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MODEEN CAEPENTKI AND JOINEKY
VOL. II
ADVANCED SERIES
BEING A COMPILATION
OF
THE VERY BEST THINGS AND MOST MODERN AND
PRACTICAL METHODS KNOWN IN THE ARTS
OF CARPENTRY AND JOINERY
PREPARED AND EDITED BY
FRED T. HODGSON
Author Practical Treatise on the Uses of "The Steel Square,"
"Modern Estimator and Contractors' Guide," "Common
Sense Stair-Builder," "Up-To-Date Hardwood
Finisher," " Practical Wood Carving," &c.
OVER 400 PRACTICAL ILLUSTRATIONS
CHICAGO FREDERICK J. DRAKE & CO., PUBLISHERS
f -■ *
Copyright 1906
BY
Frkdekick J. Drake & Co. Chicago, 111.
Copyright 1910
BY
Frederick J. Drake & Co. Chicago, III.
PREFACE
In a previous volume — Modern Carpentry — I made a fairly successful attempt to put together a series of plain and simple examples of working problems and their solution suited to the capacity of all ordinary mechanics, and owing to the fact that many thousands of that volume have found their way into the hands of carpenters and joiners in the United States, Mexico and Canada, and to the other fact that the author, as well as the publishers, have received hundreds of let- ters asking for something more on the same lines, of a higher grade, it has been determined to make another or more volumes, on the same subject, hence "Modern Carpentry and Joinery" advanced course.
In the present work I have drawn largely from ac- knowledged authorities and from workmen of recog- nized ability to which have been added the results of my own experience and observation and my knowledge of the kinks and secrets of the woodworking trades. In a work of this kind the reader must expect to find something he has met in other places, and perhaps in other adaptations, but I think that upon careful anal- ysis he will find the presentation of the cases somewhat improved, and rendered in a more simple and under- standable manner. The selections, too, I think will be found more suitable and more appropriate to the pres- ent-day practice than most of the matter found in re- cent technical literature on the subject. At any rate,
3
4 PREFACE
it has been my endeavor in the formation of the present volume to place in a handy form, instructive examples of the better class of work of the carpenters and joiners' ''Art."
It is not intended to repeat what has already been published in the first volume of Modern Carpentry, un- less such repetition will be necessary to explain or formulate some similar matter.
Young men are apt to think that because they have a fair knowledge of their trades, they know all that is required and if they peruse the first volume of this work, and have mastered its contents, they have reached the limit. This, of course, is a great mistake, as will readily be discovered by a glance over the matter of the present volume, and I can say right here, and now, that even the present volume does not by any means cover the whole subject, for, at least a half dozen other volumes could be written without touching on the other two already in the market, and the subject would not nearly be exhausted.
It is not necessary for me to quote the authorities from whom I have drawn, unless the matter is of such importance as to demand special recognition. I may say, however, that in very few books written during the last 25 years on the subject of carpentry, there has been but very little presented that had not been pub- lished before in some form or other, but the descrip- tions generally, in most cases — not in all — have been improved more or less; and the present book does not differ a great deal from most others that have pre- ceded it, only that it is more up-to-date, and more suited to 20th century requirements.
PREFACE 5
It must also be understood, that while this book goes out to the public under my name as author, I do not claim authorship, for really, it is more of a com- pilation than an original work, but, I do claim that the selections and compilations made are better and more suited to the wants of the present-day workman than can be found in any other similar work published in this or any other country.
FRED T. HODGSON, Editor and Compiler.
Collingwood,
PREFACE TO SECOND EDITION.
MODERN CARPENTRY.
(advanced series.)
VOL. II.
A preface is an introduction or "foreword" to a new book or a new discourse, but in this case, it is a second introduction to a well-known book which has re- cently had very important improvements and annexes added to the original work ; and the author finds it diffi- cult to say much of the main work other than was said in the preface sent out in the first edition of this volume ; and the recent additions are hardly bulky enough to warrant a lengthy introduction, as they may be taken as being in the same class as the general contents of the work, which is good, practical, instructive, up-to-date and equal, if not superior, to anything of the kind offered to American workmen. The most reliable indication of the merits of a book of this kind, is the opinions of the practical men who buy such books for profitable use; and the thousands of flattering letters that have been received by the author and publishers these few years bear the strongest testimony of the sterling value of the first edition. To give extracts from these letters would take up as much space as the contents of the volume re- quires, so that cannot be attempted, and may not be necessary, but, I may be pardoned for offering a few,
7
8 PREFACE
which may perhaps prove as effective as a thousand would :
"I am well pleased with your book, Vol. 2, Modern Carpentry. It is the best thing I ever saw of the kind, and has helped me wonderfully." Louis Duprey, Win- nipeg, Man.
"We use five of your books in our Trade Teaching De- partment, and find 'Modern Carpentry' the most useful we have. We make it a text-book. ' ' Rev. J. B. Gautheir, Iberville College, Iberville, Quebec.
" 'Modern Carpentry,' second volume, arrived some time ago, and I found in it just what I wanted, and this repays me more than ten fold the cost." John Wilks, Minneapolis, Minn.
"Your book, 'Modern Carpentry,' second volume, is a dandy. I have got a lot of good things out of it. I would not take $100.00 for it if I could not buy another. Good luck to you." James E. Weeks, San Francisco, Cal.
As before stated, I could fill a volume with similar ex- tracts, but my modesty forbids me from giving more un- asked.
This is my preface to this second volume of the new edition of ' ' Modern Carpentry. " It is like a ' ' Donkey 's Gallop," short and sweet, but somewhat "wobbly."
Fred T. Hodgson.
Collingwood, Ont.
PART I.
SOLID GEOMETRY.
INTRODUCTORY.
In the first volume of Modern Carpentry I gave a short treatise on plain or Carpenters' Geometry, which I trust the student has mastered, and thus pre- pared himself for a higher grade in the same science, namely, Solid Geometry: and to this end, the follow- ing treatise has been selected, as being the most simple and the most thorough available.
Solid geometry is that branch of geometry which treats of solids— i.e., objects of three dimensions (length, breadth and thickness). By means of solid geometry these objects can be represented on a plane surface, such as a sheet of paper, in such a manner that the dimensions of the object can be accurately measured from the drawing by means of a rule or scale. The "geometrical" drawings supplied by the architect or engineer for the builder's use are, with few exceptions, problems in solid geometry, and there- fore a certain amount of knowledge of the subject is indispensable, not only to the draughtsman who pre- pares the drawings, but also to the builder or work- man who has to interpret them.
As the geometrical representations of objects con- sist entirely of lines and points, it follows that if pro- jections of lines and points can be accurately drawn,
9
10
MODERN CARPENTRY
the representation of objects will present no further difficulty. A study of lines and points, however, is somewhat confusing, unless the theory of projection has first been grasped, and for this reason the sub- ject will be introduced by a simple concrete example,
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'VerUcal Projections (or Elevations) and Horizontal Projection (or Plan) of a Table In the Uldille of an Oblong Room
Fig. 1.
such, as a table standing in the middle of a room. The four legs rest on the floor at A, B, C, and D (Fig. 1), and the perpendiculars let fall from the corners of the table-top to meet the floor at E, F, G, and H. The oblong E F G II represents the horizontal projection or "plan" of the table-top; the small squares at A,
SOLID GEOIMETRY H
B, C, and D represent the plan of the four legs, and the lines correcting them represent the plan of the frame work under the top. The large oblong J K L M is a plan of the room.
The plan or horizontal projection of an object is therefore a representation of its horizontal dimensions — in other words, it is the appearance which an object presents when every point in it is viewed from a posi- tion vertically above that point.
In a similar manner an elevation or vertical projec- tion of an object is a representation of its vertical di- mensions, and also, it should be added, of some of its horizontal dimensions, — i. e., it is the appearance which an object presents when every point in it is viewed from a position exactly level with that point, all the lines of sight being parallel both horizontally and ver- tically. Thus, the front elevation of the table (or the vertical projection of the side G H) will be as shown at G' H' C' D', and the vertical projection of the side J K of the room will be J K J K. The vertical projection of the end of E G of the table will be as shown E" G" A" C" and of the end K M of the room will be K" M" K M ; the drawing must be turned until the line KM is horizontal, for these and projections to be properly seen.
By applying the scale to the plan, we find that the length of the table-top is six and a half feet, the breadth 4 feet, and the distance of the table from each wall 4% feet. From the front elevation we can learn the height of the table, and also give its length and dis- tance from the end walls of the room. From the end elevation we can ascertain the breadth of the table and
12 MODERN CARPENTRY
its distance from the sides of the room, as well as its height.
To make the drawing clearer, let us imagine that the walls of the room are of wood and hinged at the level of the floor. On the wall J K draw the front ele- vation of the table and then turn the wall back on its hinges until it is horizontal, — i. e., in the same plane as the floor. Proceed in a similar manner with the end K M, and we get the three projections of the room and table on one plane, as show^n in the diagram. To avoid confusion the end elevation will not be further con- sidered at the present.
It will be seen that the line J K represents the angle formed by the wall and floor, — in other words, it repre- sents the intersection of the vertical and horizontal "planes of projection," it is known as "the line of in- tersection," or "the ground line." If a line is let fall from G' perpendicular to J K, the two lines will meet at G, and they will be in the same straight line. Simil- arly, the perpendiculars H' h and h H are in the same straight line. Lines of this kind perpendicular to the planes of projections are known as "projectors" and are either horizontal or vertical G' g and H' h are ver- tical projectors; G g, C c, D d, and Hh are horizontal projectors.
Vertical projectors are not always parallel to one of the sides of the object represented, or, if parallel to one side, are not parallel to other sides which must be represented; thus, a vertical projector or "eleva- tion" of an octagonal object, if parallel to one of the sides of the octagon, must be oblique to the two adja- cent sides. In Fig. 2 an octagonal table is shown. The
SOLID GEOMETRY
13
plan must first be drawn, and from the principal points of the plan projectors must be drawn perpen- dicular to the vertical plane of the projection, until they cut the ground line, and from this perpendiculars must be erected to the height of the several parts of
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the table. The elevation can then be completed with- out difficulty. The side B C of the table is parallel to the vertical plane of projection, but the adjacent sides A B and C D are oblique.*
*A distinction must be made between "perpendicular" and "vertical." The former means, in geometry, a line or plane at right angles to another line or plane, whether these are hori- zontal, vertical or inclined; whereas a vertical line or plane is always at right angles to a horizontal line or plane. The spirit-level gives the horizontal line or plane, the plumb-rule gives the vertical.
14
MODERN CARPENTRY
2. POINTS, LINES, AND PLANES.
I. To determine the position and length of a given straight line, parallel to one of the planes of the pro- jection.
Let G H (Fig. 3) be the given straight line. To de- termine its position (i e., in regard to horizontal and vertical planes), it is only necessary to determine the position of any two of the extreme points.
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Fig. 3.
Let A B C D be a vertical plane parallel to the given line, and C D E F a horizontal plane. The vertical pro- jection or elevatiori of the line is represented by the line h g, and the horizontal projection or plan by the point g', the various projections being shown by dotted lines. The given line is proved to be vertical, because
SOLID GEOMETRY
15
its horizontal projection is a point; its length as meas- ured by the scale, is 6 inches; its height (gg") above the line of intersection C D is 4 inches ; and its hori- zontal distance (g' g") from the same line is 8 inches. If the illustrations are turned so that CDEF be- come a vertical plane, and A B C D the horizontal plane then GH will be horizontal line, because one of its vertical projections is a point. Other vertical projec- tions of the line can be made,— as, for example, a side elevation, — in which the projection will appear as a line and not a point:, but a line must be horizontal if any vertical projection of it is a point. ;
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-No. 1. Perspective View. No. 2. Vertical and Horizontal ProJeclionB
Fig. 4.
Let the given line GH (Fig. 4) be parallel to the vertical plane, but inclined to the horizontal plane. Then gh will be a vertical projection, and g' h' its horizontal projection or plan. By producing h g till it cuts C D at e, it will be found that the given line is inclined at an angle of 60° to the horizontal plane; its length, as measured by the scale along the vertical
16
MODERN CARPENTRY
projection gh, is 8 inches; the height of G above the horizontal plane (measured at gg") is 3 inches, and the height of H (measured at hli") is 9% inches.
II. To determine the length of a given straight line, which is oblique to both planes of projection.
Fig. 5.
Let ab (Fig. 5) be the horizontal projection and a' b' the given vertical projection of the given line. Draw the projection A B on a plane parallel to a b in the manner shown. From this projection the length of the given line will be found (by applying the scale) to be 10 inches. The height C B is of course equal to the height c' b', and the horizontal measurement A G is equal to the horizontal projection a b, and the angle
SOLID GEOMETRY
17
A C B is a right angle. It follows therefore that, if from b the line b d is drawn perpendicular to a b and equal to the height c' b', the line joining a d will be the length of the given line. To avoid drawing the horizontal line a' e' the height of a' and b' above x y are usually set up from a and b as at a" and b", the line a" b" is the length of the given line.*
III. The projections of a right line being given, to find the points wherein the prolongation of that line would meet the planes of projection.
II.
-I. Perspective Viewj II. Vertical and Horizontal Projections Fig. 6.
Let ab and a' b' (Fig. 6, II) be the given projection of the line AB. In the perspective representation of the problem it is seen that A B, if prolonged, cuts the horizontal plane in c, and the vertical plane in d, and the projections of the prolongation become e e and f d. Hence, if ab, a' b' (Fig. 6, II) be the projections
*The application of this problem to hips is obvious. Sup- pose that a b is the plan of a hip-rafter, and a' h' an eleva- tion, the length of the rafter will be equal to a d or a b.
18 MODERN CARPENTRY
of A B, the solution of the problem is obtained by pro- ducing these lines to meet the common intersection of the planes in f and e, and on these points raising the perpendiculars f c and e d, meeting a b produced in c and a' b' produced in d; c and d are the points sought.*
IV. If two lines intersect each other in space, to find from their given projections the angles which they make with each other.
Let a b, c d, and a' b' of d' (Fig.7) be the projections of the lines. Draw the projectors e' f f e', perpen- dicular to the line of intersection a' c', and produce it indefinitely towards E" ; from e draw indefinitely, e E' perpendicular to the line e' f, and make e E' equal to f" e, and draw fE'. From f as a centre describe the arc E" gE" , meeting e f produced in E", and join a E" c E". The angle a E" c is the angle sought. This problem is little more than a develop- ment of problem II. If we consider e f, e'f, as the horizontal and vertical projections of an imaginary line lying in the same plane as a e and c e, we find the length of this line by problem II to be f E' ; in other words f E' is the true altitude of the triangle a e c, a' e' c'. Construct a triangle on the base a e with an altitude f E' equal to f E', and the problem is solved. The practical application of this problem will
*The points d and c are known respectively as the vertical and horizontal "traces" of the line a b, the trace, of a straight line on a plane being the point in which the straight line, produced if necessary, meets or intersects the plane. A hori- zontal line cannot therefore have a horizontal trace, as it can- not possibly, even if produced, meet or intersect the horizontal plane; for a similar reason, a vertical line cannot have a ver- tical trace.
SOLID GEOMETRY
19
be understood if we imagine a e c to be the plan and a' e' c' the elevation of a hipped roof ; f E' gives us the length and slope of the longest common rafter or spar, and E" c is a true representation of the whole hip, i, e., on a plane parallel to the slope of the top. It will
Fig. 7.
be observed that the two projections (e and e') of the point of intersection of the two lines are in a right line perpendicular to the line of intersection of the planes of projection. Hence this corollary. — The pro-
20
MODERN CARPENTRY
jections of the point of intersections of two lines which cut each other in space are in the same right line per- pendicular to the common intersection of the planes of projection. This is further illustrated by the next problem.
•I, Vertical and Horizontal Projettions. II. Perspective V|eW
Fig. 8.
V. To determine from the projection of two lines which intersect each other in the projections, whether the lines cut each other in space or not.
Let ab, c d, ab', c d' (Fig. 8, I) be the projections of the lines. It might be supposed, as their projections
SOLID GEOMETRY 21
intersect each other that the lines themselves intersect each other in space, bnt on applying the corollary of the preceding problem, it is found that the intersec- tions are not in the same perpendicular to the line of intersection a c of the planes of projection. This is represented in perspective in Fig. 8, II. We there see that the original lines a B c D do not cut each other, although their projections a b, c d, a b' , c d', do so. Ffom the point of intersection e raisei a perpen- dicular to the horizontal plane, and it wi|ll cut the original line c D, in E, and this point therefore belongs to the line c D, but c belongs equally to- a^. As the perpendicular raised on e passes through E ^n the line c D, and [through E on the line a B, these pjoints E E' cannot b^ the intersection of the two' lines, Isince they do not touch; and it is also the same in regard to f f. Hence, when two right lines do not cut each other in space, th.6 intersections of their projections are not in the same right line perpendicular to the common in- tersection of the planes of projection.
VI. To find the angle made by a plane with the horizontal plane of projection.
Let a b and a c, Fig. 9, be the horizontal and verti- cal' traces of the given plane, i. e., the lines on which the given plane would, if produced, cut the horizontal and vertical planes of projection. Take any conven- ient point d in a b, and from it draw d e perpendicular to ab, and cutting the line of intersection xy in e, from e draw e d perpendicular to x y and cutting a c in d' . The angle made by the given plane with the horizontal plane of projection is such that, with a base d e, it has a vertical height e d' . Draw such an
22
MODERN CARPENTRY
angle on the vertical plane of projection by setting ofl from e the distance e d" equal to e d, and joining d' d". The angle d' d" e is the angle required.*
Fig. 9.
VII. The traces of a plane and the projections of a point being- given, to draw through the point a plane parallel to the given plane.
In the perspective representation (Figs. 10 and 11) suppose the problem solved, and let B C be the given
*The angle made by a plane with the vertical plane of pro- jection can be found in a similar manner. If we imagine the part above x y in Fig. 9 to be the horizontal projection and the part below x y to he the vertical projection — in other words, if Fig. 9 is turned upside down — a b becomes the vertical trace and a c the horizontal trace, and the angle d' d" € is the angle made by the given plane with the vertical plane of projection.
SOLID GEOMETRY
k23
plane, and A C, A B its vertical and horizontal traces, and E F a plane parallel to the given plane, and G F, G E its traces. Through any point D, taken at pleasure
I. Vertical and Horizontal Piojectioiis. II. Perspective View
Fig. 10.
on the plane E F, draw the vertical plane H J, the horizontal trace of which, I H, is parallel to G E. The plane H J cuts the plane E F in the line k V ,.and its vertical trace GF in 1'. The horizontal projection
24
MODERN CARPENTRY
of k' 1' is H I, and its vertical projection k' V ; and as the point D is in k 1' , its horizontal and vertical projections will be .'- and d' . Therefore, if through d be traced a line «-'- 1, parallel to A B, that line will be the horizontal projection of a vertical plane passing through the original point D; and if an 1 be drawn the indefinite perpendicular L" 1', and through d', the vertical projection of the given point, be drawn the horizontal line d' 1', cutting the perpendicular in r, then the line F G drawn through 1 parallel to A C, will be the vertical trace of the plane required; and
Fig. 11.
the line G E drawn parallel to A B, its horizontal trace. Hence, all planes parallel to each other have their projections parallel, and reciprocally. In solving the problem, let A B, A C (Figs. 10, I) be the traces of the given plane, and d d the projections of the given point. Through d draw d 1 parallel to A B, and from I draw I V perpendicular to A H. Join d d' and through d' draw d' 1' parallel to the line of in- tersection A H. Then F T G drawn parallel to A C, and G E parallel to A B, are the traces of the requir< ed plane.
