THE LIBRARY OF THE UNIVERSITY OF CALIFORNIA PRESENTED BY PROF.CHARLES A. KOFOID AND MRS. PRUDENCE W. KOFOID Aan) Alt LIBRARY ; z ¥ i w ; 2 ® 8 IR - iy 4 é # » $ % % E ; ON Poa? Df = 7 Zar Crngraocd. 4 Cer” Toren Ds 7 4 & + PETER MIC A ro A TREATISE ON PROJECTION, CONTAINING FIRST PRINCIPLES OF PLANS AND ELEVATIONS, AND THE MODES OF DELINEATING SOLIDS, AND EVERY FORM OF MECHANICAL CONSTRUCTION, SO AS TO PRESENT THE MOST STRIKING IMAGE OF THE OBJECT TO BE CARRIED INTO EXECUTION; ON ENTIRELY NEW PRINCIPLES. TOGETHER WITH A COMPLETE SYSTEM OF ISOMETRICAL DRAWING, THE WHOLE PRACTICALLY APPLIED TO ARCHITECTURE, BUILDING, CARPENTRY, MACHINERY, SHIPBUILDING, ASTRONOMY, AND DIALLING, WITH NUMEROUS PLATES. AAA ANAAAANADL ANA BY PETER NICHOLSON, AUTHOR OF “THE ARCHITECTURAL DICTIONARY,” “ THE PRINCIPLES OF ARCHITEC. TURE,” “THE BUILDERS’ DIRECTOR,” &c. &e. de. LONDON: RICHARD GROOMBRIDGE, 6, PANYER ALLEY; AND CHARLES THURNAM, CARLISLE. 1840. ~~ hk SA 507 N& Arc Li: 4 TO JOHN BUDDLE, ESQ, THIS WORK, UNDERTAKEN AT HIS SUGGESTION, IS INSCRIBED BY THE AUTHOR, AS A MARK OF ADMIRATION OF HIS GREAT PROFESSIONAL TALENTS, AND RESPECT FOR HIS PRIVATE WORTH. CONTENTS. QE A EE EE ——— PREFACE, vu savevsiras Severus wveare ries seeres a reve seve ras Vien seiniis ade ix Introduction; .......evesvsaseseessas sosess ios ‘evansnrevavers eees Mevestsnavrarseene veer Xi PART LL—_ELEMENTARY PRINCIPLES. Definitions, ....... 9s hans neva traverse sesasisore + x Stans ase wNuse ssn ser NeVeR 1 Theorems, ........... evse sees cavasasses sssnevsvo steressaseesnseare-stvastssussnarsanie son 2 PART IL—OF THE OBJECTS OF PROJECTION. Definitions of Geometrical Solids, ......c.ecevseesens wrreviuevaiivainy evveivves vemieve 9 Definitions of the Circles of the Sphere,....ccoveeenie veeraersinsennns wveeree serene . a2 Definitions of the Regular Solids, ......cccireisriensinmiiininivisnnseeienn. 13 PART IIL—PROJECTION UPON THE LOWER PLANE. Of the Projection of Points, Lines, and Plane Figures, ......... « wooceereenns 14 The Construction of Plans and Elevation of Geometrical Solids, ...ceceeees « 18 Definition of the Horizontal Plane of Projection, ..u..eceues se ssrseranesne vene 18 Definition of the Vertical Plane of Projection, ........cceeseesvessersiones caseees 18 Definition of the Ground Line denoted by ¥ Z, ....cceeereererune cunennnens wie JB Definition of a Plan, .....:... 00000. 10: satus sesetisnnashiveabimsssivtiussaws bes sseuteeirone 18 Definition of an Elevation, ........ tevenren 53s resp abeditrn sonra neis sires v ttseassi ane VG The Plan of a Right Line being given to find the Elevation, ...... «ceuvveene 18 To find the Plan and Elevation of a Cone,........e.. rerun ees Te setes verdes 20) Plans and Elevations of the Regular Solids, ... .vvssesseernsreseressasserassnsane 20 To draw the Plan and Elevation of the Tetrahedron, ... 21 To draw the Plan and Elevation of the Hexahedron, -.. 22 To draw the Plan and Elevation of the Octahedron,....uueesesressssrasres senses 22 To draw the Plan and Elevation of the Dodecahedron, 22 To draw the Plan and Elevation of the Icosahedron, ...ee....eeeccosecsssesnes . 28 Plans and Elevations of the Circles of the Sphere, ............. Vesvsehseritvers 24 To find the Plan and Elevation of a Parallel Circle,e...veeseceernesensessnsassees 24 "To projects Merldional CIrcle, .. ......cecsesssoss. sseresnsorsssnunsessramemsitae . 24 To project a Meridional Circle, under other Data, ......veevueresisevareeeninens 25 Example, shewing a Plan and Elevation of the Sphere, ... voeoeereiserns weer 25 Of the Intersection of the Surfaces of Cylinders and Cones, ....c.oeeveveeeeeenne 26 To find the Projection of the Intersection of the Surfaces of two Cylinders, 27 To find the Projection of the Intersection of the Surfaces of a Cone and a CVHIAEY, 1octrsssreesrsensiss surrevrenassss ippsbiinsniiyevuumvus swussist une yredeiivs 28 To find the Projection of the Intersection of two CONES, veuve vuseeseneenesnnne 29 On the Methods of finding the Covering of an entire Cylinder, or an entire Cone, or any Portions of these Solids, .,.e.-sssevsrvesssvusivssanernidensoy wre swe | 130 On the Methods of covering the Surfaces of Solids of Revolution, .......c.... 36 Wo find the Covering of 2 Hemisphere, ...v..vi som imisrvicismwmmrvessvsnes moos 37 Plans and Elevations of practical ObJects, ....es vereeresvresnerersnsrenses wetvanens 39 ._ ExadupLES IN MACHINERY. Projection of an Undershot Water Wheel, ........... . sottesng ota eis wweivisemengs 42 Projection of two Pinions working into each other, ..... Cy srnves 44 Projection of the Plan and Elevation of a Pinion or Spur Wheel, ............ 45 vi CONTENTS. Page Projection of the Plan and Elevation of a Bevel Wheel either right or ob- Hague, oi. Ly ders Gain ch RR GE A runs vider dn ra nar ea ewan nst se Projection of the Plan and Elevation of a square and triangular threaded SCICW, ci.erreeeeneenseivstnsiesssasoncusrssaseseiivensaan huss +oviseonnsrasesnnsisdues EXAMPLES IN ARCHITECTURE. Plans and Elevations of a Gable-ended, as also of a Hipt-roofed House, ... The Planand the Elevation of Houses which form the Perimeter of a Polygon, The Plan and Elevation of Houses forming a Crescent, ....cceoeciuneraaeennes Plan and Elevation of a circular Wall, with Windows,.. cecceveene veveenrennas Plan and Elevation of the Inside of a parallel circular-headed Window,...... Plan and Elevation of an oblique circular-headed Window, ......cceeeeuveueene. Plan and Elevation of a Building similar to the Saloon under the Dome of SE Paul, rt Is Re Fe ei Basta enas hue sees Plan and Elevation of an Octagonal Dome Ceiling, .....cccouuvuienns cavneennas Plan and Elevation of an Octagonal Ceiling having the Intrados of the Win- dows in the Tnside Conical SUITACES, . vc..v.sersstinsrnvossnsaersisesssesssarse Plan god Elevatlon of an Octagonal Ceiling, with Coves adjacent to the U1, es fie ii tie anit rans as Aras sat Ankenes neh aR Rta dlrs drs ie pembneas Plan and Elevation of a Hemispherical Dome and Niche, shewing the Joint- linesiof thie Masonry,:..cec.ii eee sss eninn: iesnsesh svasntnss savans sesh snsssnasaoente Plan and Elevation of a Dome with Coors, ...eec sees caorisssnsecsasessnsscasss Plan and Elevation of a Stair beginning and ending with Flyers, and with Windows'between the FLYers, ..ceccesssrerenssassesvasnernssessesersans coraseen Plan and Elevation of a Room which rises from a Square and terminates under the Roof in an Octagon,.......ceeeueeens Sabie ana ers Wa rhea ss sae Exercises in the Plans and Elevations of Groin Ceilings, ....ceeeuivsiniinanennns Plan and Elevation of the Earth in its Orbit, ....... Seresincsetan sesesirsesevers PART IV.—PROJECTION UPON THE OBLIQUE PLANE. Projections of Points and Lines, either in or making any given Angle with the original Plane, cide. eessse ceresaee, centtassssvissitetsenannenassiasasessraesere A useful Problem, which is the Foundation of Dialling, ........ «.eevuuinannns The Projections of three Edges perpendicular to each other at the Vertex of a Triangular Pyramid, are perpendicular to the sides of the Trian- gular Base, and intersect each other in the Projection of the Vertex,... Given the Projections of the three Edges of a Right-angled Solid Angle, and the Sides of the Sectional Triangle, to find the original Lengths of the three EBdges,.i isis ii seiviinisiisciniieivvessnsnestas soneseas Sots esetivenes Given the Dihedral Angle which any Face of the Solid Angle makes with the Plane of Projection,and the angles which the Sides ofthat Face con- taining the Right Angles makes with the intersecting Line, to find the Projections and Lengths of the three Edges, ........... siethuvirerae, Elucidation of the Principle of the directing Diagram,.....c.eeeieareuenn wennenne Definitions of Lines in the directing Diagram,........... vccoveriiniiianiininnnn, From a given Point in a given projecting Line parallel to a given Director, to cut off a part which shall be the Projection of a given Length, ...... Given any original Scale of Feet and Inches for any one of the three Edges of the Solid Angle, to find the Scale which shall give the correspond- ing Measures of the Length of the Projection, ......cccoeciiinnenns winiiii. To find the Projection of a Rectangle, ........ sThieate Sraredve estan ieaduennarie ves To find the Projection of a regular Octagon, ..... vviveeeeruiienrennes vevnnnnins To find the Projection of a Circle in a Plane parallel to any one of the three Faces of the Solid Angle, ..i...... .cedhecaiicniesiceness cinseerenes "To find the Projection of a Rectangular Prism,.. ..... «iieiiirreireeiiaennn. To find the len of a Triangular Prism, .........cecter fie Srknsameests 48 78 80 81 84 "A Table from 1 Inch to 9 Feet Isometrica CONTENTS. vii Page. "To find the Projection of a Cylinder, ... eases suren ive rserriiieiaeni i 96 To find the Projection of a Rectangular Pyramid, esdestivilisey. cxavhant hisaseiil OF "To find the Projection of 0 CONC, ve. eres ress seississnssone statessnseres coianseiic IF To find the Projection of the Frustum of a Rectangular Pyramid, ... ..... 98 To find the Projection of the Frustum of a Coneyeee..viueenes eennnnaes ceseeess 98 To find the Projection of a Sphere reposing upon a given Point, ............ 99 To find the Projection of a Circular Ring,..c...ccvveeee vureens Sesussesndedeieniy 99 To find the Projection of the Section of a Cylinder,.ce..civuiserernis vareerenens 100 To find the Projection of the Section of a Cone, .....c...eevrueururenranes essere 101 To find the Projection of the Intersection of the two Cylinders, .... ...... 102 To find the Projection of the Intersection of a Cone and a Cylinder,......... 103 To find the Projection of the Intersection of two Cones, ...... Se sseemszee ossense. JOI Projection of the Regular Solids, ..oeveriuns siiveriniiiiiininnniieiiie seine. .. 105 Examples in projecting the Tetrahedron, Hexahedron, Octahedron, and ‘the Projection of the Dodecahedron, iuis.. ec ossisssssvsiovnssneinsesinsinessins diss 107 Projection of a Regular Cylindric Octahedron circumscribing a given He- NHSPNETE, se:. cersaresnesrarnsisnssonssnssssnnrnarsh sazssosysessannesvessesheatos: duane 108 In the Directing Diagram when the Section Triangle is equilateral, the Length of any Edge, the Length of its Projection, and the Length of a Right Line drawn from the Summit of the Solid Angle to meet the in- tersecting Line perpendicularly are to one another, as V3 Vv 2, y/1.... 109 Projection of a Regular Cylindric Octahedron inscribed in a given Herder PHETE,. os uxrisernrnsiassovsas susunvivssnsvatavanussssnnnesnssseonssisnshvinnrenasans 110 To find the Projection of Sphere with twelve Meridians and four Parallel Circles,........ Ee A Ch A Ae MN CI DC SPRL SO Rl 110 Application of Projection upon the Oblique Plane to the Arts,... ceeseenee 112 Examples In Architecture, ... ci siveet vavsotisirdsnesitsrsenssvssss lanes sossbibsns ees weve 112 Isometrical Delineation of a Hip-roofed House,....covtereresrnnansn. seansseieines 114 Delineation of the same House turned round 20°,........... ves sonn vides vies 114 Delineation of the same House turned round 45° from its first Position, .,, 116 Delineation of a Pendentive Dome, with COffers, .. «vevveieererrieresinsnrnseanes 118 Uses of the Right-angled Triangle of which the Acute Angles are 60° and 30° in Isometrical DYAWING, ion. irons ve ressesmenssnsvds ss bates shiyuvetistosmanse 120 To determine the Scale of an Original Object corresponding to a given Iso- metrical 8cale,... vc. svn, vvrisincavrrasisitesnrinaispns servis sasesndarsvaabhsvesny 121 To determine the Semi-axis Major and the Semi-axis Minor of the Projec- Jection of a CHTCIe, .uuiivitidoe sins ds Bh Be a 0 iain desta e Wabteiees 121 Example of the Use of the Axes in finding the Isometrical Representation of a Circle, servi ven anys at RAR Tas A eran es ede ee Nayar a SUN in dh ees his Fake 122 + Example in Isometrical Drawing in fngoe the Projection of Spur Wheel, 124 Radius, for Andi the two Axes of the Ellipse, which is the Projection of a Circle,.. wvseeeeennreeereness... 126 Oblique Projection applied to the Representation ofa Boat, ......... soseive 127 Oblique Projection applied to Dialling, .....esssnrrssesns senses svesvesesseces seases. 128 1sometrical Drawing with Examples, ".......teueitttrersisesrasassasssssssvaesvonsies 130 Isometrical Delineation of a Trough of greater Dimensions at the open Side or Top than at the Bottom, ....... ce 1.:ssssevesvasnienions itive’ soive ve + 134 Isometrical Delineation of a Doric Cornice inverted, ........e.vuunee.ns 3 svesee 134 Isometrical Delineation ofan Ionic Cornice, ...... tenvssnessenrasine’ saesrreranse . 134 A Tower diminishing upwards in several Forms, .....c.ccevrvurasirnssnssaennan, 134 Orthographical Delineation of a double Floor, with a girder, binding, bridg- ing, and ceiling Jolstn, ..cvtviissetiorisitineesiomsresnnivnntisecssysattveietis 135 Orthographical Delineation of a Roof, upon Wall Pleas, with h Principles, Purling, Spars, King’ Post, and Struls, i ii voiveeeissriniivessansivtbiveretvaiil 3 APPENDIX. ww Projection of a regular Octagonal Prism, ...... 990000700 s0ve s00000 rilessserevierees 136 viii CONTENTS. Page. Projection of a regular Octagonal Pyramid, .....cccveniiiiiniiniisiniiieineninn. 136 Projection of the Frustum of a regular Octagonal Pyramid, .. crarvesee cussnnines | 136 GEOMETRICAL PROBLEMS. Po describe.a. regular. Octagon, [vie..ce cidesssvsane | yoisasnss sane evel ssesesaseneess (58 To describe a Curve which shall be very near Approximation to the Curve of an Ellipse, the Axes being of any given Dimensions.—See the note, 17 This Method of describing an Ellipse is applied to one or more of the Figures in each of the following Plates :—2, 3, 5, 11, 12, 16, 17, : 3s 19, 20, 21, 22, 24, 25, 29, 31, 32, 33, 34, 35, 36, 57, 38, To accommodate the general Method of fempioen an Ellipse to Isometrical DELNCation.——See the MOLE, «.«ie: reernrsassersstinsseioeranssesss, sis onsets suse 123 The Isometrical Mode of describing an Ellipse is applied to one or more of the Figures.—Plates 42, 47, 48, 49, 50. To find the Elevation of a Cylindric Octahedron, circumscribing a given Hemisphere or a given Segment of a Sphere.—See Plate 41, No. 1 ANE NU es snes. chenssusicnnson srensios: sattnsssanarsuseisesios ssesesessseessruseene. JOB DIRECTIONS TO THE BOOKBINDER. Pages. Pages. Plate Y-tofacer w..ii.cu ioe 14 and 15 | Plate 30 to face......... eri areane 95 2:80: f208:.. ovis vaniiiiis we. 16 and 17 31 to faee. iin aii. 96 3 to face vensensse: 18 80419} en B2 tO £000. Curis innnnns v1 197 ——- 4 to face w.22and 23 | rv 83.to faces....cc.s se diouns 98 tt tO BE deh unnenasgabnnns 24.and 25.1 —— B4.t0. face... cru sres steesres 99 == A to face . oii 28 ——— 35.60 faCe.0urerareiranee .e« 100 and 101 "rt F106. qecarrsnsrsarsrasses 29 —— 36 to face..ceneunnnninnnns 102 rege BLO TAGE... od eas ous tus has 32 and 33 | ——- 37 to face........ ira 103 — ety fea et a 34 and 35 | ——- 38 to face.......... ne, 104 —z 8ite faee:..... OR EAL 8 —- 39 to face.................. 106 ——— 9 to face... — 40 to face.ee.....or00.nnnnn 107 —i 10 bo: face, ——- 41 to face..... ceeuun...e .. 108 wr-z.1] ta face... ——- 42 to face... .... 110 and 111 wie [12 80 £000: 54 — 43. to face... ceieirinirese 112 w— 13 to face —- 44 to face..........ieuunene 113 HA tO TREO... eereerserisies SONA BLY co 45 £0 1000. 0sususeseerseenrer 114and 115 re=_"15 {0 face....... a varee aunens 53 teem AGO ACC as ssn he reneriss MB 8nA 11] res "1 10 fac... Sees 54 and 55 | —— 47 to face.................. 118 and 110 —— VE tO TREO ies a vives 58-and.59 —— 48 tofacell ioe 120 cnieiean JS HOTDOBc1 2s i csann roneevns 60 and 61 | —— 49 to face.......... knees 122 and 123 sb HGH OHAO Be ves . 62:20:03 |] —— 50 to Fagor. ..cociiicasanes 124 and 125 ip 20.10. T0801 cis isvtinrsnnein 62 ——= 511 10-260: .. «ese 127 —- 21 to face 66 and 67 | —— 52 to face......... eeerese. 1128.20nd 129 —- 22 to face 63 and 69 | —— 53 to face.........eeavsesse 1 131 — 23 to face 70 and Fl ——— 54101066, 0... sxc rrveseets 132 — 24 to face ... 74 and 75 | — 55 to face....uu...nnnneenne 133 —— 25 to face... «ees 76 and 77 56 to face... 134 ——- 26 to face... .. 78 and 79 | Plates 57, 58, 59, to follow 56 —- 27 to face 88 and 8Y 1 ~—— 28 to face ... 92 and 93 oie — 20 £0 F200. cucererss senrisnee: 94 —- 62 after plate 61. PREFACE. Tuk great utility of a practical work reducing mechanical drawing to fixed and scientific principles must be sufficiently obvious. To the Engineer, the Architect, the Mechanic, and to all employed in the erection of buildings, or the construction of machinery, a knowledge of the theory and practice of mechanical drawing is absolutely ne- cessary ; and, it has often, therefore, struck the author, as a circum- stance to be regretted, that the elements of a system sufficiently easy and general have not yet appeared. Draughtsmen, from the want of principles to direct their judgment, frequently find insurmountable difficulties in representing complex objects, and, in consequence, the practical mechanic, whose office it is to execute the work, experiences corresponding difficulties in ascer- taining from the drawing the intention of the designer. It has been the principal aim of the author, in the following work, to remove those difficulties, and to supply the deficiency which he conceives to exist, by furnishing an easy and obvious system of delineation, whereby every object to be represented may be reduced to drawing on geometrical principles of unerring accuracy, so that the form, position, and magnitude of all the parts may be easily under- stood, and correctly executed. b x PREFACE. The most useful kinds of mechanical drawing depend on the theory and practice of orthographical projection, which forms the principal object of the present treatise. The theory of projection is of universal application; a knowledge of this useful branch of delineation will enable the designer to instruct the workman with nearly as much ease as if he had the model before him, and to explain the effect of an imaginary object as if it really ex- isted ; this knowledge in the workman will enable him to forsee how the different parts of an object will join upon each other, to understand drawings and designs with readiness, and to execute them with accu- racy. Among many other uses to which this truly admirable science ex- tends its influence, may be mentioned the construction of the center: ings of arches and groin vaults, the formation of hand-rails and stairs, the cutting of stones for bridges and oblique arches, and the delineation of plans, and elevations of buildings and machinery. But the utility of an intimate acquaintance with the principles of this useful art is not confined to the workshop alone, a certain knowledge of these princi- ples should form a part of that stock of information which is essential to the student in the arts of design, and the rapid strides which have of late been made in other departments of the arts and sciences render it far from improbable that we shall shortly see the theory and prac- tice of projection taught in our public schools, as a necessary branch of education. Considerable attention has been paid to the arrangement of the fol- lowing treatise; and it has been attempted, in the first place, to lay down a body of definitions affording a clear explanation of the terms PREFACE. x1 which are necessary in treating this subject. Many authors, of great pretensions in the present march-of-intellect-school of mechanics, pre- tend to dispense with definitions altogether, and to explain the terms employed as they arise in the course of reading ; but these terms are generally unscientific expressions which are ill calculated to describe the properties of the objects to which they allude, and being detached from each other, the reader is deprived of the instruction he would otherwise receive by the use of definitions, admitted by general con- sent into science, and arranged in the order in which they depend on each other, so that they may be readily understood and easily refer- red to. The definitions and the elementary part of the work occupy only the first eight pages ; but to obtain a complete knowledge of the theory, the eleventh book of Euclid should be understood. In the remainder of the present volume, the reader will find a full explanation and a clear elucidation of the methods of delineating ob- jects in every position in which the plane of projection can be placed and the plates have been so arranged, that in reading the propositions, and examples, he will find the diagrams to which they refer, generally opposite the page he is reading. The engravings* are, for the most part, in outline, as it would have been impossibie, in shadowed plates, to distinguish the letters of re- ference. To avoid perplexing the student by introducing letters of the same name and form in the same diagram, it has been found necessary not * The work of Mr. Collard, of Newcastle upon Tyne. xii PREFACE. only to employ the three varieties of the English Alphabet consisting of capitals, small roman and italics, but also to make use of Greek let- ters, the nine numerals and even letters and figures with accents, and it is hoped, that this method will obviate that confusion which so often prevents the reader from finding the place referred to on the diagram. For the sake of perspicuity, a line is always referred to by two letters one at each extremity, by which it can be readily distinguished from any other line; whereas, in many modern publications where single letters are used to find the line, the reader is often at a loss to deter- mine the one referred to. The author trusts, that most of the references will be found correct, as particular care and attention have been bestowed on this portion of the work. The author cannot conclude this preface without expressing his un- feigned thanks to John Buddle, Esq., for many acts of kindness, and particularly for his munificent patronage of the present work. INTRODUCTION. Tue word projection has a general meaning, as comprising the ortho- graphical and scenographical representation of objects, but as in this treatise the orthographical is the only kind of projection used, the generic term will be employed throughout. Projection is here divided into two parts, viz., projection upon the lower plane and projection upon the upper plane. Projection upon the lower plane is when the object is projected up- on two planes, of which one of the planes is perpendicular to the other. By menns of these two projections, the length of every rec- tilineal edge of the object may be ascertained by its position to the two planes of projection ; for if such an edge be parallel to both planes of projection, the length of that edge will be the length of each of its projections, and if such an edge be parallel to one of the planes of projection, the projected length of the edge upon that plane is the length of that edge, and if such an edge be perpendicular to one of the planes of projection, the length of the edge will be the length of its projection upon the other plane only ; and if such an edge be ob- liquely situated to the two planes of projection, its length will be a right line, equal to the hypothenuse of a right angled triangle, of which one of the sides containing the right angle will be the projected length of the edge on one of the planes of projection, and the other side equal to the difference of the perpendicular distances between the ends of the projection of the line on the other plane and the inter- secting line of the planes, and thus from two projections the length of every rectilineal edge of the object may be ascertained. XIV INTRODUCTION. Projection upon the upper plane is when the plane of projection is required to make an oblique angle with the plane upon which the object is situated, and thus if every solid angle of the object be iden- tical to the solid angle of a cube, the length of every edge of the ori- ginal may be exactly ascertained. Projection upon the upper or oblique plane has the advantage over that upon the lower plane in showing the connection of the adjacent sides, and of representing the object in a pictorial form approximating to perspective. By means of a single figure, called the directing diagram, which ac- companies every plate, the oblique projection of every object, however complex, will be readily found without the trouble of constructing angles, for every line in the projection must be parallel to some line or other in the directing diagram, and the object may be drawn to a scale of any magnitude without regard to the size of this diagram. The directing diagram is no more than the projection of the three edges, and the development of the three faces of a triangular pyramid upon the plane of the base, the three edges being perpendicular to each other. By means of this useful figure, the projections of numerous objects may be found, which would be attempted in vain by any other method yet published. The perfection of oblique projection will be seen by inspecting plates 36, 37, 38, where in each plate the same object is projected in different positions upon two planes, one being parallel, and the other obliquely situated to the plane containing the axis of the two inter- secting solids : then in any one of these three plates, Figure 1 being the oblique projection is necessarily regulated by the directing dia- gram ; and if, in Figure 2, the right lines, which are the projections of the two axes, be respectively drawn parallel to the two lines con- taining the right angle of the developement of the horizontal face of the solid angle in the directing diagram, and if any two corresponding points of the two projections be brought into a right line perpendicu- lar to the intersecting line of the horizontal face of the said diagram, a right line passing through any other two corresponding points in these two projections, shall be parallel to the line passing through the first two corresponding points. Thus in plate 36, if the right line ¢ ¢ which passes through the projections ¢ ¢ of the intersection of the two INTRODUCTION. Xv axes, be perpendicular to U V in the directing diagram, and if the lines y 2, w x, Figure 2, be respectively parallel to U S, § V, in the directing diagram ; and if from the points u,v, 8, y, Figure 2, right lines be drawn parallel to # #, those lines will pass through the corres- ponding points %, v, 3, y, Figure 2, which points are the projections of the ends of the cylinders, and, in right lines, which are tangents to the curves The practice of isometrical drawing is derived from the principles of orthographical projection upon the oblique plane. The rules for delineations of this kind are in many cases more sim- ple than other modes of orthographic projection, the plane of projec- tion being equally inclined to the faces of the object to be represented, There are, however, very few objects that consist of lines which, when projected, would be altogether isometrical ; but the projections of other lines which are necessary to complete the entire projection may easily be found, provided each of these lines is the hypothenuse of a right-angled triangle, of which the sides, containing the right angles, are isometrical lines. The methods of isometrical drawing is elucidated by means of the directing diagram, and is applied to a great variety of objects difficult in their form and of frequent practical use. The isometrical protractor shown in Plate 48, Figure 5, may be applied to finding the projection of any angle in a plane parallel to any of the three faces of the solid angle. The chief utility of thisis in the projection of geological and en gineering surveys, and its application to these objects is fully detailed and illustrated by numerous examples in Mr. Sopwith’s recent work on Isometrical Drawing. * The method of describing an ellipse explained in the note, page 17, may be used in describing the projections of circles which are ellipses, as has been done by the engraver in all the diagrams of this work where * A Treatise on Isometrical Drawing, as applicable to Geological and Mining Mining Plans, Picturesque Delineations of Ornamental Grounds, Perspective Views and Working Plans of Buildings and Machinery, and to general purposes of Civil Engineering ; with details of improved methods of preserving Plans and Records of Subterranean Operations in Mining Districtt. By T. Sopwith, Land and Mine Surveyor, Member of the Institutiou of Civil Engineers, Author of “ Geological Sections of Mines,” ¢ Account of Mining Districts,” &c.