Drawing Doom Chairs. 
 
 Published by Fisher. Son & C° Caxton. London. June 23. 1827. 
 
THE CABINET M AKE R 
 UPHOLSTERER, 
 
 AND 
 
 COMPLETE DECORATOR, 
 
 WITH GEOMETRICAL SECTIONS, 
 
 AND FURNITORE COLOURED ENGRAVINGS, 
 
 AND A 
 
 PERFECT GLOSSARY OF TECHNICAL TERMS 
 USED IN CABINET MAKING , 
 
 BY 
 
 PETER NICHOLSON. 
 
 fonbmt: EISHER & CAXTON. 
 
INTRODUCTION. 
 
 The want of a complete treatise, by the aid of which all the various 
 articles of Cabinet Furniture may be readily executed by the Mechanic 
 with precision and tasteful elegance, and the designs made clearly intelligible 
 to the Gentleman and the Amateur, has been long and sensibly felt as a 
 serious inconvenience. Aware of this deficiency, the principal object of the 
 Author, in the following Essay, is to supply the desideratum, by furnishing 
 such a variety of designs as shall illustrate the principles upon which these 
 articles are constructed, and afford every requisite information on this 
 important and interesting subject. Before, however, he proceeds to 
 enumerate the contents of this work, it may be necessary for him to prepare 
 in some degree the mind of the reader for an advantageous perusal, by 
 pointing out to him the nature and extent of his scientific researches. 
 
 In the construction of objects, our first and chief consideration is utility * 
 but when our actual wants are fully supplied^ we naturally seek further 
 gratification in their varied forms and external ornaments ; yet in the choice 
 of figure, we must' nevertheless, bo always governed by certain criteria, 
 founded upon peculiar associations. It is one of the grand designs of this 
 treatise to put the reader in possession of these principles, as they will prove 
 eminently useful to him, both as an artisan and an amateur, by directing his 
 attention to classic forms and tasteful embellishments. 
 
 Every sensible object irwfche phenomena of surrounding nature, must 
 necessarily be contained under some external form and appearance. But as 
 it is desirable that the eye should be pleased, and the imagination gratified, 
 the particular figure under which a thing presents itself can never be a matter 
 of indifference; for it is a well-known fact, that while the contour of some 
 
VI 
 
 INTRODUCTION. 
 
 bodies communicate agreeable sensations, others are incapable of producing 
 any ; while many again are wholly repugnant to every possible agreement 
 with our feelings and ideas. But indeed, allowing, even in this instance, that 
 we were satisfied, the illusion would not be lasting. The charm of novelty is 
 gradually dissolved, and a repetition of the same object, however pleasing at 
 first, soon loses its effect, and in the course of time ceases altogether to 
 attract or delight. 
 
 In order, therefore, to rouse the dormant spirit of curiosity, and rekindle 
 the dying embers of exhausted pleasure, it becomes necessary to seek out 
 fresh sources of intellectual gratification; and thus the powerful and inventive 
 genius of man is called forth, and kept in constant and persevering exertion. 
 It is a trite and oft-repeated observation, that “ there is nothing new under 
 the sun;” hence many imagine that every possible variety has already been 
 discovered ; that nature at length is completely exhausted ; and that nothing- 
 further remains to be produced. This, however, is far from the truth; since 
 although the number of forms may in some degree be said to be limited, their 
 combinations are almost infinite. It is a knowledge of this interesting fact, 
 that enables us to construct new figures, and to fashion them according to our 
 desire; yet our taste must at all times be governed by certain principles, that 
 are fixed upon as sure a foundation as are the sublime propositions contained 
 in the Elements of Euclid. We shall therefore endeavour to establish some 
 specific rule by which our judgment may invariably be directed, and which 
 may serve both as a test of merit and as a standard of excellence. 
 
 A judicious selection, then, of the beautiful forms and varieties of nature, 
 and a strict adherence to curve lines produced by organical description, ought 
 to constitute our elements of taste.* 
 
 It was a knowledge of these principles that inspired the genius of the 
 illustrious artisans of Greece, and enabled them to construct those sublime 
 works, which remain, even to this day, the proud and lofty monuments of 
 unrivalled excellence and universal admiration. The labours of remote 
 antiquity would never have been held in such profound veneration, had not 
 the scientific principles, which guided the hand of the artist, been clearly 
 visible in those splendid examples that have been handed down to posterity, 
 and have hitherto withstood the destroying hand of time. 
 
 * The pleasing forms and wonderful variety of systems of curves, produced by Mr. Jopling’s 
 ingenious contrivances, originally suggested by an article in the Imperial Magazine for 1822, 
 are truly astonishing. 
 
INTRODUCTION. 
 
 VII 
 
 A caiefiil and an attentive study, therefore, of the works of the ancients, is 
 most eminently useful to such as are desirous of improvement in the art of 
 decoration ; and in older to furnish an example, we shall, in the present work, 
 endeavour to construct our figures upon geometrical principles, drawing them 
 fiom the purest classic models, selected from those precious remains of 
 antiquity which were executed when the arts and sciences were in the very 
 zenith of perfection. 
 
 Mere language alone is insufficient to convey an adequate idea of those 
 images that are conceived in the mind of one person, to the understanding of 
 anothei . for since we know that even such simple terms as a point or an 
 angle cannot be strictly or accurately defined, it may be easily inferred, that 
 oui ideas of many forms and positions, though easily conceived, cannot be 
 expressed intelligibly by words. From this acknowledged defect of written 
 desciiption, we are, therefore, compelled to have recourse to some sensible 
 mode of representation, by which the analogy, positions, and connexions of 
 the separate parts of a body, may be clearly exhibited to the eye. 
 
 Two distinct modes of representation have accordingly been invented, by 
 which the forms, positions, and measures of the original parts of a complex 
 object may be correctly ascertained. One is denominated Perspective, the 
 other Orthographical Projection, and each possesses a peculiar advantage of 
 its own. 
 
 By the aid of perspective drawing, viewed from a certain point, the object 
 is exhibited to the eye in such a manner as to produce the same image, and 
 the same effect on the imagination, as the real object itself, when viewed from 
 a corresponding point. 
 
 By means of an orthographical projection* although the object be not so 
 agreeably represented to the eye, the exact measures of all its parts can be 
 asceitained with incomparably less trouble than by perspective repre- 
 sentations. 
 
 Orthographic projection, as applied to the representations of the circles of the sphere, has been 
 long understood; but general representations are only of modern date. The French method of 
 projection, by means of co-ordinate planes and traces, is very ingenious, and is in general use 
 among them ; but a more commodious method, by means of an intersecting line, was first 
 introduced into Rees’ Cyclopaedia by the author of this work, Mr. Peter Nicholson, in the year 1813. 
 
 3 his gentleman has been successful in applying his method to many architectural solids, floors, 
 roofs, &c. (see his Builder’s Director.) Had Professor Farish been acquainted with these principles, 
 
 he never would have given so imperfect a system, under the unmeaning name of Isometrical 
 Perspective. — E d. 
 
VIII 
 
 INTRODUCTION. 
 
 A species of orthographical projection has been long understood and 
 practised by Cabinet-makers, Carpenters, Joiners, Architects, and artisans 
 concerned in Building, by means of plans, sections, and elevations. 
 
 A Plan is a representation of all the horizontal and other lines in the height 
 of the surface of an object upon a horizontal plane, by means of lines, drawn 
 perpendicularly from all, or a sufficient number of points in these lines, to the 
 said horizontal plane. A Section is a projection on a vertical plane, (dividing 
 the object into two parts,) and is formed by lines falling perpendicularly from 
 all the points in the horizontal and other lines of the object on one side of 
 the plane of projection, upon the said plane; shewing, at the same time, those 
 parts of the object that are really cut, or that coincide with the plane of 
 projection. Lastly, an Elevation is a projection, upon a vertical plane, placed 
 before the object by lines falling perpendicularly from all the points in the 
 horizontal or other lines in the surface of the object to the plane of projection. 
 Hence, the projecting rays in a section, and in an elevation, are parallel to 
 the horizon; and hence, also, when the lines of the object are parallel to the • 
 plane of projection, the lines of the projections are equal and similar to the 
 corresponding lines of the object. 
 
 Plans and elevations are eminently useful, for the purpose of enabling 
 workmen to comprehend the designs of rectangular objects, especially where 
 the object is simple, and all the faces of the three planes of projection are 
 parallel to one or other of the faces of the object. The different forms in 
 the three planes being represented by equal and similar figures, furnish an 
 easy method of finding the dimensions. Yet, how useful soever this mode 
 of representation may be to a tradesman, it does not convey to others, who 
 are unacquainted with this species of drawing, any idea of the design; nor is 
 it, after considerable study, that even proficients in new designs can connect 
 all the different parts in their imagination, such as they would be in the 
 original object. In complex, designs, many plans, sections, and elevations 
 are required to explain them. Much labour and attention are therefore 
 spared by an orthographical projection, which not only connects the several 
 parts, but also gives their due proportions. 
 
 As the construction of works, as well as every kind of delineation, requires 
 the aid of mathematical principles, we have commenced our treatise with 
 Practical Geometry, confining ourselves to such problems as may be 
 found necessary, either to direct the mechanical execution, or to represent 
 the object intended to be shewn. 
 
INTRODUCTION. 
 
 IX 
 
 In these two different species of projection, the art of perspective 
 enables us to explain our designs in the most natural and in the most 
 
 agreeable manner, both to the eye and to the understanding of others, and a 
 sufficient number of its rudiments is given and illustrated, by their application 
 to the subjects included in this treatise. 
 
 In order, however, to proceed with the utmost certainty in explaining to 
 workmen the forms and positions of all the parts of a complex object, which 
 are intended to be executed in every part, agreeably to certain dimensions in 
 feet and inches obtained from a scale ; we have deemed it requisite, without 
 referring to more than one drawing, to explain as many of the principles of 
 orthographic projection as may be required by the designer to communicate, 
 and by the workman to understand, the complete intention of the design. 
 
 After having explained the principles of representing objects, we shall next 
 proceed to the construction of Furniture; in which, from certain given parts, 
 the remaining parts are found. This is a branch of solid Geometry, the 
 knowledge of which is indispensable. It is equally useful to the Cabinet- 
 maker, the Carpenter, and the Joiner. 
 
 The three grand branches which this work is designed to embrace, are, 
 Cabinet-making, Upholstering, and Decorating. In these depart- 
 ments it is not intended to direct the Artist in his mere mechanical operations, 
 for with these it is presumed he is already fully acquainted; but to place 
 before him various scientific principles, that on all occasions may become 
 his guide. 
 
 To the Cabinet-maker will be presented certain specific and permanent 
 rules, equally applicable to dimensions and proportions in all their variety. 
 These in their result will invariably ensure order, harmony, symmetry, and 
 beauty, from which the eye receives gratification, and communicates pleasure 
 to the mind. 
 
 To the Upholsterer such directions will be given as will imprint on his 
 work the footsteps of elegance and taste; so that while fancy dictates new 
 modes of embellishment, an imperishable grandeur may associate with all 
 his performances. 
 
 The Decorator must always, in a certain degree, be exposed to the 
 freaks of caprice, and the ever varying mutability of fashionable taste. With 
 him, invention and genius are indispensable qualifications; and these, when 
 combined, give perfection to his art. But, amidst this instability, there are 
 certain principles, from which his lines, his curves, his festoons, his drapery, 
 
X 
 
 INTRODUCTION. 
 
 Vs ^ r t> 
 
 his combinations, and his arrangements, must not depart. With these 
 principles it is the design of the Author to furnish him. 
 
 The Plates which embellish this volume, being entirely from original 
 inventions and drawings, will have a particular claim upon public attention ; 
 especially as they will be numerous, and executed in a superior style of 
 elegance. But notwithstanding their originality, the explanations and 
 references which accompany each, will render the whole perfectly intelligible; 
 and their utility and advantage must be obvious to all who reflect upon their 
 import and application. 
 
 To furnish the reader with every possible information on the subjects of 
 which this volume treats, being the primary and anxious design of the Author, 
 he flatters himself, that, from the principles developed, and the incalculable 
 variety of designs to which they may be applied, it will be found a useful 
 instructor, even should a different taste prevail. 
 
 Amidst the fluctuations of tasteful elegance, he will watch the kaleidoscope 
 of fashion in all its movements, while this work is passing through the press, 
 and avail himself of every appearance that may be deemed worthy of 
 preservation. 
 
 But whatever changes the progress of time may introduce, of this truth he 
 has an indubitable assurance, that although fashions are fickle and capricious, 
 yet the scientific principles, upon which alone these fashions are built, are 
 imperishable and eternal. 
 
 London, May 1st, 1826 
 
THE 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 AND 
 
 COMPLETE DECORATOR. 
 
 GEOMETRY. 
 
 Geometry forming the basis on which our superstructure is to be 
 raised, has the first claim on our attention. This exalted science appears 
 to have been understood in some degree by the ancient Egyptians, as we are 
 informed by Herodotus: and indeed, in a country like Egypt, subjected to 
 the annual innundations of the river Nile, which destroyed all the landmarks 
 either by the impetuosity of its waters, or by the deposition of mud or sand, 
 a practical knowledge of Mensuration became necessary, in order to restore 
 to the proprietors of the lands their proper portions, which had been blended 
 into one undistinguishable surface. Hence, the word was originally limited 
 to the science of Mensuration, for it is compounded of two Greek words, yv, 
 earth , and /xerpew, to measure ; which, together, literally signify, measuring of 
 the earth. But the Geometry of the Greeks is the science which treats of 
 the properties of magnitudes, considered either as Lines, Angles, Surfaces, 
 or Solids. 
 
12 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 It is to this latter people that we are indebted for the perfection of this 
 sublime branch of knowledge, and “ it is remarkable (says Professor 
 Playfair) that the writings of Euclid, at the distance of two thousand years, 
 continue to form the most approved introduction to the mathematical 
 sciences.” 
 
 Having thus given a brief sketch of the origin of Geometry, we shall now 
 collect as many of its simple rules as will be useful to the Cabinet-maker 
 in the formation of working drawings, and shall therefore proceed to 
 
 PRACTICAL GEOMETRY. 
 
 Practical Geometry treats of the methods of drawing lines in all manner 
 of directions, and of the description of figures according to certain simple 
 properties enumerated in words called 
 
 Definitions. 
 
 Def 1. Every material substance, in the language of Geometry, is 
 termed a Solid. 
 
 Def. 2. The external appearance (or outward face) of a solid is called 
 a Surface. 
 
 Def. 3. When a surface is divided, either in reality or in imagination, 
 into two parts, the division is termed a Line. 
 
 Def. 4. When two lines cross one another, the intersection is termed 
 a Point. 
 
 Surfaces are therefore merely the conceived divisions of a solid, lines are the divisions of a 
 surface, and points are those of a line, into parts, but have no separate existence as substances. 
 The lines and points of the diagrams of the practical Geometer are, however, real surfaces, 
 because it is necessary that they should be visible ; but these are treated in the science as if they 
 were metaphysically correct. 
 
 Def 5. If while a line revolves round its two extreme points, every 
 intermediate point remains in the same place, or point of space, the line is 
 called a Straight Line. 
 
 Cor . 1 . Hence, one straight line cannot meet another straight line in 
 more than one point, unless they coincide in every point. 
 
AND COMPLETE DECORATOR. 
 
 13 
 
 Cor . 2. Hence, two straight lines cannot enclose a space. 
 
 Cor. 3. Hence, if the position of two points in a straight line be 
 determined, the position of the whole line is determined, for only one straight 
 line can pass through these points. 
 
 The definition of a straight line, in Euclid, as it Is translated from the Greek text, is obscure; 
 a straight line, or right line, is defined to be a line which lies evenly between its extreme points, the 
 word “ evenly” standing as much in need of an explanation, as the word straight or right, which it 
 is intended to define. If the word equally had been used by the translator instead of “evenly,” the 
 sense would have been quite as agreeable to the original text, and, though not explicit, would seem 
 to imply the very idea of a straight line as we have defined it; viz. “a straight line is that which lies 
 equally between its extreme points,” which means, that it is alike related on all sides of the 
 surrounding space. Professor Playfair’s definition is excellent, it is as follows : If there he two 
 lines which cannot coincide in two points without coinciding altogether , each of them is called a 
 straight line. The objection to this is, that we are under the necessity of having recourse to 
 lines instead of a line. This might have been avoided by supposing a line to be divided into 
 any two parts, then if these parts could not coincide in two points without coinciding altogether, 
 the line is a straight line; but the idea of cutting the line into two segments is not altogether 
 agreeable. 
 
 If there be any objection to the definition of a line which we have given, 
 the same must apply to the definition of solids generated by revolving a plain 
 figure about an axis, as a cone is generated by a right-angled triangle. 
 
 Def. 6. A line, in which, if any two points be taken, and if the intercepted 
 portion, however small, be not straight, such a line is called a Curve. 
 
 Def. 7. A Mixed Line is that which is compounded of straight and 
 curve lines. 
 
 Def. 8. Linear Extension is the distance between one point and another, 
 along a straight line between them, and this is also called the Distance of the 
 Points. 
 
 Def. 9. A Plane, or Plane Surface, is that which will every where coincide 
 with a straight line. 
 
 It is a curious fact, however, that though a surface may admit of two straight lines being 
 drawn in that surface through any point in it, yet the surface may not be a plane, but if the 
 surface admits of three straight lines being drawn upon it through any point, the surface is 
 necessarily a plane. 
 
 Two straight lines, which meet each other, are always in a plane. 
 
14 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 Def 10. An Angle is the opening or space contained between two lines 
 which meet each other. 
 
 Def. 11. A Rectilinear Angle is that w ich is contained by two straight 
 lines, which are not in the same straight line. 
 
 All angles henceforth expressed are supposed to be rectilinear, unless the contrary is stated. 
 
 Def. 12. The point where two straight lines, which form an angle, meet 
 each other, is called the Vertex. 
 
 Def. 13. If in two angles, in which, if the one be applied upon the other, 
 so that when their vertices and one of their containing lines coincide, the two 
 remaining lines coincide also, these Angles are said to be Equal. 
 
 Hence the length of the lines, containing an angle, do not make the angle 
 greater or less by being longer or shorter. 
 
 Def. 14. If two equal angles have a common line of separation, and the 
 other two containing lines in the same straight line, each of these angles is 
 called a Right Angle; likewise the common line is called a Perpendicular to 
 the straight line formed of the other two. 
 
 Def. 15. If the contained space of an angle be applied upon that of a 
 right angle, so that the two vertices and one of the containing lines of each 
 may coincide, and the other containing line of the angle applied fall between 
 the containing lines of the right angle, the angle thus applied is called an 
 Acute Angle. 
 
 Hence an acute angle is less than a right angle; for the right angle 
 contains the acute angle and another angle besides. 
 
 Def. 16. If the contained space of an angle be applied upon that of a 
 right angle, so that the two vertices and one of the containing lines of each 
 may coincide, while the other containing line of the angle applied falls 
 without the space contained by the two lines of the right angle, the angle thus 
 applied is called an Obtuse Angle. 
 
 Hence an obtuse angle is greater than a right angle; for the obtuse 
 angle contains not only the right angle, but likewise another angle besides 
 the right angle. 
 
AND COMPLETE DECORATOR. 
 
 15 
 
 Def. 17. If two lines, in the same plane, are so situated that they cannot 
 form an angle, however far they may be extended, these two lines are said to 
 be Parallel. 
 
 Def. 18. A Figure is that which is bounded by one or more lines, called 
 the sides or edges of the figure. 
 
 Def. 19. A Rectilineal Figure is that which is bounded by straight lines. 
 
 Here we shall observe, that all boundaries are supposed to be rectilineal, 
 unless a curve is expressed. 
 
 » 
 
 Def. 20. A Triangle is a figure, consisting of three sides, and consequently, 
 of three angles; because an angle is opposite to every side. 
 
 In respect to the Sides of a Triangle. 
 
 Def 21. If a triangle has all its sides equal, it is called an Equilateral 
 Triangle. 
 
 It is easy to prove, and it is almost self-evident, that the angles, as well as 
 the sides of an equilateral triangle, are equal. 
 
 Def. 22. If a triangle have two of its sides equal, it is called an Isosceles 
 Triangle. 
 
 Def. 23. If a triangle have no two of its sides equal, it is called a Sealine 
 Triangle 
 
 In respect of the Angles of a Triangle. 
 
 Def. 24. If a triangle have one right angle, it is called a Right-angled 
 Triangle. 
 
 Def 25. If a triangle have one obtuse angle, it is called an Obtuse-angled 
 Triangle. 
 
 Def. 26. If a triangle have its three angles acute, it is called an Acute- 
 angled Triangle. 
 
 Def. 27. A four-sided figure is called a Quadrangle, Quadrilateral, or 
 Trapezium. 
 
16 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 Def. 28. If a four-sided figure have one pair of its sides parallel, it is 
 called a Trapezoid. 
 
 j Def. 29. If a four-sided figure have both pairs of its sides parallel, it is 
 called a Parallelogram. 
 
 Def. 30. A parallelogram, of which neither the angles nor the sides are 
 all equal, is called a Rhomboid. 
 
 Def 31. A parallelogram, which has all its four sides equal, but not its 
 angles, is called a Rhombus. 
 
 Def. 32. A parallelogram, which has one of its angles a right angle, is 
 called a Rectangle. 
 
 It may be proved, and indeed it is in some measure evident, that all 
 the angles of a parallelogram are equal, and that the opposite sides are 
 equal. 
 
 Def. 32. A rectangle, which has not all its sides equal, is called an Oblong. 
 
 Def. 33. A rectangle, which has all its sides equal, is called a Square. 
 
 Def. 34. Every figure, which has more than four sides, is called a Polygon. 
 
 Def. 35. A polygon, which has all its sides equal, is said to be Equilateral. 
 
 Def. 36. A “polygon, which has all its angles equal, is said to be 
 Equiangular. 
 
 Def 37. A polygon, which is both equilateral and equiangular, if such 
 can be, is called a Regular Polygon. 
 
 There may be some objections to this definition, which is that usually given, for it is easy to 
 conceive, that all the sides of a figure may be equal, but not that the angles are equal at the same 
 time. This coincidence of the equality of sides and angles requires demonstration, therefore, in 
 order to qualify it, we have inserted the words, if such can be. 
 
 Def. 38. A regular polygon, of three sides or angles, is called a Trigon. 
 
 A trigon is, therefore, the same as an equilateral triangle, for an equilateral 
 triangle may be proved to have its angles as well as its sides equal. 
 
 Def. 39. A regular polygon of four sides, is called a Tetragon. 
 
 A tetragon is, therefore, the same as a square, for it may easily be proved, 
 that all the angles of a square are equal, as well as the sides. 
 
AND COMPLETE DECORATOR. 
 
 17 
 
 Def 40. A polygon of five, six, seven, eight, nine, ten, eleven, twelve, &c. 
 sides, is called a Pentagon, Hexagon, Heptagon, Octagon, Enneagon, Decagon, 
 Hendecagon, Dodecagon, &c. 
 
 Def. 41. A Circle is a plane figure, contained by one line called the 
 Circumference, and this line is every where equally distant from a point in the 
 figure, called the Centre of the Circle. 
 
 Def. 42. The Radius of a Circle is any straight line drawn from the centre 
 of the circle to its circumference. 
 
 Def. 43. The Diameter of a Circle is a straight line passing through the 
 centre, and terminated at each extremity by the circumference. 
 
 Corol. Hence the diameter of a circle is equal to twice the radius. 
 
 Def. 44. A Chord is any straight line terminated at the two extremities 
 by the circumference. 
 
 Def. 45. An Arc of a Circle is any part of the circumference. 
 
 Def. 46. A Segment of a Circle is a portion of a circle contained by an 
 arc and a chord. 
 
 Def. 47. A Sector of a Circle is a portion of the area contained by two 
 radii, and the arc intercepted between them. 
 
 E 
 
18 
 
 ABINET-MAKER, UPHOLSTERER, 
 
 PLATE I. 
 
 Problem I. 
 
 To bisect a given straight Line de, Figure 2, by a Perpendicular. 
 
 From the points d and e, as centres, with any equal radii, describe the 
 intersections f and g. Draw fg t intersecting de in the point f, and f is the 
 middle of the line de; and the line fg, which passes through f, is at the 
 same time perpendicular to de. 
 
 Problem II. 
 
 From a given Point c, in a given straight Line ab, Figure 1, to erect a 
 
 Perpendicular. 
 
 Mark the points c and d in the line ab, equally distant from c, and from 
 each of the points c, d , with any equal radii, greater than cc or c d, describe 
 the intersection d, and draw cd. cd is perpendicular to a b. 
 
 Problem III. 
 
 From a given Point g, Figure 3, in a given straight Line a b, to draiv a 
 
 Perpendicular. 
 
 From some convenient point i, above the line ab, asa centre, and with 
 the distance ig, describe an arc fgh, meeting ab in h. Join hi; produce 
 hi to f and draw fg, and fg is perpendicular to ab. 
 
 Problem IV. 
 
 From a given Point c, Figure 4, to draiv a Perpendicular to a given Line g h. 
 
 From the point c, describe an arc de, so as to intersect gh in the points 
 d and e. From each of the points d and e, as centres, with any equal radii 
 greater than the half of de, describe arcs intersecting each other in f, and draw 
 the line c f: cfis perpendicular to gh. 
 
/ / ut \ tic y// ( reometr } y 
 
 PISHES. SON, & C? LONDO.fr, 183* 
 
AND COMPLETE DECORATOR. 
 
 19 
 
 Problem Y. 
 
 At a given Distance e f, Figure 5, to draw a straight Line parallel to a given 
 
 straight Line ab. 
 
 From any two convenient points e,f, in the line ab, with the radius e f, 
 describe arcs at c and d , and draw the straight Line CD to touch the arcs 
 at c and d. c d is parallel to a b, as required. 
 
 Problem VI. 
 
 At a given a, Figure 6, No. 1, in the straight Line Ab, to make an Angle 
 
 equal to a given eDf, No. 2. 
 
 From the point d, with any convenient radius, describe an arc ef and 
 with the same radius, from the point a, describe the arc be. Make the arc 
 be equal to ef and draw a c. b ac is the angle required. 
 
 Problem YII. 
 
 To bisect a given Angle abc, Figure v. that is, to divide it into two equal 
 
 Parts. 
 
 Let c be any point in bc; make ba equal to bc.; and from the points 
 a and c, with any equal radii, describe two arcs so as to intersect in d, and 
 draw the line dB. d b will divide the angle abc into two equal angles. 
 This is called bisecting the angle. 
 
 Problem YIII. 
 
 To divide a straight Line a b, Figure 7, into any number of equal Parts. 
 
 Draw any line ad, and through the point b draw bc parallel to ab. 
 Run as many equal parts along the lines a d and b c, as the line a b is to 
 contain equal parts. In this example it is required to divide the line a b 
 into six equal parts. Join the point a to the last division c, and through the 
 points 1, 2, 3, &c. in ad, draw lines parallel to ac, to intersect the line a b, 
 and a b is divided, as required to be done. 
 
20 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 Problem IX. 
 
 Upon a given straight Line a b, Figure 8, to describe an Equilateral Triangle. 
 
 From each of the points a and b, as centres, and with the radius ab, 
 describe an intersection c on one side of the line. Join ca and cb, and 
 abc is the equilateral triangle required. 
 
 Problem X. 
 
 In a given Triangle abc, Figure 9, to inscribe a Circle. 
 
 Bisect any two of the angles a and b. From the point e, where the 
 bisecting lines meet, draw e h perpendicular to bc, on any one of the sides. 
 From the centre e, with the radius e h, describe the circle hik, and the circle 
 hik will touch the three sides of the triangle abc, as required. 
 
 Problem XL 
 
 Upon a given straight Line bc, Figure 10, to describe a Pentagon . 
 
 Bisect bc by the perpendicular Iiy. Draw c i parallel to hv, and make c i 
 equal to ch. Join b *, and produce b« to l, making il equal to ic. From the 
 centre c, with the distance cl, cut the line h f in f. From f, as a centre, with 
 the radius fb or fc, describe the circle abcde; repeat bc within the circum- 
 ference, so that the extremities of every two adjacent chords may meet, and 
 the lines thus drawn will divide the circumference into five equal parts. 
 
 If a pentagon be inscribed in a circle, it is very easy to inscribe a 
 decagon, by bisecting the sides of the pentagon by perpendiculars, and 
 mining the points where the perpendiculars cut the circumference to the 
 extremities of the chords of the pentagon, as in the figure aA,bB,cc,dv>,eE,a. 
 
 Problem XII. 
 
 In a given Circle , Figure 11 ,to inscribe a Triangle, a Hexagon, or a Dodecagon. 
 
 Apply the radius of a circle to the circumference as connected chords, 
 and it will contain the radius six times, as ab, bc, cd, de, ef, fa. Bisect 
 
AND COMPLETE DECORATOR. 
 
 21 
 
 each of the Arcs in the points a, b y c , d, e,f and join these points to the 
 extremities of the chords of the hexagon, and a dodecagon will be 
 obtained. 
 
 In this manner may the circle be divided into 24, 48, &c. parts, always 
 doubling the number of sides. 
 
 GENERAL OBSERVATION. 
 
 Any regular polygon, inscribed in a circle, being once obtained, we may 
 double the number of sides, and double this last number again, and so on, as 
 often as we please; and thus, if we have a triangle inscribed in a circle, we 
 may have the series of polygons, having 3, 6, 12, 24, 48, &c. sides. If we have a 
 pentagon inscribed in a circle, we may have the following series of Polygons, 
 having 5, 10, 20, 40, &c. sides. If we have an octagon inscribed in a circle, 
 we may have the series of polygons, having 4,8, 16,32, &c. sides; and 
 thus we have direct rules for dividing a circle into 3, 4, 5, 6 equal parts, 
 or any polygon, of which the number of sides is expressed by any term of a 
 geometrical progression, of which the [first term is 3, 4, 5, or 6, and the 
 common multiplier 2. Rules for other polygons are neither easy to be 
 described in words, nor to be given in figures of construction. 
 
 Problem XIII. 
 
 In a given Circle to inscribe any regular Polygon. 
 
 Figures 12, 14, 15. 
 
 Draw the diameter, and divide it into as many equal parts as the polygon 
 is to have sides. Bisect the diameter by a perpendicular, cutting the 
 opposite sides of the circumference. Produce the perpendicular, so that the 
 part without the circle may be three-fourths of the radius. From the 
 extreme point of the part produced, draw a straight line through the second 
 point of division in the diameter to meet the circumference. Join the 
 intersected point in the circumference to the extremity of the diameter, 
 whence the equal parts commence, and the line thus joined will be a 
 
 p 
 
22 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 chord of the circumference in which it may be contained, as often as 
 required, very nearly. 
 
 Thus, in Figure 12, the intention is to inscribe in a circle a polygon of 
 seven sides. The diameter gi is divided into seven equal parts. The 
 perpendicular hd is drawn from the centre h, and the radius is divided into 
 four parts, and three are set without the circle from four to seven. A line is 
 drawn from d through the second point 2 in the diameter g i, to meet the 
 circumference in a, and the line ag is the side of a polygon of seven sides; 
 and by inspecting Figures 14 and 15, it will be seen that the same principle 
 extends in all cases. 
 
 Figure 12 is a heptagon; Figure 14 an enneagon; and Figure 15 an 
 hendecagon. These methods are, however, only near approximations. 
 
 Problem XIV. 
 
 In a given Square A BCD, Figure 13, to cut out the largest regular Octagon . 
 
 Draw the diagonals ad and cd, intersecting each other at i. From the 
 points a, b, c, d, as centres, with the radius a i, equal to half one of the 
 diagonals, cut the sides of the square in the points a , b, c , d, e,fg, h. Join 
 ab, cd, ef, gh , and abcdefgh is the octagon required. 
 
 Problem XV. 
 
 To draw a Tangent to a given point c, Figure 16, in the Circumference of a 
 
 Circle, of which the Centre is a. 
 
 Join ac. Through the point c draw bd perpendicular to Ac, and bd is 
 the tangent required. 
 
 Problem XVI. 
 
 Through three given Points b, a, c, Figure 17, not in a straight line , to draw 
 
 the Circumference of a Circle. 
 
 Join the two points b and c to the middle point a. Bisect each of the 
 lines ab, ac, by perpendiculars meeting in h. From the point h, as a 
 
AND COMPLETE DECORATOR. 
 
 23 
 
 centre with the radius bh, ah, or ch, describe a circle bac, and the 
 circumference will pass through the three points b, a, c, as required. 
 
 Problem XVII. 
 
 To describe the Arc of the Segment of a Circle , of which the Chord is ac, 
 Figure 17, and the Rise of the Arc the Line d e. 
 
 Bisect the chord a c by a perpendicular hf, intersecting A c in the point 
 h. Make hb equal to de. Join ab, and bisect ab by a perpendicular hd. 
 From the point h , where the two perpendiculars meet as a centre, with the 
 distance Ab, describe the arc abc, which is that required. 
 
 Another Method , Figure 19. 
 
 Having set off the perpendicular from the middle of the chord to the 
 point c in the circumference; fasten two laths, ae, ec, together, so as to 
 intersect at c, and the edges to he on the points b and d. Brace these 
 laths together by a cross-piece bd, so that the angle bcd may be 
 immoveable. Bring the point c to b, and move the apparatus round, 
 so that the leg a c may run upon the point or pin at b, and the leg c e upon 
 the point or pin at d ; then a pencil held at c, in the act of moving, will 
 describe the curve required. 
 
 This method may be used where the radius would extend beyond the 
 space upon which the operation is required to be performed. , 
 
 Problem XVIII. 
 
 To describe a Gothic Arch to any Width , ab, Figure 20, and Perpendicular 
 Height i c in the middle of a b, so that the two Sides may meet in any 
 given Angle in the Point c. 
 
 Produce ci to g. Make the angle gcd equal to half of the vertical 
 angle. Draw a d parallel to i c. In g c make g f equal to a d, and join a f. 
 Divide af and ad each into the same number of equal parts, at the points 
 1, 2, 3, &c. From g, through the points 1, 2, 3, &c. in a f, draw lines to 
 
24 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 intersect lines drawn from c to the corresponding points 1, 2, 3, &c. in a d, 
 in the points «, b, c, d. Put pins in the points a, a, b,c, d, c, and bend a 
 slip of wood round them, and draw a curve by the edge which touches the 
 pins, and this curve will form one side of the Gothic Arch. 
 
 In the same manner we may describe the other side. But when one side 
 of a symmetrical figure is obtained by any method, the other side will be 
 found by the following process. 
 
 From the points a, b , c, d, draw lines parallel to a b, to terminate on the 
 other side of cg, as double ordinates. From the points of intersection in 
 ic, transfer the ordinates on the left of ig to the right hand side of it, and 
 they will reach to h,g,f e. Put in nails as before in the points B,h,g,f, e, c, 
 and bend the slip of wood to touch all the pins, and draw the curve as 
 before. 
 
 Problem XIX. 
 
 To describe an Ellipse , to any Length and Breadth , by means of the Instrument 
 called a Trammel , an Instrument well known to Workmen. Figure 21. 
 
 Let three pins, «, b,\c, in a straight line, be fixed to three cross-pieces 
 through which the rod passes; suppose one of the pieces which contains 
 the pin c, to be fixed to the rod, and the pieces a and b , in which the other 
 two pins are fixed, to be moveable, and the nearest point to the fixed point 
 being of steel, and the other to be a pencil. 
 
 Move tbe nearest sliding pin from the fixed pin, so that their distance 
 may be equal to half the breadth of the ellipse, and move the pencil point 
 from the fixed pin, so that their distance may be equal to half the length, 
 or to half the longer axis. Place the arms of the trammel so that the 
 grooves may coincide with the axis, the fixed point c of the rod in 
 the groove which coincides with the conjugate axis, and the middle 
 point in the groove over the transverse axis: then move the rod round 
 from the extremity a of the longer axis, and the pencil at a will describe 
 the curve. 
 
AND COMPLETE DECORATOR. 
 
 25 
 
 Figure 22, shews how points may be found in the curve upon the same 
 principle. 
 
 Let ghi be a slip of pasteboard, or a piece of ivory, straight on the edge 
 gh i. Mark the three points i, h, g on the straight edge, so that i h may be 
 equal to the semi-axis minor, and ig equal to the semi-axis major. Place 
 the point i in the axis minor, and the point h in the axis major, and mark the 
 point g upon the board or paper, which will give one point in the curve 
 Repeat this operation till a sufficient number of points are found ; stick in 
 pins or nails in all the points, bind a pliable slip of wood so as to coincide 
 with all the pins or nails, and draw a curve by the edge of the slip which 
 touches the pins, and it will be the ellipse required. 
 
 If the points are found on paper, the curve may be drawn through the 
 points with sufficient accuracy by hand. 
 
 Problem XX. 
 
 To find any number of continued Proportionals , or to divide ct given Line a c 
 into a Series of continued Proportionals , the first cd being given. 
 
 Draw a line a c, making any angle ca c, with ac. In a c, cut off a c 
 equal to Ad , and join cc. Draw db parallel to c c , meeting ac in b . From 
 ac cut off a a equal to a b , and draw ab parallel to db , and proceed in this 
 manner, and the line will be divided as required to be done. 
 
 PLATE II. 
 
 Problem XXI. 
 
 Upon a given straight Line to form a Polygon , similar to a given Polygon 
 
 abcde, Figure 1. 
 
 Make the straight line ae equal to ae. Find the point d by describing 
 arcs with the radii ad, ed, from the centres a,e ; find the point b , by 
 
 G 
 
26 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 describing arcs with the radii ab, db, from the centres a,d,c, and find the 
 point c, by describing arcs with the radii b c, dc, from the centres b and d. 
 Join ab , be, cd, de , and abode will be the polygon required. 
 
 i 
 
 Problem XXII. 
 
 To make a Square equal to the Sum of two' given Squares, h, f. Figure 2. 
 
 Draw the straight line a e and e d perpendicular to ae. Make a e equal 
 to the side of the square f, and e d equal to the side of the square h. Join 
 A d. Draw a b perpendicular to a d, and make a b equal to a d. Find the 
 point c, by describing arcs from b and d, with a radius, each equal to ad. 
 Join bc, and cd; then abcd is the square required. 
 
 Problem XXIII. 
 
 Given three straight Lines ab, 6d, be, to find a fourth Proportional. 
 
 Draw the two straight lines ba, bd, making any convenient angle with 
 each other. Make bd equal to ab, bd equal to b d, cut ba in f so that bf 
 may be equal to b f. Join ad. Cut bd in g, by drawing /g parallel to ad, 
 an 1 bg is the fourth proportional required. 
 
 Problem XXIV. 
 
 To divide a given straight Line ab, Figure 4, in the same Proportion as 
 
 another ca is divided. 
 
 Draw the two straight lines ac, ab, making any convenient angle with 
 each other. Make ac equal to ac, ab equal to a b, and join be. Cut the 
 straight line ac in the points d, e,f,g, so that ad, ae, af ag be equal to the 
 parts ad, ae, af, ag of the given line ac. Cut the line a b by drawing lines 
 through the points d, e,fg in ac, parallel to cb, and ab will be divided in the 
 same proportion as ac. 
 
AND COMPLETE DECORATOR. 
 
 27 
 
 Problem XXV. 
 
 Two straight Lines ab, cd, Figure 5, being given , to find a mean 
 
 Proportional. 
 
 In the straight line da make db , ba, respectively equal to the lines cd, 
 ab. On da, as a diameter, describe the semicircle dca. Draw be 
 perpendicular to da, and be is the mean proportional required. 
 
 Problem XXVI. 
 
 To describe an Elliptic Arch, by finding Points in the Curvature to any 
 Width bc, and Height fe, Figures 6 and 7. 
 
 Find the point a in ef, produced by making fa equal to f e. Divide 
 b f and b d each into the same number (5) of equal parts. Find the points 
 a, c, e, g, by drawing lines from the points 1,2, 3, &c. in bd bE, to meet the 
 corresponding lines drawn from a through the points 1, 2, 3, &c. in bf. 
 Through the points b, a, c, e, g, e, draw a curve by the edge of a lath or slip 
 of wood. The other half may be found in the same manner, or by transferring 
 them upon the double ordinate. 
 
 Another Method of finding Points in the Curve of an Ellipse is that 
 
 exhibited in Figure 8. 
 
 Let a b and cd be the two axes given in position and magnitude. Draw 
 i n parallel to the greater axis. Find the points a, b, c in the curve, thus 
 draw lines ad, bb', cc, &c. through the centre f. Cut the line in in a, b, c, 
 &c.by producing Fa, eb, fc, &c. Find the point e, by producing fd, and 
 making de equal to fa or fb. From e, with the radius e d, describe an 
 arc. Find the points a, b,c in the arc, by joining g e, h e, i e. Draw aa,bb,cc, 
 and through the points d, a, b, c, a draw a curve. Make Fa', f b', f c 
 respectively equal to Fa, f b, f c, See. 
 
HQ 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 Problem XXVII. 
 
 To describe a Parabola , Figure 9, by finding Points a, c, e, g in the curve . 
 
 Let a b be the base or double ordinate, and c d the height or abscissa. 
 
 Draw ae and de parallel to cd and ab. Divide ae and de each into 
 (5) the same number of equal parts. Find the points a,c,e,g, by drawing 
 lines fiom 1,2,3, &c. in ae to d, to intersect lines drawn parallel to cd 
 through 1, 2, 3, &c. in de. Draw the curve Aacegjy through the points, 
 and the other side by transferring the ordinates to those on the other side of 
 the abscissa. 
 
 Problem XXVIII. 
 
 To draw the Section of an Ovolo Moulding in the Form of a Parabolic Curve 
 A, Figuie 10 being the Point of Projection , and d the Point where the 
 End and Upper Edge of the Fillet below meet each other. 
 
 Draw ac perpendicular to the length of the moulding, and draw dc at 
 pleasure, depending on the form of the moulding. 
 
 Divide ac and dc each into the same number of equal parts (5). Find 
 the points a , b, c, &c. by drawing lines through the points 1 , 2, 3, &c. in dc 
 to the point a, to intersect lines drawn parallel to cd through the points 
 1, 2, 3, &c. in a c, and draw the curve d a b c d Af 
 
 Problem XXIX. 
 
 To describe an Hyperbola , having an Ordinate bd, Figure 11, and the 
 
 Abscissa b c. 
 
 Draw df and cf parallel to bc and bd. Produce bc to a. Take the 
 point a at a greater or less distance from c, as the curve is required to be 
 flatter or quicker at the point c. Divide bd and fd each into the same 
 number of equal parts (5). Draw lines from 1, 2, 3, &c. in b d to a, to intersect 
 lines drawn through 1,2, 3, &c. in ed to c. Draw a curve through d and 
 through the points a,c,e,g of intersection, and it will be one-half of the 
 curve required. 
 
AND COMPLETE DECORATOR. 
 
 29 
 
 Figure 12, exhibits the application of the hyperbolic curve, in drawing the 
 section of a moulding, bd being the ordinate, dc the abscissa, and the point 
 d being the upper edge of the fillet at its projection. 
 
 Problem XXX. 
 
 To describe the Section of a Cima-recta or Ogee Moulding , a, Figure 13, being 
 the Top of the Section , and b the Bottom. 
 