SOLID GEOMETRY 25
VIII. The traces A B, B C, and AD, DC, of two planes which cut each other being given, to find the projections of their intersections. ^
The planes intersect each other in the straight line A C (Fig. II, 11), of which the points A and C are the traces, since in these points this line intersects the planes of projection. To find these projections, it is only necessary to let fall on the line of intersection in Fig. II, 1, the perpendiculars A a, C c, from the points A and C, and join A c, G a. Ac will be the horizontal projection, and C a the vertical projection of G A, which is the line of intersection or arris of the planes.
IX. The traces of two intersecting planes being given, to find the angle which the planes make be- tween them. \
The angle formed by two planed is measured by that of two lines drawn.from the same point in their inter- section (one along each of the planes), perpendicular to the line formed by the intersection.. This will be better understood by drawing a straight line across the crease in a double sheet of note-paper, at right angles to the crease ;' if the two leaves of the paper are then partly closed so as to form, an angle, we have an angle formed by two planes, and this angle is the same as that formed by two lines which have been drawn perpendicular to the line of intersection of the two planes. These lines in effect determihe a third plane perpendicular to the arris. If, therefore, the two planes are cut by a third plane at right angles to their intersection the solution of the problem is ob- tained. ,
On the arris A C (Fig. 12, 11) take at pleasure any
26
MODERN CARPENTRY
I. Vertical and Horizontal Projections. II. Perspective Viev Fig. 12.
SOLID GEOMETRY 27
point E, and suppose a plane passing through that point cutting the two given planes perpendicular to the arris. There results from the section a triangle D E F, inclined to the horizontal plane, and the angle of which, D E F, is the measure of the inclination of the two planes. The horizontal projection of that tri- angle is the triangle D e F, the base of which, F D, is perpendicular to A c (the horizontal projection of the arris A C), and cuts it in the point g, and the line E g is perpendicular to D F. The line g E is necessary perpendicular to the arris A C, as it is .in the plane D E F, and its horizontal projection is g e. Now, sup- pose the triangle D E F turned on D F as an axis, and laid horizontally, its summit will then be at E", and D E" F is the angle sought. The perpendicular g E is also in the vertical triangle A C c, of which the arris is the hypothenuse and the sides A c, C C are the projections. This description introduces the so- lution of the problem.
Through any point g (Fig. 12, 1) on the line Ac, the horizontal projection of the arris or line of inter- section of the two planes, draw F D perpendicular to Ac; on A c (which for the moment must be consid- ered as a "line of intersection" or "ground line" for a second vertical projection) describe the vertical pro- jection of the arris by drawing the perpendicular c C, and then joining A C A C gives the true length and inclination of the arris. From g draw g E' per- pendicular to A C, and meeting it in E' ; from E' let fall a perpendicular E e' or A c, meeting A c in e. F e D is the horizontal projection of the triangle F E D (see No. 11) and from this the vertical projection
28
MODERN CARPENTRY
f e' D can be drawn as shown. We have now ob- tained the vertical and horizontal projections of two intersecting lines, namely, F e, e D, and f e', e' D, and by problem (4) the triangle which they make with each other can be found. It will be seen that g E' is the
I. Huiizoiital Projection. II. Vertical Projection on Plane AD- III, Vertical Projection on Plane_cj>. Fig. 13.
true altitude of the triangle f e' D, as e E is equal to e' e. Set off therefore from g towards A a distance g E equal to g E:, and join F E, E" D; F E" D is the angle made by the two intersecting planes.
SOLID GEOMETRY
29
4. Vertical and Horizontal Projections. II. Perspective View
Fig. 14.
30 MODERN CARPENTRY
X. Through a given point to draw a perpendicular to a given plane.
Let a and a' (Fig. 14, II.) be the projections of the given point, and B C, C D the horizontal and vertical traces of the given plane. Suppose the problem solved, and that A E is the perpendicular drawn through the A to the plane B D, and that its* intersection with the plane is the point E. Suppose also a vertical plane a F to pass through A E', this plane would cut B D in the line g D, and its horizontal trace a h would be perpendicular to the trace B C. In the same way a' e', the vertical projection of A E, would be perpen- dicular to C D, the vertical trace of the plane B D. Thus we find that if a line a h is drawn from a, per- pendicular to B C, it will be the horizontal trace of the plane in which lies the required perpendicular A E, and h F will be the vertical trace of the same plane. From a', draw upon C D an indefinite perpen- dicular, and that line will contain the vertical pro- jection of A E, as a h contains its horizontal pro- jection. To find the point of intersection of the line A E with the given plane, construct the vertical pro- jection g D of the line of intersection of the two planes, and the point of intersection of that line with the right line drawn through a', will be the point sought. If from that point a perpendicular is let fall on a h, the point e will be the horizontal projec- tion of the point of intersection E.
In Fig. 15, 1, let B C, C D be the traces of the given plane, and a, a' the projections of the given point. From the point a, draw a h perpendicular to B C ; a h will be the horizontal projection of a plane passing
SOLID GEOMETRY
31
vertically through a, and cutting the given plane. On a h as "ground line" draw a vertical projection as follows : From a, draw a A perpendicular to a g, and make it equal to a" a' ; from h draw h D' per- pendicular to h a, and make h D' equal to h D ; draw g D, which will be the section of the given plane by
i. Horizontal Projection. II. Vertical Projection on Plane parallel to ha". III. Vertical Projection on Plane parallel to /lu
Fig. 15.
a vertical D g, and the angle h g D' will be the meas- ure of the inclination of the given plane with the hor- ' izontal plane ; there is now to be drawn, perpendicu- lar to this line, a line A E through A, which will be the vertical projection on a h of the line required. From the point of intersection E let fall upon aha
32 MODERN CARPENTRY
perpendicular, which, will give e as the horizontal pro- jection of E. Therefore a e is the horizontal projection of the required perpendicular, and a' e' its vertical projection on the original "ground line" h a". It follows from this problem that — Where a right line in space is perpendicular to a plane, the projections of that line are respectively perpendicular to the traces of the plane.
XI. Through a given point to draw a plane perpen- dicular to a given right line.
Let a, a' (Fig. 16, 1) be the projections of the given point A, and b c, b' c' the projections of the given line B C.
The foregoing problem has shown that the traces of the plane sought must be perpendicular to the projec- tions of the line, and the solution of the problem consists in making to pass through A, a vertical plane A f (Fig. 16, 11), the horizontal projection of which will be perpendicular to b c.
Through a (Fig. 16, 1) draw the projection a f perpendicular to b c. From f raise upon K L the in- definite perpendicular f f, which will be the vertical trace of the plane a f f perpendicular to the horizontal plane, and passing through the original point A (in No. 11). Then draw through a in the vertical pro- jection a horizontal line, cutting f f in f, which point should be in the trace of the plane sought; and as that plane must be perpendicular to the vertical pro- jection of the given right line draw through f a per- pendicular to b c, and produce it to cut K L in G. This point G is in the horizontal trace of the plane sought. All that remains therefore, is from G to draw
SOLID GEOMETRY
33
34 MODERN CARPENTRY
G D perpendicular to b c. If the projections of the straight line are required, proceed as in the previous problem, and as shown by the dotted lines; the plane will cut the given line at k in the horizontal projec- tion, and at k in the vertical projection.*
XII. A rig-ht line being given in projection, and also the traces of a given plane, to find the angle which the line makes with the plane.
Let A B (Fig. 17, 11) be the original right line inter- secting the plane C E in the point B. If a vertical plane a B pass through the right line,- it will cut the plane C E in the line f B, and the horizontal plane in the line a b. As the plane a B is in this case parallel to the vertical plane of projection, its projection on that plane will be a quadrilateral figure a b of the same dimensions, and f B contained in the rectangle will have for its vertical projection a right line D b, which wnll be equal and similar to f B. Hence the two angles a b D, A B f, being equal, will e(iually be the measure of the angle of inclination of the right
*The diagram will be less confusing if the projection on a h is drawn separately, as in Fig. 16, wtiere I is the hori- zontal projection or plan, II the vertical projection or eleva- tion on a plane parallel to ha", and III the vertical projection or elevation on a plane parallel to h a. To draw No. Ill draw first the ground-line h a equal to h a on No. I, and mark on it the point g ; from h draw the vertical h d' equal to h d, and from a draw the vertical a a equal to a" a'; join d' g and a h, and from the point of intersection e let fall a perpendicular on h a, cutting it in e. a e is the actual length and inclination of the required line, and a e its horizontal length. Transfer the length a e to No. I, and from e draw e e' perpendicular to 7i a" and cutting a' h in e'. If the drawing has been correctly made, a line from e parallel to the ground-line h a" will also intersect a' h in e'.
SOLID GEOMETRY
35
line A B to the plane C E. Thus the angle a b D (Fig. 17, 1), is the angle sought.
This case presents no difficulty; but when the line is in a plane which is not parallel to the plane of
|
ay |
2 |
/ |
||||
|
, |
D |
/ |
b |
» |
||
|
a. |
./ |
h |
||||
|
c |
Fig. 17.
projection, the problem is more difficult; as, however, the second case is not of much practical value, it will not be considered.
36
MODERN CARPENTRY
3. STRAIGHT-SIDED SOLIDS.
XIII. Given the horizontal projection of a regular tetrahedron, to find its vertical projection.
Let A B C d (Fig. 18) be the given projection of the tetrahedron, which has one of its faces coincident with the ground-line. From d draw an indefinite line
dJ) perpendicular to d C, and make C D equal to C B, C A, or A B ; d D will be the height sought, which is carried to the vertical projection from c to d. Join A d and B d, and the vertical projection is complete. This problem might be solved in other ways.
SOLID GEOMETRY 37
XIV. A point being given in one of the projections of a tetrahedron, to find the point on the other pro- jection.
Let e be the point given, in the horizontal projec- tion (Fig. 18). It may be considered as situated in the plane C B d, which is inclined to the horizontal plane, and of which the vertical projection is the tri- angle c B d. According to the general method, the vertical projection of the given point is to be found somewhere in a perpendicular raised on its horizontal projection e. If through d and the point e be drawn a line produced to the base of the triangle in f, the point e will be on that line, and its vertical projection will be on the vertical projection of that line fed, at the intersection of it with the perpendicular raised on e. If through e be drawn a straight line g h, par- allel to C B, this will be a horizontal line, whose ex- tremity h will be on B d. The vertical projection of d B is d B ; therefore, by raising on h a perpendicular to A B, there will be obtained h, the extremity of a horizontal line represented by h g in the horizontal plane. If through h is drawn a horizontal line h g this line will cut the vertical line raised on e in e, the point sought. If the point had been given in g on the arris c d, the projection could not be found in the first manner; but it could be found in the second manner, by drawing through g, a line parallel to C B, and prolonging the horizontal line drawn through h, to the arris c d, which it would cut in g, the point sought. The point can also be found by laying down the right-angled triangle C d D (which is the devel- opment of the triangle formed by the horizontal pro-
38
MODERN CARPENTRY
jection of the arris C d, the height of the sohd, and the length of the arris as a hypotenuse), and by draw- ing through g the line g G perpendicular to C d, to in- tersect the hypotenuse in G, and carrying the height g G from c to g in the vertical projection. One or other of these means can be employed according to circumstances. If the point had been given in the vertical instead of the horizontal projection, the same operations inverted would require to be used.
Pig. 19.
XV. Given a tetrahedron, and the trace of a plane (perpendicular to one of the planes of projection) cut- ting it, by which it is truncated, to find the projection of the section.
First when the intersecting plane is perpendicular to the horizontal plane (Fig. 19), the plane cuts the
SOLID GEOMETRY
39
base in two points e f, of which the vertical projec- tions are e and f ; and the arris B d is cut in g, the vertical projection of which can readily be found in any of the ways detailed in the last problem. Having found g join egf g and the triangle egf is the pro- jection of the intersection sought.
When the intersecting plane is perpendicular to the vertical plane, as e f in Fig. 20, the horizontal projec- tions of the three points egf have to be found. The point g in this case may be obtained in several ways. First by drawing G g h through g, then through h drawing a perpendicular to the base, produced to the arris at h, in the horizontal projection, and then draw- ing h g parallel to C B, cutting the arris B d in g, which is the point required. Second, on d B, the hor-
40 MODERN CARPENTRY
izontal projection of the arris, construct a triangle d D B, d B being the altitude of the tetrahedron, and B D the arris, and transfer this triangle to the ver- tical projection at d d B. From g draw the horizontal line cutting B d in G ; g G is the horizontal distance of the required point in the arris, from the vertical axis of the tetrahedron as d is horizontal projection of the vertical axis, and dB the horizontal projection of the arris, it follows that the length gG, trans- ferred to d g, will give the required point g. The points e and f are found by drawing lines from e and f perpendicular to the ground-line, and producing them till they meet the horizontal projections of the arrises in e and f. The triangle e f g is the horizontal projection of the section made by the plane e f.
XVI. The projections of a tetraJbedron being given, to find its projections when inclined to the horizontal plane in any degree.
Let ABCd (Fig. 21) be the horizontal projection of a tetrahedron, with one of its sides coincident with the horizontal plane, and e d B its vertical projection ; it is required to find its projections when turned round the arris AB as an axis. The base of the pyramid being a horizontal plane, its vertical projection is the right line c B. If this line is raised to e by turning on B, the horizontal projection will be A c2 B. When the point c, by the raising of B e, describes the arc c c, the point d will have moved to d, and the per- pendicular let fall from that point on the horizontal plane will give d3, the horizontal projection of the extremity of the arris C d; for "as the summit d moves in the same plane as C, parallel to the vertical plane
SOLID GEOMETRY
41
of projection the projection of the summit will evi- dently be in the prolongation of the arris C d, which is the horizontal projection or trace of that plane. The process therefore, is very simple and is as fol- lows: Construct at the point B the angle required, c B c, and make the triangle c d B equal to c d B ;
Fig. 21.
from d let fall a perpendicular cutting the prolonga- tion of the arris C d in d3, and from c a perpendicular cutting the same line in c2 ; join B c2, A c2, B d3, A d3. The following is a more general solution of the problem: Let ABCd (Fig. 22), be the horizontal projection of a pyramid resting with one of its sides on the horizontal plane, and let it be required to raise, by its angle C, the pyramid by turning around the arris AB, until its base makes with the horizontal
42
MODERN CARPENTRY
plane any required angle, as 50°. Conceive the right line C e turning round e, and still continuing to be perpendicular to A B, until it is raised to the required angle, as at e C. If a perpendicular be now let fall from C, it will give the point C as the horizontal pro- jection of the angle C in its new position. Conceive
a vertical plane to pass through the line C e. This plane will necessarily contain the required angle. Sup- pose, now we lay this plane down in the horizontal projection thus : Draw from e the line e C
making
SOLID GEOMETRY
43
with e C an angle of 50°, and from e with the radius e c describe an arc cutting it in C. From C let fall on C e, a perpendicular on the point C, which will then be the horizontal projection of C in its raised position. On C e draw the profile of the tetrahedron C D e inclined to the horizontal plane. From D let fall a pei-pendicular on C e produced, and it will give d as the horizontal projection of the summit of the pyramid in its inclined position. Join Ad, B d, Ac, B c to complete the figure. The vertical projection of the tetrahedron in its original position is shown by a d b, and in its raised position by a, c2, d2, b, the points c2, and d2 being found by making the perpendiculars c3 c2 and d3 d2 equal to C C and d D respectively.
XVII. To construct vertical and horizontal pro- jections of a cube, the axis 1 of which are perpendic- ular to the horizontal plane.
If an arris of the cube is given, it is ea.sy to find its axis, as this is the hypotenuse of a right-angled tri- angle, the shortest side of which is the length of an
44
MODERN CAEPENTRY
arris, and the longest the diagonal of a side. Conceive the cube cut by a vertical plane passing through its diagonals E G, A C (Fig. 23), the section will be the rectangle AEGC. Divide this into two equal right- angled triangles, by the diagonal EC. If in the upper and lower faces of the cube, we draw the diagonals F H, B D, they will cut the former diagonals in the points f and b. Now, as the lines b B, b D, f F, f H, are perpendicular to the rectangular plane AEGC, fb may be considered as the vertical projection of B F and D H, and from this consideration we may solve the problem.
Fig. 24.
Let AE (Fig. 24) be the arris of any cube. The letters here refer to the same parts as those of the preceding diagram (Fig. 23). Through A draw an indefinite line, A C, perpendicular to A E. Set off on
SOLID GEOMETRY 45
this line, from A to C, the diagonal of the square of A E, and join E C, which is then the axis of the cube. Draw the lines E G, G G, parallel respectively to A C and A E, and the resulting rectangle, A E G C, is the section of a cube on the line of the diagonal of one of its faces. Divide the rectangle into two equal parts by the line b f, which is the vertical projection of the lines B F, D H (Fig. 23), and we obtain, in the figure thus completed, the vertical projection of the cube, as a c b d (Fig. 25).
Through C (Fig. 24), the extremity of the diagonal E C draw y z perpendicular to it, and let this line represent the common section or ground-line of the two planes of projection. Then let us find the hor- izontal projection of a cube of which A E G C is the vertical projection. In the, vertical projection the axis E C is perpendicular to y z, and consequently, to the horizontal plane of projection, and we have the height above this plane of each of the points which terminate the angles. Let fall from each of these points perpendiculars to the horizontal plane, the projections of the points will be found on these per- pendiculars. The horizontal projection of the axis E C will be a point on its prolongation, as c. This point might have been named e with equal correctness, as it is the horizontal projection of both the extremities of the axis, C and E. Through c draw a line parallel to yz, and find on it the projections of the points A and G, by continuing the perpendiculars A a, G g, to a and g. We have now to find the projections of the points bf (representing D B F H, Fig. 23,), which will be somewhere on the perpendiculars b b, f f, let fall
46
MODERN CARPENTRY
from them. We have seen in Fig. 23 that B F, D H are distant from b f by an extent equal half the diag- onal of the square face of the cube. Set off, therefore, on the perpendiculars b b and f f , from o and m, the distance. A b in d, b, and f, f and join da, a b, b f , f g, g f , to complete the hexagon which is the hori- zontal projection of the cube. Join f e , f e, and a c, to give the arrises of the upper half of the cube. The dotted lines d c, be, g c, show the arrises of the lower side. Knowing the heights of the points in these ver- tical projections, it is easy to construct a vertical pro- jection on any line whatever, as that on R S below.
