis is equal to the length of the chord of a circle whose radius is w’v’ and whose distance from the centre of the circle is u’w’. 1113. Problem 8. To find the developed surface in the preceding problem. Construction. The extreme visible element o’s’, Fig. 190, is shown in its true length and is the slant height of the cone. All elements are of the same length and hence, any indefinite are AA, drawn so that o’A=o’s’ will be the first step in the development. The length of the arc AA is equal to the rectified length of the base, and, as such, is laid off and divided into any convenient number of parts. Eight equal parts are shown here because the auxiliary planes were chosen so as to cut the cone into eight equal parts. If the conical surface is cut along the element OA, then o’a’ may be laid off on both sides equal to o’’a’’ in the development. The elements o’d’, o’c’ and o’b’ are not shown in their true length, therefore, it is necessary to revolve the cone about the axis so as to make them parallel, in turn, to the vertical plane. The points will ultimately reach the positions indicated by p,’ q’ and 1’, and, hence, 0b” =o’r’, o”’c’’=0'q' and o’d”=o’p’. The element o’e’ is shown in its true length and is laid off equal to o’’e”. A smooth curve through abcd’, etc., completes the required development. * See Art. 906 in this connection. 194 GEOMETRICAL PROBLEMS IN PROJECTION When frustra of conical surfaces are to be developed the elements of the surface may be produced until they meet at the vertex. The development may then proceed along the usual lines. 1114. Application of conical surfaces. As the ellipse is a symmetrical curve, the upper and lower portions of the cone may be turned end for end so that the axes intersect. The resultant shape is indicated in Fig. 191, used at times in various sheet metal designs, such as oil-cans, tea-kettles, etc. 1115. Problem 9. To find the line of inter- section of a doubly curved surface of revolution with a plane inclined to its axis. Construction. Let Fig. 192 show the elevation of the given surface of revolution.* Atten- tion will be directed to the construction of the section shown in the plan. The high- est point on the curve is shown at a, which is found from its corresponding projection a’, at the point where the plane T cuts the meridian curve that is parallel to the plane of the paper. The point b is directly under b’ and the length bb is equal to the chord of a circle whose radius is m’n’ and whose distance from the cen- tre is b’n’. Similarly, cc is found by drawing an indefi- nite line under c’ intersected Fic. 192. by op=o’p’ as a radius. As many points as are necessary are located, so as to get the true shape. One thing, however, must be observed: The plane TT *Many of these constructions can be carried to completion without actually showing the principal planes. The operations on them are per- formed intuitively. INTERSECTIONS OF SURFACES BY PLANES 195 cuts the base of the surface of revolution at h’ in the elevation and therefore hh is a straight line, which is the chord of a circle, whose radius is q’r’ and which is located as shown in the plan. The construction of the supplementary view resembles, in many respects, the construction in plan. The similar letters indicate the lengths that are equal to each other. 1116. Problem 10. To find the line of intersection of a bell- surface with a plane. Construction. Fig. 193 shows the stub end of a connecting rod as used on a steam engine. The end is formed in the lathe by turning a bell-shaped surface of revolution on a bar of a “ Q@ gest eie. ah Mees “s-?Z “of zy Ss = en Sb SSaaS ‘ ELEVATION a 1\ TEND VIEW ' e! Saree rectangular section, as shown in the end view, the shank being circular. The radius of curvature for the bell surface is located on the line ab as shown. The plane TT is tangent to the shank of the rod and begins to cut the bell-surface at points to the left of ab; hence, the starting point of the curve is at c. To find any point such as d, for instance, pass a plane through d perpendicular to the axis. It cuts the bell-surface in a circle (1024) whose radius is od’ =o’d’”’. Where the circle intersects the plane TT at d’, project back to d, which is the required point on the curve. The scheme is merely this: By passing auxiliary planes perpendicular to the axis, circles are cut from the bell- surface which pierce the bounding planes in the required points of intersection. Attention might be directed to the piont e which is on both plan and elevation. This is true because’ the planes TT and SS intersect in a line which can pierce the bell- 196 GEOMETRICAL PROBLEMS IN PROJECTION surface at only one point. The manner in which nn=n'n! is located in the plan will be seen by the construction lines which are included in the illustration. To follow the description is more confusing than to follow the drawing. 1117. Development by Triangulation. In the developments so far considered, only the cases of extreme simplicity were selected. To develop the surface an oblique cone or an oblique cylinder requires a slightly different mode of procedure than was used om °° ‘ LA i if Ye ° Fic. 194. heretofore. If the surface of the cone, for instance, be divided into a number of triangles of which the rectilinear elements form the sides and the rectified base forms the base, it is possible to plot these triangles, one by one, so as to make the total area of the triangles equal to the area of the conical surface to be developed. The method is simple and is readily applied in practice. Few illustrations will make matters clearer. 1118. Problem 11. To develop the surfaces of an oblique hexagonal pyramid. Construction. Fig. 194 shows the given oblique pyramid. It is first necessary to find the true lengths of all the elements of the surface. In the case of the pyramid, the edges alone need INTERSECTIONS OF SURFACES BY PLANES 197 be considered. Suppose the pyramid is rotated about a perpen- dicular to the horizontal plane, through o, so that the base of the pyramid continually remains in its plane, then, when oa is parallel to the vertical plane, it is projected in its true length and is, hence, shown as the line o’a’’. All the sides are thus brought para lel, in turn, to the plane and the true lengths are found. In most cases it will be found more convenient and less confusing to make a separate diagram to obtain the true lengths. To one with some experience, the actual lengths need not be drawn at all, but simply the distances a’’, b”, c’’, etc., laid off. Then from any point o’”, draw arcs o’a”’ =0'"a’”, o’b” =0'"b’”, of” = Fic. 195. of’, ete. From a’, lay off distances equal to the respective bases; in this case, ab=bc=cd, etc., and hence any one side will answer the purpose. Therefore, take that length on a divider and step off ab=a’"b’’=b'"c'"” =c'"d'” =a"'f’”, etc. The development is then completed by drawing the proper lines. The case selected, shows the triangulation method applied to a surface whose sides are triangles. It is extremely simple for that reason, but its simplicity is still evident in the case of the cone as will be now shown. 1119. Problem 12. To develop the surface of an oblique cone. Construction. Fig. 195 shows the given oblique cone. The horizontal projection of the axis is op and the base is a circle, 198 GEOMETRICAL PROBLEMS IN PROJECTION hence, the cone is not a cone of revolution because the axis is inclined to the base. Revolve op to oq, parallel to the vertical plane. Divide the base into any number of parts ab, be, cd, etc. preferably equal, to save time in subsequent operations. The element oa is parallel to the plane, hence, o’a’ is its true length. The element ob is not parallel to the plane, but can be made so by additional revolution as shown; hence o’b’ is its true length. From any point 0”, draw 0”a’’=0’a’, ob” =o’b’, oc’ =o0'c’, etc. On these indefinite arcs step off distances a’’b’”’ = b’c” =c’’d”, etc., equal to the rectified distances ab=bce=cd, etc. A smooth curve through the points a’b’c”, etc., will give Fia. 195. the development. As in previous problems, the cut is always made so as to make the shortest seam unless other requirements prevail. 1120. Problem 13. To develop the surface of an oblique cylinder. Construction. In the case of the oblique cone, it was pos- sible to divide the surface into a series of triangles which were plotted, one by one, and thus the development followed by the addition of these individual triangles. In the case of the cylin- der, however, the application is somewhat different, although two cones may be used each of which has one base of the cylinder INTERSECTIONS OF SURFACES BY PLANES 199 as its base. The application of this is cumbersome, ani the better plan will be shown. Let Fig. 196 show the oblique cylinder, chosen, for convenience, with circular bases. Revolve the cylinder as shown, until it is parallel to the vertical plane. In the revolved position, ass a plane T through it, perpendicular to the axis. The curve so cut when rectified, will develop into a straight line and the elements of the surface will be perpendicular to it. The true shape of the section of the cylinder is shown by the curve a”b’’c’’d”, etc., and is an ellipse. Its construction is indicated in the diagram. Draw any base line AA and lay off on this the rectified portions Fria. 196. of the ellipse a”’b”’=al’’b’”, b’c’=b'c’”’, etc. The revolved positions of the elements shown in the vertical plane are all given in their true sizes because they are parallel to that plane. Hence, lay off a’’o’” =a’o’, b’’p’’=b’p’, etc., below the base line. Do the same for the elements above the base line and the curve determined by the points so found will be the development of the given oblique cylinder. 1121. Transition pieces. When an opening of one cross- section is to be connected with an opening of a different cross- section, the connecting piece is called a transition piece. The case of transforming a circular cross-section to a square cross- section is quite common in heating and ventilating flues, boiler flues and the like. In all such cases, two possible methods offer 200 GEOMETRICAL PROBLEMS IN PROJECTION themselves as solutions. The two surfaces may be connected by a warped surface, the rectilinear elements of which are chosen so that the corners of the square are joined with the quarter points on the circle while the intermediate elements are distributed between them. A warped surface, however, is not developable, it is also difficult of representation and therefore commercially unsuited for application. The better method is to select some singly curved surface, or, a combination of planes and singly curved Fia. 197. surfaces since these are developable. The application, which is quite 2 common one, is shown in the following problem. 1122. Problem 14. To develop the surface of a transition piece connecting a circular opening with a square opening. Construction. Let the two upper views of Fig. 197 represent the transition piece desired. The square opening is indicated by abcd and the circular opening by efgh. The surface may be made up of four triangular faces aeb, bfc, cgd, dha, and four conical surfaces hae, ebf, feg, gdh, whose vertices are at a, b, c and d. The development is shown at D, on the same figure. The INTERSECTIONS OF SURFACES BY PLANES 201 triangle a’’b’’e”’ is first laid out so that the base a/b” =ab; and the true length of the sides are determined as found in the construction leading to the position a’e’=a”e’=b’e”’. The conical surfaces are developed by triangulation. For example, the arc he is divided into any number of parts em, mn, etc.; and the true lengths of the elements of the surface are found ce Fig. 198. by the construction leading up to the positions a’e’, a’m’, a’n’. Hence, ae” =a’e’, am’ =a'm’; and me” =me is the rectified arc of the base of the conical surface. When four of these com- binations of triangle and conical surface are laid out, in their proper order, the development is then complete. If a rectangular opening is to be joined to an elliptical open- ing, the manner in which this may be accomplished is shown in Fig. 198. 