—Published by J. Weale, London. xvi INTRODUCTION. such projections are required. The figures 1 4, 2 4, 3 A, plate 29 and 3 4, 3 B, 4 C, plate 34, show the radii of curvature at the ex- tremities of the axis-minor and major of the ellipses to which they are adapted in the plate, these figures are similar and have the same let- ters of reference as 5 G, plate 2 explained in the note above alluded to. The method of adapting the ellipse to the isometrical projection of a circle is demonstrated in Proposition LXXXI. The projection of objects upon the oblique plane, shows the con- nection between geometrical and perspective deliniation: for if, in the directing diagram, the sectional triangle be considered as the plane of the picture, and the vertex of the solid angle the point of sight, the three sides of the sectional triangle will be the vanishing lines of the three nearest conterminous faces of the cube and the an- gular points of the triangle will be the vanishing points of the squares forming the faces, Suppose, now, the vertex of the solid angle formed by the three conterminous faces to touch the centre of the picture, the cube and the point of sight being upon opposite sides of the pic- ture ; then the greater the distance the point of sight is from the pic- ture, the vanishing lines being always parallel, will be distant from the centre of the picture in the same proportion, and the perspective representation will also be greater, and if the distance from the centre to the point of sight be infinite, the three vanishing lines will be in- finitely distant from the centre, and the representation will become the orthographical projection of the cube. TREATISE ON PROJECTION. PART 1. CHAPTER I.-.ELEMENTARY PRINCIPLES. DEFINITIONS. 1. Projection is the art of drawing objects upon a plane, so that right lines proceeding, according to a given law, from the extreme boundary, and from the intermediate lineal parts of the object, may fall upon the corresponding parts of the drawing. 2. The plane upon which the drawing is made, is called the plane of projection. 3. The drawing is called the projection, image, representation, or picture. 4. The right lines which pass between the corresponding points of the object and its projection, are called projecting rays. 5. The objects to be represented, are called original objects, which may be points, lines, figures, or solids. 6. Ifa point, line, plane figure, or the face of a solid to be project- ed, be in a plane, making a given dikedral angle* with the plane of projection, the plane which thus contains such original point, line, plane figure, or the face of a solid, is called the original plane. 7. The right line, in which the original plane and the plane of pro- jection meet each other, is called the intersecting line. 8. If the projecting rays fall perpendicularly upon the plane of projection, the image or projection is called the orthographic projec- tions. 5 The angle made by a right line in an original plane and the in- tersecting line is called the angle of that line. 10. The angle made by the projection of a right line, and the in- tersecting line, is called the angle of projection of that line. * Dihedral angle is the angle formed by the meeting of two planes, and is equal to the angle contained by two right lines, one drawn in each of the planes from the same point perpendicular to the common section. +The above definitions are general for every species of projection ; but as the orthographic projection is the only one used in this work, the adjective orthogra- phic is omitted. B 2 ELEMENTARY PRINCIPLES. 11. The projection of an object upon a horizontal plane is called a plan. 12. The projection of an object upon a vertical plane is called an elevation. 13. A plane passing through any right line perpendicular to the plane of projection is called the projecting plane of that line. AxI0MS. 1. A right line which coincides with the intersecting line is, at the same time, the projection of that line. 2. The point in which a right line may meet the plane of projection is in the projection of that line. 3. Every right line which is not parallel to the plane of projection will have its intersection in the projection of the line. 4. Every right line in an original plane, not parallel to the inter- secting line, will meet its projection in the said intersecting line. 5. A right line perpendicular to the plane of projection is projected into a point. 6. The projecting plane of every line is perpendicular to the plane of projection. PROPOSITION I. THEOREM. Every right line which is not perpendicular to the plane of projection shall be projected upon a right line, For the intersection of the projecting plane, passing through the right line with the plane of projection, is a right line; but the inter- section of the projecting plane and the plane of projection is the pro- jection of the line; hence every right line which is not perpendicular to the plane of projection is projected upon a right line. PROPOSITION II. THEOREM. The projection of every right line parallel to the plane of projection shall be parallel to the original, For let A B be the right line. Draw the projecting rays A a, B b meeting the plane of projection in the points a and b, and join the points @ and 4. Then since the projection of a right line is a right line (Prop. I.), and since @ and b are the projections of 4 and B, a bis the projec- tion of A B. Now A B being parallel to 2 the plane P Q R cannot meet it: for, if possible 4 B will meet its pro- jection @ b in the plane 4 a b B; but since a b is also in the plane P Q R, the original line 4 B will meet the plane P Q R; there- fore A B cannot be parallel to the plane P @ R, unless it is parallel to its projection a b. ELEMENTARY PRINCIPLES. 3 PROPOSITION III. TuEOREM. The projection of a right line parallel to the plane of projection shall be equal in length to the original, For since, by the preceding demonstration a b is parallel to 4 B, and 4 a parallel to Bb, the figure A B b a is a parallelogram (Eu. b. i., definition following Prop. 39); hence the opposite sides are equal (Eu, b. i., p. 34); therefore a b is equal to 4 B. PROPOSITION IV. THEOREM. The projections of parallel right lines shall be also parallel, For let a projecting plane pass through each of the lines to meet the plane of projection, then these planes will be parallel to each other; hence their intersections with the plane of projection are also parallel (Eu. b. xi., p. 16), therefore the projections of parallel right lines are also parallel. PROPOSITION V. TuEoREM. The projections of parallel right lines shall have the same proportion to one another that the originals have. For let A B, C D be two parallel right lines, and P @ R the plane of pro- jection; moreover, let a b, ¢d be the projections of A B, C D then a b is pa- rallel to ¢ d (Prop. IV.). Draw the pro- jecting rays A a, B b of the points 4, B, and the projecting rays C ¢, D d of the points C, DD; moreover, draw 4 &" pa- rallel to a b, meeting B b in 4’, and 0 draw Cd’ parallel to ¢ d, meeting D d and d’. Then, because the projecting rays A a, B b, C ¢, D d are parallel, and the right lines A B, CD are parallel, and since 4 &’, C d’ are respectively parallel to a b, ¢ d, the right lines 4 &’, C & are parallel ; hence the triangles A BV, CDd are similar; therefore AB: AV :: CD : Cd (Eu. b. vip. 4), and AB:CD:: AV : Cd& (Eu b.v. p. 16); but since 4 a b b’ is a parallelogram « & is equal to 4 #, and since C ¢ d d' is a parallelogram c¢ d is equal to Cd’, hence AB: CD ::ab:0d 4 ELEMENTARY PRINCIPLES. PROPOSITION VI. THEOREM. The projection of a right line divided into any number of parts shall be divided into the same proportion as the original. Let the original line 4 D be divided D into any number of parts, 4 B, B C, C D, &c., equal or unequal, and let the B projecting rays 4 a, Bb, C¢, &c, fall upon the plane of projection P Q R, in the points a, b, ¢, &c.; then a d shall be E » divided in the same proportion as 4 D. ? “fe da For if A D be parallel to the plane of Q projection, a d is parallel to D A (Prop. IL), and @ b, b ¢, ¢ d, &c., are equal, respectively equal, to 4 B, B C, C D, &c. (Prop. IIL.) ; but if 4 D is not parallel to the plane of pro- jection; let D A and d a meet in E, then the right lines Bb, Cc, &c., being drawn parallel to the side D d of the triangle, & d D will cut the other two sides & D, FE d, proportionally (Eu. b. vi., p. 2.; hence the projection a d shall be divided in the same proportion as the original A D. Cor. If the original line 4 D be divided into equal parts, its pro- jection a d will also be divided into equal parts. PROPOSITION VII. THEOREM, The projection of a rectilineal triangle in a plane parallel to the plane of projection, shall be a triangle equal to the original. For every side of the projected triangle is equal to the correspond- ing side of the original (Prop. IIL); therefore the projected figure shall be equal to the original, PROPOSITION VIII. THEOREM. The projection of a rectilineal angle, in a plane parallel, to the plane of projection, shall be a rectilineal angle equal to the original, For draw any right line, meeting the legs of the angle, and the figure will be a rectilineal triangle, and therefore its projection (Prop. VIL.) will be a rectilineal triangle equal to the original ; hence the an- gles of the projected triangle will be equal to the corresponding angles of the original; therefore the projection of the rectilineal angle which is one of the angles of the projected triangle, is equal to the original angle. PROPOSITION IX. THEOREM. The projection of any rectilineal figure, in a plane parallel to the plane of projection is equal to the original, For the whole original figure may be divided into triangles, and ELEMENTARY PRINCIPLES. 5 since the projected triangles will be respectively equal to their corres- ponding originals, the projection of the whole figure will be equal to the original. PROPOSITION X. THEOREM. The projection of any curvilineal figure, in a plane parallel to the plane of projection, shall be equal to the original, For by drawing right lines from any point within the original figure, the projections of the angles around the assumed point will be equal to the corresponding originals, and the projections of the right lines will be equal to their corresponding originals (Prop. IIL); hence the projected figure of the curve will coincide with the original at the points in which the right lines intersect. PROPOSITION XI. TuEoREM. The projection of a circle, in a plane parallel to the plane of pro- Jection, shall also be a circle equal to the original, For any number of radii being drawn in the original circle, will be projected into equal right lines, meeting in the same common point (Prop. IIL.) ; hence the curve passing through the unconnected ends shall be a circle equal to the original. PROPOSITION XII. THEOREM. Every right line or figure in a plane perpendicular to the plane of projection, shall be projected upon a right line, For since the projecting rays are perpendicular to the plane of pro- jection, the projecting rays which pass from the line or figure will be in a plane perpendicular to the plane of projection, and, therefore, the projection of the line or figure shall fall upon the intersection of these two planes; hence the proposition is manifest. PROPOSITION XIII. THEOREM. A right line on the original plane parallel to the intersecting line, shall have its projection also parallel to the intersecting line. Let P Q R be the plane of projec- tion, and P Q AS the original plane, and let the original line 4 B be parallel to the intersecting line P Q, and let a b be the projection of A B, then a b shall also be parallel to P Q. For draw the projecting rays Aa, Bb P meeting the plane P @ R in @ and b, 4 and join @ b. Draw A ¢, B d perpendi- En cular to P Q and join ae, bd. Then the planes 4 ¢ a, Bd b will not only be parallel to each other, but at the same time perpendicular 6 ELEMENTARY PRINCIPLES. to PQ; and because 4 B is parallel to P @ the plane 4a b B will be parallel to P @; therefore the intersection a b of the plane A a b B with the plane P @Q R shall be parallel to P Q. PROPOSITION XIV. THEOREM. A right line parallel to the intersecting line shall have its projection equal in length to the original, For let the original line 4 B (see the preceding figure) in the plane P Q S be parallel to the intersecting line P @, and let @ b in the plane P @ R be the projection of 4 B. Then, because 4 B is parallel to P Q, the projection a b is pa- rallel to P @ (Prop. XIIL); but right lines parallel to the same right line are parallel to one another (Eu. b. i., p. 30), hence a b is parallel to A B; moreover, since the projecting rays 4 a, B b are parallel, the figure A a b Bis a parallelogram; hence a b is equal to 4 B (Eu. b. i. p. 34); therefore, the length of the projection is equal to that of the original. Cor. Hence a right line in the original plane, parallel to the in- intersecting line, shall have its projection parallel to the intersecting line and equal to the original line. PROPOSITION XV. THEOREM. Every right line in the original plane, not parallel to the intersecting line and its projection, shall meet each other in the intersecting line, For the right line and its projection are in the projecting plane, therefore the line and its projection will meet each other in the pro- jecting plane. Now the right line is in the original plane, and its projection is in the plane of projection ; but there is no point common to the original plane, the plane of projection and the projecting plane, except the point in which the original line meets the intersecting line ; hence the proposition is manifest. PROPOSITION XVI. THEOREM. A right line in the original plane, meeting the intersecting line per- pendicularly, shall have its projection also upon a right line, meeting the intersecting line perpendicularly in the same point, For an intermediate plane, perpendicular to the intersecting line, meets both the dihedral planes, in right lines, perpendicular to the intersecting line ; hence the right line in which the intermediate plane meets the plane of projection is the projection of the right line in which the intermediate plane meets the original plane. ELEMENTARY PRINCIPLES. 7 PROPOSITION XVII. THEOREM. A right line in the original plane, meeting the intersecting line per- pendicularly, shall be to its projection as radius to the cosine of the di- hedral angle of the planes, For the angle made by the line and its projection is the dihedral angle of the planes, and the length of the line is the hypothenuse and its projection the base of a right triangle; hence, by trigonometry, the original line shall be to its projection as radius to the cosine of the dihedral angle of the planes. PROPOSITION XVIII. THEOREM. A right line perpendicular to the original plane shall be projected into a right line perpendicular to the intersecting line, For an intermediate plane passing through the right line, perpen- dicular to the original plane and to the intersecting line, is also perpendicular to the plane of projection ; hence the projection of the line will fall upon the right line in which the intermediate plane meets the plane of projection. PROPOSITION XIX. THEOREM. A right line in the original plane meeting the intersecting line per- pendicularly, shall have its projection less than that of a right line in any other position, For the cosines of angles to the same radius are less as the angles are greater, and if the line which meets the intersecting line perpen- dicularly be radius, the cosine of the angle will be equal to the length of the projection of the line, but the angle which this line makes with its projection is greater than the angle made by any other line in the same plane with its projection ; hence a right line in the original plane meeting the intersecting line perpendicularly shall have its projection less than that of a right line in any other position. PROPOSITION XX. THEOREM. A right line in the original plane, parallel to the intersecting line, shall have its projection greater than that of a right line in any other position, For the projection of a right line making an angle with the plane of projection is less than the original. Now, every line in a plane, in- clined to the plane of projection, will form an angle with its projection, except it is parallel to the intersecting line ; therefore, the projection of every right line, which is not parallel to the intersecting line, is less than the original; but the projection of a right line, parallel to the intersecting line, is equal to the original (Prop. XIV.) ; hence, a right 8 ELEMENTARY PRINCIPLES. line in the original plane, parallel to the intersecting line, has its pro- jection greater than that of a right line in any other position. PROPOSITION XXI. THEOREM. The projection of a circle shall be an ellipse, of which the greater axis shall be parallel, and the less perpendicular to the intersecting line ; and the greater axis shall be to the less as radius is to the cosine of the dihe- dral angle of the planes, For since the diameter of the circle is bisected by its centre, the projection of every diameter shall also be bisected by the projection of the centre (Prop. VI., Cor.) Let one of the diameters be parallel to the intersecting line, and its projection shall also be parallel to the inter- secting line (Prop. XIIL); and let another diameter be perpendicular to the intersecting line and its projection shall also be perpendicular to the intersecting line (Prop. XVL); the projections of these diameters will, therefore, bisect each other at right angles; but the projection of the diameter, which is parallel to the intersecting line, is equal to the diameter of the circle (Prop. XIV.), and the diameter of the circle, perpendicular to the intersecting line, is, to its projection, as radius is to the cosine of the dihedral angle of the planes (Prop. XVII.) ; there- fore the projection of the diameter, which is parallel, shall be to the projection of the diameter, which is perpendicular to the intersecting line, as radius is to the cosine of the dihedral angle of the planes. Moreover, the projecting rays passing from the original circle to the plane of projection form the surface of a cylinder; therefore the section of the cylinder intercepted, or cut by the plane of projection, is an ellipse ; and because the projection of the diameter, which is pa- rallel to the intersecting line, is greater than that of a right line in any other position (Prop. XX.) ; and because the projection of the diame- ter, which is perpendicular to the intersecting line, is less than that of a right line in any other position (Prop. XIX.) ; the projection of the two diameters of the circle, of which one is parallel, and the other per- pendicular to the intersecting line, shall be the axes of the ellipse which is the projection of the circle. The practice of projection depends upon the situation of the object to be represented with respect to the plane of projection, which may be either of the two dihedral planes; the one being naturally placed before the other or above the other. If the object be before the plane of projection, the lower must be considered as the plane of projection, as in the construction of plans and elevation; but if the object be upon a horizontal plane, it will be necessary, or at least convenient, to sup- pose the plane of projection placed before the object, making any given dihedral angle, and any given line of intersection with the horizontal plane. GEOMETRICAL SOLIDS. PART 1. OF THE OBJECTS OF PROJECTION. The objects to be projected being points, lines, plane figures, or so- lids ; their construction requires to be understood before they can be projected. Definitions, and some of the most obvious relations of lines and plane figures, are supposed to be known to the reader ; but those of solids, and of certain lines in the surfaces of solids, which are not so generally understood, are here explained, so that a complete idea of the body to be projected can be immediately obtained without loss of time or having recourse to any other publications. CHAPTER I.—.GEOMETRICAL SOLIDS. DEFINITIONS. 1. A polyhedron is a solid figure, comprised by any number of plane figures, which are called its faces. 2. A polyhedron, comprised by four faces, which is the least num- ber possible, is called a tetrahedron ; by six, a hexahedron ; by eight, an octahedron ; by twelve, a dodecahedron ; by twenty, an icosahe- dron ; and so on. 3. The intersections of the faces of a polyhedron are called edges, or arrises. 4. A polyhedron is said to be regular when its faces are similar and regular polygons. 5. A parallelopiped is a solid figure, having six faces, of which every opposite two are parallel. 6. A rectangular parallelopiped is that which has one of its solid angles contained by three right angles. 7. A cube is a rectangular parallelopiped which has three edges ierinaied in the vertex of one of the solid angles equal to one ano- ther. 8. A prism is a solid figure, comprised by any number of faces, all of which, except two, are parallelograms, having each two opposite edges in common with the two remaining faces. 10 OBJECTS OF PROJECTION. 9. The parallel edges which join the two remaining faces are called principal edges. 10. The faces of the prism, which form the parallelograms, are cal- led the sides. Il. The two opposite faces, formed by the opposite edges of the parallelograms, are called ends or bases. 12. The altitude of a prism is the perpendicular distance between the two ends. 13. A prism is denominated by the number of edges, comprising the polygonal ends; thus when the ends are triangles, or quadrangles, or pentagons, &c., the solid is called a triangular, or quadrangular, or pentagonal, &c. prism. 14. A prism is said to be right or oblique, according as the prin- cipal edges are perpendicular to the ends, or inclined to them. 15. A regular prism is aright prism, which has for its ends two re- gular polygons. 16. The axis of a right prism is the right line which joins the cen- tres of the polygonal ends. 17. A pyramid is a solid figure, composed by any number of faces, at least four, all of which, except one, are triangles, having one com- mon vertex, and each edge opposite the vertex in common with an edge of the remaining face. 18. The edges which end in the common vertex are called prin- cipal edges. 19. The triangular faces of a pyramid are called sides. 20. The remaining face in which the principal edges meet its ver- tices is called the end or base. 21. The point in which the sides of a pyramid terminate is called the apex. 22. The altitude of a pyramid is the perpendicular distance of the vertex from the base or the plane of the base produced. 02 23. A pyramid is denominated by the number of edges comprising the polygonal end or base, and is therefore called triangular, or qua- drangular, or pentagonal, &c., according as the base is a triangle, or a quadrangle, or a pentagon, &c. 24. A regular pyramid is that which has for its base a regular polygon, and the right line which is drawn from the apex to the centre of the base perpendicular to the base. : 25. If a pyramid be divided into two parts by a plane parallel to its base, the part next the base is called a frustum of a pyramid, or some- times a truncated pyramid. 26. A cylinder is a solid figure, the surface of which is partly curved and partly plane; the plane portions being two equal and parallel cir- cles, and the curved portion such that a right line drawn through any point in the circumference of either circle, parallel to the right line joining the centres, lies wholly in the surface. The curved surface of a cylinder is called the convex surface, and the circles are called the ends or bases, and the right lines which join the centres of the ends are called the axis of the cylinder. GEOMETRICAL SOLIDS. ¥1 27. A eylinder is said to be right or oblique, according as the axis is perpendicular or inclined to the ends. 28. A cone is a solid figure, the surface of which is partly plane and partly curved; the plane portion being a circle and the curved portion such that a right line drawn through any point in the circumference of the circle and through a certain point, not in the plane of the circle, lies wholly in the surface. The curved surface of a cone is called the convex surface; the point through which the right line always passes is called the summit ; the circle is called the end or base, and the right line which is drawn from the summit to the centre of the base is called the axis. 29. A cone is said to be right or oblique according as the axis is perpendicular or inclined to the base. The slant side of a right cone is a straight line, which is drawn from the vertex to any point in the circumference of the base. 30. Ifa cone be divided into two parts by a plane parallel to the base, the part next the base is called a frustum of a cone, or sometimes a truncated cone. 31. A sphere is a solid figure, every point in the surface of which is at the same distance from a certain point within the figure which is called the centre. The distance from the centre to the surface is call- ed the radius, and a right line passing through the centre, terminated on both ends by the surface is called the diameter. A sphere may be conceived to be generated by the revolution of a semicircle about its diameter. Hence there can be no point on the surface with which some point of the semicircle will not have coincided. Hence all the sections of a sphere are circles. The right cylinder, right cone, and sphere, are sometimes styled, by way of pre-eminence, the three round bodies. They are also term- ed solids of revolution, because each of them may be conceived to be generated by the revolution of a plane figure about a fixed right line taken in its plane. Thus a sphere may be conceived to be generated by a semicircle revolving upon its diameter, a right cylinder by a rect- angle or oblong revolving upon one of its sides, a right cone by a right-angled triangle revolving upon one of the sides which contain the right angle. 