Fig. 25.
XVIII. To construct the projections of a regular octahedron, when one of its axis is perpendicular to either plane of projection.
Describe a circle (Fig. 26), and divide it into four equal parts by the diameters, and draw the lines a d, d b, be, ca; a figure is produced which serves for either the vertical or the horizontal projection of the octahedron, when one of its axis is perpendicular to either plane.
SOLID GEOMETRY
47
XIX. One of the faces of an octahedron being given, coincident with the horizontal plane of projection, to construct the projections of the solid.
Let the triangle ABC (Fig. 27) be the given face. If A be considered to be the summit of one of the two pyramids which compose the solid, B C will be one of the sides of the square base, k C B i. The base makes with the horizontal plane an angle, which is easily found. Let fall from A a a perpendicular on B c, cut- ting it in d, with the length B c as a radius, and from
Fig. 27.
d as a centre, describe the indefinite arc e f. The per- pendicular A d will be the height of each of the faces, and, consequently, of that which, turning on A, should meet the side of the base which has already turned on d. Make this height turn on A, describing from that point as a centre, with the radius A d, an indefinite arc, cut- ting the first are in G, the point of meeting of one of the faces with the square base ; draw the line G A, G d .- the first is the profile or inclination of one of the faces on the given face ABC according to the angle d A G ; the second, d G^ is the inclination of the square base,
48 MODERN CARPENTRY
which separates the two pyramids in the angle A d G. The face adjacent to the side B C is found in the same manner. Through G, draw the horizontal line G H equal to the perpendicular A d. This line will be the profile of the superior face. Draw d H, which is the profile of the face adjacent to B C. From H let fall a perpendicular on A d produced, which gives the point h for the horizontal projection of H, or the summit of the superior triangle parallel to the first, draw h i par- allel to C A, h k parallel to A B, A k and B h per- pendicular to A B, and join k C, C h, B i, and A i and the horizontal projection is complete. From the heights we have thus obtained we can now draw the vertical projection shown in No. II, in which the parts have the same letters of reference.
The finding of the horizontal projection may be abridged by constructing a hexagon and inscribing in it the two triangles A C B, h i k.
XX. Given in the horizontal plane the projection of one of the faces of a dodecahedron, to construct its projections.
The dodecahedron is a twelve-sided solid, all the sides being regular and equal pentagons. It is neces- sary, in order to construct the projection, to discover the inclination of the faces among themselves. Let the pentagon A B C D E (Fig. 28) be the side on which the body is supposed to be seated on the plane. Con- ceive two other faces, E F G II D and D I K L C, also in the horizontal plane, and then raised 'by being turned on their bases, ED, DC. By their movement they will describe in space arcs of ^circles, which will terminate by the meeting of the sides D H, D I.
SOLID GEOMETRY
49
To find the inclination of these two faces. — From the points 1 and H let fall perpendiculars on their bases produced. If each of these pentagons were raised ver- tically on its base, the horizontal projection of H and 1 would be respectively in zz- but as both are raised
0 "
Fig. 28.
together, the angles H and I would meet in space above h where the perpendiculars intersect therefore, h will be the horizontal projection of the point of meeting of the angles. To find the horizontal projection of K, pro- long indefinitely z I, and set off from z on z I the length X K in k ; from z as centre with radius z I, describe
50 MODERN CARPENTRY
the arc II cutting the perpendicuhir h I in I ; jcn z I ; then from z as centre, with the radius z k, describe an arc cutting z I produced in the point K, from which let fall on zk a perpendicular Kk, and produce it to k in X K. If, now, the right-angled triangle z k K, were raised on its base, k would be the projection of K. Conceive now the pentagon C D I K L turned round on C D, until it makes an angle equal to k z K with the horizontal plane, the summit K will then be raised above k by the height k K, and will have for its hori- zontal projection the point k. In completing the figure practically ; — from the centre o, describe two concen- tric circles passing through points h D. Draw the lines h D, h k, and carry the last round the circum- ference in m n o p r s t u : througli each of these lines draw radially the lines m C, o B, r A, t E, and these lines will be the arrises analogous to h D. This being done, the inferior half of the solid is projected. By reason of the regularity of the figure, it is easy to see that the six other faces will be similar in those already drawn, only that although the superior pentagon will have its angles on the same circumference as the in- ferior pentagon the angles of the one will be in the middle of the faces of the other. Therefore, to de- scribe, the superior half ; — through the angles n p s v k, draw the radial lines n 1, p 2, s 3, v 4, k 5, and join them by the straight lines 1 2, 2 3, 3 4, 4 5, and 5 1.
To obtain the length of the axis of the solid, observe that the point k is elevated above the horizontal plane by the height k K : carry that height to k K : the point r, analogous to h, is raised the same height as that point, that is to say h i, which is to be carried from r
SOLID GEOMETRY 51
to R ; and the line R K is the length sought. As this axis should pass through the centre of the body, if a vertical projection of the axis in O, and therefore 0 o i's the half of the height of the solid vertically. By doubting this height, and drawing a horizontal line to cut the vertical lines of the angles of the superior face is obtained, as in the upper portion of (Fig. 28), in which the same letters refer to the same parts.
XXL One of the faces of a dodecahedron being given, to construct the projections of the solid, so that its axis may be perpendicular to the horizontal plane.
Let ABED C (Fig. 29, 1) be the given face. The solid angles of the dodecahedron are each formed by the meeting of three pentagonal planes. If there be conceived a plane B C passing through the extremities of the arrises of the solid angle A, the result of the section would be a triangular pyramid, the sides of whose base would be equal to one of the diagonals of the face, such as B C. An equilateral triangle bcf (Fig. 29, 11) will represent the base of that pyramid inverted, that is, with its summit resting on the hori- zontal projection, it it required to find the height of that pyramid, or which is the same thing, that one of the three points of its base bcf, for as they are all equally elevated, the height of one of them gives the others. There is necessarily a proportion between the triangle Abe (No. 11) and ABC (No. 1), since the first is the horizontal projection of the second. A g is the hori- zontal projection of A G ; but A G is a part of A H, and the projection of that line is required for one of the faces of the solid; therefore as AG:Ag::AH:x. In other words, the length, the length of x may be ob-
52
MODERN CARPENTRY
tained by drawing a fourth proportion of the three lines it will be found to be equal to A h : or it may be obtained graphically thus : — Raise on A g at g an in- definite perpendicular, take the length AG (No. 1) and carry it from A to G (No. 11) ; g is a point in the
Fig. 29.
assumed pyramidal base b c f . Since A G is a portion of A H, A G will be so also. Produce A G, therefore, to H, making AH equal to AH (No. 1) and from H let fall a perpendicular on A g produced, which gives h the point sought. Produce H h, and carry on it the
SOLID GEOMETRY 53
length H D or H E from h to d and h to e ; draw the lines e d, be, and the horizontal projection of one of the faces is obtained inclined to the horizontal plane, in the angle HAh. As the other two inferior faces are similar to the one found, the three faces should be found on the circumference of a circle traced from A as centre, and with A d or A e as a radius. Join f A, prolong An, A o, perpendicular to the sides of the triangle f c b, and make them equal to A h, and through their extremities draw perpendiculars, cutting the cir- cumference in the points ik, Im. Draw the lines ib, k f , If, m c, and the horizontal projections of the three inferior faces obtained. The superior pyramid is sim- ilar and equal to the inferior, and solely opposed by its angles. Describe a circle passing through the three points of the first triangle, and draw within it a second equilateral triangle nop, of which the summits corres- pond to the middle of the faces of the former one. Each of these points will be the summit of a pentagon, as the points b c f . These pentagons have all their sides common, and it is only necessary therefore to determine one of these superior pentagons to have all the others. Six of the faces of the dodecahedron have now been projected ; the remaining six are obtained by joining the angular points already found, as q d, e r, t i, k s, &c.
To obtain the vertical projection (No. Ill) begin with the three inferior faces. The point A in the horizontal projection being the summit of the inferior solid angle, will have its vertical in a ; the points b c f , when raised to the height g G, will be in b c f, or simply b f. The points b g c being in a plane perpendicular to
54 ~ MODERN CARPENTRY
the vertical plane, will necessarily have the same ver- tical projection, b. The line a f will be the projection of the arris A f, and a b will be that of the arrises A b, A c, and of the line A g, or rather that of the triangle Abe, which is in a plane perpendicular to the vertical plane. But this triangle is only a portion of the given pentagonal face (No. 1), of which AH is the perpen- dicular let fall from A on the side E D. Produce a b to e, making a e equal to A H; e is the vertical projec- tion of the arris e d. This arris is common to the in- ferior pentagon, and to the superior pentagon e d q p r, which is also perpendicular to the vertical plane, and, consequently, its vertical projection will be e p, equal to a e.
This projection can now be obtained by raising a vertical line through p, the summit of the superior pentagon, and from e as a centre, and with the radius AH or AH, describing an arc cutting this line in p, the point sought. But p n o belong to the base of the superior pyramid; itherefore, if a perpendicular is drawn from n through y z to n, and the height p is transferred to n by drawing through p' a line parallel to y z, n will be the projection of the points n and o. Through n draw sua parallel to a e cutting perpen- diculars drawn through s and A in the horizontal pro- jection. Through s draw s f parallel to p e, and join a f, a p ; set off on the perpendicular from r the height of s above y z at r, and draw r t parallel to y z, cutting the perpendicular from t, and joint n t. Draw perpen- diculars from k and i throiigh y z, to k and i, make k and i the same height as e, and draw k i, and join i b, i t. The vertical projection is now complete.
SOLID GEOMETRY
55
XXII. In a given sphere to inscribe a tetrahedron, a hexahedron or cube, an octahedron and a dodecahed- ron.
Let AB (Fig, 30) be the diameter of the given sphere. Divide it into three equal parts, D B being one of these parts. Draw D E perpendicular to A B, and draw the chords A E, E B. A E is the arris of the tetra-
Fig. 30.
hedron, and E B the arris of the hexahedron or cube. From the centre C draw the perpendicular radius C F, and the chord F B is the arris of the octahedron. Divide B E in extreme and mean proportion in G, and B G is the arris of the dodecahedron. The arrises being known, the solids can be drawn by the help of the prob- lems already solved. Draw the tangent A H equal to AB; join HC and AT; A I is the arris of an icosahe-
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MODERN CARPENTRY
dron which can be inscribed in the sphere, an icosa- hedron being a solid with twenty equal sides, all of which are equilateral triangles. See Fig. 301/2-
Fig. 305i.
SOLID GEOMETRY
57
4. THE CYLINDER, CONE, AND SPHERE.
XXIII. The horizontal projection of the cylinder, the axis of which is perpendicular to the horizontal plane, being given to find the vertical projection.
Let the circle ABCD (Fig. 31) be the base of the cylinder, and also its horizontal projection. From the points A and C raise perpendiculars to the ground-line
a c, and produce them to the height of the cylinder — say, for example, a e, c f. Draw e f parallel to a c, and the rectangle a e f c is the vertical projection required. XXIV. Given the traces of an oblique plane, to de- termine the inclination of the plane to both the H. P. and the V. P.
58 MODERN CARPENTRY
Let V t and h t (Fig. 32) be the traces of the given plane. Draw the projections of a semi-cone having its axis a' b' in the vertical plane, the apex a' in the given vt and its base (a semi-circle) c e d in the hp and lying tangentially to the given h t. Then the base angle (0) of the cone gives the inclination of the plane to the h p. To determine the inclination of the plane to the V. p., draw the projections of a second semi-cone, hav- ing the axis m n in the h p, and the apex m in the
Fig. 32.
given h t, while the base is in the v p and tangential to the V t. The base angle (0) of this cone gives the inclination to the v p.
XXV. The base of a cylinder being given, and also the angles which the base makes with the planes of projection, to construct the projections of the cyl- inder.
Let the circle AGBH (Fig. 33) be the given base, and let each of the given angles be 45°. Draw the diameter AB, making an angle of 45° with the ground line or ver-
SOLID GEOMETRY
59
tical plane, and draw the line A B, making with A B the given angle ; and from A as a centre, with A B and A C as radii, describe arcs cutting A B in B and C. Then draw A D, B E perpendicular to A B and equal
Fig. 33.
to the length of the cylinder; the rectangle A E is the vertical projection of the cylinder parallel to the vertical plane and inclined to the horizontal, plane A B in an angle of 45°, Now prolong indefinitely the diameter B A, and this line will represent the projection on the horizontal plane of the line in which the generating circle moves to produce the cylinder.
60 MODERN CARPENTRY
If from B and C perpendiculars be let fall on A B, k will be the horizontal projection of B, A k that of the diameter A B, and c that of the centre C. Through c draw h g perpendicular to A B, and make c hj c g equal to C H, G G; and the two diameters of the ellipse, which is the projection of the base of the cyl- inder, will be obtained ; namely, A k and h g.
In like manner, draw D F E the lines D d, F f, E c, perpendicular to the diameter A B produced, and their intersections with the diameter and the sides of the cylinder will give the means of drawing the ellipse which forms the projection of the farther end of the cylinder. The ellipses may also be found by taking any number of points in the generating circle as I J, and obtaining their projections i j. The method of doing this, and also of drawing the verti- cal projection c f, will be understood without fur- ther explanation.
XXVI. A point in one of the projections of a cone being given, to find it in the other projection.
Let a (Fig. 34) be the given point. This point be- longs equally to the circle which is a section of the cone by a plane passing through the point parallel to the base, and to a straight line forming one of the sides of a triangle which is the section of the cone by a plane perpendicular to its base and passing through its vertex and through the given point, and of which f a g is the horizontal, and f a g the vertical projec- tion. To find the vertical projection of a, through a draw a a perpendicular to b c, and its intersection with f g is the point required; and reciprocally, a in the
SOLID GEOMETRY
61
horizontal projection may be found from a in the vertical projection, in the same manner.
Otherwise, through a, in the horizontal projection, describe the circle a d c, and draw e e or c c, cutting
Fig. 34.
the sides of the cone in e and c; draw c e parallel to the base, and draw a a, cutting it in a, the point required.
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MODERN CARPENTRY
XXVII. On a given cylinder to describe a helix.
Let abed, &c. (Fig. 35), be the horizontal projec- tion of the given cylinder. Take on this curve a series of equal distances, a b, b c c d, &c., and through each
of the points a, b, c, &c. draw a vertical line, and pro- duce it along the vertical projection of the cylinder. Then conceive a curve cutting all these verticals in the points abed, in such a manner that the height of the point above the ground-line may be in constant rela- tion to the arcs a b, b c, c d ; for example, that a may
SOLID GEOMETRY
63
be the zero of height, that bb may be 1, c c 2, d d 3, &c. ; then this curve is named a helix. To construct this curve, carry on the vertical projection on each vertical line such a height as has been determined, as 1 on b, 2 on c, 3 on d; and through these points will pass the curve sought. It is easy to see that the curve
so traced is independent of the cylinder on which it has been supposed to be traced; and that if it be isolated, its horizontal projection will be a circle. The helix is named after the curve which is its horizontal projection. Thus the helix in the example is a helix with a circular base. The vertical line f n is the axis
64 MODERN CARPENTRY
of the helix, and the Ijeiglit b b, comprised between two consecutive intersections of the curve with a verti- cal, is the pitch of the helix.
XXVIII. On a given cone to describe a helix.
Let the projections of the given cone be as shown in Fig. 36. Divide the base of the cone in the horizon- tal projection into any number of equal parts, as a b, b c, c d, &c., and' draw lines from the vertex to tlie points thus obtained. Set off along these lines a series of distances increasing in constant ratio, as 1 at b, 2 at c, 3 at d, &c. The curve then drawn through these points when supposed to be in the same plane, is called a spiral. If these points, in addition to approaching the centre in a constant ratio, are supposed also to rise above each other by a constant increase of height, a helical curve will be obtained on the vertical pro- jection of the cone.*
XXIX. A point in one of the projections of the sphere being given, to find it in the other projection.
Let a be the given point in the horizontal projection of the sphere hb c i (Fig. 38). Any point on the sur- face of a sphere belongs to a circle of that sphere. Therefore, if a is a point, and a vertical plane b c is made to pass through that point to A B, the section of
*Tlie octahedron is formed by the union of eight equilateral triangles; or, more correctly, by the union of two pyramids with square bases, opposed base to base, and of which all the solid angles touch a sphere in which they may be inscribed.
It is essential that the diagram should be clearly seen as a solid, and not as a mere set of lines in one plane. Imagine h as the apex of one pyramid on the base k c n i. and a as the apex of the other pyramid on the opposite side of the same base. The octahedron is shown to be lying on its side a b c.
SOLID GEOMETRY
65
the sphere by this plane will be a circle, whose diam- eter will be b c, and the radius consequently, d b or d c ; and the point a will necessarily be in the circum- ference of this circle. Since the centre d of this circle is situated on the horizontal axis of the sphere, and
Fig.
Fig.
as this axis is perpendicular to the vertical plane, its vertical projection will be the point d. It is evident that the vertical projection of the given point a will be found in the circumference of the circle described from d with the radius d b or d c and at that point of it wh<;)re it is intersected by the lim drawn through
66 MODERN CARPENTRY
a, perpendicular to A B. Its vertical projection will therefore be either a or a, according as the point a is on the superior or inferior semi-surface of the sphere. The projection of the point may also be found thus: Conceive the sphere cut by a plane parallel to the hori- zontal plane of projection passing through the given point a. The resulting section will be the horizontal circle described from k, with the radius k a; and the vertical projection of this section will be the straight line g e, or g e ; and the intersections of these lines with the perpendicular drawn through a, will be the projection of a, as before.
5. SECTIONS OP SOLIDS.
To draw sections of any solid requires little more than the application of the method described in the foregoing problems. Innumerable examples might be given, but a few selected ones will sulTiee.
XXX. The projections of a re^lcir tetrahedron be- ing given, to draw the section mzCe by a plane perpen- dicular to the vertical plane and inclined to ths hori- zontal plane.
Let A B C D and abed (Fig. 39) be the given pro- jections, and E F G the given plane perpendicular to the vertical plane and inclined to the horizontal plane at an angle of 30°. The horizontal projection ef g of the section is easily found as shown. To fmd the cor- reet section draw through e, f, and g, lines parallel to
SOLID GEOMETRY
67
the ground-line A C, and a c one of them as E e set off the distances E F, E G at E f and E g, and through f and g draw perpendiculars cutting the other lines in F and G'. Join E G, G F, and F E. E F G is the cor- rect section made by the plane.
Fig. 39.
XXXI. The projections of a hexagonal pyramid be- ing given, to draw the section made by a plane perpen- dicular to the vertical plane and inclined to the hori- zontal plane.
Let A B C D and a b c d e f g (Fig. 40) be the given projections, and H I J K the given plane. The hori-
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MODERN CARPENTRY
zontal projection h i j k 1 m of the section is easily- found as shown. Through m, 1, h, j, and i draw lines parallel to the ground-line A D, and on one of them as h m, set off the distances H I, H J, H K, at h M,
Fig. 40.
ih 1, h k. From h, M, 1, and k draw perpendiculars meeting the other lines in H, I, L, J, and K, and join the points of intersection. H I J Iv L M is the true section made by the plane.
