PRACTICAL TREATISE 
 
 ON 
 
 THE ART OF 
 
 MASONRY AND STONE-CUTTING. 
 
 WITH NUMEROUS DIAGRAMS. 
 
A 
 
 PRACTICAL TREATISE 
 
 THE ART OF 
 
 MASONRY 
 
 STONE-CUTTING: 
 
 ©otttatntng the Construction of 
 
 PROFILES OP ARCHES, " VERTICAL CONIC VAULTS, 
 
 HEMISPHERIC NICHES, CYLINDRO-CYLINDRIC ARCHES, 
 
 HEMISPHERIC DOMES, RIGHT ARCHES, 
 
 CYLINDRIC GROINS, OBLIQUE ARCHES, 
 
 AND GOTHIC CEILINGS : 
 
 ESSENTIAL TO ENGINEERS, ARCHITECTS, AND STONE-MASONS, 
 
 EACH ARTICLE BEING PRECEDED BY THE REQUISITE INFORMATION 
 IN PLANE AND SOLID GEOMETRY: 
 
 ILLUSTRATED BY NUMEROUS DIAGRAMS 
 ON FORTY-THREE ELEGANTLY ENGRAVED PLATES. 
 
 By peter NICHOLSON, Esq. 
 
 ARCHITECT AND ENGINEER; 
 AUTHOR OF THE « BUILDER'S NEW DIRECTOR," " THE CARPENTER'S GUIDE," 
 AND OTHER WORKS ON PRACTICAL SUBJECTS. ' 
 
 THE THIBD EDITION. 
 
 LONDON: 
 PRINTED FOR JAMES MAYNARD, 
 
 PANTON STREET, HAYMARKET, 
 
 1835. 
 
Co/VS 
 
 TH , 
 
 A/S3 
 /83S 
 
 ■ f 
 
 THE GETTY CENTER 
 LIBRARY 
 
PREFACE. 
 
 It is rather surprising that, notwithstanding the 
 number and magnitude of our public edifices exe- 
 cuted in stone, and the numerous Treatises which 
 have been published on the art of Carpentry in 
 this and other countries, no work on the Principles 
 of Masonry has yet appeared in England, except 
 one on Stone-cutting by General Vallancey, pub- 
 lished in London in 1766. The General, though 
 he has not avowed it, has copied the diagrams and 
 references from a very able French work on the 
 subject by De la Rue. This English publication 
 was intended to consist of five parts, yet only one 
 has been given to the Public; and that without 
 inserting the most valuable of the precepts and 
 observations which are in the original work. And 
 this is more surprising, when we consider that no 
 department in the art of building affords so great 
 a scope for ingenuity, or needs so much an exten- 
 sive knowledge of the best aids of Geometry, as 
 the Art of Stone-cutting. 
 
vi 
 
 PREFACE. 
 
 To be able to direct the operations of Stone 
 Masonry, taken in the full extent of the Art, re- 
 quires the most profound mathematical researches, 
 and a greater combination of scientific and prac- 
 tical knowledge, than all the other executive 
 branches in the range of architectural science. 
 
 It is not the intention of the Author in this 
 undertaking to give a complete Treatise on Ma- 
 sonry, nor even on Stone-cutting, as such a pro- 
 duction would far exceed his prescribed limits; 
 but to produce such a Work as he hopes will be 
 found of the most essential service in the art 
 treated of, in regard both to the methods employed 
 and the examples adduced ; most of those in the 
 French publications being such as the English 
 workman would never meet with in the course of 
 his practice. 
 
 As the art of Masonry requires a considerable 
 knowledge both of Plane and Solid Geometry, such 
 Problems have been introduced in Practical Geo- 
 metry, as respect the positions of lines and points ; 
 and in Solid Geometry are shewn the properties 
 arising from the intersection of lines and planes 
 
PREFACE. 
 
 vii 
 
 with each other ; and also the construction of 
 trehedrals under various circumstances. To enable 
 the workman to construct the plans and elevations 
 of the various forms of arches or vaults, as much of 
 descriptive Geometry and Projection is introduced, 
 as will be found necessary to conduct him through 
 the most difficult undertakings. 
 
 The several branches are so arranged, that 
 those subjects which are subsequent are made to 
 depend on those which have preceded them; and 
 as the most obvious methods are not always the 
 most eligible for practice, there are given in se- 
 veral instances, for the more ready information of 
 the student, the methods which are best calculated 
 to explain the principles ; and from, these, others 
 are deduced, still more convenient and expeditious 
 for practice. 
 
 The Author has availed himself of the useful 
 precepts and observations which are given in the 
 most valuable French publications on the Art of 
 Stone-cutting ; but, except in one or two instances, 
 he has neither followed their methods, nor adopted 
 their examples, for the reasons before alluded to: 
 
Vlll 
 
 PREFACE. 
 
 he also thinks it necessary to observe, that this 
 Treatise contains several constructions, and many 
 methods which are nowhere else to be found. 
 
 The numerous examples chosen are correctly 
 displayed on forty-three elegantly-engraved Plates, 
 and are not subjects of mere curiosity and specu- 
 lation, but of elegance and utility, necessary to be 
 known by all persons engaged in the construction 
 of Cathedrals, Churches, Bridges, Mansions, and 
 other large Works. 
 
 To conclude : the following Work is the first 
 and only one in English on the Art of Stone- 
 cutting ; and as such a publication has been long 
 and greatly desired, the Author indulges the hope, 
 from the experience he has had in treating prac- 
 tically, other subjects intimately connected with 
 the present, that the contents of this Volume will 
 be found important, the instructions and diagrams 
 be easily understood, and that his labours will not 
 be in vain, nor disappoint the expectations of the 
 Reader. 
 
 P. N. 
 
 1832. 
 
CONTENTS. 
 
 CHAPTER I. 
 
 PRACTICAL GEOMETRY, ADAPTED TO MASONRY AND STONE-CUTTING. 
 
 SECTION 1. 
 On the position of Lines and Points. — p. 1. 
 SECTION II. 
 
 On the species, nature, and construction of Curi e Lines. — p. 6. - 
 SECTION III. 
 
 Of the positions of Lines and Planes, and the properties arising from their 
 intersections. — p. 10. 
 
 SECTION IV. 
 
 Of the right sections of Arches or Vaults. — p. 11. 
 
 SECTION V. 
 
 On the nature and construction of Trekedrals. — p. 15. 
 
 SECTION VI. 
 
 On the projection of a Straight Line bent upon a cylindric surface, and the 
 method of drawing a Tangent to such a projection. — p. 20. 
 
vi 
 
 CONTENTS. 
 
 CHAPTER II. 
 
 THE GEOMETRY OF PLANS AND ELEVATIONS, ADAPTED TO THE CONSTRUCTION 
 OF ARCHES AND VAULTS. 
 
 SECTION I. 
 
 Preliminary principles of Projection. — p, 22. 
 
 SECTION n. 
 
 On the def elopements of the Surfaces of Solids.— p. 32. 
 
 CHAPTER III. 
 
 CONSTRUCTION OF THE MOULDS FOR HORIZONTAL CYLINDRETIC VAULTS, EITHER 
 TERMINATING RIGHTLY OR OBLIQUELY, UP01< PLANE OR CYLINDRICAL 
 WALLS, WITH THE JOINTS OF THE COURSES EITHER IN THE DIRECTION OF 
 THE VAULT, PERPENDICULAR TO THE FACES, OR IN SPIRAL COURSES. 
 
 SECTION I. 
 
 Definitions of JSIasonry, Walls, Vaults, ^x.—p. 35. 
 
 SECTION II. 
 On oblique Arches. — p. 39. 
 
 SECTION III. 
 
 A circular Arch in a circular Wall. — p. 60. 
 
 SECTION IV. 
 
 A conic Arch in a cylindric Wall. — p. 62. 
 CHAPTER IV. 
 
 CONSTRUCTION OF THE MOULDS FOR SPHERICAL NICHES, BOTH WITH RA- 
 DIATING AND HORIZONTAL JOINTS, IN STRAIGHT WALLS. P. 66. 
 
 SECTION I. 
 
 Examples of Niches, with radiating joints, in straight walls, as in 
 Plate XXIV.— p. 67. 
 
 SECTION II. 
 
 Examples of Niches in straight walls with horizontal courses, as in 
 Plate XXVII.—p. 69. 
 
CONTENTS. Vii 
 
 CHAPTER V. 
 
 CONSTRUCTION OF THE MOULDS, AND roRMATION OF THE STONES, FOR 
 DOMES UPON CIRCULAR PLANES, AS IN PLATE XXX. 
 
 On the Constri/ction of Spherical Dumes—p. 72. 
 CHAPTER VI. 
 
 CONSTRUCTION OF CIRCULAR ROOFS, OF WHICH THE EXTERIOR AND INTERIOR 
 SURFACES ARE CONICAL AND CONCENTRIC WITH EACH OTHER.— p. 85. 
 
 CHAPTER VII. 
 
 CONSTRUCTION OF THE MOULDS, AND FORMATION OF THE STONES, FOR 
 RECTANGULAR GROINED VAULTS. 
 
 Construction of Groined Vaults, with Cylindretic Surfaces p. 87. 
 
 CHAPTER VIII. 
 
 CONSTRUCTION OF THE STONES FOR GOTHIC VAULTS, IN RECTANGULAR 
 COMPARTMENTS UPON THE PLAN. 
 
 Groined Arches, springing from Polygonal Pillars.— p. 91. 
 
 CHAPTER IX. 
 
 THE MANNER OF FINDING THE SECTIONS OF RAKING MOULDINGS.— P. 95. 
 
 CHAPTER X. 
 
 CONSTRUCTION OF A LINTEL, OR AN ARCHITRAVE, IN THREE OR MORE 
 PARTS, OVER AN OPENING, AND THE STEPS OF A STAIR, OVER AN 
 AREA. P. 97. 
 
 r 
 
 CHAPTER XI. 
 
 OF WATERLOO BRIDGE.— P. 99. 
 
A 
 
 PRACTICAL TREATISE, 
 
 &c. &c. 
 
 CHAPTER I. 
 
 PRACTICAL GEOMETRY, ADAPTED TO MASONRY AND 
 STONE-CUTTING. 
 
 SECTION I. 
 ON THE POSITION OF LINES AND POINTS. 
 
 As the construction of every complex object in nature 
 consists of certain combinations of the simple operations of 
 geometry ; and as positions cannot be understood without 
 angles and parallel lines, it will be necessary to treat of the 
 practical part of this science, at least as far as the operations 
 of the positions of lines and points are concerned, in order 
 to render the constructions and the language of geometry 
 familiar to the student in their applications to the principles 
 of Masonry. 
 
 PROBLEM I. 
 
 From a given point in a given straight line to draw a 
 perpendicular. 
 
 Plate I. Let AB, fig. 1 , be a given straight line, and c the given 
 point. In AB take two equal distances, cd and ce. From <Z as a centre, 
 Avith any radius greater than cd or ce, describe an arc at /, and with the 
 same radius, from the point c, describe another arc intersecting the for- 
 mer at/, and draw/c, and fc is the perpendicular required. 
 
 R 
 
2 
 
 A PRACTICAL TREATISE ON 
 
 [CH. I. 
 
 PROBLEM II. 
 
 From the one extremity of a straight line to draw a per- 
 pendicular. 
 
 Fig. 2, Let AB be the given straight line, and let it be required to 
 draw a perpendicular from the extremity B. On one side of the line AB 
 take any convenient point c ; and from c, as a centre, with any radius 
 that will cut the line, describe an arc (ZBe, intersecting AB in the point d ; 
 through c draw the diameter, de, and join eE, and eB is the perpen- 
 dicular required. 
 
 PROBLEM III. 
 
 From a given point out of a straight line to let fall a 
 perpendicular. 
 
 Fig. 3. Let AB be the given straight line, and c the given point ; it 
 is required to draw a straight line from c perpendicular to AB. From c, 
 as a centre, with any distance that will cut the line AB, describe an arc 
 intersecting AB in the points d and e ; from d, as a centre, with any 
 radius greater than the half of de, describe an arc, and from e, with the 
 same radius, describe another arc intersecting the former in f, and draw 
 fc, and fc is the perpendicular required. 
 
 The criterion of the truth of the method of Jig. 2. is that of the angle 
 in a semicircle being a right angle. 
 
 PROBLEM IV. 
 
 At a given point in a given straight line to make an angle 
 equal to a given angle. 
 
 Fig. 4. Let CBA be the given angle, and LF the given straight line. 
 Let it be required to draw a straight line, at the point L, to make an 
 angle with the line LF, equal to the angle CBA. From the point B, 
 with any radius, describe an arc meeting BA in h, and BC in g ; and 
 from the point L, with the same radius, describe an arc ik, meeting LF 
 in i. Make ik equal to gh, and through k draw the straight line LD, 
 and FLD is the angle required. 
 
 PROBLEM V. 
 
 Through a given point / to draw a straight line parallel 
 
 to a given straight line AB. 
 
 Fig. 5. Let / be the given point, and AB the given straight line. 
 Draw any straight line Je, meeting AB in e, and draw gh, making the 
 angle /«gB equal to /eB. Make gh equal to ef. Through the points/ 
 and k draw the line CD, and CD is parallel to AB, as required. 
 
CH. I.] MASONRY AND STONE-CUTTING. 
 
 3 
 
 PROBLEM VI. 
 
 To draw a straight line parallel to a given straight line 
 
 at a given distance from the given straight line. 
 
 Fig. 6. Let AB be the given straight line ; it is required to draw a 
 straight line at a given distance from BC. In AB take any two points 
 e and /; from e, with the given distance, describe an arc gh; and from 
 with the same distance, describe another arc ik. Draw the line CD to 
 touch the arcs gh and ik, and CD is parallel to AB, as required. 
 
 PROBLEM VII. 
 To bisect a given straight line AB by a perpendicular. 
 
 Fig. 7- From the point A as a centre, with any radius greater than 
 the half of AB, describe an arc cd ; and from B, with the same radius, 
 describe another arc intersecting the former at c and d, and draw cd, in- 
 tersecting AB in e ; then AB is divided in e, as required. 
 
 PROBLEM VIII. 
 Upon a given straight line to describe an equilateral 
 triangle. 
 
 Fig. 8. Let AB be the given straight line. From the point A, with 
 the radius AB, describe an arc, and from the point B, with the radius 
 BA, describe another arc, intersecting the former in C, and draw the 
 straight lines CA and CB; then ABC is the equilateral triangle re- 
 quired. 
 
 PROBLEM IX. 
 
 Upon a given straight line to describe a triangle, of which 
 the sides shall be equal to three given straight lines, pro- 
 vided that any one of the three given lines be less than the 
 sum of the other two. 
 
 Fig. 9. Let the three given straight lines be A, B, C, and let DP 
 be the straight line on which the triangle is required to be described. 
 Make DF equal to the given straight line A. From D, with the radius 
 of the line B, describe an arc, and from F, with the radius of the line C, 
 describe another arc, meeting the arc described from D in the point E. 
 Draw ED and EF, then DEF is the triangle' required. 
 
 PROBLEM X. 
 
 Given the base and height of the segment of a circle to 
 find the centre of the circle, and thence to describe the arc. 
 
 B 2 
 
4 
 
 A PRACTICAL TREATISE ON 
 
 [CH. I. 
 
 Fig. 10- Let AC be the base bisect, AC in D by the perpendicular 
 BE; make DB equal to the height, and join the points A and B. Make 
 the angle BAE equal to ABE, and the point E is the centre required. 
 
 From the point E, with the radius EA or EB, describe the arc ABC ; 
 then ABC is the arc required. 
 
 N.B. The centre must also have been found by bisecting AB by a 
 perpendicular, which would have met BE in the point E. 
 
 PROBLEM XI. 
 