32. A segment of a sphere is any portion of the sphere which is cut off by a plane, and the circle in which the plane cuts the sphere is called the base of the segment. When the plane passes through the centre of the sphere, the two segments into which it is divided are equal to one another, and are therefore each of them called a hemisphere. The convex surface of a segment of a sphere is called a zone. 33. A double based spherical segment is a portion of the sphere in- tercepted by two parallel planes, and the circles in which these planes cut the sphere are called the bases of the segment. 34. A sector of a sphere is the solid figure comprised by a portion of the convex surface of the sphere, and the convex surface of a right 12 OBJECTS OF PROJECTION. cone, which has the same base with the spherical segment, and for its summit the centre of the sphere. 85. A spherical orb is a portion of the sphere comprised between its surface and the surface of a less sphere which has the same centre. 36. A spherical wedge, or ungula, is a portion of a sphere inter- cepted between two planes, each of which passes through the centre of the sphere. The convex surface of an ungula is called a lune. PROPERTIES OF THE SECTION OF A SPHERE. 1. Every plane section of a sphere is a circle of which the centre is either the centre of the sphere or the foot of a perpendicular drawn to the plane from the centre of the sphere. 2. Any two great circles of the sphere bisect each other. 8. If two spherical surfaces intersect each other, their common sec- tion is a circle. CHAPTER 11.—OF THE CIRCLES OF THE SPHERE. DEFINITIONS. 1. If a sphere be cut by a plane which passes through the centre, the section is called a great circle of the sphere. 2. If a sphere be cut by a plane which does not pass through the centre, the section is called a small circle of the sphere; the radius of such a section being less than that of the sphere. 3. The axis of any circle of the sphere is that diameter of the sphere which is perpendicular to the plane of the circle; and the ex- tremities of the axis are called the poles of the circle. 4. Parallel circles of the sphere are those that have their planes parallel. 5. Meridional circle, or meridian of the sphere, is a circle passing through the axis of another circle. : j 6. Any portion of the circumference of a great circle is called a spherical arc. 7. The polar distances of any circle of the sphere are the spherical arcs of a meridian between each pole, and the point in which the me- ridian intersects that circle. By the polar distance (singly) is meant the distance between the circle and the nearest pole. REGULAR SOLIDS. 13 CHAPTER 11I—OF THE REGULAR SOLIDS. DEFINITIONS. 1. A regular polyhedron, or regular solid, is a solid figure, which has all its faces regular polygons, equal to one another; the faces being either equilateral triangles, squares, or regular pentagons. The number of regular solids is five. 2. A regular tetrahedron is a solid figure, comprised by four faces, which are equilateral triangles. 8. A regular hexahedron is a solid figure, comprised by six faces, which are squares. 4. A regular octahedron is a solid figure, comprised by eight faces, which are equilateral triangles. 5. A regular dodecahedron is a solid figure, comprised by twelve faces, which are regular pentagons. 6. A regular icosahedron is a solid figure, comprised by twenty faces, which are equilateral triangles. The solid angles of a regular solid are formed by plane angles, equal each to an angle made by two sides of a polygonal face of the solid ; hence in the tetrahedron, the octahedron, and the icosahedron, each of the plane angles which form a solid angle is 60°; in the hexahe- dron, each of the plane angles which comprise a solid angle is 90°, or a right angle, and in the dodecahedron, each of the plane angles which comprise the solid angle is 108°, or three-fifths of two right angles.— In the tetrahedron, hexahedron, and dodecahedron, the solid angles are comprised each by three plane angles ; in the octahedron the so- lid angles are each comprised by four plane angles, and in the icosa- hedron, the solid angles are each comprised by five plane angles. Each of the regular solids, except the tetrahedron, has for every face an opposite and parallel face. 14 PROJECTION UPON THE LOWER PLANE. [PLATE 1. PART 111. PROJECTION UPON THE LOWER PLANE. CHAPTER I.—ELEMENTS OF PRACTICE, CONSISTING OF THE PROJECTION OF POINTS, LINES, AND PLANE FIGURES. PROPOSITION XXII. Prosrem I. Given the intersecting line and dihedral angle of the planes, to find the projection of a point. Let A (fig. 1.) be the given point. Draw 4 S perpendicular to U V, the intersecting line cutting it in 8, and make the angle « 8 § equal to the dihedral angle. From g, with the distance g 4, cut g « in E, and draw FE a perpendicular to 2 .S, meeting 2 Sin a, and a is the projection of the point A. For, suppose the plane containing the angle » 8 S, connected with the plane of projection only in the right line #8 iS, as a hinge to be revolved until it become perpendicular to the plane of projection, then the point a will be the projection of the point FE, and the right line « @ will be perpendicular to U/ V. Moreover, suppose the original plane, containing the point 4, connected with the plane of projection only in the right line U V, to be revolved upon U V, and because B « and B 4 are both perpendicular to U V, the right line 8 4 may be made to coincide with 8 x; let, therefore, 8 4 and 8 « coincide ; then because 8 4 and B KE are equal, the point 4 will fall upon Z'; hence, the point @ being the projection of Z| will be the projection of the point A. SCHOLIUM. In finding the projection of a point it is not necessary that the side 8S (fig. 2) of the angle = ¢ S be in the perpendicular, connecting the original point and its projection. Thus let B be the original point ; at any convenient point 8 in U V draw g .§ perpendicular to U V, and make the angle «~ 8 S equal to the dihedral angle ; from the same point @ draw 8 4 also perpendicular to & V, and draw B 4 parallel and B &. perpendicular to I V. Find « the projection of the Prue 1. Pageds. £¥ 4s hee Bo i% PLATE L] ELEMENTS OF PRACTICE. 15 point A (Prob. L.); draw a b parallel to UU V, and & is evidently the projection of the point B. PROPOSITION XXIII. Prosrewm II. Given the intersecting line and the dihedral angle, to find the projec- tion of a given right line. Let the right line B C (fig. 3.) meet the intersecting line U V in C'; find b the projection of the point B (Prob. I.) ; joinb C, and b C is the projection of B C. Or if the given right line do not join the intersecting line, and if it be found inconvenient to prolong it, let BB D be the position of the given line, find the projections b and d of the two extremities B D of the line (Prob. I.) ; join 6 d, and bd is the projection of the right line B D. PROPOSITION XXIV. ProprLewMm III. Given the angles which an indefinite right line and its projection make with the intersecting line and the definite length of a part of the line, to find the length of its projection. Let D F (fig. 4) be the indefinite line, and FE F its indefinite pro- jection, making the angles D F' V, EF V with the intersecting line, it is required to find the projection of the definite part 7 A of the in- definite line F' D. Draw 4 a perpendicular to U V, meeting £ F in a, and a is the projection of the point 4 upon the supposition that the planes are raised to the given dihedral angle. ScHOLIUM. In the construction of this problem, it is evident that the angle made by the original line, and its projection must be greater than the angle made by its projection and the intersecting line. If it were necessary to find the dihedral angle of the planes, con- struct a right angled triangle, of which the hypothenuse is equal to the distance 4 C of the point, and one of the sides containing the right angle equal to the distance a C of its projection ; then the angle contained by the hypothenuse, and the distance of the projection of the point A is the dihedral angle of the planes. PROPOSITION XXV. ProsrLem IV. Given the dihedral angle, and the angle which an original right line makes with the intersecting line, to find the angle which the pro- Jecting line makes with the intersecting line, and the inclination of the line to the plane of projection. Let UB A (figs. 5 or 6) be the angle which the right line makes with the intersecting line. Find a the projection of the point 4 (Prob. 1.) and join @ B. Draw a F perpendicular to « B; make a & equal to 16 PROJECTION UPON THE LOWER PLANE. [PLATE IL a E and join FF Bj; then the angle which the projection of the line makes with the intersecting line is equal to U B a or 8 B a, and the in finion of the line to the plane of projection is equal to the angle a ; . For if the planes of the three triangles 48 B, E a 3, B a F, be only connected with the plane of projection by the lines of 8 B, a 8, a B, as hinges and revolved until the points 4, E, F, meet each other; the proposition will be evident, for 4 B will fall over its projection a b. PROPOSITION XXVI1. ProBrLEM V. Given the intersecting line and the distance of the projection of one of the vertices of a given parallelogram, to find the projection of the parallelogram. Prolong the sides B 4, B C (fig. 1), of the exterior angle 4 B C, of the figure to meet the intersecting line in U and V. Draw B 8 perpendicular to U ¥, meeting U V in 8 and in B 3, make $b equal to the distance of the projection of the vortex B. Draw the right lines b U, 6 V, and find the projections a, c, of the points 4, C (Prob. III). Draw a d parallel to b ¢, and c d parallel to 4 a, and the parallelogram « b ¢ d is the projection of the parallelogram 4 B C D. ScHOLIUM. In this problem and in the next it is evident that since there are given the intersecting line, the distance of the projection of one of the vertices of a plane figure, that is, a point and its projection being given, the dehedral angle made by the planes may be found, if re- quired (see Scholium Prob. IIL); and as one or the other must be given or assumed, it is more convenient to give or assume the projec- tion of one of the vertices of an angle of the figure than to give or assume the dehedral angle. It is obvious from Prop. X VL. that every point and its projection must be in a right line perpendicular to the intersecting line ; and, since the distance of a point is to the distance of its projection as radius to the cosine of inclination, the distance of the projection of the point must always be less than the distance of the point itself. PROPOSITION XXVIII. ProsrLEm VL Given the intersecting line and the distance of the projection of one of the vertices of a given regular pentagon, to Jind the projection of the pentagon. Prolong the sides B 4, B C (fig. 2), of the exterior angle 4 R C, of the figure to meet the intersecting line in Uand V. Draw Bg perpendicular to U ¥, meeting U Vin 3, and in B 3 make $3 b equal to the distance of the projection of the vortex B. Draw the right lines b U, b V, and find the projections a, c, of the points 4, C (Prob. II1.). Draw the diagonals B E, B D; prolong B E, B D tomeet U Vin Rand N, join 4 R,b N. Draw cd parallel to 4 R, meeting 6 N in FIAT 9. oT 2 FLAT 2. o PLATE 2.] ELEMENTS OF PRACTICE. 7 d, and draw a ¢ parallel to & N, meeting 6 E in e. Join ed and the polygon a b ¢ d ¢ is the projection required. For, in the original figure, the diagonal B D is parallel to 4 E, and the diagonal B E is parallel to C D, and, because the projections of pa- ir allel lines are parallel, ¢ d is parallel to 6 cor R, and a e is parallel to orb N. : : PROPOSITION XXVIII. Prosrem VIL Given the intersecting line and the dihedral angle, to find the pro- pects of a given circle. » Having constructed the angle » 8 S (fig. 3 or 4), as in Pro- blem I, in » 8 make 8 F equal to the distance of the centre of the circle, and make 77 G' and F H equal each to the radius. Draw Gg, FHL, perpendicular to 8 § meeting it in g, f, 4, and draw g b, . fe,’ d, parallel to U V. Inf c¢ make f e equal to the distance of the centre of the, circle from 8S, and through e draw 4 d perpendi- Scularto U V.,, Thon a cand b d as axis describe the ellipse a b ¢ d,* a the ellipse. a rt is the projection of the circle required. prey br 4 oF or : JICHAPTER II.—THE CONSTRUCTION OF PLANS p . AND a OF GEOMETRICAL SOLIDS. » vo “As all figures in planes parallel to the plane of projection are pro- jected into equaliand similar figures, and as all lines and surfaces per- a i » JA very near rate curve of an ellipse of which the two axes are given, in position andmapnip tude is the following. Let d fand g e (fig. 5) be the two axes bisecting each other at right angles in % aw (in the subsidia y diagram fig. 54 e right lines j o, j n making any nient angle 0,j on each othe ye ‘made jm, j k, respectively equal to the a -axes y d, y g, OF TYLY ¢ (fig: rom J with, the distances jm, Jk,cutjn nd 7, pad id I. Draw kp, no 0, parallel t to I m meeling j n, jo, at p, o. # Es ge ( out wags 7.an 2 “make g r and ¢ equal each to j o (fig. 5 G), and in d f (fi e d w,, , each equal to j p (fig. 5 G.). fs From 7 ies 2 5) With’ the radius 7 & ihe the arc a b, from the point . with the radius Ze, describe the arc. ch, from the point w, with the ra- dius w J, describe the arc i g, and from the point : 2, W. ih the radius z J, describe the arc $2, ‘Make ‘each of these a on each side of each extremity of each axis, and each of the Sa es will nea y coincide with the elliptic curve. ~The part in the middle of eacl quarter directed by a correct € , being drawn by hand, will co plete the figure, which will be s ficiently pee to the curve of an ellipse that the one may be substituted for. the other. I & 0 yd If the curve is larg rge, four points may be found by finding. the two foci, « and © vu, drawing a double ordinate 8 v9 through eact i S 5 and making each or- dinate u @,u 8, y, v 3, equal to J » P(g p? G ans we have a point in each of the vacant Sheen. nt eg v w oe 18 PROJECTION UPON THE LOWER PLANE. CpLATE 3. pendicular to the plane of projection are projected into points and lines, the projection of aright prism, or of a right cylinder, upon a plane parallel to one of its ends will be an equal and similar figure to that end ; hence the projection of the cylinder will be a circle equal to that of the end of the cylinder. Likewise the outline of the pro: jection of a regular pyramid or of a right cone, having the end i plane parallel to the plane of projection Will be a figure equal and | lar to that end; hence the projection of the cone will be a circle - to the circular base, and the projection of the axis will be the centre of that circle. In the pyramid, as well as in the cone, as the axis is projected into a point, the projection of the principal edges owing right line drawn from that point to each of the vertices of the o of" the projection. Moreover, the projection of such a pric yl - der, pyramid or cone, having the principal edges of the prism, or the axis of the cylinder, pyramid, or cone, parallel to the plane of projec: tion, will be a figure equal and similar to a section through par allel to the plane of projection. The projections of geometrical right solids sfbosited are there- 1 fore too easy to require separate propositions : and from the little Siow 8 § he axis of oblique solids it will not be necessary in this Chapte Chpter to to a3 their projections may be found. w eg “ - "> DEFINITIONS. 1. The intersection of the vertical ft Re planes of projec- tion, is called the ground ny is erally denoted by WZ. 2. The surface of the paper upon the ag side of the ground line is called the horizontal plane of projection, and the projection,it- self, which is made upon the horizontal plane, is called the plan of the object. -” a 3. The surface of the paper upon ely side of the gr ine is called the vertical plane of 2 opectio and fame itself wh ich is made upon this vertical plane;"is ca Fo the elevation of the ob- ject. “ ! we i ¢ + PROPOSITION a ProBLEN VIL, oh ® Given the projection of a right di / in relation to one cite nes inclination in relations. the Let a b (fig 1)sbe ns nrojecti 2 ground line ¥ Draw a 4 p the angle a 7 ual t glina tio of t projection. Draw ET JP rpendicula in e, make e ¢ equal to @ fi i oin : o o> Ci 3 bats make ¢ C equal to. | joind th be is th e pr gjection the angle ¢ 6.C sion “of the 5 raion to the vertieal © plane of proj Pk : Bs t : on = *3 oe Jo wd'its projec Rigends eS. of the line pieting the the line tos the plane o ie 27 Pn o fLigline % & 1 # endic ular to , and wakeg a 8 5 4 » Lage 19 Page 19 BLATE 3-3 : ELEMENTS OF PRACTICE. 19 For suppose the plane e ¢ b to be raised upon the ground line ¥ Z, as a hinge perpendicular to the plane a e b, the plane b ¢ C raised upon b ¢ perpendicular to the plane b e ¢: and the plane 4 a b raised upon a b perpendicular to the plane @ e b; now the right line A a being always perpendicular to @ b, will be perpendicular to the plane ea b ; hence the point a will be the projection of the point 4 upon the horizontal plane e @ 6 : similarly it may be shown that ¢ will be the projection of C upon the vertical plane of projection bec; hence a b is the projection of A b upon the horizontal plane, and ¢ b is the projectiomiof C b upon the vertical plane; but 4 band C b coinciding make but one right line ; and since the planes « e b, and ¢ e b are perpendicular to each other, a 4, ¢ b, are the two projections of the line, the one on one of the planes and the other on the other plane, a b being the plan and ¢ b the elevation of the line. OF PROPOSITION XXX. ProsrLem IX. Given the projection of a right line, the distances between the extre- mities of the line and the extremities of its projection in relation to one of the planes, to find the similar relations in respect of the other plane. Let a b (fig. 2) be the projection of the line. Draw a 4,6 B, per- pendicular to @ 4; make a A equal to the distance between the end A of the line and its projection @ and b B equal to the distance be- - tween the other end'BB of the ling and its projection b. Join 4 B and A B is the length of the line. Draw a ¢ and b 4 intersecting the ground line Y Z perpendicularly in ¢ and f. Make e ¢ equal to 4 a, Jd equal to b B, and join ¢d. Then ¢ d is the projection of the line en the other plane, Draw d D and ¢ C perpendicularto dc. Make d D equal to f b, and ¢ C equal to ea. Join C D, and C D is the length of the line upon the other plane or elevation. For suppose the plane e¢ ed f to be raised upon the ground line Y Z as a hinge perpendicular to the plane 4 ef B, and the plane A a b B raised upon a b perpendicular to the plane @ e £5, and the plane cd D C raised upon cd perpendicular to the plane e cd f; then the figure comprised by the four lines e a, ec, ¢ C, a 4, will be a rectangle, e ¢ and a A being of equal length will be opposite sides, Cc and e a being equal will also be opposite sides. Moreover the figure formed by the four lines d D, d f, fb, b B, will also be a rectangle ; hence the points 4 and C, as also the points B and D, will coincide, and the right lines 4 B and C D will form one and the same right line of which its projection on the horizontal plane is a b, and on the vertical plane is ¢ d. 20 PROJECTION UPON THE LOWER PLANE. [PLATE 30 ® PROPOSITION XXXII. ProsLEM X. Given the projection of the axis of a cone, the distance between the extremities of this axis and the extremities of its projection, in relation to one of the planes of projection, to find the projection of the cone upon each plane. hy Let a b (fig. 3) be the projection of the axis of the cone upon the horizontal plane @ e fb ; here are now given the projection of a right line, the distances between the extremities of the line a the extremi- ties of its projection in relation to one of the planes, to find the similar requisites in relation to the other; therefore, as in the preceding proposition, construct the four trapezoids A a b B,ae f beef d, Cc d D, which, being done, 4 B and C D are each equal to the length of the axis of the cone, and @ &'is the rojection of the axis on one of the planes and ¢ d the projection of the same axis on the other plane of projection. Through 4 draw G' H perpendicular. to A B; make A G, A H, each equal to the radius of the circular end, "and join G B, H B: similarly through C draw K L perpendicular to C D; make C K, C L, each equal to the radius of the circular end, and join K D, L D. Then the isosceles triangles B G H, D K L, ‘are sections of the cone through its axis. ~ Prolong 4 a to g,and let % be in@,b ; draw G g, H h, perpendicular tog b; prolong A a toi: in Ai make a i, a f each equal to the ra- dius of the circular base ; upon the two axes i f;g &, describe the el- lipse i ¢ f kh, and i g f A will be the projection of the circular base. Draw two right lines from & to touch the curve of the ellipse, and b i g f will be the projection of the cone in the plane a ¢ fb. Similarly prolong d ¢ to k, and let { be in ¢ d; draw Kk, L I, per- pendicular to % d; prolong Cec to n; in C » make ¢ m, ¢ n, each equal to the radius of the circular base; upon the two axes m =, k I, describe the ellipse m % n I, and m % n I is the projection of the cir- = cular base. Draw two right lines from d to touch the curve of the ellipse, and d m %k n is the projection of the cone on the plane cedf CHAPTER I1I1.—.THE CONSTRUCTION OF THE PLANS AND ELEVATIONS OF THE FIVE REGU. LAR SOLIDS. The solid to be projected is supposed to have one of its faces in the horizontal plane of projection, and thus the face and its projection “ ~ PLATE 4.] ELEMENTS OF PRACTICE. 21 will have the same common figure. The solid is now supposed to be cut by a plane, perpendicular to the plane of the face upon which it reposes, bisecting one of the angles of this face; the plane of section will thus divide the solid into two symmetrical halves, and in the dode- cahedron and icosahedron will pass through the edge terminating in the vertex of the angle bisected, and this edge will form, with either of the two edges containing the bisected angle, two sides of a regular pentagon, of which each side will be equal to any edge of the solid. Hence the projection of the edge of the solid through which the plane of section passes will fall upon the bisecting line prolonged ; therefore if either of the edges containing the bisected angle be made the intersecting line, the length of the projection of this edge may be easily found (Prob. III). In the dodecahedron, the pentagon of which the two edges alluded to are sides, is one of the faces of the solid; in the icosahedron, this pentagon is not a face of the solid but the base of a pyramid which may easily be observed on examina- tion. gular solids parallel to the bisecting line* of the angle, and the vertical plane of projection will be parallel to the plane of section.— As the solid is divided into two symmetrical parts, the plane of section will either bisect the faces, or pass through the edges of the faces which meet in pairs, with which it comes in contact. The outline of the elevation will easily be found by describing a figure equal and si- milar to the section of the solid. & PROPOSITION XXXII. ProsrLem XI. To draw the plan and elevation of the tetrahedron. With the given side of one of the triangular faces of the solid de- scribe the equilateral triangle 4 B C (fig. 1), for the horizontal face upon which the solid repose: draw the right lines, 4 S, B §, CS, respectively perpendicular to the opposite sides B C, C 4, A B, and A B C 8, No. I, is the plan. «. Now A D being the sectional line, the section will be formed by the following lines, viz., the perpendicular represented by A D, the perpendicular represented by 0 S, and the edge represented by .S A. Hence the section and consequently the outline of the elevation will be an isosceles triangle having two sides equal each to the perpendicu- lar, and the remaining side equal to the length of the side of one of the triangular faces; hence, At any convenient distance parallel to 4 D, draw the ground line Y Z, and perpendicular to ¥ Z draw 4 a, B b, meeting it in a, b ; also perpendicular to ¥ Z draw § s. From the point a, with a distance equal to the length of one of the edges, cut S's in 5, and join sa, sb; then a bs, No. 2, is the elevation. * This is called the sectional line, or line of section. “ws The plan Being finished, place the ground line ¥ Z for all the re-: » » 22 PROJECTION UPON THE LOWER PLANE, [PLATE 4. PROPOSITION XXXIII. ProsrLEm XII. To draw the plan and elevation of the hexahedron. The plan and elevation of the hexahedron will be evidently squares, of which the sides are each equal to an edge of the solid. PROPOSITION XXXIV. ProerLEm XIII & To draw the plan and elevation of the octahedron. With the given side of one of the triangular faces of the solid, de- scribe the equilateral triangle 4 B C (fig. 2) for the horizontal face upon which the solid reposes. From the centre O describe the cir- cle A B C ; bisect the arcs 4 B, B C, C 4, at the points D, EF, F ; jomAD,DB,BEEC,CF,F A, and the regular hexagon is the outline of the plan. Join D E, FE F, FF D, and the entire plan, No. 1, is complete. a * . © Now D C, being the sectional line, the perpendicular represented by G C, the perpendicular represented by C H, the perpendicular represented by H D, and the perpendicular represented by G D, will give the outline of the elevation, which will, therefore, be an equilateral parallelogram, having each side equal to the perpendicular of one of the triangular faces: hence’ At any convenient distance parallel to D C, draw the ground line Y Z, and; perpendicular to ¥V Z draw B b, C ¢, meeting it in b, ¢; also perpendicular to ¥ Z draw D d, Ee. From the point b, with a distance equal to the length of the perpendicular of one of the trian- gular faces cut D d ind, and from the point ¢ with the same distance, cut Eeine. Join bd, de, ec, and b ¢, and No. 2 is the elevation. PROPOSITION XXXV. ProsLEm XIV. To draw the plan and elevation of the dodecahedron. With the given side of one of the pentagonal faces of the solid de- scribe the regular pentagon A B C D E (fig. 3), for the horizontal face upon which the solid reposes, and from the centre O, through the points 4, B, C, D, E, draw OF, OG, OH, OI, OJ. Perpen- dicular to C D draw B H; then C H is the projection of one of the edges of the solid on a plane perpendicular to the face 4 B CD E, bisecting the dehedral angle of two adjoining faces (Prop. IIL.). From the centre O, with the radius O H, describe the circle F G H1J ; bisect the arcs F G, G H, HI, IJ, J F, and complete the regular decagon FKGLHMINJP. Draw PO,KO, LO, MO, N 0, to meet the circumference of the circle containing the pentagon A B CD E in the points Q, R, S, T), U, and joining the points of division, the pentagon @ R S T' U is the projection of the upper face which completes the entire plan, No. I. Now F M being the sectional line, the perpendicular of a pentagon represented by A 5, the perpendicular represented by 5 HM, the side Rw PLATE £. ww... . u @ we PLATE4.] ® . & ELEMENTS OF PRACTICE. 