SOLID GEOMETRY
69
XXXII. The projections of an octagonal pyramid being- given, to draw the section made by a verti- cal plane.
Let A B C D F and a b e d e f (Fig. 41) be the given projections, and g h i j k of the section made by the plane is easily found by drawing g g, h h, i i, &.c., perpen-
\f!V
Fig. 41.
dicular to the ground-line A D. On A D produced set off the distances g h, h i, i j, and j k at G h, h i, &c., and from the points thus found draw perpendiculars to G K meeting lines drawn from h, i, and j parallel to A D, in H, I, and J. Join G H, H I, I J, and J K. G H I J K is the true section made by the plane.
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MODERN CARPENTRY
A cylinder may be cut by a plane in three different ways — 1st, the plane may be parallel to the axis; 2nd, it may be parallel to the base; 3rd, it may be oblique to the axis or the base.
In the first case the section is a parallelogram, whose length will be equal to the length of the cylinder, and whose width will be equal to the chord of the circle of the base in the line of section. Whence it follows, that the largest section of this kind will be that made by a plane passing through the axis; and the smallest will be when the section plane is a tangent — the section in that case will be a straight line.
When the section plane is parallel to the base, the section will be a circle equal to the base. When the section plane is oblique to the axis or the base, the section will be an ellipse.
XXXIII. To draw the section of a cylinder by a plane oblique to the axis.
Let A B C D (Fig. 42) be the projection of a cylin- der, of which the circle E H F K represents the base .divided into twenty equal parts at b c d e, &c., and let it be required to draw the section made by the plane
SOLID GEOMETRY 71
A C. The circular base must be drawn in such a posi- tion that the axis of the cylinder when produced meets the centre of the circle. Through the centre of the circle drav/ the diameter II K perpendicular to the axis produced. Then through the divisions of the base, bed, &c., draw lines parallel to the axis, and meeting the section plane in 1, 2, 3, &c., and through these points draw perpendiculars to A C making them equal to the corresponding perpendiculars from H K, i e, 1 b, 2 c, 3 d, &c. A curve drawn through the points thus found will be an ellipse, the true section of A B C D on the plane A C*
The Cone. A cone may be cut by a plane in five dif- ferent ways, producing what are called the conic sec- tions: 1st. If it is cut by a plane passing through the axis, the section is a triangle, having the axis of a cone as its height, the diameter of the base for its base, and the sides for its sides. If the plane passes through the vertex, without passing through the axis, as c e (Fig. 43), the section will still be a triangle, having for its base the chord c e, for its altitude the line c e, and for its sides the sides of the cone, of which the lines c e, o e are the horizontal and the line c e the vertical projections. 2nd. If the cone is cut parallel to base, as in g h, the section will be a circle, of v/hich g h will be the diameter. 3rd. When the section plane is oblique to the axis, and passes through the oppo- site sides of the cone, as m p h, the section will be an ellipse, m n h. 4th. When the plane is parallel to one of the sides of the cone, as r h, the resulting sec-
*The point i3 assumed to bs id the inferior half of the cylinder.
72
MODERN CARPENTRY
Fig. 43.
tion is a parabola r s h t u. 5th. When the section plane is such as to pass through the sides of another cone formed by producing the sides of the first be-
SOLID GEOMETRY 73
yond the vertex, as the plane q h, the resulting curve in each cone in a hyperbola.
Several methods of drawing the curves of the conic sections have already been given in Plane Geometry, Vol. 1. Here their projections, as resulting from the sections of the solid by planes, are to be considered. If the mode of finding the projections of a point on the surface of a given cone be understood, the projections of the curves of the conic sections will offer no diffi- culty. Let the problem be: First, to find the projec- tions of the section made by the plane m h. Take at pleasure upon the plane the several points, as p, »S:c. Let fall from these points perpendiculars to the hori- zontal plane, and on these will be found the horizontal projection of the points ; thus, in regard to the point p
. Draw through p a line parallel to A B : this line
will be the vertical projection of the horizontal plane cutting the cone, and its horizontal projection will be a circle, with s n for its radius. With this radius, therefore, from the centre e, describe a circle cutting, twice, the perpendicular let fall from p ; the points of intersection will be two points in the horizontal pro- jection of the circumference of the ellipse. In the same manner, any other points may be obtained in its circumference. The operation may often be abridged by taking the point p in the middle of the line mh; for then mh will be the horizontal pro- jection of the major axis, and the two points found on the perpendicular let fall from the central point p will give the minor axis.
To obtain the projections of the parabola, more points are required, such as r, 2, s, 3, h, but the mode
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MODERN CARPENTRY
of procedure is the same as for the ellipse. The ver- tical projection of the parabola u s h t u is shown at u s h t u.
The projections of the section plane which produces the hyperbola are in this case straight lines, q h, z h.
XXXIV. To draw the section of a cone made by a plane cutting both its sides, i. e., an ellipse.
Let ADB (Fig. 44) be the vertical projection of the cone A C B the horizontal projection of half its base, and E F the line of section. From the points E and F let fall on AB the perpendiculars EG, FH.
SOLID GEOMETRY 75
Take any points in E F, as k, 1, m, n, &c., and from them draw lines parallel to A B, as k p, 1 q, m r, &c., and also lines perpendicular to A B, as k 1, 12 m 3, &c. Also from p, q, r, &e., let fall perpendiculars on AB, namely, pa, q z, r y, &c. From the centre of the base of the cone, 1, with radius la, Iz, ly, &c., describe arcs cutting the perpendiculars let fall from k, 1, m, &c., in 1, 2, 3, &c. A curve traced through these points will be the horizontal projection of the section made by the plane E F. To find the true section, — Through k, 1, m, &c., draw kt, lu, m v, n w, perpendicular through E F, and make them respectively equal to the corre- sponding ordinates, 51, 6 2, 7 3, &c., of the horizontal projection G 4 H, and points will be obtained through which the half E w F of the required ellipse can be traced. It is obvious that, practically it is necessary only to find the minor axis of the ellipse, the major axis E F being given.
XXXV. To draw the section of a cone made by a plane parallel to one of its sides, i. e., a parabola.
Let ADB (Fig. 45) be the vertical projection of a right cone, and A C B half the plan of its base ; and let E F be the line of section. In E F take any number of points, E, a, b, c, e, F, and through them draw lines EH, a 61, b 7 2, &c., perpendicular to A B, and also lines parallel to A B, meeting the side of the cone in f, g, h, k, 1 : from these let fall perpendiculars on A B, meeting it in m n, o, p, q. From the centre of the base 1, with the radii Im, In, lo, &c., describe arcs cutting the perpendiculars let fall from the section line in the points 1, 2, 3, 4, 5 ; and through the points of intersec- tion trace the li^e H 12345G, which is the horizontal
76
MODERN CARPENTRY
projection of the section. To find the true section, from E, a, b, c, d, e, raise perpendiculars to EF, and make them respectively equal to the ordinates in the horizontal projection, as E r equal to E H, as equal to
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6 1, &c., and the points r s t u v w in the curve will "be obtained. The other half of the parabola can be drawn by producino: the ordinates we, v d, &c., and setting the same distances to the right of E F.
SOLID GEOMETRY
77
XXXVI. To draw the section of a cone made by a plane parallel to the axis, i. e., an hyperbola.
Let d c d (Fig. 46) be the vertical projection of the cone, d q r d one half of the horizontal projection of the base, and q r the section plane. Divide the line r q into any number of equal parts in 1, 2, 3, h, &c., and
Fig. 46.
through them draw lines perpendicular to d d. From c as centre, with the radii cl, c2, &c., describe the arcs cutting d d ; and from the points of intersection draw perpendiculars cutting the sides of the cone in 1, 2, 3, and these heights transferred to the corresponding per- pendiculars drawn directly from the points 1, 2, o, &c., in rq, will give points in the curve.
78 MODERN CARPENTRY
XXXVII. To draw the section of a cuneoid made by a plane cutting both its sides.
Let ACB (No. 1, Fig. 47) be the vertical projection of the cuneoid, and A5B the plan of its base, and AB (No. 4) the length of the arris at C, and let D E be the line of section. Divide the semi-circle of the base into any number of parts, 1, 2, 3, 4, 5, &c., and through them draw perpendiculars to AB, cutting it in 1, m, n, o, p, &c., and join CI, C m, Cn, &c., by lines cutting the section line in 6, 7, 8, 9 &c. From these points draw lines perpendicular to D E and make them equal to the corresponding ordinates of the semi- circle, either by transferring the lengths by the com- passes, or by proceeding as shown in the figure. The curve drawn through the points thus obtained will give the required section.
The section on the line D K is shown in No. 2, in which AB equals DK; and the divisions efghk in D K, &G., are transferred to the corresponding points on A B ; and the ordinates el, f m, g n, &c., are made equal to the corresponding ordinates 11, m2, n3, of the semi-circle of the base. In like manner, the section of the line G IT, shown in No. 3, is drawn.*
XXXVIII. To describe the section of a cylinder made by a curve cutting the cylinder.
Let AB D E (Fig. 48) be the projection of the cylin- der, and C D the line of the section required. On A B
*A cuneoid is a solid ending in a straight line, in which, if any point be taken, a perpendicular from that point may be made to coincide with the surface. The base of the cuneoid may be of any form; but in architecture it is usually semi- circular or semi-elliptical, and parallel to the straight line forming the other end.
SOLID GEOMETRY
79
describe a semi-circle, and divide it into any number of parts. From the points of division draw ordinates 1 h, 2k, 31, 4m, &c., and produce them to meet the line of section in o, p, q, r, s, t, u, v, w. Bend a rule or slip of paper to the line C D, and prick off on its points C,
No. 2.
No. 8.
'k
ijHk
No.«.
Fig. 47.
o, p, q, &c. ; then draw any straight line F G, and, un- bending the rule, transfer the points C, o, p, q, &c., to F, a, b, e, d, &c. Draw the ordinates al, b 2, e 3, &c., and make them respectively equal to the ordinates h 1, k2, 13,, &c., and through the points found trace the curve.
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MODERN CARPENTRY
XXXIX. To describe the section of a sphere.
Let A B D C (Fig. 49) be the great circle of a sphere, and F G the line of the section required. Then since all the sections of a globe or sphere are circles, on F G de- scribe a semi-circle F 4G, which will be the section re- quired.
Or, in F G take any number of points, as m, 1, k, H, and from the centre of the great circle E, describe the arcs H n, k o, 1 p, m q, and draw the ordinates H 4, k 3, 12, ml, and n4, o3, p2, ql; then make the ordinates on F G equal to those on B C, and the points so ob- tained will give the section required.*
*Tlie projections of sections of spheres are, if the section panes are oblique, either straight lines or ellipses, and are found as follows:
Let a 1) (Fig. 50) be the horizontal projection of the sec- tion plane. On the line of section take any number of points, as a, c, 1), and through each of them draw a line perpendicular to y X,
SOLID GEOMETRY
81
XL. To describe the section of an ellipsoid when a section through the fixed axis, and the position of the line of the required section are given.
Let A B C D (Fig. 51) be the section through the fixed axis of the ellipsoid, and F G the position of the line of the required section. Through the centre of
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the ellipsoid draw B D parallel to F G; bisect F G in H and draw A C perpendicular to F G; join B C, and from F draw F K, parallel to B C, and cutting A C pro- duced in K; and then will H K be the height of the semi-ellipse forming the section on F G.
82 MODERN CARPENTRY
Or, the section may be found by the method of or- dinates, thus: As the section of the ellipsoid on the line A C is a circle, from the point of intersection of B D and A C describe a semicircle A E C. Then on H G, the line of section, take any number of points, 1, m, p, and from them raise perpendiculars cutting the ellipse in q, r, s. From q, r, s draw lines perpendicular to A C, cutting it in the points 4, 5, 6 ; and again, from the intersection of B D and A C as centre draw the arcs 41, 5m, 6n, Co, cutting HG in 1, m, n, o; then H 0, set off on the perpendicular from H to K, is the height of the section ; and the heights H n, M m, HI, set off on the perpendiculars from 1 to 3, n to 2 and. p to 1, give the heights of the ordinates.
XLI. To find the section of a cylindrical ring: per- pendicular to the plane passing* through the axis of the ring, the line of section being' given.
Let ABED (Fig. 52) be the section through the axis of the ring, A B a straight line passing through the concentric circles to the centre C, and D E be the line of section. On A B describe a semicircle ; take in its circumference any points, as 1, 2, 3, 4, 5, &c., and draw the ordinates 1 f , 2 g, 3 h, 4 k, &c. Through the points f, g, h, k, 1, &c., where the ordinates meet the line A B, and from the centre C, draw concentric cir- cles, cutting the section line in m, n, o, p, q, &c. Through these points draw the lines ml, n 2, o 3, &c., perpendicular to the section line, and transfer to them the heights of the ordinates of the semicircle f 1, g 2, &c. ; then through the points 1, 2, 3, 4, &c., draw the curve D 5 E, which is the section required.
SOLID GEOMETRY
83
Again, let R S be the line of the required section ; then from the points t, \i, v, w, c, x, cl, &c., where the concentric circles cut this line, draw the lines t 1, u 2, V 3, &e., perpendicular to R S, and transfer to them the corresponding ordinates of the semicircle ; and through the points 1, 2, 3, 4, e, 5, f, &c., draw the curve R e f S, which is the section required.
Fig. 52.
XLII. To describe the section of a solid of resolu- tion the generating curve of which is an ogee.
Let ADB (Fig. 53) be half the plan or base of the solid, A a b B the vertical section through its axis, and E F the line of section required. From G draw C 5 perpendicular to E P, and bisecting it in m. In Em
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MODERN CARPENTRY
take any number of points, g h k, &c., and through them draw the lines g 1, h 2, k 3, &c., perpendicular to E F. Then from C as a centre, through the points g, h, k, &c., draw concentric ares cutting A B in r, s, t, u, V, and through these points draw the ordinates r 5, s 4, t 3, &c., perpendicular to A B. Transfer the heights of the ordinates on A B to the corresponding ordinates on each side of the centre of E F; and through the points 1, 2, 3, 4, 5, &c., draw the curve E 5 F, which is the section required.
Fig. 53
XLIII. To find the section of a solid of resolution, the generating curve of which is of a lancet form.
Let A D B (Fig. 54) be the plan of half the base, A E B the vertical section, and F G the line of the re- quired section. The manner of finding the ordinates
SOLID GEOMETRY
85
and transferring- the heights is precisely the same as in the last problem.
XLIV. To find the section of aji ogee pyramid with a hexagonal base.
Let A D E F B (Fig. 55) be the plan of the base of the pyramid, AabB a vertical section through its axis, and G H the line of the required section. Draw the arrises C D, C E, C F. On the line of section G H, at the points of intersection of the arrises with it, and
Fig. 55.
at some intermediate points k, m, o, q (the correspond- ing points k and q, and m and o, being equidistant from n), raise indefinite perpendicularsf. Through these points k, 1, m, n, o, p, q, draw lines parallel to the sides of the base, as shown by dotted lines; and from the points where these parallels meet the line A B, draw r 4, s 3, t 2, u 1, perpendicular to A B. These perpendiculars transferred to the ordinates n 4, m 3, o 5, 1 2, p 6, k 1, q 7, will give the points 1, 2, 3, 4, 5, 6, 7, through which the section can be drawn.
86 MODERN CARPENTRY
6. INTERSECTIONS OF CURVED SURFACES.
When two solids having curved surfaces penetrate or intersect each otheu, the intersections of their sur- faces form curved lines of various kinds. Some of these, as the circle, the ellipse, &c., can be obtained in the plane ; but the others cannot, and are named curves of double curvature. The solution of the following problems depends chiefly on the knowledge of how to obtain, in the most advantageous manner, the pro- jections of a point on a curved surface ; and is in fact the application of the principles elucidated in the sev- eral previous problems. The manner of constructing the intersections of these curved surfaces which is the simplest and most general in its application, consists in conceiving the solids to which they belong as cut by planes according to certain conditions, more or less dependent on the nature of the surfaces. These sec- tion planes may be drawn parallel to one of the planes of the projection; and as all the points of intersection of the surfaces are found in the section planes, or on one of their projections, it is always easy to construct the curves by transferring these points to the other projection of the planes.
XLV. The projection of two equal cylinders v.'hich intersect at right angles being given, to find the pro- jections of their intersections.
Conceive, in the horizontal projection (Fig. 56), a series of vertical planes cutting the cylinders parallel to their axis. The vertical projections of all the sec- tions will be so many right-angled parallelograms, sim-
SOLID GEOMETRY
87
ilar to e f e f, which is the result of the section of the cylinder from surface to surface. The circumference of the second cylinder, whose axis is vertical, is cut by the same plane, which meets its upper surface at the two points g, h, and its under surface at two corre- sponding points. The vertical projections of these points are on the lines perpendicular to a b, raised on
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each of them, so that upon the lines e f , e f , will be situated the intersections of these lines at the points g, h, and g, h and the same with other points i, k, 1, m. It is not necessary to draw a plan to find these pro- jections. All that is actually required is to draw the circle representing one of the bases (as n o) of the cylinder laid flat on the horizontal plane. Then to produce g h till it cuts the circle at the superior and
88
MODERN CARPENTRY
inferior points G, G, and to take the heights e G, e G, and carry them, upon a b, from g to g g and from h to h, h.
Fig. 57 is the vertical projection made on the line
xz.
Fig. 57.
XL VI, To construct the projections of two unequal cylinders whose axis intersect each other obliquely.
Let A (Fig. 58) be the vertical projection of the two cylinders, and h S d e the horizontal projection of their axis. Conceive in the vertical projection, the cylin- ders cut by any number of horizontal planes ; the hori- zontal projections of these planes will be rectan<?les, as in the previous example, and their sides will be parallel to the axis of the cylihders.
The points of intersection of these lines will be the points sought. Without any previous operation, six of those points of intersection can be obtained. For ex- ample the point c is situated on d e, the highest point of the smaller cylinder; consequently, the horizontal projection of c is on d e, the horizontal projection of
SOLID GEOMETRY
89
d e, and it is also on the perpendicular let fall from e, that is to say, on the line e f parallel to the axis of the cylinder S h. The point sought will, therefore, be the intersection of those lines at c. In the same way i is
obtained. The point j is on the line k 1, which is in the horizontal plane passing through the axis d e, the hori- zontal projections of k .1 are k 1, and its opposite m n ; therefore in letting fall perpendiculars from j p, the intersections of these with k 1, m n, give the points j j,
90 MODERN CARPENTRY
p p. Thus six points are obtained. Take at pleasure an intermediate point q; through this point draw a line r s parallel to a b, which will be the vertical pro- jection of a horizontal plane cutting the cylinder in q. The horizontal projection of this section Avill be, as in the preceding examples, a rectangle which is ob- tained by taking, in the vertical projection, the height of the section plane above the axis d e, and carrying it on the base in the horizontal projection from G to T. Through T is then to be drawn the line Q U perpen- dicular to G T ; and through Q and U the lines parallel to the axis; and the points in which these lines are in- tersected by the perpendiculars let fall from q u are the intermediate points required. Any number of inter- mediate points can thus be obtained; and the curve being drawn through them, the operation is completed.