 Given two converging lines through a given point in one 
 of them, to draw a third straight line, so that the angles on 
 the same side of the line thus drawn, made by this line and 
 each of the first two given lines, may be equal to each other. 
 
 Fig. 11. Let the two converging lines be AC and BD, and let A be 
 the given point. Draw AE parallel to BD; bisect the angle CAE by 
 the straight line AB ; then will the angles CAB and DBA be equal to 
 one another. 
 
 For, suppose AE to be produced from A to F, and suppose AC and 
 BD to be produced to meet in some point G, then AC would have been 
 a line falling upon the two parallel straight lines AF and BD, and con- 
 sequently making the angle at G equal to the A angle FAC ; and since 
 the three angles of every triangle are equal to two right angles, and 
 since the angles FAC, CAB, BAE, are also equal to two right angles, 
 and since FAC is equal to the vertical angle of the triangle, the angle 
 CAE is equal to the sum of the angles at the base ; and therefore, since 
 CAB is half the sum, the angle ABD must be equal to the other half. 
 
 PROBLEM XII. 
 
 Given two converging lines to describe the arc of a circle 
 through a given point in one of them, without having re- 
 course to a centre, so that the point of convergency may be 
 in the centre of the arc. 
 
 Fig. 12. Let AB and EF be the two converging lines, and A the given 
 point through which the arc is . to pass. Draw AE, making the angles 
 BAE and FEA, equal to each other. Bisect AE by the perpendicular 
 CD, and draw Ah, making the angles BAh and D^ A equal to one ano- 
 ther ; then Ah is the chord of the arc, and nh is the versed sine. Sup- 
 pose now that the three points A, h, E, are transferred to A, B, C, Jig. 
 13. Join BA and BC. Produce BA to d, and BC to e. Make the edge 
 of a slip of wood to the angle dBe. Move the edge dBe of the slip of 
 
CH. I.J 
 
 MASONRY AND STONE-CUTTING. 
 
 5 
 
 wood so that the point B may be upon A ; then move this slip again, so 
 that while the part Bd of the edge of the slip is sliding upon the pin at 
 A, and the part Be upon the pin at C, a pencil held to the angle B, will 
 describe a curve ; then this curve will be the arc required. 
 
 PROBLEM XIII. 
 Given two straight lines to find a third proportional. 
 
 Fig. 14. From any point A, draw any two straight lines BA, AC, at 
 any angle. Make AB, equal to one of the given straight lines, and AC 
 equal to the other; and in AB make Ad equal to AC. Join BC and 
 draw de parallel to BC, meeting AC in e ; then Ae is the third propor- 
 tional required. 
 
 Or if Ae be equal to one of the given straight lines, and Ad equal to 
 the other. Make AC equal to Ad. Join de and draw CB parallel to ed, 
 then AB is the third proportional. 
 
 PROBLEM XIV. 
 Given a straight line, any how divided, to divide anothe 
 
 m the same proportion. iL^^c--r^ 
 
 ! ; 1 o i X 
 
 Fig. 15. Draw the lines BA, AC as in the preceding problem, a^d ^ ^ 
 let AB be the given divided line, d and e being the poi? ts of division^; ' 
 and let AC be the line to be divided. Join BC and draw eg and c?/" pa- 
 rallel to BC, meeting AC in / and g ; then AC is divided in / and g, in 
 the same proportion as AB is divided in the points d and e. 
 
 PROBLEM XV. 
 Given three straight lines to find a fourth proportional. 
 
 Fig. 15. The angle BAC being made as before ; let Ae be equal to 
 one of the given lines. Ad equal to a second, and af equal to the third. 
 Join df and draw eg parallel to df ; then Ag is the fourth proportional. 
 
 SECTION II. 
 
 ON THE SPECIES, NATURE, AND CONSTRUCTION OF 
 CURVE LINES. 
 
 The geometrical orders of lines employed in architecture 
 in the construction of arches, are circular and elliptic, and 
 occasionally parabolic, hyperbolic, cycloidal, and catenarian 
 curves. 
 
6 A PEACTICAL TREATISE ON ^CH. I. 
 
 In houses,. the chief lines employed in the construction of 
 arches and vaults, are circular and elliptic curves, generally 
 a semi, and sometimes less, but seldom or never greater. 
 When a circular or elliptic arc is adopted, one of the axes of 
 the curve is most frequently situated upon the springing 
 line ; but is sometimes placed so as to be parallel to it. The 
 most usual portions of circular or eUiptic curves are the 
 semi; and in the pointed style of architecture, parabolic 
 and hyperbolic curves are very frequently employed. 
 
 In bridge building, besides circular and elliptic curves 
 which are the most often used in the construction of stone 
 arches, cycloidal curves may also be introduced. In chain 
 bridges, or bridges of suspension, not only the circular and 
 parabolic curves, but that of the catenarian may be em- 
 ployed. The suspending chains necessarily assume the 
 form of catenarian curves ; but the road-way may be any 
 curve line whatever ; but as all curves are nearly circular 
 at the vortex, it will be better to employ those in the con- 
 struction of works which are susceptible of the most easy 
 calculation. 
 
 > Among the numerous orders of curve lines, the parabolic 
 affords the most easy means of computing its ordinates and 
 tangents, which will be found necessary in ascertaining the 
 rise and inclination of the road- way in all points of the 
 curve, from either extreme to the centre of the bridge. 
 
 The base of an arc is that upon which the arc is supposed 
 to stand ; and the highest point of an arc is that in which 
 a straight line parallel to the base would meet the curve, 
 without the possibility of coming within the area included 
 by the curve and its base, and tliis point is called the 
 summit of the arc. 
 
 As the curves employed in building are generally sym- 
 metrical, therefore they are equal and similar on each side 
 
CH. I.] MASONRY AND STONE-CUTTING. 
 
 7 
 
 of the summit, and their areas are equal and similar on each 
 side of the perpendicular from the middle of the base. 
 
 PROBLEM I. 
 
 To describe a semi-ellipse upon the transverse axis. 
 
 Plate II. Let Ka, fig. \, be the axis major, and let BC bisecting Aa 
 perpendicularly in the point C, be the semi-conjugate axis. 
 
 Upon the straight edge mi of a rule, mark the point m at or near one 
 of its ends, and the point I at a distance ; from m equal to BC, the semi- 
 conjugate axis ; an<?. the point ^ at a distance from m, equal to AC or Ca 
 the semi-transverse axis ; the distance hi being equal to the difference of 
 the two axes. To find any point in the curve, place the point h in the 
 line BC produced, and the point I in the axis Aa ; and mark the paper 
 or plane on which the figure is to be described at the point m. Proceed 
 in this manner until a sufficient number of points are found, and draw 
 a curve through them, and the curve will be the semi-ellipse required. 
 
 PROBLEM II. 
 
 Upon a given double ordinate to describe the segment of 
 an ellipse, to a given abscissa, and to a given semi-axis in 
 that abscissa. 
 
 Figs. 2 and 3. Let Mw be the double ordinate, PH the abscissa, and 
 HC the semi-axis. 
 
 Through the centre C, draw parallel to M7/1. From either extre- 
 mity m of the double ordinate as a centre with the distance HC of the 
 given semi-axis, as radius describe an arc intersecting in r. Draw 
 mr intersecting HC in q, or produce mr, and HC to meet in q ; then mq, 
 fi/r. 2 will be the semi-transverse, and mr the semi-conjugate, and in 
 Jig. 3 the contrary will take place, mr will be the semi-transverse, and 
 mq the semi-conjugate ; the two axes being thus found, the curve may be 
 described as in the immediately preceding problem. 
 
 PROBLEM III. 
 
 Given two conjugate diameters to find any number of 
 points in the curve, and thence to describe it. 
 
 Figs. 4 and 5. Let Aa, Bb, be the conjugate diameters. Draw AD 
 parallel to BC, and BD parallel to CA. Divide AD and AC each into 
 the same number of equal parts. Through the points of division in AC 
 draw straight lines from b, and through the points of division in AD 
 draw other straight lines to the point B, meeting those dravwjji from b in the 
 
8 
 
 A PRACTICAL TREATISE ON 
 
 [CH. I. 
 
 points/, g, h. Draw a curve line through the points A,f, g, h, B which 
 will be one quarter of the whole figure. The other three will of course 
 be found in the same manner. 
 
 PROBLEM IV. 
 
 To draw a normal, or a line perpendicular to the curve 
 of an ellipse at a given point in the curve. 
 
 Fig. 6. Let the curve be ABa, and let Aa be the transverse axis, 
 and CB the semi-conjugate, and let it be required to draw a line from the 
 point 71 perpendicular to the curve. With AC the semi-axis major as a 
 radius ; from the point B describe an arc, intersecting Aa in the foci, /, 
 From the points/',/, and through the point «, draw f'd and/e, 
 and bisect the angle e ?i d, and the bisecting line nN Avill be perpendicu- 
 lar to the curve as required. 
 
 PROBLEM V. 
 
 To draw a tangent to the curve of an ellipse at a given 
 point. 
 
 Fig. 6. Let m be the given point. Draw fm, and produce fm to g, 
 and join the points/', m. Bisect the angle fmg, and the bisecting line 
 will be the tangent required. 
 
 PROBLEM VL 
 
 The curve of an ellipse being given to find the two axes. 
 
 Fig. 7. Let AMNww be the given curve within the figure; draw any 
 two parallel lines Mm, N?j. Bisect Mm in o, and N« in p, and draw 
 the straight Aopa. Bisect Aa in C, from C as a centre, with any 
 radius that will cut the curve : describe the arc rr', intersecting the 
 curve in the points r, r' , and draw the straight line rr'. Bisect rr' in h, 
 and through the points h and C draAv the line de, then de is the axis 
 major; and a line drawn through the point C at right angles to de, to 
 meet the curve on each side of C will be the axis minor. 
 
 PROBLEM VH. 
 
 With a given abscissa and ordinate to describe a parabola. 
 
 Fig. 8. Let AB be the abscissa, and BC the ordinate. Draw CD 
 parallel to BA, and AD parallel to BC. Divide CD and CB each into 
 the same number of equal parts. From the points 1, 2, 3 in CD draw 
 lines to A, and from the points 1, 2, 3 in CB, draw lines parallel to BA, 
 meeting the former lines to A in the points f,g,h. Draw the curve 
 Cy^/iA, which will be one half of the parabola, the other half will be 
 
CH. I.] MASONllY AND STONE-CUTTING. 
 
 9 
 
 found in the same manner. The radius of curvature at the point A, is 
 half the parameter. 
 
 PROBLEM VIII. 
 The curve of a parabola being given to find the para- 
 meter. 
 
 Fig. 8. Let CAN be the curve of the parabola. Bisect BC in the 
 point 2, and draw A2 and 2d perpendicular to A2, meeting AB produced 
 in d ; then BcZ is one fourth part of the parameter. 
 
 For AB : B2 :: B2 : Bd, now let AB=«, BC=b, then B2=^b, 
 
 hence a \h \ \ hh : \]) ; whence aj)—!)^ or p r= — . 
 
 PROBLEM IX. 
 
 To draw a tangent to any point M, in the curve of a 
 parabola. 
 
 Fig. 8. Draw the ordinate PM^ and produce PA to q. Make A q 
 equal to AP, and draw the straight line 9 M ; then q M will be the 
 tangent required. 
 
 For the subtangent of the curve is double to the abscissa. 
 
 PROBLEM X. 
 
 To form the curve of a parabola by means of tangents. 
 
 Fig. 9. Let AC be the double ordinate. Draw DB bisecting AC, 
 and make DB equal to the abscissa. Produce DB to and make BE 
 equal to BD. Draw the two straight lines EA and EC. Divide AE 
 and EC each in the same proportion, or into the same number of equal 
 parts at the points 1, 2, 3, &c. in each line. Draw the straight lines 1-1, 
 2-2, 3-3, &c. and their intersections will circumscribe the curve of the 
 parabola as required. 
 
 Scholium. Small portions of the curves of conic sections, near to the 
 vertices, may be described with compasses so as not to be perceptible ; and 
 thus, not only in the parabola, but in the ellipse ; and, in the hyperbola, 
 the radius of curvature at the vertices is half the parameter, which passes 
 through the focus. In the parabola, the parameter is a third propor- 
 tional to the abscissa and ordinate ; and in the ellipse and hyperbola, 
 the parameter is a third proportional to the transverse and conjugate 
 axis ; and therefore may be easily found by lines or by calculation on 
 large works, such as bridges, &c. 
 
10 
 
 A PEACTICAL TREATISE ON 
 
 [CH. I. 
 
 SECTION III. 
 
 OF THE POSITIONS OF LINES AND PLANES, AND THE 
 PROPERTIES ARISING FROM THEIR INTERSECTIONS. 
 
 A plane is a surface in which a straight line may coin- 
 cide in all directions. 
 
 A straight line is in a plane when it has two points in 
 common with that plane. 
 
 Two straight lines which cut each other in space, or 
 would intersect, if produced are in the same plane; and 
 two lines that are parallel, are also in the same plane. 
 
 Three points given in space, and not in a straight, are 
 necessary and sufficient for determining the position of a 
 plane. Hence two planes which have three points com- 
 mon, coincide with each other. 
 
 The intersection of two planes is a straight line. 
 
 Plate III. When two planes ABCD, ABFE, Jig. \, intersect, they 
 form between them a certain angle, which is called the inclination of the 
 two planes, and which is measured by the angle contained by two 
 lines ; one drawn in each of the planes perpendicular to their line of 
 common section. 
 
 Thus, if the line AF, in the plane ABEF, be perpendicular to AB, 
 and the line AD, in the plane ABCD, be perpendicular also to AB, then 
 the angle FAD is the measure of the inclination of the planes ABEF, 
 ABCD. When the angle FAD is a right angle, the two planes are 
 perpendicular. 
 
 Fig. 2. A line AB, is perpendicular to a plane PQ when the line 
 AB is perpendicular to any line BC in the plane PQ, which passes 
 through the point B, where the line meets the plane. The point B is 
 called the foot of the perpendicular. 
 
 A line AB, 3, is parallel to a plane PQ, when the line AB is pa- 
 rallel to another straight line CD, in the plane PQ. 
 
 If a straight line have one of its intermediate points in common with a 
 plane, the whole line will be in the plane. 
 
CH. I.] MASONRY AND STONE-CUTTING. 
 
 11 
 
 Two planes are parallel to one another when they cannot intersect in 
 any direction. 
 
 The intersections of two parallel planes with a third are parallel. 
 Thus in fig A, the lines AB^ CD comprehended by the parallel plane PQ, 
 RS are parallel. 
 
 Any number of parallel lines comprised between two parallel planes, 
 are all equal. Thus the parallel lines Ka, B6, Cc, . . . . , comprised by 
 the parallel planes PQ, RS are all equal. 
 
 If two planes CDEF, GHIJ, Jig. 6, are perpendicular to a third 
 plane PQ, their intersection AB will be perpendicular to the third 
 plane PQ. 
 
 If two straight lines be cut by several parallel planes, these straight 
 lines will be divided in the same proportion. 
 
 SECTION IV. 
 
 OF THE RIGHT SECTIONS OF ARCHES OR VAULTS. 
 PROBLEM 1. 
 
 To describe the arc of a circle which shall have a given 
 tangent at a given point, and which shall touch another 
 given arc. 
 
 Plate IV. Let B^, Jig. 1, be one of the given arcs, and lau the other, 
 and let it be required to describe the arc of a circle, which shall touch 
 the arc B^, in the point h, and the arc lati in some point to be found ; 
 let g be the centre of the arc B^. 
 
 Draw gk, and make kj) equal to the radius of the circle lau. Draw a 
 straight line from p to q, the centre of the arc lau, and bisect by a 
 perpendicular, meeting kg in m. Join the points m, q, and prolong mq 
 to I. It is manifest that mk and ml are equal ; therefore, from m with 
 the radius mk or ml describe an arc kl ; then kl will be the arc required. 
 