23 represented oy T,, the perpendicular represented by 7" 6, the per- pendicular represented by 6 #7, and the side represented by F' A, will form the outline of the elevation, which will, therefore, be a hexagon, having two opposite sides equal to a side, and the remaining four sides equal each to the perpendicular of one of the pentagonal faces. At any'convenient distance parallel to 7’ M, draw the ground line Y Z, and perpendicular to Y Z draw A a, Bb, C c, meeting it in a, b, e. * Also perpendicular to ¥ Z draw Ff, Rr, Tt, Mm. From the Point, with the distance AS5o0r6 T, cut Mm in m; from the point'm, with a distance equal to the side of the pentagonal faces, cut T'tin ¢; from the point @, with the same distance, cut F fin f; and from. the point f, with a distance again equal to the perpendicular of «the pentagonal face, cut Rr in». Join af, fr, rt, tm, m c. Paral- lel to Y.-Z dvaw fh, mk meeting ¢ m in 4 and fr in %, and perpen- ’ “dicular to ¥Y Zdraw Gg, LI, S s, meeting fhing, mkin land » ¢ ins. Joinbdg,gk, gl, ls, Lh, and the figure, No. 2, is the elevation. oi Ea "PROPOSITION XXXVI. ProsrLEm XV. i vw To find the, plan and elevation of the Icosahedron. A «With the given sideiof one of the triangular faces describe the equi- "lateral triangle. 4 B C (fig. 4), for the horizontal face underneath and, from the centre O of the circle, containing the equilateral tri- “angle AB C, and, through the points 4, B. C, draw O D, O E, OF. Make an angle C B N equal to 108° and make B NV equal . ito one of the sides of the triangular faces. Perpendicular to B C draw NV E; then BE is the projection of one of the sides of the base of the pentagonal pyramid (Prop. III). From the centre O, wiwith the radius O E, describe the circle D H E I F G, and with the radius bisect the.arcs D FE, IL F, F D, in H, I, G, and complete the regular hexagon D H E I F G. Now D Lbeing the line of section, the perpendicular represented "by A.5, the perpendicular represented by 5 Z, the side represented ~ by I L, the perpendicular represented by Z 6, the perpendicular re- presented by 6 D, and the side represented by D A, will form the outline of the elevation which will, therefore, be a hexagon, having two sides equal each to the side of the triangular face, and the other four sides equal each to the perpendicular of the said triangular face. ‘At any convenient distance parallel to D I, the line of section, draw the ground line ¥ Z, and perpendicular to ¥ Z draw 4 a, B b * meeting it in ¢ and b. Also perpendicular to Y Z draw D d, Kk, Ll, Ii. From the point @, with a distance equal to the side of the . . triangular faces, cut D d in d; from the point d, with a distance equal to ‘the perpendicular, cut K % in %2 ; from b, with the same distance, cut Iiin ¢, and, from the point ¢, with a distance equal to the side of the triangular face, cut L/inl. Joinad, dk kl, 1i,ib. Drawde and ¢ A parallel to ¥ Z, and draw H % and FE e perpendicular to ¥V Z. Join ha, hd, hk, hehbek,el, ei, eb, and the figure, No. 2, is the elevation. on TR a ¢ $ . Re g $ 4 24 PROJECTION UPON THE LOWER PLANE. [PLATE 5. @ B CHAPTER IV..—CONSTRUCTION OF THE+«PLANS AND ELEVATIONS OF THE CIRCLES OF THE SPHERE. ¥ wid & PROPOSITION XXXVI Prosrem XVI. Given the radius of a sphere, the elevation of the poleand the distance of a circle from the pole to find the plan and elevatio that circle. From any convenient point C (fig. 1) with the given radius, de- scribe the circle D A P K B and let the circle D A P_K B be con- sidered as a section of the great circle ofa sphere passing through thes pole perpendicular to the horizontal plane of projection, and let the diameter D K be considered as a section made by the horizontal plane of projection. Make the arc K P equal to the eleyation of the pole and join C P. Then theiradius C P is half the axis. Make * the arc P A equal to the distance of the circle from the poles draw the chord A B perpendicular to C P, intersecting CP in FE, and 4 A B is the projection of the circle in a plane perpendicular to C P. ik Draw C c¢ perpendicular to D Kj from any con nient point ¢ ‘with the given radius describe the great circle 2.4 j %, and draw the . diameter d % perpendicular to C ¢. Draw A a, FE e, B.b parallel to © Ce, meeting d % in a, e, b, ands prolong E eto bh. In E h make ¢'g, e h equal each to 4 EF or E B; with the major axis g / and the minor axis a b describe the ellipse a g b hy which is the plan of the circle required. >a - fil Let A B intersect D K in F; draw Fj parallel to C ¢, meeting ~~ the circumference ¢ d j £ in ¢ and j; then @ j is the intersecting line . of the plane of the circle of which a g ¢ 6; kis the glen. A $ PROPOSITION XXXVIIL SprobLey XVIL" . ¢ Given the radius of the sphere and the elevation of the pole, to find the projection of a meridional circle, making a given angle with the vertical meridional circle. OP i ® From any convenient point. ¢ (fig. 2), with the radius of the sphere describe the great circle d l P k, and draw the diameter dk. ‘Let d k be the intersection of the vertical meridional circle, and d % is at _ the same time its projection. Make the arc # P equal to the eleva- tion of the pole ; join ¢ P, and draw P p perpendicular to d k, meet- ing d k in p. Prolong d & to R; draw P Q perpendicular to" ¢ P, meeting ¢ BR in Q, and through Q draw 7" S perpendicular to d R. Make Q R equal to Q P, and make the angle Q R equal to the dihedral angle of the planes of the two meridional circles.— Join ¢ iS, and let ¢ § intersect the circle d! P kin kh. Prolong 4 ¢ to meet the opposite part of the circumference in /. Draw p m per- pendicular to / 2 meeting 7 2 in m. In the auxiliary (fig. 2 4) draw PLATE 5.] ELEMENTS OF PRACTICE. 25 the two right lines 7 s, s # perpendicular to each other. Maker s,s ¢ respectively to p m, p P; join 7 ¢, and the angle s » ¢ is the dihedral angle of the planes of the meridional circle to be projected, and the vertical meridional circle given by its projectiond k. Prolong r s (fig. 2, A) tog, and 7 £ to wu. Make ru equal to the radius of the sphere, and draw u g parallel to ¢s. Through ¢ (fig. 2) draw a b perpendicular to 2 I, and make c a, ¢ b equal each to r ¢ (fig. 2 A). Upon 4 land a b as axes, describe the ellipse a? b A, and this ellipse is the projec- tion of the circle required. i ; The same words explain the projections of all the meridional circles in fig. 3, similar letters of reference being affixed to one of them. PROPOSITION XXXIX. Prosrem XVIII. To find the projection of a meridional circle, making a given angle, with a meridional circle in the plane of projection. Let A P B P’ (fig. 3, No. 1) he the given meridional circle, C its centre, and P P’ the axis. Draw the diameter A B perpendicular to P P’, and prolong P P’to3. From any convenient point 8 in C3 draw 8 y parallel to 4 B, and from 3, with the radius of the sphere, describe the arc o» 3. Let 4 8 3 be considered as a quarter of the section of the sphere through 4 B. Make the angle y 85 equal to the dihedral angle of the meridional planes, and let the point 5 be in the arc 4 9, or which is the same thing, make the arc 4 5 equal to the number of degrees in this angle. Parallel to P P’ draw 5 «, meeting A B in a, and in A B make C b equal C a. With the major axis P P’, and the minor axis a b describe the ellipse P a P’ b, which is the projection of the meridional circle required. Example of a complete projection of the sphere. Fig. 3 exhibits a plan and elevation of the parallel and meridional circles, dividing the surface into spherical squares, or quadrilateral paggions. In this example the elevation of the pole is 51° 30’, be- ing the latitude of London; the parallel circles are 10° distant from each other, and the dihedral angles of the meridional circles 15°, and, consequently, the meridians being 24 in number, are hour circles. The elevation, No. 1, is a projection of the sphere upon the plane of a meridional circle. Here the parallel circles being in planes perpen- dicular to P PF’, are perpendicular to the plane of projection, and are therefore projected into right lines, which are parallel chords to the circle D A P K B, perpendicular to, and bisected by, PP’. The parallel circles are drawn by Problem XVI, and the meridional cir- cles by Problem XVIII. The plan, No. 2, is a projection of the sphere upon the plane of the horizon. The parallel circles are projected by Problem XVI, as in fig. 3: No. 2, and the meridional circles by Problem XVII, as in g. 2. In the elevation No. 1, the right line D K is the horizontal line, the arc K P is 51% degrees, A B perpendicuar to P P’ is the projection E 26 PROJECTION UPON THE LOWER PLANE. [PLATE 5. of the equatorial diameter. The arcs 4 P, A P' are each divided in- to nine equal parts, answering to 10°, 20°, 30°, &c., and the chords drawn through the points of division parallel to 4 C are the projec- tion of the parallel circles. The arc § being divided into six equal parts answering to 15°, 30°, 45° &c., and lines being drawn parallel to P P’ to meet A B gives the minor axis of all the twelve meridional circles, the major axis being 2 P’, which is equal to the real axis or diameter of the sphere. Wo CHAPTER V.—CONSTRUCTION OF THE PLANS OR OF THE ELEVATIONS OF TWO ROUND RECTI- LINEAL SOLIDS* INTERSECTING EACH OTHER. Two cylinders, or two cones, or a cylinder and a cone, may inter- sect each other; the two solids which thus intersect are supposed to be cut simultaneously by an indefinite number of planes, either pa- rallel to both the axes when the solids are cylinders, or through the two summits when the solids are cones or parallel to the axis of the one and through the summit of the other, when the former is a cylin- der and the latter a cone. Every cutting plane will thus intersect the surfaces of both solids in right lines, meeting each other in the inter- section of the two surfaces; hence it is evident that the projection of every two of the right lines in every plane will give, by their meeting, the projection of an indefinite number of points in the projection of the intersection of the two surfaces. A In order to obtain the most simple projection of two such solidgn- tersecting each. other, it will be proper to show the positions of the cylinder and cone individually, which will give the most simple pro- jections. In the right cylinder one of the most simple projections will be ob- tained when the axis is parallel to the plane of projection, for in this case its projection is a rectangle of which one dimension is equal to the length of the axis, and the other equal to the diameter of the circular ends ; another of the most simple projections will be obtained by mak- ing the axis of the cylinder perpendicular to the plane of projection ; for in this case its projection is a circle, having its diameter equal to the diameter of the circular ends of the cylinder, the circle being the projection of the curved surface, and the centre the projection of the axis of the cylinder. * By the two round rectilineal solids are understood to be the right cylinder, and the right cone, which have right lines on their surfaces in planes passing through the axis. PLATE 36.] ELEMENTS OF PRACTICE. 27 In the right cone one of the most simple projections will be ob- tained when the axis is parallel to the plane of projection, for in this case its projection is an isosceles triangle, of which the base is equal to the diameter of the circular end, and the perpendicular equal to the length of the axis. Another of the most simple projections will be obtained by placing the axis of the cone perpendicular to the plane of projection ; for in this case its projection is a circle equal in dia- meter of the circular end of the cone, the axis of the cone being pro- jected upon the centre of the circle. Therefore the most simple projection of the intersection of two cylinders, or of two cones, or of a cylinder and a cone, is when the two axes are both parallel to the plane of projection, or one parallel and the other perpendicular to it. In the solids about to be projected, the projections of a great num- ber of points cannot be made intelligible ; a few, however, will be suf- ficient to explain the principles in the most simple manner. PROPOSITION XL. Prosrem XIX. To find the projection of two cylinders intersecting each other, hav- ing their axes at right angles, and in a plane parallel to the plane of projection. Let u v e f (Plate A fig. 1) be the projection of one of the cylinders, and 8 4 & g the projection of the other, as if both solids were entire. —These two projections will be identical to the sections of the cylinders cut by a plane through the axis. Bisect » v in w, and draw w x parallel to v e or uf, meeting e fin x, and bisect 8 y in z, and draw z y parallel to » 4 or 8 g; then w x is the projection of the axis of the cylinder, of which the entire projection is « v e f and will bisect e f in x, and 2 y is the projection of the axis of the cylinder, of which the entire projection is By % g, and will bisect 2 gin y. It will be evident that the sides uf, v e of the projections of the one cylin- wil intersect Bg, v h, the projection of the other sides, in the points d, d, d, d’ of the intersection of the curved surfaces. Prolong # w to A, and from any convenient point M in x A, with a radius equal to w u or w v, or to the radius of the end of the small cylinder, describe the semicircle D A 2D’ (No.1)and draw the diame- ter D Ir parallel to « v. Between A and D take any number of points B, C'; and parallel to D D’ draw the chords B B’, C C' inter- secting 4 M in O and P. . Prolong y z to L ; from any convenient point M' in y L, with a ra- dius equal to z 8 or z 4, or to that of the end of the large cylinder, de- scribe the semicircle D L D’ (No. 2) and draw the diameter D 2 perpendicular to M ZL. Upon the radius M' L make M' N, M' O, M' P respectively equal to M 4, M O, M P (No. 1), and through the points NV, O, P (No. 2) draw the chords 4 A’, B B', C C/, pa- rallel to D D’. Parallel to z y, draw the right lines 4 a, B 0, Ce, and parallel to w z (from No. 1) draw 4 a, Bb, C ¢, then a, b, c, are points in the projection of the intersection of the curved surfaces 28 PROJECTION UPON THE LOWER PLANE. [PLATE 37. of the two cylinders. The points ¥, ¢/, d, will be found by prolonging the lines B b, C ¢, (from No. 2) to ¥/, ¢/, and by drawing B ¥, C ¢/ (from No. 1) parallel to w 2. The points thus found are in the projec- tion of the intersection of the curved surfaces of two cylinders on one side of zy. In the same manner points may be found in the projection of the intersection of the curved surfaces of the two cylinders on the other side of z y, as is evident. PROPOSITION XLI. ProsrLeEm XX. To find the projection of a cone and a cylinder intersecting each other, the two axes being at right angles, and on a plane parallel to the plane of projection. : Let ¢ s u v (figure 2) be the projection of the cylinder, and 7 7’ g that of the cone, as if both solids were entire, and these two projec- tions will be identical to the sections of the cylinder and cone respec- tively through each axis. Itis evident that the projections of the sides of the cylinder and cone intersect each other in the projection of the intersection of the two curved surfaces ; thus the points d and e the intersections of s « and ¢ v, projections of two sides of the cylinder, and 7 ¢ the projection of one side of the cone, are points in the intersec- tion of the two curved surfaces. Bisect ¢ sin y, and draw y n parallel to £ v or s « meeting « v in #, and bisect 7 7/ in £; and join fg ; then y n is the projection of the axis of the cylinder, and will bisect « v in n and f g is the projection of the axis of the cone, and will be perpen- dicular to 7 77, the right lines ¢ s v being the projections of the ends of the cylinder are equal to the diameters of the circles, and the right line 7 +’ being the projection of the base of the cone is equal to the diameter of the circle. Prolong ¢s and 7 7’ to meet each other in WW. Prolong # W to H, from any convenient point Zin W H draw Z G perpendicular to WW H, and draw g G perpendicular to WW ¢ Pro- long 2 y to meet Z G' in FF; from FF with a radius equal to the radius of the circular ends of the cylinder, describe the circle A B CD (No.1) intersecting Z G in D and E; draw G Hto touch the circle in 4, and draw F 4, meeting G' H perpendicularly in A; then A is the point of contact in the circle 4 B C D, and the right line G H. Between 4 and D take any number of inter- mediate points B, C; and draw 4 a, Bb, Cc, parallel toy a. Through the points B, C, draw G I, G J, meeting W H in 1. J, and intersect- ing the circle A B CD in B, C. Prolong ¢ Wto K; in W K, make W Yequal to W Z; draw YF perpendicular to W K. Prolong gf to meet Y F in IF; from 2” with a radius equal to that of the circular base of the cone, describe the circle A B CD (No. 2), intersecting YF" in D. In W K make YK, YL, Y M, respectively equal to ZH, Z I, Z J. Parallel to fg, draw Ao, Bp, Cq, Dr, meeting 77'ino, p, g,7, and draw og, p g, ¢ g, meeting the right lines A a, Bb, C'¢, drawn from the circumference of the-circle No. 1 in the points @, 6, ¢, and @, b, ¢, d, are points in the ® = Lo PLATE 38.] ELEMENTS OF PRACTICE. 29 intersection of the surfaces of the cone and cylinder. In the same manner points may be found in the intersection of the surfaces on the other side of the cone. PROPOSITION XLIE Prosrem XXL To find the projection of two cones intersecting each other, the axes being at right angles, and in a plane parallel to the plane of projection. Let ¢ uz (fig. 2) be the projection of the one cone, and v ¢ y that of the other, as if both cones were entire, and these two projections will be identical to the sections of the cone through the axis, the points d, d’ being the intersections of the projections # 2, # 2 of the sides of the one cone, and » y the projection of a side of the other cone, are points in the projection of the intersection of the two curved surfaces. Draw 2 g perpendicular to ¢ », meeting ¢ u in g, and draw y h perpendicular to v gq, intersecting v ¢ in %; then z g is the projection of the axis of the cone of which the projection is ¢ % z, and will bisect £% in g, and y 4 is the projection of the axis of the cone of which the projection is v ¢ y, and will bisect v ¢ in %; the right lines ¢ u and v ¢ being the projection of the bases of the cones are equal to their diameters. Through the summits y and z draw the right line w 2, and prolong ¢# and v ¢ at both ends to meet each other in WV, and to meet w 2 in w and «, and thus forming the triangle w Wa. : Prolong x W to I; from any convenient point Z in W I draw Z H perpendicular to W I, and draw w H perpendicular to Ww. Pro-" long z g to meet Z H in G', from G with a radius equal to the radius of the circular base of the cone represented by. u z, describe the circle A B C D (No. 1) intersecting Z Hin D, D’, draw H I to touch the circle, and draw G' 4 meeting H I perpendicularly in 4; then A is the point of contact of the circle and the right line # I. Between A and D take any number of intermediate points B, C; draw A 1, B 2, C3, D u parallel to g z, meeting ¢ » in the points 1, 2, 3, u, and draw the right lines 1 2, 2 z, 3 2. Through the points B, C draw H J, H K meeting I Win J and K, and intersecting the circle A B C D in B’, C". Prolong w Wto L; in W L make W Y equalto WZ; draw ¥ O perpendicular to W L, and draw 2 O perpendicular to Wa. Pro- long y 2 to meet ¥ O in G’; from G’ with a radius equal to the ra- dius of the circular base of the cone represented by v ¢ y, describe the circle 4 B C D (No. 2) intersecting ¥ O in D. In W L make YL, YM YN, respectively equal to Z 1, Z J, Z K, and draw L O, M O, N O intersecting the circle A B C D in A, B,C. Parallel to hy draw A p, B q, C r meeting v ¢g inp, gq, r; draw py, 9 y, 7 y in- tersecting the right lines 1 2, 2 2, 3 2 in the points a, b,¢, and a, b, ¢ are points in the intersection of the two conical surfaces, also the point d in which u z and v y intersect is another point in the intersection of these surfaces. In the same manner the points &’, ¢/, d’ may be found as also those upon the other side of the projection of the greater cone. we w @ 30 4. PROJECTION UPON THE LOWER PLANE. w CHAPTER VI.—PROJECTION APPLIED TO FIND. ING A PLANE FIGURE WHICH WILL COVER ANY GIVEN PORTION OF THE SURFACES OF CERTAIN SOLIDS, WITHOUT LEAVING ANY VACUITIES, AND THE CONTRARY. - > @ al Before proceeding with this subject it will be necessary to premise the following - “ eo DEFINITIONS. ow 1. A solid geligrated by any plane figure revolving upon a given fixed right line in the same plane with it, until any point whatever in the revolving edge of the figure describes the circumference of a circle, is called a solid of revolution. 2. The figure carried round to form a solid of revolution, is called the generating section. ; 3. The right line round which the generating section is carried, is called the axis of rotation. 4. Lines drawn upon the curved surface of a solid of revolution, each in the same plane with the axis of rotation, are called me- ridians. “ 5.A solid of revolution, which has a circle for its generating sec- tion, is called a eylindric annulus. Besides the cylindric annulus, and the three round solids, viz., the right cylinder, the right cone, and the sphere or globe, there are, as is evident from Defininition I. of this Chapter, an infinite variety of so- lids of revolution. > 6. The curved surface of a solid upon which a flat limber surface (such as a piece of paper), can be bent, so that every point of the surface thus applied may naturally fall upon the curved surface, is said to be developable. 7. A flat limber surface which cannot be bent upon the curved surface of a solid, so that every point of the surface applied may come in contact with the curved surface, is said to be non-develop- able. : 8. The figure containing a flat limber surface which, when bent, may exactly cover and coincide in every point with a given portion of a de- velopable curved surface of a solid, is called tke envelope of that sur- face. 9. The figure containing a flat limber surface, which, when bent, may exactly cover and coincide in every point with a portion between two meridians of a given developable surface of solid is called a gore. 10. If a point, line, &c., be given upon an envelope, and if the envelope be bent upon the curved surface, the point, line, &c., thus placed upon the curved surface is called tke point, line, &c., of envelop- "= ELEMENTS OF PRACTICE. &% 31 . ment, and the point, line, &c., on the envelope, is called the point, line, &e., of devlopment. 4 No surfaces of solids of revolution are developable unless their me- ridians are right lines. The only geometrical solids which have deve- lopable curved surfaces are the cylinder and the cone. All solids of revolution which have curved meridians have non-developable sur- faces. Thus the surface of a sphere is a non-developable surface. The surface of every cylinder or cone, whether right or oblique, is developable. ; In thischapter original points, lines, &c., are supposed to be situated upon the curved surface of a cylinder or cone, and projectedqsupon a. plane Passing through the axis, so that the otiginal of the projection thus found shall coincide with a point, line, &c, given in position upon the envelope, and as the ingerse is often required, it isiconvenient to aiviior nd the entire envelope. i KE Ast sli whether it is a cylinder or a cone, is supposed to be a solid of revolution, the plane of projection must be limited to the figure of the section passing through the axis. If the solid be a cylin= der, the plane of projection will be a rectangle, of which one dimension will be equal to the length of the axis, and the other equal to the. dia- meter of one of the circular ends, and as the curved surface which contains any original point, line, &c., is entirely upon oneside of the plane of projection, the envelope will be a rectangle of which one di- mension will be equal to the length of the axis, and the other equal to the semi-circumference* of one of the ends. If the solid bea e, the limits of the plane of projection will be confined to an socal angle, of which the base will be equal to the diameter of the circular base of the cone, and the perpendicular equal to the length of the axis. . It is evident that in the projections of points, lines, &c., situated up- * In finding the projection of a point situated upon the curved surface o cylinder or cone, it is frequently required to make one line equal to ee When a right line is required to be made equal to a given arc, or an arc equal to a given right line, or an arc equal to another given arc, the most convenient method is to divide the given line into equal parts, and with the same extent cut off as many parts of the line requiféd as the number of parts contained in the given line ; but as the chord is always less than the arc, if the given line be a curve, and the line required a right line, the length of the right line thus found will be something less than the length of the curve, or if the given line be a right line, and the length of the line required be a portion of a given curve, the length of the curve thus found will be greater than the length of the Sivan right line. It is evident, therefore, that the difference between the length of the given line and the length of the required line will be less as the number of parts into which the given line is divided is greater. There is no correct geometrical method by which a right line can be made equal to a given arc. When the diameter of a circle is 7, the circumference will be something less than 22 ; therefore if a right line be required to be made equal to the circumference of a given circle, repeat the diameter three times upon the line, and add one-seventh of the said diameter, and the sum will be something greater than the circumference ; or if a right line be required equal to the semi-circumference, divide the diameter into seven equal parts, and repeat the extent eleven times upon the line. ld “a de 32 #% PROJECTION UPON THE LOWER PLANE. [PLATE 6. w on the curved surface of a cylinder, the axis will divide the rectangle which is the plane of projection into two symmetrical rectangles, hav- ing in each one dimension equal to the length of the axis, and the other equal to the radius of the circular ends; and in the projection of points, lines, &c., situated upon the curved surface of a cone, the axis will divide the plane of projection into symmetrical right angled tri- angles, having in each the perpendicular equal to the length of the axis, the base equal to the radius of the circular base, and the hypothenuse equal to the side or meridian of the cone. Moreover, it is evident in cylindric projections, that the meridians will be projected in right lines ‘parallel to the axis, and in conical projections in right lines terminating in, or tending to, the vertex of the angle contained by theo equal sides. PRO : > 'ROPOSITION XLIII. ProBLEM XXII a a To find the envelope of the curved surface of a given cyl: “ Let ABCD (fig. 1) be the section of the cylinder through the axis, and H 7 J a section perpendicular to the axis, which will therefore be a circle. Draw A F perpendicular to 4 D, and make A F equal to the circumference of the circle Z1.J. Draw F FE parallel to 4 D, and D E parallel to A F,and 4 D E F will be the envelope required. PROPOSITION XLIV. ProsrLem XXIII. Given a point of development to find the projection of the point of envelopment for a cylindric surface. Let A BC D (fig. 1) be the plane of projection, 4 D E F the en- velope, and let Z be the point of development. Prolong D A and C B to H and J; draw H J parallel to 4 B, and upon H Jas a diameter describe the semicircle # I J, Draw L I perpendicular to 4 D inter- secting A D in k; divide L k into equal parts, and upon the semi- circular arc H IJ, with the extent of one of the parts cut off the same number of arcs from H to I. Draw I parallel to 4 D, and l is the projection of the point of envelopement. & - PROPOSITION XLV. Prosrem XXIV. Gliven the projection of a point situated on the curved surface of a cy- linder to find the development of the point. Let 7 (fig. 1) be the projection of the point. Prolong D 4 and C B to H and J; draw H J. parallel to A B, and upon H J as a diameter describe a semicircle H IJ. Draw the right line Z l intersecting A D perpendicularly in %, and draw / Z parailel to 4 D. Divide the arc H 1 into equal parts, and with the same distance upon 2 ZL, mark the same number of parts, and the extremity Z is the developement of the point of which the envelopement is the original of the point / of projection. 