XLVn. To find the intersections of a sphere and a cylinder.
Let e f c d and i k g h (Fig. 59) be the horizontal pro- jections of the sphere and cylinder respectively. Draw parallel to A B, as many vertical section planes are considered necessary, as e f, c d. These planes cut at the same time both the sphere and the cylinder, and the result of each section Avill be a circle in the case of the sphere, and a rectangle in the case of the cylin- der. Through each of the points of intersection g, h, i, k, and from the centre 1, draw indefinite lines per- pendicular to A B. Take the radius of the circles of the sphere proper to each of these sections and with them, from the centre 1, cut the correspondent perpen- diculars in g g, h h, i i, &c., and draw through these points the curves of intersection.
SOLID GEOMETRY
91
XL VIII, To construct the intersection of two rig'ht cones with circular bases.
The solution of this problem is founded on the knowledge of the means of obtaining on one of the projections of a cone a point given on the other.
Fig. 59.
Fig. 60.
No. 1. Let AB (Fig. 60) be the common section of the two planes of projection, the circles g d e f and g h i k the horizontal projections of the given cones, and the triangles d 1 f and h 1 k their vertical projec- tions. Suppose these cones cut by a series of horizon- tal planes : each section will consist of two circles, the
92 . MODERN CARPENTRY
intersections of which will be points of intersection of the conical surfaces. For example, the section made by a plane m n will have for its horizontal projections two circles of different diameters, the radius of the one being i m, and of the other 1 o. The intersecting points of these are p and q, and these points are com- mon to the two circumferences ; and their vertical pro- jection on the plane m n will be p q. Thus, as many points may be found as is necessary to complete the curve.
But there are certain points of intersection which cannot be rigorously established by this method with- out a great deal of manipulation. The point r in the figure is one of those ; for it will be seen that at that point the two circles must be tangents to each other, and it would be difficult to fix the place of the sec- tion plane s t so exactly by trial, that it would just pass through the point.
It will be seen that the point r must be situated in the horizontal projection of the line g i a perpendicu- lar i 1 equal to the height of the cone. From one raise a perpendicular and make it equal to the height of the second cone, and draw its side L i ; and from the point of intersection R let fall a perpendicular on g i meet- ing it in r; through r draw an indefinite line perpen- dicular to A B, and set up on it from A B to r the height rR. The point r can also be obtained directly in the vertical projection "by joining i g and 1 i as shown.
Bisect r g in u, and from u as a centre, with the radius u r, describe a circle, the circumference of which will be the horizontal projection of the intersec-
SOLID GEOMETRY
93
tion of the two cones. The vertical projection of this circle will be g p r q, and can be found by the method indicated above.
No. 2. Conceive the horizontal projection (Fig. 61^ No. 1) a vertical plane C D cutting both cones through their axis: the sections will be two triangles, having
Fig. 61.
the diameters of the bases of the cones as their bases, and the height of the cones as their height. And as in the example the cones are equal, the triangles will also be equal, as the triangles c e f , g f d, in the vertical projection. Conceive now a number of inclined planes, as c n m, e k p, &c., passing through the differ- ent points of the base, but still passing through the summits of the cones ; the sections which result will still be triangles (as has already been demonstrated), whose bases diminish in proportion as the planes re-
94 MODERN CARPENTRY
cede from the centres of the bases of the cones, until at length the plane becomes a tangent to both cones and the result is a tangent line whose projections are h c, h c, g e, f f. It will be observed that the circum- ferences of the bases cut each other at m and i, which are the first points of their intersections, whose verti- cal projections are the point m merely. If the pro- jections of the other points of intersection on the lines of the section planes are found (an operation present- ing no difficulty, and easily understood by the inspec- tion of the figure), it will be seen that the triangles u c m, m c o, k e p, q e r, &c., in the horizontal projec- tion, have for their vertical projections the triangles n e m, m o, k e p, &c., and that the intersections of the cones are in a plane perpendicular to both planes of projection, and the projections of the intersections are the right' lines i m, m 3. From the known properties of the conic sections, the curve produced by this plane will be a hyperbola. Fig. 61, No. 2, gives the projec- tions of the cones on the line o x.
No. 3. The next example (Fig. 62) differs from the last in the inequality of the size of the cones. Sup- pose an indefinite line C D to be the horizontal projec- tion of the vertical section plane, cutting the two cones through their axis e f. Conceive in this plane an in- definite line e f D, passing through the summits of the cones, the vertical projection of this line will be ef d: from d let fall on C D a perpendicular meeting it in D ; this will be the point in which the line passing through the summits of the cones will meet the horizontal plane; and it is through this point, and through the summits e and f, that the inclined section planes should
SOLID GEOMETRY
95
be made to pass. The horizontal traces of these phines are 0 D, G D, &e. : 0 D is then the trace of a tangent plane to the two conical surfaces 0 e, Pf; and the plane e G D cuts the greater cone, and forms by the section the triangles G e H in the horizontal, and g e h in the vertical projection; and it cuts the lesser cone, and forms the triangles If J, if j. In the horizontal
Fig. 62.
projection it is seen that the sides H e, 1 f of the trian- gles intersect in k, which is therefore the horizontal projection of one of the points of intersection ; and its vertical projection is k. In the same manner, other points can be found. It is seen at once that M, N, 1, are also points in the intersection. The curves traced through the points M k 1 N in the horizontal, and m k 1 in the vertical projection, are the projections of the intersection of the two cones.
96
MODEKN CAKFENTRY
7. COVERINGS OF SOLIDS.
1. Regular Polyhedrons. A solid angle cannot be formed with fewer than three plane angles. The sim- plest solid is, therefore, the tetrahedron or pyramid having an equilateral triangle for its base, and its other three sides formed of similar triangles.
The development of this figure (Fig. 63) is made by- drawing the triangular base ABC, and then drawing around it the triangles forming the inclined sides. If the diagram is on flexible material, such as paper, then cut out, and the triangles folded on the lines A B, B C, C A, the solid figure will be constructed.
Fig. 63.
Fig. 64.
The hexahedron, or cube, is composed of six equal squares (Fig. 64) ; the octahedron (Fig. 65) of eight equilateral triangles; the dodecahedron (Fig. 66) of twelve pentagons; the icosahedron (Fig. 67) of twenty
SOLID GEOMETRY
97
equilateral triangles. In these figures, A is the eleva- tion and B the development.
The elements of these solids are the equilateral tri- angle, the square, and the pentagon. The irregular polyhedrons may be formed from those named, by cut- ting off the solid angles. Thus, in cutting off the
Fig. 65.
Fig. 66.
angles of the tetrahedron, there results a polyhedron of eight faces, composed of four hexagons and four equilateral triangles. The cutting off the angles of the cube, in the same manner gives polyhedron of four-
98
MODERN CARPENTRY
teen faces, composed of six octagons, united by eight equilateral triangles.
The same operation performed on the octahedron produces fourteen faces, of which eight are hexagonal and six square; on the dodecahedron it gives thirty- two sides, namely, twelve decagons, and twenty trian- gles; on the icosahedron it gives thirty-two sides — twelve tentagons and twenty hexagons. This last ap- proaches almost to the globular form and can be rolled like a ball.
Pig. 67.
The other solids which have plane surfaces are the ]\vramids and prisms. These may be regular and irreg- ular: they may have their axis perpendicular or in- clined; they may be truncated or cut with a section, parallel or oblique, to their base. ^
SOLID GEOMETRY
99
II. Pyramids. The development of a right pyra- mid, of which the base and the height are given, offers no difficulty. The operation consists (Fig. 68) in ele- vating on each side of the base, a triangle having its height equal to the inclined height of each side, or, otherwise, connecting the sides, together as shown by the dotted lines.
Fig. 68.
Pig. 69.
In an oblique pyramid the development is found as follows: Let abed (Fig. 69) be the plan of the base of the pyramid, a b c d its horizontal projection, and E F G its vertical projection. Then on the side d c construct the triangle c R d, making its height equal to the sloping side of the pyramid F G. This triangle is the development of the side d P c of the pyramid.
100 MODERN CARPENTRY
Then from d, with the radius E F, describe an arc 0 ; and from R, with the radius equal to the true length e G of the arris E G, describe another arc intersecting the last at 0. Join R 0, d 0 ; the triangle d R 0 will be the development of the side a P d. In the same way, describe the triangle c R T, for the development of the side b P c. From R, again, with the same radius R 0, describe an arc S, which intersect by an arc described from 0 with the radius a b ; and the triangle 0 R S will be the development of the side a P b.
If the pyramid is truncated by a plane w a parallel to the base, the development of that line is obtained. by setting off from R on R c, and R d, the true length of the arris G a in x and 2, and on R S, R 0, and R T, the true length of the arris G w in 4, 3, 1 ; and drawing the lines 1 X, x2, 2 3, 3 4, parallel to the base of the respect- ive triangles T R c, c R d, d R 0, 0 R^S. If it is trun- cated by a plane w y, perpendicular to the axis, then from the point R, with the radius equal to the true length of the arris G w, or G y, describe an arc 1 4, and inscribe in it the sides of the polygon forming the pyramid.
III. Prisms. In a right prism, the faces being all perpendicular to the bases which truncate the solid, it results that their development is a rectangle composed of all the faces joined together, and bounded by two parallel lines equal in length to the contour of the bases. Thus, in Fig. 70, a b c d is the base, and b e the height of the prism; the four sides will form the rec- tangle b e f g, and e h i k will be the top of the prism. The full lines show another method of development.
When a prism is inclined, the faces form different
SOLID GEOMETRY
101
angles with the lines of the contours of the bases; whence there results a development, the extremities of which are bounded by lines forming parts of poly- gons.
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Fig. 70.
After having drawn the line C C (Fig. 71), which in- dicates the axis of the prism and the lines A B, D E, the surfaces which terminate it, describe on the middle of the axis the polygon forming the plan of the prism, taken perpendicularly to the axis, and indicated by the figures 1 and 8. Produce the sides 1 2, 6 5, parallel to the axis, until they meet the lines A B, D E. These lines then indicate the four arrises of the prism, cor- responding to the angles 12 5 6. Through the points
102
MODERN CARPENTRY
8 3 7 4 draw lines parallel to the axis meeting A B, D E in F H, G L. These lines represent the four arrises 8 3 7 4.*
In this profile the sides of the plan of the polygon 12 3 45678 give the width of the faces of the prism,
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and the lines AD, F H, G L, BE their length. From this profile can be drawn the horizontal projection, in the manner shown below. To trace the development of this prism on a sheet of paper, so that it can be folded
*In Fig. 75 another example is given, but as the method of procedure is the same as in Fig. 71, detailed description is un- necessary.
SOLID GEOMETRY 103
together to form the solid, proceed thus: On the middle of C C raise an indefinite perpendicular M N. On that line set off the width of the faces of the prism, indicated by the polygon, in the points 01234567 8. Through these points draw lines parallel to the axis, and upon them set off the lengths of the lines in profile, thus: From the points 0, 1, and 8, set off the length M D in the points D D D ; from 2 and 7, set off a H in H ; from 3 and 6, set off b L in L and L ; and so on. Draw the lines DD, D H L E, EE, ELHD, for the contour of the upper part of the prism. To obtain the contour of the lower portion, set off the length M A from 0, 1, and 8 to A A A, the length a F from 2 and 7 to F and F, the length b G from 3 and 6 to G and G, and so on ; and draw A A, A F G B, B B, B G F A, to complete the contour. The development is completed by making on B B and E E the polygons 12 3 4 5 6 BB, 12 3 4 5 6 EE, similar to the polygons of the horizon- tal projection.
IV. Cfiylinders. Cylinders may be considered as prisms, of which the base is composed of an infinite number of sides. Thus we shall obtain graphically the development of a right cylinder by a rectangle of the same height, and of a length equal to the circumference of the circle, which serves as its base.
To find the covering of a right cylinder.
Let A B C D (Fig. 72) be the seat or generating sec- tion. On A D describe the semicircle A 5 D, represent- ing the vertical section of half the cylinder, and divide its circumference into any number of parts, 1, 2, 3, 4,
104
MODERN CARPENTRY
5, &c., and transfer those divisions to the lines A D and B C produced ; then the parallelogram D C G F will be the covering of one half the cylinder.
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Fig. 72.
To find the edge of the covering when it is oblique in regard to the sides of the cylinder.
Let A B C D (Fig. 73) be the seat of the generating section, the edge B C being oblique to the sides A B, D C. Draw the semicircle A 5 D, and divide it into any
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Fig. 73.
number of parts as before ; and through the divisions draw lines at right angles to AD, producing them to meet B C in r s, t, u, v, &c. produce A D, and the lines 1 a, 2 b, 3 e, &c., perpendicular to D F. To these lines
SOLID GEOMETRY
105
transfer the length of the corresponding lines inter- cepted between A D and B C, that is, to la transfer the length p z, to 2 b transfer o y, and so on, by drawing the lines z a, y b, x c, &e., parallel to A F. Through the points thus obtained, draw the curved line C a b c, &c., to G ; then shall D F C G be the development of the covering of the semi-cylinder A B C D.
Fig. 74.
To find the covering- of a cylinder contained between two oblique parallel planes.
Let A B C D (Fig. 74) be the seat of the generating section. From A draw A G perpendicular to A B, and produce C D to meet it in E. On A E describe the semi- circle, and transfer its perimeter to E G, by dividing it into equal parts, and setting off corresponding di- visions on E G. Through the divisions of the semicircle draw lines at right angles to AE, producing them to
106
MODERN CARPENTRY
meet the lines A D and B C, in i, k, 1, m, &e. Through the divisions on E G draw lines perpendicular to it; then through the intersections of the ordinates of the semicircle, with the line A D, draw the lines i a, k z, 1 y^
hdfkK.
Fig. 75.
&c., parallel to A G and where these intersect the per- pendiculars from E G, in the points a, z, y, x, w, u, &e., trace a curved line G D, and draw parallel to it the curved line H C ; then will D C H G be the development of the covering of the semi-cylinder A B C D.
To find the covering" of a semi-cylindric surface bounded by two curved lines.
The construction to obtain the developments of these coverings (Figs. 76 and 77) is precisely similar to that described in Fig. 74.
SOLID GEOMETRY
107
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Fig. 7«.
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Fig. 77.
108'
MODERN CARPENTRY
V. Cones. We have considered cylinders as prisms with polygonal bases. In a similar manner we may re- gard cones as pyramids.
In right pj^ramids, with regular symmetrical bases,
as the lines of the arrises extending from the summit to the base are equal, and as the sides of the polygons forming the base are also equal, their developed sur- faces will be composed of similar and equal isosceles triangles, which, as we have seen (Fig. 78, a, b, e, d), will, when united, form a part of a regular polygon
SOLID GEOMETRY 109
inscribed in a circle, of which the inclined sides of the polygon form the radii. Thus in considering the base of the cone K H (Fig. 78) as a regular polygon of an infinite number of sides, its development will be found in the sector of a circle, MAFBM (No. 3), of which the radius equals the side of the cone K G (No. 1), and the arc equals the circumference of the circle forming its base (No. 2).
To trace on the development of the covering, the curves of the ellipse, parabola, and hyperbola, which are the result of the sections of the cone by the lines D I, E F, I G, it is necessary to divide the circumfer- ence of the base AFBM (No. 2) into equal parts, as 1, 2, 3, &c., and from these to draw radii to the centre C, which is the horizontal projection of the vertex of the cone ; then to carry these divisions to the common intersection line K H, and from their terminations there to draw lines to the vertex G, in the vertical pro- .iection No. 1. These lines cut the intersecting planes, forming the ellipse, parabola, and hyperbola, and by the aid of the intersections we obtain the horizontal projection of these figures in No. 2 — the parabola passing through M E F, the hyperbola through GIL, and the ellipse through D I.
To obtain points in the circumference of the ellipse upon the development, through the points of inter- section 0, p, q, r, &c., draw lines parallel to K H, car- rying the heights to the side of the cone G H, in the points 1, 2, 3, 4, 5, 6, 7, and transfer the lengths G 1, G 2, G 3, &c. to G 1, G 2, G 3, G 4, &c., on the radii of the development in No. 3 ; and through the points thus obtained draw the curve z D 1 X.
110
MODERN CARPENTRY
To obtain the parabola and hyperbola, proceed in the same manner, by drawing parallels to the base K H, through the points of intersection ; and transfer- ring the lengths thus obtained on the sides of the cone G K, G H, to the radii in the development.
Nos. 4, and 5 give the vertical projections of the hy- perbola and parabola respectively.
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Fig. 79.
To find the covering of the frustum of a cone, the section being made by a plane perpendicular to the axis.
Let A C E F (Fig. 79) be the generating section of
SOLID GEOMETRY
111
the frustum. On A C describe the semicircle ABC, and produce the sides A E and C F to D. From the centre D, with the radius D C, describe the arc C H ; and from the same centre with the radius D F, describe
Fig. 80.
the arc F G. Divide the semicircle ABC into any number of equal parts, and run the same divisions along the arc C H ; draw the line H D, cutting E G in G ; then shall C H G F be half the development of the covering of the frustum A C F E.
112 MODERN CARPENTRY
To find the covering of the frustum of a cone, the section being made by a plane not perpendicular to the axis.
Let A C F E (Fig. 80) be the frustum. Proceed as in the last problem to find the development of the eov- erinff of the semi-cone. Then — to determine the edge of the covering of the line E F — from the points P, q, r, s, t, &c., draw lines perpendicular to E F, cutting A C in y, x, w, v, u ; and the length u t transferred from 1 to a, V s, transferred from 2 to b, and so on, will give a, b, c, d, e, &c., points on the edge of the cover- ing. .
To find the covering of the frustum of a cone, when cut by two cylindrical surfaces perpendicular to the generating sectios.
Let AEFC (Fig. 81) be the given frustum, and A k C, E p F, the given cjdindricai surfaces. Produce A E, C F, till they meet in the point D. Describe the semicircle ABC, and divide it into any number of equal parts, and transfer the divisions to the arc C H, described from D, with the radius D C. Through the divisions in the semicircle 1, 2, 3, 4, &c., draw lines per- pendicular to A C, and through the points where they intersect A C draw lines to the summit D. Draw lines also through the points 1, 2, 3, 4, 5, &c., of the arc C H, to the summit D ; then through the intersections of the lines from A C to D, with the seats of the cylindrical surfaces k, 1, m, n, o, and p, q,r, s, t, draw lines parallel to A C, cutting C D ; and from the points of intersec- tion in C D and from the centre D, describe arcs cut- ting the radial lines in the sector D C H in u, v, w, x, y,
SOLID GEOMETRY
113
&c., and a, b, c, d, e, &c., and curves traced through the intersections will give the form of the covering.