1123. Problem 15. To develop the surface a transition piece connect- ing two elliptical openings whose major axes are at right angles to each other. Construction. Fig. 199 shows three views of the elliptical transition piece. By reference to the diagram, it will be seen that the surface Fic. 199. may be divided into eight conical surfaces, turned end for end; that is, four vertices are situated on one ellipse and four vertices are situated on the ’ 202 GEOMETRICAL PROBLEMS IN PROJECTION other. The shaded figure indicates the arrangement of the conical surfaces and the elements are shown as shade lines. In all developments of this character, the ar- rangement of the surfaces has considerable effect on the appearance of the transition piece when completed. In the case chosen, the eight conical surfaces give a pleasing result. If the intersec- tion of the eight alter- Fie. 199. nate conical surfaces is objectionable then it is possible to use a still larger number of divisions. The development of the conical surfaces is performed by the triangulation method. When finished, its appearance is that a” of” a” eo” a!” a" BZ o” a” a” Fie. 200. of Fig. 200. The similar letters indicate the order in which the surfaces are assembled. 1124. Development of doubly curved surfaces by approx= imation. Doubly curved surfaces are non-developable because successive elements cannot be made to coincide with a plane. It is possible, however, to divide the surface into a series of singly curved surfaces and then to develop: these. By dividing the doubly curved surface into a sufficiently large number of parts, INTERSECTIONS OF SURFACES BY PLANES 203 the approximation may be made to approach the surface as closely as desired. For doubly curved surfaces of revolution, two general methods are used. One method is to pass a series of meridian planes through the axis and then to adopt a singly curved surface whose contour is that of the surface of revolution at the meridian planes. This method is known as the gore method. The second method, known as the zone method, is to divide the surface into frusta of cones whose vertices lie on the axis of the doubly curved surface of revolution. The following examples will illustrate the application of the methods. 1125. Problem 16. To develop the surface of a sphere by - the gore method. Construction. Let the plan P of Fig. 201 show the sphere. Fic. 201. Pass a series of meridian planes aa, bb, cc, dd, through the sphere and then join ab, bc, cd, etc. From the plan P construct the elevation E and then pass planes through i’ and g’, perpendicular to the axis of the sphere, intercepting equal arcs on the meridian circle, for convenience. Find the corresponding circles in the plan and inscribe an octagon (for this case) and determine ef, the chord length for that position of the cutting planes through i’ and g’. This distance may then be laid off as e’f’ in the eleva- tion E and the curve n’e’a’e’m’ be drawn. For the development D, it will be noted that ab=a’b’ =a"b” 204 GEOMETRICAL PROBLEMS IN PROJECTION and ef=e'f/=e’f’. Also, the rectified distance m’g’=m’’g”, g’h’=g"h", h/i’=h’i’’, etc. A suitable number of points must be used in order to insure the proper degree of accuracy, the number chosen here is insufficient for practical purposes. By adding eight of these faces along the line a’’b’’, the development is completed. When this method is commercially applied, the gores may be stretched by hammering or pressing to their true shape. In this case they are singly curved surfaces no longer. Fie, 201. 1126. Problem 17. To develop the surface of a sphere by the zone method. Construction. Let P, Fig. 202, be the plan and E, the eleva- tion of a sphere. Pass a series of planes a’a’, b’b’, c’c’ etc., perpendicular to the axis of the sphere. Join a’b’, and pro- duce to h’, the vertex of a conical surface giving rise to the frustum a’'b’b’a’. Similarly, join b’c’ and produce to g’. The last conical surface has its vertex e’ in the circumscribing sphere. The development D is smilar to the developnent of any frustum of a conical surface of revolution. That is, with h” as a centre, draw an are with h”’a’’=h’a’ as a radius. At the centre line make a’’o”’ =ao, o’’n”’=on etc. The radius h’’b” =h’b’ so that the arc b’b’b” is tangent internally, to the similarly lettered arc. The radius for this case is g’b’=g’=b’. The development is completed by continuing this process until all the conical surfaces are developed. INTERSECTIONS OF SURFACES BY PLANES 205 Fie. 202. 1127. Problem 18. To develop a doubly curved surface of revolution by the gore method. Let Fig. 203 represent the doubly curved surface of revolution as finally approxi- mated. Through the axis ath pass a series of equally spaced meridian planes, and then draw the chords, one of which is lettered as dd. In the ele- vation E, pass a series of planes k’k’, 11’, m’m’, perpendicular to the axis. The lines b’b’, c’c’, d’d’, etc., are drawn on the resulting singly curved | surface and are shown in their true length. The corresponding projections of these lines in the plan P are also shown in their true length. To develop the surface of one face lay off the rectified dis- Fig. 203. GEOMETRICAL PROBLEMS IN PROJECTION j/k’ =j"k’, kV =k’l’, etc. At the points j’, k’, 1”, etc., lay off a’a’=a’a’’, b’b’=b’b”, etc., perpendicular to h”j’’ so that the resulting figure D is symmetrical about it. When eight of these faces (or gores) are joined together and secured along the seams, the resulting ! figure will be that shown in plan and [ |:\ » elevation. "3 @ 4 7 This surface may also be approxi- re Aba mated by the zone method. The Wy le ‘\22" | choice of method is usually governed \Y i tf Ashe by commercial considerations. The WAL ity gore method is perhaps the more Fra. 203. $ economical in material. 1. 2. oo ee RH Nb HOVONOTH oo eo 14. QUESTIONS ON CHAPTER XI State the general method of finding the piercing point of a line on a plane. Show by an oblique projection how the piercing points of a doubly curved line are found on the principal planes. . Why is the projecting surface of a doubly curved line a projecting cylindrical surface? . What is a “line of intersection” of two surfaces? . Distinguish between “line of intersection” and “section” of a solid. What is meant by the development of a surface? Explain fully. . What are the essential characteristics of a developable surface? Why is a sphere non-developable? . Why is any doubly curved surface non-developable? Why is a warped surface non-developable? . Prove the general case of finding the intersection of a right prism and a plane inclined to its axis. . Show the general method of finding the development of the prism in Question 11. . Prove the general case of finding the intersection of a right circular cylinder and a plane inclined to its axis. Show the general method of finding the development of the cylinder in Question 13. . When a right circular cylinder is cut by a plane why must the ellipse be reversible? . How is the torus approximated with cylindrical surfaces? . Prove the general case of finding the intersection of a right pyramid and a plane inclined to its axis. 18. 19. 20. 21. 22, 23. 24, 25, 27. 28. 29, 30. 31. 382. 33. 34, 41, 42, 43. INTERSECTIONS OF SURFACES BY PLANES 207 Show the general method of finding the development of the right pyramid in Question 17. Prove the general case of finding the intersection of a right circular cone and a plane inclined to its axis. Show the general method of finding the development of the right cone in Question 19. Why is the ellipse, cut from a cone of revolution by an inclined plane, reversible? Prove the general case of finding the intersection of a doubly curved surface of revolution and a plane inclined to the axis. Is the surface in Question 22 developable? Why? Prove the general case of finding the intersection of a bell-surface with planes parallel to its axis. Is the bell-surface developable? Why? . What is development by triangulation? Prove the general case of the development of the surface of an oblique pyramid. Prove the general case of the development of the surface of an oblique cone. Prove the general case of the development of the surface of an oblique cylinder. What is a transition piece? Why is it desirable to divide the surfaces of a transition piece into developable surfaces? Prove the general case of the development of a transition piece which joins a circular opening with a square opening. Prove the general case of the development of a transition piece which joins an elliptical opening with a rectangular opening. Prove the general case of the development of a transition piece which joins two elliptical openings whose major axes are at right angles to each other. . What is the gore method of developing doubly curved surfaces? . What is the zone method of developing doubly curved surfaces? . Prove the general case of the development of a sphere by the gore method. . Prove the general case of the development of a sphere by the zone method. . Prove the general case of the development of a doubly curved sur- face of revolution by the gore method. . A right octagonal prism has a circumscribing circle of 2’ and is 4 high. It is cut by a plane, inclined 30° to the axis, and passes through its axis, 1} from the base. Find the section. Develop the surface of the prism of Question 40. A cylinder of revolution is 23” in diameter and 37” high. It is cut by a plane, inclined 45° to the axis and passes through its axis 2” from the base. Find the section. Develop the surface of the cylinder of Question 42. 208 GEOMETRICAL PROBLEMS IN PROJECTION 44, 45. 46. 47, 48. 49, 50. 51. 52. 53. 54, 55. 56. 57. 58. A right octagonal pyramid has a circumscribing base circle of 2}” and is 4” high. It is cut by a plane, inclined 30° to the axis and passes through its axis 2” from the base. Find the section. Develop the surface of the pyramid of Question 44. A right circular cone has a base 2” in diameter and is 3} high. It is cut by plane inclined 45° to the axis and passes through its axis 21’ from the base. Find the section. Develop the cone of Question 46. A 90° stove pipe elbow is to be made from cylinders 4” in diameter. The radius of the bend is to be 18”. Divide elbow into 8 parts. The tangent distance beyond the quadrant is 4’. Develop the surface to suitable scale. Assume a design of a vase (a doubly curved surface of revolution) and make a section of it. The stub end of a connecting rod of a steam engine is 4’ x6”; the rod diameter is 34”. The bell-surface has a radius of 12’. Find the lines of intersection. Draw to suitable scale. An oblique pyramid has a regular hexagon for a base. The circum- scribing circle for the base has a diameter of 2’; the altitude is 33”’ and the projection of the apex is 23’’ from the centre of the base. Develop the surface of the pyramid. An oblique cone has a circular base of 24” in diameter and an altitude of 3”. The projection of the vertex on the plane of the base is 13” from the centre of the base. Develop the conical surface. An oblique cylinder has a circular base 12” in diameter; it is 23” high and the inclination of the axis is 30° with the base. Develop the surface. A transition piece is to be made joining a 3’-0” x5’-6” opening with a 4’-0’” diameter circle. The distance between openings 3’-6”. Develop the surface. Use a suitable scale for the drawing. A transition piece is to be made joining an opening having a 3’-0” <4'-0”" opening to a 2’-0” x6’-0” opening (both rectangular). The distance between openings is 4-0’. Develop the sheet and use suitable scale for the drawing. A square opening 4’’-0’ x4’-0” is to be joined with an elliptical opening whose major and minor axes are 5’-0’”’ x 3’ —0” respec- tively. The distance between openings is 3’-0’. Develop the sheet. Use suitable scale in making the drawing. A circular opening having a diameter of 3’-0” is to be connected with an elliptical opening whose major and minor axes are 4’-0” x2’-6"" respectively. The distance between openings is 6’-0". Develop the sheet. Use suitable scale in making the drawing. An opening having two parallel sides and two semicircular ends has overall dimensions of 3’-0’ x5’-6”. This opening is to be joined: with a similar opening having dimensions of 3’-6’ x4’-6”, The distance between openings is 3’-6’. Develop the sheet. Use suitable scale in making the drawing. INTERSECTIONS OF SURFACES BY PLANES 209 59. A frustum of a cone has a base which is made up of two parallel sides and two semicircular ends and whose overall dimensions are 3’-6” <6’-0’"".. The height of the vertex above the base is 8’-6”; that of the upper base is 4’-0’’ above the lower base. Develop the sheet and use suitable scale in making the drawing. 