2 Se PLATE 0. Lig ld. I Page 32 : J El AE PLATE 6.] ELEMENTS OF PRACTICE. 33 PROPOSITION XLVI. ProsrEm XXV. Given the projection of a line, to find the development of the line. Let the plane of projection be 4 B C D (fig. 2), and let ADEF be the envelope, and let H i be the given projection of the line. Up- on A B, as a diameter, describe a semicircle A G B ; then the right line A F, being equal to the arc A GB, divide the right line 4 F and the arc A G' B each into the same number of equal parts, in the points 1, 2, 8, &c. From the points 1, 2, 3, &c., in the arc AG B, draw right lines 17, 2 %, 3 /, &c., meeting A B perpendicularly in the points j, k, I, &c. Parallel to 4 D draw the right linesj ¢, k 7,1 s, &c. ; also parallel to 4 D, and from the points 1, 2, 3, &c. in A F, draw 1 Q, 2 R, 3 S, &c., and draw ¢ Q,r R, s S, &c, parallel to AF. Then H, Q, R, S, &c., are points in the development of the line ; hence, by drawing a line through these points, # Q R S...1 is the development required. PROPOSITION XLVII. Prosrem XXVI. Given the development of a line to find the projection of the enve- lopment of the line. Let A B C D (fig. 3) be the plane of projection, and 4 D F E the envelope as in the preceding proposition, and let the given de- velopment of the line be 4 E. Having divided the right line 4 FF and the semicircumference 4 G' B each into the same number of equal parts at the points 1, 2, 3, &c.; from the points 1, 2, 3, &c. in the semicircular arc 4 G' B, draw the right lines 1 7, 2 %, 3 I, &c., meeting A B perpendicularly in the points j, %, /, &c. Parallel to A D from the points 1, 2, 3, &c., in A F, draw 1 @, 2 R, 3 8, &c., meeting the line 4 £ in the points Q, R, S, &c., also parallel to A D draw j q, k 7, Ls, &c., and parallel to A B draw Q g, Rr, S's, &c., then the points g¢, 7, s, &c., are the projections of the enveloped points of which the developments are @, F, S, &c., and the line 4 gr ¢ ...C' drawn through these points is the projection of the line of which the development is 4 Q R S...E. ScHOLIUM. The given line in Problem XXV. and in Problem XXVI., may be any line whatever, either right, curved, or irregular, In Problem XXV. the given projection H i (fig. 2) is a right line, and the de- velopment H I found, is a curve of contrary flexure. In Problem XX VI. the development A E (fig. 3) is a right line, and the projection 4 C found, is a curve of contrary flexure; and the development H I (fig. 2), and the projection 4 C (fig. 3) are curves of the same species as that which is commonly called the figure of the sines ; moreover, the two curves would be identical, provided that U H (fig. 2) and A B (fig. 3) were equal to each other, as also U I and B C. The curve line A gr s5...Cis the projection of half a revolution of a cylindric spiral. F 34 PROJECTION UPON THE LOWER PLANE. [PLATE 7. PROPOSITION XLVIII. ProsrLem XXVIL To find the envelope of the curved surface of a semi-cone. Let A B C (fig. 1) be the section of the cone through the axis, and A D B be a semicircle equal to the base of the semicone ; and as the cone is right, the axis will be perpendicular to the base, and the triangle A B Cisosceles. From the point C, with the distance C' 4. describe the arc 4 FE, and make the arc 4 FE equal to the semicir- cumference 4 D B. Join C E, and C A E is the envelope required. PROPOSITION XLIX. ProsrLem XXVIII. To find the envelope of the curved surface of the frustum of a semi- cone. Let A B H F (fig 2) be the section of the frustum of the semi- cone through the axis, and 4 D B a semicircle equal to that of the greater end. Find the envelope C 4 E (Prob. XXVIL.), from C, with the radius C F, describe the arc F G, meeting ££ C in G, and A IE G F is the envelope required. : In the following propositions it will be understood that the section A B C, which is the plane of projection, and the envelope C 4 E of the semicone are given or aiready found, as in Problem XXVIL PROPOSITION L. ProsrLem XXIX. Given the development of a point situated on the curved surface of a semicone, to find the projection of the point of envelopment. Through G (fig. 3), the point of development, draw the right line C' H meeting the arc A Ein H, and make the portion 4 D of the semicircular arc equal to the portion 4 H of the arc 4 E. Draw D7 perpendicular to 4 B meeting it in Z; join 2 C; from C with the ra- dius C @ describe the arc G F meeting C Ain F; draw F g paral- lel to A B meeting C kin g, and ¢ is the projection of the point re- quired. ; PROPOSITION LI. ProsrLEm XXX. Given the projection of a point, situated on the curved surface of a semicone to find the development of the point. Let g (fig. 3) be the projection of the point. Through g draw the right line C% meeting A Bin % ; draw 42D perpendicular to 4 B, and make the portion 4 H of the arc A E equal to the portion 4 D of the semicirculararc 4 D B. Join C H; draw g F parallel to 4 B meeting C A in F; from C with the radius C F describe the arc F G meeting C Hin G and G is the development of the point.* * This might be found otherwise. (Fig 4.) Suppose the point H found as above. Draw g k parallel to C 4 ; meeting A B in k; join H C3 in H C make H G equal to k g and G is the development of the point required. PLATE. 7. Loge 3¢ PLATE 7.] ELEMENTS OF PRACTICE. 35 PROPOSITION LILI. ProsrLem XXXI. Given the projection of a line situated on the curved surface of a cone to find the development of that line. Let A G (fig. 5) be the given projection of the line. Divide the semicircle A D B, and the arc A FE each into the same number of equal parts 1, 2, 3, &c. Draw the right lines 1 4, 2 4, 3 j, &c. per- pendicular to 4 B meeting A Bink, i,j, &c ; join Ck, C1, Cj, &c., intersecting A G in the points p, ¢, 7, &c.; and join also the points in the arc 4 E with C, viz.,join C1,C 2, C 3, &c.—Draw p 7), q U, r V, &c., meeting C A in the points 7), U, V, &c.; describe the arcs TP, UQ, VR, &c, meeting C1, C2, C3, &c., in the points P, Q, R, &c.,and draw the line 4 P Q R...F, which is the development re- quired. PROPOSITION LIII. Prosrem XXXII Given the development of a line situated in the curved surfaceof a cone, to find the projection of the envelopment of the line. Let A P Q R...F (fig. 6) be the given development of the line. Divide the arc A FE, and the semicircular arc A D B each into an equal number of equal parts at the points 1, 2, 3, &c.; from the points 1, 2, 3, &c., in the arc A E draw the right lines 1 C, 2 C, 3 C, &c., intersecting the given line 4 Fin the points P, @, R, &c.; from the points 1, 2. 3, &c., in the semicircular arc, draw the right lines 1 4, 2 4,3 j, &c., perpendicular to A B, meeting A B in the points h, i, j, &c., and join C &, C i, Cj, &c. From C, with the distances CP, CQ, CR, &c., as radii, describe the arcs P 7, QU, R V, &c., meeting C A in the points 7), U, V, &c.,draw 7 p, U ¢, Vr, &c., parallel to A B, meeting the right lines C %, C i, Cj, &c., in p, ¢, 7, &c., and draw the line 4 p g r...f, which is the projection of the envelopment of the line. SCHOLIUM. In figure 6, the given line 4 P @ R...F is the proportional spi- ral of which C is the centre ; therefore the radii C4, C P, C Q, C R, &c. make equal angles with the curve; hence the line 4 p gr ..f is the projection of a spiral line drawn upon the curved surface of the cone, making equal angles with the meridians of which the projec- tions are 4 C, kh C,¢ C,j C, &c. The line 4 p gq r...f is the projection of half a revolution of the conic spiral. If the develop- ments C' 4, C P, of the two first meridians are given, the develop- ments C Q, C R, &c. may be found with a pair of proportional compasses, thus :—move the sliding centre so that the distance be- tween the points of the two longer branches may be to the distance between the points of the two shorter branches as C 4isto CP; take the distance C P between the points of the two longer branches, and make C @ equal to the distance between the points of the shorter 36 PROJECTION UPON THE LOWER PLANE. [PLATE 7. branches ; again take C Q between the points of the longer branches, and with the distance between the points of the two shorter branches from C mark C BR. Proceed in the same manner throughout all the intermediate gradations of lines until having found the point F, and the points P, Q, R,...F thus marked will be points in the propor- tional spiral. If the projection of the spiral is required to be con- tinued, repeat the operation now explained, and every repetition will give the development of a new line of which the projection will be the projection of another half revolution of the conic spiral as is shown on the figure. The distance on the line C 4 from C of the first point of the new curve is always equal to the last distance on CE from C of the curve last completed. Here we suppose the solid to be a whole cone; the first line 4 fis the projection of half a revolu- tion of the spiral on the upper surface, the second line f Z is the projection of half a revolution of the spiral on the lower half, and is shown by a dotted line. The projection of this spiral, however, - far continued, would never terminate in the point C. If, in the same plane, the curve X G be revolved upon C until X coincide with F) and the curve Z H also be revolved upon C until the point Z coincide in its turn with G'; the three curves will then be one continuous spiral and the whole being lapped closely round the conic surface will form a spiral, making one revolution and a half, and making equal angles with the meridians of the cone, and the undulated line being the projection of this spiral. ON THE COVERINGS OF SOLIDS OF REVOLUTION. Although no envelope for the whole or any portion of a surface of revolution having curved meridians can be found, yet a very near ap- proximation to the covering may be obtained, having the generating section given, by means of a number of cylindrical surfaces having the edges of every two terminating upon each other in a meridional plane, and each touching the curved surface of the solid in an inter- mediate meridional line, or otherwise by a series of pieces, every two meeting each other in the circumference of a circle, perpendicular to the axis of rotation, and each touching the curved surface of the so- lid in an intermediate parallel circle, or in both edges, according as the meridians are convex or concave on the outside. The pieces which thus form the covering will, therefore, in their enveloped state, form the curved surfaces of as many frustrums of cones as the number of pieces having their axes in common with that of the solid. If the covering be constructed by cylindric pieces, joined together at their edges, it will form a regular spherical polyhedron of a great number of sides or faces. each face being a tangent to the curved sur- face of the sphere and the middle line, which divides the face into two symmetrical parts, will fall in all its points upon a meridian. It is evi- dent that the greater the number of faces the nearer will the approxi- mation of the polyhedron be to the surface of revolution. ‘ LPIATES. PLATE 8.] ELEMENTS OF PRACTICE. 37 PROPOSITION LIV. ProsrEm XXXIIIL Given the radius of a sphere, to find a near approximate covering to the hemisphere or to the entire sphere. From any convenient point C (fig. 1) as a centre, with the radius of the sphere describe the equatorial circle D V W F, and draw the diameters 2) W and V F perpendicular to each other. Let this circle be the plane of projection. Divide the quadrantal arc D V into twice as many equal parts at the points 1, 2, 3, &c., as the number of spheric portions intended to be covered ; through the points 1, 3, 5, &c., draw the right lines C 4, CR, C 8, &c.; join C D, C2, C 4, &c, and through the points D, 2, 4, &c., draw the tangents A B, A R, R S, &c. Make D B equal to D 4, and join C B. Since the equatorial circle is the plane of projection, the meridional circles and meridional planes will be projected into right lines, terminating in the centre C ; hence the radii C' D, C2, C4, &c., are the projections of the meridional circles, and the isosceles triangles C A B, CAR, C R 8, &c., the planes of projection, or the projections of the cylindric surfaces, meeting each other upon meridional planes, and each touching the spherical surface in every point of the meridional circle between the two meridional planes upon which the edges of the cylindric surface terminates. As all these surfaces form thesurface of a spherical polyhe- dron, and as the polyhedron is regular, if the figure of development of one surface is found, each of the remaining ones will be identical. Now, here are given the radius C D of the cylinder, the two right lines C A, C B, making equal subcontrary angles with the axis, or with the line A B parallel to the axis, and the right section C D F, which is one quarter of the section of the entire eylinder, to find the develop- ment of the lines of which the projections are C A, C B. FIRST METHOD. Prolong C D to E, and make D FE equal to the arc D F. Divide the right line D E and the arc D F each into an equal number of equal parts at the points 1, 2, 3, &c., and, from the points 1, 2,3, &c., in the arc D F, draw 1 Z,2 M, 3 N, &c., intersecting C D perpendicularly in the points G, H, I, &c., and meeting A C in the points L, M, N, &c. Parallel to 4 B, through the points 1, 2, 3, &c., in DE, draw I, mm', nn, &c. Make 17,17, each equal to G L : 2 m, 2m, each equal to # M; 3n,3 #’, each equal to I NV, &c.; draw the curve lines 4 I m...E, BI w'...E, and the figure 4 I m...E... m’ ¥ B will be the development of one of the cylindric surfaces, which, when bent as required, will be contained betweeen the two meridional planes, and in contact with a meridional circle upon the spherical surface. SECOND METHOD. Draw the right line D E (fig. 2); make D E equal to the length of thearc D F (fig. 1). In DE ( fig. 2) prolonged take any convenient 38 PROJECTION UPON THE LOWER PLANE. [PLATE 8. point C, and draw C F perpendicular to C D. From the centre C, with a radius C D, describe the arc D F. Through D, draw A B parallel to C F, make D A and D B each equal to D 4 or D B (fig. 1), and join 4 C. Divide the right line D Z (fig. 2) and the arc D F each into an equal number of equal parts at the points 1, 2, 3, &c., and from the points 1, 2,3, &c., inthe arc D F,draw 1 L,2 M,3 N, &c., intersect- ing C D perpendicularly in the points @, H, I, &c., and meeting A C in the points L, M, N, &c. Parallel to 4 B, through the points 1,2,8, &c.,,in D BE, draw lI’, mm/, nn’. Make 17,1 7 each equal to GL; 2m,2m eachequal to HM ; 3n,3 x, each equal to IV, &ec.; draw the curve lines 4 I m..E, Bl w'...E, and the figure Alm...E..ow I! B will be the development of the cylindric surface which when bent as required will be contained between the two meri- dional planes and in contact with a meridional circle upon the spheri- cal surface.* THIRD METHOD. Draw the right line D E (fig. 3).; make D E equal to the length of the arc D F (fig. 1). Through D (fig. 3) draw the right line A B perpendicular to D FE; make D A, D B each equal to D 4 or D B (fig. 1); from the point D (fig. 3), with the distance D A or D B, describe the semicircle 4 Q B, and prolong E D to Q. Divide the quadrantal arc 4 Q, and the right line D Z, each into an equal number of equal parts; through the points Z, M, N, &c., of division in the arc 4 Q draw L G, M H, N I, &c., parallel to 4 B, meeting D Qin G, H, I, &c., and through the points 1, 2, 3, &c., in the right line D E, draw Il, m m/, nn’, &c., parallel to 4A B. Make 17,117, each equal to G L; 2 m, 2m’ each equal to H M; 3 n, 3 #, each equal to I N, &c.; draw the curve lines A I m...E, Bl w'...E, and the figure 4 Im...E..m' I’ B will be the development of the cylin- dric surface, as in the first two methods. ScHoLIUM. "The method here given is not confined to covering the entire sphere, in parts or gores; but any zone, whether polar or interme- diate, may be covered in the same manner. If the number of gores were 36, instead of 16, the edges which meet each other would be meridians at 10° distant, and if the length of the gores had been divided into 9 equal parts, the ZZ m m’ » »/, would have formed pa- vallels of lattitude. The covering of the entire sphere would have required the gores to have extended from the one pole to the other. The development of the surface between two meridians would form in their envelopment meridians at 221° distant from each other. The methods here employed for the development of the surface of a sphere are much more simple than those given by any other au- thor, and they are no less accurate. * It will be perceived that the last paragraph of the second method is expressed in the same words as the last paragraph of the first method. This is in conse- quence of C D (fig. 2) being divided in the same proportion as C D (fig. 1). PLATE 4.] PRACTICAL PROPOSITIONS AND RULES. 39 CHAPTER VI.—APPLICATION OF PLANS AND ELEVATIONS TO MACHINERY, ARCHITECTURE, DIALLING, GEOGRAPHY, AND ASTRONOMY. ProrosiTioNs OF FREQUENT OCCURRENCE. 1. An original plane, perpendicular to one of the planes of projec- tion, has its intersection upon the other plane in a right line perpen- dicular to the ground line, except when the original and the other plane are parallel ; in this case there is no intersection. 2. A circle in an original plane, perpendicular to one of the planes of projection, has a circle for its projection upon the other plane, when the ground line is parallel to the original plane. 3. If an original circle be in a plane perpendicular to one of the planes of projection, and inclined in any given dihedral angle to the other, the projection of the circle will be an ellipse upon that plane of projection to which the plane of the original circle is oblique ; more- over, the major axis of the ellipse, which is the projection of the cir- cle, will be in a right line perpendicular to the ground line, and the minor axis in aright line parallel to the ground line. 4. In the ellipse, which is the projection of a circle, the major axis is equal to the diameter of the circle, and the major axis is to the minor as radius is to the cosine of inclination of the planes. 5. Any surface, plane or curved, is said to be perpendicular to the plane of projection when a right line can be drawn through any given point coinciding with the surface, and, at the same time, per- pendicular to the plane of projection. In a cylinder, if the axis be perpendicular to the plane of projection, the cylindrical surface is also perpendicular to the plane of projection. 6. Every cylindrical surface, perpendicular to the plane of projec- tion, is projected into a circle, and the axis into a point which is the centre of that circle, the diameter, or radius of the projected circle, being equal to the diameter or radius of one of the ends of the cylinder. 7. All cylindrical surfaces having one common axis perpendicular to the plane of projection, are projected into circles, and the common axis into a point, which is the common centre of these circles. 8. If a plane be a tangent to a cylindrical surface, and if the axis of the cylinder be perpendicular to the plane of projection, the cylindrical surface and the plane will be projected into a circle, and a right line, which will be tangent to the circle. 9. If the axis of a cylinder be perpendicular to the plane of projection, every plane which is parallel to the axis will be projected into a right line. 40 PROJECTION UPON THE LOWER PLAN. Practicar RuLgs. 1. When the projection of a point upon a plane, and the height of the point above the plane are given, the projection of the point upon the other plane will be found by drawing a right line from the given pro- jection of the point to intersect the ground line perpendicularly, and by making a distance upwards from the point of intersection upon the line thus drawn equal to the height of the point, and the point of dis- tance will be the projection of the point upon the other plane. 2. Similarly, When the point, which is the projection of a right line perpendicular to one of the planes of projection, and the length of the line are given, the projection of the line upon the other plane will be Jound by drawing a right line from the point which is the projection of the line to intersect the ground line perpendicularly, and by mark- ing a distance upon the line thus drawn from the point of intersection upwards, equal to the length of the line, and the portion of the line intercepted between the point of intersection and the point of distance will be the projection of the line on the other plane. 3. When the projection of a right line parallel to one of the planes of projection, and the distance between the line and its projection are given, ils projection upon the other plane, when perpendicular to the line, will be found by prolonging the projection of the line to the upper side of the ground line, and by marking the distance upon the line thus prolonged from the point of intersection upwards, and the point of distance will be the projection of the line upon the other plane. 4. When the projection of a right line parallel to one of the planes of projection, and the distance between the line and its projection are given, the projection of the line upon the other plane, when it is also parallel to the line, will be found by drawing two right lines from each extremity of the given projection to intersect the ground line perpen- dicularly, and by setting the given distance from one of the points of intersection upwards, and by drawing a right line parallel to the ground line, from the point of distance to meet the other line perpen- dicular to the ground line, and the right line thus drawn will be the projection of the line on the other plane. 5. If of two faces of a rectangular parallelopiped one be parallel to one of the planes of projection, and the other parallel to the other plane of projection, the figure of each of the projections will be identical to that of the face of the solid to which it is parallel. APPLIED TO MACHINERY. 41 In the projection of machinery, as all the parts of wheels which surround the axis, excepting the arms, are comprised by surfaces of revolution, which may be either plane and cylindrical, or plane and conical, or plane conical and spherical, the intersection of any two of these surfaces is a circle. The circles of every wheel will therefore be in parallel planes. When the axis of a wheel is perpendicular to one of the planes of projection, the planes of the circles will be paral- lel to this plane of projection, and when the axis of the wheel is parallel to the plane of projection, the planes of the circles of the wheel will be perpendicular to the plane of projection. The most useful plan and elevation of a wheel is that which has the axis of the wheel perpendicular to one of the planes of projec- tion, and consequently parallel to the other ; for when the axis is per- pendicular to the plane of projection, the circles will be projected into circles equal to their originals, and the projected circles will have for their common centre the projection of the axis; and when the axis is parallel to the plane of projection, the circles of the wheel will be projected into parallel right lines equal in length to the diameters of the original circles, and the projection of the axis will bisect these parallel lines at right angles. The plan of the wheel may be either the projection in which the axis is perpendicular, or that in which the axis is parallel to the horizontal plane. The axis of a wheel may either be parallel to both planes of projec- tion, or parallel to one of the planes and perpendicular to the other, or parallel to one of the planes and oblique to the other, or oblique to both planes. If the axis of a wheel be perpendicular to one of the planes of projection and parallel to the other, the lengths of all the parts of the wheel itself may be found ; but if the axis of the wheel be parallel to both planes of projection, the plan and elevation would be identical, and therefore without a third projection upon a plane per- pendicular to the axis, the plan and elevation would not be sufficient to ascertain the dimensions of the wheel. Hence a plan and two elevations are often necessary in the construction of the object in- tended. In every case whatever, the projection of a wheel upon a plane per- pendicular to the axis will always be found necessary. It would not be eligible in working drawings to choose the plane of projection in an oblique position to the axis of the wheel ; but it some- times happens in the construction of plans and elevations of machinery that the axes of wheels are in various positions ; the axis of a wheel may have an inclination to the horizon, or to the walls of the building, or to both the horizon and the walls of the building ; the projection, there- fore, of a wheel in such a position is not a matter of choice but of ne- cessity. 