VI. Spheres, Ellipsoids, &c. The development of the sphere, and of other surfaces of double curvature,
Fig. 81,
is impossible, except on the supposition of their being composed of a great number of small faces, either plane, or of a. simple curvature, as the cylinder and the
114
MODERN CARPENTRY
/ b\ r
Fig. 82.
cone. Thus, the sphere or spheroid may be consitlered as a polyhedron, terminated, 1st, by a great number of plane faces, formed by truncated pyramids, of which the base is a polygon, as in Fig. 82; 2nd, by parts of
SOLID GEOMETRY
.115
truncated cones forming zones, as in Fig. 83, the part above AB being the vertical projection, and the part below A B the horizontal projection ; 3rd, by parts of cylinders cut in gores, forming flat sides, which dimin- ish in width, as in Fig. 84.
Fig. 83.
Fig. 84.
In reducing the spheres, or spheroid, to a polyhedron with flat sides, two methods may be adopted, which dif- fer only in the manner of arranging the developed faces.
The most simple method is by parallel circles, and others perpendicular to them, which cut them in two opposite points, as in the lines on a terrestrial globe. If we suppose that these divisions, in place of being circles, are polygons of the same number of sides, there will result a polyhedron, like that represented in Fig. 82, of which the half, A D B, shows the geometrical ele- vation, and the other half, A E B, the plan.
To find the development, first obtain the summits
116 MODERN CARPENTRY
P, q. r, s, of the truncated pyramids, which from the demi-polvhedron A D B, by producing the sides A 1, 12, 23, 34, until they meet the axis ED produced; then from the points p, q, r, s, and with the radii P A, PI, q I q 2, r 2, r 3, and s 3, s 4, describe the indefinite arcs A B, 1 b, 1 b, 2 t 2 f. e q, 3 g, 4 h, and from D de- scribe the arc 4h; upon all these arcs set off the divi- sions of the demi-polygons A E B. and draw the lines to the summits p. q, r, s, and D, from all the points so set out, as A, 1, 2, 3, 4, &c., from each truncated pyramid. These lines will represent for every band or zona the faces of the truncated pjTamids of which they consti- tute a part.
The development can also be made by drawing through the centre of each side of the polygon A E B, indefinite perpendiculars, and setting out upon them the heights of the faces in the elevation, Al 2 3 4 D, and through the points thus obtained drawing paral- lels to the base. On each of these parallels then set out the widths, h, i, k, 1, d, of the corresponding faces (e, e. e. &c.) in the plan, and there will be thus formed trapeziums and triangles, as in the first development, but arranged differently. This method is used in con- structing geographical globes, the other is more con- venient in finding the stones of a spherical dome.
The development of the sphere by reducing it to conical zones (Fig. 83) is accomplished in the same manner as the reduction to truncated pyramids, ^vith this difference, that the developments of the arrises, indicated by A 1 2 3 4 5 B in Fig. 82, are arcs of circles described from the summits of cones, in place of being polygons.
SOLID GEOMETRY
117
The development of the sphere reduced into parts of cylinders, cut in gores (Fig. 84), is produced by the second method described, but in place of joining, by straight lines, the points E, h, i, k, 1, d (Fig. 82), we
unite them by curves. This last method is used in tracing the development of caissons in spherical or spheroidal vaults.
To find the covering of a segmental dome.
In Fig. 85, No. 1 is the plan, and No. 2 the elevation of a segmental dome. Through the centre of the plan
118 MODERN CARPENTRY
E draw the diameter A C, and the diameter B D per- pendicular to A C, and produce B D to I. Let D E rep- resent the base of semi-section of the dome ; upon D E describe the arc D k with the same radius as the arc F G H (No. 2) ; divide the arc D k into any num- ber of equal parts, 1, 2, 3, 4, 5, and extend the division upon the right line D I, making the right line D I equal in length, and similar in its divisions, to the are D k : from the points of division, 1, 2, 3, 4, 5, in the arc D k, draw lines perpendicular to D E, cutting it in the points q, r, s, &c. Upon the circumference of the plan No. 1, set off the breadth of the gores or boards 1 m, m n, n 0, 0 p, &c. ; and from the points 1, m, n, o, p, draw right lines through the centre E : from E describe concentric arcs q v, r u, st, &c., and from 1 describe concentric arcs through the points D, 1, 2, 3, 4, 5, : 1 m, being the given breadth of the base, make 1 w equal to q v, 2x equal to r u, 3 y equal to st, &c. ; draw the curved line through the points 1, w, x, y, &c., to 1, which will give one edge of the board or gore to coin- cide with the line 1 E. The other edge being similar, it will be found by making the distances from the centre line D 1 respectively equal. The seats of the different boards or gores on the elevation are found by the perpendicular dotted lines, pp, oo, nn, m m, &c.
To find the covering of a semicircular dome.
Fig. 86, Nos. 1 and 2. — The procedure here is more simple than in the case of the segmental dome, as, the horizontal and vertical sections being alike, the or- dinates are obtained at once.
SOLID GEOMETRY
119
To find the covering of a semicircular dome when it is required to cover the dome horizontally.
Let A B C (Fig. 87) be a vertical section through the axis of a circular dome, and let it be required to cover this dome horizontally. Bisect the base in the point
120
MODERN CARPENTRY
D, and draw D B E perpendicular to A C, cutting the circumference in B. Now divide the arc B C into equal parts, so that each part will be rather less than
Fig. 87.
the width of the a board ; and join the points division by straight lines, which will form an inscribed polygon of so many sides; and through these points draw line parallel to the base A C, meeting the opposite sides of
SOLID GEOMETRY
121
the circumference. The trapezoids formed by the sides of the polygon and the horizontal lines may then be regarded as the sections of so many frustums of cones ; whence results the following mode of procedure, in
J c
accordance with the introductory illustration Fig. 82; —Produce, until they meet the line D E., the lines g f, f, n, &c., forming the sides of the polygon. Then, to de- scribe a board which corresponds to the surface of one of the zones, as f g, of which the trapezoid m 1 f g is a
122
MODERN CARPENTRY
section, — from the point h, where the line f g produced meets D E, with the radii h f, hg, describe two ares, and cut off the end of the board k on the line of a radius h k.
To obtain the true length of the board, proceed as in Fig. 89. The other boards are described in the same manner.
Fig. 89.
To find the covering of an elliptic dome.
Let A B C D (Fig. 90) be the plan, and F G H the elevation of the dome. Divide the elliptical quadrant F G (No. 2) into any number of equal parts in 1, 2, 3, 4, 5, and draw through the points of division lines
SOLID GEOMETRY 123
perpendicular to F H, and produced to A C (No. 1), meeting it in i, k, m, n, : these divisions are transferred by the dotted arcs to the gore b E c and the remainder of the process is as in Figs. 85 and 86.
To find the covering boards of an ellipsoidal dome.
Let A B C D (No. 1, Fig. 89) be the plan of the dome, and FGH (No. 2) the vertical section through its major axis. Produce F H indefinitely to n ; divide the circumference, as before, into any number of equal parts, and join the divisions by straight lines, as p m, ml, &c. Then, describe on a board, produce the line forming one of the sides of the polygon, such as 1 m, to meet F n in n; and from n, with the radii n m, n 1 describe two arcs forming the sides of the board, and cut off the board on the line of the radius n o. Lines drawn through the points of divisions at right angles to the axis, until they meet the circumference ADC of the plan, will give the plan of the boarding.
To find the covering of an ellipsoidal dome in gores.
Let the ellipse A B C D (Fig. 90, No. 1) be the plan of the dome, A C its major axis and B D its minor axis ; and let A B C (No. 2) be its elevation. Then, first, to describe on the' plan and elevation the lines of the gores, proceed thus: — Through the line AC (No. 1) produced at H, draw the line E G perpendicular to it, and draw BE, D G, parallel to the axis A C, cutting E G, then will E G be the length of the axis minor, on which is to be described the semi-circle E F G, rep- resenting a section of the dome on a vertical plane passing through the axis minor.
Divide the circumference of the semi-circle into any number of equal parts, representing the widths of the
124
MODERN CARPENTRY
covering boards on the line B D ; and through, the points of division 1, 2, 3, 4, 5, draw lines parallel to the axis A C, cutting the line B D in 1, 2, 3, 4, 5. Divide the quadrant of the ellipse CD (No. 1) into any num- ber of equal parts in e, f, g, h, and through these points draw the lines e a, f b, g c, h d in both diagrams, per-
■*
JVJ
Pig. 90.
pendicular to A C, and these lines will then be the seats of vertical sections through the dome, parallel to E F G. Through the points e, f, g, h (No. 1) draw lines parallel to the axis A C, cutting E G in o, n, m, k;
SOLID GEOMETRY 125
and from H, with the radii H o, H n, H m, H k, de- scribe concentric circles o 9 9 p, n 8 8q, m z y r, &c. To find the diminished width of each gore at the sections a, e, b f , c g, d h : — Through the divisions of the semi- circle, 1, 2, 3, 4, 5, draw the radii Is, 2t, 3n, 4v, 5w, 6x ; then by drawing through the intersections of these radii with the concentric circles, lines parallel to H F, to meet the section lines corresponding to the circles, the width of the gores at each section will be obtained ; and curves through these points will give the repre- sentation of the lines of the gores of the plan.
In No. 2 the intersections of the lines are more clearly shown. The quadrant E G F is half the end elevation of the dome, and is divided as in No. 1. The parallel lines 5 5, 4 4, 3 3, show how the divisions of the arc of the quadrant are transferred to the line D B, and the other parallels a h, b k, cl, d m, are drawn from the divisions in the circumference of the ellipse to the line E G, and give the radii of the arcs m, 1 o, k p, h q.
To describe one of the gores draw any line A B (No. 3), and make it equal in length to the circumfer- ence of the semi-ellipse ADC, by setting out on it the divisions 1, 2, 3, 4, 5, &c., corresponding to the divisions C h, h g, g f, &c., of the ellipse : draw through those divisions lines perpendicular to A B. Then from the semi-circle (No. 1) transfer to these perpendiculars the widths 6 5 to g n, 9 9 to f m, 8 8 to e 1, y z to d k, and X w to c h, and join Ac, d d» de, e f, f g, A h, h k, k 1, 1 m, and m n, ; which will give the boundary lines of one-half of the gore, and the other half is obtained in the same manner.
126 MODERN CARPENTRY
To describe the covering of an ellipsoidal dome with horizontal boards of equal width.
Let A B C D (No. 1, Fig. 91) be the plan of the dome, ABC (No. 2) the section on its major axis, and L M N the section on its minor axis. Draw the circumscribing parallelogram of the ellipse, namely, FGHK (No. 1), and its diagonals FHGK. In No. 2 divide the cir- cumference into equal parts, 1, 2, 3, 4, representing the number of covering boards, and through the points of division draw lines 18, 2 7, &c., parallel to A C. Through the points of divisions draw 1 p, 2 t, 3 x, &c., perpendicular to A C, cutting the diagonals of the cir- cumscribing parallelogram of the ellipse (No. 1), and meeting its major axis in p, t, x, &e. Complete the parallelograms, and inscribe ellipses therein cor- responding to the lines of the covering. Produce the sides of the parallelograms to intersect the circumfer- ence of the section on the minor axis of the ellipse in 1, 2, 3, 4, and lines drawn through these parallel to L N, will give the representation of the covering boards in that section. To find the development of the cover- ing, produce the axis D B, in No. 2, indefinitely. Join by a straight line the divisions 1 2 in the circumfer- ence, and produce the line indefinitely, making e k equal to e 2, and k g equal to 12; 1 2e k g will be the axis major of the ellipses of the covering 12 7 8. Join also the corresponding divisions in the circumference of the section on the minor axis, and produce the line
1 2 b to meet the axis produced ; and the length of this line will be the semi-axis minor, e h, of the ellipse,
2 h k, while the width f h will be equal to the division 12 in N M L.
SOLID GEOMETRY
127
Fig. 91.
128
MODERN CARPENTRY
To find the covering of an annular vault.
Let ACKGEFA (Fig. 92) be the generating sec- tion of the vault. On A C describe a semi-circle ABC, and divide its circumference into equal parts, repre- senting the boards of the covering. From the divisions of the semi-circle, b, m, t, &c., from the centre D of
Fig. 92.
the annulus, with the radii D r, D s, &c., describe the concentric circles, s q, &c., representing the covering boards in plan. Tlirough the centre D draw PI K per- pendicular to G C, indefiniteh- extending it through K. Join the points of division of the semi-circle, A b, b m, m t, by straight lines, and produce them until they cut
SOLID GEOMETRY
129
the line K H as m b n, t m ii, when the points n, u, &c., are the centres from which the curves of the covering boards mo, t v, &e., are described.
To find the covering of an ogee dome, hexagonal in plan.
Let A B C D E F (No. 1, Fig. 93) be the plan of the dome, and H K'L (No. 2) the elevation, on the di- ameter F C. Divide H K into any number of equal parts in 1, 2, 3, 4, 5, k, and through these draw per- pendiculars to H L, and produce them to meet F C
130 MODERN CARPENTRY
(No. 1) in 1, mm, n, o, p, G. Through these points draw lines J d, me, n f, &c., parallel to the side F E of the hexagon ; bisect the side F ^ in N, and draw G N, which will be the seat of a section of the dome, at right angles to the side E F. To find this section nothing more is required than to set up on N G from the points t, u, V, &c., the heights of the corresponding ordinates ql, r2, s3, &c., of the elevation (No. 2) to draw the ogee curve N 1 2 3 4 5 p, and then to use the divisions in this curve to form the gore or covering of one side E g h k M D.
PART II. PRACTICAL SOLUTIONS.
Having- taken a thorough eouryo in Solid Geometry, the student should be now prepared to solve almost any problem in practical construction, almost as soon as they present themselves. The various problems in construction, however, are so numerous, and in many cases, so intricate, that the student Avill ofte^i be con- fronted by problems which will require so many appli- cations of the rules he has been taught, in different forms, that without some helping guidance, he will fail to see exactly what to do.
The following examples, with their explanations, are intended to give him the aid necessary to solve many difficult problems, and equip him with the means of still further investigation and sure results.
It is often necessary for the workman to find the exact stretchout or length of a straight line that shall equal the quadrant or a semi-circle. '
To accomplish this: — Take AB radius, and A cen- tre; intersect the circle at C; join it and B; draw from D, parallel with CB, cutting at H; then AH will be found equal to curve A D. Fig. 1.
This method is somewhat different to that already given; both, however, are practical.
Fig. 2. To find a straight line whicli is equal to the circumference of a circle. Draw from the centre, 0,
131
132
MODERN CARPENTRY
Fig. 1.
Fig. 2.
PRACTICAL SOLUTIONS 133
any rio-ht anole, cuttino: at J and V; join JV; draw from 0 parallel with JV; square down from J, cut- ting at N ; joint it and V ; then four times N V will be found to equal the circumference.
How to find the mitres for intersecting- straight and circular mouldings.
Figure 3 shows the form of an irregular piece of framing or other work, which requires to have mould- ings mitre and properly intersect.
The usual way of doing this is to bisect each angle, or to lay two pieces of moulding against the sides of framing, and mark along the edge of each piece, thus making an interseetic or point, so that by drawing through it to the next point, which is the angle of framing, the direction of mitre is obtained. This pro- cess, however, is not the quickest and best by any means. The most simple and correct method is to ex- tend the sides A L and P H.
Now suppose we wish to find a mitre from L; take it as centre, and with any radius, as K, draw the circle, cutting at J; join it and K; draw from L parallel with J K, and we have the mitre at once.
Now come to angle on the right; here take H as centre, and with any radius, as E, draw the circle, cutting at F ; join it and E ; draw from H parallel with E F, and you will find a correct mitre.
The next question is the intersection of straight and circular mouldings.
In the present case an extreme curve is given, in order to show the direction of mitre here, which is simply on the principle of finding a centre, for three points not on a straight line. For example, ABC are
134
]\IODERN CARPENTRY
^v:-.
•'•M
"-v..
PRACTICAL SOLUTIONS 135
points; bisect AB and BC; drawn through intersec- tions thus made, and lines meeting in point D give a centre, from which strike the circular mitre as shown.
Here it may be stated that in some cases a straight line for mitres will answer ; this means when the curve is a quadrant or less.
Fig. 4 shows the intersections of rake and level mouldings for pediments.
The moulding on the rake, increases in width, and is entirely different from that on the level, yet both mitre, and intersect, the rake moulding being worked to suit the level. If the curves of Fig. 4 are struck from centres as shown, then by the same rule, the rake moulding is also struck from centres.
Take any point in the curve, as C ; square up from it, cutting at B; draw from C parallel to SL; join LK, which bisect at N ; make E D equal to A B on the right ; join L D and D N ; bisect L D, also D N ; draw through intersections thus made, and the lines meeting in F, give a point from which draw through N ; make N J equal H F ; then F and J are centres, from which strike the curve, and it will be found to exactly intersect with that of Fig. 4.
Both mouldings here are shown as solid, and of the same thickness. This is done for the purpose of mak- ing the drawing more plain and easily understood; but bear in mind that all crown mouldings are gener- ally sprung.
To find the form of a sprung' or solid moulding on any rake without the use of either ordinates or centres.
It may not be generally known, that if a level mould- ing is cut to a mitre, that the extreme parts of mitre,
136
MODERN CARPENTRY
PRACTICAL SOLUTIONS 137
when in a certain position, will instantly give the exact form of a rake moulding, and it will intersect, and mitre correctly with that of level moulding. To do -this, take the level piece which has been mitred; lay its flat surface on the drawing; make its point P at Fig. 5, stand opposite point P at Fig. 6; keep the outer edge fair with line N L. The piece being in this position, take a marker, hold it plumb against the mitre, and in this way, prick off any number of points as shown, through which trace the curve line, and the result is a correct pattern by which the rake moulding is worked.
A moment's consideration will convince us that this simple method must give the exact form of any rake moulding to intersect with one on the level.
To cut the mitres and dispense with the use of a box, this method will be found quick and off-hand. Take, for example, the back level moulding, and square over on its top edge any line, as that of F N ; con- tinue it across the back to H ; make H V equal T L above, and from V, square over lower edge H K. Now take bevel 2 from above, and apply it on top edge, as shown; mark FL; then join L V; cut through these lines from the back, and the mitre is complete.
To cut the mitre on the rake moulding, square over any line on its back, as that of HJ ; continue it across the top and lower edge ; take bevel X, shown above Fig. 5, apply it here on top edge, and mark D A ; take the same bevel, and apply it on small square at E, and mark E 2.
We now want the plumb cut on lower edge J K, and the same cut on front edge N P, shown at Fig. 6. Take
138
MODERN CARPENTRY
Fig. 5.
Fig. 6.
PRACTICAL SOLUTIONS 139
bevel W above Fig. 5, apply it here and mark 2 B ; join BA; this done, apply the same bevel on front edge NP, and mark the plumb cut, it being parallel with that of 2B here, or K J, Fig. 5 ; now cut through lines on the back, and the mitre is complete.