60. An elliptical opening having a major and minor axis of 5’-0” and 61 62 63 64 65 66 4’-0”, respectively, is to be joined with a similar opening but whose major axis is at a right angle. Develop the sheet and use suitable scale in making the drawing. . An elliptical opening having a major and minor axis of 4’-6”’ and 3’-6’”, respectively, is to be joined with a similar opening but whose major axis is at an angle of 30°. Develop the sheet and use suitable scale in making the drawing. . An elliptical opening having a major and minor axis of 5’-0” and 3’-6”, respectively, is to be joined with another elliptical opening whose major and minor axes are 4’-6” and 3’-6’’, respectively. Develop the sheet and use suitable scale in making the drawing. . A sphere is 6” in diameter. Develop the surface by the gore method: and, divide the entire surface into sixteen parts. Use a suitable scale in making the drawing. . A sphere is 6” in diameter. Develop the surface by the zone method and divide the entire surface into twelve parts. Use a suitable scale in making the drawing. . Assume some doubly curved surface of revolution and develop the surface by the gore method. . Assume some doubly curved surface of revolution and develop the surface by the zone method. CHAPTER XII INTERSECTIONS OF SURFACES WITH EACH OTHER, AND THEIR DEVELOPMENT 1201. Introductory. When two surfaces are so situated with respect to each other that they intersect, they do so in a line which is called the line of intersection. It is highly desirable in the conception of intersections to realize that certain elements of one surface intersect certain elements on the other surface, and that the locus of these intersecting elements is the desired line of intersection. , The line of intersection is found by passing auxiliary surfaces through the given surfaces. Lines will thus be cut from the given surfaces by the auxiliary surfaces, the intersections of which yield points on the desired line of intersection. The simplest auxiliary surface is naturally the plane, but sometimes cylindrical,* conical and spherical surfaces may find application. It is not necessary only to use a simple type of auxiliary surface, but also to make the surface pass through the given surfaces so as to cut them in easily determinable lines—preferably, in the elements. As illustrations of the latter, the case of a concial surface cut by a plane passing through the vertex furnishes an example. It cuts the surface in straight lines or rectilinear elements. Likewise, a plane passed through a cylindrical sur- face parallel to any element will cut from it, one or more lines which are also rectiliner elements. For doubly curved surfaces of revolution, a plane passed through the axis (a meridian plane) cuts it in a meridian curve. When the plane is perpendicular to the axis, it cuts it in circles. An interesting and valuable * A careful distinction must be made between solids and their bounding surfaces. The terms, cylinder and cylindrical surface are often used indis- criminately. The associated idea is usually obtained from the nature of the problem. 210 INTERSECTIONS OF SURFACES WITH EACH OTHER 211 property of spheres is that all plane sections in any direction, or in any place, are all circles, whose radii are readily obtain- able. When prisms or pyramids intersect, auxiliary planes aré not required to find their intersection, because the plane faces and the edges furnish sufficient material with which to accomplish the desired result. These types of surfaces are therefore exempt from the foregoing general method. 1202. Problem 1. To find the line of intersection o* the surfaces of two prisms. Construction. Let Fig. 204 represent the two prisms, one of which, for convenience, is a right hexagonal prism. The face ABDC intersects the right prism in the line mn, one line of the required intersec- tion. The face CDFE is intersected by the edge of the right prism in the point o, found as shown in the con- struction. Hence, mo is the next line of the required intersection. Similarly, the edge EF pierces the faces of the right prism at p, and op is the con- tinuation of the intersection. In this way, all points are found. The curve must be closed because the inter- Fic. 204. penetration is complete. Only the one end where the prism enters is shown in construction. The curve where it again emerges is found in an identical manner. 1203. Problem 2. To find the developments in the preceding problem. Construction. The development of the right prism is quite simple and should be understood from the preceding chapter. The points at which the oblique prism inter- sects the right prism are also shown. Thus, it is only necessary to make sure on what element the point pierces, and this may be obtained from the plan view of the right prism. 212 GEOMETRICAL PROBLEMS IN PROJECTION In developing the oblique prism, D’ of Fig. 205, the same general scheme is followed as was used in developing the oblique cylinder (1120). Revolve the prism of Fig. 204 so that the elements are parallel to the plane of projection; they are then shown in their true length. Pass a plane perpendicular to the edges and this line of intersection will develop into a straight line which is used as a base line. Find now the true section of the prism and lay off the sides perpendicular to the base line and at their proper distances apart. Lay off the lengths of the edges above and below the base line and join the points by lines to complete the development. When actually constructing this as a model, it will be well to D’ D Fig. 205. carry out the work as shown, and make the oblique prism in one piece. It’may then be inserted in the opening provided for in the right prism, as indicated. 1204. Problem 3. To find the line of intersection of two cylin- drical surfaces of revolution whose axes intersect at a right angle. Construction. Let Fig. 206, represent the given cylindrical surfaces. Through 0’, pass a series of planes a’a’, b’b’, c’c’, etc. The elements cut from the cylindrical surface by these planes interest the other cylindrical surface at the points a, e, b, d, c. The two upper views lead to the construct on of the lower view. It will be seen by this construction that the entire scheme of locating points on the curve lies in the finding of the successive intersections of the elements of the surfaces. If the cylindrical surfaces have elliptical sections, instead of INTERSECTIONS OF SURFACES WITH EACH OTHER 213 circular sections as shown, the method of procedure will be found to be much the same. 1205. Problem 4. To find the developments in the preceding problem. Construction. As the elements of the surfaces depicted in Fig. 206 are parallel to the plane of the paper, they are therefore, shown in their true length. Hence, to develop the surfaces, :e! wre". he ed Fie. 206. Fic. 207. rectify the bases-and set off the points at proper distances from the base line and also their rectified distance apart. The appearance of the completed developments are shown in Fig. 207. This problem is similar to the development of the steam dome on a locomotive boiler. 1206. Problem 5. To find the line of intersection of two cylindrical surfaces of revolution whose axes intersect at any angle. Construction. Fig. 208 shows the two cylindrical surfaces chosen. If a series of auxiliary planes be passed parallel to the plane of the intersecting axes they will cut: the cylindrical surfaces in rectilinear ele- ments. The elements of one surface will intersect the elements of the other surface in the required points of the line of inter- Fra. 208. section. The construction shown in the figure should be clear from the similar lettering for the simiar points on the line of intersection. 214 GEOMETRICAL PROBLEMS IN PROJECTION It may be desirable to note that this method of locating the required points on the curve is also applicable to the con- struction of Fig. 206, and vice versa. If the plane of the intersecting axes is not parallel to the plane of the paper, the construction can be simplified by revolving the cylindrical surfaces until they become parallel. The method of procedure is then the same as that given here. If the axes cannot be made to lie in the same plane, then the construction is more difficult. The method in such cases is to pass a plane through one cylinder so as to cut in in elements; and the same plane will cut the other cylinder in some line of intersection. The intersection of these will Yet F" gu fe" g” yield points on the desired line ; of intersection. 1207. Problem 6. To find the developments in the pre- ceding problem. Construction. Fig. 209 shows the development as it appears. If the plane of the base is perpendicular to the elements of the surface, the base will develop into a straight line. The elements will be at Fic, 209. right angles to the base line and as they are shown in their true length, they may be laid off directly from F ig. 208. Fig. 210. Fig. 211, 1208. Application of intersecting cylindrical surfaces to pipes. The construction of pipe fittings as commercially used furnish examples of intersecting cylindrical surfaces. Fig. 210 INTERSECTIONS OF SURFACES WITH EACH OTHER 215 shows two cylindrical surfaces whose diameters are the same and whose axes intersect at a right angle. The line of intersection for this case will be seen to consist of two straight lines at right angles to each other. In Fig. 211, the cylindrical surfaces have their axes intersecting at an angle of 45° and have different diameters. 1209. Problem 7. To find the line of intersection of two cylindrical surfaces whose axes do not intersect. Construction. Fig. 212 shows two circular cylindrical sur- faces which are perpendicular to each other and whose axes are offset. Pass a series of planes ab, cd, ef, etc., through 0. These planes cut the cylindrical surface in elements, which in turn, intersect elements of the other surface. As the construction lines are completely shown, the description is unnecessary. m” mo ay gi" ay f “le i" ya” em e” gi” geld Ke i c Ae - par nl D b a” L ar at Fic. 212. Fig. 213. 1210. Problem 8. To find the developments in the pre- ceding problem. Construction. First consider the larger cylindrical surface, D’, Fig. 213, which is cut along any element. With a divider, space off the rectified distances between elements; and on the proper elements, lay off b’” to correspond with b”;_ k’"d’” =k’d” le” a te, ete. For the smaller cylindrical surface, rectify the entire circle ace... la and divide into the number of parts shown. As the elements are shown parallel to the plane of projection in the side elevation, they may be directly plotted as indicated by the curve g’7i/"k'”"". . . eg’. This completes the final development. 