 PROBLEM II. 
 
 To describe an oval, representing an ellipse, to any given 
 dimensions of length and breadth, given in position. 
 
 Let Aa, Bi, Jig-.^, be the two given lines bisecting each other in 
 C ; Aa being equal to the length, and Bb to the breadth. 
 
A PRACTICAL TREATISE ON 
 
 [CH. I. 
 
 Find a third proportional to this semi-axis Ca, CB,* and make ah 
 equal to the third proportional ; also find a third proportional to CB, 
 Ca,f and make equal to the third proportional. 
 
 Make the angle Bgk equal to about 15°, and let gk meet Aa in the 
 point i. From g with the radius gB, describe an arc Bk,X and from k 
 with the radius ha, describe an arc la. Describe the arc kl by the pre- 
 
 * Thus in Jig' 3, draw the two lines GA, AH making any angle with 
 each other ; make ac equal to aC, Jig. 2, and Ad equal to Cb,fig. 2, and 
 make Ae equal to Ad. J oin cd, and draw ef parallel to cd ; then of is 
 the third proportional. 
 
 f That is in Jig. 3, make Ac equal to AG, or aC Jig. 2, and Ad equal 
 to CB or C6, Jig. 2, and make AG equal to Ac and join cd. Draw GH 
 parallel to dc ; then AH is the third proportional. 
 
 :}: Mathematicians have demonstrated that in all the conic sections, the 
 radius of curvature is equal to the cube of the normal, divided by one- 
 fourth of the square of the parameter, and that this radius at either ex- 
 tremity of the axis major, is half the parameter, which passes through the 
 focus ; and, moreover, that the radius of curvature at either extremity of 
 the axis minor, is half the parameter to this axis. Now the focal semi- 
 parameter is a third proportional to the semi-axis major and the semi- 
 axis minor, and the parameter of the axis minor is a third proportional to 
 the semi-axis minor and the semi-axis major. These parameters are 
 found geometrically, as in fig. 3, or they may easily be found by calcu- 
 lations, by finding the normal from this formula. Thus, if n represent 
 the normal, a the semi-axis major, and b the semi-axis minor, x the 
 
 abscissa from the centre, and the ordinate as usual, then « — '-^ 
 ;^^a'' — (a* — and if R represent the radius of curvature, then 
 
 R — - — by making x~o the expression for the value of the normal 
 
 4 P 
 
 would become simply b ox nz=.b ; therefore the radius of curvature at 
 
 either extremity of axis minor would be j— 2 ; and if x ~ a wo. should 
 
 4 f 
 
 b^ b^ b^ 
 
 then have nz=. — ; but a : b b : ^p, hence ^ p = — , or I = 
 
 whence R= therefore the radius of curvature at either extremity of 
 the axis minor, is a third proportional to the semi-axis minor and the serai, 
 axis major. Again, if a; a, the equation n zz tj ^a"^ — {a" — or | 
 
CH. I.] MASONRY AND STONE-CUTTING. 
 
 13 
 
 ceding problem to touch the arc Hk in k, and to touch the arc al at /, 
 and thus one quarter of the oval will be completed ; the other three will 
 be found by placing the centres in their proper positions. 
 
 Three or more points a, b, c might easily have been found in the curve. 
 Thus draw Ad parallel to 'Bb, and Jid parallel to CA. Divide Ad into 
 four equal parts, and divide AC also into four equal parts at 1,2, 3. 
 From b and through 1 , 2, 3 in C A, draw ba, bb, be, and from the points 
 1, 2, 3 in Ad, draw lines towards B, to intersect the former in a, b, c, so 
 that we may find the radius of curvature upon the sides, and at the two 
 ends, by finding the centre of a circle passing through three points at 
 each extremity, the extremity being the middle point. 
 
 Fig. 4 exhibits the use of this method of describing an oval, in finding 
 the direction of the joints of arches, so as to agree with the normals drawn 
 from the several divisions of the inner arc. The arcs are marked the 
 same as in figure 2. 
 
 PROBLEM III. 
 
 To describe a Gothic isosceles arch to any width, height, 
 and to a given vertical angle. 
 
 Plate V. Let AB, Jig. 1, be the span or width of the arch, mC per- 
 pendicular to AB, from the middle point m the height, and eCf the ver- 
 tical angle given by the tangents Ce and C/, making equal angles with 
 the line of height mC 
 
 In this example, the points e and /, the lower extremities of the tan- 
 gents, are regulated by erecting Ae and B/, each perpendicular to AB, 
 and making each equal to f of the height line ?«C. 
 
 From the point A, towards B, make Ak equal to Ae or B/", that is, 
 equal to | of ?nC ; and from the point C, the vortex of the arch, draw Ci 
 perpendicular to Ce. In Ci take C^, equal to Ak, and join kl ; bisect kl 
 by a perpendicular, di meeting Ci in the point i; join ik, and produce 
 ik to q. 
 
 From the point i, with the radius iC, describe an arc C^, meeting the 
 line iq in the point g, and from the point k, with the radius kg, describe 
 
 would become n— — ; hence R= — ' hence the radius of curvature 
 
 at either extremity of the axis major, is a third proportional to the semi- 
 axis major and the semi-axis minor ; the geometrical magnitude for R 
 
 when equal to — , and when equal to — , have been found by the con- 
 struction of figure 3. 
 
14 
 
 A PRACTICAL TREATISE ON 
 
 [CII. I. 
 
 an arc gA, and A^C will be one-half of the intrados of the Gothic arch 
 required. 
 
 Produce Cm to meet ki in the point n, and in AB make mu equal to 
 mk. Join ?iu, and prolong 7iu to t and un to o. Make no equal to ni. 
 From the centre o, with the radius oC, describe the arc Ch, meeting ut 
 in the point k, and from u, with the radius uk, describe the arc hB, and 
 BkC will be the other half of the intrados. 
 
 Upon AB, prolonged both ways to p and s, make Ap and B^ each 
 equal to the length of each one of the arch stones, in a direction of the 
 radius. 
 
 From the point k, as a centre with the radius kp, describe the arc pq ; 
 and from the point /, with the radius iq, describe the arc qr, and pqr will 
 be half of the extrados of the arch. 
 
 In the same manner will be found str, the other half of the extrados. 
 
 The arch-stones are divided upon the dotted line in the middle into 
 equal parts, and the joint lines are drawn by the centres of the intrados 
 and extrados of the arch. 
 
 REMARK. 
 
 When the height of the arch is equal to, or greater than half the span, 
 and when it is not necessary that the vertical angle should be given, the 
 curves of the intrados and extrados on the one side may be described 
 from the same centre, as also those of the other side from another centre. 
 
 The most easy Gothic arch to describe is that of which the height of 
 the intrados is such as to be the perpendicular of an equilateral triangle, 
 described upon the spanning line as a base, such is ^g. 2, and these 
 centres are the points to which the radiating joints must tend. 
 
 Gothic arches seldom exceed in height the perpendicular of the equi- 
 lateral triangle inscribed in the intrados of the aperture ; but when the 
 arch is surmounted, and the height less than the perpendicular of the 
 equilateral triangle made upon the base, draw a straight line from one 
 extremity of the base to the vertex, and bisect this line by a perpendi- 
 cular. From the point where the perpendicular meets the base of the 
 arch, and with a radius equal to the distance between this point and the 
 extremity of the base joined to the vertex, describe an arc between the 
 two points, joined by the straight line, and the curve which forms one 
 side of the intrados will be complete. In the same manner will be found 
 the curve on the other side, see Jig. 3, so that by only two centres the 
 whole of the intrados will be formed. 
 
 The curves of all kinds of Gothic arches whatever may be described 
 by means of conic parabola, to a given vertical angle, and to any given 
 
CH. I.] MASONRY AND STONE-CUTTING. 
 
 15 
 
 dimensions. Thus in Jig. 4, let Ce, Cf, be the two tangents and Ae, 
 and B/" the heights of their extremities. Divide Ae and cC each into the 
 same number of equal parts by the points 1, 2, 3, in each of these lines. 
 Draw lines from the corresponding points 1-1, 2-2, 3-3, &c. ; and the in- 
 tersections will form the curve of one side of the intrados, as we have 
 already seen. The curve on the other side will be formed in the same 
 manner. 
 
 Join BC, and bisect it in g and join gt^, intersecting the curve in /. 
 Draw hk parallel to CB, meeting gf in k.''^ Make li equal to Ik, and ih 
 joined is a tangent at k. Hence, hm perpendicular to hi, is the joint. 
 
 SECTION V. 
 
 ON THE NATURE AND CONSTRUCTION OF TRE- 
 HEDRALS. 
 
 DEFINITIONS. 
 
 Every stone bounded by six quadrilateral planes or faces 
 forms a solid, of which the surfaces terminate on eight 
 points, every three surfaces in one point. Every three planes 
 thus terminating is termed a solid angle or trehedral. 
 
 The angles formed by the intersections of the faces with 
 one another, or the three plane angles, are called sides of 
 the trehedral, and the angles of inclination are called, by 
 way of distinction from the other, simply angles. 
 
 The three sides, as well as the three angles, are each 
 called a part ; so that the whole trehedral consists of six 
 parts ; and if any three of these parts be given, the remain- 
 ing three can be found. 
 
 Therefore, in bodies constructed of stone, which are in- 
 tended to have their solid angles to consist of three plane 
 angles, the construction of such bodies may be reduced to 
 the consideration of the trehedral. 
 
 As to the remaining surface of the solid which incloses 
 the solid, completely making a fourth side to the trehedral. 
 
16 
 
 A PRACTICAL TREATISE ON 
 
 [CH. I. 
 
 it may be of any form whatever, regular or irregular, or 
 consisting of many surfaces : it or they have nothing to do 
 in the construction. 
 
 The parts of the trehedral, which may be obtained from 
 three given parts, are the very same as three parts found 
 in a spherical triangle from three given parts. This is, in 
 fact, the same as spherical trigonometry. 
 
 We shall not, however, enter into any operose analyti- 
 cal investigations, but treat the subject in the most simple 
 manner, according to the rules of solid geometry ; and 
 only those trehedrals, which have two of their planes at 
 a right angle with each other, (though there are many 
 cases in which the oblique trehedral would be necessary) ; as 
 the bounds prescribed for this work will not admit of such 
 an extension of the principles. 
 
 If the trehedral have two of its planes perpendicular to 
 each other, it is called a right angled trehedral ; each of 
 the two faces thus forming a right angle, is called a leg^ 
 and the remaining side joining each leg, is called the hy- 
 potenuse. 
 
 PROBLEM I. 
 
 Given two legs of a right angled trehedral to find the 
 hypotenuse. 
 
 Plate VI., Jigs. I, 2, 3, 4. Let PON and POR be the given legs. 
 Draw PR perpendicular to OP, and PQ perpendicular to ON. From 
 O, as a centre with the radius OR, describe an arc intersecting PQ in 
 Q, and join OQ, and QON is the hypotenuse required. 
 
 Demonstration. — Suppose the triangle POR revolved upon OP, until 
 PR become perpendicular to the plane of the triangle OPN, then the 
 plane of the triangle OPR will be perpendicular to the plane of the tri- 
 angle OPN. 
 
 Again, suppose the triangle ONQ to revolve upon ON, and let PQ, 
 or PQ produced intersect ON in m, then ?«Q will always be in a plane 
 passing through Vm, and the plane described by r«Q will be perpen- 
 dicular to the plane ?«0P ; and as PR is, by supposition, also perpendi- 
 
CH. I.] MASONRY AND STONE-CUTTING. 
 
 17 
 
 cular to the plane 7wOP, therefore PR and mQ being thus situated in 
 the same plane will meet, except they are parallel. 
 
 Let ?nQ therefore be revolved until the straight line mQ fall upon 
 the point R ; let Q then be supposed to coincide with R ; then because 
 Q, by supposition, coincides with R, and the point O is common to the 
 straight lines OQ and OR, therefore the straight lines OQ and OR 
 having two common points will coincide, and therefore mOQ will be the 
 hypotenuse required. 
 
 PROBLEM n. 
 
 Given the hypotenuse, and one of the legs to find the 
 other leg. 
 
 Figs, 1, 2, 3, 4. Let NOQ be the given hypotenuse and NOP the 
 given leg, and let these two parts be attached to each other by the 
 straight line ON. 
 
 In ON take any point tn, and through m draw PQ perpendicular 
 to ON. Draw PR perpendicular to OP. From the point O, with the 
 radius OQ, describe an arc QR and join OR ; then will POR be the 
 other leg, as required. 
 
 These four diagrams show the various positions in which the data m 
 be placed : every one will frequently occur in practice. 
 
 PROBLEM in. 
 Given the two legs of a right-angled trehedral to find one 
 of the angles at the hypotenuse. 
 
 Figs. 5, 6. Let the two given legs be PON and POR. In OP take 
 any point P, and draw PN perpendicular to ON, and PR perpendicular 
 to PO, and PK parallel to ON. Make PK equal to PR, and join NK ; 
 then PNK will be the angle at the hypotenuse. 
 
 In 5, the two legs lie upon separate parts ; and in Jig' 6, one of the 
 legs lies upon the other. 
 
 Fig. 7 exhibits the same principle applied in finding a series of bevels 
 or angles made by the hypotenuse and a leg. Thus let the two legs be 
 PON and POR. From any point m in OP draw mR perpendicular to 
 OP. On Om, as a diameter, describe the semicircle Oqm, intersecting 
 ON in q, and join qm. Make mr equal to mq, and join rR ; then PrR 
 will be the angle required. 
 
 PROBLEM IV. 
 
 Given one of the legs and the inclination of the hypote- 
 nuse to that leg, to find the other leg. 
 
 c 
 
18 
 
 A PRACTICAL TREATISE ON 
 
 [CH. I. 
 
 Figs. 8 and 9. Let NOP be the given leg. In ON take any point m, 
 and draw vii perpendicular to ON» Make imp, equal to the angle which 
 the leg NOP makes with the hypotenuse. Through any point i, in mi, 
 draw Pp parallel to ON, and PQ, perpendicular to OP. Make PQ, 
 equal to ip, and join OQ, and QOP will be the other leg. 
 
 PROBLEM V. 
 
 Given one of the legs and the angle which the hypote- 
 nuse forms with that leg to find the hypotenuse. 
 
 Fig. 10 and 11. Li NO, take any^point m, and draw ?w» perpendicular 
 to ON. Make nmp equal to the angle which the hypotenuse makes with 
 the leg NOP. From the point m as a centre with any radius, mn describe 
 an arc np. Draw pP, reQ parallel to NO, and PQ perpendicular to NO, 
 and join OQ ; then NOQ is the hypotenuse required. 
 
 General applications of the Trehedral to Tangent Planes. 
 Example I. 
 
 Given the inclination and seat of the axis of an oblique 
 cylinder or cylindroid, to find the angle which a tangent 
 makes at any point in the circumference of the base, with 
 the plane of the base. 
 
 Figs. 1, 3, Plate VII. Let AEBO be the base of the cylinder or 
 cylindroid, CB the seat of the axis, and let BCD be the angle of inclina - 
 tion, and let O be the point where the tangent plane touches the curved 
 surface of the solid. 
 
 Draw ON a tangent line at the point O in the base, and draw OP 
 parallel to CB. Make the angle POR equal to BCD, and draw PR per- 
 pendicular to PO. 
 
 Then, if the triangle POR be conceived to be revolved round the line 
 PO as an axis, until its plane become perpendicular to the plane of the 
 circle AEBC, the straight OR will, in this position, coincide with the 
 cylindrical surface, and a plane touching the cylinder or cylindroid at O, 
 will pass through the lines ON and OR. Here will now be given the 
 two legs POR and PON of a right angled trehedral to find the angle 
 which the hypotenuse makes with the base. Draw PQ perpendicular 
 to ON, intersecting it in m, and draw PS perpendicular to PQ. Make 
 PS equal to PR, and join /«S ; then PmS is the angle required. 
 