 Join a b, and divide a b into two equal parts at the point c. Divide c b 
 into six equal parts, and divide bc, ca, each by a perpendicular cf and gd. 
 Cut the line cf by an arc described from b, with the radius of five equal parts. 
 Make dg equal to cf. From the point f with the radius/B, describe the arc 
 bc, and from the pointy, with the radius gc, describe the arc ca, and bca 
 will be the moulding required 
 
 Problem XXXI. 
 
 To describe a Semi-reversa or Ogee Reverse , Figure 14, the Points a and b 
 being the Ends of the Section of the Moulding. 
 
 Divide ab into two equal parts in c, and bisect bc, ca, each by the 
 perpendiculars cf, dg. From the middle points e and d make ef and dg, 
 each equal to Be or ec. From f, with the radius^/B, describe the arc bc, 
 and from g, with the radius gc, describe the arc ca; then bca is the semi- 
 reversa required. 
 
 Figures 15 and 16, shew the application of the method of describing an 
 ellipse, by points to an ovolo moulding, db and da being semi-conjugate 
 diameters, from which the curve may be described, similar to figures 6 and 7, 
 from the axis. 
 
30 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 PLATE III. 
 
 Problem XXXII. 
 
 To Draw the Section of aCima-recta , having theProjecture of its Extremities 
 
 at a and d. Figure 1. 
 
 Produce the upper edge of the lower fillet from a to c. Draw dc and a b 
 perpendicular to ac. Bisect bd in f and draw fg parallel to ab; then 
 fg will bisect ac as well as bd. Bisect fg, and through the point of 
 bisection draw the straight line 4-4, meeting ab in 4, and dc in 4. Divide 
 the distance between the point of bisection and each point at 4, into any 
 number of equal parts, for example into 4, and mark the points of division 
 1, 2. 3 ; also, divide the distance between the point of bisection, and each of 
 the points /and g, into the same number of equal parts, viz. four. From the 
 points a, b, cl, c, draw r lines through the points of division 1, 2, 3, each line to 
 intersect as in the figure, and through the points of intersection draw a curve, 
 and the curve will be the cima-recta required. 
 
 In the cima-reversa, figure a, the point a is given as the point of greatest 
 projecture, and the sloping direction fb, in the middle of the moulding, is 
 given in position; acdb and dfgh are parallelograms, in which cd and dh 
 are equal; also db and df are equal to each other. Produce dh to m, and 
 make hm equal to hd. Produce dc to e, and make ce equal to cd. Divide 
 hg’ gf> a ^ so a ac ’ eac h into the same number of parts; then proceed to 
 find the points of intersection, and draw the curve as before. 
 
 Problem XXXIII. 
 
 To describe a Semiovolar Curve, compounded of Circular Arcs, so as to 
 represent the Section of a Semi-Elliptic Arch to any Length and Height. 
 
 METHOD FIRST FIGURE 2. 
 
 Let a b be the length or greater axis. Bisect a b in f, by the perpendicular 
 c r, and make F c and f r each equal to the height of the arch. Draw a n 
 
FI 3 
 
 Geometry 
 
AND COMPLETE DECORATOR. 
 
 31 
 
 parallel to c r, and cn parallel to ab. Bisect the line an in the point 1, 
 and draw the lines lc and rn, intersecting each other in the point l. Bisect 
 cl by the perpendicular d d, meeting cr in d. Draw de parallel to ab. 
 From the point d, as a centre with the radius dc, describe the arc kcpe. 
 Join e b, and produce e b to meet the circular arc cpe in p. Join p d, 
 intersecting fb in i. Make fh equal to fi, and join dh, which 
 produce to k. 
 
 From the point h, with the radius h k, describe the arc k a, and from the 
 point i, with the radius i p, describe the arc p b. Then the compound curve 
 a kc pb will resemble the section of an elliptic arch as required. 
 
 METHOD SECOND, STILL NEARER — FIGURE 3. 
 
 Let a b be the length or greater axis. Bisect a b in d, by the perpendicular 
 ce. Make dc and de, each equal to the height as before. Draw a 3 
 parallel to dc, and c3 parallel to ba. Divide ad and a 3, each into three 
 equal parts. From e, through the points 1, 2, in ad, draw the lines ef,eg, 
 and from c through the points 1,2, in a 3, draw the lines cg, cf. Produce 
 c e to h. Bisect g c by the perpendicular b h, and draw the straight line g h, 
 intersecting ab in w. In db make dn equal to b m, and from h, through n, 
 draw hp. Bisect fg by the perpendicular d i, meeting gh in i. Draw i k 
 parallel to a b. From i, with the radius i f, describe the arc fk. Produce 
 ka joined to f; join fi, intersecting da in p. In db make d^ equal to Dp. 
 Produce l# joined to n. From h, with the radius h c, describe the arc gcp. 
 From i, with the radius ig, describe the arc gf. From l, w'ith the radius 
 lp, describe the arc pn. From p, with the radius p¥, describe the arc fa ; 
 and lastly, from q, with the radius q n, describe the arc nb. 
 
 Problem XXXIV. 
 
 Given three straight Lines , to find a fourth Proportional. Figure 4. 
 
 Draw any angle abd. In ba make ba equal to the first of the given lines, 
 and be equal to the second. In bd make bf equal to the third line. Join 
 cf, and draw ad parallel to cf then bd is the fourth proportional. 
 
.32 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 Problem XXXV. 
 
 Given two straight Lines, and one oj them divided into any number oj' Parts, 
 to divide the other into the same number of Parts, and in the same ratio 
 as those of the divided Line. Figure 5. 
 
 Let a b be the divided line, and c d the undivided line, and let e be a 
 point of division in ab ; also, let hgn be an equilateral triangle. From one 
 of the vertices g-, and upon the adjacent sides, make ga and gb equal to the 
 divided line ab, and draw ab, then ab will be equal to the divided line a b. 
 In ab make ae equal to ae, and join eg. Again, from the vertex g, and 
 upon the adjacent sides, apply the undivided line cd, and draw cd, intersect- 
 ing eg in f, then cd, equal to cd, will be divided in J in the same ratio 
 which a b is divided in e. 
 
 scholium. 
 
 It very frequently happens that segments, which have chords of very great 
 extent, but which have a very small height, are required by the Cabinet- 
 maker. It would be very inconvenient in such cases to find a rod of sufficient 
 length to describe the arc, and in others altogether impossible, since the 
 length of the radius would exceed any limited space. It will, therefore, be 
 useful to shew how flat arcs may be described by means of a sufficient number 
 of points being found in the curve, so that it may be drawn by the vertical face 
 of a pliable slip of wood bent round pins placed in the points. 
 
 We may observe, that, in such cases it is of very little consequence 
 whether the arc be that of a real circle, or that of any other curve, since all 
 curves approach to the arc of a circle near the vertex. Therefore since an 
 error cannot be made visible to the eye by using any curve indifferently, the 
 easiest method must be preferred. The following Problems may be 
 considered as various modes of effecting the object thus circumstanced. 
 
AND COMPLETE DECORATOR. 
 
 33 
 
 Problem XXXVI. 
 
 To describe the Segment of a Circle by means of Points, having the Chord 
 and the height of the Arc given. Figure 6. 
 
 Let ab be the chord. Bisect ab by a perpendicular meeting ab in d. 
 Make dc equal to the height of the segment. Draw fe parallel to ab. 
 Join ca and cb. Draw af perpendicular to ca, and be to cb. Also, 
 perpendicular to ab, draw ag and bh, meeting fe in the points g and h. 
 Divide fc, ce, da, db, ag, bh, each into the same number of equal parts, 
 for example, each into four. Join the corresponding points of division in the 
 lines ab and fe. Draw lines from the points 1 , 2, 3, in g a to c, intersecting 
 the corresponding lines drawn from the points in f e to those in a b, in the 
 points a, b, c, d, proceed with the other side in the same maimer ; then fixing 
 pins or nails in the points, draw the curve acb by means of a slip bent round 
 the points, and the curve thus drawn will be the true arc of a circle. 
 
 Problem XXXVII. 
 
 To describe the Curve of a Parabola , having a double Ordinate and the 
 
 Abscissa given. Figure 7. 
 
 METHOD FIRST — BY POINTS. 
 
 Let a b be the length or double ordinate. Bisect ab by a perpendicular 
 cd, meeting ab in d. Divide da, db, each into the same number of equal 
 parts, viz. four. Draw e e parallel to ab; also draw a e and b f perpendicular 
 to ab. Through the points 1,2,3, in da and db, draw as many lines 
 perpendicular to a b. Divide ae and bf into the same number of equal 
 parts, viz. four. Through the points 1,2,3, in ae and in bf, draw lines to c, 
 intersecting the perpendiculars, then draw the curve as before. 
 
 METHOD SECOND — BY TANGENTS. 
 
 Let a b be the chord, and de the height in the middle of ab. Produce 
 de to c, and make ec equal to ed. Join ca, cb. Divide ac and cb, each 
 into the same number of equal parts, and in the same order. Lines being drawn 
 through the corresponding points of division will form the curve required. 
 
 i 
 
34 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 ORTHOGRAPHICAL PROJECTION. 
 
 The most useful methods of representing objects, depend upon the 
 principles of Orthographical Representation, which, as before observed, 
 enables the designer of Cabinet Furniture to exhibit a representation of the 
 object which he conceives ; to shew how the different parts will be connected 
 together; and to foresee their just effect on the eye. By its means the 
 workman comprehends the design intended; as it furnishes him with the 
 angles of position and proportion of the parts, and thus enables him to bring 
 it into actual existence, by forming, joining, and fixing the parts immoveably 
 together. 
 
 D EFINITIONS. 
 
 1. Orthographical projection, is a method of representing objects on a 
 plane, by lines corresponding to those of the object itself, in such a manner, 
 that a straight line drawn from any point of a line in the object to meet the 
 plane perpendicularly, will fall in the corresponding line of representation ; 
 and reciprocally, a straight line drawn perpendicular to the plane from any 
 point of a line, will fall upon the line it represents in the original object. 
 
 Corol. Hence a perpendicular drawn from the extremity of a line of the 
 object to the plane of projection, will fall in the corresponding extremity of 
 the projection of that line. 
 
 2. The plane, on which the representation is made, is called the Plane of 
 Projection, and the representation is called the Projection, as well as the 
 Representation. 
 
 The straight lines, drawn from corresponding lines of the object and of 
 the plane of projection, are called the Projecting Rays. 
 
 3. When the projection is made on a horizontal plane, it is called a Plan ; 
 when on a vertical plane, not cutting the object, it is called an Elevation ; and 
 when it cuts the object, it is called a Section. 
 
AND COMPLETE DECORATOR. 
 
 35 
 
 4. A primitive or original plane, is a plane belonging to the object itself; 
 any line or point in such a plane, is called a primitive line or point, in contra- 
 distinction to its representation, which is also called the projection of that 
 line or point. 
 
 Though it would not be consistent with a practical treatise to enter into 
 any detail of the theory, yet for the use of the designer and workman, it will 
 be proper to enumerate such principles as will be easily understood, which 
 being fixed in the mind, will greatly facilitate the practice of delineation. 
 
 Propositions. 
 
 1 . Every primitive straight line is projected into a straight line. 
 
 2. Every primitive straight line, parallel to the plane of projection, is 
 projected into an equal straight line. 
 
 3. Any primitive straight line, not parallel to the plane of projection, is to 
 its projection as radius is to the cosine of inclination. 
 
 4. The parts of a divided primitive straight line, have the same ratio 
 among themselves, as the parts of the projected line have among themselves. 
 
 5. Any curve line in a plane, parallel to the plane ot projection, is 
 projected into an equal and similar line. 
 
 6. Any figure in a plane, parallel to the plane of projection, is projected 
 into an equal and similar figure. 
 
 7. A straight line, perpendicular to the plane of projection, is projected 
 into a point. 
 
 8. An angle in a plane, parallel to the plane of projection, is projected 
 into an equal angle. 
 
 9. The area of any plane figure is to the area of its projection, as radius 
 is to the cosine of inclination of the plane. 
 
 10. The projection of a circle is an ellipse, of which axis major is to the 
 axis minor, as radius to the cosine of inclination of the plane of the circle to 
 the plane of projection. 
 
 Hence the axis major is equal to the diameter of the original circle, and 
 it is parallel to the intersecting lines of the two planes. 
 
36 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 A primitive object is a point, line, or plane figure, of which the projection 
 is required. 
 
 In order to have a comprehensive idea of orthographical projection, we 
 must first shew the projections of a point, and of lines and figures situated in 
 a plane, and as the intersecting line is the only line that fixes the position of 
 the object to the projection, and its projection must always be given* when 
 the primitive object is given to find its projection. 
 
 In Plate iv. the intersecting line is understood to be given, as well as the 
 angle of inclination of the primitive plane, and the plane of projection; 
 therefore as these are constant in the data, it will not be necessary to 
 enumerate them, but merely to suppose, that their being given is understood. 
 In the six diagrams of this plate, the intersecting line is denoted by 1 1. 
 Hence, when it is said, “ Given the position of a point, or of a line,” it must 
 be understood, that the position of such a point or line is as it regards the 
 intersecting line ; and when the position of a plane figure is said to be “ given,” 
 the meaning is, that the angle which one of the sides, or of some principal 
 line, makes with the intersecting line, is given. 
 
 It is evident, that the projection of a point is in a straight line drawn 
 through the point perpendicular to the intersecting line ; for suppose that the 
 primitive plane revolves round the intersecting line as an axis, the point will 
 describe the circumference of a circle in a plane, perpendicular to the 
 intersecting line; hence, the intersection of the plane of the circle with the 
 plane of projection will be perpendicular to the intersecting line. 
 
 The seat of a point is the foot of the projecting ray of that point, viz. the 
 point where the projecting ray meets the plane of projection. 
 
 * The Author of this treatise was the first person in this country who introduced the science of 
 orthographical projection in general, and that by means of an intersecting line. He first discovered 
 the principle in 1796, and prosecuted this method with unremitted vigour, as was well known to 
 that able and ingenious engraver, Mr. Wilson Lowry, who recommended him shortly after to 
 Longman and Co. to write the article “ Projection” for Rees’ Cyclopaedia, which was published 
 in the year 1813, see plates 1 and 2. 
 
Tig . 3. 
 
 Fig . 6 
 E 
 
 Published by H Fisher Son & C° Cseton London June 24,1826 
 
 C 1 rtiuHjraplvicdi Projection . 
 
 B 
 
 Fuf . 2 . 
 
 M~d.u>7ic?wLscn del. 
 
 \liwcnizd by 2 Ihcholsor^ . 
 
AND COMPLETE DECORATOR. 
 
 37 
 
 Angle of projection, of a plane figure, is the angle of inclination of the 
 primitive plane and the plane of projection. 
 
 The straight line, which is drawn through a primitive point, perpendicular 
 to the intersecting line, is called the perpendicular locus of that point. 
 
 A straight line in the plane of projection, parallel to the intersecting line, 
 which contains the seat of a point, is called the parallel locus of that point. 
 
 In order to preserve perspicuity in referring to the following diagrams, 
 when the primitive object is given in position and magnitude, to find the 
 projection; for the sake of conveniency, the primitive plan is supposed to be 
 one side of the intersecting line, and in the same plane with the plane of 
 projection, which is situate on the other side. Hence, the angle of inclination 
 of the primitive plane, and the plane of projection, must be supposed to be 
 acute. By this means the confusion of lines is avoided, which must 
 necessarily happen when the plane of projection is covered by the piimitive 
 plane, and the object and its representation are completely separated. 
 
 When the separation of delineation is completed, the primitive object is 
 obliterated, being supposed to be drawn with pencil. 
 
 The shortest distances of a primitive point, and its projection, are to each 
 other, respectively, as radius is to the cosine of the angle of projection. 
 
 PLATE IV. 
 
 Problem I. 
 
 To find the Projection of a Point b. 
 
 Draw b b, Figure 1, perpendicular to the intersecting line n, crossing 11 
 at e. At any point e, in u, draw e n perpendicular to n. Make the angle 
 n e h equal to the angle of the inclination of the planes. Make e b equal to 
 es, Draw bb parallel to ii, and the point b is the projection of the point b. 
 
38 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 Problem II. 
 
 Two primitive Points , a and b, being given, and the Projection a of one of the 
 Points, to find the Projection of the other. 
 
 Draw a straight line Ac, to meet the intersecting line n in c, at oblique 
 angles. Join a c. Draw b d parallel to n, meeting ac in Draw de 
 perpendicular to 1 1, meeting ac in e. Lastly, draw eb parallel, and b b 
 perpendicular to n, and the point b is the projection of the primitive 
 point b. 
 
 N. B. This Problem is given as an exercise to construct the diagram, by 
 assuming an intersecting line, and two primitive points a and b ; hence the 
 figure is not exhibited in the Plate. 
 
 By this Problem we may find the projection of as many points as we 
 please, having the projection of one of them given, without liav'ng the angle 
 of the inclination oa the planes. And thus the projection of any curve line 
 may be found. 
 
 Problem III. 
 
 To find the Projection of a Rectilineal Angle. 
 
 Let abc be a rectilineal angle. Produce the sides ba, bc, to meet the 
 intersecting line n in the points i\ i 2 . Find the projection b, of the point b, 
 by Problem 1 . Draw the straight lines b» M , b i 2 , and i 1 bP will be the 
 projection of the angle required. 
 
 Problem IV. 
 
 Given the Projection f one Extremity of a Line, to find the unlimited 
 
 Projection of the Line. 
 
 Case 1 . When the primitive line is perpendicular to the intersecting line. 
 Draw a line through the given point, perpendicular to the intersecting line, 
 and the line thus drawn is the unlimited projection required. 
 
AND COMPLETE DECORATOR. 
 
 39 
 
 Case 2. When the primitive is parallel to the intersecting line. 
 
 Draw a straight line through the projection of the given extremity, 
 parallel to the intersecting line, and the line thus drawn is the unlimited 
 projection required. 
 
 Case 3. When the primitive line is inclined to the intersecting line in a 
 given angle. 
 
 Produce the primitive line at the other extremity, of which the projection 
 is not found, to meet the intersecting line. Join the point thus found to the 
 projection of the extremity given, and the line thus joined will be the 
 unlimited projection required. 
 
 Problem V. 
 
 Given the Projection of one Extremity of a Line, and the unlimited 
 Projection of the Line, to find the Determinate Length of the Projection 
 of that Line. 
 
 Through the extremity of the primitive, of which its projection is not 
 found, draw a line perpendicular to the intersecting line to meet the unlimited 
 projection of the line, and the line comprehended between the two points, of 
 which the one is now found, and the other given, is the projection of the line 
 required. 
 
 But if the unlimited projection of the line be perpendicular to the 
 intersecting line, and if the projection of one extremity be given, the projection 
 of the other extremity will be found thus. Draw two lines parallel to each 
 other, through each extremity of the primitive line, to meet the intersecting 
 line at oblique angles. Join the intersecting point of the line that passes 
 through the extremity of which the projection is given. Through the other 
 intersecting point draw a line parallel to the line joining the first intersecting 
 point, and the given point of projection, to meet the perpendicular, and the 
 part of the perpendicular comprised between this point and the given point of 
 projection, will be the projection of the line required. 
 
/ 
 
 40 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 Problem VI. 
 
 To find the Projection of a Parallelogram. 
 
 Find the projection of one of its angles. Produce each of the other two 
 opposite sides of the figure to be projected to meet the intersecting line. 
 Draw lines from each of these two points in the intersecting line, parallel to 
 the lines which contain the projected angle, to meet each of these lines, and 
 the parallelogram, formed in the plane of projection, will be the projection of 
 the parallelogram required. 
 
 Example . — Let it be required to find the projection of the parallelogram 
 
 ABC D. 
 
 Find b the projection of b, by Problem 1. Produce ba and bc to meet 
 ii in i l i 2 , and join b* 1 , hi 2 . Produce ad and cd to meet 1 1 in i 3 , t 4 . 
 D raw i 3 a parallel to hi 2 , and « 4 c parallel to b«’, and the parallelogram ahcd 
 will be the projection of the parallelogram abcd. This example is an 
 illustration of Problems 1, 2, and 5. 
 
 Hence, if the projection of an angle be given, the projection of any point 
 in either of the sides, containing that angle, may be found. Thus, let 1 b i 2 
 be the projection of the angle a bc, the projection of the point A maybe 
 found; for, draw a* 3 , parallel to bc, meeting 1 1 in i 3 . Draw i 3 a parallel to 
 hi 2 , meeting bs 1 in a, and a is the projection of a. 
 
 Problem VII. 
 
 To find the Representation of any Rectilineal Figure whatever. 
 
 Find the projection of the vertex of one of the angles by Problem 1. 
 Draw the diagonals of the primitive figure, from the vertex of which the 
 projection is found to meet the intersecting line. Draw lines from each point 
 so found in the intersecting line, to the point found, which is the projection 
 of the vertex, and the lines thus drawn are the corresponding diagonals of the 
 projection. Draw lines through each of the vertices of the primitive figure, 
 except that of which the projection is found, perpendicular to the intersecting 
 
AND COMPLETE DECORATOR. 
 
 41 
 
 line, to meet the corresponding lines. Join the unconnected points thus found 
 in the same order as in the sides of the primitive figure, and the resulting 
 figure will be the projection of the rectilineal figure or polygon required. 
 
 Example 1. Figure 2. Let the primitive figure abcd be a pentagon. 
 Find the projection i l a i 2 of the angle eab, by Problem 3. Draw the 
 diagonals a i 3 , aj 4 , to meet the intersecting line in i 3 , t 4 . Draw ai 3 , ai 4 , and 
 these lines are the corresponding diagonals of the figure to be projected to the 
 diagonals a i 3 , a* 4 of the original figure. Draw b&, c c, orf, Ee perpendicular 
 to ii, meeting ai 2 , ai 3 , ai 4 , ai 1 in the points b, c, d, e, and lastly, join be, cd, de, 
 and the pentagon a b c d e is the projection of the pentagon abcde. 
 
 Example 2. Let the primitive figure be that of a hexagon abcdef, fig. 3. 
 Find a the projection of a. Produce a b, a c, a d, a e, a f to meet 1 1 in i 2 , i 5 , i\ 
 i 3 , i 1 . Draw the lines ai 2 , ai 5 , ai 4 , ai 3 , ai find the points b, c, d,e,f as in 
 Example 1. Join be, cd, de, ef, and a be def will be the projection of the 
 primitive figure abcdef. 
 
 remark. 
 
 In the examples here given, if the primitive plane be revolved upon the 
 intersecting line, until it makes an angle equal to the given angle contained by 
 the two planes ; then lines drawn through the vertex of every angle of the 
 primitive figure, perpendicular to the plane of projection, will fall upon the 
 corresponding vertices of the projected figure. Thus suppose, in figure 1, 
 that the angle b e n is made to revolve on the side n e, until its plane become 
 perpendicular to the plane Pbi 2 ; and suppose then, that the primitive plane 
 i l Bi 2 , revolve on the intersecting line, the lines joining the corresponding 
 points a, a; b ,b ; c, c ; d ,d will be perpendicular to the plane of projection. 
 
 Problem VII. 
 
 To find the Projection of a Circle. 
 
 Find the projection of the two diameters, of which one is parallel, and the 
 other perpendicular, to the intersecting line, by Problem V. Upon the two 
 
 L 
 
42 
 
 BINET-MAKER, UPHOLSTERER, 
 
 given lines thus found as axes, describe an ellipse, and it will be the 
 representation of the circle required. 
 
 EXAMPLE. 
 
 Through the centre e, figure 4, draw Bb perpendicular to the intersecting 
 line, crossing it in f. Make the angle nfb , as in Problem 1, equal to the 
 angle of inclination of the two planes. On the side fb , of the angle, make fe 
 equal to f e, and fb equal to f b. Draw ec and 6 b parallel to the intersecting 
 line, and let ec intersect Bb in e, and 6b in b. IneF, make ed equal to eb. 
 Draw a a parallel, b b meeting ec in a. Make ec equal to ea; then ac is the 
 projection of the diameter, which is parallel, and bd the projection of the 
 diameter, which is perpendicular to the intersecting line. Upon ac and bd, 
 as axes, describe an ellipse. 
 
 Problem VIII. 
 
 Knowing only the Angle of Projection, to find the Ratio between a primitive 
 straight Line and its Projection , when the primitive Line is perpendicular 
 to the primitive Plane. 
 
 Let nfb, Figure 4, be the angle of projection. In fb, one of the sides 
 of the angle, take any point h, and draw hi perpendicular to the other side, 
 meeting it in i; then any primitive line in the primitive plane is to its 
 projection, when the said primitive line is perpendicular to the intersecting 
 line, as fh to hi, that is, as radius is to the cosine of the angle of the 
 inclination of the planes. 
 
 But if fh be the length of any primitive line, perpendicular to the 
 primitive plane, hi will be the length of its projection. 
 
 Hence, the ratio of the projections of the two diameters of a circle, one of 
 which is parallel to the intersecting line, and the other perpendicular to the 
 primitive plane, are to one another as radius to the sine of the angle of 
 projection. 
 
AND COMPLETE DECORATOR. 
 
 43 
 
 PROJECTION OF SOLIDS. 
 
 In case of rectangular prisms, and even in parallelepipeds, one intersecting 
 line will be sufficient to find the projection of the entire solid; but in 
 projecting solids, of which its solid angles have more than three faces 
 terminating in one vertex, it will be necessary to find an intersecting line for 
 each face ; this, however, we shall not attempt in this place, as such objects 
 are seldom to be met with in the construction of Cabinet Furniture.* 
 
 Problem IX. 
 
 To find the Representation of a Cube, one of its Faces being given in Position 
 
 and JSIagnitude. 
 
 Let the square abcd, figure 5, be the given face, which we shall suppose 
 to be the top of the solid in the primitive plane, the body being underneath. 
 
 Find a bed, the projection of the face abcd. Draw the straight lines 
 dg, ae, bf perpendicular to the intersecting line. Make ae equal to the sine 
 of the angle of projection, to a radius equal to the length of one of the 
 edges of the cube. Draw eg- parallel to a d, and ef parallel to a b, and 
 bedgef is the projection required. 
 
 OR THUS, 
 
 Having found a bed the projection of the face abcd, which is given in 
 the primitive plane, let eva be the angle of inclination of the two planes ; and 
 let Fa be equal to fa. Draw «e perpendicular to f a, and make a e equal to 
 the length of one of the edges. Complete the parallelograms a dge, a bfe, 
 and we shall have the solid as before. 
 
 In the same manner the projection of the prism, figure 6, of which one of 
 the ends is abcd, will be found. 
 
 * Those who are desirous of obtaining a knowledge of the projection of the regular solids, will 
 rind ample instruction in Rees’ Cyclopaedia, under the article “Projection,” see Plate I. by the 
 author of this work. 
 
44 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 PLATE V. 
 
 Problem X. 
 
 Given the Projections da, db, dc, Figure 1. of the three Edges of the solid 
 Angle of a rectangular Prism, and the intersecting Point a, of one of the 
 Edges represented by d a, to find the intersecting Lines of the three Faces. 
 Draw a b and a c respectively perpendicular to dc and d b, meeting d b 
 and dc in b and c. Join cb, and ab, bc, ca are the three intersecting lines 
 of the faces of the solid angle. 
 
 REMARK. 
 
 If the point a be not an intersecting point, the plane of the triangle abc 
 will be parallel to the plane of projection, and the three lines ab, bc, ca will 
 be each parallel to the plane of projection. 
 
 Problem XI. 
 
 Given the three intersecting Lines ab, bc, ca, Figure 1, of the Faces of a 
 solid Angle of a rectangular Prism , to find the Projections of the 
 three Edges. 
 
 From the vertices a, b, c, of the triangle abc, formed by the three 
 intersecting lines, draw ad, bd, cd respectively perpendicular to the sides 
 bc, ca, ab, and ad, bd, cd are the projections of the three edges, as 
 required. 
 
 Problem XII. 
 
 Given the three intersecting Lines ab, bc, ca, Figure 2, of the Faces of a 
 solid Angle of a rectangular Prism , to find the Distance of the Vertex 
 of the solid Angle from the Plane of Projection. 
 
 Find ad, bd, cd, the projections of the three edges. Produce any one of 
 them, as a d, to meet the opposite side in h. On a h, as a diameter, describe 
 the semicircle a gh. Draw vg perpendicular to Ah, and ng is the distance 
 of the angle required. 
 
AND COMPLETE DECORATOR. 
 
 45 
 
 Problem XIII. 
 
 Given the three Projections of the Edges of a solid Angle of a rectangular 
 Prism, and the intersecting Point a, Figure 2, of one of them, to find the 
 Distance of the Vertex from the Plane of Projection. 
 
 Find the three intersecting lines ab,bc,ca, by Problem I. Produce ad 
 to meet the opposite side bc in h. On a h, as a diameter, describe the 
 semicircle a gh. Draw Dg perpendicular to a h, and Dg will be the 
 
 distance required. 
 
 To explain this principle to workmen, to an artisan, or to an amateur : 
 let ad, bd, cd be the projections of three of the edges of a solid angle of a 
 rectangular prism, and let ab, bc, ca be the three intersecting lines of the 
 faces. On each of the three sides ab, bc, ca, as diameters, describe 
 respectively the three semicircles a/b, Aec, eg a. From the point d draw 
 v>f De, Dg respectively perpendicular to ab, bc, ca. Draw AffB ; Be, ec; 
 eg, g a. Then the triangle a/b, being revolved on the line ab, the triangle 
 Bee revolved on the line bc, and the triangle eg - a revolved on the line ca, 
 as axes, until the three vertices f, e,g coincide, it is evident that the solid 
 angle of a rectangular prism will be formed by the three right angles, and 
 that the point d will be the projection of its vertex. This may be done 
 with a stiff piece of paper or pasteboard, by drawing a point, not quite 
 sharp, along the lines ab, bc, cd, with a moderate degree of pressure, so 
 as not to cut the paper or pasteboard, in order to make the three triangles 
 turn more freely upon the said three lines. By uniting the three primitive 
 planes in this manner, it is obvious that any one of them, and its projection, 
 are both in a plane perpendicular to the plane of projection; and it is also 
 obvious, that if the three intersecting lines be equal in length, the three 
 projections of the edges must be also equal in length, and will divide the 
 circumference of a circle, described from the projection d of the vertex, into 
 three equal parts, see figure 4, so that each of the angles must be 120°, 
 therefore, when this happens to be the case, the projections of the three 
 
46 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 edges of every solid angle of a rectangular prism must either be 120° or 60°; 
 that is, they will either be, as in figure 5, where each is 120°, or, in figure 6 , 
 where the two lines a b and a c make 60° each, with the middle line a d. 
 
 This case has been most absurdly called isometrical perspective. It has 
 nothing to do with perspective; since no point, however distant, can ever 
 project the object on a plane, by lines radiating from that point, and passing 
 through the lines of the object, to meet corresponding lines in the plane of 
 projection; yet the eye is not only talked of, but a line of sight also. The 
 question anay be fairly asked, In what point of the line of sight must the eye 
 be placed in order to project an object isometrically? If the eye be placed 
 in any point of the diagonal of a cube, and if the plane of projection be 
 perpendicular to this diagonal, the edges of the solid angle will certainly be 
 projected, so as to make each of the angles they contain 120°; but this equa- 
 lity will not obtain in more than one of the solid angles of a parallelopiped, this 
 being the form in which framed work and machinery are usually arranged. 
 
 Projection is a general name, that not only applies to the representation 
 of objects formed by perpendiculars from the object to the plane of projection, 
 but also to radiating lines passing through the object to the plane of projection, 
 or radiating lines intercepted by the plane of projection. When a new science 
 is invented, it must have a new name. New names to inventions, previously 
 known by other names, convey an erroneous idea of the thing intended. And 
 as this case of orthographical projection is founded upon the common 
 principles of the science, and is the same in practice as w hen the three lines 
 w hich form the projections of the angles of a cube are unequal, and since it 
 offers nothing in point of facility to the delineator, except that of taking the 
 measures from one scale, it is not a separate species of projection, but a 
 particular case of orthographical representation, and therefore it is not 
 entitled to any name distinct from that to which it belongs. 
 
 Ihere wall, however, be no impropriety in applying a single word of 
 qualification, in order to save the explanation of a circumstance which would 
 otherwise require several words ; we shall, therefore, make use of the word 
 
AND COMPLETE DECORATOR. 
 
 47 
 
 Isometrical adjectively, in order to accompany the substantive Line, when this 
 line is one of the three edges of a solid angle of a cube, or of a rectangular 
 prism, which are equally inclined to the plane of projection. 
 
 Problem XIV. 
 
 To find the Inclination of any One of the three Edges to the Plane of 
 
 Projection. 
 
 Draw any straight line ab, figure 8, and in ab set off three equal parts of 
 any length from a to b, marking the divisions c and /. From the middle 
 point e , of a b, with the radius a, or eB , describe the semicircle adb. Draw' 
 cd perpendicular to ab, and join bd; then abd will be the angle of 
 inclination required.* 
 
 OR THUS. 
 
 In any straight line bc, figure 8, set four equal parts of any length each, 
 and let ce be a fourth of cb. Draw cd perpendicular to c b from e , with the 
 distance es, describe an arc at d ; join bd, and cbd is the angle required. 
 
 * In order to shew that ac, figure 7, is one-third of the perpendicular ab, we see that the two 
 right-angled triangles eac and f ac arc equal to each other, for ef is bisected in a, and the angles 
 aec and afc opposite the common perpendicular ac, are each 30°, and each of the angles eca,fca 
 are therefore 60°; therefore the two triangles ace and acf might be formed into one equilateral 
 triangle, by uniting the legs ae and af, so that the point e might coincide withy, and a with a, 
 and placing the right angles adjacent in the same plane, then a c would be half one of the sides, viz. 
 equal to the half of ce or cf; that because ce, cf, and cb are equal, ac would be the half of cb, 
 and therefore one-third of a b. Hence ac, cb, and a b are respectively as 1, 2, 3. 
 
 This is the foundation of the above construction. The method of finding the angle, which 
 each of the edges makes with the plane of projection, is preferable to any other method hitherto 
 given. This angle is the principal thing required by the draughtsman, he has no need of number; 
 however, as the numerical ratios are very easily derived from the construction, we shall shew how 
 this may be done ; for this purpose, cd is a mean proportional between ca and cb; therefore, cd 
 rr V2, ac being 1, and c62; then, with reference to figure 8, we have bc=2 and CDn^/2; 
 hence BDr^B^* x c d 5 ) = V(4 + 2)=6 ; therefore CD, cb, bd are as V 2, 2, >/6, and 
 multiplying these by the >/2, they become 1, 2, \/3. 
 
 As to the angle of inclination of the primitive line, and the plane of projection, it is nearly 
 equal to 35° 16°, for considering radius equal 1, we shall have b d to D c, as radius to the sine of 
 the angle of projection ; that is, since bd = \/6, and DC— <\/2, \/6 : \/2 ; : 1 : sine of the angle of 
 projection — V-j = •5773+ or more nearly ’5774 — which answers to 35° 10'. 
 
48 
 
 C A BI NET-MAKER, UPHOLSTERER, 
 
 A REASON FOR THE TWO PRECEDING METHODS. 
 
 These methods are evident by considering figure 7, of which the equilateral 
 triangle bef may be supposed to be formed, by the intersections of the three 
 planes which constitute the solid angle. Since any edge of a solid angle of a 
 rectangular prism, is always perpendicular to the plane passing through the 
 other two; therefore, the primitive line, represented by be, is perpendicular 
 to the primitive plane represented by ecf; and consequently every section of 
 the solid passing through the edge, of which be is the projection, will form a 
 right-angled triangle, of which the primitive vertex of the right angle is 
 represented by c ; and in the present case of equal inclinations of the edges 
 to the plane of projection, if the plane of section be considered as perpen- 
 dicular to the said plane of projection, the perpendicular ba, from any angle 
 b to the opposite side ef must be the base of the triangle. 
 
 Problem XV. 
 
 Given any Isometrical Length, to find the Primitive Length. Find abd, 
 Figure 9, the Angle of Inclination. 
 
 In one of the sides ba, containing the angle abd, make Be equal to the 
 length of the projection of the line. Draw ef perpendicular to ba, meeting 
 b d, or bd produced iny.* then By is the length of the primitive line. 
 
 Problem XVL 
 
 Given the Isometrical Length oj the side oj a Rhombus, representing a 
 
 Square, to describe the Rhombus. 
 
 From any point b, in the straight line bd, figure 10, with a radius equal to 
 the length of the line given, describe the arc adc. From the point d, with 
 the same radius, describe an arc on each side of the line b d, each intersecting 
 the former arc in the points a and c. Join da and dc; then abcd is the 
 rhombus required. 
 
 .9 
 
AND COMPLETE DECORATOR. 49 
 
 Problem XVII. 
 
 The Rhombus which represents a Square being given , to find the Axis of 
 the Ellipse which represents the Circle inscribed in that Square. 
 
 Find the angle cbd, figure 11, No. 1, of inclination of the plane of 
 projection, and the original plane. Let fghi, No. 2, be the given rhombus 
 which represents the circumscribing square. In bc, No. 1. make b k equal 
 to the half of any one of the four sides; that is, because the rhombus 
 consists of two equilateral triangles, equal to the half of the shorter diagonal, 
 viz. equal to ei or eG, No. 2. Draw hi, No. 1, perpendicular to bc, meeting 
 bd in /. Then bI is the semi-axis major, and Ik the semi-axis minor of the 
 ellipse which represents the primitive circle inscribed in the square, 
 represented by the rhombus fghi. 
 
 Having thus shewn the modes of constructing the projections of points, 
 lines, surfaces, and solids, from their primitives, and the primitive figures from 
 their projections ; we shall now proceed with practical examples, which will 
 at once shew the utility, the superiority, and the generality of our principles 
 above those limited descriptions which may be elsewhere found. 
 
 PLATE VI. 
 
 Figure 1, exhibits the projection and development of a cube by the three 
 right-angled triangles. The squares of the cube, which constitute the solid 
 angle at the summit, are so placed that one of the right angles of each square 
 may coincide with the right angle of each of the triangles ; if these right- 
 angled triangles be revolved on their hypothenuses ab , bc, ca, until the 
 vertices d, e,g coincide, three sides to be projected of the solid, would then 
 be brought into their real situation. Then the triangle a bc, of which the 
 sides are the three intersecting lines, being equilateral, the figure airlpn, of 
 the contour of the projection, would be a regular hexagon, divided internally 
 by the projections of the three edges of the solid angle into three equal 
 rhombuses. In this case, all the edges are projected into equal straight 
 
 N 
 
50 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 lines, which are equal to one another, and any primitive line is to its projection 
 as V3 is to V2. 
 
 Figure 2, exhibits the development of the sides of a rectangular prism 
 upon the three intersecting lines of their planes, supposed to be extended to 
 meet the plane of projection in the straight lines a b, bh , ha , in the same 
 manner as has been shewn for the cube, figure 1, except that the lengths are 
 not projected into equal measures. In this example the ends are squares, 
 hence the sides are equal rectangles. Because the plane angles, which form 
 the faces of the solid angle, are right angles, they are all contained in a 
 semicircle. Therefore having the three intersecting lines given, in order to 
 prepare this development, we must first find the projection f of the vertex of 
 the solid angle, by drawing lines from each of the three angles perpendicularly 
 to their opposite sides. On each of the said three sides, as a diameter, 
 describe a semicircle. Draw a straight line from f perpendicular to each 
 diameter, to meet the semicircular arc ; from this point of meeting, which is 
 the vertex, draw a line to the extremity of each diameter. Place the end of 
 the solid, which is a square in the right angle acb, and the side adjacent to 
 this angle in the next vertex, so that the two edges of equal length may 
 coincide, then the three sides being turned up as before, will exhibit the three 
 sides of the primitive solid which are to be projected. 
 
 EXAMPLE i. 
 
 Find the representation of a prism, with a mortise cut from the upper end, 
 according to given dimensions, by means of isometrical lines. 
 
 Make a scale p q, of feet and inches, to any convenient size, as exhibited 
 on the right-hand side of Plate VI, between figures 5 and 8, and let this scale 
 be that from which the measures of the parts of the several diagrams of this 
 Plate are taken. 
 
 Draw a straight line a b, figure 3, to represent one of the arrises of the 
 prism, in the direction of its height. From a , as a centre, with any radius, 
 
AND COMPLETE DECORATOR. 
 
 51 
 
 describe an ar c pq, intersecting ab in r. From r , as a centre, with the same 
 radius ra, describe an arc cutting the former at p on one side of ab, and at 
 q on the other ; and draw ap and aq , which produce if necessary. Suppose 
 the length of the solid to be five feet nine inches, in the breadth three feet 
 one inch, and the thickness one foot seven inches. From the scale p q, take 
 five feet nine inches, and apply the extent from a to b; take three feet one 
 inch from the scale, and apply this extent in the line ap, or ap produced from 
 a to e, and take one foot seven inches from the scale, and apply this distance 
 from a to f. Having now the two sides ab, ae, complete the parallelogram 
 abce; having the two sides ab, af, complete the parallelogam abdf; and 
 having the two sides be, bd, complete the parallelogram cbdg , which will 
 complete the representation of the prism. 
 
 To exhibit the Mortise. — Suppose the mortise to be nine inches wide in 
 the middle of the thickness ; then as the thickness is one foot seven inches, 
 by deducting nine inches, there will remain ten inches for the thickness of the 
 two substances on each side of the mortise ; hence the thickness of each is 
 five inches. In the line bd make bl and dm each equal to the extent of five 
 inches from the scale; draw the lines Ir and ms parallel to the side be. 
 Suppose now the length of the mortise to be in the middle of the breadth of 
 the prism, and to be one foot five inches, now the breadth of the prism being 
 three feet one inch, deducting one foot five inches, there will remain one foot 
 eight inches, which allows ten inches of substance at each end of the mortise ; 
 therefore, in the edge be make bh and ci , each ten inches. Draw lit and is 
 parallel to bd or eg, and the parallelogram rstu will be the upper end of the- 
 mortise. Draw sv parallel to ba, meeting ru, and the cavity will be 
 exhibited by the two sides which join in sv. 
 
 Having gone through all the particulars in the construction of figure 3, a 
 complete idea of that of every other prismatic solid, or any solid composed of 
 prisms, may be easily conceived : we shall therefore only give the dimensions, 
 and leave the construction as an exercise to the learner. 
 
52 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 EXAMPLE II. 
 
 Construct a cross, having the transverse parts of a prismatic form, the one 
 passing through a mortise in the other at right angles. The piece which 
 contains the mortise to stand upon a square base, each side of the base to be 
 seventeen inches ; the length six feet seven inches ; the distance of the arms 
 from the base to be three feet three inches ; the two arms of the cross to 
 project equally on each side, and to be of the following dimensions, viz. the 
 length of both four feet six inches ; the breadth one foot two inches ; and the 
 thickness in the breadth of the solid, which contains the mortise, to be eight 
 inches, and the arm to be fixed in the middle of the breadth which contains 
 the mortise. Also, to represent a mortise in the other side of the same piece 
 which contains the mortise already spoken of, at the same height as the 
 former, and in the middle of the breadth, so that the breadth of this mortise 
 may be equal to the breadth of the other, through which the transverse piece 
 which forms the two arms passes, proceed as in Example I. 
 
 EXAMPLE III. 
 