216 GEOMETRICAL PROBLEMS IN PROJECTION 1211. Intersection of conical surfaces. Ifa plane be passed through the vertex of a conical surface it cuts it in rectilinear elements if at all. When two conical surfaces intersect it is possible to pass a plane through the vertices of both. Thus, the plane may be made to intersect both surfaces in rectilinear elements, the intersections of which determine points on the required line of intersection. Fig. 214 shows this pictorially. If the vertices m and n be joined by a line and its piercing point o be found, then, any line through o will determine a plane. Also, if od be the line chosen, the plane of the lines mo and od will cut the conical surfaces in Fig. 214. the elements ma, mb, nc and nd. These elements determine the four points e, f, gh. Any other line through o will, if prop- erly chosen, determine other elements which again yield new points on the line of intersection. The following problem will bring out the details more fully. 1212. Problem 9. To find the line of intersection of the surfaces of two cones, whose bases may be made to lie in the same plane, and whose altitudes differ. Construction. Let A and B, Fig. 215 be the vertices of the two cones in question, whose bases are in the horizontal plane. If the bases do not lie in the principal planes, a new set of prin- cipal planes may be substituted to attain the result. Join A and B by a line and find where this line pierces the horizontal INTERSECTIONS OF SURFACES WITH EACH OTHER 217 plane atc. Any line, through c, lying in the horizontal plane, in addition to the line AB will determine a plane, which may cut the conical surfaces in elements. Let, for instance, cd be such a line. This line cuts the bases at d, e and f, the points which will be considered. The elements in the horizontal pro- jection are shown as db, eb and fa; in the vertical projection, they are d’b’, e’b’ and f’a’. The element f’a’ intersects the element d’b’ at g’, and the element e’b’ at h’, thus determining g’ and h’, two points on the vertical projection of the required line of intersection. The corresponding projections g and h may also be found, which determine points on the horizontal projection of the same line of intersection. Consider, now, the line ci, tangent to the one cone. Its element is ib in the horizontal projection and i’b’ in the vertical projection; the corresponding element cut from the other cone is ja in the horizontal projec- tion and j’a’ in the vertical projection. The points k and k’ are thus found, but they represent only one point K on the actual cones. This process is continued until a sufficient number of points are determined, so that a smooth curve can be drawn through them. The cones chosen, have complete interpenetration, as one cone goes entirely through the other. Thus, there are two distinct curves of intersection. That, near the apex B, is found in an identical way with the preceding. 218 GEOMETRICAL PROBLEMS IN PROJECTION Attention might profitably be called to the manner in which points are located so as to miminize confusion as much as possible. To do this, draw only one element on each cone at a time, locat- ing one, or two points, as the case may be; then erase the con- struction lines when satisfied of the accuracy. Several of the prominent points of the curve may be thus located and others estimated if considerable accuracy is not a prerequisite. If the surfaces are to be developed and constructed subsequently, however, more points will have to be established. There is nothing new in the development of these cones, the case is similar to that of any oblique cone and is therefore omitted. One thing may be mentioned in passing however, and that is, while drawing the elements to determine the contour of the base, the same elements should be used for locating the line of inter- section as thereby considerable time is saved. 1213. Problem 10. To find the line of intersection of the surfaces of two cones, whose bases may be made to lie in the same plane and whose altitudes are equal. Construction. Let A and B, Fig. 216, be the desired cones. Join A and B by a line which is chosen, parallel to the planes of projection. Hence, this line cannot pierce the plane of the bases, and the preceding method of finding the line of intersection is thus inapplicable. It is possible, how- ever to draw a line cd parallel to ab. These lines therefore determine a plane which passes through both vertices and intersects the surface in rectilinear elements. The comple- Fie. 216. tion of the construction becomes evident when the process which leads to the finding of F and G is understood. The cases of intersecting cones so far considered have been so situated as to have bases in a common plane. This may not always be convenient. Therefore to complete the subject, and to cover emergencies, an additional construction will be studied. INTERSECTIONS OF SURFACES WITH EACH OTHER 219 1214. Problem 11. To find the line of intersection of the surfaces of two cones whose bases lie in different planes. Construction. Let A and B, Fig. 217, be the vertices of the two cones. The cone A has its base in the horizontal plane; B has its base in the plane T, which is perpendicular to the vertical plane. Join the vertices A and B by a line, which pierces the horizontal plane at c and the plane T in the point d’ horizontally projected at d. Revolve the plane T into the vertical plane, and, hence, the trace Tt becomes Tt’, The angle t’Tt’” must 6 eee emssoosase be a right angle, because it is perpendicular to the vertical plane, and, therefore cuts a right angle from the principal planes. The piercing point of the line AB is at d” in the revolved position, where d’d” is equal to the distance of d from the ground line. In the revolved position of the plane T, draw the base of the cone B and from it accurately construct the horizontal projec- tion of the base. Draw a line d’e’”’ tangent to the revolved position of the base of the cone B at f’’. From it, find f, the horizontal, and f’, the vertical projection of this point, and draw the elements fb and f’b’. Lay off Te’ =Te, and draw ce. The lines CD and DE determine a plane which cuts the horizontal 220 GEOMETRICAL PROBLEMS IN PROJECTION projection of the base of the cone A in g and h, and thus, also, the two elements ag and ah which are horizontal projections, and a’g’ and a’h’ the corresponding vertical projections of the elements. The element BF intersects the elements AG and AH in the points M and N shown, as usual, by their projections. M and N are therefore two points of the required curve, being prominent points because the elements are tangent at these points. To obtain other points, draw a line d’’i” and make Ti’ =Ti. Where ci cuts the cone A, draw the elements as shown, also the correspond- 0@%------------ Fig, 217. ing elements of the cone B. The points of the curve of intersec- tion are therefore as indicated. One fact is to be observed in this and in similar constructions. Auxiliary planes are passed through the vertices of both cones, cutting, therefore, elements of the cones whose intersection determine points on the curve. All planes through the vertices of both cones must pass through c in the horizontal plane and through d” in the plane T. Also, such distances as Te” must equal Te because these auxiliary planes cut the horizontal plane and the plane T on their line of intersection Tt; the revolution of the plane T into the vertical plane does not disturb the location INTERSECTIONS OF SURFACES WITH EACH OTHER 221 of any point on it. Hence such distance as Te will be revolved to Te” where Te=Te”. When applying this problem to a practical case, it would be better to select a profile plane for the plane T, as then Tt” would coincide with the ground line and all construction would be simplified. It has not been done in this instance, in order to show the generality of the method, and its adaptation to any condition. 1215. Types of lines of intersection for surfaces of cones. When the surfaces of two cones are situated so that there is com- Fig. 218. Fig. 219. plete interpenetration, the line of intersection will appear as two distinct. closed curves. Fig. 215 is an example. If the surfaces of the cones are such that the interpenetration is incomplete, only one closed curve will result. Figs. 216 and 217 are examples of this case. When the surfaces of two cones have a common tangent plane then the curve is closed and crosses itself once. An example under this heading is given in Fig. 218. It is possible to have two cones the surfaces of which have two common tangent planes. In this case there are two closed curves which cross each other twice. This illustration appears in Fig. 219. 222 GEOMETRICAL PROBLEMS IN PROJECTION 1216. Problem 12. To find the line of intersection of the surfaces of a cone and a cylinder of revolution when their axes intersect at a right angle. Construction. Let Fig. 220 represent the cone and the cylinder. Through the cone, pass a series of planes perpendicular to its axis. If the planes are properly chosen, they will cut the conical surface in circles and the cylindrical surface in rectilinear elements, the intersection of which determine points on the line of intersection. A reference to Fig. 220 will show this in con- struction. As a check on the accuracy of the points on the curve, it is possible to draw elements of the conical surface through some point; the corresponding projections of the elements must con- Fia. 220. tan the corresponding projections of the points. In the illustration, the elements OA and OB are drawn through the points E and J respectively. 1217. Problem 13. To find the line of intersection of the surfaces of a cone and a cylinder of revolution when their axes intersect at any angle. Construction. The given data is shown in Fig. 221. In this case it is inconvenient to pass planes perpendicular to the axis of the cone, since ellipses will be cut from the cylinder. A better plan is to find m, the intersection of their axes, and use this as a centre for auxiliary spherical surfaces. The spherical surfaces intersect the surfaces of revolution in circles (1025). INTERSECTIONS OF SURFACES WITH EACH OTHER 223 The details of the construction are shown in Fig. 221. To check the accuracy of the construction it is possible to draw an element of one surface through some point and then find the corresponding projection of the element; the corresponding projection of the point must be located on the corresponding projection of the element. Two elements OF and OG are shown in the figure. The points determined thereby are shown at D. 1218. Problem 14. To find the line of intersection of the surfaces of an oblique cone and a right cylinder. Construction. Let Fig. 222 illustrate the conditions assumed. C. ¢ @ Fie. 221. Fig. 222. To avoid too many construction lines, the figures have been assumed in their simplest forms. Pass any plane through o, perpendicular to the horizontal plane; it cuts elements DO and CO from the cone and the element from the cylinder which is horizontally projected at f and vertically at f’g’. In the vertical projection, the elements c’o’ and d’o’ intersect the element f’g’ at g’ and h’, two points of the required curve. The extreme element OE determines the point i by the same method. There are two distinct lines of intersection, in this case, due to complete interpenetration. The points on the line of intersection near the vertex are located in a manner similar to that shown. 224 GEOMETRICAL PROBLEMS IN PROJECTION 1219. Problem 15. To find the developments in the pre- ceding problem. Construction. The oblique conical surface is developed by triangulation. In Figs. 222 and 223, the lines o’a’ and o’b’ are shown in their true length in the vertical projection; hence, lay off o’a’=0"a”, and o’b’=o0"b”’. If the element oe be revolved so that it is parallel to the vertical plane, the point i’, in the vertical projection, will not change its distance above the horizontal plane during the revolution. Hence it moves from i’ to k’, the revolved position. Accordingly, lay off o”i’=o’k’ and one point of the intersection on the development is obtained. Other points, of course, are found in absolutely the same manner. Fic. 223. Extreme accuracy is required in most of these problems. Con- structions like this should be laid out to as large a scale as con- venient. The development of the cylinder is also shown and is perhaps clear without explanation. 