 The hypotenuse will be easily constructed at the same time, thus — 
 
CH. I.] MASONRY AND STONE-CUTTING. 
 
 19 
 
 make mQ, equal ta mS, and join OQ, then NOQ will be the hypotenuse 
 required. 
 
 In Jig. 1, the method of finding the angle which the tangent plane 
 makes with the base and the hypotenuse is exhibited at four different 
 points. In the two first points O from A in the first quadrant, the tan- 
 gent planes make an acute angle at each point O ; but in the second 
 quadrant, they make an obtuse angle at each point O. 
 
 Fig. 2 is the second position of the construction from the point A, for 
 finding the angle which the tangent plane makes with the base, and for 
 finding the hypotenuse enlarged ; in order to show a more convenient 
 method by not only requiring less space, but less labour. It may be thus 
 described, the two given legs being P'O'R' and P'O'N'. 
 
 Draw P'm' perpendicular to O'N', meeting ON in m'. In P'O', make 
 P'«' equal to P'm', and draw the straight line v'R', then PVR' will be 
 the inclination of the tangent plane at the point O. 
 
 Again in O'P', make O't' equal to O'm', and draw t'u' parallel to P'R'. 
 From O', with the radius O'R', describe an arc meeting i'u' in ti, and 
 draw the straight line O'm' ; then t'O'u' is the hypotenuse. 
 
 For since P'S' is equal to P'R', and PV equal to P W, and the angles 
 w'P S', and vP'R', are right angles ; therefore the triangle ?/P'R', is 
 equal to the triangle m'P'8', and the remaining angles of the one, equal 
 to the remaining angles of the other, each to each ; hence the angle F'v'R' 
 is equal to the angle P'w'S'. 
 
 Again, because O't' is equal to O'm, and O'Q' is equal to O'R', and 
 O'm' is also equal to O'R' ; therefore O'm' is equal to O'Q', and since 
 the angles O'^'m' and O'tw'Q' are each a right angle, therefore the two 
 right angled triangles have their hypotenuses equal to each other, and 
 have also one leg in each, equal to each other ; therefore the remaining 
 side of the one triangle is equal to the remaining side of the other, and 
 therefore also the angles which are opposite to the equal sides are equal ; 
 hence the angle P'O'm' is equal to N'O'Q' . 
 
 By considering this construction by the transposition of the triangles, 
 the whole of the angles which the tangent planes make at a series of 
 points O in figures 1 and 3, and their hypotenuses may be all found in 
 one diagram, as in figure 4. 
 
 Thus, in Jig. 4, if the angles AGO, AGO', AGO ", AGO'", be re- 
 spectively equal to AGO, AGO', AGO", AGO'", .fig. 1, and in Jig. 4, the 
 semicircle AO'B be described, and if CD be drawn perpendicular to 
 AB, and the angles GAD, GBD, be made equal to BGD,/g-. 1 ; then 
 each half of Jig. 4, being constructed as in Jg. 2 ; the angles at vi, m, 
 
 c 2 
 
20 
 
 A PRACTICAL TREATISE ON 
 
 [CH. I. 
 
 7n", in", will be respectively equal to the angles PyreS, P'tw'S', Q"?w"S", 
 Q"m"'S", mjig. 1. 
 
 Also, in Jig. 4, the angles CAE, CAg, CAh, CBi, CBk, CBF will be 
 the hypotenuses at the point A, O, O', O", O'", B mjig. 1. 
 
 We may here observe, 1, that the angles which the tangent planes 
 make with the plane of the base in the first quadrant are acute ; and 
 those in the second quadrant are obtuse ; and those in the second qua- 
 drant are the supplements of the angles P«?S ; and, moreover, that all 
 the angles which constitute the hypotenuses of the trehedral, are all 
 acute, whether in the first quadrant or second quadrant of the semi- 
 circle AOB. 
 
 SECTION VI. 
 
 ON THE PROJECTION OF A STRAIGHT LINE BENT UPON 
 A CYLINDRIC SURFACE, AND THE METHOD OF 
 DRAWING A TANGENT TO SUCH A PROJECTION. 
 
 PROBLEM 1. 
 
 Given the developement of the surface of the semi-cylin- 
 der, and a straight line in that developement, to find the 
 projection of the straight on a plane passing through the 
 axis of the cylinder, supposing the developement to en- 
 case the semi-cylindric surface. 
 
 Fig. 5. Let ABC be the developement of the cylindric surface, BC 
 being the developement of the semi-circumference, and let AC be the 
 straight line given. 
 
 Produce CB to D, making BD equal to the diameter of the cylinder. 
 On BD, as a diameter, describe the semicircle BED, and divide the semi- 
 circular arc BED, into any number of equal parts, at 1,2, 3, &c. ; and its 
 developement BC into the same number of equal parts, at the points /, g, 
 hy &c. Draw the straight lines fh, gl, hm, &c. parallel to BA, meeting 
 AC at the points k, I, m, &c. ; also parallel to BA, draw the straight 
 lines lo, ^q, &c. and draw ko, Ip, mq, &c. parallel to CD ; and the 
 points o, p, q, &c. are the projections or seats of the points k, I, m, jScc 
 in the developement of the straight line AC. 
 
 The projection of a screw is found by this method : BD may be con- 
 sidered as the diameter of the cylinder from which the screw is formed ; 
 
CH. I.] 
 
 MASONRY AND STONE-CUTTING. 
 
 21 
 
 and the angle BAG, the inclination of the thread .with a straight line on 
 the surface ; or BCA the inclination of the thread with the end of the 
 cylinder. The same principle also applies to the delineations of the 
 hand-rails of stairs, and in the construction of bevel bridges, of which we 
 shall treat in a subsequent part of this work, 
 
 PROBLEM II. 
 
 Given the entire projection of a helix or screw, in the 
 surface of a semi-cylinder, and the projection of a circle in 
 that surface perpendicular to the axis, upon the plane pass- 
 ing through the axis, to draw a tangent to the curve at 
 a given point. 
 
 Fig. 6. Let BPK be the projection of the helix or screw, and BA 
 the projection of the circumference of a circle, and since this circle is in 
 a plane perpendicular to the plane of projection, it will be projected into 
 a straight line AB, equal to the diameter of the cylinder. 
 
 On AB as a diameter, describe the semicircle ArB, and draw Pr per- 
 pendicular to, and intersecting AB in q, join the points e, r, and produce 
 er to f. 
 
 Produce AB to C, so that BC may be equal to the semicircular arc 
 Br A. Draw CD perpendicular to BC, and make CD equal to AK, 
 and draw the straight line BD ; then BD will be the developement of 
 the curve line BPK. 
 
 Draw P« parallel to AC, meeting BD in u, and draw ut perpendicular 
 to BC. Draw rg perpendicular to er, and make rg equal to Bf. Draw 
 gn perpendicular to AC, meeting BC in n, and draw the straight line wP ; 
 then nP will touch the curve at the point P. 
 
 Or the tangent may be drawn independent of BCD thus : Draw Pr 
 perpendicular to AB, and rg a tangent at r. Make rg equal to the deve- 
 lopement of rB, and draw gn perpendicular to BC, meeting BC in w, and 
 join nP, which is the tangent required. 
 
22 A PRACTICAL TREATISE ON [CH. II. 
 
 CHAPTER II. 
 
 THE GEOMETRY OF PLANS AND ELEVATIONS, ADAPTED 
 TO THE CONSTRUCTION OF ARCHES AND VAULTS. 
 
 SECTION I. 
 PRELIMINARY PRINCIPLES OF PROJECTION. 
 
 If from a point A', fig. 1 in space, a perpendicular A'a 
 be let fall to any plane PQ whatever, the foot a of this 
 perpendicular is called the projection of the point A' upon 
 the plane PQ. 
 
 If through different points A', B', C', D' figs. 2, 3, 4, 
 
 of any line A'B'C'D' ..... whatever in space, perpendicu- 
 lars A'«, B'Z», Cc, D'f/, be let fall upon any plane PQ 
 
 whatever, and if througli a, b, c, d, the projec- 
 tions of the points A', B', C, D', in the plane PQ 
 
 a line be drawn, the line thus drawn will be the projection 
 of the line A'B'C'D' ... .given in space. 
 
 If the line A'B'C'D' fig. 3, be straight, the pi o- 
 
 jection abed .... will also be a straight line ; and if the 
 line A'B'C'D . . . .fig. 2, be a curve not in plane perpen- v 
 dicular to the plane PQ, the curve abed . .) . which is the 
 projection of the curve A'B'C'I^ .... in space, will be of 
 the same species with the original curve, of which it is the 
 projection. Hence, in this case, if the original curve 
 A'B'C'D .... be an ellipse, a parabola, hyperbola, &c., the 
 projection abed .... will be an ellipse, a parabola, an hy- 
 
CH. II.] MASONRY AND STONE-CUTTING. 
 
 perbola, &c. The circle and the ellipse being of the same 
 species, the projected curve may be a circle or ellipse, 
 whether the original be a circle or ellipse, as in Jig. 4. 
 
 The plane in which the projection of any point, line, or 
 plane figure is situated, is called the plane of projection, 
 and the point or line to be projected is called the pri- 
 mative. 
 
 The projection of a curve will be a straight line when 
 the curve to be projected is in a plane perpendicular to 
 the plane of projection. Hence the projection of a plane 
 curve is a straight line. 
 
 If a curve be situated in a plane which is parallel to 
 the plane of projection, the projection of the curve will be 
 another curve equal and similar to the curve of which it 
 is the projection. 
 
 The projection upon a plane of any curve of double 
 curvature whatever is always a curve line. 
 
 In order to fix the position and form of any line what- 
 ever in space, the position of the line is given to each of two 
 planes which are perpendicular to each other ; the one is 
 called the horizontal plane and the other the vertical 
 plane ; the projection of the line in question is made on 
 each of these two planes, and the two projections are 
 called the two projections of the line to be projected. 
 
 Thus we see in Jig. 5, where the parallelogram UVWX 
 represents the horizontal plane, and the parallelogram 
 UVYZ represents the vertical plane, the projection ab 
 of the line A'B' in space upon the horizontal plane 
 UVWX, is called the horizontal projection, and the pro- 
 jection AB of the same line upon the vertical plane UVYZ, 
 is called the vertical projection. 
 
 The two planes, upon which we project any line what- 
 ever, are called the planes of projection. 
 
24 
 
 A PRACTICAL TREATISE ON 
 
 [CH. II. 
 
 The intersection UV of the two planes of projection, is 
 called the ground line. 
 
 When we have two projections ah, AB of any Une A'B' 
 in space, the line A'B' will be determined by erecting to 
 the planes of projection the perpendiculars «A', ^B'. . . . , 
 
 A A', B B' through the projections 3, ; 
 
 A, B, of the original points A', B', . . . . of the line 
 
 in question. For the perpendiculars rtA', A A' erected from 
 the projections a, A of the same point A' will intersect each 
 other in space in a point A', which will be one of these in 
 the line in question. It is clear that the other points must 
 be found in the same manner as this which has now been 
 done. 
 
 When we have obtained the two projections of a line in 
 space, whether immediately from the line itself, or by any 
 other means whatever, we must abandon this line in order 
 to consider its two projections only. Since, when we de- 
 sign a working drawing, we operate only upon the two 
 projections of this line that we have brought together upon 
 one plane, and we no longer see any thing in space. 
 
 However, to conceive that which we design, it is abso- 
 lutely necessary to carry by thought the operations into 
 space from their projections. This is the most difficult part 
 that a beginner has to surmount, particularly when he has 
 to consider at the same time a great number of lines in 
 various positions in space. 
 
 The perpendicular A'«, Jig, 5, let fall from any point A' 
 whatever in space upon the plane XV of projection, is 
 called the projectant of the point A' upon this plane. 
 Moreover, the perpendicular distance between the point 
 A' and the horizontal plane XV, is called the pro- 
 jectant upon the horizontal plane, or simply the horizon- 
 tal projectant ; and the perpendicular distance A'A between 
 the original point A' and the vertical plane UY, is called 
 
CH. II.] MASONRY AND STONE-CUTTING. 
 
 25 
 
 the projectant upon the vertical plane, or simply the vertical 
 projection. 
 
 We shall remark, so as to prevent any mistake, that the 
 horizontal projectant A'a, is the perpendicular let fall from, 
 the original point upon the horizontal plane, and that the ver- 
 tical projectant is the perpendicular let fall from that point 
 upon the vertical plane. Hence the horizontal projectant 
 is parallel to the vertical plane, and is equal to the distance 
 between the original point and the horizontal plane ; and 
 the vertical projectant is parallel to the horizontal plane, 
 and is equal to the distance between the original point 
 and the vertical plane. 
 
 We may remark, that if through Jig. 6, the horizontal 
 projection of the point A' we draw a perpendicular aa to UV 
 the ground line, this perpendicular aa will be equal to the 
 measure of the vertical projectant A'A ; consequently the 
 distance of the point A' to the vertical plane is equal to the 
 distance between a, its horizontal projection, and UV the 
 ground line measured in a perpendicular to UV. In like 
 manner, if through A, the vertical projection of the point 
 A', we draw a perpendicular Aa to UV the ground line, 
 this perpendicular will be equal to the measure of the 
 horizontal projectant Aa ; consequently, the distance of this 
 point A' to the horizontal plane, is equal to the distance 
 between A its vertical projection, and UV the ground line 
 measured in a perpendicular to UV. 
 
 To these very important remarks we shall add one 
 which is not less so. Two perpendiculars, aa,^g. 6, Aa, 
 being let fall from the projections a, A to the same point 
 A', upon the ground line UV, will meet each other in the 
 same point a, of the said ground line UV. 
 
 If we now wished the two projections of a point A', 
 Jig. 6, or of any line A'B' whatever be upon one or 
 the same plane, it is sufficient to imagine the vertical 
 
26 
 
 A PRACTICAL TREATISE ON 
 
 [CH.II. 
 
 plane UVYZ to turn round the ground-line UV, in such 
 a manner as to be the prolongation of the horizontal plane 
 UVWX ; for it is clear that this plane will carry with it 
 the vertical projection A or AB of the point, or of the line 
 in question. Moreover we see, and it is very important 
 that the hues Aa, Bb, perpendicular to the ground-line 
 UV will not cease to be so in the motion of the plane 
 UVYZ ; and as the corresponding lines aa, bh, are also 
 perpendiculars to the ground-line UV, it follows that the 
 lines aa', hb', will be the respective prolongation of the lines 
 aa, bh. 
 
 Hence it results, when we consider objects upon a single 
 plane, the projections A of a point A' in space are ne- 
 cessarily upon the same perpendicular Aa to the ground- 
 line UV. 
 
 It is necessary to call to mind that the distance Aa mea- 
 sures the distance from the point in space to the horizontal 
 plane, (the point A being the vertical projection of this 
 point,) and that the line aa measures the distance from the 
 same point in space to the vertical plane. 
 
 It follows, that if the point in space be upon the hori- 
 zontal plane, its distance with regard to this last-named 
 plane will be zero or nothing, and the vertical Aa will be 
 zero also. Moreover, the vertical projection of this point 
 will be upon the ground-line at the foot a of the perpendi- 
 cular aa let fall upon the ground-line, from the horizontal 
 projection a of this point. 
 
 Again, if the point in space be upon the vertical plane, 
 its distance, in respect of this plane, will be zero, the hori- 
 zontal aa will be zero, and the horizontal projection of 
 the point in question will be the foot a of the perpendicular 
 Aa let fall upon the ground-line from the vertical projec- 
 tion A of this point. 
 
 In general, we suppose that the vertical projection of 
 
CH. II.] MASONllY AND STONE-CUTTING. 
 