 Represents a cylinder, five feet three inches in length, and three feet in 
 diameter. 
 
 Let cad, figure 8, be the angle of projection. In ac , make ad equal to 
 three feet of the scale pq. Draw be perpendicular to ac, meeting ad in e; 
 then ae is the primitive measure of ab. In figure 8, draw ac, bd, at right 
 angles, intersecting each other in k. Make kA, kc each equal to the half of 
 ae, figure 8, and make k b, k d, figure 5, each equal to the half of be, figure 8. 
 Upon the lines a c and b d, as axes, describe the ellipse abcd. Draw k l 
 perpendicular to a c. Make k l equal to five feet three inches. Through l 
 draw the lines eg, fh parallel to ac and b d. Make Ze, Z g each equal to 
 k\ or kc, and i h, i f each equal to k b or k d, and describe the ellipse efgh. 
 Draw the tangents ae and cg, which will complete the representation of the 
 solid. 
 
AND COMPLETE DECORATOR. 
 
 53 
 
 EXAMPLE IV. 
 
 Represent a right cone, so that the diameter of the base may be three feet, 
 and the altitude six feet. — Find the base and height as in the cylinder. 
 Draw lines from the vertex to touch the ellipse representing the base. 
 
 example v. 
 
 Represent a table, with a rectangular top and rectangular legs, framed to 
 bearing rails ; the top of the table being six feet in length, two feet ten inches 
 in breadth, and two inches in thickness ; the legs being three inches square, 
 two feet ten inches in height under the table, making the entire height three 
 feet ; the bearing rails being six inches high, framed flush, with the legs on 
 the outside ; the outer surface of each side of the frame being four inches 
 within each side of the table. — 
 
 We may commence the representation, figure 7, by making the angles 
 ball , cali } as in Example I ; or since the angles hah, hac are constantly the 
 same, we may draw two contiguous angles of 60° each, any where on the 
 paper out of the way, then draw the lines a b, ah, ac parallel to the sides of 
 these angles : then, as we would wish the length to be represented on the 
 left or on the right hand, we must place the length on ab or ac accordingly; 
 the one being fixed, will determine the other. Let us suppose the length to 
 be on the right, therefore make ac equal to six feet, and ab equal to two feet 
 ten inches, and ah three feet. With the two sides ah and ac, complete the 
 parallelogram hacl ; with the two sides ah, ab, complete the parallelogram 
 habi ; and with the two sides hi, hi, complete the parallelogram ihlk , 
 which will form the upper surface of the table. In the line ac make af, cp, 
 each equal to four inches, and fs, pq, each equal to three inches. In the line 
 ab make ad, br, each equal to four inches, and de, rs, each equal to three 
 inches. Through the points d,e,r,s, draw lines parallel to ac, and through 
 the points fg,p, q, draw lines parallel to the side ab. The four parallelo- 
 grams formed by the intersections of the lines parallel to ac, ab, are the 
 
 o 
 
54 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 lower ends of the feet which stand on the floor. From the three angular 
 points of these parallelograms nearest the bottom, draw lines parallel to ah, 
 which will form the vertical faces of the legs. Make mn equal to two feet 
 ten inches, and draw two lines parallel to a b and ac, to meet the two most 
 remote lines of the legs, and these two lines thus drawn will represent the 
 under lines of the bearing rails, and thus the representation of the table will 
 be complete. 
 
 On the top of the table we have represented a cup or bason, as also a box 
 at the farther end ; the principles of our projection, by means of intersecting 
 lines, will shew how this box may be drawn, according to given dimensions 
 and in a given position. 
 
 Let the side of the box be placed accidentally upon the lineMC, the corner 
 being upon u. We shall suppose its form that of a rectangular prism as usual. 
 Draw mu perpendicular to the horizontal planes, viz. parallel to the arrises of 
 the legs of the table, and draw i l perpendicular to mu, intersecting each other 
 in x. Draw mi parallel to the edge hi of the top of the table, and make xv 
 equal to xi. Produce uv to l, and join lu. Draw u z perpendicular to lu, 
 meeting il in z, and join uz. Suppose now that the length of this box is 
 seventeen inches, the breadth nine inches, and the height four inches. Take 
 seventeen inches, nine inches, and four inches from the scale pq, and in figure 
 8, set off seventeen inches in the line ac, from a to f, draw fg perpendicular 
 to ac, meeting ad in g ; also, in ac set off nine inches from a to h, and draw h i 
 perpendicular to ac, meeting a d in i. Transfer the distances ai, ag to the lines 
 vz, ul, from u to t and from u to v. Draw Tt and \ v perpendicular to il, 
 meeting the lines u z and ml in the points i and v. In mu make uw equal the 
 height hm, from the scale pq. Having thus obtained the lengths of the three 
 ridge lines, the outline of the solid will be completed by drawing parallel lines. 
 
 REMARK. 
 
 This Example will shew the great extent and facility of our principles, 
 that they are capable of representing lines and planes at any given angles with 
 each other, and solids of all dimensions and proportions. The limited and 
 
AND COMPLETE DECORATOR 
 
 55 
 
 common place observations which were published for the use of practical men 
 in 1825, and which are pretended to be new, apply only to lines in the 
 direction of the three edges of the solid angles. They will not enable the 
 student to find the proportions of the primitive lines, from their projections in 
 any other direction. Our principles, on the contrary, not only shew how this is 
 to be done, but direct us how to find the angle which is made, when one line is 
 drawn to another lin5 given in position, in order to represent a perpendicular, or 
 to represent any given angle whatever ; as also, how to draw a line parallel to a 
 given line in projection, so that the distance of the two lines may represent a 
 given distance. These positions are not only effected in the three planes 
 which represent the solid angle of a rectangular prism, but in every other 
 plane of which the position is given to any one of these three faces. Thus, if 
 we know the inclinations of all the planes of a solid, however complex, and 
 the figures of the several faces, we shall be enabled to represent that solid ; 
 and, on the contrary, if we have an orthographical representation, we may 
 construct a solid similar to the original or primitive solid. 
 
 We shall not proceed any farther in the projection of objects by equal 
 measures, since, though useful in the construction of mathematical diagrams, 
 they form the angles which represent those of the solid much too acute in 
 order to exhibit an agreeable representation, and since in point of facility 
 nothing is gained by this case of orthographical projection. We shall, there- 
 fore, proceed with a general method, which does not require to be confined to 
 any particular angles, as they may be assumed at pleasure. A scale may be 
 readily found to each of the three directions, keeping still in view the 
 projection of a solid angle, of which any one of the faces is perpendicular to 
 the line of concourse of the other two. 
 
 Though the principles of these general methods have been long given by 
 the author, he has never, till now, had an opportunity of exemplifying them in 
 practice, except those two examples published in 1823, mentioned in the second 
 note of the Introduction; he therefore trusts, that the superiority of his original 
 methods to those more recently published, will be evident by a careful perusal. 
 
56 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 If two angles of a projection of a right-angled solid angle be equal, and if 
 any distance be taken upon the line between these equal angles, from the 
 point where the three lines meet, and if upon this distance, as a diagonal, a 
 rhombus be described, then the following proposition will obtain. 
 
 Either semi-axis of the ellipse, inscribed in the rhombus, is the hypothenuse 
 of a right-angled triangle, of which each of the two legs is one-fourth of the 
 diagonal, of which this semi-axis is a part. * 
 
 This proposition is of the greatest use under the circumstance now men- 
 tioned ; for the rhombus being given, the two axes will be found instantly, 
 without knowing the angle of inclination of the line which passes through the 
 centre perpendicular to the plane of the circle, by drawing a right angle, and 
 setting a fourth (say of the shorter diagonal) from the vertex upon each side 
 of the angle, and transferring the distance between the two points of extension 
 from the centre, upon the line intended for the minor axis, and the distance 
 of the two remotest points will be the minor axis. 
 
 The following is another proposition, which will be found to be extremely 
 useful. 
 
 The sine and cosine of the angle of inclination of any face of a right- 
 angled solid angle to the plane of projection, are respectively equal to the 
 cosine and sine of the line of concourse of the other two faces, to the said 
 plane, and to the same radius, in both cases. 
 
 The number of propositions or theorems might have been extended, but as 
 our work is designed for the use of practical men, and not for the speculative 
 geometer, what are given will be sufficient to answer the purpose intended. 
 
 Before we commence our examples of representing solids in the form of 
 rectangular prisms, by lines in given positions, the following observation will 
 be found to be of great use to the student, viz. if the projection of one of the 
 plane angles of a solid angle of a rectangular prism, and the intersecting line 
 of that plane, be given, the intersecting lines of the other two planes may be 
 readily found. Thus in figure 2, suppose we have the angle afb , which 
 represents the projection of the right angle acb, and the intersecting line ab 
 
AND COMPLETE DECORATOR. 
 
 57 
 
 given ; the intersecting lines of the other two faces will be found thus : draw 
 bh perpendicular to af and ah perpendicular to bf then the intersecting 
 lines of the other two planes are ah and bh. Or, if we have the intersecting 
 line of one of the faces of a solid angle of a rectangular prism, and the pro- 
 tection of that face, we may find the projection of the other two faces as 
 before ; for since four of the edges of the solid are always perpendicular to 
 two opposite planes, the projections of these edges must also be perpen- 
 dicular to the intersecting line of either of these planes ; but the length of 
 a straight line, perpendicular to a plane, is to the length of its projection as 
 radius is to the sine of the angle of projection ; therefore having the repre- 
 sentation of one face of a rectangular prism, and the intersecting line, we 
 have only to draw lines through the upper three points of each angle perpen- 
 dicular to the intersecting line, then set off on the perpendicular which passes 
 through the vertex of the angle at the summit, a length equal to the sine of 
 the angle of projection, the primitive length being taken as radius. Or, 
 suppose we assume the intersecting line ah , and the angle afb as the 
 projection of one of the right angles at the summit, we may find the represen- 
 tation of the whole solid without using the angle of projection. Thus on ab, 
 figure 2, as a diameter, describe the semicircle acb. Draw fe perpendicular 
 to ab, and join ac and be. In ca and cb make co and cm equal to the sides 
 of the first face. Draw og and mn, meeting/a and fb in g and n. Find 
 the intersecting ah as before. On ah describe the semicircle avh. Draw' fv 
 perpendicular to ah. Join vh. In vh make vtv equal to the height of the 
 solid. Join fh. Draw wl perpendicular to ah, meeting fh in l: by com- 
 pleting the parallelograms we have the representation required. 
 
58 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 Examples in which a different Scale is used for each of the three Edges . 
 
 PLATE VII. 
 
 Example 1 , figure 1 . Represent a writing desk, with drawers, &c. 
 
 Draw ax as a director for the representation of the perpendicular lines, 
 and draw a b, ac, making any angles with ad that may be agreeable to the 
 eye, for the directors of the lines which represent the horizontal lines of the 
 object. To find the scales agreeable to these three assumed lines, on any 
 convenient part of the paper, as at figure 2 : draw qa, qb, qc respectively 
 parallel to a b, ac, ax, figure 1. Then proceed in figure 2, as follows. From 
 any point a, in qa, draw ab perpendicular to qc, and ac perpendicular to qb, 
 and join be. Then if the three lines qa, qb, qc represent the vertical solid 
 angle of a rectangular prism, ab, be, ca will be the intersecting lines of the 
 three faces. Produce aq to meet the opposite b c in p. On ap, as a diameter, 
 describe the semicircle adp. Draw qd perpendicular to aq; then qd is the 
 height which the apex of the cube is raised above its projection or seat q; 
 therefore draw qh perpendicular to qb, and qf perpendicular to qc. Make 
 qh and qf each equal to qd. Join ad, bh, and cf. Then will the three lines 
 ad, bh, cf be the primitive lines, of which their projections are aq, bq, cq. 
 Therefore we may assume in any one of the three lines aq, bq, cq, a scale of 
 such a length as may be convenient for the size of the drawing. Let, there- 
 fore, in the line qc, the distance qo be taken to represent twelve inches, or 
 one foot. Draw og perpendicular toe q, meeting cf in g ; then fg is the 
 length of a foot of the real object itself. In da make de equal to fg, and in 
 hb make hi also equal to fg. Draw eo perpendicular to aq, meeting it in o; 
 then qo is the length of a foot in the line qa; moreover, draw io perpendi- 
 cular to qb, meeting qb in o ; then qo is the length of the scale in the line qb. 
 In each of the three lines aq, bq, cq, divide the distance oq into twelve equal 
 parts as shewn, and repeat the distance of a foot as often as may be found 
 
AND COMPLETE DECORATOR. 
 
 09 
 
 necessary, and thus we have the three scales required in the three 
 different directions. The method of applying these scales will be evident 
 to most of our readers; but as there are some who do not comprehend 
 so readily as others, we shall accompany our reader a little farther, and 
 transfer these scales to a, b, c, parallel to the lines of which their lengths 
 are to be taken. 
 
 Suppose the length of the flap or lid to be three feet three inches, the 
 breadth two feet eight inches, and the height of the legs two feet, and the face 
 of the rails to be one inch within the edges of the flap ; the legs to be two 
 inches square. As it will not be necessary to state the dimensions of all the 
 parts, we shall now shew how the lines are to be represented as far as the 
 dimensions are given. 
 
 Take three feet three inches from the scale b, and apply it on the line ab, 
 figure 1, from a to b. Take two feet eight inches from the scale c, and apply 
 it on the line ac, from a to c. In the lines ab, ac, make ad, a f, b/i, each equal 
 to one inch. Also, make de,fg , hi, c/i, each equal to two inches. Draw 
 the lines fr, gt parallel to ab, and the lines div, eu parallel to ac ; proceed thus, 
 and form the foot of the legs rst,plq, uvw. Draw Id parallel to ax, and make 
 the height of Id equal to two feet from the scale a. Draw Dm parallel to ab, 
 and an parallel to ac. Having conducted the reader thus far in the delineation 
 of this object, we shall now leave him, as it is presumed that he is sufficiently 
 familiar with the method of proceeding to enable him to finish the drawing. 
 
 Example 2. Represent three circles perpendicular to three straight lines 
 as axes, of which every two form a right angle; the "representation of the 
 three lines being de, df, dg, figure 3, parallel to qa, qb, qc, figure 2. 
 Supposing the plane of the circle upon the axis represented by de to be one 
 foot one inch and a half from d, and the radius ten inches ; also the plane 
 of the circle upon the axis represented by df to be two feet two inches from 
 d, and the radius of this circle six inches, and the plane of the circle upon the 
 line represented, dg, to be one foot two and a half inches from d, and the 
 radius of this circle eight inches. 
 
60 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 In the line de make d a equal to one foot one inch and a half from the 
 scale b, or from that of qa, figure 2, which is parallel to B. Through a draw 
 be perpendicular to de. Make a b, a c, each equal to ten inches, taken from the 
 oblique scale o-12, No. 1, figure 2, parallel to ad. Place the foot of the compass 
 in the same scale at 10, and fix the other foot at such a distance, as, by 
 making it describe an arc of a circle, the arc will just touch the line qa; this 
 may be done instantly by trial, without drawing a perpendicular to aq, or 
 taking the length of the sine : this may also be otherwise expressed, by taking 
 the shortest distance between the point 10 and the line qa. Transfer the 
 distance thus obtained to figure 3, in the line de, from a to e, and from a to d. 
 Then the ellipse described upon the two axes be and de will represent the 
 circle required. In the same manner the ellipses at Nos. 2 and 3 may be 
 drawn. 
 
 The reader is requested to find the dimensions of a portable desk repre- 
 sented by figure 4. 
 
 The method of finding the measure of a foot is shew n in figure 2, on e&cli 
 of the three directors, in order that the original measures may be brought into 
 their real situation by turning up the three triangles of which the bases are 
 the lines qa, qb, qc. This is certainly the most evident method of finding the 
 proportions of the scales; but the following method is the most convenient. 
 
 Having drawn the three lines qa, qb, qc, figure 2, parallel to the three 
 lines ab, ad, ac, figure 1, and the sides ab, be, ca, figure 2, respectively per- 
 pendicular to the three lines qc, qa, qb, and having found the height qd from 
 the vertex of the solid angle to its reach ; proceed as follows : — 
 
 In figure 5, draw any straight line be, and draw ba perpendicular to be. 
 Make ba equal to qd, figure 2. In figure 5, on the straight line be, make 
 be, b d, b c respectively equal to qc, qa, qb, figure 2. In figure 5, join ae, 
 a d, ac. Draw a/ parallel to be. Set off a f equal to twelve inches, 
 or one foot, of the scale qc. Draw fi perpendicular to a f, meeting ae 
 in the point*. From a, with the radius a*, cut the lines Ad, ac in the 
 
 points k and l. Cut the line A^in the points g and h, by the lines kg, 
 
O t‘ t/wfl /‘dp /i ic&L /'ro/ c<‘( ion 
 
 FI cS. 
 
 r M 
 
 V-o 
 
 M ■ ■ . I*. ■ ■ ,-J *- 
 
 1 30 
 
 
 I'Yq.4-. 
 
 Fiy. 
 
 <( 
 
 D/ A 
 
 /•»/. t> . 
 
 J. 'I 
 
 >OiT & C° LOJTDOJI. 
 
 TISHEP.. 
 
AND COMPLETE DECORATOR. 
 
 61 
 
 lh perpendicular to a /,* then a / being the scale in the direction qc , 
 figure 2, ao' and a h will be the respective scales in the directions qa, qb, 
 figure 2. 
 
 We have here in figure 5, preferred the line a / to be , in order that the 
 points h, g, f might not interfere with the points c and d, as otherwise they 
 might have had some chance, or fall very near them. 
 
 This method is not only convenient in finding the lengths of the scales, 
 but also in finding the two axes of the ellipse upon any one of the three 
 directions. 
 
 Thus, suppose it were required to find the ellipse described round the, 
 point k, figure 3, as a centre, to represent a circle, so that the primitive line 
 of dg may pass through the centre of the circle as an axis. To do this in 
 the easiest manner, it would be better to have a scale of the original measures 
 as well as of the three directors, and this we have in any one of the three 
 sloping scales, figure 2. Or the same may be obtained by dividing any one 
 of the three radii a i, a k, a l, figure 5, into twelve equal parts. We may now 
 recollect in example 2, that the radius of the original circle upon the line 
 represented by dg , figure 3, was eight inches. Having drawn fg, figure 3 , 
 through k, perpendicular to dg, take eight inches from the scale aI, figure 5, 
 and in figure 3 make kf and kg equal to it; set the distance kf or kg, 
 figure 3, or eight inches in the line ea , figure 5, from e to m. Draw the sine 
 mn, and transfer mn to figure 3, from k to h, and from k to i; then h i is 
 the semi-axis minor 
 
 PLATE VIII. 
 example . 
 
 Represent a semicircular arch in a straight wall, of which the thickness 
 is four feet, the distance of the arch from the corner of the wall three feet and 
 a half, the breadth of the arch fourteen feet, the height from the bottom of the 
 w r all to the springing fifteen feet. 
 
 Q 
 
CABINET-MAKER, UPHOLSTERER, 
 
 f>2 
 
 In figure 3, draw the lines ca , cb, cd in any directions that may he agree- 
 able to the eye, or which the subject to be delineated may require, the 
 directors of the horizontal lines being ca, cb , and that of the vertical lines 
 being cd. In this example the right angles of the solid angle are respectively 
 143°, 108°, and 30°, the horizontal right angle in front being represented by 
 an angle of 143°, and each of the right angles in the two vertical planes by 
 71°, 30', which is the half of 143°. In figure 3, the angle acb is therefore 
 143°, and each of the angles acf, bcf, is 71°, 30'. To find the scale upon 
 each of the lines ca, cb, cd, we may assume ten feet upon any one of these 
 th ree lines, of such length as may be judged proper for the size of the drawing 
 to be made. Suppose, then, in the line cd, which is the director of the ver- 
 tical lines, that co is taken to represent ten feet ; we have now to find ten feet 
 in the lines ca and cb, or only in one of them, as they are equal. If the 
 reader recollects the operation of finding the scales of the writing desk, figure 
 1, Plate vn. he will be at no loss to find the scales in this example ; but if 
 he does not, he must return to page 58, where the explanation of figure 1, 
 Plate vn. is given. Let, therefore, co, in the line ca, be ten feet according 
 to co, the assumed ten feet in the line cd; the sloping lines 0-10 and 0-10 
 being their originals. 
 
 Having fixed the position of the directors of the lines that are to represent 
 the solid angles of the primitive object, and the scales in these three directions, 
 we may now proceed to the delineation of the object. 
 
 In figure 1, draw j or parallel to its director ca, figure 3, and make pq, qr 
 respectively three feet and a half, and fourteen feet from the scale ca. Draw 
 r s, figure 1, parallel to the director cb, figure 3, and make rs equal to four 
 feet from the same scale. In figure 1, draw qg, rf si parallel to the vertical 
 director cd, or rf, figure 3. In figure 1, make rf equal to fifteen feet from the 
 scale cd, figure 3. In figure 1, draw/* parallel to rs, and fg parallel to rp. 
 Bisect fg in the point e. We must now recollect that each of the three faces 
 of a solid angle of a rectangular prism is perpendicular to the line of con- 
 course of the other tw o ; and that when an ellipse is represented in that face, 
 
AND COMPLETE DECORATOR. 
 
 03 
 
 the axis minor is parallel to the line of concourse, and the axis major perpen- 
 dicular to it. Therefore, through e draw ab parallel, and cd perpendicular 
 to rs ; then the axis major will be in the line cd , and the axis minor in the 
 line ab. We have also to recollect that the axis major is to the axis minor, 
 or the semi-axis major to the semiaxis minor, as radius is to the sine of the 
 angle of the line of concourse of the two planes just mentioned to the plane 
 of projection. Now cb, figure 3, is the representation of this line of concourse; 
 but as ca is equal to cb , we must take our measures from ca instead of cb. 
 Take, therefore, seven feet of the original scale of ca , figure 3, and transfer it 
 to figure 1, in the line cd, from e to c, and from e to d. Now take the sine 
 of the angle of projection to radius seven feet; that is, vw, figure 3, and 
 transfer it in the line ab, from e to a and from e to b. Having now the two 
 axes ab and cd, in position and magnitude, we may describe the ellipse or 
 the part f dbg, which will represent the face of the arch. To represent the 
 arch in the opposite face of the wall, draw kl, mn, & c. parallel to fi; make 
 kl, mn, &c. each equal to fi; through the points i,l, n, &c. draw the cmwe 
 iln, &c. to intersect the curve fc tbg in the points. 
 
 A building, with arched apertures, requires more attention than one which 
 has only rectangular apertures ; for in the former case we are under the 
 necessity of finding the original measures ; but in the latter we only require 
 the measure from the three scales of the edges of a solid angle. As every 
 good draughtsman has such a command of hand as to describe a curve, through 
 a few points, with sufficient exactness, and perhaps would not be at the 
 trouble to describe portions of the ellipse required to represent the semicir- 
 cular arch from the axis; and as he has now the four points f c , b,g given, 
 he may readily find the fifth point thus, which will enable him to trace it with 
 great exactness. Draw ct parallel to rf, and make et equal to seven feet, 
 from the scale c d, fig. 3, and the point t will be the fifth point required. 
 
64 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 EXAMPLES FOR PRACTICE. 
 
 Figure 5, exhibits the projection of a flight of steps, and figure 6, repre- 
 sents a series of return steps. These two examples are drawn on the same 
 principles as the others, and therefore need not be explained. 
 
 Figure 7, exhibits the projection of a line, in a plane perpendicular to the 
 primitive plane, and the projection of the intersection of that perpendicular 
 plane, cut by another plane, of which its intersection and angle of inclination 
 to the primitive plane are given. 
 
 Let i l be the intersecting line of the primitive plane, and the plane of 
 projection; also, let ac be the intersection of the primitive plane, and the 
 vertical plane. At any point b, in a b, draw bc perpendicular to a b, and 
 draw cd' perpendicular to bc. Make the angle cbd' equal to the inclination 
 of the inclined plane to the primitive plane. Draw c d perpendicular to a c, 
 and produce cd to e. Make the angle cae equal to the inclination of the 
 line in the perpendicular plane, insisting upon ac. Find abc the projection 
 of abc. Draw ce perpendicular to i l. Make cd, ce equal to the sines of 
 the angle of projection respectively to radius cd, c.e ; Join ad and ae: then 
 ae is the projection of the line, and ad that of the intersection required. 
 
 This Problem is the same as finding the projection ae of a ray of light, 
 and to find the intersection ad of a vertical plane, passing along that ray {ae) 
 with a plane at the given angle cbd. 
 
 Hence, if ab represents one side of a building, and bc the other side, ab 
 and bc will represent a right angle : and because cd represents a line perpen- 
 dicular to the plane represented by abc, the line cd represents a line 
 perpendicular to the line represented by cb and by ca. Hence, the angle 
 bad represents the plane angle which forms the hypothenuse, and bac, cad 
 represent the legs of a right-angled trehedral. de represents the elevation of 
 the straight line c e above the inclined plane. Thus the use of this problem 
 in the projection of shadows is manifest. We shall therefore leave its appli- 
 cation, in the formation of certain shadows, to the reader. 
 
AND COMPLETE DECORATOR. 
 
 65 
 
 With regard to the scales or lines of measure in the three directors, the 
 measure, representing any length in the line opposite the greatest angle, is 
 greater than the length taken upon either of the other two lines representing 
 the same measure. On the contrary, the measure representing any length 
 upon the line opposite the least angle, is less than the length taken upon either 
 of the other two lines representing the same measure. 
 
 Hence, the three scales will be easily distinguished from each other 
 wherever they are placed; or, in other words, it will be easy to see to which of 
 the three directors any one of the three measures applies. 
 
 The three scales will be sufficient for any representation, of which all the 
 parts of the edges of the original are straight lines, parallel and perpendicular 
 to each other ; but where circles are to be represented upon any of these 
 lines as an axis, it will be necessary to find the inclination of each of the three 
 edges of the solid angle, in order to find the two axes of the ellipse which is 
 to represent the circle. 
 
 Ihe most convenient method of finding the three inclinations is, perhaps, 
 the following : draw a straight line, as a base and a perpendicular, from any 
 
 point to that straight line ; set the height of the vertex of the solid angle upon 
 the perpendicular, from the point of concourse of the two lines forming the 
 right angle, also from the point of concourse set the lengths of the three lines 
 which represent the three lines of concourse of the solid angle, upon the line 
 representing the base ; then join each of the three points to the top of the 
 perpendicular, and the acute angle, which either of these lines makes with the 
 base, is the angle of inclination. 
 
 N. B. The least inclination belongs to the director which is opposite to 
 the greatest angle, and the greatest inclination belongs to the director 
 opposite the least angle, representing a right angle of the solid angle. 
 
 R 
 
 & 
 
66 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 Part II.— ORTHOGRAPHICAL PROJECTION. 
 
 On Shadows. 
 
 We must here have the seat of the sun’s rays and their angle of inclination 
 to the horizon, to find the shadow of any vertical line, and the shadows of 
 any line perpendicular to a vertical plane. 
 
 In order to find the shadow on the vertical plane, the seat of the sun’s 
 rays must be found on this plane. 
 
 The plane of the horizon is here represented by the primitive plane. 
 
 Let abcdefghi, Plate vm. figure 2, represent a penthouse or lean-to, 
 with the ends of several timbers projecting from the front wall, also, with two 
 stalks of chimneys, of which their planes are perpendicular to the horizon, and 
 parallel and perpendicular to the plane of the front. All the walls, except the 
 front, are higher than the sloping plane of the roof. The tops of the chimneys 
 and the back wall terminate in horizontal planes. Moreover, the two gable 
 walls terminate in planes parallel to the sloping side of the roof. It is required 
 to find the shadow of the timbers projecting perpendicularly from the plane of 
 the front, the shadow of the stack of chimneys, and the shadow of the gable 
 wall on the inclined plane of the roof. 
 
 Let the line a l, figure 4, represent the seat of the sun on the horizon- 
 Draw a a parallel to ef, the lower edge of the plane of the front, figure 2, 
 and in al, figure 4, take any point l, and draw La parallel to ed, the lower 
 edge of the gable, figure 2. Draw the line lc, figure 4, parallel to dc, ea, 
 or fg, figure 2. Draw the angle lac, figure 4, to represent the angle which 
 a ray of the sun makes with its seat ; then a c will represent a ray of the sun. 
 Complete the parallelogram a bcl, and join ab, and ba will be the seat of 
 the sun on the vertical plane. 
 
 To find the shadows of the timbers projecting from the front of the pent- 
 house, figure 2. Let gf be the arris or edge of one of the timbers, perpen- 
 dicular to the plane aefg of the face of the front wall, g being the point 
 
AND COMPLETE DECORATOR. 
 
 67 
 
 where the line gf meets the plane of the front. Drawee and fe respectively 
 parallel to ba and c a, figure 4, then the straight line ge is the shadow of gf 
 In the same manner the projection of the shadows of the other three perpen- 
 dicular lines may be found. Now v being the shadow of the point v, and 
 w the shadow of the point w, ]omge, ev, vie, and tvx, then gevwx is the 
 shadow of the end of the timber as required. 
 
 N. B. The edge ev of the shadow is parallel to fv the vertical edge, and 
 vw to vw the horizontal edge of the timber; therefore if we find the seat of 
 a ray of the sun to each of the three lines of the projecting end of the timber 
 that throws the shadow, and having found the point e, in order to complete 
 the shadow, we have only to draw ev parallel to fv, and vw parallel to vw. 
 
 In the intersection of the sloping plane of the roof and the face of the 
 wall, take any length db. Draw da parallel to ed, and ba parallel to al, 
 figure 4; then the triangle ab d represents a plane parallel to the horizon. 
 Draw ac parallel to dc, ea, or fg. and draw dc parallel to ab or gh, and 
 join be; then be is a section of a vertical plane, passing through the sun in 
 the sloping plane of the roof. 
 
 To find the shadow'upon the roof by the ledge GZe h, first find the shadow 
 of the vertical line yz thus, draw yfl parallel to be, and xfl parallel to ca, 
 figure 4; then fl will be the shadow of the point z. Draw fly parallel to z«, 
 and fly is the shadow of the sloping line ze . 
 
 To find the shadow of either of the stalk of chimneys. To do this, it is only 
 necessary to shew how one of the quoin lines of two adjacent vertical surfaces 
 may be found ; therefore let this vertical line be m t. Draw m n parallel 
 to be, and tn parallel to a c, figure 4; then mn is the shadow of the vertical 
 edge mt of the stalk of chimneys. In the same manner the shadow of each of 
 the other vertical lines may be found ; then by joining all the parts, we shall 
 have the shadow complete. Or if we had found the lines h i, mn, and os, and 
 only the extremity of the shadow of the point k, produce hm and kt to meet 
 each other in l ; draw il, intersecting mn in n; draw ns parallel to tu, meet- 
 ing os in s; then liinso is the shadow of the chimney shaft as required. 
 
63 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 It is now expected that the reader is in full possession of the principle, 
 and therefore he cannot be at a loss to form the shadows of all objects, how- 
 ever varied in form from one another; we shall, therefore, close this subject 
 by shewing how the ray and seat of the sun may be found, so as to cause all 
 parallel planes, which are perpendicular to the surface of the wall, in vertical 
 as well as in horizontal positions, to throw shadows equal to their breadth. 
 In order to accomplish this, in figure 8, draw the cube abcdef. Draw the 
 diagonal fc of the base cdeA; also, draw the diagonal fc of the vertical 
 plane bcef ; then ec will be the seat, and fc a ray of the sun on the hori- 
 zontal plane. Likewise, draw the diagonal a c, and ac is the seat of the sun 
 on a vertical plane parallel to the plane abch. This position of the cube is 
 adapted to the example, figure 2. 
 
 Having thus discussed the shadows of objects, projected orthographically, 
 we shall now 7 give a few of the principles for the shadows of bodies geome- 
 trically represented in planes, elevations, and sections, which in certain cases 
 are indispensable to the draughtsman, and to fix those principles they will be 
 illustrated by a select number of useful examples. 
 
 Problem XVIII. 
 
 The Seal and Angle of the Sans Hays being given on any Plane , and the Line 
 of Intersection of a Plane, perpendicular to the first mentioned Plane, to 
 find the Seat of the Suns Rays on the perpendicular Plane. 
 
 Let ac, figure 1 , Plate VIII. be the intersection of the perpendicular plane, 
 and be the seat of the sun’s rays. Make the angle bed equal to the angle of 
 inclination of the sun’s rays. Draw bd perpendicular to be, and be perpen- 
 dicular, intersecting ac in the point a. Make ae equal to bd, and join ec . 
 D raw ef perpendicular to ec, make ef equal to bd, and join fc. 
 
 In order that the reader may have some idea of the Problem, draw bg and 
 ah perpendicular to ba, and make bg and all each equal to bd, and join g li. 
 
 Conceive now the rectangle abgh to be turned upon ab, the triangle cbd 
 to be turned upon cb, and the triangle cae to be turned upon ca, so that the 
 
AND COMPLETE DECORATOR. 
 
 69 
 
 planes of these three figures may be each perpendicular to the plane cab, 
 then will bg coincide with bcl , and ae with ah. These three planes being 
 perpendicular to the plane cab, turn the plane cef round the line ce till it 
 becomes perpendicular to the plane cae ; then will the line ef coincide with 
 hg, and the line cf will coincide with the line cd; and because the plane cbd 
 is perpendicular to the plane cab, and cb is the foot of the plane c b d; there- 
 fore cb is the projection or seat of the line cd on the plane cab. In the same 
 manner it may be shewn that ce is the seat of the line cf, that is, of cd in the 
 plane cbe. 
 
 Problem XIX. 
 
 Given the Intersection cb, Figure 2, of the Plane of Projection, and the Seat 
 ca of a Line, the Angle which the Plane of Projection mahes with the 
 given Plane , and the Angle which the Line mahes ivith its Seat, to find 
 the Seat and Angle of Inclination of the Line on the Plane not given. 
 
 Draw ae perpendicular to be, intersecting be in b, ad perpendicular to ac, 
 and ah perpendicular to ab. Make the angle adg equal to the angle of 
 inclination of the two planes, and the angle acd equal to the inclination of 
 the line to its seat. Also, make ah equal to ad. Draw hg perpendicular to 
 bg, and make be equal to bg, and join ec. Draw ef perpendicular to ce; 
 make ef equal to gh, and join cf; then ce is the seat of the line, and ecf is 
 the angle of inclination. 
 
 Problem XX. 
 
 Given be, Figure 3, the Seat of a Ray of Light, and the Angle of Inclination 
 of the Ray to its Seat, also a Line ac parallel to a Line in Space, to find 
 the Angle of Inclination of a Plane of Shade. 
 
 Draw be perpendicular to ac, intersecting ac \n a. Make the angle bed 
 equal to the inclination of the ray with its seat. Draw bcl perpendicular to 
 be, and bf perpendicular to ba. Make bf equal to bd, and join af, then 
 baf is the angle required. 
 
70 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 Problem XXI. 
 
 To find the Shadow of a Solid upon a Wall, of which Solid its Planes are 
 all parallel and perpendicular to the Plane of the Wall, given the Plan 
 and Elevation of the Solid. 
 
 Let bchg, figure 4, represent the elevation of the solid, the lines bc, ad, 
 ef, gh, ch, and bg, represent the planes perpendicular to the wall, and the 
 rectangles a d f e, abcd, efhg represent the planes parallel to the wall. 
 Also, let abcdef gh represent the plan on a plane perpendicular to the plane 
 of the elevation. In this plan, ah represents the w r all line bc, de , fg; the 
 planes abcd, adfe, efhg. 
 
 Draw b p, a r, e s, g t, h u, parallel to the seat of the sun’s rays on the 
 elevation, and draw am, dn, eo and gi parallel to the seat of the sun’s rays 
 on the plan intersecting ah in m, n, and fg in o. Draw mp, nr perpen- 
 dicular to ah, and bg rs parallel to it ; sw parallel to mp or nr, wt parallel 
 to bg, and tu parallel to gh ; then Bp qrswtua is the shadow of the solid 
 required. 
 
 From the description here given of figure 4, it will hardly be necessary to 
 enter into any particular explanation of figures 5, 6, and 7. In figure 5, 
 .hough only two of the planes are perpendicular, and the other are inclined 
 to the plane on which the shadow is thrown, yet the lines of intersection of 
 all the planes are either parallel or perpendicular to the plane on which the 
 shadow is thrown; the principle then depends upon this Problem, “ the plan 
 and elevation of a point being given, and the line of elevation, to find the 
 shadow of the point in the elevation .” Figure 6, exhibits the shadow of a 
 cross ; here all the lines are parallel and perpendicular to the plane of eleva- 
 tion. Figure 7, exhibits the shadow of any thing with a circular bow in front, 
 s f and o p are the straight parts of the shadow, and are parallel to the lines 
 bc and de, which projects them; fghilno is the shadow of the circular 
 edge, and as the shadow of a circle is either a circle or ellipse, so the shadow 
 
AND COMPLETE DECORATOR. 
 
 71 
 
 of the segment of this circle is ail ellipse, when the tw r o planes are oblique 
 to each other. 
 
 With regard to figure 5. continue the directions ed , df of the lines that run 
 through the shadow to meet the wall line ih, and find the elevations of the 
 same points, and these points will give the directions of the shadows mn, po 
 on the elevation. 
 
72 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 PERSPECTIVE. 
 
 Definitions, &c. 
 
 Def. 1 . Perspective is the art of representing a rectilineal object upon a 
 plane surface by lines, so that when the representation is placed in a certain 
 position between the eye and the object, every line in the representation will 
 intercept the view of the corresponding line of the object. 
 
 To render the principles of Perspective easy of comprehension, it is ne- 
 cessary to consider the faces of an object, as composed or formed of different 
 planes, and one of those faces coincident with a plane indefinitely extended 
 on all sides : this plane is called the original plane, ox geometrical plane, and 
 the face of the object which is coincident with the original plane, is called the 
 original figure. 
 
 The point for viewing the object is given in position to the original plane; 
 that is, some point is fixed upon in the original plane, and the eye at a given 
 distance from that point in a perpendicular drawn from the said point. 
 
 Def. 2. The point where the eye is placed, is called the point of sight. 
 
 Def 3. The foot of the perpendicular, is called the station point. 
 
 As the faces of an object may have all manner of positions, an original 
 plane may in consequence have any inclination to the horizon ; therefore an 
 original figure may be in a horizontal plane, below or above the eye ; a 
 vertical plane on the right or on the left of the spectator; or on an inclined 
 plane in any position to the horizon ; but in whatever plane an original figure 
 is situated, the method of representing the object is the same. 
 
 Def. 4. T he surface on which the object is represented, is called the plane 
 of the picture, or simply, the picture. 
 
 In order to have a comprehensive notion of the principles of perspective, 
 the student must consider, that a picture drawn in the utmost degree of per- 
 fection, and placed in its real position, ought to appear to the spectator, the 
 same as the real object itself, when seen from the point of sight. In order 
 
AND COMPLETE DECORATOR. 
 
 73 
 
 to produce this effect, it is necessary that two visual rays, from any two 
 points of the picture, should form the same angle, at the eye, as the visual 
 rays from the two corresponding points of the original object; and that every 
 ray, from the picture to the eye, ought to proceed with the same degrees of 
 light, shade, and colour, as from the corresponding points of the original 
 object. And there can be no doubt, that if these circumstances were 
 attended to in the execution, the deception would be complete; so that if we 
 were looking at the picture from its point of sight, our senses would be so 
 far deceived, as to suppose that we saw the object itself, instead of its repre- 
 sentation on a plane surface. This observation is completely verified by the 
 Diorama now exhibiting in the Regent’s Park. 
 
 Def. 5. That point in the picture, cut by a line drawn from the point of 
 sight perpendicular to the picture, is called the centre of the picture. 
 
 Def 6 . The distance between the spectator’s eye and the centre of the 
 picture, is called the distance of the picture , or principal distance. 
 
 Def. 7. The point where any original line cuts the picture, is called the 
 intersecting point of that line. 
 
 Def. 8. The line where any original plane cuts the picture, is called the 
 intersecting line of that plane. 
 
 Def. 9. The point where a line, drawn from the eye of the spectator 
 parallel to an original line, cuts the picture, is called the vanishing point of 
 that original line. 
 
 Corol. 1 . Hence it is evident, that original lines, which are parallel to 
 each other, have the same vanishing point; for only one line can be drawn 
 through the eye of the spectator that will be parallel to them all. 
 
 Corol. 2. Those lines which are parallel to the picture have no vanishing 
 points ; because no line can be drawn through the eye parallel to any one of 
 them, that will ever meet the plane of the picture. 
 
 Corol. 3. The lines that generate the vanishing points of two original lines 
 meeting together, form the same angle at the eye of the spectator, as the two 
 original lines themselves. 
 
 T 
 
74 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 Def. 10. The line formed in the picture, by a plane passing through the 
 point of sight parallel to an original plane, is called the vanishing line of that 
 original plane. 
 
 Corol. 1 . Hence original planes, that are parallel, have the same vanishing 
 line, for only one plane can pass through the eye of the spectator, that will be 
 parallel to them all. 
 
 Corol. 2. All the vanishing points of lines, in parallel planes, are in the 
 vanishing line of these planes ; for the lines which produce those vanishing 
 points are all in the plane that produces the vanishing line, by this definition. 
 
 Corol. 3. The planes which produce the vanishing lines of two original 
 planes, have their common section in a line passing through the eye of the 
 spectator parallel to the line of common section of the two original planes, 
 and are inclined to each other in the same angle as the two original planes. 
 
 Corol. 4. Hence the vanishing point of the line of common section of two 
 planes, is the intersection of the vanishing line of those planes. 
 
 Corol. 5. The vanishing and intersecting lines of the same original plane 
 are parallel to each other, since they are generated by parallel lines. 
 
 Def. 11 . The point in a vanishing line, cut by a perpendicular drawn from 
 the eye of the spectator to that vanishing line, is called the centre of the 
 vanishing line. 
 
 Corol. A line drawn from the centre of the picture to the centre of a 
 vanishing line, is perpendicular to that vanishing line. 
 
 Def. 12 . The distance between the eye of a spectator and the centre of a 
 vanishing line, is called the distance of that vanishing line. 
 
 Corol. The distance of a vanishing line is the hypothenuse of a right- 
 angled triangle, of which one leg is the principal distance, and the 
 other is the distance between the centre of the picture and the centre 
 of the vanishing line. 
 
 Def 13. A plane passing through the eye of the spectator, parallel to the 
 picture, is called the directing plane. 
 
 Def. 14. The representation of any figure, is called its projection . 
 

AND COMPLETE DECORATOR. 
 
 75 
 
 Plate I. — Perspective. 
 
 In order to comprehend the sense of these definitions more readily, let deh, 
 fig. 1 , be the original plane, and let o be the point of sight, Def. 2, p the station 
 point, j Def. 3, and abl the plane of the picture, Def. 4 : then if o c be perpen- 
 dicular to the plane abl, meeting it in c, c is the centre of the picture, Def. 
 5, and oc its distance, or the principal distance, Def. 6 . If ki be an original 
 line, and produced to meet the picture in b, b is the intersecting point of the 
 line ki , Def. 7. If o v be drawn parallel to f, meeting the picture in v, v is 
 the vanishing point of f g, Def. 9. If v l be drawn through v, parallel to the 
 original plane deh,vl is the vanishing line of that plane, or of de h, Def. 10. 
 If o s be drawn perpendicular to v l, to cut it in s, s is the centre of the 
 vanishing line v l, Def. 11 , and o s its distance, Def. 12, which is, therefore, the 
 liypothenuse of the right-angled triangle ocs. If a plane, ode, be parallel 
 to the plane of the picture, ode is called the directing plane, Def. 13. 
 