42 PROJECTION UPON THE LOWER PLANE [PLATE 9. ExAMPLES IN MACHINERY. EXAMPLE FIRST. Projection of an undershot water-wheel, consisting of two rings and float boards, supported upon brackets morticed into the rings, the arms being omitted. No. 1 is a geometrical projection of the wheel upon a plane perpendicular to the axis. The outer edges of the float boards, and the outer and inner surfaces of each of the two rings being cylindrical surfaces perpendicular to the plane of pro- jection, are each projected into a circle, and as all the cylindrical sur- faces have the same common axis, the circles, which are their projections, have one common centre. As two faces of each float board are in parallel planes, and the one of these faces upon which the water acts in a plane passing through the axis, and as the axis is perpendicular to the plane of projection, the two opposite faces of each float board will be projected into parallel right lines, of which one will tend to the common centre of the three circles. As the float boards are fixed to the brackets, one side of each bracket and the back face of the attached float board are in the same plane ; the face of each bracket opposite to that attached to the float board being in a plane parallel to the axis is projected into a right line. As the stress of'each bracket is greatest where it begins to enter the ring, the one end next to the ring is made thicker than the other at the extremity in the circumference of the outer cylinder. The thick- ness of the ring is, therefore, shown by two concentric circles. The ends of the float boards being in planes parallel to the plane of pro- jection, are each projected into a rectangle, of which one dimension is the breadth and the other the thickness of the float board. The thickness of the float board, and that of its supporting bracket, are shown by three right lines, the middle one being common to the board, and the bracket is that which radiates to the centre of the cir- cle. The directions of the outer, or oblique lines of the brackets, are found in the following manner : having drawn one of them agreeably to the thickness in circumference of the outer circle, and at the root where it enters the ring; prolong the line sufficiently within the inner circle; draw a right line from the common centre of the three circles to meet the line thus prolonged perpendicularly ; from the centre, with the length of the perpendicular, describe a small cir- cle ; having marked the thickness of the ends of the brackets in the circumference of the outer circle, draw right lines from the points thus marked out for the back of the brackets to touch the small circle, and the portions of these lines comprised between the extremities of the outer ends of the backs of the brackets, and the circle representing the outer cylindric surface of the rings, are the projections of the rec- tangular surfaces of the backs of the brackets. : To obtain the projections of the float boards, the circumference of the outer circle must be divided into as many equal parts as the float boards are in number. The whole circumference is here divided into 32 equal parts. enorme mm em eT : # PLATE 9.] APPLIED TO MACHINERY. 43 No. 2, is a projection of the wheel upon a plane parallel to its axis. The ends of the float boards, two faces of each ring, and two parallel faces of each bracket, being in planes perpendicular to the axis, are projected into right lines perpendicular to the right line which is the projection of the axis; and the edges of the float boards which are in the cylindric surfaces, and which are parallel to the axis of the wheel, are projected into right lines parallel to the projection of the axis. Draw F P perpendicular to V Z; from any convenient pointoin # P draw g % perpendicular to FP, and make o g, o £, each equal to the ra- dius of the wheel. Draw g i, % j, parallel to o P; make g ¢ equal to the length of the float boards, and draw é j parallel to g¢ 2. In g ¢ make gk, is, each equal to the distance between the outer face of each ring from the ends of the float boards ; make % m, s u, each equal to the breadth of the rings; draw % /, m n, uv, s ¢, parallel to g 4, meet- ing &jinl, n, v,¢; then the projection of the two faces of the one ring is comprised between the lines 2 /, m n, and the projection of the two faces of the other ring between the lines v, s¢. Again, in g i, make w y equal to the breadth of the brackets fixed to one ring, and make « y equal to the breadth of the brackets fixed to the other ring, so that w y and = y may be in the middle of m % and % s; draw wx, Y 2 af, yd parallel to gk, and the lines w 2, y 2, as also « 8. y 3, will comprise the breadths of the brackets in the planes perpen- dicular to the axis. In the geometrical projection of the wheel, No. 1, let 4 D E B re- present a bracket, 4 D being its thickness at the root, and £ B that in the circumference of the outer edges of the wheel, also let B C be the thickness of the float board in the same circumference. Draw D1, E 2, B3 parallel to F P, intersecting y z in 1, 2, 3, and w @ in d, e, f’; then 1 d e 2, will represent the back of a bracket, and 2 e¢ f 3 the end in the outer circumference. In a similar manner the projec- tions of all the other brackets will be found. Draw 4 p, B ¢, Cr, parallel to I P, intersecting g % in @, b,c, and ij in p, ¢, 7; then a b q p, is the projection of the back of one of the float boards, and b ¢ r g the projection of the outer edge. The projections of the re- maining float boards may be found in the same way. The rings only appear in these spaces that are vacant between two projections of adjacent float boards. Only the ends of the brackets appear upon one side ; but upon the other both the back and the ends appear. The geometrical projection, No. 1, is always an elevation ; but No. 2 may either be considered as a plan or an elevation, according as the plane of projection is horizontal or vertical. When the axis of the wheel is parallel to the horizontal plane of projection, No. 2 will be the plan and No. 1 the elevation, and when the axis of the wheel is parallel to the vertical plane of projection No. 2 will be the eleva- tion. 44 PROJECTION UPON THE LOWER PLANE (pLATE 10. EXAMPLE SECOND. Exhibits the geometrical projections of two wheels or pinions work- ing into each other. » In two pinions working into each other, the radius of either is that of a circle described from a certain intermediate point between the bottom and the extremity of the teeth. This circle is called the pitch line. 1f two pinions work upon each other, and if the number of teeth in each wheel and the diameter of one of them be given, the diameter of the other may be found; for the diameters will be as the number of teeth. In the present instance, one wheel, of 12 feet dia- meter, has 28 teeth; then the diameter of the wheel, which has 19 teeth, will be found (by 28 : 19 :: 12 : x) to be 8-1 feet, nearly. With radii, equal to those of the pitch lines (say 6 ft. and 4 ft.), describe two circles to touch each other in the point a. Divide the circumference of the greater circle, beginning at «, into 28 equal parts, and the points of division, a, b, ¢, &c., are the centres of the teeth. Divide the arc, @ b, into 4 equal parts, at 1, 2, 3; add the chord of one of the equal parts to the radius of the pitch line, and, with the radius thus lengthened, describe a circle, s ¢ #; with the ra- dius of the pitch line, diminished by something more than the chord of one of the small arcs, describe another circle, w 2. From the point b, towards ¢, make b 5 equal to 6 3 ; from a, with the radius a 5, describe the arcs % 4, f g meeting the circle s ¢ %, in the points 4, f; and the circle v w «x, in the points 7, g; from the centre 4, with the same ra- dius, describe the arcs £ /, m n meeting the circle s ¢ %, in the points k, m, and the circle v w @, in the points /, #; from the centre ¢, with the same radius, describe the arcs o p, ¢ » meeting the circle s ¢ %, in the points o, ¢, and the circle » w @, in the points p, 7. Proceed in the same manner with the remaining centres. As to the inner circle ¥ 2 a, it may be described with such a radius as will give a substance of sufficient strength. The point a, being that in which the two pitch lines meet each other; from a, divide the pitch line of the smaller wheel into 19 equal arcs, and the points of division will be the centres of the intervals. Divide each of these arcs into 2 equal parts, and the points of bisec- tion will be the centres of the teeth. Proceed similarly, as was done in the greater wheel. : ; In dividing the teeth of wheels, which are sometimes very nume- rous, geometrical rules seldom apply. All equal divisions of prime numbers above 5 must be found by trial. The division into equal parts will not be difficult if the whole number of teeth can be divided by 4. In general, if the number of teeth be a composite number, re- solve it into its prime factors: divide the whole circumference into as many equal parts as one of the factors contains units, then each of these arcs into as many equal parts as one of the remaining factors contain units, and so on. PLATE 1). - PLATE. PLATE 11.] APPLIED TO MACHINERY, 45 EXAMPLE THIRD. The two projections of a pinion or spur wheel. Draw the right line @ 6 (No. 1) in the required position to the ground line ¥ Z, and make @ b equal to the greater diameter. Perpendicular to a b draw a d, b ¢ ; make b ¢ equal to the thickness of the wheel, and draw ¢ d parallel to a b. Bisect @ b in e, and through e draw E F perpendicular to a b.— From any point FE, in E F, with the distance e a or e b describe the circle G H I for the extremities of the teeth, and complete the geo- metrical projection G' H I, as here exhibited. To save trouble the sides of the teeth are represented by right lines radiating to the cen- tre. Through E, perpendicular to BE F, draw H G, meeting the circle J K P in which the bottoms of the teeth terminate in the points J, K, and the circle Z M @Q of the inside of the ring in the points ZL, JM. Parallel to EF draw Ks, M t, L u, J v, meeting a b in the points $4 UT, Perpendicular to YZ draw e ¢, intersecting ¥ Z inm/. Inm'¢ make 7’ 0" equal to the height of the axis above the horizontal plane of projection, and through o' draw & d' parallel to ¥ Z. In m' ¢ make o' @, 0 ¢ each equal to E G or EE H. Perpendicular to Y Z draw, al, bd, and with the major axis & ¢, and the minor axis’ d, de- scribe the ellipse a’ &' ¢ d'. Again in wm’ ¢ make o’ ¢, 0’ ¢, each equal to ££ J or El K; perpendicular to ¥ Z draw sf’, v I/ meeting & d’ in f’s #/, and with the major axis ¢’ ¢’, and the minor axis f' #/, describe the ellipse ¢’ f” g’ i/. Again in m/ ¢/ make o’ ¢, o’ &’ each equal to E L or BM ; perpendicular to ¥Y Z draw ¢j/, u I, meeting I’ d' in j* U and with the major axis # #/, and the minor axesj” / describe the el- lipse ' j” 2 ¢’. The three ellipses now described are the projections of the three circles in one face of the wheel ; the other circles in the opposite face of the wheel will be found in a similar manner. In the geometrical projection G' H I of the face of the wheel, let ij k Ibe the projection of one of the teeth; parallel to ZZ F draw in, jp, k 7, intersecting « b in m, 0, ¢, and d ¢ in =, D7, perpendicular to ¥' Z draw o g, q y, intersecting the ellipse a’ &’ ¢ d’ in the points §, y; parallel to ¥ Z draw 83, y ¢; and again perpen- dicular to ¥ Z draw pd, re. Towards the centre o/ draw 2 « and y 2, meeting the ellipse ¢ f” ¢' #' in the points «, 2; draw 2-3 parallel to ys and ¢ 3 parallell to y 2 and « 8 v 2-3 ¢ 3 is the projection of the teeth on the elevation corresponding to the projection m2 p 7 q 0 in the plan No. 1, and to the projection i; % of the geometrical projec- tion comprised by G' H I. The semicircle 2 § 7), of which the diameter R 7 is perpendicu- lar to ¥ Z, and of which the radius is equal to that of the circle round the extremities of the teeth, is here used for the purpose of obtaining greater accuracy in the division of the teeth, No. 2, on each side of the minor axis &’ d’, the semicircle being divided into equal parts corres- ponding to those in the geometrical projection No. 1. 46 PROJECTION UPON THE LOWER PLANE [PLATE 12. EXAMPLE FOURTH. Exhibits the two projections of a bevel wheel. The principle of the action of two bevel wheels is similar to that of two isosceles cones, touching each other at their bases and apices, the diameter of the base of the one cone being to the diameter of the base of the other, as the number of teeth in the one wheel is to the number of teeth in the other, and the angle contained by the axis of the two cones equal to the angle contained by the axis of the two wheels, and the one cone being turned upon its axis will also turn the other cone upon its axis. It is evident, that as the summits of both cones coincide, and as the sides of both are equal to each other, the circumference of the circles of the two bases must be in the surface of a sphere of which the centre is the point of coincidence of the two apices. A bevel wheel, which has no geometrical projection. The projection which is made on a plane perpendicular to the axis is here the ele- vation, ¥ Z being the ground line, and » « the height of the projection of the centre above Y Z. Draw « M, making any convenient angle with » ». Place a sec- tion of the wheel made by a plane passing through the axis, so that the axis of the section may coincide with » M. It is evident that both sides of the section must be symmetrical, and that » M is the line of symmetry. In this general section 7 £ G' K and J F H L are sections through the ring, and 4 I K Cand B J L D are sections through the teeth or projections of the sides of the teeth. As the teeth of two level wheels which work into each other project from conical surfaces, all the lines which form the edges of the teeth will terminate in the sum- mit of the cones ; therefore, in the section, the right lines 4 C, I K, as also the right lines B D, J L meet the line of symmetry in A. As the ends of two bevel wheels, which work upon each other, ought to be in the surface of a sphere of which the centre is the apex of the two cones, the sections 4 FZ, B F, ought to be the arcs of circles of which the centre is M, and the radius equal to that of the sphere ; but as right lines are more easily projected than arcs of circles, the lines A E, BF, are respectively perpendicular to Z M, J M; the ends C G, D H, of the sections of the teeth are also right lines, re- spectively parallel to 4 E, B F. For the more easy projecting of the teeth, it will be proper to observe that the right lines 4 E, B F meet « M in N, and that the right lines CD, K L, G H, as also the right lines 4 B, IJ, E F, are bisected by N JM respectively in the points P,Q, RB; 0,8, T. From the point « (No. 1), with the distance 4 O or O B, describe the circle a b ¢ d; from wu, with the distance S Z, or S J, describe the circle e fg k; from u, with the distance P C, or PD, describe the circle i j £ Z; from wu, with the distance @ K or Q L, describe the circle m n o p ; from u, with the distance RZ G, or R H, describe the circle ¢ r s &. The projections of the tops of the teeth are comprised between the circles ea bcd, i j k I; and the projections of the bot- PLATE 12.] APPLIED TO MACHINERY. 47 tom of the teeth between the circles e f g &, m n 0 p; the projec- tions of the visible ends of the teeth on the inside are comprised be- tween the circles ¢ 7 k J, m n o p, and the projections of the invisible ends of the teeth between the circles a bc d, ef g Divide the circle a 6 ¢ d into 40 equal parts, 20 of these will com- prise the breadths of the teeth, and the other 20 the breadths of the intervals. The teeth being marked out as exhibited in No 1, » By 3 exhibits the projection of the top of a tooth, and thus the elevation No. 1 is complete. To find the plan of the wheel, No. 2; perpendicular to VY Z, draw the right line » #». In any convenient place, No. 2, place the section of the wheel so that the line M NV may be parallel to # n. Draw the right line 4 a, intersecting v » perpendicularly in 0, and make o a, 0 b, each equal to O Aor O B. Parallel to A a, draw M m, N =, intersecting © n in m, n. Join a m, bm, as also a n, b n. Parallel to A a, draw Cc, intersecting a m in ¢, and b m in d, K k intersecting amin k,and bm in I; Ii intersectinga n ind, and b nin j,and E e intersecting @ » in ¢, and b » in f. Then the projections of the tops of the teeth are comprised between the lines a b, ¢ d; the bottoms between the lines £ J, ¢ 7, and the visible ends between the lines a b, ij. Let «, 3,7, ¢, No. 1, be points in the division of the teeth in the cir- cle a, b, ¢, d ; parallel to v n, draw « «, 8 8,7 7, c 0, meeting a b, No. 2, in the points «, 8,7, 0; then the projection of the teeth being marked out in No. 2, as here shewn, « By 3 is the projection of a tooth corresponding to « 8 y 3, No. 1. Figure 2 exhibits one of the projections of a bevel wheel, the axis of the wheel being oblique to the plane of projection. Here, instead of right lines or circles, the circles of this wheel are projected into ellipses, which will be found by Proposition XX VIIL., and the distances of the teeth, by drawing lines parallel to vn, from the points of division on the a 4, No. 2, fig. 1. Thus suppose m = (fig. 2) to be parallel to mn (fig. 1) No. 2, and the section of the wheel to be put in such a position as will make the axis M N form a given angle with the plane of projection. Perpendicular to m » draw 4 a, Bb, Q gq, intersectiug m » in a’, b, g ; prolong Q g tor; in Q r make ¢ r, ¢ s, each equal to @ A or @Q B, with the major axis 7 s and the minor axis @ describe an el- lipse a r bs, which is the projection of the circle of which the dia- meter is 4 B. The projections of the remaining circles will be found in the same manner. The points m, #, which are the projections of M, NV, the vertices of the cone are the points for drawing the projections of the edges of the teeth on the conic surfaces; and so on with the rest. 48 PROJECTION UPON THE LOWER PLANE exAmpLE FIFTH (Fig 1.) Exhibits the projection of a square-threaded screw. The principle upon which this projection is founded is the same as that of a line upon the surface of a cylinder, the line being a right line when developed upon a plane. It is evident that if the edge of a limber surface (a piece of paper for instance) be cut straight, and the surface tightly enveloped upon a cylinder, the edge which was straight will become a curve, which will either be a circle or a spiral, according as the straight edge is applied perpendicularly or obliquely to the meridians. This curve will evidently intersect the meridians at equal angles, and make as many revolutions as the number of times it comes to the same meridian, and this meridian will be divided into as many equal parts as the revolutions are in number. The method of projecting such a line has already been shown (Prob. xxvi.); but more particularly as follows :— From any convenient point J, with a radius equal to that of the cylinder, of which a portion of the curved surface is the outside sur- face of the screw, describe the circle 4 C EF G ; and from the point Z, with a radius equal to that of the cylinder, of which a portion of the curved surface is the receding surface between the threads of the screw, describe the circle J LZ N. Divide the circumference A CE G into any number of equal parts A B, B C, C D, &c. (as eight), and draw the radii 47, BI, CI, &c., intersecting the circumference J L N in the points J, K, L, &c. Perpendicular to Y Z draw Z 7; in Z T make Z Q, Q R, RS, &c., cach equal to the distance which the spiral makes upon the meridian in one revolution ; divide Z @ in- to as many equal parts at the points 1, 2, 3, &c. as are in the circum- ference of the circle A CE G (viz., into eight); parallel to Y Z draw 1b, 2 ¢, 3d, &c., and parallel to Z T'draw 4 a, Bb, C ¢, Dd, &c., and through the points @, b, ¢, d, &c. draw a curve abc def gk, which is the projection of the first revolution. The second revolu- tion indicated by a’ ¥' ¢ d ef" ¢' /' is found in the same manner, and so on for as many revolutions as may be required. The cylindric surface on the outside of the thread of the screw being compressed by two indentical spiral lines intercepting equal portions of the meridians, the projections of these spirals will also intercept equal portions of the projections of the meridian; therefore, the projection of the remain- ing spiral will be found by setting equal distances from the projection of the spiral already found upon the meridians downwards. The pro- jections of the two spirals forming the recess of the screw will be found in a similar manner, viz.: the first from the inside circle J Z NN, and the line Z 7, and the second by setting the distance between the threads downwards as before. The projection of the upper spiral of the interior cylindric surface is shown by the curve line j 2 7m n... The triangle & V W is the development of the upper spiral of the outside cylindric surface, and the triangle « By is the development of . the upper spiral of the inside cylindric surface. » APPLIED TO ARCHITECTURE. 49 As all buildings stand upon a horizontal plane ; the faces of walls in order to insure their stability, must be vertical, and the surfaces of the floors, in order to be easy for walking upon, must be horizontal. As all rectangular floors are more convenient for furniture than those of other figures, and as a rectangular parallelopiped can be divided into others withcut loss of space, this figure is the most economical form of building, and being less expensive, is more generally adopted. Moreover, solids of this form, having all their planes parallel to the three faces of a right-angled solid angle, are much more easy to pro- ject than other figures of solids. The roofis a covering to the building extending over all the parts of the interior in order to protect the inhabitants from depredation or incle- ment weather, the face or faces of the roof ought to consist of one or more planes inclined to the horizon at a sufficient angle to discharge the rain and snow water. Every inclined plane should rise from a horizonal line parallel to the face of the adjacent wall. The lower edge of the inclined plane of a roof is called the eave. When the roof consists of more than one inclined plane, the eaves should be in one horizontal plane. When two adjacent inclined planes meet each other their line of meeting is called the Zip line. When two inclined planes, rising from opposite parallel walls, meet each other, their line of meet- ing is called a ridge line. Roofs of rectangular buildings which have two inclined planes rising from opposite walls, have one ridge; and roofs which have four inclined planes, have four hip lines and one ridge line, or none, according as the sides of the building are unequal or equal. As the inclined planes of the same roof make equal angles with the horizon, the ridge line, if any, will be parallel and equi-distant from the two horizontal lines forming the tops of two exterior parallel faces of opposite walls; moreover, a vertical plane passing through any hip line will bisect the right angle contained by the two horizontal lines of the tops of the exterior faces of two adjacent walls, the vertex of the angle being the foot of the hip line. From what has been now explained it is evident that the plan of a house of which the roof consists of two equally inclined planes is a rectangle, of which the breadth is bisected by the projection of the ridge line; and that the plan of a house of which the roof consists of four equally inclined planes is a rectangle, of which the angles are bi- sected by the projections of the hip lines, and the ends by the projec- tion of the ridge line, of continued. When the roof rises from twe opposite parallel walls, the house is said to be gable ended ; and when the inclined planes rise from two or more adjacent walls, the roof is called a Zip roof, and the house is said to be Aip-roofed. The right lines on the plan of a house, which are the projections of the faces of walls, are called wall lines. The heights of the walls, the ridge and chimney shafts, may be set upon any convenient right line standing upon the ground line perpendicularly, as also the heights of windows. 50 PROJECTION UPON THE LOWER PLANE. [PLATE 14. Ficures 1, 2, 3, 4, Exhibit the plans and elevations of two rectangular houses, one being gable-ended and the other hip-roofed, making, with regard to their position to the vertical plane of projection, four different exam- ples. In order to save room, these houses are all of the same dimen- sions. No. 3 is a common tranverse section, cut by a plane perpen- dicular to the ridge line. Each of these houses is symmetrical, either in respect to a vertical plane passing through the ridge, or perpendi- cular to the ridge. ConsTrUCTION OF THE TRANSVERSE SECTION (No. 3). Draw the right line 1-4, and through the point 1 draw » 8 perpen- dicular to 1-4. In 1-4 make 1-2 equal to the height of the walls, and through the point 2 draw « 6 perpendicular to 1-4. Make 1 «, 1 B each equal to half the breadth of the house, and parallel to 1-4 draw « 3, 8 meeting ¢ § in the points 3, y. Make 2 ¢, 2 0 each equal to half the breadth, and 2-3 equal to the height of the roof, and join 3 ¢ 30. Make 3-4 equal to the height of the chimnies above the ridge of the roof, and through the point 4 draw V J parallel to « g. Make 4 V, 4 W equal each to half the breadth of the chimney, and parallel to 1-4 draw V U meeting 3 :in U, and W X meeting 3 6 in X. Then U V W X is the elevation of the chimney shaft parallel to the plane of the section, «8 3 is a transverse section of the roof, » J a sec- tion of the exterior face of the wall, and 8 y a section of the exterior face of the opposite wall. CONSTRUCTION OF THE PLAN oF THE GABLE-ENDED HoUSES (Figures 1 and 3). Draw the right line 4 B, and make A B equal to the length of the house. Draw 4 D and B C perpendicular to 4 B, and in 4 D make A P, P D, each equal to 2¢, or 2 ¢ No. 3, and draw P Q, D C, parallel to A B, meeting B C in the points Qand C. Againin 4D make P G, P O, each equal to 1 « or 13, No. 3, and draw the wall lines G H, O I parallel to A B, meeting B Cin H, I. Again in 4 D make P V, P6, each equal to 4 V or 4 W, No.3, and draw V K, 6 Z, parallel to 4 B, meeting B Cin K, L. In V K make V W, RJ each equal to the thickness of the chimney shaft, and describe the pa- rallelograms VV W X 6, J K L 7. The plan of the house being now complete, A B, D C, are the projections of the eaves, E F terminated by the chimnies, the projection of the ridge, VW X6, KJ 7 L the projections of the chimnies, 4 D, B C, the wall lines of the exterior faces of the end walls and gables, G H, O I, wall lines of the front and rear. ConsTRUCTION OF THE PLAN oF THE Hrip-rRooFED HOUSES (Figures 2 and 4). Draw the right line A B, and make 4 B equal to the length of the house. Draw A D and B C perpendicular to 4 B, and in 4 £) make : PL ATE. [7 oS [1 ~~ 2 ™~ N Nol. 1 L 7 i tytucddls oo bye dea PLATE IZ No:3, ii] PLATE 14.] APPLIED TO ARCHITECTURE. ol A P, P D, each each equal to 2 ¢, or 2 8, No.3, and draw P Q, D C parallel to 4 B, meeting B Cin the point Q. In P Qmake P IL, Q F, each equal to P Aor P D, and jom EA, ED, F B, FF C. Then EA, ED, F B, F C, are the projections of the hip lines, £ I the projection of the ridge line, 4 B, B C, C D, D A, the projections of the eaves. In P @Q, make P 3, Q 4, each equai to ¢ 3, or y ¢ (No. 3) ; through the points 3, 4, draw G' O, H I parallel to 4 D, meet- ing EAin G, ED in O, F Bin H,and F Cin I, and join G H, OI; then GH, HI, I O, O G, are the wall lines. In E F make E N, F R, each equal to the distance of the outside face of each chimney shaft, from each end of the ridge, and through the points IV, R, perpendicular to 4 B, draw the right lines V6, K L. Make R K, R L, each equalto 4 V,or 4 W, No.3; draw V K, 6 L, parallel to A B; make V W, K G each equal to the thickness of the chimney shaft, and describe the parallelograms V W X 6, J K L7, which are the projections of the chimnies. EXAMPLE FIRST (Fig. 1). The plan of the gable-ended house being given, and the vertical plane of projection parallel to the ridge line to find the elevation. Draw the ground line Y Z parallel to the ridge line & F'; draw A a and B t perpendicular to ¥ Z, meeting it in ¢/, ; in g’ « make 9’ p and ¢' a respectively equal to 1-3 and 1-2, No. 3, and draw P 9 ab, parallel to ¥ Z, meeting B ¢in g and b. Make p « equal equal to 3-4 (No. 3), and in @ » make x v equal to V U, or W X, No. 3. Draw v & parallel to Y Z, meeting B ¢ in £, and perpendicu- lar to ¥ Z, draw W x, J s intersecting « ¢ in 2, sand v % in w, j, and the parallelograms » w x a, kj st, are the projections of the chimnies ; e f being the projection of the ridge, « b that of the eave belonging to the front. EXAMPLE sEcoND (Fig. 3). The plan of the gable-ended house being given, and the ridge line making a given angle with the vertical plane of projection to find the elevation. Place the plan in such a position that the ridge line £Z F may make the given angle with the ground line ¥ Z. Draw Q 4 perpendicular to Y Z, intersecting it in 1; in the right line 1-4, make 1-2, 2 ¢ re- spectively equal to 1-2, 2-3, No. 3, and through 2 draw « c¢ parallcl to ¥ Z. Perpendicular to ¥ Z draw 4 a, B b, C ¢, intersecting acin b. Join b ¢, ¢ ¢, and complete the parallelogram a b ¢ p. Draw G g, H hk, Ii, perpendicular to V Z, intersecting @ ¢ in g, h, t.