 27 
 
 a point is above the ground-line, and that the horizontal 
 projection is below; but from 'what has been said, it is 
 evident that if the point in space be situated below the 
 horizontal line, its vertical projection will be below the 
 ground-line ; for the distance from this point to the hori- 
 zontal plane, cannot be taken from the base-line to the top, 
 but from the top to the base with respect to its plane. 
 
 So if the point in space be situated behind the vertical 
 plane, its horizontal projection will be above the ground- 
 line ; from which we conclude — 
 
 1st. If the point in question be situated above the hori- 
 zontal plane, and before the vertical plane, its vertical pro- 
 jection will be above and its horizontal projection below 
 the ground-line. 
 
 2!nd. If the point be situated before the vertical plane, 
 and below the horizontal plane, the two projections will be 
 below the ground-line. 
 
 3rd. If the point be situated above the horizontal plane, 
 but behind the vertical plane, the two projections will be 
 above the ground-line. 
 
 4th. Lastly. If the point be situated above the horizon- 
 tal plane, and behind the vertical plane, the vertical pro- 
 jection will be below, and the horizontal projection above, 
 the ground-line. 
 
 The reciprocals of these propositions are also time. 
 If a line be parallel to one of the planes of projection, its 
 projection upon the other plane will be parallel to the 
 ground-line. Thus, for example, if a line be parallel to a 
 horizontal plane, its vertical projection will be parallel to 
 the ground line ; and if it is parallel to the vertical plane, 
 its horizontal projection will be parallel to the ground - 
 line. 
 
 Reciprocally, if one of the projections of a line be parallel 
 to the ground line, this line will be parallel to the plane of 
 
28 
 
 A PRACTICAL TREATISE ON 
 
 [CH. II. 
 
 the other projection. Thus, for example, if the vertical pro- 
 jection of a line be parallel to the ground-line, this line will 
 be parallel to the horizontal plane, and vice versa. 
 
 If a line be at any time parallel to the two planes of pro- 
 jection, the two projections of this line will be parallel to 
 the ground-line ; and reciprocally, if the two projections of 
 a line be parallel to the ground-line, the line itself will be 
 at the same time parallel to the two planes of projection. 
 
 If a line be perpendicular to one of the planes of projec- 
 tion, its projection upon this plane will only be a point, and 
 its projection upon the other plane will be perpendicular to 
 the ground-line. Thus, for example, if the line in question 
 be perpendicular to the horizontal plane, its horizontal pro- 
 jection will be only a point, and its vertical projection will 
 be perpendicular to the ground-line. 
 
 Reciprocally, if one of the projections of a straight line be 
 a point, and the projection of the other perpendicular to the 
 ground-line, this line will be perpendicular to the plane of 
 projection upon which its projection is a point. Thus the 
 line will be perpendicular to the horizontal plane, if its pro- 
 jection be the given point in the horizontal plane. 
 
 If a line be perpendicular to the ground-line, the two 
 projections will also be perpendicular to this line. The re- 
 ciprocal is not true ; that is to say, that the two projections 
 of a line may be perpendicular to the ground-line, without 
 having the same line perpendicular to the ground-line. 
 
 If a line be situated in one of the planes of projection, its 
 projection upon the other will be upon the ground-line. 
 Thus, if a line be situated upon a horizontal plane, its ver- 
 tical projection will be upon the ground-line ; and if this 
 line were given upon the vertical plane, its horizontal pro- 
 jection would be upon the ground-line. 
 
 Reciprocally, if one of the projections of a line be upon 
 the ground-line, this line will be upon the plane of the 
 
CH. II.] MASONRY AND STONE-CUTTING. 29 
 
 other projection. Thus, for example, if it be the vertical 
 projection of the line in question which is upon the ground, 
 this line will be upon the horizontal plane ; if on the con- 
 trary, it were upon the horizontal projection of this line 
 which was upon the ground-line, this line would be upon 
 the vertical plane. 
 
 If a line be at any time upon the two planes of pro- 
 jection, the two projections of this line would be upon the 
 ground-line, and the line in question would coincide with 
 this ground-line. Reciprocally, if the two projections of a 
 line were upon the ground-line, the line itself would be 
 upon the ground-line. 
 
 If two lines in space are parallel, their projections upon 
 each plane of projection are also parallel. Reciprocally, if 
 the projections of two lines are parallel on each plane of 
 projection, the two lines will be parallel to one another in 
 space. 
 
 If any two lines whatever in space cut each other, the 
 projections of their point of intersection will be upon the 
 same perpendicular line to the ground-line, and upon the 
 points of intersection of the projections of these lines. Re- 
 ciprocally, if the projections of any two lines whatever cut 
 each other in the two planes of projection, in such a man- 
 ner that their points of intersection are upon the same 
 perpendicular to the ground-line, these two lines in question 
 will cut each other in space. 
 
 The position of a plane is determined in space when we 
 know the intersections of this plane with the planes of pro- 
 jection. 
 
 The intersections AB, AC, of the plane in question, with 
 the planes of projection, are called the traces of this plane. 
 
 The trace situated in the horizontal plane is called the 
 horizontal trace, and the trace situated in the vertical plane 
 is called the vertical trace. 
 
30 
 
 A PRACTICAL TREATISE ON [CH. II. 
 
 A very important remark is, that the two traces of a 
 plane intersect each other upon the ground-line. 
 
 If a plane be parallel to one of the planes of projection, 
 this plane will have only one trace, which will be parallel 
 to the ground-line, and situated in the other plane of pro- 
 jection. Reciprocally, if a plane has a trace parallel to 
 the ground-line, this plane will be parallel to the plane of 
 projection which does not contain this trace. Thus : — 
 
 1st. If a plane be parallel to the horizontal plane, this 
 plane will not have a horizontal trace, and its vertical trace 
 will be parallel to the ground-line. Likewise, if a plane 
 be parallel to the vertical plane, this plane will not have 
 a vertical trace, and its horizontal trace will be parallel to 
 the ground-line. 
 
 2d. If a plane has only one trace, and this trace pa- 
 rallel to the ground-hne, let it be in the vertical plane; 
 then the plane will be parallel to the horizontal plane. So 
 if the trace of the plane be in the horizontal plane, and 
 parallel to the ground-line, the plane will be parallel to the 
 vertical plane. 
 
 If one of the traces of a plane be perpendicular to the 
 ground-line, and the other trace in any position whatever, 
 this plane will be perpendicular to the plane of projection 
 in which the second trace is in. Thus, if it be a horizontal 
 trace which is perpendicular to the ground-line, the plane 
 will be perpendicular to the vertical plan of projection ; 
 and if, on the contrary, the vertical trace be that which is 
 perpendicular to the ground-line, then the plane will be per- 
 pendicular to the horizontal plane. 
 
 Reciprocally, if a plane be perpendicular to one of the 
 planes of projection without being parallel to the other, its 
 trace upon the plane of projection to which it is perpendi- 
 cular will be to any position whatever, and the other trace 
 will be perpendicular to the ground-line. Thus, for exam- 
 
CH. II.] MASONRY AND STONE-CUTTING. 
 
 31 
 
 pie, if the plane be perpendicular to the vertical plane, the 
 vertical trace will be in any position whatever, and its 
 horizontal trace will be perpendicular to the ground-line. 
 The reverse will also be true, if the plane be perpendicular 
 to the horizontal plane. 
 
 If a plane be perpendicular to the two planes of projec- 
 tion, its two traces will be perpendicular to the ground-line. 
 Reciprocally, if the two traces of a plane are in the same 
 straight line perpendicular to the ground-line, this plane 
 will be perpendicular to both the planes of projection. 
 
 If the two traces of a plane are parallel to the ground- 
 line, this plane will be also parallel to the ground-line. 
 Reciprocally, if a plane be parallel to the ground-line, its 
 two traces will be parallel to the ground-line. 
 
 When a plane is not parallel to either of the planes of 
 projection, and one of its traces is parallel to the ground- 
 line, the other trace is also necessarily parallel to the ground- 
 line. 
 
 If two planes are parallel, their traces in each of the 
 planes of projection will also be parallel. Reciprocally, if 
 on each plane of projection the traces of the two planes are 
 parallel, the planes will also be parallel. 
 
 If a line be perpendicular to a plane, the projections of 
 this line will be in each plane of projection, perpendicular 
 to the respective traces in this plane. Reciprocally, if the 
 projections of a line are respectively perpendicular to the 
 traces of a plane, the line will be perpendicular to the 
 plane. 
 
 If a line be situated in a given plane by its traces, this 
 line can only intersect the planes of projection upon the 
 traces of the plane which contains it. Moreover, the line in 
 question can only meet the plane of projection in its own 
 projection. Whence it follows, that the points of meeting 
 of the right line, and the planes of projection are respec- 
 
32 
 
 A PRACTICAL TREATISE ON 
 
 [CH. II. 
 
 tively upon the intersections of this right line, and the traces 
 of the plane which contains it. 
 
 If a right line, situated in a given plane by its traces, is 
 parallel to the horizontal plane, its horizontal projection will 
 be parallel to the horizontal trace of the given plane, and 
 its vertical projection will be parallel to the gi'ound-line. 
 Likewise, if the right line situated in a given plane by its 
 traces is parallel to the vertical plane, its vei'tical projection 
 will be parallel to the vertical line of the plane which con- 
 tains it, and its horizontal projection will be parallel to the 
 ground-line. 
 
 Reciprocally, if a line be situated in a given plane by its 
 traces, and that, for example, let its horizontal projection be 
 parallel to the horizontal trace of the given plane, this line 
 will be parallel to the horizontal plane, and its vertical pro- 
 jection will be parallel to the ground-line. Likewise, if the 
 vertical projection of the line in question be parallel to the 
 vertical trace of the given plane, this line will be parallel to 
 the vertical plane, and its horizontal projection will be 
 parallel to the ground-line. 
 
 SECTION II. 
 
 ON THE DEVELOPEMENTS OF THE SURFACES OF 
 
 SOLIDS. 
 
 PROBLEM I. 
 
 To find the developement of the surface of a right semi- 
 cylinder. 
 
 Fig. 1. Let ACDE be the plane passing through the axis. On AC, 
 as a diameter, describe the semicircular arc ABC. Produce CA to 
 and make AP equal to the developement of the arc ABC. Draw FG 
 parallel to AE, and EG parallel to AF ; then AFGE is the develope- 
 ment required. 
 
CH. II.] MASONRY AND STONE-CUTTING. 
 
 33 
 
 PROBLEM II. 
 
 To find the developement of that part of a semi-cylinder 
 contained between two perpendicular surfaces. 
 
 Figs, 2, 3, 4. Let ACDE be a portion of a plane passing through the 
 axis of the cylinder, CD and AE, being sections of the surface, and let 
 DE and GF be the insisting lines of the perpendicular surface ; also let 
 AC be perpendicular to AE and CD. On AC, as a diameter, describe the 
 semi-circular arc ABC. Produce CA to H, and make AH equal to the 
 developement of the arc ABC. Divide the arc ABC, and its develope- 
 ment, each into the same number of equal parts at the points 1, 2, 3. 
 
 Through the points 1, 2, 3, &c. in the semi-circular arc, and in its 
 developement, draw straight lines parallel t o AE, and let the parallel lines 
 through 1, 2, 3, in the arc A, B, C, meet FG in p, q, r, &c. and AC in 
 k) I, m, &c. Transfer the distances Ap, Iq^ mr, &c. to the developement 
 upon the lines 1«, 26, 3c, &c. Through the points F, a, b, c, &c. draw 
 the curve line Fcl. In the same manner draw the curve line EK ; then 
 FEKI will be the developement required. 
 
 PROBLEM IIL 
 To find the developement of the half surface of a right 
 cone, terminated by a plane passing through the axis. 
 
 Fig. 5. Let ACE be the section of the cone passing along the axes 
 AE ; and CE the straight lines which terminate the conic surface, or the 
 two lines'which are common to the section CAE and the conic surface ; 
 and let AC be the line of common section of the axal plane, and the 
 base of the cone. 
 
 On AC as a diameter describe a semi-circle ABC. From E, with the 
 radius EA, describe the arc AF, and make the arc AF equal to the semi- 
 circular arc ABC, and join EF ; then the sector AEF, is the develope- 
 ment of the portion of the conic surface required. 
 
 PROBLEM IV. 
 
 To find the developement of that portion of a conic sur- 
 face contained by a plane passing along the axes, and two 
 surfaces perpendicular to that plane. 
 
 Fig. 6. Let ACE be the section of tlie cone along the axis, and 
 let AC and GI be the insisting lines of the perpendicular surfaces. 
 Find the developement AEF as in the preceding problem. Divide the 
 semi-circular arc ABC, and the sectorial arc AF, each into the same num- 
 ber of equal parts at the poinis 1, 2, 3, &c. From the points 1, 2, 3, &c. 
 
 D 
 
34 
 
 A PRACTICAL TREATISE ON 
 
 [CH. II. 
 
 in the semi-circular arc draw straight lines Ik, 21, 3m, &c. perpendicular 
 to AC. From the points k, I, m, &c. draw straight lines ArE, ZE, wE, 
 &c. intersecting the curve AC in p, q, r, &c. Draw the straight lines 
 pt, qu, rv, &c. parallel to one side, EC meeting AC in the points 
 t, u, V, &c. Also from the points \, 2, 3, in the sectorial arc AF, 
 draw the straight lines IE, 2E, 3E, Sec. Transfer the distances pt, qu, 
 rv, &c. to la, 2b, 3c, &c. ; then through the points A, a, b, c, &c. draw 
 the curve AcF, and AcF is one of the edges of the developement, and 
 by drawing the other edge, the entire developement, AGHF, will be 
 found. 
 
CH. III.] MASONRY AND STONE-CUTTING. 
 
 35 
 
 CHAPTER III. 
 
 CONSTRUCTION OF THE MOULDS FOR HORIZONTAL CY- 
 LINDRETIC VAULTS, EITHER TERMINATING RIGHTLY 
 OR OBLIQUELY, UPON PLANE OR CYLINDRICAL 
 WALLS, WITH THE JOINTS OF THE COURSES EITHER 
 IN THE DIRECTION OF THE VAULT, PERPENDICULAR 
 TO THE FACES, OR IN SPIRAL COURSES. 
 
 SECTION I. 
 
 DEFINITIONS OF MASONRY, WALLS, VAULTS, &C. 
 
 Stone-cutting is the art of reducing stones to such 
 forms that when united together they shall form a deter- 
 minate whole. 
 
 In preparing stones for walls, of which their surfaces 
 are intended to be perpendicular to the horizon, nothing 
 more is necessary than to reduce the stone to its dimen- 
 sions, so that each of its eight solid angles may be con- 
 tained by three plane right angles. 
 
 Moreover, in working the stones of common straight 
 right cyhndretic vaults, where the planes of the sides of the 
 joints terminate upon the intrados or extrados of the arch 
 or vault, in straight lines parallel ruler lines of the cyhn- 
 dretic surface, there can be no difficulty ; for if one of the 
 beds of the stone be formed to a plane surface, and if this 
 side be figured to the mould, and the opposite ends squared, 
 
 D 2 
 
36 
 
 A PRACTICAL TREATISE ON 
 
 [CH. III. 
 
 and, lastly, the face or vertical moulds applied upon the 
 ends thus squared, and their figures drawn, these figures 
 will be the two ends of a prism, consisting of equal and 
 similar figures, and will be similarly situated ; and therefore 
 we have only to form tiiis prism, in order to form the arch- 
 stone required. 
 