 PROPOSITION I. — THEOREM. 
 
 The representation of a line, is part of a line passing thruogh the inter- 
 secting and vanishing points of the original line. 
 
 Figure 1. For the visual rays io, ko, &c. which produce the represen- 
 tations i, k of the points i, k, &c. of the original line i k, by their intersec- 
 tions with the picture, are all in a plane passing through the original line ik, 
 and the spectator’s eye is at o. But the point b, which is the intersection of 
 the original line i k, is in that plane, because it is in the line l k ; and the line 
 o v is in the same plane, because it is parallel to the original line i k, Def. 5 ; 
 wherefore the line vb is the intersection of the plane iok with the picture; 
 and consequently ik, which is the representation of the line ik, is part of the 
 line k b which passes through the vanishing point v, and the intersection b of 
 the original line ik. 
 
 Corol. 1. Hence the representation of any number of lines that are parallel 
 to each other, but not to the picture, will pass through the same point. For 
 
CABINET-MAKER, UPHOLSTERER, 
 
 76 
 
 they all pass through the same vanishing point by this theorem, and by 
 corollary 1, Def 9. 
 
 Corol. 2. But if the original lines are parallel to the picture, as well as to 
 each other, their representations will be parallel to each other, and to the 
 originals. For the line passing through the spectator’s eye, which lin*e in 
 other cases produces the vanishing point by its intersection with the picture, 
 is, in this case, parallel to it, and therefore produces no vanishing point, so 
 that the representations can never meet each other, nor that of the line pass- 
 ing through the spectator’s eye, and consequently they are parallel to each 
 other, to that line, and to the originals. 
 
 Corol. 3. Hence it appears, that the representation of plane figures, 
 parallel to the picture, are similar to their originals. For in figure 2, the 
 picture being ab, the original figure cdef being parallel to it, and the repre- 
 sentation being cdef. If the original figure he resolved into triangles, by 
 diagonals as d f, the representation will he resolved into triangles by corre- 
 sponding diagonals df. Whence all the lines in the representation being 
 parallel to all the lines in the original, every triangle edf will he similar to 
 the corresponding original triangle edf; and consequently, all the lines in 
 the representation cdef will he in the same proportion to one another as the 
 corresponding lines in the original figure cdef. 
 
 PROPOSITION II. THEOREM. 
 
 Any line in the representation of a figure parallel to the picture, is to its 
 original line, as the principal distance is to the distance between the specta- 
 tor’s eye and the plane of the original figure. 
 
 Let og, figure 2, be perpendicular to the original plane and to the picture, 
 cutting them in g and g: the original plane being parallel to the picture, 
 therefore all the visual rays oc, od, oe, &c. are cut in the same proportion, 
 by the points c, d, e, &c. as og is by the point g ; and dc being parallel to 
 dc the triangle, ode is similar to the triangle odc; therefore dc: dc : : od: 
 od; but from what has here been observed, od: od : : og: og whence by 
 
AND COMPLETE DECORATOR. 
 
 77 
 
 equality dc : dc : : og : og, that is, representation cd is to its original line 
 c d, as the principal distance o g is to the distance between the spectator’s 
 eye and the plane of the original figure. 
 
 PROPOSITION III. THEOREM. 
 
 The distance between the vanishing point of a line and the representation 
 of any point in it, is to the distance between the vanishing point and the inter- 
 section, as the distance between the directing point and the intersection, is to 
 the distance between the directing point and the original point. 
 
 Figure 3. Let b be the intersecting point, and g the directing point, of 
 the original line ik, and v is its vanishing point; then k being the represen- 
 tation of k, and v k parallel to og. Now the triangles okv and kog are 
 similar ; wherefore & v:og::ov:kg, but o g is equal to v b, and o v equal 
 to gb; therefore & v:vb::gb:kg. 
 
 PROPOSITION IV. — THEOREM. 
 
 The radial or parallel of an original line, producing the vanishing point, 
 makes the same angle with the vanishing line as the original line makes with 
 the intersecting line. 
 
 For the intersecting line, the vanishing line and the parallel of the eye are 
 parallel to each other, and the radial, producing the vanishing point of an 
 original line, is parallel to the original line ; therefore whatever inclination the 
 original has to the intersecting line, the radial of that line has the same 
 inclination to the parallel of the eye ; and because the vanishing line is paral- 
 lel to the parallel of the eye, and the radial joins the parallel of the eye and 
 the vanishing line ; the acute angle, formed by the radial and the parallel of 
 the eye, is equal to the acute angle formed by the same radial and the vanish- 
 ing line, being alternate angles ; therefore the acute angle, formed by the 
 radial and the vanishing line, is equal to the acute angle formed by the origi- 
 nal line and the intersection. 
 
 u 
 
78 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 PROPOSITION V. — THEOREM. 
 
 If from any point b, fig. 4, taken at pleasure in the vanishing line ab, a 
 straight line be drawn to any two points d and e in the intersecting line, so as to 
 cut the indefinite representation a i of any original line in the points d and e, 
 then will the radial o a be to the distance a b, between the vanishing point a 
 and the assumed point b, as the original of d e is to d e. 
 
 Produce the indefinite representation a i to meet the intersecting line in 
 i : draw i d parallel to o a, and draw d d and e e parallel to o b, and through 
 e draw ef parallel to gi. 
 
 Now because ob is parallel to dD and eE, and because ob meets the 
 vanishing line in b, b is the vanishing point of the lines dD and eE ; but d b 
 and e b are the indefinite representations of d d and eE ; therefore the repre- 
 sentations of the original points d and e are in the lines ds and eB. Again, 
 because o a is parallel to the original line d e, and because o a meets the 
 vanishing line a b in a, a is the vanishing point of the original line de, and 
 since i is its intersecting point, a i is the whole indefinite representation of 
 d e ; therefore the points d and e are represented in the line a i ; but since 
 they are also represented in d b and e b, the points d and e, the intersections 
 of these three lines, are the representations of d and e. 
 
 Then because of the similar triangles oab and dep,oa:ab::de:ef; 
 but ef is equal to ed, and de is the original of de, whence o a is to ab, as 
 the original of de is to d e. 
 
 Corol. Hence if the length of the original of de be known, de will be 
 found, and consequently de, without having d e in its real position. 
 
 PROPOSITION VI.- — THEOREM. 
 
 If the angle eof, fig. 5, formed by the radials of any two indefinite represen- 
 tations a e and b f, be bisected, and if from the point g in the vanishing line, 
 cut by the bisecting line, be drawn lines g a, g c, g b, and g d, cutting each 
 indefinite representation in two points, and any line drawn parallel to the 
 
AND COMPLETE DECORATOR. 
 
 79 
 
 vanishing line in four points a, c, b, d, then the originals of the two intercepted 
 distances ac, bd, in the indefinite representations, will be to each other as the 
 corresponding intercepted distances ac, bd, on the line drawn parallel to the 
 vanishing line. 
 
 Take o q in the one radial at pleasure, and make o s on the other equal to 
 oq, and join qs, which will be bisected in t by on. Now in the two right- 
 angled triangles otq and ots, the base to is equal to the base ts, and the 
 perpendicular t o is common, therefore the hypothenuse o q is equal to the 
 hypothenuse os. Draw qp and sr parallel to o g, and through o draw pr 
 parallel to the vanishing line e f ; then p r will also be bisected in o ; pro- 
 duce or to any point u, and draw uv parallel to og, cutting o f in v. 
 
 Now by the preceding proposition oe is to eg as the original of ac is to 
 ac, and o f is to f g as the original of bd is to bd ; but by the similar triangles 
 o eg and qop, oe:eg::oq:op; therefore the original of ac is to ac as 
 OQ is to op: again, in the similar triangles ofg and sor, of:fg::os : 
 or; therefore the original of bd is to bd as os is to or : and because u v is 
 parallel to s R, the triangles o u v and o r s are ^similar ; whence o v : o u : : 
 os : o r : but op is equal to or, and oq equal to os ; whence ov : ou : : 
 oq : op, hence, alternately oq:ov::op:ou; therefore the original of ac 
 is to the original of bd , as op to ou, or or to ou; whence if op or or 
 be equal to ac, and ou equal to bd, oq and ov will be the originals of 
 ac and bd. 
 
 „ PROPOSITION VII. — PROBLEM. 
 
 Given the centre and distance of the picture, and the seat of a point on 
 the picture, and its distance from it, to find the representation of that point. 
 
 Let c, figure 6, be the centre of the picture, s the seat of the given point. 
 Through c and s draw at pleasure c o and s p parallel to each other ; make 
 co equal to the distance of the picture, and sp equal to the distance of the 
 original point from its seat s. Draw op and sc, cutting each other in p, 
 which will be the representation of the original point required. 
 
80 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 For imagine the eye of the spectator to be placed at o, and p to be the 
 original point in their proper places ; then will o p be a visual ray, projecting 
 the point p at p ; and because of the similar triangles p s p and p c o, s p : : 
 pc : : s p : co ; therefore sp + pc :pc : : sp + co : co, or cs : cp : : sp + 
 co : co : now cs, sp + co and co are constant quantities; therefore the 
 distance cp is invariable, and consequently the point p must also be 
 invariable. 
 
 Hence the point p may be found by calculation thus : — s p + c o : c o : : 
 cox c s 
 
 cs : cp= -- - - - c o which is the distance of the point p from c, the centre of 
 the picture. 
 
 PROPOSITION VIII. — PROBLEM. 
 
 Given the centre and distance of the picture, the intersecting point of an 
 original line, and the seat of the line on the picture, and the inclination of the 
 line to its seat, to find the vanishing point of the original line and its inde- 
 finite representation. 
 
 Let c, figure 7, be the centre of the picture, i the intersecting point of the 
 original line, and i d its seat. 
 
 Make the angle dia equal to the angle which the original line makes with 
 its seat id; draw cv parallel, and co perpendicular, to id, and make co 
 equal to the distance of the picture. Draw o v parallel to ia, and join i v ; 
 then v will be the vanishing point, and i v the indefinite representation of the 
 original line. 
 
 Suppose the triangles cvo and dia to be turned round the lines cv and 
 id; c v i being turned towards the front, and d i a on the back, of the picture, 
 so that each two adjoining planes may be perpendicular to each other, then 
 the point o will be in the point of sight, and i a will be the original line ; then 
 o v will be still parallel to a i ; therefore o v will be the radial of a i, v its 
 vanishing point, and i v its indefinite representation. 
 
AND COMPLETE DECORATOR. . 
 
 81 
 
 PROPOSITION IX. — PROBLEM IV. 
 
 Prom a given point in the indefinite representation of a straight line, to 
 cut off a part of that indefinite representation that shall represent a given 
 portion of the original. 
 
 METHOD FIRST. 
 
 Let a v, fig. 8, be the given indefinite representation ; it is required to cut off 
 fi om a v a part which shall represent a line of a given magnitude, vl being the 
 vanishing line, Plate x. ; o the point of sight, and in the intersecting line. 
 
 Through o draw co parallel to vl ; make od equal to the original line. 
 In vl take any point l, draw ol and dc parallel to ol. Draw La, cutting 
 
 in in a; make ab equal to oc, and join Lb, cutting av in b, then ah is the 
 part required. 
 
 This method is excellent when only one or two distances are required to 
 be ascertained; but when many are wanted, the following method will be 
 found to be productive of very great facility. 
 
 The principle and methods of performing this operation are quite new, 
 and entirely free from the defects of the method first introduced by Dr. 
 Brook Taylor, and in practice in every recent publication founded upon his 
 principles. 
 
 METHOD SECOND. 
 
 The same data being referred to by the same letters, will be known by the 
 preceding diagram. 
 
 Diaw oc, figure 9, parallel to vl; make oc equal to the distance be- 
 tween the vanishing line vl and intersecting line in; then the point l 
 being taken as before, draw o l and c d parallel to o l. Make the dis- 
 tance between the vanishing line vl and the parallel line ab equal to od. 
 Through l draw La, which produce to meet ba in a, and make ab equal 
 to the length of the original line, and join bL, cutting a v in b, then ab is 
 the line required. 
 
 x 
 
82 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 This depends upon the same principle as the foregoing method. 
 
 These methods have the peculiar advantage, that the measuring centre 
 may be placed on any point of the vanishing line, and consequently may be 
 brought immediately over the representation of the line to be divided ; by this 
 means the dividing lines cut the representation given more directly or less 
 obliquely than the usual process. 
 
 In the common method the measuring point cannot have any change, and 
 as it is always situated at a distance from the centre of the vanishing line, the 
 intersections made in this indefinite representation are so very oblique, that 
 the points required must, in fact, be guessed at. Taking given portions of 
 the distance of a vanishing point, as J, J, f, &c, is very troublesome, and 
 does not always obviate the inconvenience. 
 
 PROPOSITION X. — PROBLEM V. 
 
 To divide the representation of a line, whose vanishing point is given, into 
 any number of parts, so that their originals shall have a given ratio among 
 themselves. 
 
 Let ab, figure 10, be the given representation of the line to be divided, 
 and v its vanishing point. 
 
 Draw vo at pleasure, and draw ab parallel to vo. Through any point 
 o, in the line v o, draw o a and o b, cutting ab in a and b : divide a b in c, d, 
 &c. in the given ratio, and draw oc, od, &c. cutting a, b in c, d, &c. then a, b 
 will be divided as required. 
 
 For c v being parallel to ab, a b may be considered as the original line, 
 and ov its parallel; and consequently o as the point of sight, and oa, oc, 
 o d, &c. visual rays, projecting the points a, c, d, &c. 
 
 PROPOSITION XI. — PROBLEM VI. 
 
 From a given point in a given indefinite representation of a line, to cut off 
 a part, of which the original shall have a given ratio to the original of another 
 given part of that indefinite representation. 
 
AND COMPLETE DECORATOR. 
 
 83 
 
 Let b v, figure 11, be the given indefinite representation of the line, v its 
 vanishing point, be the given part; it is required to cut off from a a part 
 of which the original shall have a given ratio to the original of the part 
 represented by be. 
 
 Through v draw v l at pleasure, and a b parallel to v l, and at any distance 
 from v l draw l c and l b, which produce to meet a b in c and b ; also draw l a, 
 cutting ab at a : make Ad equal to bc, in the ratio required, and join dL, 
 cutting b v at d, then the original of ad is to the original of bc as Ad is to bc. 
 
 This is evident by considering it upon the same principle as that in the 
 preceding proposition. 
 
 The above Problem very frequently occurs in the following case, viz : — 
 supposing ab to be given in perspective, it is required to cut off a part, ad, 
 from a, so that the originals of ad and ab may have a given ratio. The 
 description is shortly thus — 
 
 PLATE XI. 
 
 Draw l a, l b, figure 1, cutting ab at A and b; make Ad equal to a b in 
 the ratio required; join dL, cutting the indefinite representation \b as 
 required. Or if a d is given, ab may be found exactly in the same manner. 
 
 PROPOSITION XII. — PROBLEM VII. 
 
 At a given point in the indefinite representation of an original Line, to 
 represent a Line equal to, or in a given Ratio to, the original of another 
 given Line, the vanishing Line, and the proposition of the Rye, being 
 given. 
 
 CASE FIRST. 
 
 Let a b, figures 2, 3, and 4, be the vanishing line, o the point of sight, 
 ab the given indefinite representation, a its vanishing point ; c b the indefinite 
 representation from which the part is required to be taken, and c the point 
 at one of its extremities, and let b be the vanishing point of cb. 
 
84 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 Parallel to ab draw any straight line ad; oin ao and bo, and draw 
 op, cutting ah in p, so that the angle aob may be bisected. Draw 
 p a, p b, pc, cutting ad in a, b, c. Make cd toab in the ratio required, 
 and join pd, cutting cb in «?, then the original of ab will be to the original 
 of cd as ab is to cd. 
 
 In figures 3 and 4, as ab is parallel to the vanishing line, the radial oa 
 is also parallel to the vanishing line. In figure 4, the angle aob being a 
 right angle, and being bisected, bp will be equal to bo, the distance of the 
 picture. 
 
 CASE SECOND. 
 
 Suppose the points b and c to coincide, then the points b and c will also 
 coincide. In this case let b become c, and b will also become c ; the opera- 
 tion will still continue to be the same, except there is only one line to produce 
 nstead of the two p b and pc; but that the method may be clearly under- 
 stood, we shall shew the construction according to this simplification, parti- 
 cularly as this case more frequently occurs than the other. 
 
 Parallel to ab, figure 5, draw any straight line ad. Draw p a, p c, meet- 
 ing ad in a and c ; make cd equal to ca, in the ratio of the two original lines, 
 and join pd, cutting cb in d. Then the original of ab will be to the original 
 of cd, as ca is to cd. 
 
 This method obtains the result, whether the two lines lie on both sides of 
 the line pc, or on the same side of it. 
 
 Hence it is evident, that if ca and cd be equal to each other, the original 
 lines, represented by ca, cd, ce, and cf, will be all equal to each other. 
 
 But here it may be observed, supposing d a and f b not to be parallel, 
 that is, to incline to each other, that we should not have had more trouble in 
 representing the triangle abc, figure 6, formed by producing da and fb to 
 meet each other in c ; so that to find the representation of a triangle in 
 the original plane, we have only to produce the three sides, and find the 
 indefinite representations of these three sides, as the triangle abc i« the repre- 
 sentation of the original triangle abc. 
 
AMD COMPLETE DECORATOR. 
 
 85 
 
 This same idea will extend to all kinds of right-lined figures, by merely 
 finding the indefinite representations of the sides of the original figure, without 
 resolving it into triangles ; and their mutual intersections w ith each other, will 
 form the representation of that figure, whatever it may be. 
 
 This method is so evident, that more examples will not be necessary, and 
 more particularly as the figures, being drawn out in their real places, would 
 be attended with great inconveniences, and would require more surface to 
 draw r upon than can generally be found. 
 
 The principle will be clearly understood by turning the plane y upon the 
 plane of the picture towards the eye, and the plane x behind the picture, so 
 that the planes x and y may be parallel to each other ; then the picture and 
 the other two planes being supposed to be transparent, and tbe eye placed at 
 o, each indefinite representation will intercept the view of its original, whether 
 point or line. 
 
 PROPOSITION XIII.- -PROBLEM VIII. 
 
 Given the vanishing and intersecting Lines of a Plane , the Centre and Dist- 
 ance of the vanishing Line , to find the Projection of any Line , or Side of 
 a Figure , given in the original Plane. 
 
 Let gh, figure 6, Plate xi. be the vanishing line, s its centre, and let the 
 intersecting line be d e. 
 
 Draw so perpendicular to g h, and make so eeual to the distance of the 
 picture. 
 
 Then to find the representation of the straight line a b, in the original 
 plane x, folded on the plane of the picture, as well as the vanishing plane y. 
 
 Draw og parallel to a b, meeting the vanishing line g h in g, and produce 
 a b to meet the intersecting line in e, then e is the intersecting point of ab; 
 joinGE; and g e is the indefinite representation of ab. Now, to find the 
 representation of any point in the original plane, we must make two lines 
 pass through that point, and find the indefinite representations of these two 
 lines, and their intersection will give the representation of that point. With 
 
86 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 respect to the representations of the points a, b, which mark the extremities 
 of the original line ab; a common indefinite representation g e has been 
 found for each of the points a, b ; then as one indefinite representation has 
 been found for each point, we have only to find another for each of these 
 points; therefore, draw any line ad, and find its indefinite representation dh, 
 also any line be, and find its indefinite representation fi, then the intersec- 
 tions a and b , of these two indefinite representations, mark the extremities of 
 the line ab. which is the representation of a b as required. 
 
 PROPOSITION XIV. — PROBLEM IX. 
 
 Given the vanishing and intersecting Lines of a Plane , and the Centre 
 and Distance of the vanishing Line, to find the original of a given 
 Repi esentation. 
 
 Let it be proposed to find the original of the triangle abc , fig. 6. Produce 
 the sides ba,ac,bc, of the given representation to their vanishing points g, h, i, 
 and to their intersecting points e, d, f, and draw r the radials go, ho, io; 
 draw ea parallel to og, dc parallel to oh, and fc parallel to oi, then the 
 triangle abc, formed by the intersections of these three parallels, will be the 
 original of the triangle abc as required. 
 
 This is evident from the last proposition being only the reverse of it. 
 
 PROPOSITION xv.' — PROBLEM X. 
 
 Given the vanishing and intersecting Lines of a Plane, and the Centre and 
 Distance of the vanishing Line, to find only the Length of the Original 
 of a Projection given. 
 
 Let g h, fig. 7, and a b, be the intersecting and vanishing lines, s the centre 
 of the vanishing line gh,o the point of sight placed as usual, and ab the given 
 projection or indefinite representation. 
 
 Produce ba to its vanishing point g ; join g o, and make g h equal to go. 
 Draw h« and ub, cutting the intersecting line ab in a and b ; then ab will 
 represent a distance equal to the original of ab. 
 
AND COMPLETE DECORATOR. 
 
 87 
 
 Demonstration. Let c be the intersection of the given representation a b. 
 Now g h being equal to go, the distance of the vanishing point g, and ac 
 being parallel to g h, the point h may be considered as the point of sight, and 
 a bc as the original line, ha and hb, as visual rays, producing the projection 
 or indefinite representation a b. 
 
 PROPOSITION XVI. PROBLEM XI. 
 
 (riven the vanishing Line of aP lane , the Centre and Distance of that vanish- 
 ing Line , and the vanishing Point of a Line in that Plane , to find the 
 vanishing Point of another Line in the said Plane , so that the two 
 original Lines shall form a given Angle. 
 
 Let g h, figure 8, be the vanishing line, o the point of sight, and g the 
 given vanishing point situated as before. 
 
 Join g o. Draw o h, making the angle goh equal to the given angle, then 
 h is the vanishing point required. 
 
 If the sides of the original angle produced are not subtended by the inter- 
 secting line, make the angle goi, not subtended by the vanishing line, equal 
 to the original angle, and produce i o to meet the vanishing line in h, then h 
 is the vanishing point required. 
 
 This is evident, since the radials make the same angle at the point of 
 sight that the originals make with each other. 
 
 PROPOSITION XVII. PROBLEM XII. 
 
 Given the vanishing Points of an Angle , and the Representation of the Angular 
 
 Point , to represent that Angle. 
 
 Let g and h, figure 8, be the vanishing points, and b the representation of 
 the angular point ; then if the original angle is subtended by the intersecting 
 line, draw bo, and b h, and g6h will be the representation required. 
 
 But if the original angle is not subtended by the intersecting line, produce 
 Gb or nb, according to which side it may lie upon; suppose h b produced to 
 c, then the angle g bc, or a bc, will represent the angle required. 
 
88 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 PROPOSITION XVIII. — PROBLEM XIII. 
 
 Given the vanishing Line of a Plane , the Centre and Distance of the vanishing 
 Line , and the Representation of one Side of a Triangle of a given Species 
 in that Plane , to find the Projection of the whole Triangle. 
 
 Let vl, figure 9, Plate xi. be the vanishing line, s its centre, and o the 
 point of sight as before, and ah the representation given. 
 
 Produce ah to its vanishing point v, or let v be the vanishing point of ah; 
 and let ah subtend the angle, which is also subtended by the intersecting line. 
 Make the angle v ok equal to the original of the angle at h, the side ok cut- 
 ting v l in k ; draw k he, and suppose the original of the angle at a , not sub- 
 tended by the intersecting line : make the angle vop equal to the original of 
 the angle a, and produce po to its vanishing point l, and draw l ac, then ahe 
 is the triangle required. 
 
 OR THUS. 
 
 Draw a b, fig. 9, parallel to v o. Then on a b describe a triangle similar and 
 similarly situated to the original triangle. Draw o l parallel to a c, and o k 
 parallel to bc; then the intersections of these lines give the vanishing points 
 l and k ; then the angles, subtended by the intersecting line, will be the 
 opposite angles to those subtended by the vanishing line ; therefore the same 
 angles of the triangle abc, that are subtended by the vanishing v l, will have 
 their representations also subtended by the vanishing line v l ; and the same 
 angles in the triangle abc, of which the contained sides would meet the 
 vanishing line when produced to form an opposite angle, will have their 
 representation also subtended by the vanishing line; thus the angle ach 
 represents an angle equal to acb, and the angle c ha represents an angle 
 equal to cba ; because ab and cb would meet the vanishing line the same 
 as a h and ch. 
 
AND COMPLETE DECORATOR. 
 
 89 
 
 PROPOSITION XIX. — PROBLEM XIV. 
 
 Given the Vanishing Line of a Plane , and the Centre and Distance of the 
 Vanishing Line , and the Representation of any Side of a Figure in that 
 Plane, to find the Representation of the whole Figure. 
 
 Resolve the original figure into triangles, by means of diagonals, and find 
 the representations of these triangles one after the other, by Proposition xvm. 
 beginning with the triangle which has the line given, then every triangle, as it 
 is represented, will give a new line for the next to be represented. 
 
 EXAMPLE. 
 
 Given the Vanishing Line of a Plane, the Centre and Distance of the 
 
 Vanishing Line, and the Representation of one Side of a Square , to find 
 
 the Representation of the entire Figure. 
 
 Let vl, figure 10, be the vanishing line, s its centre, o the point of sight 
 
 as usual, and let ah be the representation of the given side. 
 
 Produce ah to its vanishing point v, if not already done ; join vo, and 
 
 draw o l perpendicular to v o. Draw o m, bisecting the angle vol, and meet- 
 
 *» . 
 
 ing the vanishing line v l in m. Also, draw a m and l h, cutting each other in c : 
 again, through a and c draw a l and \c, cutting each other in d ; then ahcd 
 is the representation of the square required. 
 
 The representation of a square is described with great ease, knowing 
 that the diagonal being drawn, bisects the opposite angles ; and by attending 
 to the foregoing instruction, the representations of figures with more sides 
 may be executed with equal facility. 
 
 PROPOSITION XX. — PROBLEM XV. 
 
 Given the Vanishing Line of a Plane, the Centre and Distance of the 
 Vanishing Line, and a Line representing a Perpendicular drawn 
 through the Centre of a Square to one of its Sides, to describe the entire 
 Representation of the Square. 
 
 ' Let v l, fig. 1 1, be the vanishing line, s the centre of the vanishing line, and o 
 
 the point of sight as usual. Also, let a be the representation of the centre of 
 the square, and ah the representation of the perpendicular. 
 
90 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 Produce ba to its vanishing point v ; then having drawn vo and ol per- 
 pendicular to vo : draw o m, bisecting the angle vol, and cutting the vanish- 
 ing line v l in m ; then m is the vanishing point of the diagonal of the square. 
 Draw m a and l 6, cutting each other in d; also draw mb and a l, cutting 
 each other in c. Through c draw vc, cutting d l at e, and du at f ; lastly, 
 draw l / and dv, cutting each other at^; then defg is the representation of 
 the square required. 
 
 This method is evident, since a line drawn perpendicular from the centre 
 to one of the sides of a square makes half a right angle with either of the 
 diagonals; therefore 6 m must cut ac, to represent a line equal to the line 
 represented by ab. 
 
 This operation is also evident from the next Proposition. 
 
 PROPOSITION XXI. — PROBLEM XVI. 
 
 Given the Vanishing Line of a Plane , the Centre and Distance of the 
 Vanishing Line , and a Line representing a Perpendicular drawn 
 through the Centre of an Octagon to one of its Sides, to describe the 
 entire Representation of the Octagon. 
 
 Complete the square defg , fig. 12, as in the last Proposition. Draw n h 
 parallel to v l. Produce m b and m d to meet n h at b and d. Draw b q per- 
 pendicular to n h, and make bq equal to bd, and join dq. Make dH and d n 
 each equal to dq ; bp equal to bH, and do equal to dp. Draw m n, cutting 
 the sides gf and gd of the square at m and n; and draw m h, cutting ed and 
 e/at h and i; draw om, cutting gf and gdoXl and o ; and, lastly, draw pM, 
 cutting ef and e d at k and p. Draw Ik and op , then hiklmnop is the 
 representation of the octagon required. 
 
 This method is evident from the construction of the real octagon, (see the 
 adjacent figure,) for it is well known that each side of a regular octagon is 
 equal to the difference between the side and diagonal of a square having the 
 diameter of its inscribed circle equal to that of the octagon. Now dr is half 
 
AND COMPLETE DECORATOR. 91 
 
 the side of the square, and since r s is equal to dr or r e, d s is 
 therefore equal to half the diagonal; and Rh, which is half the side 
 of the octagon, is equal to the difference between the half diagonal and 
 half side of the octagon. 
 
 PROPOSITION XXII. — PROBLEM XVII. 
 
 Given the Vanishing Line of a Plane , the Centre and Distance of the 
 Vanishing Line , and the Representation of the Centre and Radius of a 
 Circle , to complete the whole Representation of that Circle. 
 
 It is known that a circle in perspective is an ellipsis, for it is demonstrated 
 by writers on conics, that the section of a cone is an ellipsis. m Now the repre- 
 sentation of a circle is the section made by the plane of the picture intercept- 
 ing the rays which proceed from the original circle to the eye, and which form 
 a conic surface, consequently the representation is an ellipse. Therefore in 
 order to form the curve correctly, we must find its axis, or such diameters 
 that it may be described by continued motion. 
 
 But though the difficulty of doing this is not very great, yet to understand 
 the principle thoroughly requires more geometrical knowledge than most people 
 are possessed of. We shall, therefore, shew how to find a convenient number 
 of points, or of points and lines representing as many points of the original 
 circle ; which, when ascertained, the curve must be drawn through them by 
 hand, observing the elliptic figure, which will greatly contribute to the accu- 
 racy of the representation. 
 
 method first. 
 
 Let a £, figure 13, represent the given radius, the vanishing line and its 
 centre and distances being as before. 
 
 Find the representation hiklmno of the octagon, as in the preceding 
 Problem. Produce ah to cut the opposite side ml in v, and produce ca to 
 cut the side no at w: let the diagonal fd cut the two sides Ik, op in r and s; 
 draw the diagonal ge, cutting mn and hi in t and u , then through the eight 
 points h, u , c , r, v, t, w, s, draw the curve of an ellipse ; which may be very 
 
92 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 correctly done, since the circle is supposed to be inscribed in the octagon, its 
 representation will touch the representation of the octagon; therefore the 
 eight points of representation are as many points of contact, which are equi- 
 valent to sixteen points, without tangents. 
 
 When the representation is very small, we need not be at the trouble to 
 represent the octagon, the four points b, c, v, w, in the representation of the 
 sides of the square, will be sufficient; for these being points of contact, the 
 curve may be drawn with considerable accuracy. However, in most cases it 
 would be eligible to find the sides mn and hi of the representation of the 
 octagon, as also the points of contact t and u, which would limit the length 
 of the curve witfi much greater accuracy than would otherwise be obtained, 
 and this may be done without increasing any confusion of lines, even though 
 the curve should be ever so narrow, or ever so near the vanishing line. 
 
 This is one of the best practical methods that has ever been invented. In 
 short, the operation is not only performed with fewer constructive lines, but 
 with much more facility, and the curve described with greater accuracy than 
 has been done by any other method hitherto published. 
 
 PLATE XII. 
 
 METHOD SECOND. 
 
 The data, fig. 1, being as before, produce the representation ab of the given 
 radius to its vanishing point v. In the vanishing line v l, take any number 
 of vanishing points k, 1,l, &c.: and draw the radials vo, ko, lo, lo, &c. : 
 through the representation of the centre a draw the indefinite radii \b, k», 1A, 
 L g, &c. Make af to represent a part equal to the part represented by a A 
 by Proposition xi. Bisect the angles v o k, ko 1, lo l, &c. and let the bisecting 
 lines meet the vanishing line in m, n, p, &c. 
 
 Through one extremity f, of the first perspective diameter fb, draw mf 
 cutting the second in i ; through the other extremity b draw m b , cutting the 
 second diameter in e. 
 
AND COMPLETE DECORATOR. 
 
 93 
 
 Having now obtained the second diameter ei , we may find the third from 
 the second, the same as the second was found from the first; and so on, from 
 one to another, till the whole are found. 
 
 For if the indefinite representations of any two lines intersect each other, 
 and if the two radials be bisected, a line drawn from the vanishing point of 
 the bisecting line, cutting the two indefinite representations, will intercept two 
 parts between the points of section and the point where these indefinite 
 representations cut each other, of which the originals will be equal. 
 
 PROPOSITION XXIII. PROBLEM XVIII. 
 
 Given the Centre and Distance of the Picture , the Intersection and Inclina- 
 tion of an original Plane to the Plane of the Picture , to find the Vanish- 
 ing Line of that Plane , and the Centre and Distance of the Vanishing 
 Line. 
 
 Let a b, figure 2, be the given intersecting line of the original plane, and 
 c the centre of the picture. 
 
 Draw co parallel, and cs perpendicular, to ab; make co equal to the 
 distance of the picture. Draw o s, so that the angle o s c may be equal to 
 the inclination of the original plane to the picture. Draw sd parallel to ab ; 
 then will sd be the vanishing line, s its centre, and o s its distance. 
 
 For, imagine the triangle o cs to be raised on the picture, so that oc may 
 be perpendicular to it ; then the point o will be the point of sight, and s d 
 being parallel to a b, a plane passing through the line s d and the point o, 
 will be the parallel of the original plane, passing through a b, and inclined to 
 the picture in the angle o s c ; therefore sd is the vanishing line, o s the 
 distance of the vanishing line from the point of sight o, and s the centre of 
 the said vanishing line. 
 
94 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 PROPOSITION XXIV. — PROBLEM XIX. 
 
 Given the Centre and Distance of the Picture , and the Vanishing Line 
 of a Plane , to find the Vanishing Point of Lines perpendicular to 
 that Plane. 
 
 Let a b, figure 3, be the vanishing line given, and c the centre of the 
 picture. 
 
 Through c draw ad perpendicular, and co parallel, to ab. Make co 
 equal to the distance of the picture; join oa, and draw od perpendicular to 
 oa ; then d will be the vanishing point required. 
 
 When the vanishing line a b passes through the centre of the picture, then 
 co will coincide with ab, and as od is drawn perpendicular to ca, both a d 
 and od will be perpendicular to ab; or, ad and od being parallel, the 
 vanishing point d will be at an infinite distance. See figure 4. 
 
 In this case the representations of lines, perpendicular to the plane pro- 
 posed, will all be perpendicular to their vanishing line a b. This is actually 
 the case when we would represent perpendiculars to a plane perpendicular 
 to the picture, or in a vertical picture, upon a plane perpendicular to the 
 horizon. 
 
 Demoyistration. — Figure 3. Imagine the triangle ao d to be turned round 
 a d until it is perpendicular to the picture ; then o will be brought into the 
 point of sight ; therefore the plane passing through o and a b, will be parallel 
 to the original plane, and the line o d, passing through the point of sight o, 
 will be perpendicular to it, and as od meets the picture in o, o will be the 
 vanishing point of all original lines parallel to o d, or perpendicular to the 
 original plane. 
 
AND COMPLETE DECORATOR. 
 
 95 
 
 PROPOSITION XXY. — PROBLEM XX. 
 
 Given the Centre and Distance of the Picture , and the Vanishing Point of a 
 Line , to find the Vanishing Line of a Plane, perpendicular to the origi- 
 nal of the Line whose Vanishing Point is given, and the Centre and 
 Distance of the Vanishing Line. 
 
 Let c, fig. 3, be the centre of the picture, and d the vanishing point given. 
 
 Through c draw da and co perpendicular to da, and make co equal to 
 the distance of the picture: draw do and oa perpendicular to do; and, 
 lastly, draw ab parallel to co; then ab is the vanishing line required, and a 
 its centre. 
 
 This construction here is the reverse of that in the preceding Problem, 
 therefore the same observations apply here, and the same demonstration as 
 in the Problem alluded to. 
 
 PROPOSITJON XXVI. — PROBLEM XXI. 
 
 Given the Centre and Distance of the Picture, and the Vanishing Line of a 
 Plane, through a given Point, to draw the Vanishing Line of another 
 Plane, perpendicular to the Plane whose Vanishing Line is given, and 
 to find the Centre and Distance of that Vanishing Line. 
 
 Let c, figure 4, be the centre of the picture, and a b the given vanish- 
 ing line. 
 
 Find the vanishing point d of a line perpendicular to a plane whose 
 vanishing line is ab, by Problem xix ; join d e, and draw c i perpendicular 
 to d e, cutting d e in f ; then d e is the vanishing line, and f its centre. 
 
 Its distance may be thus found ; let c o, parallel to a b, be the distance of 
 the picture ; from d, as a centre with the radius d o, describe an arc, cutting 
 c i at i ; then f i is the distance of the vanishing line d e. 
 
 Demonstration. — Because the plane, whose vanishing line was required, 
 is perpendicular to the other plane, its vanishing line must pass through the 
 point d of lines perpendicular to that other plane ; but as this vanishing line 
 
96 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 must also passf through the point e, de will be the vanishing line required : 
 and, because a line drawn through the centre of the picture, perpendicular to 
 a vanishing line, gives by its intersection the centre of that vanishing line ; 
 the point f is the centre of the vanishing line de. The rest will be evident 
 by turning up the triangle dao perpendicular to the plane of the picture, 
 then turning the triangle d i f round d e, until g falls upon o, then d g will 
 coincide with d o. 
 
 Corollary 1. -If fc be produced till it meet the vanishing line given in 
 the point b, b will be the vanishing point of lines perpendicular to the 
 original plane of the vanishing line de: for that vanishing point is in the line 
 fc, by the construction of Proposition x. and it is in the vanishing line given 
 by the demonstration of the present Proposition. 
 
 Corol. 2.— If the vanishing lines ab and de meet in g, the points b, d, 
 and g, will be the vanishing points of the three plane angles forming the solid 
 angle of a cube ; these planes being perpendicular to each other, and if d b 
 be joined, bg.gd, and d b will be the vanishing lines of the three planes that 
 contain that solid angle. 
 
 Corol. 3. — The distance of the vanishing line dg is also equal to the line 
 f i ; the point i being in the semi-circumference described upon dg as a 
 diameter. 
 
 Corol. 4.— In the construction of the vanishing lines forming the solid 
 angle, when the point of sight is found for each vanishing line, and the radials 
 of the common intersections of the planes drawn, every two radials at the 
 point of sight will be at right angles to each other, and every two of these 
 radials, which terminate in the intersection of two vanishing lines, will be 
 equal to each other. Hence in every two vanishing lines, if the centre 
 and direct radial be drawn in the one, the direct radial can be found in 
 the other, for the point of sight in this other will be in the circumference 
 of a circle described from the intersection of the two vanishing lines, 
 with a radius equal to the radial joining that intersection and the point 
 of sight given. 
 
AND COMPLETE DECORATOR. 
 
 97 
 
 Thus let G b and gd be the vanishing lines of two planes, and let A and 
 f be their centres, and let h be the point of sight for the vanishing line g b. 
 Draw fi perpendicular to gd; from the centre g, with the radius gh, de- 
 scribe an arc, cutting f i in I ; then f i will be the direct radial. 
 
 Corol. 5. — If the point of position of the eye for each of two of the vanish- 
 ing lines be found, the third will also be found ; for it must be in the inter- 
 section of two circles described from the two vanishing points of the third 
 vanishing line, with radii which are respectively equal to the two adjacent 
 radials. 
 
 Corol. 6 . — The length of any direct radial is a mean proportion between 
 the segments of the vanishing line, on each side of the centre ; because the 
 plane angle, formed at the eye, is a right angle; and hence, as before observed, 
 if we draw the perpendicular from the centre of the vanishing line, and de- 
 scribe a semicirtcle on the said line, the point where it intersects the perpen- 
 dicular will be the point of sight. 
 
 PROPOSITION XXVII. — PROBLEM XXII. 
 
 Given the Centre and Distance of the Picture , and the Vanishing Point of 
 the common Intersection of two Planes , inclined to each other in a given 
 Angle , and the Vanishing Line of one of them , to find the Vanishing 
 Line of the other. 
 
 Let c, figure 5, be the centre of the picture, b g the given vanishing line 
 of one of the planes, and b the vanishing point of their common section, and 
 h the angle of their inclination to one another. 
 
 Find the vanishing line gd of planes perpendicular to the lines whose 
 vanishing point is b, (by Proposition xxv.) and let that vanishing line cut the 
 vanishing line given in g. In gd find the vanishing point e of lpies, making 
 the given angle H with the lines, whose vanishing point is g ; that is, in bcf, 
 perpendicular to g fd, take f p equal to the distance of the vanishing line g d. 
 If c o be parallel to g d, and equal to the distance of the picture, make 
 fp equal to fo; join pg, and draw pe, meeting gd in e, so that the 
 
 2 B 
 
93 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 angle gpe may be equal to h, and draw b e, then b e will be the vanish- 
 ing line required. 
 
 Demonstration. — For imagine the triangle gpe to be turned on the line 
 gd, so that the point p may fall in the perpendicular drawn from c to the plane 
 of the paper or picture ; then p will be the point of sight, and the apex of 
 three imaginary planes, bpg, g p d, d p b, supposed to be raised on the vanish- 
 ing lines bg, gd, db. Then also the three planes, thus supposed to be 
 raised, will be parallel to the three original planes which have bg, g d, d b, 
 for their vanishing lines. Now the plane, whose vanishing line is dg, being 
 perpendicular to the other two by construction, because it is perpendicular to 
 their common section, whose vanishing point is b, will measure the inclination 
 of these two other planes ; therefore, by joining be, be will be the vanishing 
 line required. 
 
 PROPOSITION XXVIII. — PROBLEM XXIII. 
 
 Having the Centre of the Picture , the Vanishing Lines of two original 
 contiguous Planes , and the Centre and direct Radial of one of the 
 Vanishing Lines , to find the Centre and direct Radial of the other 
 Vanishing Line. — The reader to supply the figure himself 
 Let c be the centre of the picture, a b and b d the two vanishing lines, h 
 the centre of the vanishing line, ab and ho its direct radial. 
 
 Draw c m perpendicular to b d, cutting bd at l ; make b m equal to bo; 
 that is, with the radius b o describe an arc, cutting c m at m : then l is the 
 centre and lm the distance of the vanishing line lm. 
 
 PROPOSITION XXIX. — PROBLEM XXIV. 
 
 Given the Centre and Distance of the Picture, the Vanishing Line of a 
 Plane, the Inclination of a Line to that Plane , and the Angle which the 
 Seat of the Line upon that Plane males with the Intersecting Line , to 
 find the Vanishing Point of that Line. 
 
 Let c, figure 5, be the centre of the picture, b g the given vanishing lines 
 
AND COMPLETE DECOKATOK. 
 