It has already been shown, that we dispense with making or using a box for mitreing sprung mouldings.
In this case, the front edge or upper member, stands parallel with face of wall, so that bevel X being ap- piled, gives the plumb cut; then the cut on top edge is square with face of wall. This shows, that we have only to find the direction of a cut on the back of mould- ing to make the mitre.
To do this, take any point as R ; draw from it square with rake of gable. Now mark sections of moulding, as shown, its back parallel with RF; draw from D square with E N ; extend the rake to cut line from D at K ; this done, take any point on the rake, say L ; draw from it parallel with R F, cutting at K; take it as centre and L as radius, and draw the arc of a circle ; with same radius return to K on the right; take it as centre, and draw the arc L H ; make the first arc equal it ; then draw from H parallel with L C, cutting at C J ; draw from it square with rake, cutting at C, and join C K. This gives bevel W for cut on back of mould- ing.
A most perfect illustration of this may be had by having the drawing on card-board, and cutting it clear through all the outer lines, including that of the mould- ing on lines F D N E, making a hinge by a slight cut on line RF; also make a hinge of line RA, by a slight cut on the back, and in like manner make front edge
140
MODERN CARPENTRY
work on a hinge by a slight cut on line F V. This cut is made on top surface. Perform the same operation
Jlm/^MoulcUiy for a, &ai^
Fig. 7.
on the left. All the cuts being made, raise both sides on hinges A R and A 2 ; push the sections of mould- ings on right and left from you; make front edge rest
PRACTICAL SOLUTIONS
141
on FD. Now bring mitres together, and we have a practical illustration of mitreing sprung mouldings on the rake. (See Fig. 7.)
PROJECTION OF SOLIDS.
The following illustrations will be found by the stu- dent almost indispensable in the construction of vari- ous objects, and they will open the door to many more.
Fig. 8 shows the projection of a solid. This means the section of anything that is cut by a plane not paral- lel to its base ; or, to put this in a more practical way — take a square bar of wood and cut it in the direction of BE; the section it makes is shown by BEFH; simple as this is, it still gives the idea of what is mean/" ^y projection.
142
MODERN CARPENTRY
Fig. 9 shows the section of a square bar which has been cut by two unequal pitches, say in the direction of bevels J and H ; the line C B is called the seat ; from it all measurements are taken and transferred to lines that are square with the pitch AB; this pitch may be called a diameter, because it and the ordinate A E are at right angles.
Fig. 9.
Fig. 10 shows the sides of a square bar which may be any length. The bar is to stand perpendicular, and pass through a plank that inclines at A B ; the learner
PRACTICAL SOLUTIONS
143
is now required to show on the surface of plank the shape of a mortise that shall exactly fit the bar.
Fig. 10.
To solve this the student is left to exercise his own intelligence.
The problem may be clearly demonstrated by a card-board model; it being cut, and the parts pro-
144 MODERN CARPENTRY
jected from the flat surface will represent the plank, and show the mortise in it standing directly over the square, Fig. 10.
Fig. 11. It is here required to mark two unequal pitches on two sides of a square bar — then to have a piece of plank or board cut so that it shall exactly fit both pitches on the bar. The inclination of plank may be assumed as A B, and height of both pitches as H A ; let E F be the seat from which all measurements are taken, and transferred to lines that are on the surfaces of plank and square with AB; thus giving points to direct in drawing line C D ; and D E produced.
Could we apply the bevel J to points C and K, and have plumb lines on the edge of plank, then by cutting through these and those already marked on the sur- face, the problem would at once be solved, by making both upper and under surface of plank fit the two pitches as required. But in practice this would be in- admissible on account of the great waste of material.
The proper method is to take any point, E, and cut through the plank square with E D, and at C, cut through the plank square with C D ; here it will be noticed that line A D on the surface makes a very different pitch to that of A B on the edge of plank, and that C D differs from both.
To understand these points thoroughly is the true secret of the nicest element in the joiner's art — hand railing. It being clear that two bevels are required one for each pitch — proceed to find them. Make N L equal one side of the square. Take N as center, and strike an arc touching the line E D ; with same radius, and L centre, make the intersection in S ; join it at L,
PRACTICAL SOLUTIONS
145
Fig. 11.
146 MODERN CARPENTRY
which gives bevel R for joint at E. Again take N as centre, and strike an are touching line D C ; with same radius, and L centre, make the intersection in W; join it and L, which gives bevel P for joint at C ; the cuts being made by these bevels.
PANELLED CEILINGS IN WOOD AND STUCCO BRACKETS, AND SIMILAR WORK.
It is no part of the dutj' of this work to show designs for wooden ceilings, but in order to illustrate the method or methods, of constructing a wooden ceiling I deem it proper to show a design in this style the better to convey to the student the reason for the various steps taken to reach the desired result.
Fig. 12 shows a section of a roof and the mode of constructing a bracketed ceiling under it, but with slight modification the same arrangement can be adopted for ceilings under floors. The rafters A A are 18 inches from centre to centre. On these, straps a a, 3X1^ 'inches, are nailed at 16 inches apart, and similar straps a a, 1^X1 inch, are nailed to the ceiling joists. To the straps are nailed the brackets bbb, shaped to the general lines of the intended ceiling, and also placed 14 inches apart. The laths are nailed to the brackets for the moulded parts, and to straps for the flat parts of the ceiling. The brackets and the straps for ceiling should not be more than one inch in thick- ness, for where the brackets and straps occur the plaster cannot be pressed between the laths to form a key. If the brackets and straps are made thicker,
PRACTICAL SOLUTIONS
]47
148 MODERN CARPENTRY
parts of the ceiling are apt to be weak and irregular. The lathing shonld be well bonded and have no end joints longer than twelve inches on the ceiling, and twenty-tonr inches on the Avails and partitions. No joints of laths shonld be overlapped, as the plaster would thereby be made thinner at a part where it forms a no key, and would thus be liable to crack from the vibration of roof, floor or other causes.
Fig. 13 is a section on a larger scale at right angles to that shown in Fig. 12. Here the ceiling bracket b is shown affixed by hooks to the wall, and to the ceiling- joists by the strap a a. The ceiling having been plas- tered, and the mouldings of the cornice run on the lathed brackets prepared to receive them, the plaster enrichments marked c d e f g are then applied. No. 3 on Fig. 2 shows a curtain-box in section with its cur- tain-rod.
Fig. 14 is a section through a window-head, meeting- rails and sill of a window. The safe lintel is placed about 10 inches above the daylight of the window, for the purpose of allowing Venetian blinds to be drawn up clear of the window, and leave the light unobstructed. The framing of the window is carried up to the lintel, and between it and the upper sash a panel is -set in, and the blinds hang in front of it.
Fig. 15 is a cross vertical section, and Fig. 16 a plan, looking up, of a skylight on the ridge of- a roof, suitable for a staircase or corridor. The design can be adopted to suit various widths. The skylight is bracketed for plaster finishings, in the same manner as the ceiling already described. The framing and mouldings at b are carried down the side of the light at the same slope
PRACTICAL SOLUTIONS
149
Km,,
$^W;
wi
If!,;.
mm
ill ,
1 I
150
MODERN CAEPENTRY
as the sasli, till they butt against the sill and bridle d and e, forming a triangular panel having for its base the cornice c, which is carried round the aperture horizontally, and finishing flush with the ceiling, per- mits the cornice on the corridor to be continued with-
Fig. 16.
out interruption, Observe in Fig. 16 the lower cornice finishes the walls, and the upper mouldings 0 marked c in Fig. 15 are carried round the well of the skylight. Pig. 17 shows a plan of a panelled ceiling. The loAver members or bed mouldings are carried- round the walls of the room, and tend to build together, and give an
PRACTICAL SOLUTIONS
151
appearance of support to the several parts of the ceil- ing. The best way of making panelled ceilings is to cover the floor with boarding, and lay down the lines of the ceiling on the temporary floor thus formed. Then build and lath the ceiling on these lines. When it is completed it will be quite firm, and can be cut into sections suitable for being lifted up and attached to the joisting. The same lines will serve as' guides for the plasterer setting the moulds for running the cor- nices and for preparing the circular ornament.
Fig. 17.
Fig. 18 shows in plan and section a centre suited for this ceiling. By fixing the central portion of it one inch from the surface of the finished ceiling and con- necting it by small plaster blocks placed about one inch apart, it forms an excellent ventilator. A zinc tube can be led from this centre into a vent, and an
152
MODERN CARPENTRY
air tight valve put on the tube to prevent a down draft when the vent is not in use. F is an iron rod fixed to
Fig. 18.
a strut or dwang between the joists for the purpose of securing a gas pendant or electrolier.
Let CAB (Fig, 19) be the elevation or the bracket of a core, to find the angle-bracket.
PRACTICAL SOLUTIONS
153
First, when it is a mitre-bracket in an interior angle, the angle being 45° : divide the curve C B into any
Fig. 19.
number of equal parts 12 3 4 5, and draw through the divisions the lines Id, 2e, 3f, 4g, 5c, perpendicular to
154 MODERN CAEPENTRY
A B, and cutting it in d e f g c ; and produce them to meet the line D E, representing the centre of the seat of the angle-bracket : and from the points of intersec- tion h i k 1 c draw lines h 1, i 2, k 3, 1 4, at right angles to D E, and make them equal — h 1 to d 1, i 2 to e 2, &c. ; and through F 1 2 3 4 5 draw the curve of the edge on the bracket. The dotted lines on each side of D E on the plan show the thickness of the bracket, and the dotted lines u r, v s w t, show the manner of finding the bevel of the face. In the same figure is shown the manner of finding the bracket for an obtuse exterior angle. Let D I K be the exterior angle : bisect it by the line I G, which will represent the seat of the centre of the bracket. The lines I H, m 1, n 2, o 3, p 4, c 5, are drawn perpendicular to I G, and their lengths are found as in the former case.
To find the angle-bracket of a cornice for interior and exterior, otherwise reentrant and salient, angles.
Let AAA (Fig. 20) be the elevation of the cornice- bracket, E B the seat of the mitre-bracket of the in- terior angle, and H G that of the mitre-bracket of the exterior angle. From the points A k a b c d A, or wherever a change in the form of the contour of the bracket occurs, draw lines perpendicular to A i or D C, cutting A i in e f g h i and cutting the line E B in E 1 m n o B. Draw the lines E G, G L B H, and H K, representing the plan of the bracketing, and the par- allel lines from the intersection 1 m n o, as shown dotted in the engraving ; then make B F and H I per- pendicular to E B and G H respectively, and each equal to i A, o u to h d, n t to g c, m s to f b, 1 r to e a, 1 p to e k, and join the points so found to give the con-
PRACTICAL SOLUTIONS
155
tours of the brackets required. The bevels of the face are found as shown by the dotted lines x v y w, &c.
■Angle-bracketB for Moulded Cornices
Fig. 20.
To find the ang-le-bracket at the meeting' of a concave wall with a straight wall.
Let A D E B (Fig. 21) be the plan of the bracketing on the straight wall, and D M GE the plan on the cir-
156
MODERN CARPENTRY
cular wall, CAB the elevation on the straight wall, and G M H on the circular wall. Divide the curves
Angle-bracket at Meeting of Straight and Concave Walls
Fig. 21.
C B, G IT into the same number of equal parts ; through the divisions of C B draw the lines CD, 1 d h, 2 e i, &c., perpendicular to A B and through those of G H draw
PRACTICAL SOLUTIONS
157
the parellel lines, part straight and part curved 1 m h, 2 n i, 3 o k, &c. Then through the intersections h i k 1 of the straight end curved lines draw the curve D E, which will give the line from which to measure the ordinates hi, i 2, k 3, &c.
Angle-bracket at Meeting ot Coves ot Equal Height but Unec|ual Projection
Fig. 22.
Fig. 22 shows the method of finding the angle-bracket at the meeting of coves of equal height but unequal projection. The height C B is equal to G H, but the projection B A is greater than H I.
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MODERN CARPENTRY
Fig. 23, Nos. 1, 2, 3, shows the curb and ribs of a circular opening (CBA, No. 2), cutting in on a slop- ing ceiling. No. 1 is a section through the centre B D, No. 2 and E F I, No. 3. The height L K is divided into
Curb and Ribs ol Circular Arch Cutting into Sloping Ceiling Fig. 23.
equal parts in c, f, g, h, i, and the same heights are transferred to the main rib in No. 2 at A, 1, 2, 3, 4, 5, B. Through the points A, 1, 2, 3, 4, 5, in No. 2 lines are drawn parallel to the axis BE I; and through the points e, f, g, h, i, in No. 1 lines are drawn parallel to
PRACTICAL SOLUTIONS
159
the slope K H. The places of the ribs 1, 2, 3, 4, 5, in the latter, and their site on the plan. No. 3, and also the curve of the curb, are found by intersecting lines in the manner with which the student is already ac- quainted. ' ,
abc
N03
JC opfs abc
Spherical Niches on Semicircular and Segmental Plans
ON NICHES.
To describe a spherical niche on a semi-circular plan.
The construction of this (Fig. 24) is precisely like that of a spherical dome. The ribs stand in planes, which would pass through the axis if produced. They are all of similar curvature. No. 2 shows an elevation
160 MODERN CARPENTRY
of the niche, and No. 3 the bevelling of the ribs a, b, against the front rib at D on the plan ; a b is the bevel of a, and b e of b.
Let HBK (Fig. 24) be the plan. It is obvious that the ribs m p, nr, will be parts of the quadrant GF (No. 3\ Transfer the lengths lo, m p, n r, and r s to the line G E, as shoAvn at o p r s, and raise perpendicu- lars from these points to the quadrant ; G p is the rib m p, and o p is the bevel ; G r is the rib nr, and r s is the section of the front rib at the crown; the vertical projection of the upper arris of this rib will be a sem'- circle with radius s s or s H.
The niche of which both the plan and elevation are segments of a circle.
No. 25 is the elevation of the niche, being the seg- ment of a circle whose centre is at E. No. 1, ABC, is the plan, which is a segment of a circle whose centre is 1), Having drawn on the plan as many ribs as are required, radiating to centre D, and cutting the plan of the front rib in a, b c, d e; then through the centre D draw the line GH parallel to AC; and from D de- scribe the curves m 1, AG, C H, cutting the line GH; and make D F equal to E 0, No. 2. From F as a centre describe the curves 1 p 1 and G I H for the depth of the ribs; and this is the true curve for all the back ribs.
To find the lengths and bevel of the ribs : — From the centre D describe the quadrant and arcs a f, b g, c g, d h, &c., and draw f f , g g, h h perpendicular to D H, cutting the curve 1 p 1, and the lines of intersection will give the lengths and bevels of the several ribs.
PRACTICAL SOLUTIONS
161
'.•j}\.
•Segmental Niche on Segmental Plas Fig. 25.
162
MODERN CARPENTRY
^/ / / //\\ ! / /
£IIiptical Niche on Sei^B«aMl PUo Fig. 26.
PRACTICAL SOLUTIONS 163
Let D in the plan (No. 1, Pig. 26) be the centre of the segment. Through D draw E F parallel to A C, and continue the curve of the segment to E F. Then to find the curve of the back ribs :— From k 1 m n, any points in the curve of the front rib fNo. 2) let fall per- pendicular to the line A B, cutting it in a b c d. Then from D as a centre describe the curves a e, b f, c g, e h, d h, and from the points where they meet the line E F draw the perpendiculars e k, f 1, g m, h n, ho, and set lip on e k the height ek of the elevation and the cor- responding heights . on their other ordinates, when klmno will be the points through which the curve of the radial ribs may be traced. The manner of find- ing the lengths and bevels of the ribs is shown at
t U U V V.
A niche semi-elliptic in plan and elevation.
Let No. 1, Fig. 27, be the plan, and No. 2 the eleva- tion of the niche. The ribs in this case radiate from the centre D, and with the exception of m g (which will be the quadrant of a circle) they are all portions of ellipses, and may be drawn by the trammel, as shown in No. 4, which, gives the true curve of the rib marked d in No. 2 and b D in No. 1. The rib c, in the eleva- tion, is seen at a D in No. 1 ; the bevel of the end h i is seen at A a in No. 3, and that of the end e f at b c.
To draw the ribs of a regular octagonal niche. ^ Fig. 28.— Let No. 1 be the plan, and No. 2 the eleva- tion of the niche. It is obvious that the curve of the centre rib H G will be the same as that of either half of the front rib A G, F G. In No. 3, therefore, draw ABODE, the half -pi an of the niche, equal to A B C H G, No. 1, and make D G E equal to half the
164
MODERN CARPENTRY
12 ? 6 3 O
Xliiiiiii i-i-U-
Elliptical Niche on Elliptical Plan Fig. 27.
J^^
PRACTICAL SOLUTIONS
165
il^".3
|
A |
|
|
-'V/ a |
|
|
^\ |
|
|
^y |
|
|
^.■"""^p-^ |
4- V r...- _ 3 |
Regular Octagonal Niche
Fig. 28.
front rib. Divide D G into any number of parts 12 3 4, &G., and through the points of division draw lines par- allel to A G, meeting the seat of the centre of the angle
166
MODERN CARPENTRY
-Irregular Octagonal Niche
Fig. 29.
PRACTICAL SOLUTIONS 167
rib C E iu i k 1 m n 0. On these points raise indefinite perpendiculars, and set up on them the heights a 1 in i 1, b 2 in k 2, and so on. The shaded parts show the bevel at the meeting of the ribs at G in No. 1.
To draw the ribs of an irregular octagonal niche.
Fig. 29.— Let No. 1 be the plan, and No. 2 the eleva- tion of the niche. Draw the outline of the plan of the niche at ABCDEF (No. 3), and draw the centre lines of the seats of the ribs B G, H I, &c. ; draw also G L F equal to the half of the front rib, as given in the elevation No. 2, and divide it into any number of parts 1234. Through the points of division draw dl, c 2, b3, a 4, perpendicular to G F, and produced to the seat of the first angle rib G E. Through the points of intersection draw lines parallel to the side E D of the niche meeting the second angle rib D G ; through the points of intersection again draw parallels to D C, and so on. The curve of the centre rib is found by setting up from n 0 p q G the heights d 1, c 2, &c., on the par- allel lines which are perpendicular to K G. The curve of the rib B G or E G is found by drawing through the points of intersection of the parallels perpendiculars to the seat of the rib, and setting upon them, at h m r I G, the heights dl, c2, &c. No. 4 shows the rib C G, and No. 5 the intermediate rib H I.
DOUBLE CURVATURE WORK.