 But the formation of the stones in the angles of vaults, 
 and in the courses of spheretical niches and domes, are much 
 more difficult, and require more consideration. In such 
 constructions various methods may be employed, and some 
 of these, in particular instances, with great advantage, both 
 in the saving of workmanship and material, as we shall 
 have occasion to show. In general, however, previous to 
 the reducing of a stone to its ultimate form for such a si- 
 tuation, it will be found convenient to reduce the stone to 
 such a figure as will include the more complex figure of the 
 stone required, so that any surface of the preparatory figure 
 may either include a surface or arris of the stone required 
 to be formed, or be a tangent to their surface. 
 
 Surfaces are brought to form by means of straight and 
 curved edges, always applied in a plane perpendicular to 
 the arris-lines, so that, when a surface is thoroughly formed, 
 the edge of application may have all its points in contact 
 with the surface in its whole intended breadth. 
 
 A wall, in masonry, is a mass of stones or other material, 
 either joined together with or without cement, so as to have 
 its surfaces such that a plumb-line, descending from any 
 point in either face, will not fall without the solid. 
 
 One of the faces of a wall is generally regulated by the 
 other, and the regulating surface is called the principal 
 face. 
 
 The line of intersection of the principal face of a wall, 
 and a horizontal plane on a level with the ground, or as 
 
CH. III.] MASONRY AND STONE-CUTTING. 
 
 37 
 
 nearly so as circumstances will permit, is called the base- 
 line. 
 
 A horizontal section of a wall, through the base-line, is 
 called the seat of the wall. 
 
 The other side of the seat of a wail, opposite to the base- 
 line, is called the rear-line. 
 
 In exterior walls the outer surface is always the princi- 
 pal face, and the base and rear-lines are generally so situ- 
 ated, that normals drawn to the base-line, between the 
 base and rear-lines, are all equal to one another. This 
 uniformity most frequently takes place . also in partition or 
 division walls ; but, in some instances, on account of a room 
 being circular or elliptical, while the other faces are plane or 
 curved surfaces, this equality of the normals cannot subsist. 
 
 If a wall be cut by a plane perpendicular to the base- 
 line, or if the base-line be a curve perpendicular to a tan- 
 gent through the point of contact, such a section is called a 
 right section. 
 
 Hence, according to this definition, since the base-line is 
 always in a horizontal plane, every straight line and every 
 tangent to a base-line, when it is a curve, will be a hori- 
 zontal line, therefore the right section must be in a vertical 
 plane. 
 
 Walls are denominated according to the figure of their 
 base-line. When the base-line is straight, the wall is said to 
 be straight. Hence, if the figure of the base be an arc or the 
 whole circumference of a circle, or a portion or the entire 
 curve of an ellipse, the wall is said to be circular or elliptical. 
 Other forms seldom occur in building. 
 
 Walls are more strictly defined by the joint consideration 
 of the figures of their bases and right section. 
 
 When the base and the right section of a wall are each a 
 straight line, and all the horizontal sections straight lines, 
 
 INSTITU 
 
38 A PRACTICAL TREATISE ON [CH. III. 
 
 the face of the wall is called a ruler surface, and if all the 
 right sections have the same inclination, the wall is called a 
 straight inclined wall ; if they are all vertical, the wall is 
 called an erect straight wall, or a vertical straight wall. If 
 the right sections vary their inclination, the wall is called a 
 winding wall. 
 
 When the base line is the circumference or any arc of a 
 circle, and the right section a straight hne perpendicular to 
 the horizon, the wall is said to be cylindric. If the right 
 sections of a wall be all equally inclined to the horizon, the 
 wall is said to be conic ; and thus a wall takes also the 
 name by which its surface is called ; hence a straight wall, 
 which has its right sections either vertical or at the same 
 inclination, is called a plane wall. 
 
 A wall in tallus, or a battei^ing wall, is that of which the 
 vertical section of the principal face is a straight line not 
 perpendicular to the horizon. This vertical section is called 
 the tallus-Une. 
 
 The horizontal distance between the foot of the tallus- 
 line and the plumb-line, passing through its upper extremity, 
 is called the quantity of batter ; and the plumb line, from 
 the top of the tallus-line to the level of its foot, is called 
 the vertical of the batter. 
 
 The interstices between the stones, for the insertion of 
 cement or mortar, in order to connect the stones into one 
 solid mass, are called Joints, and the surfaces of the stones 
 between which the mortar is inserted, are called the sides 
 of the Joints. 
 
 When the sides of the joints are everywhere perpen- 
 dicular to the fece of a wall, and terminate in horizontal 
 planes upon that face, such joints are called coursing Joints ; 
 and the row of stones between every two coursing joints, is 
 called a course of stones. 
 
 An arch or vault, in masonry, is a mass of stones sus- 
 
CH. III.] MASONRY AND STONE-CUTTING. 
 
 30 
 
 pended over a hollow, and supported by one or more walls 
 at its extremities, the surface opposed to the hollow being 
 concave, and such that a vertical line, descending from any 
 point in the curved surface, may not meet the curved surface 
 in another point. 
 
 The concave surface under the arch or vault, is called the 
 intrados of that arch or vault ; and if the upper surface be 
 convex, this convex surface is called the e.vtrados. 
 
 Those joints which terminate upon the intrados in hori- 
 zontal lines, are called coursing joints, and the coursing 
 joints will either be straight, circular, or elliptic, accordingly 
 as the horizontal sections of the intrados are straight, cir- 
 cular, or elliptic. 
 
 Whether in walling or in vaulting, the joints of the stones 
 should always be perpendicular to the face of the wall, or to 
 intrados of the arch, and the joints between the stones 
 should either be in planes perpendicular to the horizon, or 
 in surfaces which terminate upon the face of the wall or 
 intrados of a vault in horizontal planes; these positions 
 being necessary to the strength, solidity, and durability of 
 the work. 
 
 Walls and vaults being of various forms ; viz. straight, 
 circular, and elHptic, depending on the plan of the work ; 
 hence the construction will depend upon the simple figure 
 or upon the complex figure when combined in two. 
 
 SECTION II. 
 ON OBLIQUE ARCHES. 
 
 PROBLEM I. 
 
 To execute an oblique cylindroidic arch, intersecting each 
 side of the wall in a semi-circle, the imposts of the arch 
 being given. 
 
40 
 
 A PRACTICAL TREATISE ON 
 
 [CH. III. 
 
 Let Jig. 1, Flaie X. be the elevation^ and in fig. 2, let ABCD, 
 EFGH, be the two imposts which are equal and similar parallelograms, 
 having the sides AB, FE one of each in a straight line, and the sides 
 DC and GH in a straight line- 
 Join GC, and on GC as a diameter, describe the semi-circle GIG, 
 which, if conceived to be turned upon the line as an axis, until its 
 plane become perpendicular to the seat BCGF of the soffit of the arch, 
 it will be placed in its due position. Divide the semi-circular arc CIG, 
 into as many equal parts as the ring-stones are to be in number. We 
 shall here suppose there are to be nine ring-stones. From the points 
 of division, 1, 2, 3, &c. draw ordinates perpendicular to GC, meeting GC 
 in the points p, q, r, &c. Perpendicular to CB, the jamb line of the im- 
 post, draw the lines, g2, ?-3, &c. ; from the point C as a centre, 
 with the chord of one-ninth part of the semi-circular arc, CIG', describe 
 an arc intersecting ;)1 CB at 1 ; from the point 1, with the same radius 
 describe an arc intersecting the line ^2, in the point 2 ; from the point 2 
 as a centre with the same radius, describe an arc intersecting the line rS, 
 in the point 3 ; and so on. Join the point C and 1 ; 1 and 2, 2 and 3, 
 &c. and thus form the entire edge CKL, of the developement of the 
 semi-circular arc CIG. 
 
 Through the points 1, 2, 3, &c. in CKL, draw the lines 1/3, 2y, 3^, 
 &c. parallel to CB, and make 1/3, 2y, 3^, &c. each equal to CB ; and join 
 I^A fiy, 7^, &c. ; then CB/31 is the soffit of the first ring-stone ; I/3y2, 
 is the soffiit of the second ring-stone ; 2yo3, the soffit of the third ring- 
 stone, and so on. 
 
 Perpendicular to GF draw FJ ; produce CB to J ; and parallel to CJ, 
 draw ps, qt,ru,&c. Intersecting FJ in the points v, w, x, &c., make r*, 
 rvt, xu, &c. respectively equal to pi, ^2, ?-3, &c. Join J and s, * and t, 
 t and u, &c. ; and complete the polygonal line JmF. Through the points 
 
 t, u, &c. draw the joint lines sy, iz, uQ, radiating to the point c; 
 then will the angles of inclination of the beds and soffits be NJ.y, ysJ, 
 the first ring-stone ; yst, zts, for the second ring-stone ; zlu, But for tie 
 third ringstone ; and so on. 
 
 From any point B in EC, Jg. 3, make the angle CBA equal to tie 
 angle ABC, of the impost Jig. 2. Prolong CB toE. From B as a cen- 
 tre, with any radius describe the semi-circular arc CDE ; and on BC as 
 a diameter, describe another semi-circular arc C^B. Divide the semi-cir- 
 cular arc CDE, in the points 1, 2, 3, Sec. into nine equal parts, equal to 
 the number of ring-stones, and draw the radials IB,2B, 3B, &c. intersec- 
 ting the semi-circular arc C^B in the points y, g, h, &c. Draw CA pei- 
 pendicular to BC ; and in BA as a diameter, describe the same circular 
 
CH. III.] MASONRY AND STONE-CUTTING. > 41 
 
 arc BCA. From the point B, with the radii B/, B^, B^, &c. describe the 
 arcs fi; gk, hi, &c. meeting the serai-circular arc BCA, in the points i, k, I, 
 &c. and draw the straight lines Bi, Hk, Bl, &c. Then, ABC being the 
 angle of the impost, ABj will be the angle of the joints at the junction 
 of the first and second ring-stones ; AB/t the angle of the joints at the ■ 
 junction of the second and third ring-stones ; AB/ will be the angle of 
 the joints at the junction of the third and fourth ring-stones, &c. 
 
 To apply the moulds for cutting any one of the ring-stones, or to form 
 the solid angles made by the face, the two beds and the soffit of the 
 stone, which being done will form that ring-stone. —For instance, let it 
 be required to form the third ring-stone :— We have given the plain angle 
 2ya, figure 2, which is a side, and the plane angle AB/t, Jig. 3, 
 another adjacent side ; also the angle ztu,^\vhich. is the inclination ^ 
 of these two sides, to construct the solid angle. This can be easily done 
 by working the bed of the stone corrresponding to the joint 2y on the 
 ^oftitfig. 2 ; then work the narrow side of the stone, from which the soffit 
 is to be formed, first as a plane surface, making an angle ztu, with the 
 bed first wrought ; place the surface of the mould abed. Jig. 4, upon the 
 narrow side of the stone, which is to form the soffit, so that the edge ab 
 may be upon the arris of the stone ; then by the edge be, draw a line ; 
 again upon the wrought side which is intended for the bed, apply the 
 angle ABk,^g. 3, so that the line AB may be upon the arris, and the 
 point B upon the same point that b was applied ; then by the leg Bk, 
 Avhich is supposed to be upon the surface of the bed, draw a line ; we 
 have only to cut away the superfluous stone on the outside of the two 
 lines on the bed and on the soffit ; and thus we shall form a complete 
 trehedral; the plane soffit of the stone being gauged to its breadth, 
 and the mould 2ed3, Jig. 1, being applied upon the last wrought side, so 
 that the points d, e, may be upon the points of the stone to which b and 
 c were applied ; then drawing a line by the edge f?3, and cutting away 
 the superfluous stone between the two lines on the front, and on the plane 
 of the soffit, will form the upper bed of the stone. 
 
 This will be made sufficiently evident by a developement of the soffit, 
 the two beds, and the front of the ring-stone. Make an equal and simi- 
 lar parallelogram abed, Jig. 4, to that of 2y33,^g-. 2. Make the angles 
 abe, dcg,fig. 4, respectively equal to the angles ABk, ABl,Jg. 3 ; then be 
 being equal to de,Jig. 1, apply the mould 2de3, so that the points d, e may 
 be upon bc,Jig. 4, and draw the front of the stone bcki,Jg. 4, and simi- 
 larly draw ad7nl. Make be equal to bi, eg equal to ck, and draw e/and 
 gh parallel to ba or cd, and this will complete the developement. 
 
 A complete model of the stone will instantly be formed, by revolving 
 
4^ 
 
 A PRACTICAL TREATISE ON 
 
 [CH. III. 
 
 the four sides abef, bcki, cdhg, dalm, upon the four lines ha, be, cd, da as 
 axes, until e coincide with i, k with g, h with m, and I with/. 
 
 We have here made use of the developement of the intrados in the 
 construction of the solid angles, as being easily comprehended. The ring- 
 stones might, however, have been formed by the angle of the joints, which 
 is one side of a trehedral ; one of the angles of the face mould, which is 
 the other adjacent side ; and the inclination of these two sides ; so that we 
 shall have here also two sides and the contained angle, to construct the 
 solid angle of the trehedral. As an example, let it again be required to 
 construct the third ring-stone. To find the angle which the face of the 
 third ring-stone makes with the bed in the second joint : we have here 
 given the two legs ABC, CB2,/g'.3, of a right-angled trehedral, to find the 
 angle which the hypotenuse makes with the side CB2 : this being found, 
 will be the inchnation of the face-mould, 2ded, Jig. 1. and ABk,Jig. 3. 
 Therefore, in this case work the bed of the stone first, then the face, to 
 the angle of inclination thus found. Upon the arris apply the leg AB of 
 the joint-mould A'Bk,^fig. 3, so that the side Bk may be upon the bed, 
 and draw a straight hue on the bed by the edge Bk ; next apply the 
 mould 2de3, so that the arc d2 may be upon the arris, and the 
 point d upon the same point of the arris to which the point B was ap- 
 plied, and the chord de upon the face ; then draAV a line on the face of the 
 stone, by the leg de ; and work off the superfluous stone, and the face will 
 be exhibited. Fig. 5, shows the stone as wrought. 
 
 From Avhat has been said, it is evident that if one of the solid angles 
 of a ring-stone be formed of an oblique arch in a straight wall, the re- 
 maining solid angle may be formed without the use of the trehedral. 
 Thus, for instance, suppose the solid angle which is formed be made by 
 the surface of the soffit, the bed, and the face of the arch — we have only 
 to guage the soflit to its breadth, and apply the head-mould upon the face 
 of the stone ; then by working off the superfluous stone between these 
 lines, another solid angle will be formed by the surface of the soflit, the 
 upper bed of the stone, and the face of the arch. 
 
 And since the angle of the joints is the same in the lower and upper 
 beds of any two ring-stones that come in contact with each other, the 
 same angle of the joints will do for both, so that in fact, if this be carried 
 from one ring-stone to another, the arch may be executed without any 
 joint mould. 
 
 This mode would, however, not only be inconvenient, but liable to very 
 great inaccuracy. It would be inconvenient, as it is necessary to work one 
 stone before another, so that only one workman could be employed in the 
 construction of the arch. It would be liable to inaccuracy when the 
 
CH. III.] MASONRY AND STONE-CUTTING. 
 
 43 
 
 number of ring-stones are many, for then any small error would be liable 
 to be multiplied or transmitted from one stone to another. Besides, it is 
 satisfactory to have a mould to apply, in order to examine the work in its 
 progress. 
 
 What has been now observed, with regard to the oblique arch in a 
 straight wall, and with respect to the angle on the edges of the point, 
 will apply to every arch of which the intrados is a cylindric or cylindroi- 
 dic surface. 
 
 In the construction of any object it is always desirable to have two 
 different methods, as one may always be a proof or check to the other. 
 Besides, though these methods may be equally true in principle, one of 
 them may be often liable to greater inaccuracy in its construction than 
 the other. 
 
 PROBLEM II. 
 