 99 
 
 Find d, the vanishing point of lines perpendicular to planes whose 
 vanishing line is eg, and let dc cut bg in s, which is the centre of the 
 vanishing line bg : make s i equal to the distance of the vanishing line bg ; 
 that is, if co, parallel to bg, be equal to the distance of the picture, make si 
 equal to so. Draw ik parallel to bg; and make the angle kig equal to 
 the angle which the seat of the line makes with the intersection. Join dg. 
 Find the centre f and distance fp of the vanishing line d g ; that is, draw 
 p f perpendicular to d g, cutting dg in f ; make g p equal to gi; and 
 draw pe, meeting dg in e, so that the angle gpe may be equal to the 
 inclination of the line to the plane, then e will be the vanishing point of 
 the line required. 
 
 This construction is evident from the foregoing demonstration, and more 
 particularly from the las t. 
 
 When the vanishing line b g passes through the centre of the picture, the 
 original plane and that of the picture will be perpendicular to each other. In 
 this case the vanishing point d will be infinitely distant from the centre of the 
 picture, and consequently the vanishing line d g will be perpendicular to the 
 vanishing line b g ; and because s and c are supposed to coincide, the points 
 f and g will also coincide : the construction will, therefore, be simplified as 
 follows : — ■* ' 
 
 Draw c i, fig. 7, perpendicular to b g ; make c i equal to the distance of the 
 picture. Draw i k parallel to bg, and make the angle kig equal to the 
 angle which the seat of the line makes with the intersecting line. Draw d g 
 perpendicular to bg; make g p equal to g i, and draw p e, meeting dg in e, 
 so that the angle gpe may be equal to the inclination of the line to the plane; 
 then e will be the vanishing point of the line required. 
 
100 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 PROPOSITION XXX. — PROBLEM XXV. 
 
 Given the Vanishing Line of a Plane , the Centre and Distance of the 
 Picture , the finite and indefinite Representations of two Lines in that 
 Plane ; from a given Point in the indefinite Representation , to cut off a 
 Part , which shall represent a Line equal to the Line represented by the 
 
 finite Representation. 
 
 Let vl, figure 8, be the vanishing line, c the centre of the picture, ab the 
 finite representation of a line, and c l the indefinite representation of another; 
 it is required to cut off from c a portion of the line c l, of which the original 
 shall be equal to the original of the line represented by a b. 
 
 Draw c o perpendicular to v l, intersecting vl in the point s, the centre 
 of the vanishing line. Draw c e parallel to v l, and make c e equal to the 
 distance of the picture. Join se, and make s o equal to se. Join o v and 
 ol. Draw the radial om, bisecting the angle vol. 
 
 Draw the straight line bd parallel to vl. From the vanishing point m, 
 ana through the points a and b , draw the straight lines ma, mb, meeting the 
 line bd in the points a and b ; also, from the same vanishing point m, and 
 through the point c, draw the straight line me, meeting b d in c. In bd, 
 make the line cd to ab, as the original of cd is to the original of ab. Draw 
 the straight line dm, intersecting cl in d, and cd is the line required. 
 
 SCHOLIUM. 
 
 This Problem is useful in the representation of solids ; for it sometimes 
 happens, after we have finished the representation of one plane, that we have 
 another to represent, and having already determined the vanishing line from 
 the circumstances of the Problem, and having the centre and distance of the 
 picture, which were necessary in the representation of that plane, we have 
 the same data still, and consequently we may, by these means, represent 
 lines, of which the originals may have any ratio to each other, and thus find 
 the representation of any figure : in short, perform any of the foregoing 
 Problems. 
 
AND COMPLETE DECORATOR. 
 
 101 
 
 Part II.— PERSPECTIVE. 
 
 Of the Representation of Solids. 
 
 All objects whatever, that are formed by art, come under the denomination 
 of some geometrical solid, as a prism, a pyramid, a cylinder, a cone, a sphere, &c. 
 Thus the body of a house is generally a rectangular prism, and the roof may 
 either be a right triangular prism, as in pediment roofs, or the frustum of a 
 pyramid, as in hip-roofs. Or, if the body of a building be a cylinder, the 
 roof may either be a cone or a segment or zone of a sphere, as in domes. 
 The surfaces of these solids are, therefore, triangles, rectangles, circles, & c. 
 Therefore to represent any object of Cabinet Furniture, or building, it will be 
 necessary to represent the several kinds of geometrical solids ; but more 
 particularly the rectangular parallelopiped. And as most objects stand per- 
 pendicular to the horizon, they will be represented as if they stood upon a 
 plane perpendicular to the picture. A cube, or rectangular prism, is 
 perpendicular to the horizon when one of its faces coincides with a 
 horizontal plane; a right cylinder, or cone, is said to stand perpendi- 
 cular to the horizon when the axis is perpendicular to the horizon ; there- 
 fore in representing such objects, the axis will be parallel to the picture, 
 and will therefore be represented by straight lines perpendicular to the 
 vanishing line. 
 
 Having treated of the representations of various kinds of plane figures, 
 nothing now remains to be done in order to represent a solid bounded 
 by a plane figure, but to combine the faces according to their angles of 
 inclination. 
 
 To enable the student to execute this part in the most general manner, 
 he has only to know from having given the centre and distance of the picture , 
 the angle of inclination of two planes , the vanishing point of their common 
 
 2 c 
 
102 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 section , and the vanishing line of one of the planes , in order to he able to find 
 the vanishing line of the other , as also its centre and distance * 
 
 If he has studied the preceding propositions with care, he is already in 
 possession of the principle for accomplishing this object: but as this gene- 
 rality in the method of representing solids is seldom required in perspective, 
 and less perhaps by the designer of cabinet work than by the architect or 
 builder, we shall here trace the principle as applied to a parallelopiped 
 to that of a rectangular prism, according to the various positions of the faces 
 of the solid. 
 
 We must consider, that as a rectangular object is bounded by six faces, 
 which are plane surfaces, and since every plane has its parallel, every two 
 planes may have a vanishing line, since the six faces may be inclined to the 
 picture. 
 
 The number of vanishing lines, and the number of vanishing points, can 
 never exceed three of each kind. 
 
 Moreover, since the solid cannot have more than two of its faces parallel 
 to the picture, it cannot have fewer vanishing lines than two, nor fewer van- 
 ishing points than one, and in this case, where two of the planes are parallel 
 to the picture, the other four planes are perpendicular to the same ; therefore, 
 the vanishing line of each pair of these four planes, must pass through the 
 centre of the picture. 
 
 When two of the faces of the solid are parallel to the picture, the other 
 four faces being perpendicular, have their four edges also perpendicular to 
 the picture ; therefore, these four edges, being parallel to the intersection of 
 the two planes, which passes through the eye parallel to the four planes 
 perpendicular to the picture, have only one vanishing point in the centre of 
 the picture. 
 
 * It may be proper to mention in this note, a simplification of a proposition which leads to the 
 solution of the above Problem, perhaps, in a more simple manner. The proposition which we 
 allude to is this : — “ Given the Centre and Distance of the Picture , and the Vanishing Point of a 
 Line, in a Plane perpendicular to the Picture, to find the Vanishing Point of another Line in that 
 Plane , perpendicular to the Line whose Vanishing Point is given.' 9 
 
AND COMPLETE DECORATOR. 
 
 103 
 
 The remaining two faces, being parallel to the picture, are represented by 
 equal and similar rectangles. In this case, since a line which is parallel to 
 the picture, has its representation parallel to the vanishing line of the plane, 
 passing through the original line ; the eight edges of the solid, which are 
 parallel to the picture, are in the four perpendicular planes to the picture 
 therefore, the representations of these eight lines are parallel to the vanishing 
 lines of these four planes. 
 
 When four of the edges of the solid are parallel to the picture, and per- 
 pendicular to the horizon, and the planes between these edges inclined to 
 the picture, the remaining two planes are perpendicular to the plane of the 
 picture, and parallel to the horizon, but the edges of these two planes 
 are inclined to the picture; therefore, these eight edges being thus inclined, 
 and in planes perpendicular to the picture, have two vanishing points in the 
 vanishing line of the two planes which are perpendicular to the picture, and 
 therefore in the horizontal line ; and as straight lines, perpendicular to a 
 plane which is perpendicular to the picture, are represented by straight 
 lines perpendicular to the vanishing line of that plane, the represen- 
 tations of these four edges are perpendicular to the vanishing line of the 
 horizon. 
 
 Having thus traced the different positions of the representations of lines, 
 frpm the different positions of the faces of the rectangular prism to the plane 
 of the picture; it will be of the greatest use to trace the same positions of 
 representation from the angles of inclination of the planes of the solid, which 
 are all right angles, and from the properties of their vanishing lines, in order 
 to trace the principle through the various stages of position. 
 
 When the centre of the picture falls in the centre of the given vanishing 
 line, we shall find the vanishing point of the four lines perpendicular to the 
 planes, of which the vanishing line is given at an infinite distance, and in a 
 line passing through the centre of the picture perpendicularly to the vanish- 
 ishing line given of the two planes, which are perpendicular to the plane of 
 the picture. 
 
104 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 This indicates that the representations of the four lines are perpendicular 
 to the vanishing line. 
 
 , In this case the edges of the two planes, of which the vanishing line is 
 given, may either have one or two vanishing points; if they have two vanish- 
 ing points, four of the edges of the solid will be parallel to the picture, and 
 the faces terminated by these edges inclined to the picture. 
 
 In the case where there are three vanishing lines, which will form a 
 triangle, and the three vanishing points will be in the vertices of the tri- 
 angles, and consequently there will be two vanishing points in each vanish- 
 ing line. 
 
 Since the vanishing point of lines perpendicular to a plane, of which the 
 vanishing line is given, always passes through the centre of the picture ; there- 
 fore, when one of the vanishing lines is given, and the centre of the picture is 
 in the centre of the vanishing line, the vanishing point of these lines will be 
 in the perpendicular to the vanishing line, at an infinite distance from the 
 centre of the picture ; the other two vanishing lines, which join the vanishing 
 point, and the two vanishing points in the vanishing line given, must be also 
 perpendicular to the same vanishing line given. 
 
 For in performing this operation, we at first draw a straight line from the 
 centre of the picture, perpendicular to the vanishing line given, and set the 
 distance of the picture from its centre, upon the perpendicular ; then the 
 other extremity of this distance is the point of sight. 
 
 When this is done, we draw the radial which joins the vanishing point of 
 the common section of the two original planes and the point of sight, and 
 from the point of sight we draw another radial perpendicular to the first, 
 and consequently perpendicular to the vanishing line given, and then a line 
 through the vanishing point found by the second radial perpendicular to the 
 vanishing lines ; and this last perpendicular will be the vanishing line of a 
 plane perpendicular to the line of which the vanishing point is given, and the 
 point of intersection of the two vanishing lines, will be the centre of this last 
 vanishing line, and the length of the last radial is its distance; so that the 
 
AND COMPLETE DECORATOR. 
 
 105 
 
 point of sight for this new vanishing line will be in the first vanishing line, as 
 the two vanishing lines intersect each other at right angles. 
 
 We now proceed to set off the angle of inclination of the two planes, at 
 this new point of sight, for the second vanishing line, so that the first vanish- 
 ing line may be one side of the angle ; and by producing the line forming the 
 other side of the angle to meet the second vanishing line, we shall have the 
 vanishing point of all those lines which are perpendicular to the common 
 section of the two original planes, and which form their angle of inclination. 
 
 This circumstance will enable the artist to represent the lid of a box at 
 any angle of inclination with the top ; to represent a ladder in any position to 
 the horizon, or to represent the rafters of a building perpendicularly to the 
 wall-plates, and making a given angle with the horizon, or top of the 
 building. 
 
 But if all the faces of a solid are planes, which are at right angles to each 
 other, it will not be necessary to find more vanishing lines than that which is 
 already given ; for in this case, though we may have a second vanishing line, 
 it would not be used in the operation; for the angle of inclination of the 
 planes being a right angle, and when this right angle is set off from the second 
 point of sight, so that the perpendicular radial, which is part of the first 
 vanishing line, may form one side of the angle, the other side of the angle 
 must therefore be parallel to the second vanishing line, and consequently 
 perpendicular to the first. This therefore indicates that the vanishing 
 point is at an infinite distance, and therefore the representation of all lines, 
 perpendicular to the plane of which the vanishing line is given, are also 
 perpendicular to the same vanishing line. So that in representing a rect- 
 angular solid, which can have only three vanishing points at most, and of 
 which one of the vanishing points is not in the centre of the picture, two of 
 them must be in the vanishing line given ; and the third series of lines, which 
 represent the edges, are perpendicular to the said vanishing line, 
 i These preliminary observations will be evident in the examples of repre- 
 senting solids in perspective. 
 
 2 D 
 
106 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 Observation, 
 
 With respect to general methods of practice in the representation of 
 objects comprehended under the forms of rectangular parallelopipeds, 
 or rectangular prisms, deduced from the immediately preceding 
 observations. — 
 
 The first consideration is, to fix on the point of view ; this will determine 
 the centre and distance of the picture ; therefore, in what follows, we shall 
 always suppose the centre and distance of the picture to be given, the 
 determination of these will therefore form no part of the operation in repre- 
 senting rectangular objects in perspective. 
 
 There are two positions, in respect to the picture, of representing rectan- 
 gular bodies, commonly called rectangular parallelopipeds, or rectangular 
 prisms ; the one is, by considering two of the faces parallel to the picture ; 
 and consequently, the other four perpendicular to the same. Hence, in this 
 case, the vanishing point of the four parallel lines, in the meeting of the four 
 planes, which planes being perpendicular to the plane of the picture, will be 
 in the centre of the picture ; the outlines of the figures representing the other 
 two planes, which are parallel to the picture, are rectangles, which have their 
 sides parallel and perpendicular to the vanishing line of the four planes which 
 are perpendicular to the picture. 
 
 The common sections of those two planes, of which the vanishing line is 
 given, and of the two planes which are parallel to the picture, are represented 
 by straight lines parallel to the vanishing line, and consequently, the other 
 two edges of the planes, which are parallel to the picture, are represented by 
 lines which are perpendicular to the vanishing line. 
 
 The other position of the solid is, by considering two of the planes to be 
 perpendicular to the picture, and each pair of the other four inclined. In 
 representing such a solid, only one vanishing line, as we have already seen, 
 will be necessary, and this vanishing line will be that of the two faces of the 
 solid which are perpendicular to the picture, therefore each vanishing point 
 will be the vanishing point of four lines, viz. those of the opposite parallel 
 
AND COMPLETE DECORATOR. 
 
 107 
 
 planes. Therefore, eight of the edges or arrises of the solid have their 
 vanishing points in the vanishing line; because the vanishing line is the inter- 
 section of a plane with the picture passing through the eye, parallel to the 
 two perpendicular planes. The arrises of the four planes, which are inclined 
 to the picture, being perpendicular to the other two planes, which are 
 perpendicular to the picture, are therefore parallel to the picture, and 
 are consequently represented by lines which are perpendicular to the 
 vanishing line. 
 
 There are, therefore, three positions of the object in respect of the picture: 
 one is, when all the sides are inclined to the picture ; a second is, when two 
 parallel faces are perpendicular to the picture ; and the third, when two 
 parallel faces are parallel, and consequently the other four perpendicular to 
 the picture. 
 
 The first of these is called the triple inclined position , or simply the 
 inclined position of the picture ; the picture being inclined to all the three 
 pairs of parallel planes, or parallel faces of the object. The second is called 
 the single perpendicular position of the object , or only the perpendicular 
 position, the picture being perpendicular only to one pair of parallel sides of 
 the object. The third is called the parallel position , or double perpendicular 
 position of the picture , the picture being parallel to one pair only of the faces 
 of the object, or perpendicular. to two pair of faces. 
 
 The art of representing rectangular bodies having one pair of its faces 
 parallel to the picture, has been denominated parallel perspective . 
 
 In all positions of the picture, its centre and distance must always be a 
 part of the data; and where the object is inclined, the inclination of one of its 
 sides to the plane of the picture must be given, or some other equivalent. 
 
 Practice in representing a Rectangular Object according to the Triple 
 
 Inclined Position. 
 
 As the position of the object to the picture depends upon choice, to save 
 a great deal of useless trouble, it is customary to assume the vanishing line of 
 
108 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 one pair of its parallel faces, and the centre of the picture, as also to assume 
 the vanishing' point of a pair of sides of the rectangles of these two parallel 
 faces of the object; this assumed point must therefore be in the assumed 
 vanishing line ; for the vanishing points of all lines in the same plane are in 
 the vanishing line of that plane. This vanishing point is, therefore, the 
 vanishing point of the line of concourse of two faces of the object that are 
 reciprocally perpendicular to each other, viz. of one of the faces of which its 
 vanishing line is assumed, and one of the four perpendicular faces to that of 
 which the vanishing line was assumed. 
 
 We have now given the centre and distance of the picture, and the vanish- 
 ing line of a plane, to find the vanishing point of lines perpendicular to that 
 plane ; and this being found with the vanishing point of the line of concourse 
 of two of the faces, will give two of the vertices of the triangle formed by the 
 three vanishing lines of the three pair of parallel faces of the object. 
 
 In order to find the third vertex of the vanishing triangle, find the point 
 of position of the eye to the vanishing line ; join this point of position to the 
 assumed vanishing point, and from the point of position draw a radial perpen- 
 dicular to the line thus joined, and the vanishing point made by this radial is 
 the vanishing point of the vertex of the remaining angle of the vanishing 
 triangle. These three vertices are the vanishing points of the twelve arrises 
 or lines of concourse of the faces of the solid, there being four parallel lines 
 for each of the three vanishing points. 
 
 Therefore, by drawing a straight line from each of the two vanishing 
 points in the vanishing line given, to the third vanishing point found, we shall 
 have the three vanishing lines ; two of these are, however, only necessary. 
 
 But in order to represent the three faces which constitute the solid angle 
 next to the eye of the observer, it will be necessary to find the centre and 
 distance of one of the last drawn vanishing lines, which being done, find the 
 representation of each rectangle, of which the vanishing line, its centre and 
 distance, as well as the vanishing points of the sides of each, are both now 
 given. The representation of the remaining face or rectangle will be found 
 

 
 
 To 
 
 Fublisked br H.Fislier. Son k V. Caxtan.Iandon. Ju2y.l835. 
 
AND COMPLETE DECORATOR. 
 
 109 
 
 by only drawing lines from the ends of the lines in the adjacent edges of the 
 trapeziums, which represent the two first rectangles to the vanishing points 
 of this last rectangle. 
 
 To represent or find the image of a rectangular solid, its position in 
 respect of the eye, and the picture being determined ; by knowing the distance 
 of the eye and the distance of the vertex of the nearest solid angle, each from 
 the picture, and having given in position in the plane of the picture, its centre 
 and the vanishing line of one of the faces. — This single example very nearly 
 embraces the whole of Perspective ; we shall therefore, in order to be distinct, 
 number the propositions. 
 
 ^ • We hav e given the distance of the picture, c its centre, l the 
 vanishing point of a line ; to find the vanishing line of a plane, perpendicular 
 to the line of which the vanishing point l is given. (See Prop, xxv.) 
 
 Through the points L and c draw the straight line ls 2 , and cn perpendicular to ls 2 . • Make 
 cn equal to the distance of the picture, and join ln. Draw ns 2 perpendicular to ln, and 
 through s 2 draw vm perpendicular to ls 2 , and vm is the vanishing line here required. 
 
 In this part of the operation we have also found s 2 the centre, and a line s 2 n equal to the dis- 
 tance of the vanishing line vm, as also v the vanishing point of the line of intersection of the 
 planes, of which vm and vl are the vanishing lines. 
 
 This vanishing line being now found, will be enumerated among the data 
 in the following Problem, as one of the given parts. 
 
 No. 2.— Given in the picture in respect of two lines in a plane, making a 
 given angle with each other, vm the vanishing line of their plane, s 2 its 
 centre, and a line s 2 n equal to its distance, to find the vanishing point of the 
 other line. 
 
 Draw s 2 p 2 perpendicular to vm, (or, since cs 2 is already perpendicular to vm, prolong, if neces- 
 sary, cs 2 to p 2 .) Join vp 2 , and make the angle vp 2 m equal to the angle which the two 
 lines make with each other; and as this angle in the present case is a right angle, draw p 2 m 
 perpendicular to vp 2 , and m is the point here required. 
 
 This vanishing point being found, it will make a part of the data of the 
 following Problem. 
 
 No. 3. — Given in the picture of two straight lines in a plane, their vanish- 
 ing points l and m, to find the vanishing line of the plane. 
 
 2 E 
 
no 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 Join the points L and M, and lm will evidently be the vanishing line required; for the vanishing 
 points of all lines in the same plane are in the vanishing line of that plane. 
 
 By these operations, one grand Problem, which embraces the whole, is solved. The Problem which 
 we now allude to is as follows : — 
 
 Given with respect to two planes, making a given angle with each other, the centre and distance 
 of the picture, the vanishing line of one of the planes, and the vanishing point of their line of 
 concourse, to find the vanishing line of the other plane. 
 
 This Problem has been solved in finding the vanishing line lm, therefore vl and vm are the 
 vanishing lines of the two planes of the solid, and v the vanishing point of their line of con- 
 course, or common section. 
 
 But in the solution of this Problem we found vm the vanishing line of a plane perpendicular to the 
 line of concourse of the two planes, of which the vanishing lines are vl, lm. But since iu 
 each solid angle of a rectangular solid, any one of the three faces is perpendicular to the line 
 of concourse of the other two, therefore vm is the vanishing line of the third plane. 
 
 The object being rectangular, supersedes the necessity of finding the third vanishing line. In con- 
 sequence of v M being the vanishing line of the third plane, y and M are the vanishing points of 
 the remaining two arrises which are common to the third plane and each of the two first. 
 
 In these operations we have found the radials P 2 v and p 2 m of the sides of the rectangles in the 
 two planes of the solid, which have v M for their vanishing line ; w r e must now find the radials 
 of the plane which have vl for their vanishing line. We shall, therefore, proceed to enume- 
 rate the Problem. 
 
 No. 4. — Given in the picture with respect to a rectangle, the vanishing 
 points of the sides, and the lengths vr 2 and ln of the two radials, to find the 
 radials in position to the vanishing line v l. 
 
 From the point L, as a centre with the distance ln, describe an arc; and from the point v, as a 
 centre, with the distance vp 2 , describe another arc, intersecting the former in p. Join pv 
 and pl, then p v and pl are the two radials which were to be found. 
 
 It is evident, that in the triangle formed by the three vanishing lines, a line drawn from any 
 one of the vertices, through the centre of the picture, will be perpendicular to the opposite side of 
 the triangle, and this can very easily be proved from the geometry of solids. 
 
 For if a straight line be drawn from the vertex of any angle of a triangle, perpendicular to the 
 opposite side, dividing this side into two segments ; and if a semicircle be described on the perpen- 
 dicular as a diameter, and if the mean proportional between the segments be applied as a chord in 
 the semicircular arc, so that the point of division may be one extremity of the chord, and if from 
 the other extremity of the chord a perpendicular be let fall upon the diameter, a straight line 
 drawn from any one of the three vertices, through the foot of the perpendicular, will be perpen- 
 dicular to the opposite side of the triangle. 
 
 Hence, with respect to a plane, and a line perpendicular to that plane, if we have the centre 
 of the picture, the vanishing line of the plane, and the vanishing point of the line, a straight line 
 drawn through the vanishing point of the line, and through the centre of the picture, will meet or 
 intersect the vanishing line in its centre. 
 
AND COMPLETE DECORATOR. 
 
 Ill 
 
 Having thus obtained the three vanishing lines of the planes of the nearest solid angle, we 
 shall now proceed first to find the image of the vertex. 
 
 No. 5. -Given the distances from the eye, and from a point each to the 
 plane of the picture, and in this plane, its centre, and the seat of the point, to 
 find the image of the point. 
 
 Join cv, and the representation of the point will be in the line Cv; and here we might proceed 
 with the operation, but to prevent the confusion of lines which would arise in the course of so 
 many operations, the reader is referred to the separate Diagram, No. 2; where, by first 
 drawing cv, and making cv equal to cv in the figure, he may proceed with No. 2 as follows: 
 draw c n' and vd' parallel to each other, making any convenient angle with vc'. Take cW 
 equal to the distance of the picture or any part of it, and take vd’ equal to the distance of the 
 point from its seat, or the same part of this distance that c'n' is of the distance of the picture. 
 As we have sufficient room on the surface we have to draw upon for this subsidiary operation, 
 we shall employ the whole distance, therefore draw n'cT, intersecting v'c' in a. Transfer the 
 distance va to the figure of the solid, to he represented from v to a in the straight line vC. 
 We have now the image of the vertex of the nearest solid angle, and by joining the point now 
 found with each of the vertices v, l, m of the triangle formed by the vanishing lines, we shall 
 have a L, «v, and am, the indefinite images of the three lines of concourse in the plane 
 of the picture. 
 
 We have, therefore, all the vanishing lines, the vanishing points, and the radials of these 
 points for two of the vanishing lines ; as also the image of the vertex of the nearest solid angle, 
 and theiefoie we have for each of two adjacent rectangular faces of the solid, the vanishing line of 
 toe indefinite images of the two sides of the rectangle of that face ; but for want of an intersecting 
 line, we cannot cut off from the two indefinite images a portion from each, so that the original of 
 the parts may be the length of the arrises or line of concourse of the adjacent faces of the solid. 
 
 We must, therefore, proceed to find an intersecting line, or line on which the measures may be 
 applied. There is no occasion, however, to find the real intersecting lines, for if we have the 
 indefinite image of a line perpendicular to the picture, the intersecting point of the line, and the 
 image of a point in this line, we can find such measures as may be applied upon a line drawn 
 through the indefinite image of the point, parallel to the vanishing line ; and as would be found by 
 the real intersecting line in a more convenient manner, from this consideration that the whole 
 distance between a plane passing through the eye parallel to the picture, and a point in space is to the 
 distance between the eye and the picture, as the length of a line parallel to the picture passing through 
 that point is to the length of a line passing through its image parallel to the vanishing line. There- 
 fore if the line which passes through the original point be equal to any one of the three dimensions 
 of the solid, the length of the line found, as the fourth term of the proportion, will be the length of 
 the side to be applied on the parallel passing through the image of the point. 
 
 To do what has now been explained, is equivalent to finding the image of a similar solid and 
 similaily situated, having the vertex of its nearest solid angle in the point a, from dimensions which 
 would be those found by the proportion just stated. Or by calling the perpendicular distance 
 between the eye and a plane passing through the vertex of the nearest solid angle parallel to the 
 picture, the entire distance, and calling any line from the intersecting point before the picture an 
 
312 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 entire indefinite line ; the distance between the intersecting and vanishing points, the entire inde- 
 finite image ; and moreover calling that part of the entire indefinite image of a straight line 
 between the image of a point and the vanishing point, the indefinite part of the entire image from 
 the image of the point. 
 
 Then any dimension of the solid proposed to be delineated, and the corresponding dimensions 
 of the similar solid which touches the picture, will be to one another in the ratio of the entire 
 indefinite image of the line passing through the vertex of the nearest solid angle perpendicularly to 
 the picture, and the indefinite part from the image of the vertex ; that is, in the ratio of cv to c a. 
 Let the lengths of the three arrises between every two faces of the solid angle be respectively 
 equal to the three lines r', s', t'; transfer the three lengths r', s', t' to No. 1, upon the line 
 v d! respectively, from v to r , from v to s', and from v to t'. Draw a! w parallel to v d' , and 
 join dr' , ds , dw, intersecting a w in the points u , v , w , then the lengths corresponding to 
 r', s', t are respectively a'u, a'v', and a'w'. 
 
 These are the dimensions of the body which answer to the intersecting line in the figure, 
 passing through a parallel to the vanishing line. 
 
 Hence we may suppose, that a'u is the length of the arris to be represented in the line a L, a'v the 
 length of the arris to be represented in the line a v, and a'w the length of the arris to be 
 represented in the line am. 
 
 No. 6. — Given in the picture of two lines meeting in a point, one finite, 
 and the other indefinite ; the two representations and their radials to cut 
 off a part from the indefinite image, so that the originals of the finite images 
 may be to one another in a given ratio. 
 
 In the radials pl and p v, from p, the point of position of the eye to the vanishing line vl, make 
 P h and P j respectively equal to the lines r', s', or in any ratio of these lines, as |, J, J, &e. 
 and complete the rectangle p hij. Draw the diagonal pi, and prolong it to meet the vanish- 
 ing line v L in li; find q in the same manner with regard to the vanishing line vm; that is, 
 make p 2 /, P 2 o in the ratio of the lines s' and t' ; complete the rectangle p 2 lmo, and draw 
 thediagonal p ~m, and produce it to q. 
 
 In the vanishing line lv, make L d' equal to the radial lp; and through a draw au parallel to vl. 
 Make au equal to a'u' in the subsidiary figure; join du, intersecting aL in d; join dv and 
 ak, intersecting each other in c; join lc, and produce lc to meet av in the point b; join 6 m 
 and aq, intersecting each other in g; and lastly, join xg, and prolong xg to meet aM in f; 
 then abcefg is the rectangular prism required. 
 
 The perspective representation of any parallelopiped would have been 
 found in the same manner, by finding the vanishing lines so as to answer the 
 inclination of the planes, instead of the right angles in the construction. 
 
AND COMPLETE DECORATOR. 
 
 113 
 
 PRACTICAL EXAMPLES OF SOLIDS. 
 
 The diagram 5, plate xn, exhibits the representation of a cube. 
 
 Having by the preceding principles found ab, a c, the perspective representations of the two front 
 lines of the base of the cube ; prolong one of them as ab, to meet the intersecting line PQ in 
 the point H. Draw Hi perpendicular to PQ and make hi equal to the height of the solid, 
 that is equal to the side of the cube, and as v is the vanishing point of ab, v will also be the 
 vanishing point of the opposite and parallel side: therefore join hv, and ad and be perpen- 
 dicular to P Q, meeting I v in the points d and e. Again, because L is the vanishing point of 
 the side a c, L will also be the vanishing point of the side opposite and parallel to a c, there- 
 fore join a L and draw eg parallel to H I meeting cZl in g. Draw g v and e L meeting each 
 other in the point/, then baegfe will be the representation of the solid. 
 
 Figure 6, plate xn, gives the representations of two rectangular parallelo- 
 pipedons attached to each other, the one being higher than the level of 
 the vanishing line, and the other lower. 
 
 Having put the front lines of the bases into perspective as before, prolong one side of each 
 base to the intersecting line, and draw a perpendicular from each intersecting point to the inter- 
 secting line : call each of these perpendiculars a line of heights. Set the height of each solid 
 upon its respective line of heights, and complete the apparent sides and the top of each solid as 
 in figure 5, and we shall have the representation of both solids thus attached. 
 
 These solids may represent two parts of a building at different heights, as is frequently the 
 case in complex objects, or it may represent a piece of furniture of the same forms. 
 
 Fig. 9, represents a kind of forum building of the most simple con- 
 struction; the walls are all rectangular parallelopipedons, and consequently 
 represented as in figures 5 and 6. 
 
 The main building, as well as the barn and other offices, are roofed in the same manner with 
 gables, and all at the same angle of inclination, therefore all the gables which stand parallel 
 will have the same two vanishing points ; and as the walls are at right angles to one another, and 
 perpendicular to the horizon, the vanishing points of the gables, which are in one set of parallel 
 planes, will have their vanishing points in the line drawn through the vanishing point of the level 
 lines of these walls, perpendicular to the vanishing line, and those in the other series of parallel 
 walls will have their vanishing points in a line drawn through the vanishing point of the level 
 lines of these walls, perpendicular to the vanishing line. 
 
 Let v L be the vanishing line, and P Q the intersecting line of the horizontal plane, upon which 
 the objects are supposed to be placed, also let c be the centre of the picture, and c o its 
 distance, o being the place of the eye. 
 
 2 F 
 
114 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 To find the vanishing points of those gables which are in the planes that have their vanishing 
 lines on the right hand of the centre of the picture. Take the distance between the vanish- 
 ing point L and the eye at o, and transfer this distance upon the vanishing line vl: from 
 the point of distance make an angle with v L equal to the inclination at the plane of the 
 roof to the plain of the horizon, and the point where the other side of this angle meets 
 the vertical vanishing line, passing through L, will be the vanishing point of one side of 
 the roof. 
 
 By setting the distauce between the vanishing point L, and the vanishing point now found above 
 L, downwards from L upon the vertical vanishing line passing through l, we shall have the 
 vanishing point of the remote lines of all the gables in the planes of which their vanishing 
 line is the perpendicular to v l, passing through the vanishing point l. 
 
 Again, transfer the distance v o of the vanishing point v, and the eye o upon the vanishing line 
 vl, and from this point of distance make an angle with the vanishing line l v, equal to the 
 angle which the sloping sides of the pediments in the w'alls, at right angles, to the former, 
 makes with the horizon, and let the other side of this angle meet the perpendicular to v l, 
 passing through the point v; and as the angle thus constructed, may be upon either side of 
 the vanishing line tl, the point where the side of this angle meets the perpendicular drawn 
 through v to the line v L, will be above or below the point v accordingly, so that upon which- 
 ever side of v this happens, the other vanishing point must be set upon the contrary side. 
 
 By means of the two vanishing points found on the right hand, we shall 
 have the vanishing point of all the gables on the right hand, and by means of 
 the two vanishing points found on the left hand, we shall have the vanishing 
 points of all the gables on the left hand, of which the two sloping sides of 
 each, and the sloping sides of every two, make the same angle with the 
 horizon. 
 
 The roof of the pent-house on the right hand has only one sloping side; 
 this sort of roof is called a shed roof. As the plane of the two ends of this 
 roof of the two sloping walls are parallel to the planes of the gables of the 
 dwelling-house, the vanishing point of the sloping top will be in the vertical 
 vanishing line drawn through l ; the vanishing point of this sloping side is 
 found exactly in the same manner as either of the sloping sides of the roof of 
 the dwelling-house, or of the other gables which have the same inclination. 
 
 The upper vanishing point in the vertical vanishing line, is the vanishing 
 point of the two sloping sides of the visible plane of the roof of the dwelling- 
 house, and of the small triangular front which terminates the door over the 
 stable ; and the vanishing point at the same distance below l, is the vanish- 
 
AND COMPLETE DECORATOR. 
 
 115 
 
 ing point of the two sloping edges of the side of the roof which is not visible, 
 also the vanishing point of the other sloping line which terminates the door 
 
 over the stable. 
 
 The vanishing point of the two equal sloping lines which terminate the 
 gable of the stable, are in the vertical vanishing line passing through v, 
 equally distant from v . 
 
 As two of the planes of the chimney shaft of the dwelling-house are 
 parallel to the gable of the same house, these planes will be intersected by the 
 inclined plane of the roof; and the line of intersection, and the plane of the 
 horizon, will form the same angle as each of the two sloping sides which ter- 
 minate this inclined side of the roof, therefore the intersection of the visible 
 planes of the chimney shaft, and inclined side of the roof, will have the same 
 vanishing points as the two lines which terminate the visible inclined side of 
 
 the roof. 
 
 As the vanishing points of all lines in the same plane are in the vanishing 
 line of that plane, if the vanishing points of any two lines in any plane be 
 given, the vanishing line of their plane will be found by drawing a straight 
 line through the two vanishing points. 
 
 To apply this observation to the present purpose, the vanishing point v of 
 the line w r hich terminates the top of the wall, is also the vanishing point of a 
 line in the visible inclined plane of the roof, and the vanishing point of the 
 line which terminates this visible inclined side of the loof, and the visible 
 gable, is therefore another vanishing point in the visible inclined plane of the 
 roof; therefore as the inclined planes of the roofs run in two different direc- 
 tions, they will have two different vanishing lines. Let these two vanishing 
 lines be found, and the point where they intersect each other will be the 
 vanishing point of the common intersection of the two inclined planes, to 
 which the two vanishing lines belong. This vanishing point will be the 
 vanishing point of the visible side of the roof of the dormer, in the front over 
 the stable, with the visible side of the root of the stable. 
 
116 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 PLATE XIV. 
 
 Pig. 1.— Shews the method of representing a cone, p being the point of 
 sight, s the station point, v 1 v 2 the vanishing line, 1 1 the intersecting line. 
 In this diagram two cones are represented ; that on the right hand is done by 
 means of the plan, and that on the left by means of the measures applied 
 upon the ground line. 
 
 The first of the representations is to find the perspective projection of the plan, and the 
 perspective of the centre of the circle ; from this point draw a perpendicular to the vanishing 
 fine, then the height of this perpendicular being found, we shall have the perspective repre- 
 sentations of the axis ; then join the upper extremity of this axis to each extremity of the curve 
 <>f the ellipse, which represents the base of the cone, as a tangent, and these two tangents and the 
 
 convex part of the curve intercepted between them, will be the perspective representation of 
 the cone. 
 
 big. 2. Exhibits the method of representing a pyramid operating in the 
 same manner as in the cone. In this, as in the former examples of a cone, 
 there are two representations of a pyramid, the one with and the other with- 
 out the plan ; the base is first represented, the centre of the base next, then 
 the summit of the pyramid, then join this summit to the points of concourse 
 of the sides of the trapezium representing the base, and we shall thus have 
 the perspective representation required of the pyramid. 
 
 The representation of the summit of the pyramid, or that of the cone, may be found thus : 
 Draw a straight line from any convenient point in the intersecting line perpendicular to the 
 vanishing line, then upon this perpendicular set the height of the axis. Join the foot of the 
 perpendicular to the foot of the axis, and prolong the line thus joined to meet the vanishing 
 me. Join the upper extremity of the perpendicular end, and the point found in the vanishing 
 me, and the line thus joined will cut the axis in the summit. 
 
 Fig. 3. Exhibits the method of putting an isosceles roof in perspective 
 intersected by a cylindrical wall. The vertical planes of the wall are here 
 found by means of the plan: v 1 v 2 is the vanishing line of the horizontal 
 planes , v v the vanishing line of the vertical parallel planes, two of which 
 are the gables ; v 3 is the vanishing point of one of the sloping sides, which 
 terminate the top of each gable. 
 
AND COMPLETE DECORATOR. 
 
 117 
 
 wf perpendicular to the intersecting line is the height of the ridge line of the roof, the 
 point w being found by producing the ridge line on the plan below. The line gf is the inter- 
 section of the plane of the roof with the plane of the picture. The circular base of the wall 
 is here drawn on the plan. The circle on the plan is here supposed to be divided into eight 
 equal parts ; first into four by two diameters, each parallel to one of the sides of the plan of 
 the building, the one being parallel to one side of this plan, and the other parallel to the other 
 side of the same, and consequently these two diameters are at right angles to each other ; then 
 a square is circumscribed about the circle, so that the sides may be parallel to the plans of the sides 
 of the wall, and the diagonals of the square being drawn will divide the circumference of the 
 circle also into four equal parts : so that the whole will be divided into eight equal parts. 
 
 Through the point where one of the diagonals intersects the circumference, draw a line 
 parallel to the side line, to meet the intersection or ground ; the line thus drawn being in the 
 visible plane of the roof. In the same visible plane of the roof draw a tangent to the circle 
 parallel to the ridge line, to meet the same intersecting or ground line. 
 
 From these two points in the intersecting line draw perpendiculars to the vanishing line 
 to meet the intersection gf of the visible side of the roof. From the points b, d, where these 
 perpendiculars meet the intersection gf of the sloping visible side of the roof, draw lines to the 
 vanishing point v 1 . Also from each of the visible extremities of the side of the square next to 
 the front in the visible side of the roof, draw a line to the station point to meet the intersecting 
 line; then draw two lines perpendicular to the intersecting line, to intersect the line du in the 
 points r and u. 
 
 From the points r and u draw lines to v 3 to meet the line drawn from f towards v 1 repre- 
 senting the ridge, and thus we have a trapezium, in which the portion of the ellipse which 
 represents the intersection of the cylindric surface and the plane of the roof is inscribed. 
 Draw the two diagonals su and sr, and draw a line from v 3 through s to meet the line ur, 
 and this point in Mr is one of the points through w'hich the curve of the ellipse must pass; 
 two other points are in the line representing the ridge, and two other points are in the diagonals 
 su and sr, so that we have five points in the representation of the visible part of this inter- 
 section of the cylindric surface, and the plane of the roof. Draw a curve through these five 
 points, and the representation of this intersection will be complete. 
 
 Then drawing the representation of the horizontal circle which terminates the top of the 
 cylindric surface, and drawing the two vertical tangents to this circle, we shall have the repre- 
 sentation of the visible part of the cylindric wall above the roof. 
 
 To give the most clear idea of this subject, the two elevations are given below at E and F, 
 with the intersections of the principal points. To those who have a clear idea of the subject, 
 these elevations may be dispensed with. 
 
 Fig. 4. — Exhibits the process of finding the perspective representation 
 of the two diagonal ribs of a cylindric ground vault. This is found by 
 applying the measures to the intersecting line. — Fig. 5, exhibits the perspec- 
 tive representation of a cylindric ground vault, the external and internal 
 cylindric surfaces being both concentric cylindrical surfaces. 
 
 2 G 
 
118 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 PLATE XV. 
 
 Fig. 1. — Exhibits the method of finding the perspective representation 
 of a dome supported upon pendentives. Fig. 1, No. 2, is the plan of the 
 dome, and fig. 4, No. 1, the section. From this the several heights to be 
 applied to the perspective drawn, fig. 1, are found, the corresponding 
 points are marked, 1, 2, 3, 4, 5, both in the section and in the representation. 
 In the perspective drawing, the heights are applied upon the intersecting 
 lines, and by this means the representation of the circles which terminate 
 the pendentives are found. 
 
 As to the base of the dome, it is a horizontal circle, and is therefore put in perspective 
 in the usual manner of representing a circle in such a position. One of the faces of the object 
 being parallel to the picture, the objects in this face, when represented in perspective, are similar 
 to the figures in the original face. * 
 
 Fig. 2. — Exhibits the representation of a dome divided into coffers or 
 pannels. Fig. 2, No. 3, is the plane of the same. The base of this dome 
 is represented by the dotted circle below. 
 
 Fig. 3. — Shews the representation of square posts at equal distances 
 from each other. Fig. 1, No. 1, exhibits the heights and distances of the 
 posts, r is the vanishing line of the tops of each row, found as usual by 
 bringing down the distance of the eye on the vanishing line, and making the 
 angle equal to that in No. 1. The measures in the base of No. 1, are 
 applied on the intersecting line. 
 
 PLATE XVI. 
 
 In the twelve following diagrams let v v be the vanishing line, 1 1 the inter- 
 secting line, c the centre of the picture, c p its distance, and p the point of 
 sight in relation to the vanishing line v v : here observe, that wherever s is 
 placed in p c the distance p s is equal to the height of the picture ; so that 
 the distance from s to the intersecting line is equal to the distance of 
 the picture. 
 
AND COMPLETE DECORATOR. 119 
 
 Hence from what has now been observed, c p is perpendicular and 1 1 
 parallel to v v. 
 
 In order to make one connected operation, the planes of each of the twelve diagrams are 
 placed contiguous to each other, so that when each operation is performed, the vanishing plane, 
 the plane of the juncture, and the original or ground plane, may be all brought into their real 
 positions, and the eye being placed in the point of sight, the spectator would see the originals as 
 he would in nature, or as they would appear as described in the definition of perspective. 
 
 The reader is to observe once for ail, in the twelve diagrams above alluded to, that as the 
 lines vl, n, and cp, and the points c and p, are given in position, it will not be necessary to 
 mention them in the enunciations of the problems. 
 
 PROBLEM. 
 