1220. Problem 16. To find the line of intersection of the surfaces of an oblique cone and a sphere. Construction. Let Fig. 224 represent the cone and sphere in question. The general scheme is to pass planes through the vertex of the cone perpendicular to the horizontal plane. These planes cut the cone in rectilinear elements and the sphere in circles; the intersection of the elements and the circles so cut will determine points on the curve. Thus, through 0, draw a plane which cuts the cone in a and b and the sphere in d and e. The elements cut are shown as 0’a’ INTERSECTIONS OF SURFACES WITH EACH OTHER 225 and o’b’ in the vertical projection. Attention will be confined to the determination of g’, one point on the lower curve, situated on the element OA. The circle cut from the sphere by the plane through oa and ob has a diameter equal-to de. If pc is a perpendicular to de from p the centre of the sphere, then c is the hori- zontal projection of the centre of the circle de, and cd andce are equal. If the cutting plane is revolved about a perpendicular through o to the hori- zontal plane, until it is parallel to the vertical plane, a will move to a” and o’a’”” will be the revolved position of this element. The centre of the circle will go to h in the horizontal projec- tion andh’ in the vertical projection, because, in this latter case, the distance ' of h’ above the horizontal plane does Fic. 224. not change in the revolution. The element and the centre of the circle in a plane parallel to the vertical plane are thus determined. Hence, with cd as a radius and h’ as acentre, describe an arc, cutting o’a’” inf’. On counter revolution, f’ goes to g’ on o’a’, the original position of the element. The point g’ is therefore one point on the curve. Every other point is found in the same way. 1221. Problem 17. To find the line of intersection of the surfaces of a cylinder and a sphere. Construction. Fig. 225 pic- tures the condition. Pass a series of planes, through the cylinder and the sphere, perpendicular to the plane of the base of the cylinder. One position of the cutting plane cuts the cylinder in a’b’ and the sphere in c’d’. Construct a supplementary view S with the centre of the sphere at o’’ asshown. The elements appear as a”a” and bb”. The diameter of the circle cut from the sphere isc’d’ and Fig. 225, 226 GEOMETRICAL PROBLEMS IN PROJECTION with c’e’ as aradius, (equal to one-half of c’d’), draw an are cutting the elements at f’gh” andi”. Lay off f’j’ =fj; gj” =gi, etc., and the four points f, g,h, i are determined on the required view. These points are on the required line of intersection. 1222. Problem 18. To find the line of intersection of two doubly curved surfaces of revolution whose axes intersect. Construction. Let Fig. 226 represent the surfaces in ques- tion. With o, the intersection of the axes, as a centre, draw a series of auxiliary spherical surfaces. One of these spherical curfaces cuts the surface whose axis is on in a circle whose diameter ‘3 cd. This same auxiliary sphere cuts the surface whose axis is Fic. 226. om in a circle projected as ef. Hence cd and ef intersect at a, a point on the required line of intersection. In the view on the right, only one of the surfaces is shown. The location of the corresponding projections of the line of intersection will be evident from Art. 1027. When the axes do not intersect then the general method is to pass planes so as to cut circles from one surface of revolution and a curve from the other. The intersections determine points onthecurve. Itis desirable in this connection to chose an arrange- ment that gives the least trouble. No general method can be given for the mode of procedure. 1223. Commercial application of methods. In practice, it is always desirable as a matter of time to turn the objects so that the auxiliary surfaces may be passed through them with the least effort. Frequently, many constructions may be car- ried out without any special reference to the principal planes. INTERSECTIONS OF SURFACES WITH EACH OTHER 227 There is no harm in omitting the principal planes, but the student should not go to the extreme in this ommission. With the principal planes at hand, the operations assume a familiar form, which will have a tendency to refresh the memory as to the basic principles involved. All the operations in the entire subject have a remarkable simplicity in the abstract; the confusion that sometimes arises is not due to the principles involved, but solely to the number of construction lines required. It, therefore, seems proper to use such methods as will lead to the least con- fusion, but it should always be borne in mind that accuracy is important at all hazards. QUESTIONS ON CHAPTER XII 1. State the general method of finding the intersection of any two ~ surfaces. 2. When the surfaces are those of prisms or pyramids, is it necessary to use auxiliary planes as cutting planes? Why? 8. Prove the general case of finding the line of intersection of the sur- faces of two prisms. 4, Show the general method of finding the development of the prisms in Question 3. 5. Prove the general case of finding the line of intersection of two cylindrical surfaces of revolution whose axes intersect at a right angle. 6. Show the general method of finding the development of the surfaces in Question 5. 7. Prove the general case of finding the line of intersection of two cylindrical surfaces of revolution whose axes intersect at any angle. 8. Show the general method of finding the development of the surfaces in Question 7. 9. Prove the general case of finding the line of intersection of two cylindrical surfaces whose axes do not intersect. 10. Show the general method of finding the development of the cylin- drical surfaces in Question 9. 11. State the general method of finding the line of intersection of two conical surfaces. 12. Prove the general case of finding the line of intersection of the sur- faces of two oblique cones whose bases may be made to be in the same plane and whose altitudes differ. 13. Show the general method of finding the developments of the surfaces of the cones in Question 12. 14. Prove the general case of finding the line of intersection of the sur- faces of two oblique cones whose bases may be made to lie in the same plane and whose altitudes are equal. 228 15. 16. 17. 18. 19. 20. 21, 22. 23. 24, 25. 26. 27. 28. 29, 30. 31. 32. 33. 34, 35. 36. 37. 38. 39. GEOMETRICAL PROBLEMS IN PROJECTION Show the general method of finding the developments of the surfaces of the cones in Question 14. Prove the general case of finding the line of intersection of the sur- faces of two oblique cones whose bases lie in different planes. Show the general method of finding the developments of the cones in Question 16. When the surfaces of two cones have complete interpenetration, discuss the nature of the line of intersection. When the surfaces of two cones have incomplete penetration, discuss the nature of the line of intersection. When the surfaces of two cones have a common tangent plane, discuss the nature of the line of intersection. When the surfaces of two cones have two common tangent planes, discuss the nature of the line of intersection. Prove the general case of finding the line of intersection of the sur- faces of a cone and a cylinder of revolution when their axes inter- sect at a right angle. Show the general method of finding the developments of the surfaces in Question 22. Prove the general case of finding the line of intersection of the sur- faces of a cone and a cylinder of revolution when their axes inter- sect at any angle. Show the general method of finding the developments of the sur- faces in Question 24. Prove the general case of finding the line of intersection of the sur- faces of an oblique cone and a right cylinder. Show the general method of finding the developments in Question 26. Prove the general case of finding the line of intersection of the sur- faces of an oblique cone and a sphere. Show the general method of finding the development of the cone in Question 28. Develop the surface of the sphere in Question 28 by the gore method. Prove the general case of finding the line of intersection of the sur- faces of a sphere and a cylinder. Show the general method of finding the development of the cylinder in Question 31. Develop the surface of the sphere in Question 31 by the zone method. Prove the general case of finding the line of intersection of two doubly curved surfaces of revolution whose axes intersect. Develop the surfaces of Question 34 by the gore method. Develop the surfaces of Question 34 by the zone method. State the general method of finding the line of intersection of two doubly curved surfaces of revolution whose axes do not intersect. What items are to be considered when applying the principles of intersections and developments to commercial problems? Is it always necessary to use the principal planes when solving prob- lems relating to intersections and developments? 45, 46. 47, 48. 49, 50. 51. 52. 53. 5p. 55. 56. 57. 58. 59. 60 . Develop the surfaces of Question 42. . Two cylinders of 2” diameter and 13’ diameter INTERSECTIONS OF SURFACES WITH EACH OTHER 229 . Find the intersection of the two prisms shown in Fig. 12-A. Assume suitable dimensions. . Develop the prisms of Question 40. . Two cylinders of 2” diameter intersect at a right angle. Find the line of intersection of the surfaces. Assume suitable dimensions for their lengths and position with respect to each other. : ER intersect at a right angle. Find the line of NX > Wf intersection of the surfaces. Assume suitable 1 dimensions for their lengths and position with respect to each other. Lys \ Develop the surfaces of Question 44. c Two cylinders of 2’’ diameter intersect at an angle yg joa. of 60°. Find the line of intersection of the sur- faces. Assume suitable dimensions for their lengths and position with respect to each other. Develop the surfaces of Question 46. Two cylinders of 2’ diameter intersect at an angle of 45°. Find the line of intersection of the surfaces. Assume suitable dimen- sions for their lengths and position with respect to each other. Develop the surfaces of Question 48. Two cylinders of 2” diameter intersect at an angle of 30°. Find the line of intersection of the surfaces. Assume suitable dimen- sions for their lengths and position with respect to each other. Develop the surfaces of Question 50. Two cylinders of 2’ diameter and 12” diameter intersect at an angle of 60°. Find the line of intersection of the surfaces. As- sume suitable dimensions for their lengths and position with respect to each other. Develop the surfaces of Question 52. Two cylinders of 2” diameter and 13” diameter intersect at an angle of 45°. Find the line of intersection of the surfaces. Assume suitable dimensions for their lengths and position with respect to each other. Develop the surfaces of Question 54. Two cylinders of 2’’ diameter and 12” diameter intersect at an angle of 30°. Find the line of intersection of the surfaces. Assume suitable dimensions for their lengths and position with respect to each other. Develop the surfaces of Question 56. Two cylinders of 2” diameter intersect at a right angle and have their axes offset 4”. Find the line of intersection of the surfaces. Assume the additional dimensions. Develop the surface of one of the cylinders of Question 58. . A 2” cylinder intersects a 12” cylinder so that their axes are offset #”" and make a right angle with each other. Find the line of intersection of the surfaces. Assume the additional dimensions. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. GEOMETRICAL PROBLEMS IN PROJECTION . Develop the surfaces of Question 60. . Two 2” cylinders have their axes offset 3’’.. The elements intersect at an angle of 60°. Find the line of intersection of the surfaces. Assume the additional dimensions. . Develop the surfaces of Question 62. . Two 2” cylinders have their axes offset }’. The elements intersect at an angle of 45°. Find the line of intersection of the surfaces. Assume the additional dimensions. . Develop the surfaces of Question 64. . Two 2” cylinders have their axes offset 3”. The elements intersect at an angle of 30°. Find the line of intersection of the surfaces. Assume the additional dimensions. . Develop the surfaces of Question 66. . A 2” cylinder intersects a 12’ cylinder at an angle of 60°. Their axes are offset 1’. Find the line of intersection of the surfaces. Assume the additional dimensions. . Develop the surfaces of Question 68. 4 tee xe Fia. 12-B. Fig. 12-C, A 2” cylinder intersects a 12” cylinder at an angle of 45°. Their axes are offset 4’. Find the line of intersection of the surfaces. Assume the additional dimensions. Develop the surfaces of Question 70. A 2” cylinder intersects a 13’ cylinder at an angle of 30°. Their axes are offset 3’. Find the line of intersection of the surfaces. Assume the additional dimensions. Develop the surfaces of Question 72. Find the line of intersection of the surfaces of the two cones shown in Fig. 12-B. Assume suitable dimensions. Develop the surfaces of Question 74. Assume a pair of cones similar to those shown in Fig. 12-B, but, with equal altitudes. Then, find the line of intersection of the surfaces. Develop the surfaces of Question 76. Assume a cone and cylinder similar to that shown in Fig. 12-C. Then, find the line of intersection of the surfaces, Develop the surfaces of Question 78. 80. 81. 82. 83. 84, 85. 86. 87 88, 89, 90. 91. 92. 93. 94. 95. INTERSECTIONS OF SURFACES WITH EACH OTHER 231 Assume a cone and cylinder similar to that shown in Fig. 12-C, but, have the cylinder elliptical. Then, find the line of inter- section of the surfaces. Develop the surfaces of Question 80. Assume a cone and cylinder similar to that shown in Fig. 12-C, but, have the cone elliptical. Then, find the line of intersection of the surfaces. Develop the surfaces of Question 82. Assume an arrangement of cone and cylinder somewhat similar to that of Fig. 12-C, but, have both cone and cylinder elliptical. Then, find the line of intersection of the surfaces. Develop the surfaces of Question 84. Assume a cylinder and cone of revolution, whose general direction of axes are at right angles, but, which are offset as shown in Fig. 12-D. Then, find the line of intersection of the sur- faces. Develop the surfaces in Question 86, = Aer Fig. 12-D. Fig. 12-E. Assume an elliptical cylinder and a cone of revolution, whose general direction of axes are at right angles to each other, but, which are offset as shown in Fig. 12-D. Then, find the line of inter- section of the surfaces. Develop the surfaces of Question 88. Assume a circular cylinder of revolution and an elliptical cone, whose general direction of axes are at right angles to each other, but, which are offset as shown in Fig. 12-D. Then, find the line of intersection of the surfaces. Develop the surfaces of Question 90. Assume an elliptical cylinder and an elliptical cone, whose general direction of axes, are at right angles to each other, but, which are offset, as shown in Fig. 12-D. Then, find the line of inter- section of the surfaces. Develop the surfaces of Question 92. Assume a cylinder and cone of revolution, similar to that of Fig. 12-E, and make angle «=60°. Then, find the line of intersection of the surfaces. 3 Develop the surfaces of Question 94. 232 GEOMETRICAL PROBLEMS IN PROJECTION 96. Assume a clyinder and cone of revolution, similar to that of Fig. 12-E, and make angle «=45°. Then, find the line of inter- section of the surfaces. 97. Develop the surfaces of Question 96. 89. Assume a cylinder and a cone of revolution, similar to that of Fig. 12-E, and make angle « =30°. Then, find the line of intersection of the surfaces. 99. Develop the surfaces of Question 98. 100. Assume an elliptical cylinder and a cone of revolution arranged similar to that of Fig. 12-E, and make the angle « =60°. Then, find the line of intersection of the surfaces. 101. Develop the surfaces of Question 100. 102. Assume a cylinder of revolution and an elliptical cone, arranged ‘similar to that of Fig. 12-E, and make the angle «=45°. Then, “ find the line of intersection of the surfaces. 103. Develop the surfaces of Question 102. Aer Fig. 12-E, Fig. 12-F. Fie. 12-G. 104. Assume an elliptical cylinder and an elliptical cone arranged similar to that of Fig. 12-E, and make the angle «=60°. Then find the line of intersection of the surfaces. 105. Develop the surfaces of Question 104. 106. Assume a cylinder and cone of revolution as shown in Fig. 12-F, and make angle «=60°. Then, find the line of intersection of the surfaces. 107. Develop the surfaces of Question 106. 108. Assume a cylinder and a cone of revolution as shown in Fig. 12-F, and make angle «=45°. Then, find the line of intersection of the surfaces. 109. Develop the surfaces of Question 108. 110. Assume a cylinder and a cone of revolution as shown in Fig. 12-F, and make angle «=30°. Then, find the line of intersection of the surfaces, 111. Develop the surfaces of Question 110. 112, Assume an elliptical cylinder and a cone of revolution, arranged similar to that shown in Fig. 12-F, and make angle «=60°. Then find the line of intersection of the surfaces. INTERSECTIONS OF SURFACES WITH EACH OTHER 233 113. Develop the surfaces of Question 112. 114. Assume a- cylinder of revolution and an elliptical cone, arranged similar to that shown in Fig. 12-F, and make angle «=45°. Then, find the line of intersection of the surfaces. / ( | Fig. 12-H. Fre. 12-I. 115. Develop the surfaces of Question 114. 116. Assume an elliptical cylinder and an elliptical cone, somewhat: similar to that shown in Fig. 12-F, and make angle «=60°. Then find the line of intersection of the surfaces. 117. Develop the surfaces of Question 116. 118. Assume a cone and cylinder as shown in Fig. 12-G. Then, find the line of intersection of the surfaces. 119. Develop the surfaces of Question 118. 120. Assume a cylinder and a sphere as shown in Fig, 12-H. Then, find the line of intersection of the sur- faces. 121. Develop the surface of the cylinder of Ques- tion 120. 122. Develop the surface of the cylinder of Ques- tion 120, and, also, the surface of the sphere by the gore method. 123. Assume a cylinder and sphere as shown in Fig. 12-I. Then, find the line of intersec- tion of the surfaces. 124. Develop the surface of the cylinder in Ques- tion 123. Fig, 12-J. 125. Develop the surface of the cylinder in Ques- tion 123, and, also, the surface of the sphere by the zone method. 126. Assume a cone and a sphere as shown in Fig. 12-J. Then, find the line of intersection of the surfaces. 127. Develop the surface of the cone in Question 126. 128. Develop the surface of the cone in Question 126, and, also, the sphere by the gore method. 234 GEOMETRICAL PROBLEMS IN PROJECTION 129. Assume two doubly curved surfaces of revolution, whose axes intersect. Then, find the line of intersection of the surfaces. 130. Develop the surfaces of Question 129 by the gore method. 131. Develop the surfaces of Question 129 by the zone method. 132. Assume two doubly curved surfaces of revolution whose axes do not intersect. Then, find’ the line of intersection of the surfaces. Fia. 12-K. 133. Develop the surfaces of Question 132 by the gore method. 134. Develop the surfaces of Question 132 by the zone method. 135. Assume a frustum of a cone as shown in Fig. 12-K. A circular cylinder is to be fitted to it, as shown. Develop the surfaces. Hint.—Find vertex of cone before proceeding with the devel- opment, PART Jl PRINCIPLES OF CONVERGENT PROJECTING-LINE DRAWING CHAPTER XIII PERSPECTIVE PROJECTION 1301. Introductory. Observation shows that the apparent magnitude of objects is some function of the distance between the observer and the object. The drawings made according to the principles of parallel projevting-line drawing, and treated in Part I of this book, make no allowance for the observer’s posi- tion with respect to the object. In other words, the location of the object with respect to the plane of projection has no influence on the size of the resultant picture, provided that the inclina- tion of the various lines on the object, to the plane of projec- tion, remain the same. Hence, as the remoteness of the object influences its apparent size, then, as a consequence, parallel projecting-line drawings must have an unnatural appearance. The strained appearance of drawings of this type is quite notice- able in orthographic projection wherein two or more views must be interpreted simultaneously, and is less noticeable in the case of oblique or axonometric projection. On the other hand, the rapidity with which drawings of the parallel projecting-line type can be made, and their adaptability for construction purposes, are strong points in their favor. 1302. Scenographic projection. To overcome the foregoing objections, and to present a drawing to the reader which cor- rects for distance, scenographic projections are used. To make these, convergent projecting lines are used, and the’ observer is located at the point of convergency. The surface on which 235 236 CONVERGENT PROJECTING-LINE DRAWING scenographic projections are made may be spherical, cylindrical, ete., and find their most extensive application in decorative painting. 1303. Linear perspective. In the decoration of an interior of a dome or of a cylindrical wall, the resulting picture should be of such order as to give a correct image for the assumed loca- tion of the eye. In engineering drawing, there is little or no use for projections on spherical, cylindrical, or other curved surfaces. When scenographic projections are made on a plane, this type of projection is called linear perspective. A linear perspective, therefore, may be defined as the drawing made on a plane surface by the aid of convergent projecting lines, the point of convergency being at a finite distance from the object and from the plane. The plane of projection is called the picture plane, and the posi- tion of the eye is referred to as the point of sight. 1304. Visual rays and visual angle. If an object—an arrow for instance—be placed a certain distance from the eye, say in the position ab (Fig. 227) and c be ; the point of sight, the two extreme rays of light or visual rays, ac and = ab, form an angle which is known as d a the visual angle. If, now, the arrow be Fig. 227. moved to a more remote position de, the limiting visual rays de and ec give a smaller visual angle. The physiological effect of this variation in the visual angle is to alter the apparent size of the object. If one eye is closed during this experiment, the distance from the eye to the object cannot be easily estimated, but the apparent magnitude of the object will be some function of the visual angle. In binocular vision, the muscular effort required to focus both eyes on the object will, with some experience, enable an estimate of the distance from the eye to the object; the mind will automatically correct for the smaller visual angle and greater distance, and, thus, to an experienced observer, give a more or less correct impression of the actual size of the object. This experience is generally limited to horizontal distances only, as very few can estimate correctly the diameter of a clock on a church steeple unless they have been accustomed to making such observa- tions. As the use of two eyes in depicting objects in space causes PERSPECTIVE PROJECTION 237 slightly different inpressions on different observers, the eye will be assumed hereafter as a single point. 