— Draw Vv perpendicular to ¥ Z intersecting a p in v, and draw v k pa- rallel to ac, intersecting b ¢ in k, Draw W w, J j, perpendicular to Y Z, intersecting vk in w,j. Prolong V v to «, and make v « equal to U V No.3. Draw ou parallell to ac ; draw X 8 perpendicular to Y Z, intersecting p ¢ in 5, and 2 « in B, and prolong W w to 52 PROJECTION UPON THE LOWER PLANE. [PLATE 14. meet « in a. Prolong Jj, Kk tomeet w wins, ¢ and draw ZL u per- pendicular to ¥ Z, intersecting ¢ ¢ in /. EXAMPLE THIRD (Fig. 2). The plan of the hip-roofed house being given, and the ridge line pa- rallel to the vertical plane of projection, to find the elevation. Draw the ground line ¥ Z parallel to the ridge line Z F. At a distance from ¥ Z, equal to 1-2, No. 3, draw the right line a b paral- lel to ¥ Z, and at a distance equal to 2-3, No. 3, the height of the roof draw ef parallel to @ 4. Perpendicular to ¥ Z draw 4 a, E e, Ff, Bb, and join @ e and b f. Perpendicular to ¥ Z draw G g, H k, intersecting a b in g, hand ¥ Zin g’ &'. Then g’ g kk is the pro- jection of the face of the wall, and a & fe that part of the roof that is seen. Perpendicular to ¥ Z draw K ¢, intersecting e fin 7 : make 7 ¢ equal to 3-4, No. 3, and in K ¢, make ¢% equal to U V, or W X, No.3. Through ¢ draw « # and through % draw v % parallel to @ b.— Draw V v, Ww, J j, perpendicular to ¥ Z, meeting v & in v, w, j, and prolong Vv, Ww, Jj, to meet wx tino, x, s, and v w xo, kj s ¢, are the projections of the chimney shafts. ExAMPLE FouRTH (Fig. 4). The plan of a hip-roofed house being given, and the inclination of the ridge to the vertical plane of projection to find the elevation. Place the plan in such a position that the ridge line may make the given angle with the ground line. At a distance from Y Z, equal to the height of the walls, draw the right line a ¢ parallel to ¥ Z, and at a distance from a ¢, equal to the height of the roof, draw e f paral- lel to @ ¢. Draw 4 a, Bb, C ¢c, perpendicular to Y Z, intersecting a c¢ in b, as well as ina and ¢; perpendicular to ¥ Z draw E'e, Ff, and join ae, bf, cf. Perpendicular to YZ draw G g, H k, I i, meeting acing, h,t, and YZ in g, /, 7. Perpendicular to Y Z draw R 7 in- tersecting ef in 7. Having prolonged Z K on the plan to meet 4 B “in 5; draw 5-5 perpendicular to Y Z, meeting a ¢ in 5, and join 5 7. Draw Kk perpendicular to Y Z, to meet 5 » in 2. Prolong K % to ¢, and make % ¢ equal to U V, or W X, No.3. Through ¢ draw « u parallel to @ c, and draw % v also parallel to @ c. Perpendicular to Y Z draw Vv, Ww, Jj, meeting k vin v, w, j. Also perpendicular to Y Z draw L u, and X g intersecting e fin, ¢, and prolong V v to «, as also W w to meet « % in , and prolong L I to u, J j to s.— Draw w § parallel to 2 r, meeting e fin 0. Parallel to Ww draw X 3, intersecting « u in f3, and ef ing, andv wb: Bx a,y kriut s are the chimney shafts. The breadths of the windows and of the piers are divided upon the wall lines G H, and the heights of the roof and chimnies, as well as the heights of the windows, may be set upon a line perpendicular to ¥ Z. PE } Po PLOT x Gooonn | |0000O0 Init 000008) IRIE, [7.000000 anuonn {mm il I TPA pO a Rp A TOOT LOnnng _ 0000 0H PLATE 15.] APPLIED TO ARCHITECTURE. 53 exampLE FIFTH (Fig. 1). Exhibits the two projections of a series of houses regularly arranged in the perimeter of a polygon.* The plans and elevations of the houses are delineated in a similar manner to those in plate 14, being individually of the same form, and need, therefore, not be again de- scribed It so happens, however, that the inclined faces of the roof are tangent planes to the surface of'a right cone upon meridian lines at regular distances from each other, and that vertical planes passing through the meridian lines would divide each chimney shaft into the symmetrical parts; therefore, the projection of the intersection of the faces of the chimney shafts, and the inclined planes of the roofs, will be parallel to the meridians of the conic surface. Hence, in order to find the projections of these meridians, conceive the circle inscribed within the polygon, formed by the eaves upon the convex side of the houses, to be the base of the cone, and let @ b in the elevation be the projection of this base, and C upon the plan be the projection of the centre of the circle. By this means the projections of the intersec- tions of the faces of the chimney shafts with the inclined planes of the roofs may be found in a more accurate and more expeditious manner than can be obtained from the projection of single or unconnected objects. Draw C ¢, bisecting the ground line Y Z, perpendicularly in d ; make the angle d a ¢, or d be, equal to the number of degrees con- tained by the planes of the roof and the horizon, and the point ¢, which is the projection of the apex of the cone, is the directing point to which all the meridians tend. Or, if the height of the roof be a certain equal part of the breadth building, as here one-fourth, divide @ binto four equal parts, and make d ¢ equal to one of them, and thus the point ¢ will be found. EXAMPLE SIXTH. (Fig. 2.) Exhibits the plan and elevation of a crescent, or of a series of houses having their principal fronts arranged in the arc of a circle. In this figure the faces of the walls and those of the roofs are surfaces of re- volution, the faces of the walls being cylindrical and those of the roofs conical. The elevation exhibits the concave surface of the building ; hence, the apex of the cone will be under the base. Let a b in the elevation be the projection of the base of the cone; bisect ab in d, and draw d ¢ perpendicular to @ 6. Make the angle da c, or db ¢ equal to the inclination of the roof, and the point ¢ is the directing point for drawing the meridians. _ The windows, in both figures, are divided upon the wall line, which is here a dotted line, in a circle concentric with that of the eaves. * The polygon, in Somer’s Town, London, h is a series of houses arranged in the perimeter of a polygon. 54 PROJECTION UPON THE LOWER PLANE. [PLATE 16. EXAMPLE SEVENTH (Figure 1.) Exhibits the plan and elevation of a circular portion of a wall pro- truding from a straight wall, the circular part having three windows, one in the middle, and also in the middle of the solid between the other two windows, the vertical plane of projection being parallel to the straight wall. : Draw the right line S 7), and let ¢ be a point in § 7’ representing the middle of the chord. In S 7"make ¢ O, : W, each equal to half the chord of the circular part of the wall; perpendicular to O W draw ¢ , and through the three points O, ,, W, describe the arc O y W, and let « be the centre of the circle. In the arc O 4 W, make o 8 and y 2 each equal to the distance between the points in the middle of the breadth of each window, and from the point » with a radius less by the thickness of the wall than the radius of the arc O W, describe the arc UP Q R V, and draw the radii « £2, « y, « 5 intersecting the arc UP Q R V in the points P, Q, R. In the arc O y W make 8 E, B F equal to each other, so that the distance E F may be equal to the breadth of a window. Draw FE G and FF H parallel to 8 «; make FE G equal to the recess of the face of the win- dow from the exterior face of the wall, and through G from the centre « describe the arc IJ intersecting /* H in H. Make GI, HJ equal te each other, so that the distance Z J may be equal to the breadth of the sash frame ; draw I K, J L parallel to 8 «; make I K equal to the thickness of the sash frame, and from the centre « de- scribe the arc K L. Inthearc UP Q R V make PM, P N equal to each other, so that the distance J NV may be equal to the intended breadth of the opening of each window in the inside of the room, and join KM, L N. Then M KL N is the recess of the sash frame from the concave face of the wall under the window ; the part KZ J I re- presents the bottom of the sash frame, and the part A B CD repre- sents the part of the stone sill protruding from the convex surface of the wall. Perpendicular to the ground-line draw E e intersecting ¥ Zin ¢’; make ¢’ e equal to the height of the bottom of the window above Y Z; through e draw 1-3 parallel to Y Z, and draw Ff perpendicular to Y Z intersecting 1-3 in f. Draw Bb, Dd perpendicular to Y Z; prolong Bb to 1, and D d to 3; make 1 4 equal to the height of the sill, and draw b d parallel to 1-3. Draw C ¢ perpendicular to Y Z, meeting b d in ¢, and prolong Cc to meet 1-3 in 2. Then 14 ¢ d 3-2 is the elevation of the sill of the window. What remains to be drawn in order to complete the entire elevation of the circular protuberance is evident from what is already done. PLATE]G. PLATE 16.] APPLIED TO ARCHITECTURE. 55 FORMATION OF A SEMICIRCULAR ARCH-HEADED WINDOW. The intrados of the arch being a surface of revolution, the window placed in a straight wall, the axes of rotation perpendicular to the planes of the faces of the wall, the surface of revolution partly conic and partly cylindric, the conic and cylindric portions separated from each other by an annular plane, it is required to draw the plan of the window. CONSTRUCTION OF THE PLAN oF THE WINDOW. (Figures 2 and 3.) Let G' H be a portion of the wall line of the exterior face, and 4 B a portion of the wall line of the interior face, these two lines being pa- rallel. Make G' H equal to the breadth of the window on the out- side ; bisect G' H by a perpendicular 7 R; draw G' E, H F parallel to I R, and make G E equal to the recess of the window from the exterior face of the wall. Through E draw C D parallel to G' H, in- tersecting A Fin F, and I R in §; make § C, § D, equal to each other, so that C D may be equal to the diameter of the conic surface joining the annular plane, and make R 4, R B, equal to each other, so that 4 B may be equal to the diameter of the circle, which is the in- tersection of the conic surface and the interior face of the wall, and join C A,D B. Then G ECA BDF His the plan of the head of the window. EXAMPLE EIGHTH (Fig. 2). The axis of rotation being perpendicular to the vertical plane of projection, and the plan of the window being given, to find the eleva- tion. Draw I 7, intersecting ¥ Z, perpendicularly, and at a distance from Y Z, equal to the height of the springing beds of the arch, draw a b pa- rallel to Y Z, intersecting Rrinz; perpendicular to Y Z, draw A a, B b, and from ¢, with the radius ¢ @, or ¢ b, describe the semicircle a » b. Again perpendicular to Y Z draw C ¢, D d, intersecting @ b in ¢ and d, and from , with the radius 7 ¢, or 7 d, describe the semicircle ¢ s d ; and again perpendicular to Y Z draw FE e, F f, intersecting ¢ b in e and f, and from ¢, with the radius ¢ e or ¢ f, describe the semicircle etf. Divide the arc a » b into as many equal parts as the arch stones are in number, which ought always to be odd, so that one of the arch stones may stand over the centre of the arch, and from the points of division draw right lines on the outside of the curve to point towards the centre ¢, and complete the remaining joints as they appear in the figure, and the elevation No. 2, will be complete. 56 PROJECTION UPON THE LOWER PLANE. PLATE 16.] EXAMPLE NINTH (Fig. 3). The axis of rotation being oblique to the vertical plan of projection, and the plan of the window being given, to find the elevation. Draw the wall line 4 B, to make an angle with the ground line Y Z, equal to the inclination which the face of the wall has to the vertical plane of projection. Draw i b at a height above Y Z, equal to that of the springing of the arched head. Upon 4 B describe a semi- circle B-1-2-3-A ; divide the semicircular arc into as many equal parts B 1, 1-2, 2.3, &c. at the points 1,2, 3, &c. as the arch stones are to be in number, and draw the right lines 1 J, 2 K, 3 L, &c. to meet A B perpendicularly in the points J, K, L, &c. Perpendicular to Y Zdraw Bb, Jj; Kk, Ll, &c., intersecting 2 6 in the points 1, 2, 3, &e.; make 17, 2 &, 3 [, &c., above the line ¢ b respectively, equal to the ordinates J 1, K 2, L 3, &c., of the semicircle, and through the points b, 7, , I, &c., describe the semi-ellipse @ » b. Perpendicular to Y Z, draw I 7, and the point ¢ is the elevation of the vertex of the cone. Again, perpendicular to Y Z draw Ce, Dd, § s intersect- ing 4 b in the points ¢, d, # ; make u s equal to § C or SD, and with u s the semi-axes major, and ¢ d the axis minor, describe the semi- ellipse ¢ s d,and ¢ s d is the projection of the semi-circle, which is the in- tersection of the conic surface and the plane of the masonry upon which the window frame adjoins. Again, perpendicular to ¥ Z draw Ee, Ff; in u s make ut equal to SE or § F, and with ¢ the semi-axis major, and ef the axis minor, describe the semi-ellipse e ¢f, and eZ f is the projection of the circle, which is the intersection of the cylin- dric surface and the plane of the masonry upon which the window frame adjoins. The elevation of the circle, which is the intersection of the exterior face of the wall, and the cylindric surface will be found in a similar manner. Let R r intersect « b in v, then v is the projection of the centre, and bj % [...a the projection of the circumference of the circle which is the intersection of the conic surface and the interior face of the wall, and from the same points j, &, /, &c., draw right lines on the outer side of the curve, pointing to » for the joints of the arch stones in the face of the walls, and from the same points j % / draw right lines to intersect the curve d sc pointing to v; from the points of inter- section in the curve d s ¢ draw right lines pointing to « to intersect the curve e ¢ f; and draw H A perpendicular to Y Z, intersecting i b in A. Make each of the parallel lines, from the curve e ¢f; equal to hf, and a curve drawn through the points, as shown, is the projection of the circle, which is the intersection of the cylindric surface and the exterior face of the wall, and by drawing the joints of the stones, as shown, the result is the elevation of the window. PLATE 17.7] APPLIED TO ARCHITECTURE. 57 EXAMPLE TENTH. To find the plan and elevation of the portion of a building which has a saloon in the centre. The interior faces of the walls of the sa- loon rise from the sides of a regular octagon, described on the level of the floor, and continue upwards to meet a spherical surface so as to form semicircular intersections; from the termination of the spherical surface a cylindric surface rises to a certain height, and the cylindric surface is surmounted by a spherical surface of the same radius as that of the cylinder; the spherical surface is terminated by a circular aperture for the purpose of giving light to the interior. The centres of the two spherical surfaces, and the axis of the cylinder, are in the vertical line passing through the centre of the octagon. Each of the walls on the sides of the octagon is pierced with an aperture for pas- sage. The aperture has its jambs or sides in vertical planes, and its soffit in a cylindric surface. The planes of the jambs, and the axis of the cylindric surface, are perpendicular to the surface from which they recede. The cylindric surface intersects the plane of the wall in a semicircle, of which the diameter terminates upon the two vertical lines, which are the intersections of the jambs, and of which the centre is also the centre of the semicircle formed by the intersection of the vertical face of the wall and the spherical surface. These apertures will, therefore, form arcades, which are eight in number, and all of the same width. The cylindric part of the saloon is pierced with windows, of which the bottoms terminate in one horizontal plane, and the tops in another. The meridians of the cylinder, which divide the windows into two symmetrical parts, divide the circumference into as many equal parts as there are windows, and as the widths of the win- dows are all equal, the breadths of the piers are also equal. Each of the arcades in four of the walls, which are not adjacent, opens into two aisles or passages, and each of the arcades on the remaining four sides opens into one aisle, so that the eight arcades open into twelve pas- sages in such a manner that of two vertical planes intersecting each other in a line passing through the centre of the octagon, either one divides the whole construction into two symmetrical parts. Both sides of each of six of the aisles are parallel to one of the planes, and both sides of each of the remaining six are parallel to the other; hence the directions of three of the aisles upon one side are at right angles to those of the three aisles upon the adjacent side, and in right lines with those of the three aisles upon the opposite side. The four aisles through which the planes pass have equal widths, but are wider than the other eight which have also equal widths. Each of the four arcades through which the intersecting planes do not pass, opens into a vestibule, of which the surface of the wall is cylindric, and that of the ceiling spherical ; and each of two of the aisles at right angles to each other enters from the vestibule, intersects the cylindric surface in two vertical lines, and the spherical surface in a semicircle, which has 58 PROJECTION UPON THE LOWER PLANE [PLATE 17. its centre in the middle of the line, joining the upper end of the two vertical lines and the diameter at right angles to these lines.* Let 1 be the centre of the octagonal plan. Through 1 draw Z J in any convenient position, and draw K ZL perpendicular to 7 J.— Make 1 7,1 K,1 J, 1 L, each equal to (30 ft.) the distance of the centre of the octagonal base from the middle of each side; through the points Z, J, draw M C, P N, parallel to K ZL, and through the points K, L, draw M P, O N, parallel to IJ, and the quadrilateral figure O M P Nis a square. Draw the diagonals O P, M N, which will intersect each other in 1, and from JZ, with the radius M I (30 ft.), cut M 1 ine From IZ, with the distance : 1 (equal to 12ft. 5in.{}), cut M O in 4, H; from K, with the same radius cut M P, in G, F; from J, with the same radius, cut P Nin D, FE, and from Z, with the same radius, cut O Nin B, C. Join AB, C D, E F, G H, and the figure A B C D EF G H is a regular octagon, of which the sides (each equal to_24ft. 10in.) are the wall lines of the interior faces of the saloon. Parallel to 7 J, draw the ground line ¥ Z, and perpendicular to Y Z, draw the right line 1 w, intersecting ¥ Z in 2. In 2 w, make 2 f equal to (R4ft.) the height of the walls, and through g draw a d pa- rallel to Y Z. Parallel to 2 w draw 14, Jj, intersecting a dina, d; make a ¢ equal to (12ft. din.) half the side of the octagonal base, that is equal to the heights of the semicircular arcs, which are the intersections of the vertical faces of the walls and the spherical surface above, the ra- + Such a building as has now been described is similar to the grand saloon of St. Paul’s Cathedral, London. On each side of the octagon, which are portions of the sides of the square, are three avenues, of which the middle one is the nave, and the two, one upon each side of the nave, are the aisles; three of the avenues are ar- ranged in the length of the building, and three, which form the transepts, in the breadth. The sides of the naves, and those of the aisles, are formed into beautiful ranges of arcades. The usual practice of decorating arcades is to form an archivolt surrounding the intrados of the arches, to terminate either upon the entablature of a complete order of architecture, with pilasters returning upon the jambs, or upon imposts surmounting vertical architraves on the des. x + It is evident from the construction that the side of the octagon is equal to the difference between the side and diagonal of the square, and, therefore, half the side of the octagon equal to the difference between half the side and half the diagonal of that square, and that MM I 1 is an isosceles right angled triangle ; therefore, MI=o/(TT1° + T12)=y/ (2X 1° )=+/ (2X30?) =+/(1800) = 4242 ft. From 42ft. 5in.( =42-42) subtract 30ft., and the remainder 12ft. in. is equal to half the side of the octagon, being also equal to the heights of th icircular arc above the springing line. 4 : Tn he a triangle 8 a i, the side « 8 is equal to half of i j, or half of IJ (equal to 30 ft.) and the height a i equal to the height of the semicircular arcs ; therefore, 3i=+/(a #2 +71%)= v/ (30% + 12422 ) = 32-46ft. =32ft. 53in. which is the radius of the spherical surface. From this the radius of the cylindric surface above may be found; for, suppose @ s to be joined, and 2 I prolonged, to meet s ¢ in 0; then, 3 s ¢ is a right-angled triangle, the sides containing the right-angle being 8 0, 0 s: now 8 s=8i 32:46 and £8 0=8 47 6=1242+2=14-42; there- fore, s §=A/(B+2—B 82)=V (32162 —14-427 )=29-08 feet, which is the radius of the concave cylindiic surface of the wall. LLATE TT. FATE 17. r PLATE 17.] APPLIED TO ARCHITECTURE. 59 dius of the semicircular arcs being equal to half the side of the octagon. Draw 4 parallel to ¥ Z, and join 8¢, 8 j. From the point g, with the radius g ¢, or Bj (32ft. 5lin.) equal to the radius of the spherical surface, describe the arc i s 5 ¢ j, intersecting 2w in 5. At a given height (say 2ft.) above ¢j draw s ¢ parallel to ¢ 7, intersecting 2 w in ¢, and draw s wu, ¢ v, parallel to 8 w. Make s u equal to (25ft.) the height of the cylindric wall, and draw » v parallel to s &. Upon u v, as diameter, describe the semicircle # z w y v, and draw the chord zy parallel to « v, so that # y may be equal to (18ft.) the diameter of the cylindric aperture for the passage of light. Perpendicular to ¥ Z draw Rr, Ll, Q ¢, meeting i jin r,1, q, also intersecting a d in 4, 3, 4, and perpendicular to ¥ Z draw B b, C ¢, meeting a d in b, c. From the point 8 with the radius 8 & or 8 ¢ describe the semicircle b lc; with the semi-axis major « 7 and the minor axis « b, describe the semi-ellipse @ 7 b. Similarly describe the semi-ellipse ¢ ¢ d. Then the semicircle B I cis the projection of the intersection of the face of the wall upon B C and the spherical surface ; being the projection of a semicircle in a plane parallel to the plane of projection ; the semi- ellipse a b is the projection of the intersection of the face of the wall upon A B and the spherical surface, being the projection of a semi- circle in a plane perpendicular to the horizontal plane of projection ~and oblique to the vertical plane of projection. The same must be observed with regard to the semiellipse ¢ ¢ d. The projections of all the remaining semicircles which are formed by the intersections of the horizontal semicylindric surfaces over the openings, and the faces of the walls are projected in the same manner, being either in the planes rising from 4 B, B C, C D. or in planes to which these lines are parallel. The projections of the semicircles standing over 4 H, D I, being in planes perpendicular to both planes of projection are right lines perpendicular to the ground line,and thus si arblcqgdj¢ is the projection of the concave portion of the spherical surface between the interior faces of the walls and the vertical concave cylindric sur- face, and if from the centre 1 on the plan with a radius ¢ s or 4 ¢ the circle 8 & 7 be described, the space between the polygon A B C, &c., and the circle S' & 7 is the projection of the spherical surface on the plan. The projections of the windows on the elevation will be found from the plan by dividing the circumference of the circle S 7, &c., into parts or chords, which contain the breadths of the windows and the breadths ‘of the piers; the breadths of the windows being eal to one another, and the breadths of the piers equal to one ano- ther. 60 PROJECTION UPON THE LOWER PLANE [rLATE 18. EXAMPLE ELEVENTH. Find the plan and elevation of an octagonal building, the ceiling being a spherical polyhedron. Every side of the building is pierced with a window in the middle. The plane of the window is parallel to the interior face of the wall. In order to admit of an uninterrupted passage for the light, a recess from each window to the interior sur- faces is formed by four vertical and two horizontal surfaces, one of the horizontal surfaces being the bottom of the recess and the other the ceiling, which is a figure contained by three right lines perpendicular to each other and a semicircular arc concave towards the middle right line. The middle right line is adjacent to the plane of the window, and the extremities of the other two, which are not adjacent to the plane of the window, terminate in the interior face of the wall. The four ver- tical surfaces are, therefore, three plane figures and the concave sur- face of a cylinder of which the axis must ih consequence be vertical. The surface of the spherical polyhedron which forms the ceiling, is formed of eight quadrantal portions of the surface of the same cylinder, so that a vertical plane passing through the axis of the octagonal walls perpendicular to any one of the eight faces of the building, will cut the axis of the cylinder perpendicularly, and the surface of the ceiling so as to make a semicircular section. It is evident that the axis of the cylindric surface, adjacent to any interior face of the build- ing, will be horizontal and parallel to that face, and that every two cy- lindric surfaces will meet each other in a vertical plane, bisecting the dehedral angles formed by the interior vertical faces of the walls, and that the figure in which they meet will be a quadrantal portion of an ellipse comprised between the two semi-axes, and that since the ellip- tic curves are in vertical planes, their horizontal projections are right lines, being halves of the diagonals which pass through the centre of the octagon, and their vertical projections quadrants of an ellipse, hav- ing ene common vertical semi-axis major equal to the radius of the cy- linder or to the semicircular section. Now in referring to the plan 5-6'-7-8-9'-10'~11'-12" is the octa- gon formed by the wall lines of the interior faces, 56, 6-7, 7-8, &c., being the sides. The figure is described as in the preceding ex- ample, X being the centre. Join 5X, 6 X, 7'X, &c., and these lines are the projections on the plan of the meetings of the cylindric sur- faces, two and two, which form the ceiling. Through X draw WY’ parallel to the side 6-7’ of the octagon, meeting the side 5-12’ in W, and the side 8-9’ in 17, and parallel to W ¥” draw the ground- line Y Z. Perpendicular to V Z draw Ww, Xz, Y'y, and parallel to V 7 at any convenient height above Y Z draw w y, intersecting X x in r. From », with the radius » w or 7 y, describe the semicircle w z y, which is a section through the ceiling upon the line # Y” in the plan. In the semicircular section 2 y, the one half w a is also the vertical pro- jection of the intersection of the two cylindric surfaces of which 5 X is the horizontal projection, and the other half @ y is the vertical projec- _ PLATES. - d hi a = Soi o : > : = os Ls ” ho / o / . t 3 / il Rm ATT PLATES. Ll a jp “Zi eg, PLATE 18.] APPLIED TO ARCHITECTURE. 