 To find the representations of lines perpendicular to the intersecting 
 line. 
 
 METHOD FIRST. 
 
 Let ba and rfc be two original lines: cut n in the points e and g, by lines drawn towards s 
 from the points b and a, and again cut n in k by producing ba. Join kc, and cut Ac in a' and 
 b' by the lines e a', gb gb' perpendicular to n, and a' b' will be the representation of a b. In the 
 same manner c' d' , the representation of cd, will be found. 
 
 » , 
 
 METHOD SECOND, FIG. 2. 
 
 Cut the line ii in k by producing ba, and in e and g by making ke equal to ka , and kg equal 
 to kb. Cut the vv mn by drawing p n parallel to ac. Join kc. Find the points a' U in up, by 
 joining en and gn and a'b', will be the representation of ah; c' d’ found in the same manner, will 
 be the representation of cd. 
 
 PROBLEM. 
 
 To find the representation of lines that are obliquely situated to the inter- 
 secting line. 
 
 METHOD FIRST, FIG. 3. 
 
 In ii find the point e by producing ba, and the points m and n by the lines as, b s. 
 
 In vv find the point n by drawing pn parallel to ba. Join en; and find a'b' by the lines 
 ma , nb perpendicular to n, and a b' will be the representation of ab. In the same manner, and 
 by means of the same vanishing point, c' d' the representation of erf will be found. 
 
 METHOD SECOND, FIG. 4. 
 
 In ii find the points mn, by drawing am and bn perpendicular to ii, and the points e and g 
 by making me equal to ma, and ng equal to nb. In vv find the point r by making E r equal to 
 cp. Find the point a' by joining me and er, and the point b' by joining nc and gr: then 
 a' b' joined is the representation of a b. In the same manner, and by the same vanishing point r, 
 c'rf', the representation of the line erf is found. 
 
120 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 PROBLEM. 
 
 To find the representation of a triangle abc, fig. 5. 
 
 Find the points in the intersecting line, where the prolongation of the accessible sides would 
 meet it; find the vanishing points of these sides and their indefinite representations, by joining the 
 intersecting and vanishing points. In each indefinite representation, find the extremity of each end 
 of the remaining side, and join the two points thus found, and the triangle will be the representa- 
 tion of the given triangle. 
 
 Thus, in n find ef, by producing cb and ab. Find n and r, the vanishing points of ab, be by- 
 drawing p n, p r, parallel to ab, cb. Find the indefinite representations fn and er by joining/n 
 and er, and the points a! and c, by drawing as and cs> gives the triangle a'b'c the representation 
 of a be. 
 
 The triangle inay also be found as in fig. 6, in the same manner as has been shewn in fig. 4. 
 
 PROBLEM. 
 
 To find the representation of a square, fig. 7, having its sides perpendicular 
 and parallel to the intersecting line. 
 
 In II find the points e,f, by producing ba and cd, and the points g, h i, by drawing s a, s d, 
 sc. Join ec and/c. In ec and /c find the points a' d' c by drawing g a, lid and ic perpendicular 
 to II. Find the point b‘ by drawing c' b’ parallel to 1 1 ; then the trapezoid a'b'c d' will be the 
 representation of the square abed. 
 
 PROBLEM. 
 
 To find the representation of a square obliquely situated to the intersecting 
 line, see the same figure, viz. fig. 7. 
 
 In the intersecting line il, find the points e, r, o,p by drawing ae, br, do, cp each perpen- 
 dicular to 1 1. 
 
 In II find the points c, A, D, B by making ec, ca, cd, eB respectively equal to the per- 
 pendiculars od, ea, pc, rb. 
 
 In vv find the point n by making c n equal to CP. Join ec, and in ec find the points f,a,g , h 
 by drawing the lines cn, An, Bn, Bn, the point a is the representation of the point a. Find the 
 points b' c d' by drawing hb' gefd parallel to 1 1 , and by joining r c, p c, and o c, and a V c' d' will 
 be the representation of the square abed. 
 
 If it be found inconvenient to have the plan on the picture, we may proceed in the following 
 manner, see fig. 8. 
 
 Let the plan be drawn on one piece of paper, and posited to the intersecting line, as shewn 
 at fig. 7. 
 
 Find the points e, r, o, p, as in fig. 7, and transfer them to the intersecting line 1 1, fig. 8, from 
 e to the points r, o, p. Draw the lines r c, o c, p c. In 1 1 , fig. 8, make ea, od, pc, rb equal to 
 the perpendiculars ea, od, pc, rb, fig. 7. In fig. 8, draw the lines an, bn, cn, dn, intersecting 
 
AND COMPLETE DECORATOR. 
 
 121 
 
 the line e c, r c, p c, o c, in the points a',b',c',d'. Join ab',b' c', c d',d' a', and the trapezium 
 a' U c d' is the square required ; and similarly that which is parallel to the picture on the left hand 
 may be represented. 
 
 In the representation of rectangular solids, the plans must be first repre- 
 sented; but before we proceed to any general method of representation, it is 
 advisable to familiarize them by a few examples. 
 
 In the following solids, the rectangular faces, which join the two ends, 
 are called sides of the solid, and the faces to which the sides are joined are 
 called the ends ; the end which coincides with the original plane is called 
 the base, and the other end is called the top of the solid ; the line in which 
 any two faces meet is called an edge, or the edge of those two faces. 
 
 Since the sides of the solid stand upon the base, we shall, first of all, 
 shew how the several bases are found in perspective. 
 
 Fig. 9, the base of the solid to be represented, is a square, aide. In the 
 intersecting line n, find the points g and /by producing ad and db, the two 
 sides which terminate in the nearest angular point to 1 1 . 
 
 Find the vanishing points n, m of the lines a b and db. Find the representation of the nearest 
 point b of the base, by drawing gn,fm. In x i, find the points e and h, by making g e equal to 
 g a, and fli equal to f d. In v v, find the points A and b, by making ota.mb, respectively equal 
 to the radials m p, n p. Find the points amg n and dmfm by joining e B and h A ; and as the 
 base is a square, this is as much as will be required, the other two edges being covered by the sides 
 of the solid. 
 
 The same process is employed in fig. 10, in finding the representations a’ d', d' c', of the two 
 edges da, dc of the square abed. 
 
 In fig. 11, the point c which represents the nearest point of the base, is 
 found as in fig. 5, and the points a and b are found as in fig. 6. 
 
 As it has already been observed, that it is often very inconvenient to have 
 the plan and the perspective drawing on the same paper, fig. 12 is supposed 
 to be done without the plan attached to it ; a preparation, however, as in 
 fig. 11, must be first made, and the intersecting points of the perpendiculars 
 and their lengths must be transferred to the intersecting line n, fig. 12, as 
 shewn by the dotted circles ; from these the representations of the two edges 
 
 2 H 
 
122 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 of the base, which terminate in the nearest point to the intersecting line, 
 will be found. 
 
 Having found the edges of the visible sides in the plan, we shall now proceed to the elevations 
 of the representations of the solids. Produce to the intersecting line one of the representations of 
 the two edges of the base, which terminate in the point of the base which is nearest the intersecting 
 line ; from the point where the prolongation meets the intersecting line, draw a line perpendicular 
 to the vanishing line ; call this perpendicular the line of heights ; on the line of heights set up the 
 height of the solid. Draw a line from the upper point to the vanishing point of the side produced. 
 Thus, in fig. 9, produce d b' to g, and draw gm perpendicular to v v. Make gm equal to the 
 height of the solid. Join mn. In mn find the points o and r, by drawing b' o, dr. Find the 
 point t by joining o to the vanishing point m, and drawing d t perpendicular to v v. Lastly, find 
 the point s in fogs g and 10, by joining t and r to the vanishing points n, m, and thus we have the 
 complete representations, figs. 9 and 10. Figs. 11 and 12 are completed by joining iu. 
 
 PLATE XVII. 
 
 DIAGRAMS ILLUSTRATING THE PRINCIPLES OF PERSPECTIVE. 
 
 \ 
 
 In all the diagrams of this Plate, let ps be the point of sight, s p the 
 station point, nw the directing line, or the parallel of the eye, and vl is 
 the vanishing line. 
 
 Fig. 1 is the representation of straight lines in three different planes, two 
 being horizontal and one parallel to the picture; these planes in all the six 
 diagrams are represented as if they were five of the sides of a room, the 
 sixth side being left open for the purpose of seeing into the interior, and one 
 of the five sides, which is the plane of the picture, is supposed to be transpa- 
 rent ; one of the two horizontal planes, which is the perpendicular to the 
 picture, represents the floor, and the other the ceiling. 
 
 In fig. 1, the lines He and lo on the floor, are parallel, and therefore have 
 a vanishing point f found by drawing a line psf through the point of sight 
 p-s parallel to these lines ; the same observation applies to the lines vr and sn; 
 all these lines are here represented on the plane of the picture by the same 
 letter. In every one of the planes in each of the six diagrams, the lines and 
 
AND COMPLETE DECORATOR. 
 
 123 
 
 points in the picture have the same letters of reference as the corresponding 
 lines and points of the objects themselves. 
 
 In fig. 2, a triangle is exhibited as if lying on the floor, and another on 
 the ceiling. 
 
 In fig. 3, a pentagon abode is placed on the side of the room which is 
 parallel to the picture. The method of representing this object is by drawing 
 lines from all the vertices of the figure perpendicular to the floor, and 
 drawing lines from the points where those lines meet the floor to the station 
 point, and again drawing lines on the plane of the picture from the points 
 where the floor and the picture intersect perpendicularly to the line of inter- 
 section ; then these lines, intersected by visual rays drawn from all the vertices 
 of the polygon upon the opposite wall, will give the vertices of the perspective 
 representation. The trapezium upon the plane of the wall, which is perpen- 
 dicular to the picture, is represented in a similar manner. 
 
 In fig. 4, a table is here represented as if standing on the floor. This is 
 done by drawing lines from the feet of the table upon the floor to s-p the 
 station point, and drawing lines on the plane of the picture from the points of 
 intersection perpendicular to the intersecting line, and by drawing lines or 
 visual rays from the extremities of all the lines of the object to the point of 
 sight, and the points where these lines intersect the former in the plane of the 
 picture will be the perspective representations of the corresponding points of 
 the object thus drawn from. In this figure a hatchway is represented in the 
 ceiling. In each of these objects the sides are parallel and perpendicular 
 to the plane of the picture, therefore the lines which are perpendicular to 
 the picture vanish in its centre. 
 
 The method of representing the objects in figs. 5 and 6, is similar to that 
 now explained, and therefore need not be repeated. Fig. 7 is the plan of 
 fig. 5, and fig. 8 the plan of fig. 6 ; these plans exhibit the positions of the 
 objects, the station point, and the manner of drawing the lines towards it 
 from all the points of the object, so as to give their intersecting points on the 
 plane of the picture. 
 
124 
 
 CABINET-MAKER, UPHOLSTERER 
 
 PLATE XVIII. 
 
 ILLUSTRATION OF VARIOUS PROBLEMS. 
 
 i , 
 
 Figs. 1 and 2, exhibit the method of finding the perspective of an octagon, 
 both horizontal, and inclined to the horizon at a given angle. The repre- 
 sentation, fig. 1, is made from the plan, and that in fig. 2, is found from the 
 measures being applied upon the horizontal vanishing line. 
 
 Figs. 3 and 4, exhibit the method of finding the perspective representation 
 of a semicircle, both in a horizontal and in an inclined plane, as if they were 
 connected together by the same common diameter. Both these representa- 
 tions are found by means of the measures of the object applied to the inter- 
 secting line. In fig. 3, the diameter which connects the two semicircles, is 
 parallel to the picture ; but in fig. 4, it is oblique. In fig. 3, the centre c of 
 the picture is the vanishing point of the lines that are perpendicular ; the 
 other parallel series of lines, being parallel in the object, are represented by 
 lines parallel to the intersecting line. In fig. 4, neither of the two series of 
 parallel lines of the object being parallel to the intersecting line, the vanishing 
 point of one series will be on one side of the centre, and that of the other series 
 on the other side of the centre. The manner of finding the intersecting points, 
 and the angle which the two semicircles make with each other, is exhibited 
 above, fig. 3. 
 
 Figs. 5, 6, 7, exhibit the perspective representation of a room where the 
 floors are laid with various kinds of pavement. In fig. 2, the lines of the pave- 
 ment being perpendicular to the plane of the picture, the vanishing point of 
 these lines is in the centre ; the breadths are regulated by the point of 
 distance v 2 . In fig. 6, the lines are regulated by two vanishing points, 
 equally distant from the centre of the picture. Fig. 7, is a combination 
 of the two. 
 
USHER, SOU, & C° LONDON", 1833 . 
 
AND COMPLETE DECORATOR. 
 
 125 
 
 THE METHOD OF DRAWING LINES TO INACCESSIBLE VANISHING POINTS. 
 
 The drawing of lines to a point out of reach was long a desideratum, and 
 numerous complex schemes were introduced to accomplish this desirable 
 object: at length, Mr. J. Fary and Mr. P. Nicholson produced instruments 
 nearly at the same time, which would completely effect the purpose. In 
 May, 1814, the Society of Arts voted to Mr. P. Nicholson the sum of twent\r 
 guineas, for the invention of the centrolinead. Mr. Fary produced his instru- 
 ment for drawing lines converging to an inaccessible point, very shortly after, 
 and was also rewarded by the Society of Arts. The principle of these 
 instruments, though they had the same object to accomplish, was very dif- 
 ferent. The one invented by Mr. Nicholson, in performing, moved in the 
 manner of a parallel ruler; but the other, invented by Mr. Fary, consisted of 
 three bars, so that two of them being moved to any given angle, the edge 
 of the third bar always besected those of the other two, and the motion of the 
 instrument was regulated by sliding these two last edges each upon a pin, in 
 the same manner and upon a similar principle to that of describing the seg- 
 ment of a circle by means of an angle. However, as Mr. Nicholson had 
 previously shewn another instrument, upon the same principle as Mr. Fary’s, 
 to several of the members of the Society of Arts, he was encouraged by those 
 gentlemen to bring forward his second invention of the centrolinead, which 
 he produced accordingly, and was rewarded with the silver medal of the 
 Society in the year 1815. 
 
 This last instrument has been found to be of the greatest use to artists. 
 It is much more simple in its construction than Mr. Fary’s, which required 
 him to perform a geometrical operation, before the pins, upon which the 
 instrument moved, could be fixed. Mr. Nicholson’s last instrument for 
 drawing lines to a vanishing point, as presented to the Society of Arts, was 
 simply constructed upon the principle of a double bevel, in which each 
 angle was independent of the other, and might be set to any two angles less 
 than 180°. 
 
126 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 This instrument was made to shift to the right or left by means of screws 
 and brass plates. Shortly after his reward, Mr. Nicholson published a small 
 tract on the use of the centrolinead. In this little work he exhibited and 
 explained the manner of making an immoveable or solid centrolinead, to 
 answer the drawing of lines which would converge to a point at any given 
 distance. He now makes three solid or fixed centrolineads, at one-fourth of 
 the expense, to answer every degree of convergency, and these he applies 
 with the greatest expedition ; but before the centrolinead can be used, it will 
 be necessary to shew how the first two converging lines on the right or left 
 hand sides of a building, or piece of furniture, of which the plan is a rect- 
 angle, are found. 
 
 Let a be, fig. 1 , Plate xvill. for instance, be the plan of two sides of the frame of a table, or 
 two sides of a building, and let 1 l be the sectional line of the picture, placed in the position 
 required by the draughtsman. 
 
 In order to avoid distortion, every picture should be viewed in the middle, and the horizontal 
 visual angle should be given ; that is, the distance of the eye ought to bear some ratio or proportion 
 to the length of the picture. We shall here suppose that the picture is viewed at a distance which 
 will be two and a half times its length ; that is, supposing the picture to be two feet long, as in 
 large drawings, the distance of the eye would be five feet; therefore, from any convenient point k, 
 in the line l L, draw km perpendicular to I L. In I L, take the point i at any convenient distance 
 from k, and in I L make kl equal to hi. In km set off il two and a half times from k to m, and 
 join m i and m l. In I l cut off the length of the picture de, by drawing ad parallel to m l, and 
 ce parallel to mi. The picture must now be viewed at a distance equal to two and a half times 
 de; that is, about ten inches and five sixteens. Now, this distance would exceed the extension of 
 the surface which we have here to work upon ; but if it can be shewn how the drawing may be 
 completed equally well, without extending the lines ad and ce till they meet, we should be most 
 happy, to avail ourselves of the information, in avoiding so great an inconvenience and so much 
 more labour. 
 
 Bisect, then, de in the pointy, and draw y^r perpendicular to I L. Divide df into five equal 
 parts. Cut the lin efg in g, by drawing lig through the fourth point h of division parallel to ad 
 or m l, then will y <7 be a fifth part of the real distance of the picture; and if the drawing or paper 
 had not admitted of this distance, df must have been divided into a greater number of equal parts; 
 for f g would be shorter in proportion to the greater number of equal parts which are contained in 
 df; so that if df had been divided into six equal parts, fg would have been one-sixth part of the 
 distance of the picture, and so on. 
 
 In 1 l, cut the points n and c, by drawing gn parallel to b a, one side of the plan, and g c 
 parallel to b c, the other side. 
 
 Let v L, fig. 2, be the vanishing line, either on the same paper, or on another, as may be con- 
 venient. In v L assume the point p for the centre of the picture. 
 
AND COMPLETE DECORATOR. 
 
 127 
 
 In vl cut off the distances p r and pu respectively, equal to fn and fc in the line i L, fig. 1. 
 In fig. 2, draw the lines r s, p F, and uy perpendicular to v L. Set the points of the dividers at any 
 convenient distance from each other, and set off as many equal parts in p F, and as many equal 
 parts wanting one in r s and uy, as df, fig. 1, was divided into: now, df was divided into five 
 equal parts, therefore p F contains five, rs and uy contain each four of such parts as the dividers 
 were set to. 
 
 Draw the straight lines fs and F y. Produce FstoE. hipcd represents the instrument, 
 H i, i p are the two straight edges that run upon the pins, and I D is the edge of the instrument by 
 which the lines are drawn. 
 
 There are here given two straight lines E F and v L, which would tend to the proper vanishing 
 point, if produced. The distance in which they will meet the line v L produced on the left, is five 
 iines the distance nf, fig. 1. 
 
 Suppose, now, the edge CD of the centrolinead to be placed on the vanishing line v L, put a 
 pin in the paper in any convenient point A of the running edge I H, and draw a line I P on the paper 
 by the other running edge I p. Shift the instrument, so that the edge I D of the blade may coincide 
 with the line ef, and the edge i h still rest against the point A; then draw a new line ip. Put 
 another pin in the intersection B of the two lines I p, I p. 
 
 Lastly, move the centrolinead until the edge c D fall upon any given point between the lines 
 ef and v L, so that the edge ih may rest against the point A, and the edge ip against the point 
 B, and draw a straight line by the edge i D, and the straight line thus drawn will tend to the same 
 point in which the other two given lines E D and v l would meet each other, if produced. 
 
 In this manner lines tending to a vanishing point, may be drawn with as 
 much expedition, as perpendicular lines could be drawn by a common 
 T-square, or parallel lines by a parallel ruler; and more lines may be drawn 
 by the instrument in five minutes, than could be drawn in twenty four hours 
 by finding each line separately by a geometrical problem. 
 
 THE PERSPECTIVE OF SHADOWS. 
 
 The shadows of objects are the dark parts on the surface of a body or 
 plane, occasioned by the intervention of another body which intercepts the 
 rays of light. 
 
 In projecting shadows by the sun, there are three different positions of 
 the luminary to be considered; viz. it must either be in the plane of the 
 picture, or on the same side of the picture with the spectator, or behind the 
 picture 
 
12P 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 In order to determine the shadows of objects on the plane of the picture, 
 the situation and altitude of the sun must be determined ; then, whether the 
 sun is before or behind the picture, the ■vanishing point of the rays of light 
 must either he found or assumed ; but when the sun is in the plane of the 
 picture, the rays will be parallel to the picture, and therefore the representa- 
 tions will be parallel. 
 
 When the sun is on the same side of the picture as the spectator, it is 
 obvious that the vanishing point of the rays of light will be below the hori- 
 zontal line ; and if on the other side, above it. 
 
 When a straight line obstructs the sun’s rays, it is evident that the rays 
 thus intercepted are all in a plane passing through the luminary and through 
 that line. The opaque surface occasioned by the obstruction of the sun’s rays, 
 is called a plane of shade. It is therefore evident, that the plane of shade 
 passes through the luminary and through the straight line. Hence the 
 indefinite shadow of a line upon a plane is the intersection of that plane and 
 the plane of shade. 
 
 If therefore we have the vanishing point of a line that throws a shadow 
 upon a plane, and the vanishing point of the sun’s rays, we have the vanish- 
 ing points of two lines in the plane of shade, and therefore the vanishing line 
 of the plane of shade ; and if we have the vanishing line of the plane of shade, 
 and the vanishing line of the plane on which the shadow is thrown, we shall 
 have the vanishing point of the shadow. For the shadow itself is the inter- 
 section of the plane of shade, and the plane on which the shadow is thrown; 
 therefore the vanishing point of the shadow is the intersection of the two 
 vanishing lines of these planes. 
 
 If the vanishing line of the plane of shade be parallel to the vanishing line 
 of the plane on which the shadow is thrown, it is evident the shadow of the 
 line will be parallel to the vanishing line of the plane on which the shadow is 
 thrown, or parallel to the vanishing line of the plane of shade ; for these two 
 vanishing lines can never intersect each other. 
 
 If the line which throws a shadow be parallel to the plane on which the 
 
AND COMPLETE DECORATOR. 
 
 129 
 
 shadow is thrown, the line and its shadow will have the same vanishing 
 point. 
 
 When the line which throws a shadow on a plane is perpendicular to that 
 plane, and when the plane is parallel to the picture, the vanishing point of 
 the line will be in the centre of the picture ; therefore the vanishing line of the 
 plane of shade will be the line joining the vanishing point of the sun’s rays, 
 and the centre of the picture; and since in this case the shadows of lines 
 thrown upon any plane, by lines perpendicular to that plane, are all parallel 
 among themselves, the shadows of these lines will be all parallel among 
 themselves. 
 
 The shadows of objects projected upon horizontal planes, bylines inclined 
 at any given angle to these planes, will have their vanishing points in the 
 vanishing line of horizontal planes. 
 
 MOST USEFUL PRACTICAL OBSERVATIONS. 
 
 With respect to the shadows of perspective drawings of Cabinet Furniture, 
 and Architectural objects, as the objects are placed upon the horizon, and 
 since generally the surfaces are planes, and rise perpendicular to the base, 
 the edges of every two adjacent surfaces thus rising from the base will be 
 straight lines perpendicular to the horizon: and as the picture is generally 
 perpendicular to the horizon, the perpendicular lines of the object will be 
 parallel straight lines to the plane of the picture, and will therefore have their 
 perspective representations perpendicular to the vanishing line of the hori- 
 zontal planes. 
 
 Again, as every two planes of the object are generally at right angles to 
 each other, those planes which are perpendicular to the horizon will have 
 their two vanishing lines perpendicular to the vanishing line of the horizontal 
 planes; and because the base on which the object stands, and all the oppo- 
 site planes above the base, are parallel to the horizon, the lines which terminate 
 these planes will have their vanishing points in the vanishing line of the 
 
 2 K 
 
130 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 horizontal planes ; and as all the horizontal lines of the object belong also to 
 the vertical planes of the same object, the horizontal lines will also have their 
 vanishing points in the two vanishing lines, which are perpendicular to the 
 horizontal vanishing line. 
 
 Hence, the points of intersection of the two vanishing lines of the planes 
 which are perpendicular to the horizon, and the vanishing line of the hori- 
 zontal planes, will be the vanishing points of all the horizontal lines of the 
 object. 
 
 The vanishing point on the one side is the vanishing point of every line 
 which is perpendicular to any one of the parallel vertical planes, of which the 
 perpendicular passing through the other vanishing point is their vanishing 
 line. Or these two vanishing points may be considered the vanishing points 
 of two lines in a horizontal plane at right angles to each other. 
 
 With regard to the effect which the position of the planes of such objects 
 have upon the shadows, we shall observe, that whatever position the luminous 
 point has in respect of the three vanishing lines now alluded to in these 
 observations, as the vertical lines of the object are represented by lines which 
 are perpendicular to the vanishing line of the horizontal plane, these lines 
 may be considered as having their vanishing point in a line drawn through 
 the centre o the picture perpendicular to the vanishing line of the horizontal 
 planes, and at an infinite distance in the line thus drawn. 
 
 And since the vanishing line of the plane of shade of a straight line pro- 
 jecting a shadow on a given plane, is a straight line passing through the 
 vanishing point of the sun’s rays, and through the vanishing point of the line 
 which throws the shadow ; therefore, with regard to the perspective repre- 
 sentation of the shadows of the lines which are perpendicular to the horizon 
 upon any horizontal plane, as these lines have their vanishing points at an infi- 
 nite distance from the centre of the picture, and in a perpendicular to the vanish- 
 ing line of the horizontal planes, a line drawn from the vanishing point of the 
 sun’s rays to this infinitely distant vanishing point, must be perpendicular to 
 the vanishing line of the horizontal planes, the conclusion is, that — 
 
AND COMPLETE DECORATOR. 
 
 131 
 
 The vanishing line of a plane of shade of a line perpendicular' to a hori- 
 zontal plane, is a straight line drawn through the vanishing point of the suns 
 rays perpendicular to the vanishing line of the horizontal planes. 
 
 And because the vanishing point of the shadow of any line meeting a 
 plane upon that plane, is in the vanishing line of that plane — 
 
 The vanishing point of the shadow of a line perpendicular to a horizontal 
 plane, projected upon any horizontal plane whatever, is that point in the 
 vanishing line of the horizontal planes, where a straight line drarvn from the 
 vanishing point of the suns rays perpendicular to this said vanishing line of 
 the horizontal planes, meets it. 
 
 With regard to the representation of the shadows of the lines of the object 
 which are perpendicular to the vertical planes upon these vertical planes, as 
 the vanishing point of the lines which throw the shadow are one of the two 
 vanishing points in the vanishing line of the horizontal planes — 
 
 The vanishing line of the plane of shade of a straight line perpendicular 
 to a vertical plane, projecting a shadow on this vertical plane, is a straight 
 line drawn from the vanishing point of the line to the vanishing point of the 
 sun's rays, the vanishing j ' of the line which throws the shadow being one 
 of the two vanishing points in the horizontal vanishing line. 
 
 But as the plane on which the shadow is thrown is vertical, the vanishing 
 line is the line drawn through the other vanishing point perpendicular to the 
 vanishing line of the horizontal plane ; hence — 
 
 The vanishing point of the shadow of a straight line perpendicular to a 
 vertical plane upon this plane, is in the perpendicular drawn through the other 
 vanishing point, which is not the vanishing point of the line to the vanishing 
 line of the horizontal planes. 
 
132 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 PLATE XIX. 
 
 APPLICATION OF THE THEORY OF SHADOWS TO REAL PRACTICE. 
 
 b c d e, fig 3, represents a wall in perspective. In this example the sun 
 is supposed to be in the plane of the picture, therefore the perspective repre- 
 sentation of every ray of light makes the same angle with the vanishing line 
 of the horizontal planes, as the altitude of the sun makes with the horizon ; 
 and consequently, the sun's rays have no vanishing point but what is in- 
 finitely distant. 
 
 Now the vanishing line of the plane bcde is the line v A drawn through the left hand v per- 
 pendicular to vv, the vanishing line of the horizontal planes, and the vanishing point of the lines 
 which throw the shadow is the right hand point v in the vanishing line vv of the horizontal 
 planes. 
 
 Therefore as v is the vanishing point of the lines which throw the shadow, the vanishing line 
 of the plane of shade must pass through v, making an angle with vv, the vanishing line of the hori- 
 zontal planes, equal to the angle which a ray of the sun makes with the horizon ; therefore draw 
 the sloping line VA through the right-hand vanishing point v, making the angle vva equal to the 
 angle which a ray of light makes with the horizon, and the sloping line v A will be the vanishing 
 line of the plane of shade. But v A perpendicular tow is the vanishing line of the plane on 
 which the shadow is thrown, therefore the point A in which the sloping vanishing line v a of the 
 plane of shade, and the vertical vanishing line v A of the plane on which the shadow is thrown, meet 
 each other, is the vanishing point of the shadow. 
 
 From what has been said, the perspective representations of the rays are all parallel to the 
 sloping line v A, and the indefinite shadow' of any line perpendicular to the plan represented by 
 bcde w r ill pass through the point A, and through the foot of the lines, and we have the law for 
 drawing the indefinite perspective shadow of the line and the perspective of the ray of light which 
 passes through the unconnected extremity of the line. 
 
 Therefore to find the shadow of any such line, we have only to draw a line from A the vanish- 
 ing point of the shadow, and through the foot of the line, and to draw another line through the 
 unconnected extremity parallel to the sloping line v A, and the point where this line intersects the 
 indefinite perspective shadow will terminate the shadow required, or the portion of the indefinite 
 shadow between the foot of the line which throws the shadow and the point of intersection of the 
 perspective representation of the indefinite shadow of the line, and the perspective of the ray of 
 light drawn through the unconnected extremity, will be the shadow of the line. 
 
 Fig. 4, is the plan of a body projecting from the face of the wall, consist- 
 ing of several faces which are plane surfaces, every two adjacent planes being 
 prependicular to each other, and those which are adjacent to the wall 
 
AND COMPLETE DECORATOR. 
 
 133 
 
 perpendicular to the plane of the wall; hence all the straight lines formed 
 by the intersection of the planes perpendicular to the wall, will also be per- 
 pendicular to the wall, and therefore parallel to the horizon; hence their 
 Vanishing point will be on the horizontal vanishing line, and consequently, in 
 the right-hand point v, which is the vanishing point of lines perpendicular to 
 the plane represented by b c d e, of which the vanishing line is v a, perpen- 
 dicular to the horizontal vanishing line v v ; and as these planes are either 
 perpendicular or parallel to the horizon, the intersections of the planes which 
 are parallel to the wall, with the planes which are perpendicular to the wall, 
 but at the same time parallel to the horizon, will be both parallel to the wall 
 and to the horizon; and the intersections of all those planes which are 
 parallel to the wall, with those which are at the same perpendicular to the wall 
 and to the horizon, are all perpendicular to the horizon, and consequently 
 parallel to the picture. 
 
 Therefore, all the lines which are perpendicular to the horizon and 
 parallel to the picture will be represented by lines perpendicular to the 
 vanishing line of the horizon, and the representation of the shadows of these 
 lines will also be perpendicular to the vanishing line v v of horizontal planes 
 
 Fig. 1 is an end elevation of the plan fig. 4, and fig. 2 is a front elevation of the same project- 
 ing body. H I projecting from the plane represented by BCD E is one of the upper edges, L being 
 the upper extremity, and H the foot of the line LH. 
 
 To find the shadow of the line LH. From the vanishing point A of the sun’s rays, and through 
 the foot H of the line, draw ai, and draw Li parallel to Avthe vanishing line of the plane of 
 shade ; then H I is the shadow of the line, h l and i the shadow of the point L. 
 
 The edge of the vertical line under L will be found by drawing a line through I downwards 
 perpendicular to the vanishing line vv of the horizon. 
 
 In the same manner the si a ow of the opposite side of the rectangle will be found, as also 
 the shadows of all the lines which are perpendicular to the plane represented by BCD e. 
 
 We now come to the shadows of those solids which are situate upon the 
 plane of the horizon, and of which their surfaces are planes at right angles to 
 one another, and parallel and perpendicular to the horizon. In this case the 
 perspective representation of the rays of light will still be found by drawing 
 lines parallel to the sloping vanishing line v a of the plane of shade. 
 
 2 L 
 
134 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 But now the vanishing point of lines perpendicular to the horizontal 
 plane, being in a perpendicular drawn through the centre of the picture, and 
 being infinitely distant from this centre, therefore the line which is thus 
 parallel to the sloping line av drawn, this infinitely distant point can never 
 reach the vanishing line vv of the plane on which the shadow is thrown; 
 hence the vanishing point of the shadow must be in the line v v, infinitely 
 distant from the centre of the picture ; therefore the perspective representa- 
 tion of all lines perpendicular to the horizon, will be parallel to the vanishing 
 line v v of the horizontal plane. 
 
 Let hi represent one of the edges of a solid of which each pair of the adjacent sides are per- 
 pendicular to each other, and let this solid be so situate that one of its planes may coincide with 
 the plane of the horizon, then calling this plane the base, the other four planes which join the base 
 will be perpendicular to the horizon ; and consequently, the four edges formed by every two of 
 these planes will be perpendicular to the horizon, and all of an equal height above it. 
 
 The line h i being one of these four perpendicular edges, the shadow of the line h i will be 
 found thus : Draw h k parallel to v v, and i k parallel to the sloping line A v, then hk will be the 
 perspective shadow of the line represented by hi. In the same manner the shadow of the other 
 edges will be found ; then to complete the shadow, we have only to join the corresponding points. 
 
 PLATE XX. 
 
 Fig. 1, exhibits a prismatic solid, with a projecture in front: o is the 
 vanishing point of the sun’s rays. As the shadows are to be thrown upon a 
 horizontal plane by lines perpendicular to the horizon, the vanishing point of 
 the perspective representation of the shadow will be the point in the vanish- 
 ing line of the horizontal plane, cut by a perpendicular to the said horizontal 
 vanishing line, drawn through the vanishing point of the sun’s rays. Hence, 
 by drawing o l perpendicular to v l, we have the vanishing point of the 
 sun’s rays. 
 
 Thus now, to find the shadow of any line, as a b, draw a L and bo, intersecting each other in 
 the point c, thus ac is the shadow of the line a b. In the same manner the shadow df of ‘he line 
 de will be found, and joining cf, cf will be the shadow of the line be, and acfd will be the 
 shadow of the plane abed. In the same manner we shall find the shadow of the other planes. 
 
/Y . ,. 9 . 
 
 Shadows . 
 
 Mlisl^a. ty Pisnei . San & C? Caxton. London. 1834 
 
 J>icA sc. 
 

 A 
 
 % 
 
 AND COMPLETE DECORAT 
 
 135 
 
 In fig. 2, after having found c,f, n, r , t , the representations of the shadows 
 of the upper extremities a, b , e, k, Z, m, p, s of the perpendiculars ba, de , 
 gk , &c. in order to complete the shadow, there is nothing more to do than 
 
 4 
 
 to trace a curve through the points c,f, i, n , r, /. 
 
 Fig. 3, exhibits an obelisk projecting a shadow' upon a cylinder, and upon 
 the plane of the horizon : the obelisk stands with its base perpendicular tc 
 the horizontal plane, and with its edges equally inclined to the axis which is 
 perpendicular to the horizon. The perspective representation of the square 
 on the top must be also found in the base; by this means the perpendiculars 
 from the upper extremities will be found in the same manner as in the two 
 preceding examples, and by connecting the extremities the whole will be 
 found complete. The shadow of the cylinder upon the horizontal plane is 
 found by making a sufficient number of sections through the cylinder, and in 
 the plane passing through the line that throws the shadow, and the sun. 
 
 We have now finished our article upon Perspective. Its application to 
 the following elegant subjects will explain more of its use to the designer and 
 cabinet-maker than volumes of words could explain. — The subjects to which 
 it is applied throughout this work are the following : 
 
 Four-post beds 
 Sofa 
 
 Occasional table 
 
 Side-board 
 
 Library table 
 
 Bed in the Grecian style 
 
 Couch in the Grecian style 
 
 Drawing-room chairs 
 Card-tables 
 Music-stool 
 Circular table 
 
 Chair in which the Duke of York 
 died 
 
 Book-case. 
 
 In the design of the Grecian couch, the vanishing line is out of the 
 picture ; but as it is now supposed that the reader is in possession of the 
 principles, he can very easily supply its place in imagination. The measures 
 are applied along the ground line as exhibited at the bottom, and the per- 
 spective lengths are cut off by means of the points of distance ; these points 
 
136 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 of distance being the distance between each vanishing point and the eye 
 transferred upon the vanishing line. 
 
 SCHOLIUM. 
 
 The use of perspective representation is to shew objects in the most 
 natural manner, and to the greatest advantage, so as to give a correct idea 
 of the combinations of their various parts ; and though by having the 
 centre and distance of the picture, and knowing that the angles of the 
 object are right angles, the correct length of every line might be obtained ; 
 or, if the centre and distance of the picture be given, and the two vanishing 
 points, we shall be able to obtain the original lengths and angles of all the 
 parts of the object; but the centre and distance of the picture are seldom 
 given ; and if they were, the reverse process of finding the original lengths and 
 angles of the lines would be so tedious, as to exhaust the patience of the 
 draughtsman, or persons employed to execute the work 
 
 A perspective drawing may answer to innumerable angles; but one thing 
 we should observe, that when the distance of the picture is a mean propor- 
 tional between the distance on the vanishing line from the centre of the 
 picture, and each vanishing point; and if the object have only two vanishing 
 points, the angles are all right angles. 
 
 It is necessary to mention these circumstances, that the reader may see 
 the whole extent of the use of perspective, which will save his time by pre- 
 venting him from making useless researches 
 
 Once for all, in perspective, as far as regards the cabinet-maker: — Though 
 a plan of the object is often drawn for the use of the workman, yet he or the 
 draughtsman generally applies the measures upon the ground line, and they do 
 not draw from a station point, as architectural draughtsmen are accustomed. 
 
 Orthographical representation gives the same connexion of parts as 
 perspective representations; and though it does not shew the object quite so 
 natural, it has this great advantage, that all the parts can be measured by 
 means of a scale, and therefore the object can be readily put into execution 
 
AND COMPLETE DECORATOR. 
 
 137 
 
 Indeed, the difference between perspective representation taken at a very long 
 distance, and an orthographical representation, is so trifling, as hardly, in 
 many instances, to be perceived, as will be easily seen from the following 
 objects, which are all orthographically represented : 
 
 Escritoir Sofa 
 
 Lady’s wardrobe Chiffonier 
 
 Dressing-table and dressing-glasses. 
 
 We cannot say more than we have already observed under the article of 
 orthographical projection, and shall only repeat, that the three directions 
 of the solid angle may be assumed by any three lines which may be agreeable 
 to the eye ; the scale may be found for lines in each of the three directions. 
 In order that the reader may know which of the scales belongs to any given 
 direction, the scale is always applied parallel to the lines which are measured 
 by it ; that is, vertical lines by the vertical scale, oblique lines by the oblique 
 scale, parallel to these lines. 
 
 In point of drawing, orthographical representation admits of much more 
 
 rapid execution than perspective representation. 
 
 Throughout the whole of “ The Cabinet-maker,” wherever the sides of 
 furniture are geometrically represented, we have always affixed a scale, in 
 order to obtain the dimensions of the parts, so that any of the designs may be 
 put into execution. The designs thus represented are — 
 
 W indo w-curtains 
 
 Candelabra 
 
 Firescreen 
 
 Chairs 
 
 French bed 
 
 Tea-poy 
 
 Tripod and Stand 
 
 Sofa 
 
 Stools 
 
 Pier-tables 
 
 Window-seats 
 
 Side-board 
 
 Side-board and Celeret 
 Cabinet and looking-glass 
 Looking-glass 
 
 Circular-table , * 
 
 Drapery for three windows 
 Ladies’ work-tables 
 Hall-seat, bason-stands, and pot- 
 de-chambre recess 
 
 2 M 
 
138 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 Card-tables 
 Couch 
 Music-stools 
 Circular library-table" 
 Plan of a circular-table 
 
 Wine-coolers 
 
 Upright piano-forte in the Grecian , 
 style 
 
 Upright piano-forte in the Gothic 
 style. 
 
 PLATE XXL 
 
 Finding the angle ribs for canopies upon square, rectangular, polygonal, 
 and circular plans ; describing the forms of their veneers, mitering trays in 
 the form of the frustum of a pyramid ; method of enlarging and diminishing 
 moulding, and of tracing the sections and mitres of raking mouldings. 
 
 PROBLEM. 
 
 To describe the ribs of a canopy upon a square plan, the section perpen- 
 dicular to the front being given. 
 
 Let a, b, c, d, fig. 1, be the plans of the posts upon which the canopy is 
 supported, and let the given rib be q 1 2 3 4 5 6, placed with its base perpen- 
 dicular to the side d c. The curve q 1-2-3-4-5# being of any form required. 
 In this example it is the quadrantal arc of a circle ; e h being perpendicular 
 to e q. 
 
 Draw the mitre lines between the inner points of the plans a, c, and d, b, 
 crossing each other at e. 
 
 First, to find the angle rib. — In the curve of the given rib, take the inter- 
 mediate points I, 2, 3, 4, 5 at equal distances or at pleasure. Draw the lines 
 1 #, 2 r, 3 g, 4 e, 6 c, meeting the mitre line in the points t, r, g, e, c , and cross- 
 ing the base p e in the points v, o, n , m, l. Draw the lines tu, r s, g h , ef,cd, 
 e b perpendicular to the mitre line on the plan. Transfer the perpendiculars 
 p 1 , o2, n3, m4, 1 6 between the base and the curve of the given rib to the 
 
AND COMPLETE DECORATOR. 
 
 139 
 
 perpendiculars t a, r s, g h, ej \ c d, & b raised upon the diagonal or mitre line, 
 and through the points d, u, s, h,f d , b draw a curve which will give the 
 outer edge of the angle rib. The inner edge will be found in the same 
 manner. 
 
 Fig. 1, No. 1, fig. 2, No. 2, and fig. 2, No. 3, exhibits the method of backing 
 the ribs of figs. 1 and 2. 
 
 The diagonal ribs in fig. 2, are found in the same manner as in fig. 1 ; but 
 in order to find the form of the covering or cloth upon each side, divide the 
 curve line e 1 2 3 4 5 into any number of equal parts, the more, the truer 
 the operation will be ; we shall, however, suppose that five parts are sufficient, 
 then extend these parts upon a straight line b 5 perpendicular from the point 
 b to ac , and through the points 1, 2, 3, 4, 5 in the line b 5, draw the lines 
 1 a, 2 a, 3 a, 4 a parallel to a c \ Make 1 a, 2 a, 3 a, 4 a, respectively equal to 
 the lines e n, e n, e n, &c. between the lines l b and e b on the plan which are 
 parallel to le ; and through the points a, a, a, a, &c. draw the curve a aaaab, 
 and in the same manner draw the curve c 5 on the other side, then a c 5 will 
 be the entire covering of the canopy. 
 
 In the same manner the covering a 6 c, fig. 3, will be found, and likewise 
 the spherical canopy exhibited in fig. 4. 
 
 These bod'ies not only exhibit the coverings of canopies, but the figure of 
 a veneer to any square, rectangular, or circular plan. 
 
 Fig. 5, shews another method besides that exhibited in fig. 4, for finding 
 the figure of a veneer upon a spherical surface. 
 
 Let abc be a vertical section of the spherical surface. Divide the curve 
 into the equal parts b h, fig, gffe, &c. Let a bisect the line a b. Draw 
 a c perpendicular to ab . Prolong a c and fig to meet each other in d, and 
 from d with the distances d fi, dg, describe the arcs ho and gn, and these 
 arcs will be the edge of the veneer, in order to cover the spheric surface of 
 which g h is a section. 
 