 To construct the moulds for a cylindretic oblique ar^\ 
 terminating upon the face of a wall in a plane at obliq|p>:' STlT^ 
 angles to the springing plane of the vault, so that the cours- / ,, 
 ing joints may be in planes parallel to the ruUer lines of the < 
 intrados of the vault. 
 
 Let the vertical plane of projection be perpendicular to the axis of the 
 intrados, and it will therefore be also perpendicular to all the joints of 
 which their planes are parallel to the axis : hence 
 
 The vertical projection of the intrados will be a curve equal and simi- 
 lar to the curve of the right section of the intrados. 
 
 The vertical projections of the coursing joints will bo radiant straight 
 lines, intersecting the curve lined projection of the intrados. 
 
 The vertical projections of all the joints which are in vertical planes pa- 
 rallel to the axis, will be straight lines perpendicular to the ground line. 
 
 The vertical projection of all the joints m, horizontal planes, will be 
 straight lines parallel to the ground-line. 
 
 Moreover the vertical projections of the intersections of planes which 
 are parallel to the axis will be points. 
 
 The horizontal projections of the planes of the coursing joints, and of all 
 the intersections of the planes of all joints which are parallel to the axis, 
 will be straight lines perpendicular to the ground-line. 
 
 And because the axis of the archant is perpendicular to the vertical 
 plane, the vertical projections of the intrados, and of the joints which are 
 parallel to the axis, will have the same position to one another, as the 
 curve and other lines in the right section which are formed by the joints 
 in planes parallel to the axis. 
 
A PRACTICAL TllEATISE ON [CH. III. 
 
 All sections which are perpendicular to the horizon, will have straight 
 lines for their horizontal projections. 
 
 The length of any inclined line will be to the length of its projection, as 
 the radius is to the cosine of the line's inclination to the plane of projection. 
 
 We shall suppose that the stones which constitute the intrados of the 
 archant, have not fewer than three, nor more than four, of their faces that 
 intersect the intrados. The stones which form the face of the archant, 
 when they do not reach the rear of the vault, have three of their faces 
 which intersect the intrados, and three at least which intersect the face. 
 
 We shall call all these surfaces ^vhich intersect the intrados, or face of 
 the archant, the retreating sides of joints of the stones; and the surface of 
 any stone which forms a part of the intrados, the douelle of the stone. 
 
 When the stones do not reach from the front to the rear of the intrados 
 of an archant, they are arranged in rows, in such a manner, that the 
 stones which constitute any one of the rows, have as many of their re- 
 treating sides as there are stones in the row, in one continued surface, 
 and the opposite retreating sides of all the stones in another continued 
 surface, while the heads form a portion of the intrados extending from 
 front to rear of the vault, and the remaining retreating sides of the stones 
 either come in contact, or are connected together by mortar. 
 
 Every such row of stones is called a course of vaulting. 
 
 One course may be joined to another by bringing their adjacent conti- 
 nued surfaces in contact ; but they are generally cemented with mortar, 
 which is called the coursing Joint, and as this cementing substance should 
 be as thin as possible, and of an equal thickness, we shall suppose that 
 the coursing joints intersect the intrados in lines, extending from front to 
 rear of the vault, we shall call these lines the coursing lines of the in- 
 trados. 
 
 In this example, as the vertical projection of the intrados, and of the 
 joints which are in planes parallel to the axis, are identical in all respects 
 to the lines of the right section, the dimensions between every two cor- 
 responding points being equal in both, we may therefore substitute at 
 once the right section for the vertical projection, placing the right sec- 
 tion upon the ground-line UV. 
 
 Plate XI. Let No. 1 be the right section placed in the situation of 
 the vertical plane projection upon the ground-line UV, the curve-line 
 COC being the vertical projection of the intrados, AD, BF, CH, the 
 projections of the vertical projection coursing joints, meeting the projec- 
 tion of the intrados in the points A, B, C. Of these radiant lines CH 
 is the projection of the springing. The line BF meets the line FG pa- 
 rallel, and EF perpendicular to the ground-line UV. The extradns 
 
CH. III.} MASONRY AND STONE-CUTTING. 
 
 45 
 
 ZEDY of this Bection is a straight line parallel to the ground-Una. As 
 this right section of this vault is symmetrical, we shaO only describe one 
 half, the other will be understood by the same rules. 
 
 Let rs, No. 2, be the trace of the vertical face of the wall on the hori- 
 zontal plane of projection, making a given angle with the ground-line 
 UVj and let uv and rs, be the traces of the inclined face of the wall ; 
 the inclination of this face being given by a right section of the wall. 
 
 Let FAAn, No. 3, be the right section of the wall, of which All, 
 the base, is equal to the shortest distance between the two traces uv and 
 rs. No. 2, of the faces of the wall. The line IIF of this section, is the 
 section of the vertical face, and AA, that of the inclined face of the wall. 
 
 This section FAAIl, No. 3, is so situated, that the base line AH is 
 perpendicular to the traces uv, rs, of the faces of the wall. No. 2, the 
 point n being in the line rs or sr prolonged, therefore the point A in 
 the line uv, or in vu prolonged, and Fir being perpendicular to AH will 
 be in the same straight line with the horizontal trace rs of the vertical 
 face of the wall. 
 
 In order to obtain the projection of the intersection of the intrados 
 and of the joints which are in planes parallel to the axis of the intrados 
 witii the inclined face of the wall ; we must find the projection of every 
 line in this inclined face made by the intersection of a horizontal plane 
 passing through every point in the right section which is formed by 
 every two lines in its construction. 
 
 For this purpose it will be necessary to find the horizontal projection 
 of eTery point of the lines where the intersections of the planes parallel 
 to tihe axis meet the inclined face of the wall. To proceed : — 
 
 Take all the heights of the points of the right section, and apply them 
 respectively from the point FT in the line ITF No. 3 ; through these 
 points draw lines parallel to AIT, so that each line may meet the sloping 
 line A A. From each of the points in the line A A draw lines parallel to 
 the horizontal trace uv, No. 2, and lines being drawn from the corre- 
 sponding points of the right section will give the points required by the 
 intersection of the two systems of parallel lines- 
 
 Thus to find the horizontal projection of the intersection of any parti- 
 culair line which is parallel to the axis with the inclined face of the 
 wall, this line being given by its intersecting point in the right section. 
 No. 1 ; this point being the intersection of one of the coursing lines, 
 viz. the first A from the middle of the section. No. 1. 
 
 Draw Aa perpendicular to the ground-line, and transfer the height 
 KA of the point A, No. 1, upon the line IIF, No. 3, from II to ]. 
 Draw 1-2 parallel to IIA, AA meeting PQ in 2. From 2 draw 2a parallel 
 
46 A PRACTICAL TREATISE ON [CH. III. 
 
 to either of the horizontal traces uv, or rs. No. 2, and the point a (No. 
 2) is the horizontal projection of the extremity of the coursing line of 
 the intrados which passes through the point A of the right section. 
 
 In the same manner may be found the projections 6 and c of the in- 
 tersections of the coursing joints of the intrados, with the face of the 
 archant, and also those of the intersections of the planes parallel to the 
 axis; the projections of these points being exhibited by Italic letters 
 corresponding to these of the Roman in the right section. 
 
 To find the developement of the intrados or soffit of the 
 arch. 
 
 Parallel to the ground-line in No. 2, draw the regulating line in the 
 horizontal plane of projection^ intersecting the projections aa', bb' ,cc', &c. 
 of the coursing joint-lines in the points a, fi, y, &c. 
 
 In any convenient situation, No. 4^ draw the line VW, and in VW 
 take any convenient point o. In oV make oa equal to OA, No. \, 
 the half-chord of the arc of the section of the key-course ; and in No. 4, 
 make a^, fiy, &c. equal to the succeeding chords AB, BC, &c. No. I, 
 of the sections of the courses in intrados. 
 
 Through the points a, (i,y, No. 4, draw the lines aa', bb', cc', perpen- 
 dicular to YW, and make aa, (ib, yc, respectively equal to aa, fib, yc. 
 No. 2, as also aa', fib', yc, No. 4, equal to aa', fib', yc', No. 2. In No. 4, 
 join ab, be, on the one side, and db', b'c, on the other ; then aa'b'b, 
 bb'c'c, will be the chord-planes of the soffits of the courses of the stones 
 on each side of the key-course. The figures of the chord-planes of the 
 right-hand side of the arch being found in the same manner, will give 
 the entire developement of the intrados by joining the corresponding ends 
 of the chord plane of the key- course. 
 
 Through any convenient point V, No. 4, in the line VW, draw ac' per- 
 pendicular to VW, and prolong WV to D. Make VD equal to AD, 
 No. 1, and through D, No. 4, draw dd' parallel to ac. In ac, No. 4, 
 make Va, Va', respectively equal to aa, aa', No. 2, and make Dd, Dd', 
 No. 4, respectively equal to \d, \d' , No. 2. Join ad, ad', then will aa'd'd. 
 No. 4, be the side or figure of the coursing joint corresponding with the 
 line AD, No. 1. In the same manner the remaining figures bb'f'f, cc'h'h, 
 Avill be found, as also the remaining figures of the coursing-joints on the 
 right-hand side. 
 
 Then the figures of the moulds for the course of stones, of which 
 the right section is a figure equal and similar to ABFED, No. 1, are 
 No. 1, and adb'b, aa'd'd, bb'f'f. No. 4. All the stones are wrought to 
 the form of right prisms before the heads in the front and rear of the 
 
CH. III.] MASONRY AND STONE-CUTTING. 
 
 47 
 
 arch are formed, then the moulds of the upper and lower beds are ap- 
 pliedj and their figures are drawn upon the surfaces of the coursing- 
 jointSj so as to give the intersections of the coursing-joints with the face 
 of the arch. 
 
 In the course of stones, on the left hand next to the key-course adb'b. 
 No. 4j is the chord-figure of the intrados, aa'd'd. No. 4, the upper bed, 
 and bb'f 'f the lower bed. ; 
 
 To find any point in the oblique face of the arch. Let 
 the point to be found be the point corresponding to the 
 point A. 
 
 The place of the point A in the oblique line AA, No. 3, is at the point 
 2, and its place upon the projection No. 2, is at a. Draw AY perpendi- 
 cular Av, or to uv, and in AY make A2, equal to A2 in A A. From the 
 point 2, in AY, draw 2p parallel to uv, and draw ap perpendicular to uv, 
 and the point p will in the curve of the oblique face of the arch. 
 
 In the same manner will be found the points i, q, See in the curve of 
 the oblique face of the arch, as also all other points, by first finding 
 their projections as at No. 2, and the heights of these points upon the ob- 
 lique line AA, No. 3, and then transferring the points thus found upon 
 the perpendicular AY. Through the points found in the perpendicular 
 AY, draw lines parallel to uv, to intersect with lines drawn perpendicu- 
 lar to uv from the projections of the points to be found in No. 2, and the 
 points of intersection of every two lines, will be the points in the oblique 
 face of the arch, corresponding to those in the section, No. 1. 
 
 The curve thus found in the oblique face of the arch will be an oblique 
 curve ; therefore the line uv will not be an axis, but a diameter. 
 
 To find the direction of any joint in the oblique face of 
 the arch, the plane of the joint being perpendicular to the 
 springing plane of the arch. 
 
 Suppose, for instance, the plane passing through LT in the elevation 
 No. 1, perpendicular to UV. Find the projection t and I in the hori- 
 zontal plane of projection of the points represented by T and L in the 
 vertical plane of projection, and find the point i in the curve of the oblique 
 face of the arch, corresponding to the point T in the vertical plane of 
 projection ; then joining the points I and i, the straight line li will be 
 the position of the joint in a plane perpendicular to the springing plane 
 of the intrados of the arch. 
 
 Plate XII. The diagram in this plate exhibits the construction of an 
 arch of the same species as that in the preceding plate ; but here the 
 
4>S 
 
 A PRACTICAL TREATISE ON 
 
 [CH. III. 
 
 figure of the curve in the oblique face of the arch, is a given symmetrical 
 figure, and therefore the right section of the arch is an oblique curve, 
 which is exactly the reverse of that in the immediately preceding plate. 
 
 PROBLEM III. 
 
 To construct an oblique arch for a canal with a cylin- 
 dric intrados, so that the sides of the coursing joints may be 
 in planes which intersect each other in straight lines per- 
 pendicular to the two faces of the arch, and parallel to the 
 horizon, and that the planes of the coursing joints may 
 make equal angles with each other : — 
 
 Plate XIII. Let ABCD be the plane of the arch ; AD and BC being 
 the plans of the faces, and AB, DC, the plans of the springing lines of 
 the intrados of the arch parallel to the line of direction of the canal. 
 
 Find the middle point e of the parallelogram ABCD, and draw ef per- 
 pendicular to AD or BC. Through any convenient point f in ef draw 
 GH perpendicular to ef, and from the point / with a radius equal to 
 half of AD or BC, describe the semi-circumference ikl meeting GH in 
 i and /. Divide the circumference ikl from i into as many equal parts as 
 the coursing joints are intended to be in number : for example, let it be 
 divided into nine equal parts, il, 12, 23, &c. Draw the tangent QR pa- 
 rallel to GH, and from /, and through the points 1, 2, 3, &c. of division, 
 draw the straight lines, fm,fn,fo, fp, &c. meeting QR in the points m, 
 n, 0, p. 
 
 Through e draw st parallel to AB or DC, and draw ms, nu, ow, py, 
 perpendicular to GH, meeting si in the points s, u, w, y. Make ez, ex, ev, 
 et, equal respectively to ey, ew, eu, et. Prolong CD to meet ef in y, and 
 prolong/e and AB to meet each other in the point /3 ; then with the two 
 diameters st and /3y describe the ellipse sfity, and with the two diameters 
 uv and fty describe the ellipse vfivy, and so on ; then the portions of these 
 curves comprised between the lines AD and BC, will be the plans of the 
 coursing joints. 
 
 The method which has now been shown for finding the joint lines of 
 the intrados of the arch is quite satisfactory as to the principle, since it 
 exhibits the plans of the complete sections of the cylinder by the cutting 
 planes of the joints to the several angles of inclination. We shall show 
 how the joint lines of the intrados themselves may be found, as depend- 
 ing upon the plans of the joints. 
 
 To find the plane curves for the joints of the intrados : 
 
 Having found the conjugate diameter fty, and the semi-conjugate cs, 
 
CH. III.] MASONRY AND STONE-CTUTTING. 49 
 
 as also the semi-conjugate diameters eu, erv, eyi, plate XIV, as has been 
 shown in the immediately preceding plate, procteed in the following man- 
 ner. Draw at, uv, wx, yz, perpendicular to es, and make st, uv, wx, yz, 
 each equal to the radius of the semi-circle ikl. Join et, evj'ex, ez. Draw 
 ss, nu', WW, yy , perpendicular to /3y or /3/ ; and from the point e as a 
 centre, with the radii et, ev, ex, ez, describe the arcs is', vu', xw' , zy'. 
 Join es, eu', ew, ey . 
 
 With the diameters es , eit, etv, ey', and with their common conjugate 
 /3y, describe the semi-ellipsis ^s'y, fiu'y, (3w'y', fiy'y, &c. then the por- 
 tions of these curves contained between the lines BC and AD will be 
 the curve lines of the joints required. 
 
 In order to describe the curve lines of the joints of the intrados, the 
 conjugate diameters of the plans must be founcd. The operation by the 
 following method is very convenient in finding the plans of the joint- 
 lines of the intrados, but it does not afford the nneans of finding the joint- 
 lines themselves, and therefore is but of little mse in the construction of 
 the moulds. 
 
 Let ABCD;,^o-. 2, plate XIII, be the plan, wihich is a parallelogram as 
 before. Divide AB into any number of equal psarts, as, for example, into 
 four, at the points 1, 2, 3, and draw the lines la, 2/3, 3y, parallel to BC 
 or AD, meeting DC in the points a, 13, y, and let % be the ground-line 
 of the elevation ; then AD, \a, 2/3, 3y, BC, a.re the plans of semi-cir- 
 cular sections of the intrados, and are each parallel to the ground-line 
 hg, the elevations of these plans will be semi-cir-cles. 
 