61 tion of the intersection of the two cylindric surfaces of which 8’ X is the horizontal projection. Perpendicular to ¥ Z draw 6'-6, 7-7, meeting wy in 6, 7- With the semi-axis major » 2, and the semi-axis minor r 6, describe the quadrantal part 6 x of the curve of an ellipse, and with the semi-axis major » x, and the semi-axis minor » 7, describe the qua- drantal part 7 x of the curve of'an ellipse. Then the curve 6 x is the vertical projection of the intersection of the two cylindric surfaces of which 67 X is the horizontal projection, and the curve 7 x is the verti- cal projection of the two cylindric surfaces of which the horizontal pro- jection is 7” X. So that w x 6,6 27, and 7 x y are the projections on the elevation of three of the cylindric surfaces ofwhich 5 X 6/, 6 X7,7 X8, are the projections on the plan. Now let A K on the plan be the breadth of a window in the side '-6’ of the octagon, ZL. W the breadth of a window in the side 6’-7, and let A K and Z W be equal to each other. Draw X . bisecting L W perpendicularly in «, and from the point » with the radius a Z or a W, describe the semicircular arc L © W intersecting « X in Q. From the centre X with the radius X » describe the arc « ¢, and draw Qo perpendicular to = X. In the arc « 4 take any number of points 3, v, 3; & and perpendicular to » X draw 8 1, 5 2, 33, ¢ 4, meet- « X in the points 1, 2, 3, 4. Perpendicular to » X draw 1 M, 2 N, 3 0, 4 P, meeting the semicircular arc Z @ W in the points M, N, 0, P. Inr x make rs, rt, ru, rv, 7 gq, respectively equal to 18, 2+, 33, 4¢ Q¢, and parallel to Y Zdraw bs,¢ ¢,d u,e v, f gq. Perpen- dicular to Y Z draw Ll, Mm, Nn, Oo, P Ps Qq, intersecting w y, bs, ct, du, ev, fq, respectively in, m, n, 0, p, ¢q, and draw the curve I mn o pq, and this curve is half the projection of the intersection of the vertical cylindric surface of the middle window, and the cylin- dric surface of the ceiling of which the axes is horizontal. Upon A K, as a diameter, describe the semicircle 4 F K; make the arcs AB, A C, AD, &c. respectively equal to the arcs LZ M, LL N, I 0, &c., and draw the chords B J, C I, D H, &c. parallel to A K. Draw Aa, Bb Cc, Dd, Ee, F f, &c., perpendicular to V Z, meeting wy, bs, ct, du, ev, fq, in the points a, b, ¢, d, e, f, &c., and draw the curve a b ¢ d ef....k which is the projection of the intersection of cylin- dric surfaces in the oblique face. The whole vertical projection of the window in the oblique face, and half of the vertical projection of the window in the parallel face, on the left hand side of the vertical line X 2, being now obtained, and, since both sides of the elevation are symmetrical, the entire elevation will be found by transferring the curves to the right band side of the line of gymmetry X z. The same is to be understood in every symmetrical elevation in the following ex- ample. 62 PROJECTION UPON THE LOWER PLANE. [PLATE 19. EXAMPLE TWELFTH. Find the plan and elevation of a building having as in the last ex- ample the interior faces of the walls in the form of an octagonal prism, the ceiling a corresponding spherical polyhedron, and the same num- ber of windows. The windows have semicircular arched heads, formed each by surfaces of revolution, so that each axis of rotation may pass through the centre of the spherical polyhedron perpendicu- larly to the face of the wall through which the aperture of the window passes. The surface of revolution of each window is terminated at one end by the plane of the exterior face of the wall, and at the other by the concave adjacent cylindric surface of the ceiling; the exterior face of the wall will, therefore, be in a plane, perpendicular to the axis of rotation. The surface of revolution consists of two concave sur- faces, separated from each other by an annular plane figure, upon which the vertical plane of the window rests. The concave surface next to the exterior face of the wall is cylindric, and that next to the ceiling conical, the greatest diameter being next to the interior of the building. Thus the intrados of each window is identical in its general form to the semicircular arched headed window, page 55. The ceil- ing being the same form as the spherical polyhedron in the last exam- ple, in which its projection is shown, it will only be necessary to find the projection of the windows as regards their plan and elevation in the interior or concave side of the building. In order to find the projec- tion of the intersection of the conic surface of each window, and the adjacent cylindric surface, it is proper to observe that the axis of ro- tation of the window, and that of the cylindric surface, are in the same horizontal plane, and perpendicular to each other, Let X be the projection of the centre of the spherical polyhedron, and X will, therefore, be the point of intersection of the axes of the eight cylindric surfaces of the ceiling ; and since the interior plan of the building is a regular octagon, the axes will form angles of 45 de- grees, and will bisect the sides of the octagonal plan. Let ¢ ;" be the interior breadth of a window, and let 4 7 4-5” be the recess from the inside to the face of the window. Draw X 6’, bisecting ¢ j/ perpen- dicularly in «, and the right line X 6’ will divide the figure gj’ 4-5’ into two symmetrical parts. Prolong 5’ ¢ from the one extremity to and from the other extremity to meet X 6’ in the point 6, and pro- long 7 4 to 6. From X, with the radius X «, describe the arc « s, and draw ¢ P parallel to 47’, intersecting X 6 in 7. From 7, with the radius 7's, describe the semicircle ¢ O P, intersecting X « in 0; divide the quadrantal arc P O into any number of equal parts PY, 1-2, 2-3, 3’ P, and draw the right lines 1’ Q, 2’ R, 3’ §, pa- rallel to O 7, meeting P T'in the points @, R, S. Parallel to P 7. through the points 1’, 2, 3/, draw right lines intersecting O 7, from the points of intersection ; and from 7) as a centre, describe arcs in- tersecting « 7’, and from the points of intersection draw right lines to the point 6/, intersecting the arc « ¢ respectively in 8, , 3. Parallel to ¢ j/ draw BX, y U, ¥m/, intersecting « Tin %, v w; and join Q 6, PLATED. M) @. PLATE 19. PLATE 19.] APPLIED TO ARCHITECTURE, 63 R 6, S 6, intersecting 8 &, y I, & m/, in the points kU mw, and draw the curve j& ¢ m/ T...9, which is half the projection of the curve of the intersection of the conic intrados and the corresponding cylindric por- tion of the ceiling. Make each of the curves ABC D EF G H1, J KL M N...& so that each half may be symmetrical in respect of the right line bisecting the breadth of the window perpendicularly and identical to the curve j/ # I m' T; that is, so that if the curve J # Um T be applied to the curve 4 B C D EF G H I, the points 7s ®, Us m', T, may fall upon the points 4, B, C, D, E, and being re- versed, the points j/ 2 I T, may fall upon the points 7, H, G, F, E, and also that the curve j/ # 7 m/ T being applied upon the curve JK L M N, the points j', k, I, m, T, may fall upon the points J, K, LM, N. Parallel to J & draw the ground line ¥ Z, and perpendicular to Y Z draw X n. Parallel to Y Z draw @ 0 meeting X 7 in o, and in 0 » make 0-1, 0-2, 0-3, 0 n, respectively equal to ug, v y, w 8 Te. Parallel to Y Z draw 1, ¢ 2, d 3, e n, and parallel to X » draw 4 a, Bb, Cec, Dd, Ee, &ec.,and draw the curve a bed e fg h i, which is the projection of the intersection of the conic surface of the intrados of the window, and the corresponding cylindric surface of the ceiling in the oblique face of the walls. Moreover, perpendicular to Y Z draw J j, Kk, L I, M'm, intersecting a 0,5 1,¢ 2,d 3, en, in the points Fk l,m, n, and draw the curve j kl mn, which i is half the projection of the curve of the conic intrados of the window in the side of the building parallel to the vertical plane of projection. As the right line X n divides the elevation symmetrically, the projection of the other half and that of the remaining window may be easily found. 64 PROJECTION UPON THE LOWER PLANE. [PLATE 20. - EXAMPLE THIRTEENTH. Find the plan and elevation of an octagonal building, having a ceil- ing partly coved and partly plane. The coved part consists of eight concave cylindric surfaces, comprised each between two meridians; one of the meridians of each concave surface coincides with a right line in the plane of the inner face of the wall, and the other with a right line in the plane of the ceiling, and thereby forming two regular octagons, each in a horizontal plane; one upon the ceiling, and the other upon the inner faces of the wall. The windows are the same in number and form as in the last example. The curved surfaces of the window and the ceiling which intersect each other have a common tangent plane. The projections of the curves, which are the intersections of the cylindric surfaces, are here found upon the plan and elevation in a si- milar manner to those shewn in example eleventh. Let 4 7 be the breadth of the window in the inside, and let 45” 4-5 ; be the recess of the window, and let 1-2-3-4-5-6-7-8 be the projection of the octagon surrounding the plane part of the ceiling. Draw the right line O 6’ bisecting 07 in 2 and the side 6-7 of the octagon in the point &, and from & with the radius & « describe the arc = «. Let point 6” be the apex of the conic intrados of the window. Upon 6/ & as a diameter, describe a semicircular arc intersecting the arc « ¢ in ¢, and join 6’ ¢, which must fall upon the side 5’ ¢ of the window. Pa- rallel to ¢ j/ draw-¢ P, intersecting O 6” in 7, and join 7'¢, 75. From 7", with the radius 7 ¢, describe the semicircle : O P, and di- vide the quadrant O P into any number of equal parts, P 1, 1-2, 27-3, 8 0. Draw l' Q, 2’ R, 3’ §S intersecting : P perpendicularly in Q, R, S, and join Q 6’ R 6’ S 6, intersecting ;’ 7 in the points kK, I m'. Parallel to ¢ P, draw &'B, I, m'D, intersecting « 7 in the points %, v, w, and the arc « ¢ in the points 8, y, 3 Form the plan A B CD E... I of the oblique window, and the plan J KL M N... Y’ of the parallel window identical to the plan j” 2’ m’ T...0. Parallel to J Y’ draw the ground line ¥ Z, and perpendicular to Y Z draw the right line N n. Parallel to ¥Y Z, at the proper height of the level of the springing of the windows and the coves of the win- dows, draw a y intersecting N z in 0 ; make heights upon on respec- tively equal to u 8, v4, w 3, and through the points of division parallel to a y draw b k,c I, dm, e n. Perpendicular to ¥ Z draw 4 a, B J, Cece, Dd, Ee &ec,as also Jj, Kk LI, Mm, and the curves abedefghi,jklmn..ybeing drawn will complete the elevations of two windows. PLATE 20 PLATE 21.] APPLIED TO ARCHITECTURE. 65 EXAMPLE FOURTEENTH. Plan and elevation of a hemispherical dome, and a semi-hemisphe- rical niche recessed from the dome, showing the joint lines of the ma- sonry of the two spherical surfaces, the centres being on the same horizontal plane as the springing lines of the concave surfaces of the dome and the niche, and the vertical plane of projection being parallel to the plane in which the two spherical surfaces intersect each other. It is a general principle in masonry, at least in constructing walls or in building vaults formed by entire surfaces of revolution which have vertical axes, to form the joints so that aright line perpendicular to the exterior surface, whether plane or curved, passing through the meet ing of any two stones in that surface, may coincide with both surfaces which thus form their joint, and that the joints may be in sueh posi- tions as to cause one series of joint lines to be in horizontal planes, and the other in vertical planes. Therefore, if the exterior surface be a surface of revolution, and the axis of rotation be vertical, the joint lines in the horizontal planes will be parallel circles, and the joint lines in the vertical planes will be in meridional lines, or in circles, intersect- ing each other in the axis. The series of joints, which form the pa- rallel circles, will be conic surfaces, having their vertices in the axis of rotation. If the surface of the revolution be spherical, the conic surfaces of the joints will have one common apex in the centre of the spherical surface. As the intersection of two spherical surfaces is a cirele, and as the right line passing through their centres passes perpendicularly through the centre of this circle, and as the vertical plane of projection is paral- lel to the plane of the circle, and consequently the horizontal plane of projection perpendicular to it, the projection on the plan of this semi- circle will be a right line, bisected by the projection of the right line which passes through the centres of the two spherical surfaces. In fact, the projection of the intersection of the spherical surfaces on the plan will be a right line joining the two points of intersection of the two circles described, from centres at a distance from one another, equal to the distance of the centres of the two spherical surfaces with radii respectively equal to those of the spheres; moreover, the pro- jection of this semicircle, in the elevation, will be an equal semicircle. The axis of rotation of the spherical surface of the niche is in the right line which joins the centres of the two spherical surfaces, and is, therefore, perpendicular to the vertical plane of projection, and the planes of the beds of the stones and the axis of rotation are in the same plane ; hence the beds of the stones are in planes perpendicular to the vertical plane of projection, and having equal inclinations with one another, will be projected in the elevation into right lines, making angles with one another, equal to the angles of inclination of the beds of the stones, and the radiating lines will divide the semicircle, which is the projection of the intersection of the two spherical surfaces into as many equal parts as there are beds. The joint lines of the niche will thus radiate upwards, so as to meet the joints in the meridional arcs and those of the parallel circles. K 66 PROJECTION UPON THE LOWER PLANE [PLATE 21. Let @ be the projection of the centre of the spherical surface of the dome, and ¢’ the projection of the centre of the spherical surface of the niche, the points @, ¢’ being distant from one another, equal to the distance between the centres of the two spherical surfaces. From the point @, with the radius of the spherical surface of the dome, describe the circle « 4 O ¢' R, and from the point ¢/, with the radius of the spherical surface of the niche. describe the circle 4 £ ¢/, and through the intersections A, ¢, of these circles, draw the com- mon chord 4 ¢’, which is bisected by ¢’, and consequently 4 P ¢' is a semicircle. Parallel to 4 ¢’ at any convenient distance, draw the ground line Y Z, and prolong Q ¢ to intersect the semicircular arc 4 P ¢’ in P, and the ground line ¥ Zin ¢. From the point ¢, with the radius of the sphere surface, describe the semicircle ¥ ¢ Z, which is the pro- jection of the meridional circle, in the plane parallel to the vertical plane of projection. Parallel to Q ¢, draw 4 a meeting ¥ Z in a; from ¢, with the radius ¢ a, describe the semicircle a e ¢, and from the same point ¢, with the radius ¢ s, equal to the radius of the cen- tral stone of the niche, describe the semicircle s w 8.% Upon the semicircular arc Y'¢ Z, make the chords Y 1, 1-2, 2-3, 3-4, &c., each equal to the breadth of one of the courses of stone of which the joint lines are in the parallel circles, and draw the right lines 1-1, 2-2, 3-3, 4-4, &c., parallel to ¥ Z, meeting the arc Z : in the points 1, 2, 3, 4. As the number of stones, forming the head of such a niche, is al- ways odd, in order that one of the stones may be in the middle, divide the semicircle @ e ¢ into as many equal odd parts as the stones are in number (being ninet in this example) at the points a, b, ¢, d, e, &c., and draw the radiating lines 5 ¢, j , Iv, n w, &c., intersecting the semi- circular arc @ e ¢ in the points b, ¢, d, e, &c. and meeting the semicircular arc s w 4 in the points ¢, u, v, w, &c. and the projections 1-1, 2-2, 3-3, 4-4, &c., of the coursing joint lines in the points 5, 7, /, n, &c. The radiating lines between the two semicircles @ e @, s w ¢, are the pro- jections of the joints formed upon the concave spherical surface of the niche, being the projections of great circles intersecting each other in a right line perpendicular to the vertical plane of projection. As all the vertical joints of the stones of the dome intersect the spherical surface in quadrantal arcs of circles, and thus form meri- dians which have a vertical axis, these meridians, being projected into right lines upon the plane, will divide the circumference of the circle of the base of the concave hemispheric surface of the dome into equal parts, and will be projected upon the elevation into quadrants of ellipses, of which the semi-axis major will be common, and the semi- axis minor will be found upon Y Z, between the centre ¢ and the * This central stone in the form of the frustum of a cone has no particular English name ; but in the French language it is called Trampillion. + The thickness of the stones which form the head of the niche must be less than the thickness of the courses of the dome which meet the concave surface in parallel circles, so as properly to intersect each other. Bi (PLATE. 2] | 3 1 | 3 PLATE 21. » k PLATE 21.] APPLIED TO ARCHITECTURE, 67 point cut by a right line drawn parallel to Q ¢, from the end of the horizontal projection of the meridian in the circle of the base. Thus, let F Q be the projection of one of the meridians upon the plan; draw Ff parallel to @ ¢, meeting Y Z in f; then with the semi-axis major ¢s and the semi-axis minor ¢ f; describe the elliptic quadrantal curve fs, which is the elevation of the meridian standing over FZ Q upon the plan, and is the curve line in which the projections of the alternate joints fall. In the same manner the pro- jections of the remaining joints may be found. a Let fg, h 4,7 k Im, &c., be the vertical projections of the joint lines of the stones of the dome upon the vertical beds, and let 5¢ fu, lv, nw, be the vertical projections of the stones of the niche upon the radiating beds. On the plan draw the diameter « R parallel to the ground line Y Z, and from the points 1, 2, 3, 4, &c., in the arc Z ¢, perpendicular to Y Z, draw 1-1/, 2-2/, 3-3/, 4-47, &c., meeting o R in the points 1, 2, 3, 4, &c. From the centre Q, with the radii Q 1’, Q 2, Q 3, Q 4, &c., describe the arcs G H 1’, IJ2, KL3, MN¥, &c., which are the horizontal projection of the parallel circles. Perpendicular to YZ draw g G, hk H, i Lj J, RK,1L,m M,n N, and from F, H,J, L, draw F G, HI, JK, L M, radiating towards ©. Divide the semicircular arc below the diameter « R on the plan into nine equal parts as in the niche, and through the points of division, draw right lines, meeting « R perpendicularly in 8, vy, 3 ¢&, &c. With the semi-axis major @ O, which is common, and the semi-axes minor @Q 3, Qy, @ 3, Q ¢, respectively describe the elliptic curves g HB 0, JC O,3 LD O,: NE O. Perpendicular to Y Z, draw b B, ¢ C, dD, ¢ E, &c., meeting 4 ¢' in B, C, D, E, and with the semi-axis major ¢/ P, which is common, and the semi- axes minor ¢ B, ¢ C, ¢' D, ¢' I, respectively, describe the elliptic curves BT P, CUP, D VP, EWP. Perpendicular to Y Z draw s §, meeting the semicircular arc 4 P ¢’ in S, and through § draw a chord. Again, perpendicular to YZ draw ¢ 7, wu U, v V, w W, &c. Then in the two projections of the joints of the niche, as connected with the joints of the dome 8S AF G H B 7 is the plan, and sa fg hbt the elevation of the first stone above the springing surface, TB H IJ C U the plan, and ¢b k ij c u the elevation of the second stone, U CJ KL D V the plan, and ucj kl dv the elevation of the third stone, V.D L M N E W the plan, and v d ! m n e w the elevation of the fourth stone. 68 PROJECTION UPON THE LOWER PLANE [PLATE 22. EXAMPLE FIFTEENTH. Plan and elevation of a dome, of which both the exterior and inte- rior surfaces are surfaces of revolution, having the same axis of rota- tion, which, of course, is in a vertical position. Both the exterior and interior surfaces are spherical. The interior surface is that of a he- misphere, and the exterior is that of a segment of a sphere, less than a hemisphere, but of a greater radius than that of the interior surface, in order that the substance of the masonry comprised between the convex and concave surfaces may continually diminish in thickness from the bottom towards the pole or summit, The dome is lighted by a cylindric aperture through the top. The interior is decorated with coffers, in the form of spherical squares, receding from the con- cave hemisphere surface. These coffers are comprised by meridional and parallel circles, so as to form both horizontal and ascending rows. Each coffer has three distinct recesses, which are each in the form of a spherical square. The recesses diminish in breadth so as to leave equal margins all round. The coffers are divided from each other by zones, which are por- tions of the spherical surface, each being comprised by two parallel circles, and the ascending rows are divided from each other by lunes, which are portions of the spherical surface, each being comprised by two meridional circles. The coffers are all equal in any horizontal row, but in the ascending rows their breadth becomes less and less as they approach the pole or summit of the hemispheric surface. The breadth of each line is to the breadth of each coffer in the hemispheric surface in the ratio of 1 to 3. It will be evident, from the particulars explained, that the horizon- tal projections of the lunes, zones, and coffers, are comprised by radia- ting lines and circles, drawn from the centre, which is the projection of the centre of the sphere; the radii of these circles being respective- ly equal to the radii of the parallels of which they are the projections, and the vertical projections of these lunes, zones, and coffers, are com- prised by right lines drawn parallel to the ground line and quadrantal elliptic curves, the radiating lines on the plan being the horizontal projections of the meridians, and the quadrantal elliptic curves the vertical projections of the meridians ; moreover, the circles upon the plan being the horizontal projections of the parallel, and the lines drawn parallel to the ground line in the elevation the vertical projec- tions of the parallels. : 3 In this example the number of coffers in the circumference is twenty. Let B upon the plan be the projection of the centre of the hemispheric surface, and let ¥ Z be the ground line. From B, with the radius of the hemispheric surface, describe the circle 4 C D E, which is the horizontal projection of the circle at the bottom of the concave surface. Perpendicular to ¥ Z draw B b, intersecting the citcle A CDE in C and Y Zin ¢; draw the diameter 4 D per- pendicular to B C, and proiong C' B to E. Divide each quadrant A C, CD, D E, E F into five equal parts, and the points of division ATES BN PLATE 22.] APPLIED TO ARCHITECTURE. 69 will mark the centres of the coffers upon the circle. All the coffers being thus marked out, so as to correspond to the particulars descri- bed, proceed thus. Parallel to B b draw A a, meeting ¥ Z in a, and from ¢, with the radius ¢ a, describe the semicircle @ b d. Let a b, fig. 2, be the de- velopment of the arc a b on the elevation, and through a draw i n perpendicular to @ b; make a i, a n, each equal to CZ, C' N on the plan, and find the figure ¢ z b of development according to any one of the three methods shown in Prob. xxxiii. In Fig. 2, make s¢, uv, respectively equal to .S 7, U V on the plan, so that ¢ s may be equal to» ¢, and ¢ » equal to n v, and draw curves to b so as to divide all the lines parallel to ¢ # in the same proportion. Upon the middle line a 5 make ae equal to the breadth of the zone be- low the first horizontal row of coffers, and through e draw j o parallel to ¢ », meeting the curves i, » b in 7, 0. Join io; draw o %, making an angle of 45 degrees, with oj to meet the curve ¢ b in %, and draw k p parallel to j 0, meeting the curve n bin p. Draw % g parallel to i 0, meeting the curve nb in ¢; draw ¢ m parallel to o %, meeting the curve ¢b inm ; draw ¢ / and m r parallel to j 0, meeting the curve DESCRIPTION AND USE OF AN IMPROVED LEVELLING STAVE. By T.SopwiTH. The improved Levelling Stave may be had as above, Price £2. 10s. each, ac- companied with a Letter-press Description. FOURTEEN DAY USE i RETURN TO DESK FROM WHICH BORROWED | This book is due on the last date stamped below, or on the date to which renewed. Renewed books are subject to immediate recall. General Library LD 21-100m-2,’55 : } . . ’ University of California (B139s22)476 rkeley