 In the same manner will be found f g ml, effci, &c. the figures of the 
 succeeding veneers. 
 
140 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 To find the mitres at the angles of a knife-box. — Let abcd, fig. 6, be 
 the plan at the top or wide part, and abdc the plan of the bottom or narrow 
 part, a a, d c, c d, b b, being the plans of the diagonals, and let ef be the 
 elevation. 
 
 Through any point a in the plan a a of a diagonal, draw p n perpendicular 
 to a a, meeting the sides a d and a b in p and n. From the point a in the 
 lin e pn make am equal to the height of the elevation, and join a m. Draw 
 ag perpendicular to Am, meeting Am in g, and from a in the line a a make 
 ah equal to ag. Join hn by drawing a straight line from h to n , and nha 
 will be the angle of the mitre required. The same bevel will answer each of 
 the other three mitres, as is evident. 
 
 To diminish a cornice in any given proportions. — Let ck be the given 
 cornice, consisting of a fillet cima-recta, corona fillet, and cima-reversa. Upon 
 the height line be make b f equal to the height of the required cornice. Join 
 c c, and draw fa parallel to c c, meeting be in a. At the point c draw the 
 straight line c a, making any angle c a with c b. Make c a equal to ba, and 
 join b a. 
 
 Through the points in the height c b, where the horizontal lines of the given 
 cornice cut the line cb, draw lines parallel to b a, meeting the line c a. 
 Draw a b perpendicular to a c , and make a b equal to ba. Draw r c d perpen- 
 dicular to cb, the height of the" given cornice, and from all the points in the 
 projection of the given cornices draw lines to meet cd perpendicularly. 
 Draw b d perpendicular to ba, and draw cd 1 , making any convenient angle 
 with cd. Join d d, and through the points where the perpendiculars meet 
 c d draw lines parallel to d d to meet c d. Through the points in the height 
 cb cut by the horizontal lines, draw straight lines parallel to bA to meet the 
 line ca. Through the points in ca draw straight lines parallel to ab, 
 and through the points in cd draw lines parallel to ca, and the points 
 where these lines meet will be the points of projecture in the cornice required. 
 
 To find the angle-bars of return glass frames. — In fig. 8, b is the given 
 bar, a is an angle-bar placed in a right angle, and c is an angle-bar placed in 
 
AND COMPLETE DECORATOR. 
 
 141 
 
 an obtuse angle, all the three bars being made to the same thickness. Draw 
 parallel lines to the sides to meet the mitre lines of each angle-bar w, two 
 from each point. Take the line 1 1 in a, and make 1 1 in f the section b 
 equal to 1 1 in a, and draw the oblique line 1 1 in b ; also in b draw 22, 33, &c. 
 parallel to the oblique line 1 1 ; then in tlhe section a make 22, 33, &c. respec- 
 tively equal to 22, 33 in the given section b, and thus we shall find the curves 
 on both sides of the section a, and in th*e same manner those of the section c 
 will be found. 
 
 Given a horizontal moulding, in a vertical plane, to find the section of a 
 moulding upon the rake to meet the horizontal moulding, and the section 
 of another horizontal moulding to meet the raking moulding on a line parallel 
 to the direction of the horizontal moulding, so that the three mouldings shall 
 be in three vertical planes at right angles to each other. 
 
 In fig. 9, let the cavetto 1234 be th<e given horizontal moulding. Draw 
 lines parallel to the rake, and through the points 1, 2, 3, 4, draw lines parallel 
 to the horizon, meeting the perpendicular line from the lower end of the 
 profile of the moulding. Make the proj ections of the section at right angles 
 to the rake, equal to those of the horizontal moulding. To find the section 
 or profile of the return horizontal moulding at the top, set the projectures of 
 the horizontal moulding at the bottom upon a horizontal line, and from the 
 points in the extremities of the projectures, draw perpendiculars to the 
 horizon to meet the raking ordinates, and the points where the perpen- 
 dicular and raking lines meet each other will be in the section of the moulding 
 required. 
 
 Fig. 10, exhibits a level horizontal moulding in a plane parallel to the 
 plane in which the raking moulding is, this moulding being connected to the 
 raking moulding by another horizontal moulding in a plane perpendicular to 
 the other two planes and to the horizon. 
 
142 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 SPIRAL LINES. 
 
 We have now shewn some of the most useful processes which are neces- 
 sary to the workmen, in order to be able to construct various pieces of 
 cabinet furniture ; and though fancy is in a great measure regulated by the 
 eye, the designer must frequently resort to hand-sketching. In this, however, 
 there are many cases in long sweeping lines regularly bending on the same, 
 or in different sides, where he would find it difficult to form such a composi- 
 tion as would be agreeable to the discerning eye, which must be the criterion 
 of every natural line thus produced. 
 
 Geometry, which he has found useful in the construction of plans and 
 elevations, and in projection and perspective, will here also be found not only 
 to strengthen his imagination in the designing of beautiful forms, but will 
 even lend her assistance in producing the most elegant curves that can be 
 conceived. 
 
 If the human mind were capacious enough to retain a knowledge of 
 mathematical lines of all orders, there is not a curve belonging to any species 
 which would not be of use to the designer of cabinet furniture, as there is not 
 one but what has its peculiar degree of elegance ; but since memory is 
 not sufficiently retentive, we shall only introduce such curves as may easily 
 be recollected, and which will be found to be of the most essential service in 
 the art of design. We shall therefore proceed to explain the law of their 
 formation. 
 
 THE NATURE AND DESCRIPTION OF SPIRAL LINES. 
 
 Definitions. 
 
 DeJ. 1. If round a fixed point another be supposed to move, either con- 
 tinually approaching or continually receding from the fixed point according 
 to a given law, the line thus described is called a Spiral. 
 
AND COMPLETE DECORATOR. 
 
 143 
 
 Def. 2. The fixed point is denominated the centre of the spiral. 
 
 Def. 3. The moving point is denominated the describant. 
 
 Def. 4. If the describant has gone once round the fixed point, the spiral 
 is said to have one revolution; and as often as the describant has moved 
 round the fixed point, so many revolutions the spiral is said to have. 
 
 Def 5. Any straight line drawn from the centre of the spiral to meet the 
 describant, or any point in the curve, is called an ordinate. 
 
 Spirals are of many different species, of these the most useful are the 
 spiral of Archimedes, the reciprocal or hyperbolic spiral, and the logarithmic 
 or proportional spiral. 
 
 Spiral of Archimedes. 
 
 Def. 6. If in the description of a spiral, both the angular space and the 
 distance of the describant from the centre increase uniformly, the spiral is 
 called the spiral of Archimedes . 
 
 Def. 7. If the construction of a spiral be such, that when the curve is 
 divided into parts by any succession of ordinates making every two adjacent 
 angles equal, and if in any two adjacent angles the separating or dividing 
 line be a mean proportion between the other two, which separates each of 
 these two angles from each of those angles on each side of them, the spiral 
 is called the proportional spiral , or otherwise it is called the logarithmic 
 spiral. 
 
 Therefore, when the ordinates of a proportional spiral divide the space 
 round the centre into equal angles, in every three succeeding ordinates, the 
 third will always be a third proportional to the two first; that is, as the first 
 is to the second, so is the second to the third: as the second is to the third, 
 so is the third to the fourth : as the third is to the fourth, so is the fourth to 
 the fifth ; and so on. From the definition and explanation here given, we may 
 find as many points in the logarithmic spiral as will be necessary to con- 
 struct it. 
 
 By means of spiral lines, the designer of cabinet-work is enabled to form 
 these elegant terminations and symmetrical forms which are to be seen in the 
 
144 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 various parts of the designs in this work, under the heads of Ornaments, 
 Grecian Ornaments, Elements of Foliage, Chairs, and in almost every species 
 of furniture. In fact, there are very few embellished compositions where 
 such lines cannot be introduced with effect. The characteristic of the Ionic 
 order, and the elegance of the Corinthian capital, are derived from the grace- 
 ful flowing lines of this species of lines. 
 
 In many kinds of curve lines, the curves can only be formed by means 
 of points ; this is also the case with the curve lines of the spiral species, and 
 thus to produce a continued line, an infinite number of points would be 
 required ; but instead of this unlimited multitude of points, we must be con- 
 tented with a sufficient number, and either connect the successive points by 
 hand agreeably to the eye, or join them by circular arcs made to the 
 radius of curvature at the points of junction. 
 
 This last is the most eligible method ; for though no two parts of a curve 
 taken in succession have exactly the same degree of curvature, yet when the 
 points are not very distant, and succeed each other according to a regular law, 
 the curve formed by means of circular arcs will not differ essentisdly from the 
 true spiral, and the deviation will scarcely be perceived by the most keen 
 observer. 
 
 OF VOLUTES, OR SCROLLS. 
 
 Definitions. 
 
 Def. 1. A volute is an ornament in Cabinet-work and in Architecture, 
 consisting of two or more spiral lines of the same species of curve. Such 
 ornaments are mostly used in vertical planes, as in the Ionic capital, or in 
 vertical surfaces upon a curved plan, as in the volutes of the Corinthian and 
 Composite capitals. Sometimes in these capitals the volutes, besides their 
 convolution round the centre spring outwardly, being in this case regulated 
 by the surface of a cone, and terminating in the apex of the conic surface as 
 the centre. 
 
AND COMPLETE DECORATOR. 
 
 145 
 
 Sometimes, between the spiral lines, the intermediate surfaces are formed 
 into mouldings, as in the Temple of Erechtheus and Minerva Polias at 
 iVthens. These mouldings increase the effect of the simple outline of the 
 volute to extreme elegance, particularly when the volute is large in propor- 
 tion to the diameter of the column. 
 
 Def. 2. Cathetus of a volute is a straight line drawn through the centre 
 of the eye, perpendicular to the horizon, to meet the uppermost spiral line. 
 
 PLATE XXII. 
 
 TO DRAW THE VOLUTE, FIG. 1, THE HEIGHT, B E, BEING GIVEN. 
 
 Divide the perpendicular height b e into eight equal parts, the points of 
 division being l', 2', 3', 4', 5', 6', 7', proceeding from the bottom upwards. Bisect 
 the distance between the points 3' and 4’ in the point 3, and from the point 3 
 thus found as a centre, with the radius 3 3’ or 3 4 , describe the circumference 
 of a circle which will form the eye of the volute. Through the centre of the 
 eye draw the straight line c f perpendicular to b e, and bisect the right angle 
 c 3 e formed at the centre of the volute by the straight line a d. 
 
 Divide the part n l of the line a d intercepted by the opposite parts of the 
 circumference, which contains the eye of the volute, into six equal parts. 
 From the point l, with the radius lb, describe the arc bcd; from the point 
 n, with the radius nd, describe the semicircular arc d e fg; from the centre 5, 
 with the radius 5 g, describe the semicircular arc gh; from the centre 1, with 
 the distance 1 h, describe the semicircular arc hi; from the centre 4, with the 
 radius 4 i, describe the semicircular arc i k ; from the centre 2, w'ith the radius 
 2 k, describe the semicircular arc k m, and lastly describe the arc mn; then 
 will the curve line abcdefghikmn be one of the spirals of the volute 
 required. 
 
 2 o 
 
146 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 A spiral described in this manner, with the centres all in one straight 
 line, is not exactly circular, the outlines being somewhat elliptical. If the 
 spiral had been set upon the line cf instead of bc, we should have had 
 the same curve as that given by W. Newton in his translation of Vitruvius, 
 for the Vitruvian volute. The species of this curve is not of any of the 
 mathematical forms which we have described; its figure makes a rude 
 approach to the proportional or logarithmic spiral, and though not so eligible 
 in the rich orders of Architecture, it may nevertheless be used in many 
 fanciful compositions with considerable effect ; the chief thing to recommend 
 it is the simplicity of the operation by which it is described : its application 
 
 is partial, since it can only be described to a particular point; but to 
 
 form an elliptical volute, with compasses, of a graceful figure, to any 
 length and height, is a problem which involves considerable difficulty in its 
 construction ; we shall, however, give a figure which will approach to the 
 truth of such a method. 
 
 PROBLEM I. 
 
 To describe an elliptical volute to any given height and projection, from 
 the centre, on one side of the centre very nearly. — 
 
 Let d f, fig. 2, be the proposed height. Divide df into eight equal parts, 
 
 as before. Bisect the fourth part from the bottom, or the fifth part from the 
 
 top, in the point e ; so that if the whole line f d were divided into sixteen equal 
 parts, the part e f of the line d f below e, would contain seven of these parts, 
 and the part ed would contain nine of the same parts. 
 
 From the point e as a centre, with the radius equal to one of the sixteen 
 equal parts, describe the circumference of a circle which will have the same 
 diameter as the eye of the volute. To render the diagram as clear as 
 possible, this circle is described with a dotted line instead of a continued 
 line. Through the centre e draw the straight line co perpendicular to df, 
 and draw the straight line o h, bisecting the angle deg, and intersecting 
 
AND COMPLETE DECORATOR. 
 
 147 
 
 the dotted circle in the point l : again, through e draw i k perpendicular to 
 
 0 h, intersecting the dotted circle in the points i and k. 
 
 In the line cg make ea equal to the projection of the volute, and from 
 the point l, as a centre with the radius ld, describe the arc da, meeting cg 
 in the point a. Divide ba into six equal parts, and make ac equal to one 
 of the six equal parts, and in oh make lp equal to bc, so that ep is equal to 
 the sum of half the height of the eye and the distance b c. Again, in oh 
 make eq equal to ep, and through the points i, q, k, p, describe the rectangle 
 12 34, of which the sides 14 and 23 are parallel to pq, and the sides 43 and 
 
 1 2 are parallel to i k. Draw the diagonals 1 3 and 2 4 of this rectangle, and 
 divide each of them into six equal parts. Describe the rectangles 56 7 89 
 and 9-10-11-12. Prolong 4 1 to a, 2 1 to b, 2 3 to c, 4 3 to d , 8 5 to e, 6 5 to/, 
 6 7 to g, 8 7 to h , 12-9 to 
 
 From the centre 1, with the distance id, describe the arc a b; from the 
 centre 2, with the distance 2 5, describe the arc bc; from the centre 3, with 
 the distance 3c, describe the arc cd; from the centre 4, with the distance 
 4 d, describe the arc de; from the centre 5, with the distance be, describe the 
 arc ef; from the centre 6, with the radius 6f, describe the arc f g; from 
 the centre 7, with the radius 7 g, describe the arc g h; from the centre 8, 
 with the radius 8 h, describe the arc hi; from the centre 9, with the radius 
 9 i, describe the arc ik; from the centre 10, with the radius \0k, describe the 
 arc kl; from the centre 11, with the distance 1 11, describe the arc Im; from 
 the centre 12, with the distance 12 m, describe the arc mn. 
 
 Thus we have one of the elliptic spirals which will form a part of the volute. 
 Lastly, round the point e, as a centre, describe an ellipse, of which the two 
 axes are in the same ratio as the height of the volute is to its width, so that the 
 axis may be upon the lines d f and c g, and that the same axis major may be 
 equal to the distance e n. Then another spiral being described in the same 
 proportions, will complete the volute. 
 
 This is a singular example, and it shews what may be accomplished by 
 the powers of Geometry. The joinings are not quite perfect, as the two 
 
148 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 adjacent angles at the beginning of every revolution have not their centres in 
 the straight line passing through them and the point of junction. 
 
 PROBLEM II. 
 
 To describe a volute to any given height, and to terminate at the end 
 of any given number of revolutions in the circumference of a circle whose 
 diameter is given. 
 
 Divide the difference between the diameter of the eye and the height of 
 the volute into four times as many equal parts, together with two more parts, 
 as the number of intended revolutions is to consist: that is, if the spiral 
 is to have one revolution, the remainder must be divided into six equal parts; 
 if it is to have two revolutions, the remainder must be divided into ten 
 equal parts ; if three revolutions, into fourteen equal parts, and so on ; for 
 four limes 1 with 2 makes 6, four times 2 with 2 makes 10, four times 3 
 with 2 makes 14, and so on. 
 
 Then take half the number of these parts and one more, together 
 with half the diameter of the eye, and set this distance from the top of 
 the volute downwards, and the lower extremity will be the centre of the 
 eye ; or take half the number of those parts made less by one, with half 
 the diameter of the eye, and set this distance on the perpendicular line up- 
 wards, and the upper extremity will give the centre of the volute as before. 
 
 Construct a square of which the centre is the centre of the volute, and 
 two of its sides are parallel to the perpendicular, and consequently the other 
 two at right angles to each side are equal to one of the equal parts of the 
 number into which the difference between the diameter of the eye and 
 the height of the volute is divided. Draw' the two diagonals of the 
 square which will cut each other in the centre of the volute, and divide 
 the half of each diagonal into as many equal parts as the intended number 
 of revolutions are, which w ill give the centre of the spiral. 
 
 Join the point of division, so as to make as many squares w'ith that first 
 constructed as there are to be revolutions, the sides of these squares being 
 
AND COMPLETE DECORATOR. 
 
 149 
 
 all parallel and perpendicular to each other. The centres of the first revolu- 
 tion are in the vertices of the angles of the first square ; the centres of the 
 second revolution are in the vertices of the angles of the second or next 
 square, and so on ; the first centre of every revolution being in the same half 
 diagonal, and in the first vertex of every square ; every centre is the centre 
 of the quadrantal arc of a circle, terminated upon two straight lines drawn 
 from the centre of this arc, of which straight lines, one is either upon the height 
 line, or parallel to the height line, and the other is perpendicular to the 
 height line. 
 
 In describing the spiral, it is obvious, that if we begin next to the centre, 
 we shall continually expand; or if we begin at the most remote point, we 
 shall continually diminish, and at last touch the eye of the volute. It will be 
 preferable to begin drawing the spiral at the most remote point. In beginning 
 the volute, it must be considered whether the volute is to be upon the right 
 or left side ; if upon the right side, the first quadrant will have its centre on 
 the left extremity of the upper side of the first square, and this side of the 
 square prolonged from the other extremity will terminate the other extremity 
 of the arc. 
 
 In general, the point from which an arc which forms a portion of a spiral 
 is drawn, as also the points in which it terminates, is in a straight line, 
 either in the height line, or in a line parallel to the height line, or in a line 
 perpendicular to the height line of the volute; the point in which any arc 
 begins is in the prolongation of the side of the square from the centre of the 
 arc, and the point in which any arc terminates is in the prolongation of 
 the adjacent side of the square from the remote extremity to the centre. 
 
 The methods of describing spirals by means of squares limited to three 
 revolutions, may be seen in the oldest Italian authors ; but the generalization 
 of all the known principles of describing spiral lines by means of compasses, 
 was first accomplished by Mr. P. Nicholson, and published in his ‘Principles 
 of Architecture, 1796,’ as dotted on the Plate. 
 
150 
 
 CABINET-MAKER, UPHOLSTERER, 
 
 Example. 
 
 Let it be required to describe a volute with four revolutions: let the 
 point c be the given centre. From the point c, with the semidiameter of the 
 eye, describe the circumference of a circle. Through c draw la parallel to 
 the height line, intersecting the circumference of the circle in the point r. 
 Divide a r into two equal parts at the point u, and divide ur or ua into one 
 part more than the number of intended revolutions, and set one of these parts 
 from the centre each way upon the perpendicular, and the line drawn through 
 the centre at right angles to the perpendicular. Through these points draw 
 the square 1 234, of which the sides 2 1, 3 4 are perpendicular to, and the 
 sides 2 3, 1 4 parallel to, the height line. 
 
 D raw the diagonals 2 4, 13, which will intersect each other in the centre c 
 of the circle, and divide each half cl, c 2, c3, c 4, into four equal parts, 
 because four revolutions are intended in the volute. Through these points 
 draw squares, of ^hich the sides are parallel to the sides of the first square 
 1 2 3 4, then 1, 2, 3, 4, are the centres of the outer spiral of the first revolution 
 of the volute 5, 6, 7, 8, the centres of the outer spiral of the second revolu- 
 tion, 9, 10, 11, 12, the centres of the outer spiral of the third revolution, and 
 13, 14, 15, 16, the centres of the outer spiral of the fourth revolution. These 
 centres are clearly exhibited, fig. 3, No. 1. 
 
 Prolong the line 1 2 to m, 2 3 to n, 3 4 to o, 4 1 to p, and so on. From 
 the centre 1, with the radius 1 a, describe the arc a m; from the centre 2, with 
 the radius 2 m, describe the arc m n; from the centre 3, with the radius 3 n, 
 describe the arc n o ; from the centre 4, with the radius 4 o, describe the arc 
 o p, and so on, completing the first revolution. 
 
 By proceeding in the same manner, we shall find a second series of centres 
 for the second spiral, and thus the volute will be completed. 
 
AND COMPLETE DECORATOR. 
 
 151 
 
 PROBLEM III. 
 
 To describe the logarithmic spiral to any given height, so that the centre 
 of the volute may divide that height in a given proportion. — 
 
 Figure 4. 
 
 Through the centre of the spiral draw a straight line perpendicular to the 
 height, and find a mean proportional between the two parts of the height, and 
 mark a distance from the centre upon the line drawn through the centre 
 perpendicular to the height, then the extremity of this perpendicular will be 
 a point in the spiral as well as at the extremity of the height. Bisect each 
 of the right angles, and find the bisecting lines a mean proportional between 
 the next two lines, one on each side of it, then the ordinates will be at 45° 
 distant from each other: by this means we have found one half of the first 
 revolution. In the same manner we shall find the other half of this revolu- 
 tion. But if the ratio between any two adjacent ordinates be given, we shall 
 be able to find all the succeeding ordinates ; suppose, for example, the first 
 two ordinates are given, 0-20 and 0-18-3, then every third ordinate will be a 
 third proportional to the two given ones; say, as 20: 18-3:: 183 : 16-7, 
 18*3 : 16’7 : : 16 7 : 15*3, 16 7 : 153 : : 15 3 : 14, and so on, so that by pro- 
 ceeding in this manner we shall have the points 1 2*8, 11*7, 9'8, &c. and by 
 drawing a curve through them, we shall have the logarithmic spiral, as 
 described. 
 
 The arcs of circles may be described through every three given points, so 
 as to form a spiral in the following manner. Take the length of the middle 
 ordination as a radius, then from each extremity of each extreme ordinate 
 describe the arc of a circle to meet each of the extreme ordinates. Thus, in 
 fig. 4, the lines or distances 0-20, 0-18-3, 0-16-7, are three consecutive ordi- 
 nates; then taking the middle ordinate 0-18-3 as a radius from the points 20 
 and 16'7, describe an intersection. From this point of intersection, with the 
 same radius, describe an arc 20-18'3-16*7. 
 
152 
 
 CABINET-MAKER, UPHOLSTERER, &C. 
 
 Again, 147, 153, and 14, are the three next consecutive ordinates, the 
 middle one being 153, and the extreme ordinates 167, and 14. Take the 
 middle ordinate 15 3 as a radius, and from each of the points 16-7, and 14, * 
 describe an intersection; and from the point of intersection as a centre 
 describe the arc 16 7- 15*3-14, and thus two quadrantal arcs are described, 
 which together form half a revolution of the exterior spiral of the volute : in 
 the same manner the whole outward spiral may be formed 
 
 If any point be given in the curve of a second spiral, and in the greatest 
 ordinate, the second spiral may be described to the same proportion in all its 
 ordinates as the first, by saying, as the first ordinate is to the second in the 
 first spiral, so is the first ordinate to the second in the second spiral. Again, 
 as the second ordinate is to the third ordinate in the first spiral, so is the 
 second ordinate to the third in the second spiral, and so on; then when all 
 the points are found, draw a curve which will complete the second spiral, 
 and thus the entire volute, according to the principles of the proportional 
 spiral. 
 
 This method of describing the proportional or logarithmic spiral, was 
 discovered by the author of the present work, in the year 1811, for the 
 express purpose of instructing his pupils, and was first published very shortly 
 after in Rees’ Cyclopedia; since that time, the method has been copied into 
 several other publications. 
 
( 1 ) 
 
 GLOSSARY OF TECHNICAL TERMS 
 
 CABINET-MAKING; 
 
 WITH A DESCRIPTION OF APPROPRIATE FURNITURE ADAPTED TO THE VARIOUS SPECIES OF 
 ROOMS IN DWELLING-HOUSES, AND OF THE OPERATIONS TN CABINET-MAKING. 
 
 BAN 
 
 ANGLES, in Cabinet-work, are those two 
 surfaces of a piece of furniture which meet and 
 form a corner. See Block. 
 
 ANIMAL Forms, were introduced as embel- 
 lishments into furniture by the ancients: of 
 
 these we have many Greek examples, both real 
 and chimerical. 
 
 ANTE ROOM Furniture, should contain some 
 appropriate sculpture, and a few' paintings, with 
 a suitable number of marble tables, witli massive 
 carved plain frames, and a convenient number 
 of seats of a simple but elegant form, so 
 as to admit of two or three persons to sit. 
 There might also be narrow tables attached 
 to the walls by hinges, supported by scroll 
 brackets. 
 
 ARM-CHAIR, a chair with resting-places for 
 the arms, and hence they are frequently called 
 Elbow-chairs. 
 
 B. 
 
 BAMBOO, a species of cane which grows in 
 India to a large size. Bamboo is employed in 
 chairs and tables: of this material, bed-room 
 chairs are frequently constructed. 
 
 BANDING, is a narrow veneer used as a 
 border, or part of a border, either round the 
 solid wood, or round a large veneer in the centre. 
 When the part surrounded is solid, a rebate 
 must be sunk for the banding. Cabinet-makers 
 employ three kinds of banding. When the wood 
 is cut lengthwise of the grain, it is called 
 Straight Banding; when the wood is cut across 
 the grain, it is called Cross Banding; and when 
 it is cut at an angle between two, it is called 
 Feather-edged Banding. From the various 
 species of coloured woods, a most astonishing 
 variety of elegant combinations may be pro- 
 
 B I L 
 
 duced. The grain of the wood at the joints 
 should be w'ell matched. 
 
 BASIN-STAND, a piece of furniture placed 
 in bed and dressing rooms, to hold a vessel or 
 basin with water for washing. 
 
 BASSO-RELIEVO, a carving in which the 
 projection of figures is less than its real propor- 
 tion, or less than the true projection, supposing 
 the figures imitated to be of the real size. 
 
 beAd, a semicylindric moulding. Beads are 
 of two kinds; when the moulding touches the 
 surface from which it is made, and the remain- 
 ing part within the surface, the bead is called a 
 Recess Bead; but when the semicylindric sur- 
 face of the bead projects beyond the surface 
 to which it is attached, the bead is called a 
 Cocked Bead. 
 
 BED-ROOM Furniture, are bedsteads, dress- 
 ing-tables, drawers, washing-stands, dressing- 
 glasses, clothes-horses, night conveniences; as 
 also various small articles, so well known as not 
 to require enumeration. 
 
 BEDSTEAD, a frame of wood or metal, upon 
 which a bed is laid. The metal generally 
 employed in the construction is either iron bars 
 or brass tubes. 
 
 BEECH, a hard close wood much employed 
 in chairs, bedsteads, sofas, &c. It is very soon 
 injured by worms. 
 
 BIDET, a seat containing a pan of japanned 
 tin or earthenware, most commonly made in 
 the form of a stool, having a loose cover. 
 
 BILLIARD-RDOM Furniture, should not 
 have much embellishment ; and whatever orna- 
 ments may be employed, they should have some 
 analogy to the use of the room. 
 
 BILLIARD-TABLE, a table to play at bil- 
 
 2 Q 
 
BOO 
 
 ( 2 ) 
 
 B U H 
 
 liards upon. The top is framed with small 
 pannels, and covered with cloth, in order to 
 prevent the whole, or any part, from warping ; 
 the pannels ought to be cut so that the annular 
 plates should be perpendicular to the apparent 
 surface of the work. 
 
 BLIND, a screen placed on the inside of a 
 window, to prevent the rays of the sun from 
 entering the interior, or persons without from 
 seeing into the room. High blinds, or those 
 that extend the height of the window, are used 
 as sun-blinds ; and low blinds are used in 
 parlour windows, to prevent people from seeing 
 the interior or its contents: for that purpose 
 wire gauze fixed in mahogany frames is much 
 used. 
 
 BLOCK, a piece of wood glued into an angle 
 where tw r o pieces of wood meet, in order to 
 secure them more firmly than the mere joining 
 of the two parts by glue or otherwise. 
 
 BOASTING, is the first process used in 
 carving, by cutting away such of the superfluous 
 parts of the wood as do not require any great 
 attention, leaving the remaining portion for the 
 more minute part of the operation. The carver 
 who is best skilled in the application of drawing 
 is the most fit to be employed in this part 
 of the operation, as he is a better judge of 
 the quantity of sinking that is requisite. The 
 general outline is shaped by the use of saws and 
 gouges. 
 
 BOOK-CASE, a kind of press or cupboard 
 fitted up with shelves, at appropriate distances, 
 for the reception of books, maps, prints, and 
 other things adapted to scientific or literary 
 amusement. A book-case may either be a 
 piece of insulated furniture, or fitted up as 
 a portion of the room which is destined for its 
 reception. 
 
 The most common kinds of book-cases are 
 made of mahogany, but those of a superior cast 
 are made of oak, with ^arved roses and other 
 ornaments. Sometimes rose-wood is used, with 
 gilded ornaments, and inlaid with brass. 
 
 The shelves of book-cases are generally 
 moulded on the edge, which is also sometimes 
 made of a flat-round, with a brass bead put in 
 the middle. The shelves may either be made to 
 rest on racks, or on round pins of hard wood of 
 about an inch in diameter, inserted on the end 
 about three-eighths of an inch, and projecting as 
 much. The half portion of the pin which pro- 
 jects is cut away, and the other half is let into 
 the other side of the shelf, each shelf being 
 supported by four pins. 
 
 Four square-headed brass pins are sometimes 
 used for supporting each shelf, instead of 
 
 wmoden pins, with the square part or head let 
 into the under side of the shelf, the holes for 
 receiving them being in narrow brass plates, the 
 circular part of the pin about one-fourth of an 
 inch in diameter. 
 
 The distances and breadth of the shelves must 
 be regulated by the sizes of the books. In order 
 that the same cell or recess may be made to 
 answer to a uniform row of books, all of a given 
 size, the racks and holes for the pins should be 
 adjusted, so as to alter to the required dimen- 
 sions of the four following species of books, 
 viz. twelves, octavo, quarto, and folio, the ex- 
 
 treme sides being 
 
 
 
 Twelves . 
 
 7i in. 
 
 high, and 4i in 
 
 Small octavo . 
 
 9 
 
 . . . 5i . 
 
 Imperial octavo 
 
 lli 
 
 . ... 8i . 
 
 Small quarto . 
 
 11 
 
 . . . 9 . 
 
 Imperial quarto 
 
 1 6i 
 
 . . . Hi . 
 
 Small folio 
 
 15 
 
 . . . lli . 
 
 Imperial folio 
 
 23 
 
 . . . . 16 . 
 
 The lower recess should always be deep enough 
 for the largest folios. 
 
 BOOK-CASE DWARF, a book-case, all the 
 shelves of which can be reached without steps. 
 
 BORDERS, a display of foliage, fret-work, 
 or inlaying round a plane or uniformly decorated 
 surface. See Inlaying. 
 
 BRACKET, a support fixed against a wall, 
 for placing any useful article or ornament upon. 
 The shelf supported by the bracket may either 
 be fixed or moveable by means of hinges. 
 
 BRASS, Inlaying with. See Inlaying. 
 
 BRASS-WORK, cleaning of. When veneers 
 are inlaid with brass, the brass should be filed 
 level with a fine single-cut file, then rubbed with 
 pumice-stone, next with fine powdered tripola 
 mixed with linseed oil upon a piece of felt or 
 hat, until the desired effect is produced. When 
 the ground is of ebony, or dark rose-wood, the 
 brass may be brought to a better surface with 
 char finely powdered. 
 
 BRONZE, a compound metal consisting of 
 tin and copper, used for cast ornaments, &c. 
 The term is also applied to a green colour made 
 in imitation of the colour which this metallic 
 composition acquires by age. 
 
 BUHL-WORK, a kind of inlaid work em- 
 ployed in patterns, borders, &c. which took its 
 rise about the middle of the 17th century, and 
 w as carried to great perfection in France by the 
 ingenuity of many able artists, among these 
 the most distinguished was the famous Buhl, 
 from whom the name of this species of work is 
 derived. 
 
 Th is kind of inlaying was introduced into 
 other countries, but declined for want of encou- 
 
B U H 
 
 CAS 
 
 ( 3 ) 
 
 ragement. It has lately been revived in England, 
 where it has risen to superior elegance, and has 
 superseded those gaudy imitations of animals, 
 fruit, flowers, &c. which were so much employed 
 about ten years ago. 
 
 In joining this ornament with that to which 
 it is to unite, the part for the ornament and that 
 for the ground must be glued together, drawn 
 upon the one, and laid upon another such piece 
 joined in the same manner, but placed in reverse 
 order ; then the design being drawn upon the 
 one, and the whole substance making one thick- 
 ness, is cut through by means of a fine bow- 
 saw, thus producing four parts joined in twos 
 reversed to each other. 
 
 The plates of brass, or other metal, shoidd be 
 about the usual thickness of a veneer, previously 
 roughed on both sides with a coarse file or 
 smoothing-plane, as also the veneers or other 
 substance to be combined with them. In the 
 act of joining, or connecting them together, 
 the plates and veneers should be previously 
 heated, a thin coat of glue being passed over 
 one of the metal plates, and covered with a sheet 
 of thin paper, coat the paper also with glue, and 
 cover it with the veneer. Pinch them together 
 by means of a screw-press, or otherwise by first 
 placing them between smooth and even boards, 
 to prevent the injury which they would sustain 
 by the pressure. This being done, the work 
 must remain till it becomes dry, the parts will 
 then be found to adhere together with sufficient 
 firmness for cutting the pattern, which should 
 be drawn on the veneer, or, if the lines should 
 not be sufficiently distinct, on a piece of paper 
 pasted on the veneer. The pattern should be 
 cut with a very fine narrow bow-saw, and, in 
 order to join the parts so as to produce two 
 complete ornaments, the one consisting of a 
 veneer inlaid with metal, and the other of metal 
 inlaid with a veneer ; place a piece of paper of 
 a sufficient size, with the veneer over it, upon a 
 plane surface, and insert the metal in the veneer 
 with strong glue, rubbing it well down with the 
 veneering hammer and with glue. Cover the 
 whole over with another piece of paper, then 
 having warmed and greased it over with tallow, 
 pinch them by means of a press. When properly 
 dry, the work will separate from the boards; 
 the paper being taken away, it may be laid in 
 its place as a veneer : a caul is usually employed 
 in preference to the hammer. The reverse pat- 
 tern should be prepared in the same manner. 
 The process is the same whether metal and 
 w'ood, or metal and tortoise-shell, or two woods 
 of different colours, are employed. When 
 tortoise-shell is used, its underside is sometimes 
 
 coloured with red, yellow, or some bright colour, 
 and is sometimes gilt. 
 
 BUREAU, a desk. 
 
 BUST, an imitation of a man or woman 
 downwards as far as the breast, carved of any 
 material, or cast from a mould. 
 
 C. 
 
 CABINET-MAKING, the art of constructing 
 in wood every species of furniture useful for the 
 conveniences of man. 
 
 CABLING, an imitation of the twisted strands 
 of a rope. Flutes are frequently filled with 
 cables, which produce a very good effect. 
 
 CADDY, a small box for keeping tea. 
 
 CANDELABRA, a candlestick supported 
 from the floor, and used in various situations. 
 Sometimes one is placed at each end of a side- 
 board. 
 
 CARD-TABLE, a table to play at cards upon ; 
 the top is generally covered with green cloth, 
 and made in parts so as to fold together. 
 
 CARVING, the art of reducing any species of 
 material similar to the form of a given object. 
 
 With respect to Cabinet-making, the material 
 upon which this operation is performed is wood. 
 The first part of the work is to form the object 
 roughly, leaving the surface to be completed by 
 a subsequent part of the labour. The first part 
 of the process is called Boasting : the next part 
 of the process consists in forming the principal 
 directing lines, so as (o be a guide to the finish- 
 ing of the work. — In carving valuable wood, the 
 workman must contrive to put the ornament 
 together in parts, so that the joints may be the 
 least visible when the work is complete. The 
 combination of carved and turned works with 
 annular leaves, produces a most beautiful effect, 
 particularly when the turned part is ornamented 
 with flutes or reeds. 
 
 CAST, the figure of a required object, formed 
 in a mould by a material in a liquid state. 
 
 CASTING Ornaments or Moulding to re- 
 semble wood. Make a very clear cement with 
 five parts of Flanders glue, and one part of 
 isinglass, by dissolving the two kinds separately 
 in a large quantity of water, then, after having 
 separated the heterogeneous parts which could 
 not be dissolved by straining them through fine 
 linen, mix them together. The glue thus pre- 
 pared must be so much heated, that the finger 
 can but only endure the temperature ; by this a 
 little water is evaporated, and thus the glue 
 acquires more consistence. Mix raspings of 
 wood, or saw-dust, passed through a fine sieve, 
 with the glue, forming it into a paste. Having 
 rubbed the plaster or sulphur-mould with linseed 
 or nut oil, as in plaster-casts, put in the paste. 
 
C H A 
 
 C L E 
 
 ( 4 ) 
 
 and press the parts by hand, so that no vacuity 
 may remain ; and in order that the whole may 
 acquire a perfect form, cover it over with an 
 oiled board, place a weight above it, and when 
 the casting is dry remove the rough parts, and 
 if any inequalities remain, they are smoothed; 
 the ornament thus prepared may be fixed with 
 glue to the thing intended. 
 
 CASTERS, rollers which revolve round 
 sockets affixed to the legs of tables, chairs, 
 &c. for the more easily removing them from 
 one place to another. 
 
 CAUL, a kind of mould, made of solid wood, 
 used in veneering; it is formed to the surface to 
 be veneered, and having been well heated, oiled, 
 and greased, it is screwed down upon the ve- 
 neer, and its heat and pressure forces the glue 
 out of the joint, and therefore causes the veneer 
 to bed closely upon the ground. Cauls for 
 curved surfaces are sometimes made of thin 
 wainscot ; being heated, they are made to bend to 
 the surface. Cauls for card-table tops are gene- 
 rally made of inch deal, and keyed across with 
 wainscot, to keep them straight. 
 
 CEDAR, a name given to two kinds of wood ; 
 the one, which has a very strong scent, is called 
 Pencil Cedar; the other is simitar in appearance, 
 but without any perceptible smell. 
 
 CELLARET, a wine-cooler, or wine-cistern. 
 The most convenient situation for a cellaret is 
 under the centre of a sideboard when it is not in 
 use. The smallest size should be made to hold 
 nine bottles, and for larger roonvs the dimensions 
 of the cellaret may be increased so as to hold 
 twelve or fifteen bottles. The depth should not 
 be less than 13 inches, and the space for each 
 bottle is usually 3| inches square. 
 
 CEMENT-STOPPING, a composition made 
 with clear glue and fine saw-dust, used for 
 stopping up the holes or defects of hard wood. 
 When the surface is to be painted, whiting may 
 be used instead of saw-dust; and the stopping 
 must be allowed to be thoroughly dry, before 
 anv attempt can be made to finish the surface. 
 
 CHAIR, the well-known article for sitting and 
 resting ourselves upon. 
 
 CHAIR-MAKING : the tenons should always 
 be in the direction of the grain of the wood ; 
 and as the framing is of a trapezoidal form, the 
 mortises should be made obliquely, to receive 
 the tenons ; they are easily put in a vertical 
 position, by means of a proper saddle made to 
 the bevel : the greatest care should be taken 
 that the sides of the mortise and the sides of the 
 tenons should be parallel, and as exactly fitted 
 to each other as possible ; by this means, when 
 the parts are properly glued, and the work is 
 
 forced together by means of a cramp, the parts 
 wiil unite in the strongest manner. — In cliaii- 
 making, the beauty of the work depends on 
 correct forms of the curves and scrolls, and in 
 being agreeable to the eye. 
 
 CHAISE LOUNGE, a kind of sofa. 
 
 CLEANING of Work, is the art of reducing 
 the surface to the required degree of smooth- 
 ness. The finishing of the surface of work is 
 also called Polishing. 
 
 To clean the surface of solid work — Smooth 
 it by means of a finely set smoothing-plane, and 
 reduce the ridges by a scraper; then rub the 
 surface with glass paper, finishing.it with the 
 finest kind : but if the grain of the wood be open, 
 we must proceed farther in addition to the 
 process now described. Let the surface be 
 uniformly wetted with a wet spunge, and when 
 it becomes dry, rub it a second time with glass 
 paper, till it has acquired the requisite degree 
 of smoothness. 
 
 This last or additional part of the process 
 might have been accomplished in the following 
 manner. While the surface is wet, rub it with 
 a piece of pumice-stone in the direction of the 
 fibres; and when it becomes dry, wet it again, 
 and the grain will be raised in a much less 
 degree than in the former wetting ; then the 
 process of rubbing being repeated, the surface 
 will be found to be still more compact, and will 
 therefore receive a better polish. 
 
 To clean a veneered surface — Having taken or 
 scraped away the glue, tooth the surface in an 
 oblique direction to the fibres, and in proportion 
 as the surface acquires a regularity, set the plane 
 finer. The final part of the operation of planing 
 is accomplished by a finer tooth-plane. Remove 
 the marks of the tooth-plane by a scraper, and 
 finish the surface with glass paper, or pumice- 
 stone and glass-paper. Veneers, being of a 
 closer texture than solid wood can always be 
 found, do not require so much labour as open- 
 grained solid wood. 
 
 To clean veneers laid with brass— File the 
 brass level with a fine single cut file, and then 
 rub it over with pumice-stone, next with fine 
 powdered tripoli mixed with linseed oil, until 
 the desired polish is acquired. 
 
 CLEANING old Furniture, is the art o 
 giving the surface an appearance of newness 
 or at least of freshness, which may be accom- 
 plished by means of the following operation ; — 
 
 When furniture has been finished by means of 
 the French polish, grease or dirt may be re- 
 moved from the surfaces hy rubbing it quickly 
 over with a little spirit of turpentine, which will 
 not soften the varnish. 
 
COL 
 
 ( 
 
 5 
 
 COM 
 
 When furniture has been finished with wax 
 composition, the polish may be renewed by 
 repeating the original process of the wax compo- 
 sition with a small quantity, carefully rubbed off. 
 
 To remove stains in tables — Wash the surface 
 over with stale beer or warm vinegar, and the 
 stains will be removed by rubbing them with a 
 rag dipped in spirits of salt, then the surface 
 may be repolished in the same manner as new 
 work is done. 
 
 When the work is not stained, wash the sur- 
 face with clean spirits of turpentine, and repolish 
 it with furniture oil. 
 
 To clean lacquered brass-work in furniture — 
 Wash it over with a soft linen or muslin rag 
 wetted in warm w'ater ; and if the spots cannot 
 be removed by this method, they must be re- 
 lacquered. 
 