1305. Vanishing point. When looking along a stretch of straight railroad tracks, it may be noted that the tracks appar- ently vanish in the distance. Likewise, in viewing a street of houses of about the same height, the roof line appears to meet the sidewalk line in the distance. It is known that the actual distances between the rails, and between the sidewalk and roof is always the same; yet the visual angle being less in the distant observation, gives the impression of a vanishing point, which may, therefore, be defined as the point where parallel lines seem to vanish. At an infinite distance the visual angle is zero, and, hence, the parallel lines appear to meet in a point. The lines themselves do not vanish, but their perspective projections vanish. z 1306. Theory of perspective projection. The simplest notion of perspective drawing can be obtained by looking at a distant house through a window-pane. If the observer would trace on the window-pane exactly what he sees, and locate all points on the pane so that corresponding points on the house are directly behind them, a true linear perspective of the house would be the result. Manifestly, the location of each successive point is the same as locating the piercing point of the visual ray on the picture plane, the picture plane in our case being the window- pane. 1307. Aerial perspective. If this perspective were now colored to resemble the house beyond, proper attention being paid to light, shade and shadow, an aerial perspective would be the result. In brief, this aerial perspective would present to the eye a picture that represents the natural condition as near as the skill of the artist will permit. In our work the linear perspective will alone be considered, and will be denoted as the “‘ perspective,” aerial perspective finding little application in engineering. 1308. Location of picture plane. It is customary in per- spective to assume that the picture plane is situated between the eye and the object. Under these conditions, the picture is smaller than the object, and usually this is necessary. The eye is assumed to be in the first angle; the object, however, is generally in the 238 CONVERGENT PROJECTING-LINE DRAWING second angle * for reasons noted above. The vertical plane is thus the picture plane. A little reflection will show that this is in accord with our daily experience. Observers stand on, the ground and look at some distant object; visual rays enter the eye at various angles; the mean of these rays is horizontal or nearly so, and, therefore, the projections naturally fall on a vertical plane interposed in the line of sight. The window- pane picture alluded to is an example. 1309. Perspective of a line. Let, in Fig. 228, AB be an arrow, standing vertically as shown in the second angle; the point of sight C is located in the first angle. The visual rays pierce the picture plane at a” and b”, and a”b’’, in the picture plane, is the perspective of AB in space. The similar condition by ersped- tive, i ts Object Picture Plane ! | 1 1 ' ¢ Fig. 228. Fig. 229, of Fig. 228 in orthographic projection is represented in Fig. 229. The arrow and the point of sight are shown by their pro- jections on the horizontal and vertical planes. The arrow being perpendicular to the horizontal plane, both projections fall at the same point ab; the vertical projection is shown as a’b’; the point of sight is represented on the horizontal and vertical planes as c and c’, respectively. Any visual ray can likewise be represented by its projections, and, therefore, the horizontal projection of the point of sight is joined with the horizontal pro- * The object may be located in any angle; the principles are the same for all angles. It is perhaps more convenient to use the second angle, as the lines do not have to be extended to obtain the piercing points as would be the case for first-angle projections. PERSPECTIVE PROJECTION 239 jections. Similarly, the same is true of the vertical projections of the arrow and point of sight. It is now necessary to find the piercing points (506,805) of the visual rays on the picture plane (vertical plane); and these are seen at a’ and b”. Therefore, ab” is the perspective of AB in space. 1310. Perspectives of lines perpendicular to the hori-= zontal plane. In Fig. 230, several arrows are shown, in projec- tion, all of which are perpendicular to the horizontal plane. Their perspectives are a’’b’”’, d’e”’, gh’. On observation, it will be noted that all the perspectives of lines perpendicular to the horizontal plane are vertical. This is true, since, when a plane f By d''g? gh ° ; ” Tal nig! a ee | i eo Toy SG] Q’ e6 Le \] an Ww # a, ; 16 | rs om ab i | h i 1 1 1 | 1 \ . c Fie. 230. . Fig. 231. is passed through any line and revolved until it contains the point of sight, the visual plane (as this plane is then called) is manifestly perpendicular to the horizontal plane, and its vertical trace (603) is perpendicular to the ground line. 1311. Perspectives of lines parallel to both principal planes. Fig. 231 shows two arrows parallel to both planes, the projections and perspectives being designated as before. This case shows that all perspectives are parallel. To prove this, pass a plane through the line in space and the point of sight; the vertical trace of the plane so obatined will be parallel to the ground line* because the line is parallel to the ground line. * If the line in space lies in the ground line, the vertical trace will coincide with the ground line. This is evidently a special case. 240 CONVERGENT PROJECTING-LINE DRAWING 1312. Perspectives of lines perpendicular to the picture plane. Suppose a series of lines perpendicular to the vertical plane is taken. This case is illustrated in Fig. 232. The per- spectives are drawn as before and designated as is customary. It may now be observed that the perspectives of all these lines & x ee ss |e eee NS, I c Fig. 232. (or arrows) vanish in the vertical projection of the point of sight. This vanishing point for the perspectives of all perpendiculars to the picture plane is called the centre of the picture. The reason for this is as follows: It is known that the visual angle ce Vv e \ S Paes | Lk TSS i 1 d a € a +o b I eg a e a | oN | c Fig. 233. Fic. 234. becomes less as the distance from the point of sight increases; at an infinite distance the angle is zero, the perspectives of the lines converge, and, therefore, in our nomenclature vanish. 1313. Perspectives of parallel lines, inclined to the picture plane. Assume further another set of lines, Fig. 233, PERSPECTIVE PROJECTION 241 parallel to each other, inclined to the vertical plane, but parallel to the horizontal plane. Their orthographic projections are parallel and their perspectives will be seen to converge to the point V. Like the lines in the preceding paragraph, they vanish at a point which is the vanishing point of any number of lines parallel only to those assumed. The truth of this is established by the fact that the visual angle becomes zero (1304) at an infinite distance from the point of sight and, hence, the perspectives vanish as shown. To determine this vanishing point, draw any line through the point of sight parallel to the system of parallel lines; and its piercing point on the vertical or picture plane will be the required vanishing point. It amounts to the same thing to say that the vanishing point is the perspective of any line of this system at an infinite dis- tance. A slightly different condition is shown in Fig. 234. The lines are still parallel to each other and to the horizontal plane, but are inclined to the vertical plane in a different direction. Again, the perspectives of these lines will converge to a new van- ishing point, situated, however, much the same as the one just preceding. 1314. Horizon. On observation of the cases cited in Art. 1313, it can be seen that the vanishing point lies on a horizontal line through the vertical projection of the point of sight. This is so since a line through the point of sight parallel to either sys- tem will pierce the picture plane in a point somewhere on its vertical projection. As the line is parallel to the horizontal plane, its vertical projection will be parallel to the ground line and, as shown, will contain all vanishing points of all systems of horizontal lines. Any one system of parallel lines will have but one vanishing point, but as the lines may slope in various directions, and still be parallel to the horizontal plane, every ‘system will have its own vanishing point and the horizontal line drawn through the vertical projection of the point of sight will be the locus of all these vanishing points. This horizontal ‘line through the vertical projection of the point of sight is called the horizon. As a corollary to the above, it may be stated, that all planes parallel to the horizontal plane vanish in the horizon. As before, 242 CONVERGENT PROJECTING-LINE DRAWING the visual angle becomes less as the distance increases, and, hence, becomes zero at infinity. There are an unlimited number of systems of lines resulting in an unlimited number of vanishing points, whether they are on the horizon or not. In most drawings the vertical and hori- zontal lines are usually the more common. Thelines perpendicular to the picture plane are horizontal lines, and, therefore, the centre of the picture (vanishing point for the perspectives of the per- pendiculars) must be on the horizon. 1315. Perspective of a point. The process of finding the perspective of a line of definite length is to find the perspectives 4 \ a | ee Fig. 235. Fia. 236. of the two extremities of the line. When the perspectives of the extremities of the line are found, the line joining them is the perspective of the given line. In Fig. 235, two points A and B are chosen whose perspectives will be found to be a” and b”. When the point lies in picture plane it is its own perspective. This is shown in Fig. 236, where a’ is the vertical projection and a is therefore in the ground line; hence, the perspective is a’. In the construction of any perspective, the method is to locate the perspectives of certain points which are joined by their proper lines. The correct grouping of lines determine certain surfaces which, when closed, determine the required solid. 1316. Indefinite perspective of a line. Let AB, Fig. 237, be a limited portion of a line FE. Hence, by obtaining the PERSPECTIVE PROJECTION 243 piercing points of the visual rays AC and BC, the perspective a’’b’’ is determined. Likewise, DE is another limited portion of the line FE and its perspective is d’e’’. If a line be drawn through the point of sight C, parallel to the line FE, then, its piercing point g’ on the picture plane will be the vanishing point of the perspectives of a system of lines parallel to FE (1313). Also, as FE pierces the picture plane at f’, then f’ will be its own perspective (1315). Therefore, if a line be made to join f’ and g’, it will be the perspective of a line that reaches from the picture plane out to an infinite distance. And, also, the perspective of any limited portion of this line must lie Fig. 237. on the perspective of the line. From the construction in Fig. 237, it will be observed that a’’b” and d’e’” lie on the line f’g’. Consequently, the line f’g’ is the indefinite perspective of a line FE which reaches from the picture plane at f’ out to an infinite distance. Thus, the indefinite perspective of a line may be defined as the perspective of a line that reaches from the picture plane to infinity. Before leaving Fig. 237, it is desirable to note that g’ is the vanishing point of a system of lines parallel to FE. It is not located on the horizon because the line FE is not parallel to the horizontal plane. ‘ 244 CONVERGENT PROJECTING-LINE DRAWING 1317. Problem 1. To find the perspective of a cube by means of the piercing points of the visual rays on the picture plane. Construction. Let ABCDEFGH, Fig. 238, be the eight corners of the cube, and let S be the point of sight. The cube is in the second angle and therefore both projections are above the ground line (315, 317, 514, 516). The horizontal projec- tion of the cube is indicated by abcd for the upper side, and efgh for the base. The vertical projections are shown as a’b’c’d’ for the upper side, and e’f’g’h’ for the base. The cube is located in the second angle so that the two projections overlap. The point of sight is in the first angle and is shown by its horizontal projection s and its vertical projection s’. Join s with e and v’ Vv Fig. 238. obtain the horizontal projection of the visual ray SE in space; do likewise for the vertical projection with the result that s’e’ will be the visual ray projected on the picture plane. The point where this pierces the picture plane is E, and this is one point of the required perspective. Consider next point C. This is found in a manner similar to point E just determined. The horizontal projection of the visual ray SC is sc and the vertical projection of the visual ray is s’c’ and this pierces the vertical or picture plane in the point C as shown. In the foregoing man- ner, all other points are determined. By joining the correct points with each other, a linear perspective will be obtained. It will be noticed that CG, BF, DH, and AE are vertical because the lines in space are vertical (1310), and hence their perspectives are vertical. This latter fact acts as a check after PERSPECTIVE PROJECTION 245 locating the perspectives of the upper face and the base of the cube. It will be noticed, further, that the perspectives of the horizontal lines in space meet at the vanishing points V’ and V (1313, 1314). 