To Obtain to Soffit Mould for marking the veneer (see Fig. 35), divide the elevation of lower edge of the head (Fig. 30) into a number of equal parts, as A, B, C, D, E, S; and drop projectors from these points into
168 MODERN CARPENTRY
the plan cuttii^ the chord line A" S" in A", B", C", D", E", S". Draw the line s' s', Fig. 35, equal in length to the stretch out of the soffit in the elevation (the length of a curved line is transferred to a straight one by taking a series of small steps around it with the compasses, and repeating a like number of the straight line), and transfer the points ABC, &c., as they occur thereon, repeating them on each side of the centre line. Erect perpendiculars at the points, and make each of these lines equal in length to its cor- despondingly marked line in the plan, as A' a' a' Fig. 35, equal to A" a a Fig. 32, these letters referring re- spectively to the chord line and the inside and outside edges of the head. Draw the curves through the points so found. As will be seen by reference to Fig. 35, the mould is wider at the springing than at the crown; this is in consequence of the pulley stiles being parallel. If they were radial their width would be the same as the width of the head at the crown, and the head would be parallel; the gradual increase in width from the crown to the springing is also apparent in the sash- head and the beads, as indicated by the line 0 0, Fig. 35, which is the inside of the sash-head, and the out- side of the parting head; this variation in width ren- ders it impossible to gauge to a thickness from the face or the groove in the head from its edges.
To Form the Head. Having prepared the cylinder (Fig. 33) to the correct size, prepare a number of staves to the required section, which may be obtained by drawing one or two full size on the elevation, as shown to the right in Fig. 36. The staves should be dry straight-grained yellow deal, free from knots and
PRACTICAL SOLUTIONS
169
Fig. 30.
Fig. 31.
Fig. 32.
170
MODERN CARPENTRY ^ ^(/>
sap, and not be so wide that they require hollowing to fit. If the veneer is pine, it will probably bend dry, but hardwood will require soft- ening with hot water. One end
Ul
o
Fig. 33.
should be fixed as shown m Fig. 34, by screwing down a stave across it. Then the other end is bent gently over initil tlic crown is reached,
Fig. 35.
Fig. 34.
when another stave is screwed on, and the bending continued until the veneer is well down all round, and a third stave secures it until it is thoroughly dry, when the re-
PRACTICAL SOLUTIONS
171
mainder maj' be glued on. It is as well to interpose a sheet of paper between the cylinder and the veneer, in ease any glue should run under, which would then ad- here to the paper instead of the veneer. The head should not be worked for at least twelve hours after glueing. If a band saw is at hand, the back should be roughly cleaned off and the mould bent round it, the shape marked, and the edges can then be cut vertically with the saw, by sliding it over the cylinder sufficiently
ULM
Fig. 36.
for the saw to pass. When cut by hand, the mould is applied inside and the cut is made square to the face, the proper bevel being obtained with the spokeshave, and found by standing the head over its plan and trying a set square against it. When fitting the head to the pulley stiles the correctness of the joints is tested by cutting a board to the same sweep as the sill, with
172
MODERN CARPENTRY
tenons at each end. and inserting it in the mortises for the pulleys, as shown at B, Fig. 36. A straight-edge applied to this and the sill will at once show if the head is in the correct position, and if the edges are vertical as they should be.
To Obtain a Developed Face Mould. Make the line i h, Fig. 37, equal to the stretch out of the plan of the face of sash-heads, viz., I H, Fig. 32. Transfer the di- visions as they occur, and erect perpendiculars thereon.
Make these equal in height to the corresponding ordi- nates over the springing line in the elevation, and draw the curves through the points so found. The groove for the parting bead can be marked by running a % in. piece around the inside of the sash-head, this bead being generally put in parallel. It is sometimes omitted altogether, the side beads being carried up until thev die off on the head.
PRACTICAL SOLUTIONS 173
To Find the Mould for the Cot Bar. Divide its cen- tre line in the elevation into equal parts, as 0, P, Q, R, T. Drop projectors from these into the plan, cut- ting the chord line 1 1 in o, p, q, r, t. These lines should be on the plan of the top sash, but are produced across the lower to avoid confusion with the projectors from the other bars. Set out the stretch out of the cot bar
xj-rrrn
M R Q' P 0' P' Q'
Fig. 38.
on the line M' 0' T', Fig. 38, and erect perpendiculars at the points of division, and make them equal in length to the correspondingly marked lines in the plan. The cot bar is cut out in one piece long enough to form the two upright sides as well as the arch. The straight parts are worked nearly to the springing, and the bar which is got out wider in the centre is then steamed and bent around a drum, and afterwards cut to the mould (Fig. 38) and then rebated and moulded. The arched bar should not be mortised for the radial bars, but the latter scribed over it and screwed through from inside.
To Find the Mould for the Radial Bars. Divide the centre line of the bar into equal parts, as 1, 2, 3, 4, 5, Fig. 30, and project the points into the plan, cutting the chord line J J in 1', 2', 3', 4', 5'. Erect perpendicu- lars upon the centre line from the points of division, and make them equal in length to the distance of the corresponding points in the plan from the chord line, and draw the curve through the points so obtained.
174
MODERN CARPENTRY
The other bar is treated in the same maimer, the projectors being marked with full lines in the plan.
Fig. 39.
Fig. 40.
The soft moulds for the head linings are obtained in similar manner to those for the head mould, the width being gauged from the head itself
PRACTICAL SOLUTIONS
175
A Frame Splayed Lining with Circular Soffit as
shown in part elevation in Pig. 39, plan Fig. 40, and section Fig. 41. The soffit stiles are worked in the'solid
Pig. 41.
in two pieces joined at the crown and springings. The rails are worked with parallel edges, their centre lines radiating from the centre of the elevation. Edge
1,76
MODERN CARPENTRY
moulds only are required for the stiles, and a- developed inould for the panels. The method of obtaining these has been shown on a separate diagram, Fig. 42.
Method of Obtaining Moulds for Circular Splayed Linings. Fig. 42.
To Obtain the Face Moulds for the Soffit Stiles. Draw the line E E, Fig. 42, and produce the faces of
PRACTICAL SOLUTIONS 177
the jambs until they meet in point C. From C draw the line C D perpendicularly to E E. This line will contain the centres of the various moulds, which are lo- cated by producing the plans of the edges of the stile across it e, 1, 2, 3 being the respective centres, and the inside and outside faces of the jambs affording the necessary radii for describing the arcs A a and B b.
To Apply the Moulds. Prepare the stuff equal in thickness to the distance between the lines e 1 and 2 3, Fig. 42. Apply the mould A to the face of the pieces intended for the front stile, and cut the ends to the mould, and square from the face. Set a bevel as at F, Fig. 42, and apply it on the squared ends, working from the lines on the face, and apply the mould a at the back, keeping its ends coincident with the joints, and at the points where the bevel lines intersect the face. The piece can then be cut and worked to these lines, and the inside edge squared from the face. The outside edge is at the correct bevel, and only requires squaring slightly on the back to form a seat for the grounds. The inside stile is marked and prepared sim- ilarly.
The Development of the Conical Surface of the SofRt is shown on the right hand of the diagram. Fig. 42, and is given to explain the method of obtaining the shape of the veneer, but it is not actually required in the present construction, as the panel being necessarily constructed on a cylinder, its true shape is defined thereon, and its size is readily obtained by marking direct from the soffit framing when the latter is put together. Let the semi-circle E D E, Fig. 42, represent a base of the semi-cone, and the triangle E C E its ver-
178
MODERN CARPENTRY
tical section. From the apex C, with the length of one of its sides as radius, described the arc E F, which make equal in length to the semi-circle E D E by step- ping lengths as previously described. Join F to C, and E C F is the covering of the semi-cone. The shape of the frustum, or portion cut off by the section line of the linings, is found by projecting the inner edge of the lining upon the side C E, and drawing the con-
Fig. 43.
centric arc IJ; then EIJE represents the covering of the frustum. Any portion of this, the panels for instance, is found in the same way. To draw the rails, divide their centre lines equall}^ on the perimeter, and draw lines from the divisions to the centre as II C; make the edges parallel with these lines.
A Window With a Splayed Soffit and Splayed Jambs is shown in part elevation, Fig. 43, plan Fig. 44; and
PRACTICAL SOLUTIONS
179
section Fig. 45. The linings are grooved and tongued together, as shown in the enlarged section, Fig. 46. To obtain the correct bevel required for the shoulder of the jamb and the groove in the soffit, the lining must
Fig. 44.
be revolved upon one of its edges until it is parallel with the front, when its real shape can be seen. This operation is shown in the diagram on the right-hand
Fig. 45.
half of the plan and elevation. Draw the line a b, rep- resenting the face of the jamb in plan, at the desired angle. Project the edges into the elevation, and in- tersect them b}^ lines c d, projected from the top and bottom edges of the soffit in section. This will give the projection of the linings in elevation. Then from point
180
MODERN CARPENTRY
a as centre, and the width of the lining a b as radius, describe the arc b b', bringing the edge b into the same plane as the edge a. Project point b' into the elevation, cutting the top edge of the soffit produced in of . Draw a line from c' to d, and the contained angle is the bevel for the top of the jamb. When the soffit is splayed at the same angle as the jambs, the same bevel will answer for both; but when the angle is different, as shown by the dotted lines at e, Fig. 45, then the soffit also must be turned into the vertical plane, as shown at e' and a line drawn from that point to intersect the projection of the front edge of the jamb in F ; join this point to the intersection of the lower edges, and the contained angle is the bevel for the grooves in the soffit. (See Fig. 46.)
n^
Fig. 46.
Fig. 47.
The Enlarging and Diminishing of Mouldings. The design of a moulding can be readily enlarged to any desired dimensions by drawing parallel lines from its members, and laying a strip of paper or a straight-edge of the required dimensions in an inclined direction be- tween the boundary lines of the top and bottom edges;
PRACTICAL SOLUTIONS 181
and at the points where the straight-edge crosses the various lines, make marks thereon which will be points in the new projection, each member being increased proportionately to the whole. Projections drawn at right angles to the former from the same points will give data for increasing the width in like manner.
To Diminish a Moulding. The method to be ex- plained, which is equally applicable to enlargement, is based upon on& of the properties of a triangle, viz., if one of the sides of a triangle is divided into any number of parts, and lines drawn from the divisions to the opposite side of the triangle, any line parallel Avith the divided side will be divided in corresponding ratio. See Fig. 47, where ABC is an equilateral tri- angle, the side A B being divided into six equal parts, and lines drawn from these to the apex C. The two lines DE and PH, parallel to A B, are divided into the same number of parts, and each of these parts bears the same ratio to the whole line that the corresponding part bears to A B, viz., one-sixth; the application of this principle will now be shown. Let it be required to reduce the cornice shown in Fig. 48 to a similar one of smaller proportions. Draw parallel projectors from the various members to the back line A B, and upon this line describe an equilateral triangle. Draw lines from the points on the base to the apex, then set off upon one of the sides of the triangle from C a length equal to the desired height of the new cornice as at G or H, and from this point draw a line parallel to the base line. At the points where this line intersects the iuclined division lines, draw horizontal projectors cor- responding to the originals. To obtain their length
182
MODERN CARPENTRY
or amount of projection, draw the horizontal line b E, Fig. 48, at the level of the lowest member of the cor- nice, and upon this line drop projectors at right angles to it from the various members. Describe the equilat- eral triangle b i E upon this side, and draw lines from the divisions to the apex i. To ascertain the length
Fig. 49.
Fig. 48.
that shall bear the same proportion to b E that the line GH bears to A B, place the length of b E on the line B A from B to F, and draw a line from F to C : the portion of the line G H cut ofl' from J to H is the pro- portionate length required. Set this length off parallel to b E within the triangle, as before described, and also draw the horizontal line L H, Fig. 49, making i^
PRACTICAL SOLUTIONS
183
eqna] in length to a a, Fig. 48. Upon this line set off the divisions as they occur on a a, noting that their direction is reversed in the two figures. Erect per- pendiculars from these points to intersect the previ- ously drawn horizontals, and through the intersections trace the new profile. The frieze and architrave are reduced in like manner, M B, Fig. 48, representing the height of the original architrave, and NH the reduc- tion. The cornice can be enlarged similarly by pro-
Fig. 50.
ducing the inclined sides of the triangle, as shown bj' the dotted lines on Fig. 48, sufficiently to enable the required depth to be drawn within it parallel to A B. One member has been enlarged to indicate the method, which should be clear without further explanation.
Raking Mouldings. Fig. 50 shows the method of finding the true section of an inclined moulding that is required to mitre with a similar horizontal moulding
184
MODERN CARPENTRY
at its lower end, as in pediments of doors and windows. The horizontal section being the more readily seen, is usually decided first. Let the profile in Fig. 50 repre- sent this. Divide the outline into any number of parts, and erect perpendiculars therefrom, to cut a horizontal line drawn from the intersection of the back of the moulding v/ith the top edge, as at point 7. With this point as a centre and the vertical projectors from 1 to 7 as radii, describe arcs cutting the top of the inclined mould, as shown. From these points draw perpendicu-
Fig. 51.
lars to the rake, to meet lines parallel to the edges of the inclined moulding drawn from the corresponding points of division in the profile, and their intersections will give points in the curve through which to draw the section of the raking mould. When the pediment is broken and a level moulding returned at the top, its section is found in a similar manner, as will be clear by inspection of the drawing. If the section of the raking moulding is given, that of the horizontal mould can be found by reversing the process described above. Fig. 51 s^bws the application of the method in finding the'section of a'return bead when one side is level and
PRACTICAL SOLUTIONS
185
the other inclined, as on the edge of the curb of a sky- light with vertical ends.
Sprung Moulding-s. Mouldings curved in either ele- vation or plan are called "sprung," and when these are used in a pediment, require the section to be deter- mined as in a raking moulding. The operation is sim- ilar to that described above up to the point where the back of the section is drawn perpendicular to the in- clination, but in the present ease this line E x. Pig. 52,
Fig. 52.
is drav/n radiating from the centre of the curve, and the. projectors are drawn parallel to this line. The parallel projectors, 1, 2, 3, 4, 5, are also described from the centre until they reach the line E x, when per- pendiculars to this line are raised from the points of intersection to meet the perpendicular projectors.
Mitreing Straight and Curved Mouldings Together. If a straight mitre is required, draw the plan of the mouldings, as in Fig. 53, and the section of the straight mould at right angles to its plan as at A. Divide its
186
MODERN CARPENTRY
profile into any number of parts, and from them draw parallels to the edges intersecting the mitre line. From these intersections describe arcs concentric with the plan of the curved mould, and at any convenient point thereon draw" a line radial from the centre. Erect perpendiculars on this line from the points where the arcs intersect it. and make them equal in height to the corresponding lines on the section of the straight moulding A, and these will be points in the profile of
Fig. 53.
the curved moulding B. "When it is required that the section of both mouldings shall be alike, a circular mitre is necessary, and its true shape is obtained as shown in Fig. 53. Draw the plan and divide the profile of the straight moulding as before, drawing parallels to the edge towards the seat of the mitre. Upon a line drawn through the centre of the curved moulding set off divisions equal and similar to those on the straight part, as 1 to 8 in the drawing. From the centre of the
PRACTICAL SOLUTIONS
187
curve describe ares passing through these points, and through the points of intersection of these arcs with the parallel projectors, draw a curve which will be the true shape of the mitre. Cut a saddle templet to this shape, and use it to mark the mouldings and guide the chisel in cutting.
Fig. 54.
To Obtain the Section of a Sash Bar Raking in Plan. Fig. 54 represents the plan of a shop front sash with bars in the angles. On the left hand is shown the sec-
188 MODERN CARPENTRY
tion of the stile or rail into which the bars have to mitre. Divide the profile of moulding into a number of parts, as shown from 1 to 6, ajid from these points draw parallels to the sides of the rails intersecting the centre line of the bar. Also draw perpendiculars from the same ponts to any line at right angles to them, as at A. DraAv a line at right angles to the centre line of the bar, and on it set off the divisions from 1 to 6 as at A. Draw projectors from these points parallel to the centre line of the bar, and where they intersect the correspondingly numbered lines drawn parallel with the sides of the sash will be points in the curve of the section of the bar. It will be noticed that there is no fillet or square shown on the bar, and that in trans- ferring the points from the line A they must be re- versed on each side of the centre. Should a fillet be required on the bar, additional thickness must be given for the purpose. Three methods of forming the rebates in the bar are shown, the screwed saddle beads being the best for securing the glass.
Interior Shutters Include Folding-, Sliding-, Balanced, and Rolling-, Exterior Consist of Hanging, Lifting, Spring, and Venetian Shutters.
FOLDING, or as they are frequently called, BOX- ING SHUTTERS, because they fold into a boxing or recess formed between the window frames and the walls, are composed of a number of narrow leaves, framed or plain, as their size may determine, rebated and hinged to each other and to the window frames. They should be of such size that when opened out they will cover the entire light space of the sash frame and a margin of a ^4 i^i- ^^ addition. Care must be taken
PRACTICAL SOLUTIONS 189
to make them parallel, or they will not swing clear at the ends; and as a further precaution, they should not be carried right from soffit to window board, but have clearance pieces interposed at their ends about % in. thick. The outer leaf, which is always framed, is termed a shutter j the others are termed flaps. It is not advisable to make the shutters less than 1^4 in- thick, and flaps over 8 in. wide should be framed ; those less than 8 in. may be solid, but should be mitre clamped to prevent warping. In a superior class of work the boxings are provided witli cover flaps which conceal the shutters when folded, and fill the void when they are opened out. The sizes and arrangements of the framing are determined by the general finishings of the apartment, but it is usual to make the stiles of the front shutters range with those of the soffit and elbow linings. When Venetian or other blinds are used inside, provision is made for them by constructing a block frame from 2]^2 to 3 in. thick inside the window frame and the shutters are hung to this.
The leaves are hung to each other with wide hinges called back flaps, that screw on the face of the leaves, there not being sufficient surface on the edges for butt hinges. In setting out the' depth of boxings, at least i/g in. should be allowed betweeii each shutter to pro- vide room for the fittings; the shutters are fastened by a flat iron bar hung on a pivot plate fixed on the inner left-hand leaf, and haying a projecting stud at its other end which fits into a slotted plate, and it is kept in this position by a cam or button. Long shutters are made in two lengths, the joint coming opposite the meeting rails of the sashes; these are sometimes re- bated together at the ends.
190
MODERN CARPENTRY
Fig. 55.
PRACTICAL SOLUTIONS
191
Fig. 55 is a sectional elevation of a Window Fitted with Boxing- Shutters having a cover flap and spaces for a blind and a curtain. One-half of the elevation shows the shutters opened out and the front of the finishings removed, showing the construction of the boxings, &c. The plan, Fig. 56, is divided similarly, one half showing the shutters folded back, with por^
Fig. 56.
tions of soffit cornice, &c. ; the other half gives the plan and sections of the lower parts, the dotted lines show- ing the Mandow back.
Fig. 57 is a vertical section and Fig. 58 an enlarged section through the boxings. The framed pilaster cov- ering the boxing is cut at the level of the window board, and hung to the stud A', this being necessary for the cover to clear the shutters when' open (see dotted lines on opposite half). The cover flap closes into rebates at the top and bottom, as shown in sec-
192
MODERN CARPENTRY
tion in Fig. 55. The window back is carried behind the elbows, and grooved to receive the latter. The rails of the soffit must be wide enough to cover the boxings, and should have the boxing back tongued
Fig. 57.
Fig. 58.
into it,