 COLOURED WOODS, are those which have 
 a natural colour, or such as may be produced 
 by the process of dying. Wood of the natural 
 colour is always the most valuable ; but as it can 
 never be obtained in sufficient quantities to form 
 large surfaces without striking defects, a com- 
 bination of different coloured woods, or of metals 
 arid shells with woods, has been resorted to. 
 These combinations are effected by coloured 
 bands and ornaments of wood, or by beads, 
 lines, or ornaments of brass, which are fre- 
 quently enriched by engraved lines on the 
 surface. In such combination, care must be 
 taken to harmonize the colours, so as to give 
 sufficient relief. 
 
 There are three primitive colours by w hich all 
 the colours in nature may be produced ; these 
 are red, yellow, and blue. Green is produced 
 by the mixture of yellow and blue, orange by 
 the mixture of red and yellow, and so on. 
 
 Every two colours in the six following, which 
 produce the most opposite effect, may be found 
 thus — Divide the area of a circle into six equal 
 sectors, by straight lines drawn from the cir- 
 cumference to the centre ; then suppose the 
 first to be made blue, the third red, the fifth 
 yellow, then the second will be purple, the 
 fourth orange, and the sixth green ; and the 
 colours which are in opposite sectors are those 
 which are the most opposed to each other in 
 their nature : thus, orange being opposite to 
 blue, have the most opposite quality ; and it is 
 worthy of observation, that the intermediate 
 colour between two adjacent primitive ones, is 
 the composition which arises from their mixture ; 
 thus, between yellow and blue we shall have 
 green, between blue and red we shall have pur- 
 ple, and so on. 
 
 Again, by various combinations of the six 
 
 ) 
 
 colours thus produced, an infinite variety of 
 gradations may be produced. If any two co- 
 lours in opposite sectors be mixed, a neutral 
 tint or brown will be the result ; thus, orange 
 and blue make brown, purple and yellow make 
 brown, red and green make brown. 
 
 The illustration of the theory of colours may 
 be farther extended, by forming three equal 
 sectors of circles, of which the sides or radii are 
 the sides of an equilateral triangle, the arcs 
 being described from the vertices, so as to meet 
 every two sides of the triangle, and to divide its 
 area into seven parts, of which one will be in 
 the centre : then if the three entire sectors be 
 painted with the three primitive colours, the 
 intermediate one upon each side of the triangle 
 will be the colour which arises from the mixture 
 of those in the two adjacent quadrilaterals, 
 which terminate the angles of that side; and 
 the colour in the trilateral figure which occupies 
 the centre, being that which is produced from 
 mixing the three together, is brown. 
 
 Painters divide colours into warm and cold ; 
 the warm colours are yellow, orange, and red ; 
 and the cold colours are purple, blue, and green. 
 In each class of colours the middle one is that 
 which has the greatest intensity ; therefore in 
 the three warm colours the orange is the 
 warmest, and in the three colours the blue is 
 the coldest. 
 
 We shall only remark, that when any rich 
 colour is applied, very little variety need be 
 introduced; but if the colour be light and deli- 
 cate, the addition of others will be desirable in 
 order to give a more lively effect by contrast, 
 which, being skilfully managed, will give force 
 and lustre to the ground; but, on the contrary, 
 opposition would entirely destroy its natural 
 harmony. 
 
 It is of the greatest advantage to the cabinet- 
 maker to have a clear idea of the result arising 
 from the combination of two or more colours, 
 in order to give the best effect to his work, and 
 to avoid incongruous associations of them. 
 
 COMMODE, a low wardrobe, or a piece of 
 drawing-room furniture, of which the lower part 
 is often enclosed with doors, and the upper part 
 is furnished with book-shelves. 
 
 COMPOSITION ORNAMENT, is a species 
 of ornament made of whiting and thin glue of 
 the consistence of glaziers’ putty, and pressed 
 into a mould which has been previously oiled 
 with sweet oil. It is then taken out, and if 
 applied upon a plane surface, it is suffered to 
 dry; but if applied upon a curve surface, it 
 should be bent while it is soft, to the proper 
 form. 
 
 2 R 
 
 ft 
 
DIN 
 
 EAT 
 
 ( 6 ) 
 
 CONSOLE, an ornament with a double scroll 
 one turning inwardly and the other outwardly, 
 for supporting a cornice. Consoles are also 
 used as modillions. 
 
 CONSOLE TABLE, a table supported at 
 the wall upon consoles. 
 
 CONTOUR, the most extreme boundary of 
 a body. 
 
 CONTRAST, the difference of colour, or 
 figure, or proportion. 
 
 CONVERSATION CHAIRS, those that are 
 made to keep the body in an easy posture during 
 conversation. See the various designs in this 
 work. 
 
 COPAL, a varnish made of the juice of an 
 American tree: it is a concrete, hard, shining, 
 and odoriferous; the colour is transparent, like 
 that of citron. This substance is such as will 
 neither dissolve in spirit of wine nor essential 
 oils, except by a certain process. It may, how- 
 ever, be dissolved by digestion in linseed oil. 
 Copal varnish is superior to any other. 
 
 CORNICE, an ornament consisting of one or 
 several mouldings situated at the top of a piece 
 of furniture. 
 
 COUCH, a long seat on which persons occa- 
 sionally lie down to rest. 
 
 D. 
 
 DECORATION, every species of ornament is 
 called a decoration. 
 
 DEFECTS, are those parts of a piece of wood 
 where the material is wanting, as a hole, shake, 
 or the like. 
 
 DESK, a piece of furniture so constructed as 
 to make it convenient for writing upon, and to 
 contain small articles and valuable papers. 
 Those boxes used by travellers for writing upon, 
 consisting of two parts hinged together, are 
 called Portable Desks. 
 
 DINING-ROOM CHAIRS, are of course 
 those chairs adapted to the use of dining ; their 
 construction and ornament should depend on 
 the character of the architectural finish dis- 
 played in the embellishments of the room. 
 
 DINING ROOM FURNITURE, should, as 
 we have observed in chairs, depend on the 
 species of architecture introduced. As the walls 
 cannot, in the Grecian style, admit of being 
 entirely covered with architectural ornaments, 
 a few plain portraits may be introduced in the 
 large spaces. 
 
 DINING -TABLES, those used in dining- 
 rooms. Dining-tables are of various kinds; 
 some have frames and legs, and others have 
 pillars and claws : those with legs have the 
 advantage of being steady, and made in parts, 
 so as either to be extended or contracted at 
 
 pleasure. Tables with pillars and claws are 
 more elegant than those constructed with frames, 
 and in the act of dining they are entirely free of 
 the sitters at table; whereas in the other the 
 legs are in the way ; but they are much more 
 expensive in point of workmanship. Dining- 
 tables should be about 28 inches in height. 
 The tops of dining-tables are exposed when the 
 dessert comes to be set: the w r ood should be 
 uniform in its texture, even, and hard ; Spanish 
 mahogany is very well adapted to the purpose. 
 
 DOVE-TAILING, a mode of framing two 
 parts together by means of prismatic pins, and 
 recesses cut in their right section in the form of 
 dove-tails, so as to leave no open space between 
 the joints. There are several kinds of dove- 
 tailing, as common dove-tailing, lap dove-tail- 
 ing, and mitre dove-tailing, used on different 
 occasions. 
 
 DRAWERS, a deep frame consisting of 
 several shelves or partitions, with sliding-boxes 
 called Drawers, open at the top for holding 
 linen, ladies’ dresses, &c. The drawers, as well 
 as the frame or drawer-case, are always dove- 
 tailed. 
 
 DRAWING-ROOM, or Withdrawing-room, 
 a room in which company collect before dinner, 
 and to which they retire after. It is appropriated 
 to the reception of company and formal visits ; 
 hence the style of the furniture should be 
 elegant. The plain surfaces of the walls are 
 frequently embellished with large glasses upon 
 the solid parts between the windows, and still 
 more generally over the chimney-piece. The 
 more extensive surfaces should be ornamented 
 with a select choice of paintings adapted to the 
 size of the room. 
 
 The furniture of drawing-rooms consists of 
 commodes, pier -tables, fire-screens, vases, 
 bronzes, tables, sofas, couches, chairs, seats, 
 foot-stools, loo-tables, &c. Drawing-room 
 tables are made of the best mahogany, the most 
 richly figured British oak, rose-wood, &c. ; 
 sometimes foreign woods are employed ; the 
 tops are frequently inlaid with brass in orna- 
 mental borders. 
 
 DRESSING-ROOM FURNITURE, such 
 furniture as is appropriate to the use of the 
 dressing-room. This may consist of drawers, 
 dressing-table, writing-tuble or secretary, cabi- 
 nets, sofas, chairs, stools, wardrobe, &c. 
 
 E. 
 
 EATING-ROOM FURNITURE, should be 
 side-boards, dining-tables, chairs, side-tables, 
 waiters, &c. The decorations and style of the 
 furniture should be more plain than those of 
 the withdrawing-room. 
 
F R E 
 
 FUR 
 
 ( 7 ) 
 
 EBONY, a fine compact wood, much used in 
 mouldings which surround pannels. 
 
 ENTRANCE HALL, a room or passage used 
 for waiting. The furniture should consist of 
 marble tables and a convenient number of seats 
 and chairs. 
 
 The chairs of entrance halls usually have the 
 crest of the owner. The seats and chairs are 
 generally made of the same material, and are 
 often constructed of oak. The frames of the 
 marble tables should be massive, and they may 
 be either plain or carved. The style of the 
 furniture should bear some analogy to that of 
 the other rooms. 
 
 F. 
 
 FLUTES, a series of equal hollows sunk into 
 a plane, or cylindric, or conoidal surface. The 
 right sections of flutes are most commonly semi- 
 circular or semi-elliptic arcs. When they are 
 applied to a conic or a conoidal frustum, they 
 are made to diminish in proportion to the cir- 
 cumference of the body at every part of its 
 extent or height. Flutes sometimes come in 
 contact with each other, and sometimes a por- 
 tion of the surface from which they recede 
 remains between them ; these surfaces are called 
 fillets, and are generally much less in breadth 
 than the breadth of the flutes. The objects to 
 which flutes are applied are turned legs of tables, 
 chairs, &c. When flutes come in contact with 
 each other, the circular or elliptic must be 
 always less than a semi, as it would be im- 
 possible to preserve the arris; but though a 
 sharp arris is the perfection of the workmanship, 
 it is frequently taken away in order to preserve 
 it from injury. 
 
 FOLIAGE, ornaments consisting of leaves, 
 fruit, flowers, frequently connected with stalks, 
 making agreeable turnings in lines of contrary 
 curvature, and terminating in elegant spiral 
 lines. 
 
 FOOT-STOOLS, those for placing the feet 
 upon in order to give greater ease to the person 
 who sits. 
 
 FRENCH POLISH, is a new and durable 
 mode of polishing or varnishing, by which means 
 it is not so much necessary to polish the surface 
 of the wood itself. This mode consists in ap- 
 plying a considerable thickness of transparent 
 gum-lac over the surface, so that the surface of 
 the gum appears as if it were the surface of the 
 wood. 
 
 This polish is a kind of spirit-varnish, consist- 
 ing principally of shell-lac dissolved in spirits of 
 wine; and in order to render its colour paler, it 
 is mixed up with gum-mastich and gum-san- 
 darach in the following proportions : — Shell-lac, 
 
 3 parts ; gum-sandarach, 1 part; gum-mastich, 
 1 part; spirit of wine, 40 parts. — Thegum-sau- 
 darach and gum-mastich must be first dissolved 
 in the spirit of wine, and the shell-lac next. 
 Put these in a bottle loosely corked, and place 
 the bottle in a vessel of water heated to the 
 boiling point of spirit of wine, or a little below 
 173°, until the solution is effected. 
 
 The surface to be polished should be placed 
 opposite to the light, in order that the effect 
 produced by the polishing may be better seen ; 
 prepare a rubber formed of the flat coil of thick 
 woollen cloth torn off the piece, so that the face 
 of the rubber formed of the torn edge may be 
 soft and pliant; then, after having rolled it into 
 form from one to three inches diameter, secure 
 it with a thread coiled round it. The face of 
 the rubber being covered with a soft linen cloth 
 doubled, the remainder of the cloth at the other 
 end of the rubber forms a handle to hold it by. 
 The face of the cloth must be moistened with a 
 little raw linseed oil, and, if required, may be 
 coloured with alkanet-root, applied upon the 
 finger to the middle. The varnish is enclosed 
 in the double fold of the cloth. Then rub over 
 the surface quickly and lightly in circular direc- 
 tions, confining the operation to a space not 
 exceeding ten or twelve inches square, until the 
 area is gone over, and until the space is nearly 
 dry. Proceed in this manner with equal areas 
 or portions, until the whole surface has been 
 gone over: the operator must be careful to join 
 every two adjacent surfaces. 
 
 In giving the surface a second coat, the rubber 
 may be moistened with the varnish w'ithout oil, 
 and applied as directed. A third coat may be 
 laid on in the same manner, and then a fourth 
 with a little oil. The process must then be 
 repeated in a similar order, viz. with two others 
 without oil, and so on, repeating the coats until 
 the varnish has acquired its proper thickness. 
 Then the inside of the rubber being moistened 
 with a little spirit of wine, and having wetted it 
 with the varnish, rub the surface quickly and 
 uniformly over every part, which will greatly 
 contribute in giving it uniformity, and the polish 
 required. Next wet the cloth with a little spirit 
 of wine and oil without varnish, and rub over 
 the surface until it is nearly dry ; then the effect, 
 of the operation will shew whether the process 
 is complete or not; if not complete, the opera- 
 tion must be repeated. 
 
 FURNISHING, Styles of. The style of 
 Furnishing ought to correspond to the style 
 of architecture of the building, and more parti- 
 cularly that of the rooms in which the furniture 
 is placed. The style may, therefore, be Egyp- 
 
FUR 
 
 (jr ii O 
 
 ( 8 ) 
 
 tian, Greek, Roman, Gothic, &c. We have also 
 modern styles, as French and English. In point 
 of form, French and English furniture has each 
 a peculiar character ; but with respect to decora- 
 tion, the ornaments are imitated from the ancient 
 Egyptian, Greek, or Roman. There is no style of 
 architecture that affords such a variety, and so 
 many elegant forms, as the Greek. The articles 
 of furniture were, however, not very numerous 
 among the Greeks ; they consisted chiefly of 
 couches, chairs, tables, tripods, and vases; and 
 have been chiefly collected from Winkelman, 
 Count Caylus, and Stewart. From the exam- 
 ples thus found Mr. Hope formed a complete 
 system of Cabinet furniture. But though the 
 articles of Greek furniture were not very nume- 
 rous, the variety of ornaments which their 
 architecture have presented has afforded the 
 means of producing a multitude of elegant forms 
 and splendid embellishments. The couches of 
 the Greeks were of different forms; the plan 
 was sometimes three sides of a square, and 
 sometimes a semicircle ; the legs were often 
 formed into elegant shapes, and occasionally 
 highly ornamented. 
 
 The seats of the Greeks were adapted for one 
 or more persons ; those intended for more than 
 one were probably used for conversation, as 
 they were often of a semicircular form, which 
 was extremely convenient, as it brought the 
 persons opposite to each other. The table was 
 so situated as to be surrounded on three sides 
 by the couch. 
 
 In religious ceremonies, „ripods were some- 
 times used ; and a brazier, for warming the 
 apartments, was generally supported by a tripod 
 or stand. These tripods were generally of an 
 elegant form, consisting of three legs supporting 
 a ring to receive the brazier. The legs were 
 sometimes formed in the shape of the legs of 
 animals ; other decorations, as paws, heads, &c. 
 were also used. 
 
 Of all the kinds of furniture, the Greeks be- 
 stowed the most splendid ornaments upon their 
 vases. Of these vases some were used for do- 
 mestic purposes, others contained the ashes of 
 the dead, others were for ornament; but to what- 
 ever purpose they w'ere destined, they were 
 generally formed of the most elegant figure, and 
 decorated in the most splendid manner. 
 
 The ruins of Herculaneum and Pompeii have 
 recently furnished us with numerous examples 
 of the Roman style. The tops of the tables of 
 the Romans were sometimes supported by a 
 pillar, with a flat base or pedestal ; others by 
 a pillar rising from a base supported upon three 
 or four low turned feet. The legs sometimes 
 
 stood on a flat base, which was frequently 
 raised with turned feet and the claws of animals. 
 
 The furniture of the Romans was of a similar 
 description to that of the Greeks, but was greatlv 
 inferior in point of form and decoration ; they, 
 however, bestowed great attention on the form 
 and embellishment of their candelabra, which 
 were sometimes six feet in height. 
 
 The style of Egyptian furniture may be chiefly 
 collected from Denon, who travelled in the 
 Egyptian expedition with Buonaparte. 
 
 G. 
 
 GALLERY FURNITURE consists of seats to 
 admit of two or three persons, narrow tables 
 supported by scroll brackets, &c.: the walls are 
 decorated with the paintings of the great masters. 
 
 GERMAN STYLE of Carving, is a combina- 
 tion of fantastic chimerical subjects, introduced 
 by the Germans, but now antiquated. 
 
 GILDING, a glittering surface made by means 
 of gold leaf. In order to produce a golden eff ect, 
 the parts to be gilt should be chosen with care, 
 and not over-abundant, which would produce a 
 taw'd ry effect. 
 
 GLUE, a strong cement for wood, made of 
 the skin, ears, &c. of animals; first formed into 
 hard cakes, then steeped in w'ater until all the 
 parts have become soft. It is then dissolved by 
 heating it in water; and while in the fire, it 
 should be carefully stirred, to prevent it from 
 sticking to the vessel or glue-pot, until the w'ater 
 boils and the pieces of glue are thoroughly 
 melted. The joints to be glued must be per- 
 fectly close, so as to prevent seeing between the 
 surfaces, which ought to come in contact, though 
 the work may be opposed to the light. The 
 hotter the glue is, the more force it will exert in 
 keeping the two parts together ; therefore, in all 
 large and long joints, the glue should be applied 
 immediately after boiling. Glue loses much of 
 its strength by being often melted ; therefore 
 glue that is newly made is preferable to that 
 which has been used. 
 
 GOTHIC FURNITURE, is that which is 
 constructed in the Gothic style, in order to 
 correspond to the style of architecture of the 
 room where it is placed. 
 
 GREEK FURNITURE, is not only such as 
 has been used by the ancient Greeks, and which 
 has been transmitted to the present times by 
 means of the sculpture of their architectural 
 edifices ; but all sorts of furniture, whether used 
 by those ancient people or not, that is decorated 
 by Grecian ornaments. 
 
 GROOVES, are those channels which are 
 sunk into the substance of a board or other 
 work, in order to insert the edge of another 
 
LIB 
 
 M U S 
 
 C 
 
 board or piece of work, so as to fasten the two 
 boards or piece of work together. Grooves are 
 of two kinds, viz. either plain or dovetail. 
 Dovetail grooves are those that are wider at the 
 bottom than at the surface to which the work is 
 parallel ; by this means the piece or board which 
 has its inserted edge cut in the same form, can- 
 not be drawn out by any force acting perpendi- 
 cularly to the surface of the other board into 
 which it is inserted, without breaking the edge; 
 but can only be taken out before it is glued in 
 the direction of the groove. 
 
 GROUPE, a collection of individual things, 
 as a groupe of figures, &c. 
 
 H. 
 
 HALL FURNITURE, in splendid houses, 
 may consist of marble tables with massive frames, 
 and a convenient number of chairs and seats, 
 which ought to be decorated suitably to the 
 rooms to be entered. In grand mansions, the 
 walls may be embellished with paintings, and 
 the interior furnished with a sufficient quantity 
 of appropriate sculpture. 
 
 I. 
 
 INLAYING, an expensive mode of ornament- 
 ing furniture, by inserting other materials of a 
 richer colour, or better quality, either of natural 
 or dyed woods, metals, or shells. 
 
 The art of inlaying is very old. It was prac- 
 tised in ancient times by the people of Asia. 
 These nations were subdued by the Romans, 
 who imitated and acquired the art from the 
 spoils of the cities which they had conquered ; 
 and from the Romans, the nations of modern 
 Europe received their instructions, and became 
 expert proficients in the art. 
 
 JARDINIERE, a support for a small room- 
 garden, a flower- stand. 
 
 K. 
 
 KEYS, very thin slips of wood let into saw- 
 kerfs, and fixed with glue at the angles where 
 two parts of a piece of w ork are joined together. 
 These slips should not be in parallel planes, but 
 should incline towards one another in the man- 
 ner of dovetails. 
 
 L. 
 
 LADIES’ DWARF WARDROBE, a wardrobe 
 for ladies, lower than the room in which it is 
 placed. 
 
 LAP DOVETAILING, is a species of con- 
 cealed dovetailing, where the front is made to 
 lap over the side, so that the end of the w'ood 
 of the front piece is exhibited as if it were in 
 the same surface as the return part. 
 
 LIBRARY FURNITURE, may be of a similar 
 description to that of the drawing-room ; but 
 the decorations may be of a plainer kind. 
 
 9 ) 
 
 j LOO-TABLES, are contrived so as to be 
 convenient for playing at the game of loo, 
 
 M. 
 
 MAHOGANY, a well-known species of foreign 
 wood, much used in common furniture. Logs of 
 mahogany may be cut into planks of such amazing 
 breadth, as to afford table tops of immense w idth. 
 
 Mahogany is of several species, which have 
 different degrees of hardness ; some are as soft, 
 and may be as easily wrought, as deal ; other 
 kinds are hard and compact, and cannot be 
 made smooth w'ithout the use of the scraper. 
 One sort comes from Jamaica, another from 
 Yucatan, called Honduras Mahogany; but that 
 of Jamaica is much more durable and beautiful, 
 and appears as if the pores were filled with 
 chalk. In furniture, when first made, it is of 
 a light yellowish colour; but it darkens and 
 reddens gradually with age. 
 
 MITRE DOVETAILING, is a joint in which 
 the strength consists in the dovetailing, but 
 which only exhibits a plain mitre, the dove- 
 tailing being concealed. 
 
 MODILLION, a kind of console. See Con- 
 sole. 
 
 MORTISING, in the construction of cabinet- 
 w'ork, is the method of cutting prismatic recesses 
 in one part, in order to receive another in a 
 different direction, and to join and fix them 
 together, so as to become one undivided whole, 
 or part of a whole, if again to be connected with 
 other parts. The part which fills the mortise is 
 called a tenon. It is obvious that the less the 
 cavity between the mortise and tenon is, the 
 more firmly will the glue unite the two parts 
 together; therefore the mortise must be very 
 exactly cut, and the tenon nicely fitted. 
 
 MOSAIC WORK, inlaying with coloured 
 marbles. The tops of tables are often beauti- 
 fully ornamented with Mosaic work. 
 
 MOULDINGS, a prismatic surface, of which 
 the profile consists of a series of curves and 
 straight lines. The profiles of Grecian mould- 
 ings consist of all the curves of the conic 
 sections, but more particularly that of the el- 
 lipse, under varied proportions of the axis; and 
 those profiles of mouldings used by the Romans 
 consist only of the arcs of circles, and therefore 
 do not produce that pleasing variation of light 
 and shade which Grecian mouldings afford. The 
 profiles of Gothic mouldings have a much nearer 
 affinity to those of the Greek than to the Ro- 
 man, as the degrees of curvature are much more 
 varied ; but there are certain peculiarities of 
 ridge-lines in the meeting of Gothic curves that 
 are characteristic of their species 
 
 MUSIC-ROOM FURNITURE, is that which 
 
 2 s 
 
O R M 
 
 P U M 
 
 ( 10 ) 
 
 ought to be adapted to the character of the 
 Muses. The embellishments should, therefore, 
 be emblems of music, antique instruments, &c. 
 and the furniture should have considerable affi- 
 nity to that of the drawing-room. 
 
 O. 
 
 OAK, a species of wood of a rich umber- 
 brown colour. The wood of old trees with 
 excrescences, stunted trees of a slow growth, 
 and pollard trees, have been much used in fur- 
 niture, from the richness which the surface 
 presents when polished. The deep-coloured 
 woods are well adapted for moulding, as the 
 contrast between the ornamental and plane parts 
 has a fine effect. 
 
 OLD ENGLISH FURNITURE, that which is 
 made to imitate the style of furniture used in the 
 old Gothic mansions. We must, however, ob- 
 serve, that the Gothic furniture which is now so 
 much used, is that which was constructed in the 
 decline of cathedral architecture, and is a com- 
 position of ecclesiastic and castellated archi- 
 tecture. It w'as introduced during the reign of 
 Elizabeth, and is easily recognized by an osten- 
 tatious display of fantastic decorations. The 
 peculiar embellishments should be taken from 
 the screens, niches, canopies, pews, seats, 
 and monuments of ancient cathedrals and 
 churches. 
 
 ORNAMENTS, are all those embellished 
 parts of furniture which vary from smooth or 
 polished surfaces, in order to produce an agree- 
 able succession of parts. Ornaments may con- 
 sist of various combinations of surfaces, as in 
 Mosaic works, carvings in imitation of animals 
 or foliage, and mouldings of every form, or any 
 combinations of those which will be productive 
 of good effect. 
 
 Furniture is sometimes constructed, so that 
 every visible part is covered with ornament ; but, 
 as proportion and relief are necessary, the planes 
 or general curved surfaces which form the de- 
 sign, should be embellished with a repetition of 
 some beautiful figure, in order that the whole 
 may appear to advantage ; and as it is light and 
 shade that produce relief, care should be taken 
 to give carved work and mouldings their proper 
 projections. 
 
 The composition of ornaments is best acquired 
 from the study of the works of antiquity, as no 
 written rules can ever direct the taste of the 
 artist in this field of display; and his talent 
 must be acquired by long practice in the effect 
 of various combinations. 
 
 OR-MOULU, a species of gilding by means 
 of mercury, which produces a fine spotty bril- 
 liancy to very plain wood-work. This process 
 
 is accomplished by means of leaf-gold, of which 
 the leaves are thicker than those used in other 
 kinds of gilding. Having cleaned and freed the 
 surface from the acid used, cover it with an 
 amalgam of gold and mercury, spread on with a 
 brush of brass w ires, which being done, cover it 
 with a double coat of leaf-gold, pressing it close 
 with a piece of cotton, in order to prevent the 
 gold from collecting in hollows ; next place 
 the work in such a manner as to permit the 
 mercury to escape, and let it remain for about 
 an hour, until it is entirely drained ; then place 
 it over a charcoal fire for about a minute or two, 
 turning it until it becomes bright; having taken 
 it from the fire, strike it with the wire-brush, in 
 order to drive the gold into the receding parts : 
 place it again on the fire for about two minutes, 
 until the mercury evaporates, and leaves a colour 
 like that of box-wood : having cooled and washed 
 it in water, apply the amalgam of mercury a 
 second time, and a third, if necessary, but with- 
 out the leaf-gold. To remove the box-wood 
 colour, brush the work with a wire-brush in a 
 weak solution of vinegar, and rub it over with 
 a linen rag ; put it upon the fire on an iron grate, 
 and permit it to dry; having covered it with a 
 saline powder, and having again put it over the 
 charcoal fire for about half a minute on each 
 side, wash, and dry it in the air, and lastly 
 make completely dry over the charcoal. It is 
 from this species of gilding that French furniture 
 derives its brilliancy. 
 
 OTTOMAN, a seat in the form of a sofa, a 
 kind of footstool. 
 
 P. 
 
 PATENT DINING TABLES, tables which 
 may be extended into different lengths by means 
 of leaves concealed in a case under them. The 
 patent has now expired. 
 
 PATERAS, or Rosettes, ornaments made in 
 imitation of roses. 
 
 PEDESTAL, the lower support of an order, 
 applied in cabinet w'ork to side-boards, either 
 forming parts or insulated. 
 
 PUMICE-STONE, a very porous kind of 
 stone, used for polishing the surface of a piece 
 of wood which is not very close-grained. This 
 is done by spungingthe surface with water, and 
 rubbing it while wet with a flat piece of fine 
 pumice-stone, in the direction of the grain, and 
 permitting the wood to dry, and again repeating 
 the operation of wetting, rubbing, and drying. 
 By this means the surface will be found much 
 more condensed, and will have acquired the 
 desired degree of smoothness, by attentively 
 observing the effect in every stage of the pro- 
 gress during the operation. 
 
SID 
 
 S T A 
 
 ( 11 ) 
 
 R. 
 
 REEDING, a series of cylindric or conical 
 staves, arranged upon any surface, but nioie 
 particularly round the surface of a cone cylinder, 
 and in this* application they appear like a bundle 
 of rods. 
 
 ROMAN FURNITURE, is that which is 
 constructed in imitation of the Roman style. 
 The information required in the execution of this 
 stvle is chiefly collected from publications which 
 exhibit, in plates, representations of the ruins 
 of Herculaneum and Pompeii. The most splendid 
 article of furniture used by the Romans were 
 their Candelabra. Their tables were generally 
 sustained by a pillar, some having a low pedes- 
 tal, others with a base supported by three, and 
 sometimes four short turned legs, or ornamented 
 with the paws of animals. In the decoration of 
 Roman furniture, chimerical figures were used 
 in very great abundance; garlands and festoons 
 were also employed occasionally. In point of 
 taste, the Roman* furniture was stiff, and often 
 too much overcharged with ornament; it wanted 
 the fine-flowing outline, simplicity, acd chastity 
 of Greek furniture. 
 
 ROSE-WOOD STAIN, a stain given to chairs, 
 sofas, and the like, constructed chiefly of beech, 
 in order to produce a similar effect to that of 
 rose-wood. For the process, see Staining. 
 
 S. 
 
 SALOON FURNITURE, ought to be of a 
 similar description to that of halls and galleries. 
 
 SCRUTOIKE, a case of drawers to contain 
 writings. 
 
 SEAT, a bench adapted for sitting upon. 
 Under this general term are included chairs, 
 sofas, couches, but it is more particularly ap- 
 plied to a kind of form. Stools are small nar- 
 row seats without ledges. 
 
 SECRETAIRE DESK AND BOOK-CASE, a 
 convenient repository for writing, keeping valu- 
 able papers and choice books in, &c. 
 
 SHRINKAGE, a contraction of wood in a 
 transverse direction to the fibres, which takes 
 place in expelling the natural sap. — No work 
 will stand unless the wood has been sufficiently 
 seasoned ; for if thus not properly prepared, the 
 joints will be liable to loosen, the surfaces will 
 be in danger of warping and splitting, by which 
 means the strength of the work and the beauty 
 of its appearance is destroyed. 
 
 SIDEBOARD, a piece of furniture used in a 
 dining-room, for containing such articles as are 
 employed in the use of the table ; their height 
 is generally about 37 inches, and their breadth 
 from 30 to 33 inches; their length depends upon 
 the size of the room, and may extend from 5 to 
 
 10 feet. Sometimes they are constructed with 
 cases for holding wine; but when of a large 
 size, the celaret, or wine-cooler, is most con- 
 venient when separate, and its situation is gene- 
 rally under the middle of the length of the 
 sideboard. 
 
 SIDEBOARD, RUNNING, a sideboard with 
 supports for table articles at various heights. 
 
 SOFA, a seat or couch on which several per- 
 sons may sit at once, or on which one person 
 may sit or lie down, as may be found most con- 
 venient. 
 
 SOFA-TABLE, a long table to stand before 
 a sofa. 
 
 STAINING, is a method of imitating various 
 species of wood by dying the surface. This 
 effect is chiefly employed in chairs, sofas, and 
 bedsteads, and when done, the surface is po- 
 lished and varnished. The woods on which this 
 operation is performed are holly, pear-tree, and 
 beech. — An Ebony stain is effected by a mixture 
 of galls and logwood in the proportion of 12 to 2, 
 by giving the surface one coat; again, by add- 
 ing one part of verdigris to the mixture, give 
 it another; then adding one part sulphate of 
 iron, apply another; and repeat this operation, 
 if necessary. — Rose-wood may be imitated by a 
 mixture formed in the following manner: Boil 
 16 parts logwood in 64 parts water, till the 
 liquor acquires a deep red colour for the ground ; 
 then having added one part potash, apply the 
 hot mixture to the surface; repeat the operation 
 of coating, till the required rose-wood ground is 
 acquired. In order to granulate the surface, 
 heat the ebony stain before described for the 
 last coat, and with a graining-brush, like that 
 used by painters, mark the dark veins on the 
 surface of the work in imitation of the streaks 
 of the natural wood. — To imitate Mahogany, rub 
 the surface with a diluted solution of aqua-fortis; 
 dissolve by heating one ounce of dragon’s blood 
 in a pint of spirit of wine, and add one ounce of 
 carbonate of soda to the mixture, which being 
 filtered, lay it on with a soft brush in two coats, 
 and the surface of the wood will have the ap- 
 pearance desired. — The same effect is, however, 
 produced at a less expense, by rubbing the sur- 
 face with a diluted solution of aqua-fortis, and 
 then, if the article is small, it may be heated 
 upon a fire, but if large, it must be placed in an 
 oven, and covered with hot sand. — Another 
 method of imitating Mahogany is by mixing 
 Indian red with a small proportion of yellow- 
 ochre, which ought to be varied to answer the 
 grain and colour of the wood. — Oak may be 
 imitated by a due proportion of umber and 
 yellow-ochre. 
 
T R I 
 
 ( 12 ) 
 
 STAIRCASE FURNITURE, should be of a 
 similar description to that of halls, galleries, 
 ante-rooms, &c. 
 
 STOPPING, a method of making good the 
 defects in fine work, so as to render the surface 
 continuous, by stopping up the ca\ities with 
 fine saw-dust formed into a paste with clear 
 glue. If the surface is to be painted, the work 
 should be first primed : in this case whiting may 
 be used instead of saw-dust; and when the com- 
 position is dry, the protuberant part may be 
 taken away, and the surface smoothed; but 
 putty should never be used when the defects are 
 extensive. Defects are also stopped with a 
 composition made to imitate the colour of wood, 
 consisting of resin and bees-wax ; the colouring 
 matter is then added. These may be respec- 
 tively in the proportion of 1, 4, 1, the entire 
 composition being 6. 
 
 Another composition may be made by grind- 
 ing the required colour in spirits of turpentine 
 to the consistence of a thick paste ; and having 
 softened this paste with turpentine- varnish, stop 
 the defects with it, then cause the turpentine to 
 evaporate by putting it in a warm place; and 
 when it has remained about a day, the stopping 
 will be very hard, and will have the advantage 
 of not being softened by heat. — The stopping 
 used in common work is glazier’s putty, mixed 
 with colour to imitate that of the wood 
 T. 
 
 TABLE, a well-known piece of furniture for 
 containing food, writing, or any other articles 
 on its surface, so as to be convenient for the 
 purpose intended, without obliging those per- 
 sons who surround it to stoop. The height 
 should be just sufficient for a person sitting near 
 its edge to rest his arms without suffering any 
 inconvenience. 
 
 TEA POY, a tea chest supported upon a pil- 
 lar, with feet of various forms. 
 
 TENT BEDSTEAD, a bedstead in the form 
 of a tent. 
 
 TRACERY, the ornamental divisions between 
 pannels in Gothic architecture. 
 
 TRIPOD, a seat with three legs. This was 
 used by the Greeks for supporting a kind of 
 bason called a Brazier, which contained fire for 
 warming their apartments. 
 
 TRIPOLI, a soft silicious stone, of a gray 
 colour inclining to yellow, used for polishing 
 wood, metal, stones, varnish, and glass. 
 
 WIN 
 Y. 
 
 VALANCE, narrow drapery hanging round 
 the cornices, testers, steads, and curtains of 
 beds, and also accompanying window-curtains 
 to give them a fulness and completion. 
 
 VARNISH, a kind of liquid, which, being 
 applied to the surface of wood, becomes ex- 
 tremely hard, produces a beautiful shining lustre, 
 and gives the effect of a very high polish. 
 
 VASES, an ornament in the form of a solid 
 of revolution with its axis vertical. Vases were 
 anciently used as an article of furniture by the 
 Greeks. We cannot now ascertain to what use 
 these vases were applied. Their contour was 
 exceedingly graceful, and was much varied in 
 its outline, and their surfaces were highly deco- 
 rated with the most delicate and fine ornaments. 
 
 VENEERING, a method of covering the 
 surface of a piece of wood called the Ground, 
 with a thin plate of wood called Veneer, so that 
 the two surfaces may be brought in complete 
 contact, and fastened together. The object of 
 this is to make the surface appear as if the 
 material had been of the finest and most rich 
 qualify; for this reason the ground maybe of 
 inferior wood, and the veneer of the most 
 superior kind. 
 
 W. 
 
 WARPING of WOOD, is the position which 
 a piece acquires in expelling the natural sap, 
 and this is owing to an unequal contraction of 
 the fibres. It is therefore improper to construct 
 articles of furniture of unseasoned wood, since 
 the surfaces will not only be liable to shake 
 and split, but also to twist, and consequently 
 to take a very different postiion to that in- 
 tended. 
 
 WASH-HAND BASON STAND, a stand for 
 holding a bason for washing. These stands are 
 of various forms. 
 
 WASH-HAND TABLE, a table with conve- 
 Jiiencies for washing. 
 
 WAX POLISHING, is a method of polishing 
 by rubbing the surface with hard bees-wax, and 
 completing the operation of polishing by rubbing 
 with corks, brushes, and linen rubbers. This 
 kind of polish is used in various stained articles, 
 as also for chairs and bedsteads. 
 
 WINDOW CORNICES, the cornice to which 
 the curtains of a window are hung. 
 
 WINGED WARDROBE, a wardrobe with 
 wings. 
 
 London: h. fisher, printer to his majesty. 
 
Published *by !!.!■" is her. Son Sc C°. Caxtoru : .ondov Ja.lv, 1835. 
 
- - --- 
 
Published, by Fisher. Soil .1' C": Coxton, London , JaivT 1835. 
 
 ■yjsoj 
 
& C? LONDON, 1833 . 
 

LotVQ'.SC. 
 
 
 CamieLibri . 
 
 Zvy. 3. 
 
 ' fiery nel by l\f.A yuhjac n . 
 
 PnMielujl bv Pisher. San. ic C« Caxton..XctB.aaiL,Miy.l835. 
 
 
 
 Tip. 4 . 
 
 Tip. 2. 
 
 Tip.l. 
 
JaMi-shel by Fialier , San. Sc C ° Ca3±on, lanAcnx. 182 6. 
 
 (rumvufj 
 
Published. hy Fisher. Son k C° Cwrton. London, Nov 1 18. 1826 
 
 Drawing /ioom Chav 
 
FISHER, SON, &. C? LOUDON. 1.333 
 
 4 A 'fall Chair. A Drawing Room Chau: 
 
' vorjfyswf'jM nq v -.v.tj y wi/m 
 
irm Chair. 
 
 Cj an ./«'/ 
 
 jLiUri. sc 
 
 LublisTied. Fiskei. SonSc C* C ax ton, London., J'anjel.lS26 
 
 
 
Published. "by Fisher, Son X- C? Caxton, London, April 1835. 
 
 (/n/ij/ ryi ty.&jayrp-nfefa ■/bpvg^g-jjr yzyrrm-jj 
 
 Z'-' 7 ?’ V'oaJo *yn(7 m ,/rm/y m/j 
 
Drawing Room Gothic Commode. 
 
' 
 
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 'VJJ/OOJ M// (J 
 
Published, by Fisher, Son ic C° Canton , London Fcc r 30 1634 
 
 j£ Coach 
 
b 
 
 A Couch in the Grecian Styh 
 
— — 
 
J.SenshaH . sc . 
 
— — 
 
tVblished by IT. Fisher >a Sc C°Casi on, London. Oct? 16.26. 
 
 Curtains for two Wi / / d c iv s. 
 
Published by Pf.Pisher So- &C°Caxton London ISS'l' . 
 
 Draper 1/ for three Windows 
 
Greek Faleras. 
 
 Drawn, by M A Itfichj l- on . 
 
 Fig.X 
 
 Fy.3. 
 
 Published hy Fisher. Son & C° Caxton, London.. 1834. 
 
•gzgrx 3 ™!! Txo-pio'x T&xprao ;o=?-o°S /a T^STMU-I 
 
 Elements of T\ uiae/e . 
 
— ----- - 
 

FISHES-, SON, &. C9 LONDON, 1835. 
 
 
 
 
 
 ' ' ^ 
 
 - ; -- 
 
 Grecian 0 rnament. 
 
LookiiKj Glass. 
 
 Published by H. Fi'sber.Scm & C?CaxtOE.Io3lon, 1834. 
 
 
mmmt 
 
 t a oin e ? 
 
 and 
 
 1,0 
 
 
 Published dry Fisher. Son 8c C° Caxton. Loudon 182 T- 
 
 S’y.J 
 
 IMl! 
 
( haul Dress inn -Glass . 
 
 i7 
 
 Pubhsned "b^H. Fisher. Son k CPCaxton. London. Sep t 30 1834^. 
 
•3zar>2 kc-th h a i wm* 
 
 Chair legs and Sofa kgs with part of the dads. 
 
S idc Board L egs & Rails. 
 
 Published i>y PL. Fisher. Son Sc CPCazt on. London, Sen*; 9,1826 , 
 
TUTU r~ - . i 
 
l i mm 
 
 Mod 6 Min 
 
 id Me (1 6 Min 
 
 'Entablature JJlod. 
 
 Pedestal 7 Mod. 
 
 18 j Modules or . 9 ^ Miametm 
 
 zMcd 6Min 
 
 17 Mod 12 Min 
 
 Cornice 
 
 rriejje 
 
 ~Pfitatdature i Mott ' 1 
 
 Pedestal 7 Jhod. 
 
 21 Modules or w<f Diameters. 
 

■ 
 
 
M-A.Nicholson . cut 
 
 Published. "by Fisb.er . Sou. iC C? C ax ton.. I oinLoix , 1835 
 
 " 
 
 Fire Screens . 
 
 lFtfot 
 
 Fzy. Z. 
 
 Fy. 1. 
 
Window Seats. 
 
 -Prihhis'hed. Try Fisher , Son Sc C? CaztoiL London. June .1.1832. 
 
Published try Fisher. £on &• C° Caxton, London. 182 G 
 
 ‘p/vog- ?/>?$ 
 

 S ideboard. 
 
 Published by Fisher, Son Sc C° Caxton . London . Jan. 29 , 1834 . 
 
Gothic SidcBoard. 
 

 Side Board and C el r reft. 
 
Published by Fisher. Son Sc C°. Caxton London. 1834. 
 
 'xy jD f7 joyr 
 
Dcst.jncA fcJlra.U'n bu UfAMchciscn. 
 
PaVJislietL Tj j ifislLeT Sarnie C? CaxtoiL. L cmxkm. May 1826 . 
 
'?*? '“e'j-'V-'urrn 
 

 
Published by Fisher. Son k C? Caxton. London, 1835. 
 
Music Stools. 
 
Published by Fisher. Son Sc C e Caxton Londoa Jan. 1834 
 
Circular Table 
 
 Published by Ti Fisher. So:: & C° Canton. London. June 17. 1826 
 
 
: — 
 
L ibrary Table . 
 
Card Tables 
 
 Published "b j Fisher, Son & C° Caxton. London. Dee. 1834 
 
 Z>uJc jculf . 
 
 
’jpofvwT __ _ ■ w 'uonvimmi 
 
 Card Tab Us. 
 


' V? uosio-ipwvjZ' 
 
 O ccasioital Table/. 
 
■ <fr?9T '-&K ■xsjpnO’x-TiOEzs^ ;3=f ttos • zatp-rj £a ^sttcjs