1318. Perspectives of intersecting lines. Instead of locat- ing the piercing point of a visual ray by drawing the projections of this ray, it is sometimes found desirable to use another method. For this purpose another principle must be developed. If two lines in space inter- oe sect, their perspectives inter- sect, because the perspective of a line can be considered as being made up of the perspec- tives of all the points on the line. As the intersection is a point common to the two lines, a visual ray to this point should pierce the picture plane in the intersection of the perspectives. Reference to Fig. 239 will show that such Fic. 239. is the case. CG is the visual ray in space of the point of intersection of the two lines, and its perspective is g’’, which is the intersection of the perspectives ab” and d’e”’. 1319. Perpendicular and diagonal. Obviously, if it is desired to find the perspective of a point in space, two lines can be drawn through the piont and the intersection of their perspectives found. The advantage of this will appear later. The two lines generally used are: first, a perpendicular, which is a line perpendicular to the picture plane and whose perspective therefore vanishes in the vertical projection of the point of sight (1312); and second, . a diagonal which is a horizontal line making an angle of 45° with the picture plane, and whose perspective vanishes somewhere on the horizon (1313). As the perpendicular and diagonal drawn through a point are both parallel to the horizontal plane, these two intersecting lines determine a plane which, like all other horizontal planes, vanish in the horizon. Instead of using a diagonal making 45°, any other angle may be used, provided 246 CONVERGENT PROJECTING-LINE DRAWING it is less than 90°. This latter line would be parallel to the picture plane and therefore would not pierce it. It can also be observed that there are two possible diagonals through any point, one whose perspective vanishes to the left of the point of sight and the other whose perspective vanishes to the right of the point of sight. It is further known that the perspectives of all parallel lines vanish in one point and therefore the perspectives of all parallel diagonals through any point must have a common vanishing point. With two possible diagonals through a given point in space, two vanishing points are obtained on the horizon. 1320. To find the perspective of a point by the method of perpendiculars and diagonals. Let a and a’, Fig. 240, Horizon ce! Vv Fia. 240. represent the projections of a point in the second angle, and let c and c’ be the projections of the point of sight. By drawing the visual rays ac and a’c’ the piercing point a” is determined by erecting a perpendicular at eas shown. Although the perspective a’”’ is determined by this method, it may also be determined by finding the intersections of the perspectives of a perpendicular and a diagonal through the point A in space. Thus, by drawing a line ab, making a 45° angle with the ground line, the horizontal projection of the diagonal is found. As the diagonal is a horizontal line, its vertical projection is a’b’. This diagonal pierces the picture plane at b’, which point is its own perspective. Its perspective also vanishes at V, the vanishing point of all horizontal lines whose inclination to the picture plane is at the 45° angle shown and whose directions are parallel to each other. The point V is found by drawing a line PERSPECTIVE PROJECTION 247 cv parallel to ab, c’V parallel to a’b’, and then finding the piercing point V. The horizontal projection of a perpendicular through A is ad; its vertical projection is evidently a’, which is also its piercing point on the picture plane and as it lies in the picture plane, it is hence its own perspective. The vanishing point of the perspective of the perpendicular is at c’, the centre of the picture. Since the perspective of a given point must lie on the per- spectives of any two lines drawn through the point, then, as b’V is the indefinite perspective of a diagonal, and a’c’ is the indefinite perspective of a perpendicular, their intersection a’’ is the required perspective of the point A. The fact that a” has already been determined by drawing the visual ray and its Vv Horizon c' y' all BS b, a ys 1 Vv b ve dad Fia. 241. piercing point found shows that the construction is correct, and that either method will give the same result. In Fig. 241, a similar construction is shown. The points V and V’ are the vanishing points of the left and right diagonals respectively. The perspective a’ may be determined by the use of the two diagonals without the aid of the perpendicular. For instance ab and a’b’ are corresponding projections of one diagonal; ad and a’d’ are corresponding projections of the other diagonal. Hence b’V’ is the indefinite perspective of the right diagonal (since cv’ is drawn to the right) and d’V is the indefinite perspective of the left diagonal. Their intersection determines a’, the required perspective of the point A. The indefinite per- spective of the perpendicular is shown as a’c’ and passes through the point a” as it should. 248 CONVERGENT PROJECTING-LINE DRAWING Any two lines may be used to determine the perspective and should be chosen so as to intersect as at nearly a right angle as possible to insure accuracy of the location of the point. The visual ray may also be drawn on this diagram and its piercing point will again determine a”. 1321. To find the perspective of a line by the method of perpendiculars and diagonals. Suppose it is desired to construct the perspective of an arrow by the method of perpendic- ulars and diagonals. Fig. 242 shows a case of this kind. AB is an arrow situated in the second angle; C is the point of sight. A perpendicular through the point of sight pierces the picture plane in the vertical projection c’. The two possible diagonals whose horizontal projections are given by cm and cn pierce the picture ; plane in V’ and V respectively. Fig. 242. The perspective of any diagonal drawn through any point in space will vanish in either of these vanishing points depending upon whether the diagonal is drawn to the right or to the left of the point in question. Likewise, the perspective of a per- pendicular drawn through any point in space will vanish in the centre of the picture. In the case at issue, the horizontal projection of a diagonal through b is bo and its vertical projection is b’o’; the piercing points is at o’. As the perspectives all lines parallel to the diagonal have a common vanishing point, then the perspective of the diagonal through B must vanish at V and the perspective of the point B must be somewhere on the line joining o’ and V. Turning attention to the perpendicular through B, it is found that its vertical projection corresponds with the vertical pro- jection of the point itself and is therefore b’. The perspectives of all perpendiculars to the vertical or picture plane vanish in the centre of the picture; and, hence, the perspective of B in space must be somewhere on the line b’c’. It must also be somewhere on the perspective of the diagonal and, hence, it is at their intersection b’. This can also be shown by drawing PERSPECTIVE PROJECTION 249 the visual ray through the point of sight C and the point B. The horizontal and vertical projections of the visual ray are cb and c’b’ and they pierce the picture plane at b’’, which is the same point obtained by finding the indefinite perspectives of the perpendicular and the diagonal. By similar reasoning the perspective a” is obtained and, therefore, a’’b’’ is the perspective of AB. 1322. Revolution of the horizontal plane. The fact that the second and fourth angles are not used in drawing on account ‘of the conflict between the separate views has already been con- sidered (315, 317). This is again shown in Fig. 238. The two views of the cube overlap and make deciphering more difficult. To overcome this difficulty the horizontal plane is revolved ce A\ Vv “oh Horizon fe ' a’ i t I Ant ‘a b 1 \ 1 WA : Pp O71 b \ \ \ y % Fig. 248. 180° from its present position. This brings the horizontal projections below the ground line and leaves the vertical pro- jections the same as before. Now the diagonals through any point slope in the reverse direction, and care must therefore be used in selecting the proper vanishing point while drawing the indefinite perspective of the diagonal. 1323. To find the perspective of a point when the hori= zontal plane is revolved. Let Fig. 243 represent the conditions of the problem. The vertical projection of the given point is a’ and its corresponding horizontal projection is at a, below the ground line due to the 180° revolution of the horizontal plane. The vertical projection of the point of sight is at c’ and its cor- responding horizontal projection is at c, now above the ground line. The conditions are such that the given point seems like a first angle projection and the point of sight seems like a second 250 CONVERGENT PROJECTING-LINE DRAWING angle projection, whereas the actual conditions are just the reverse of this. Through a, draw ab the horizontal projection of the diagonal; the corresponding vertical projection is a’b’ with b’ as the piercing point. As ab is drawn to the right, the indefinite perspective of the diagonal must vanish in the left vanishing point at V, Hence, draw Vb’, the indefinite perspective of the diagonal. i c 7 2 { Horizon e bo | \ a! AU \ \ \ iv \ \ \ o\ \ b ! ' | 1 L Bk-— \ \ \ y %. Fia. 243. The indefinite perspective of the perpendicular remains unchanged and is shown in the diagram as a’c’. Therefore the intersection of Vb’ and a’c’ determine a’, the required perspective of the point A. The accuracy of the construction is checked by drawing the visual ray ca and then erecting a perpendicular at 0, as shown, which passes through a” as it should. 1324. To find the perspective of a line when the hori= zontal plane is revolved. The point B in space will again be located in Fig. 244 just as was v’ c” v_ donein Fig. 242. The horizontal eee projection of the diagonal is bo Sa 0" and its vertical projection is b’o’; | the piercing point, therefore, is J n 0’. The horizontal projection of the point of sight c is now above * / the ground line instead of below ey it, due to the 180° revolution of Fra. 244, the horizontal plane; the orig- inal position of this point is marked c”.