 These elevations being described, let efg be the elevation to the plan 
 BC, Mm the elevation to the plan 2/3 in the miiddle, between the plans 
 BC and AD of the semi-circular sections of the cylinder. Let c be the 
 centre of the semi-circular arc Mm, and divide ithe semi-circular arc Mm 
 into as many equal parts as there are intended to be courses in the arch ; 
 for example, let the number of courses be nine, and therefore the semi- 
 circular arc Mm must be divided into nine eq[ual parts, in the points 
 1,2, 3, &c. - /'Z'J 
 
 From the centre Ij-St 3, &c. and through the points of division r , draw 
 lines which will be the elevations of the joints, ai,nd let pt be one of these 
 lines, intersecting the five semi-circles in the points p, q, r, s, t. Draw 
 the lines pu, qv, rw, sx, ty, perpendicular to the g;round-line hg, intersect- 
 ing the plans AD, 1«, 2/3, 37, BC, in the poin ts u, v, w, x, y, and the 
 line uvrvxy being drawn, will be the correct plan of the joint required. 
 
 In the same manjier the plans of the remaining joints may be found. 
 
 E 
 
50 
 
 A PRACTICAL TREATISE ON [CH. III. 
 
 In the Supplement to the Encyclopedia Britannica, arti- 
 cle Stone Masonry, (page 569, Plate CXVIII. fig. 28. T. 
 Tredgold delin.) a method in order to effect the finding 
 of the plan here shown, is there intended to be accomplish- 
 ed ; whence the following description and diagram are verb- 
 ally copied. 
 
 " When a road crosses a canal in an oblique direction, the bridge is 
 often made oblique. When the angle does not vary more than 10 or 12 
 degrees from a right angle, the arch-stones may be formed as already de- 
 scribed; but in cases of greater obliquity, a different principle of con- 
 struction is necessary. These cases should, however, be avoided when- 
 ever it is possible ; as however solid the construction of an oblique arch 
 may be in reality, it has neither the apparent solidity nor fitness which 
 ought to characterise a useful and pleasing object. 
 
 " An oblique arch may be constructed on the principle of its being a 
 right arch of a larger span, as is shown in^^. 2, plate VI ; or inj/%. 1, 
 plate XV, of this work. 
 
 " Let ABCD be the plan, and EFGH the corresponding points in the 
 elevation : in this elevation the dotted lines show the parts which would 
 not be seen. 
 
 " The joints of the arch are supposed to be divided upon the middle 
 section, and therefore drawn to the mean centre K, which corresponds to 
 the point I on the plan. 
 
 " Divide AD into any number of equal parts, as at 1, 2, 3, &c. and 
 transferring these points to the elevation, describe the arch belonging to 
 each point, and also draw the parallel lines 11, 22, &c. on the plan. 
 
 " To find the mould for the arris of any joint, as a, draw ab parallel 
 to the base line EF : and from a, as a centre, transfer the distances of 
 the points, where the arches cut the joint, to the line ab. Then let fall 
 perpendiculars from the points in the line ab, to the lines 11, 22, &c. in 
 the plan, whence we find a, m, n, o, p, in the curve of the mould for the 
 arris of the joint a. The mould for any other joint may be found in the 
 same manner. The ends of the arch-stones will be square to the j9ints ; 
 and pcde wiU be the mould for one end, and actZ/the mould for the other 
 end. It will be of some advantage in working the arch-stones to ob^rve, 
 that the arch-stone being in its place, the soffit should be every where 
 perfectly straight, in a direction parallel to the horizon." 
 
 Whoever is the author of this article on stone-masonry 
 
CH, III.] MASONRY AND STONE-CUTTING. 
 
 51 
 
 just now quoted from the work alluded to, he is chargeable 
 with a description and representation which will lead the 
 reader to an erroneous construction of finding the plans, or 
 the arrises of the joints as he calls them. We will defy the 
 writer, or even Mr. Tredgold who affixes his name as 
 draughtsman to the plate, to demonstrate the truth of the 
 method there described. 
 
 After all, this method only divides the curve of the mid- 
 dle section, which is parallel to the front and rear faces, into 
 equal parts, and the plane passing through the semi-circular 
 arc into straight lines perpendicular to the curve ; but in 
 order to have the surfaces which form the sides of the 
 joints in the front and rear faces perpendicular to the curve, 
 and at the same time perpendicular to each face, proceed 
 according to the following method, 2, plate XV. 
 
 Let lad be the plan of one pier, and ycfthe plan of the other pier, ad 
 and c/ being the plans or horizontal sections of tHe springing lines of the 
 intrados ; also, let LF be the ground-line parallel to the planes of the 
 front and rear elevations. Describe the live semi-circles in the elevation 
 as before, ABC being that in the front, DEP that in the rear, and 
 GHI that belonging to the middle section. 
 
 Divide the semi-circular arc GHI into the number of equal parts re- 
 quired, and let the points of division be 1, 2, 3, &c. Through the points 
 1, 2, 3, &c. draw the straight lines \o, 2*, 3U, &c. radiating to the cen- 
 tre of the semi-circular arc ABC intersecting the curve ABC in the 
 points N, R, T, and the lines NO, RS, TU, will be the joint lines of the 
 face, and will be perpendicular to the curve line ABC. 
 
 In the straight line ac, which is the plan of the face of the arc, take 
 a part zn for the joint in the direction NO of the elevation, and let the 
 hues IN, 2R, 3T, intersect the semi-circular arc between the parallel 
 sections ABC and DEF in the points a, /3, y, &c. Let the points u and 
 V be in the straight line ac. Make nu and uv respectively equal to 
 Na, al, and draw uw and vx perpendicular to zv. 
 
 Divide ad into as many equal parts as the thickness of the arch is 
 divided into equal parts by the planes of the semi-circular arcs which are 
 parallel to the planes of the front and rear faces ; that is, divide ad into 
 four equal parts, and let ak, ag, be two of those parts in succession, and 
 
 E 2 
 
52 
 
 A PRACTICAL TREATISE ON 
 
 [CH. III. 
 
 draw ktv and gx parallel to oc ; then, n, w, x, will be three points in the 
 curve, which is the intersection of the plane of the curving joint and the 
 cylindric surface, forming the intrados ; and thus we might find as many 
 points as we please, by increasing the number of equi-distant sections. 
 This gives the first joint next in succession to the springing AD. 
 
 In the same manner all the other coursing joints will be found as at 
 No. 2, No. 3, No. 4, &c. 
 
 Observations on the preceding methods : — 
 
 The most simple construction of an oblique arch with a cylindrical in- 
 trados, is that where the sides of the coursing joints are in planes inter- 
 secting the intrados perpendicularly in straight lines, as in the first 
 example ; but when the arch is very oblique, the coursing joints inter- 
 sect the planes of the two vertical faces in very oblique angles. 
 
 It has been shown that when the sides of the coursing joints are 
 in planes perpendicular to the front and rear faces, these planes cut the 
 intrados very obliquely, except at the middle section, or in the best 
 method in the curve of the front and rear. It therefore appears, that 
 in an oblique arch, in order that the surfaces of the coursing joints may 
 intersect both the intracjps, and the face of the arch perpendicularly, the 
 sides of the coursing joints cannot be in planes. 
 
 In order that every arch may be the strongest possible, a straight line 
 passing through any point of the surface of a joint perpendicularly to 
 the intrados, ought to have all its intermediate points between the point 
 through which it passes ; and the intrados, in the surface of the side of 
 the coursing joint ; and in order that the stones may be reduced to their 
 form in the easiest manner possible, the surfaces should be uniform ; and 
 the forms of the stones should be similar solids, and the solids similarly 
 situated. 
 
 To obtain these desirable objects will not be possible where the faces 
 of the arch are plane surfaces ; however, even in this case, the joints 
 may be so formed by uniform helical surfaces, that they will intersect 
 the intrados perpendicularly in every point, and the faces of the arch 
 perpendicularly in two points of the curve. 
 
 This mode of executing a bridge renders the construction much 
 stronger than when the joints of it are parallel to the horizon. Since in 
 this last case, the angles of the beds and the faces are so acute upon 
 one side, that the points of the ring-stones are very liable to be broken, 
 or even to be fractured in large masses. 
 
 For, though the gravitating force acts perpendicularly to the horizon ; 
 yet, notwithstanding, when one body presses upon the surface of another. 
 
 I 
 
CH. III.] MASONRY AND STONE-CUTTING. 
 
 53 
 
 the faces act upon eaclh other in straight lines perpendicularly to their 
 surfaces. Hence a riglut-angled solid will resist equally upon all points 
 of its surface. 
 
 From this consideration, we are induced to give a preference to the 
 construction with spirad joints, though attended with greater difficulty 
 in the execution. 
 
 PROBLEM IV. 
 
 To execute a briidge upon an oblique plan, with spiral 
 joints rising nearly perpendicular to the plane of the sides. 
 
 Fig. \, plate XVI, is the plan of a bevel bridge ; Jig. 2, the elevation of 
 the same, as the two faces of the obtuse angle are shown ; the joints of the 
 intrados descend from thie face of the arch in such a manner, that supposing 
 the lines ab, a'b', a"b",^^g. 1, to be the joints of the intrados, meeting the 
 curve of the intersectiom of the face of the arch and intrados in the points 
 b, b', b", &c. then the jo)ints ba, b'a', b"a", &c. are as nearly perpendicular 
 to the curve bb'b"b"' as ptossible for the construction to admit of, supposing 
 the joints to be all paralilel to each other. By making the joints of the 
 intrados all parallel to (each other, all the intermediate arch-stones will 
 have the same section when cut by a plane at right angles to the arris- 
 line of the bed and intr-ados of the arch ; therefore, if the intermediate 
 arch-stones are equal in length, the upper and lower beds must be the 
 same winding surfaces, and consequently must all coincide with each 
 other, and all the end-jorints must be equal and similar surfaces, and thus 
 all the arch-stones may be equal and similar bodies. 
 
 The most considerable; obliquity of the joints in the intrados is at those 
 two parts of the curve where it meets the horizon. The obliquity of 
 the intradosal joints, at the crown of the arch, is considerably less than 
 at the horizon ; but in the middle of that portion of the curve, between 
 the crown and the hori2zon on each side, the intradosal joints are exactly 
 perpendicular to the horiizon. 
 
 Had it not been for tluese deviations, the execution of this arch would 
 have been extremely eaisy, and very few constructive lines would have 
 been necessary. 
 
 This arch, however, nnight be executed so that all the intradosal joints 
 would be perpendicular tto the curve-line of the face and intrados ; but 
 this position would hawe caused such a diversity in the form of the 
 stones as to increase the Uabour in a very great degree, and, consequently, 
 to render the execution very expensive ; and not only so, but as the joints 
 would have been out of ;a parallel, their effect would have been very un- 
 
54s 
 
 A PRACTICAL TREATISE ON [CH. III. 
 
 sightly. A succession of equal figures, similarly formed, has a most im- 
 posing eifect on the eye of the spectator. The laws of perspective pro- 
 duce on the imagination a most fascinating variety, the figure only 
 varying by imperceptible degrees, which yet in the remote parts produces 
 a great change. 
 
 There is still another method in which the greater part of the diffi- 
 culty may be removed without impairing the strength of the arch ; this 
 manner is to form the ring-stones so that the joints in the intrados may 
 be perpendicular to the curve forming its edge ; the intermediate por- 
 tion of the intrados to be filled in with arch-stones, which have their 
 soflSt-joints parallel to the horizon. This disposition of the joints might 
 not be so pleasant to the eye, but, if well executed, it could not be dis- 
 agreeable. 
 
 If the ends were made to form spirals, as in Jig. 3, and a wall erected 
 above the arch, as this wall could only be made to coincide in three 
 points at most with the face of the arch, no regular form of work could 
 be introduced so as to connect the wall to the ring-stones. 
 
 To form the developement of the intrados of the oblique 
 arch, with spiral or winding joints, and thence to find the 
 plan of the developement or intrados. 
 
 Let AC, plate XVII, be the inner diameter of the face of the ring- 
 stones; upon AC describe the semi-circular arc ABC, and find its deve- 
 lopement upon the straight line AD. Draw the straight lines AG and 
 DI perpendicular to AD. 
 
 In AG take any point M, and draw ML, making the angle AML equal 
 to the angle of the bevel of the bridge, meeting CH in the point L. Draw 
 La perpendicular to AG', meeting AG in a. Prolong La to meet DI in 
 Q, and draw ON parallel to LM, so that the distance between LM and 
 ON may comprise the breadth of the bridge. Let ON meet CH in O, 
 and AG in N ; then will LMNO be the plan of the bridge. Find the 
 developement MPQSRN upon the straight line AG', the curve MPQ 
 being the developement of the arc insisting on ML, and NRS the deve- 
 lopement of the curve line upon NO. 
 
 Draw MQ, and divide MQ into as many equal parts as there are in- 
 tended to be arch-stones, which we shaU here suppose to be fifteen ; 
 hence there will be a ring-stone in the middle, and the number of ring- 
 stones will be equal on each side of the middle one ; let P be the middle 
 point of the line MQ, and let a, b, c, &c. be the points of division on one 
 side of P, and d,b',c, &c. the points of division on the other side. 
 
CH. III.] MASONRY AND STONE-CUTTING. 
 
 55 
 
 Through the middle point P draw the straight line WX. Through 
 the points a, b, c, &c. draw the lines do, ep, fq, &c. parallel to WX, 
 meeting the curve MQ in the points k. I, m, &c. and the curve NS in 
 the points o, p, q, &c. ; also through the points d, b', c, &c. draw the 
 lines d'd, e'p, fq, &c. parallel to WX, meeting the curve MQ in the 
 points U, I, Tti, &c., and the curve NS in the points o, p, q, &c. ; then 
 ao, Ip, mq, &c. ; also do', I'p, ffi'g", &c. will be the joint lines on the in- 
 trados of the arch ; the heading joints are marked on the developement 
 at right angles to these joints. The curves on the plan are projected by 
 means of Problem 1, Ch. I. Sec. VI. 
 
 Thus, d g r'lB the seat of the developement do; e h s the seat of the 
 developement e p, &c. 
 
 Now as all the intermediate arch-stones are equal and similar, it 
 will only be necessary to show how one of the stones may be formed. 
 For this purpose, let uvwx be the developement of the soffit. Draw 
 vy parallel to MN or QS. Run a straight draught vt/ diagonally upon 
 the intrados of the stone, making an angle uyv with the edge uy, or ux, 
 of the soffit. Draw ua and w/3 perpendicular to vy. 
 
 Make two moulds Z,Z to the arc ABC, so that their chords may be 
 equal ; then cut two draughts ua and fim so as to coincide with the 
 convex edges of the two moulds Z,Z, while the straight edges of the two 
 moulds Z,Z are out of winding. 
 
 That is, apply the moulds Z,Z at the same time; the one upon the 
 line ua, and the other upon the line and sink a cavity or 
 
 draught under each line ; so that, after one or more trials, the convex 
 edge may coincide with the bottom of each draught ; and that the point 
 marked upon each circular edge may coincide with the bottom of the 
 draught vy ; and that the two chord-lines of each circular mould may 
 be in the same plane, that is in workman's terms out of winding or out 
 of twist. 
 
 The remaining superfluous part may be worked off as directed by two 
 straight edges, and thus the cylindric surface of the soffit of the stone 
 will be formed. 
 
 The longitudinal spiral joints may be formed by means of the bevel at 
 y, where it is applied to the section of one of the arch-stones : but before 
 the heading joints and beds are wrought, a pliable or flexible mould 
 uvtvx must be made, and bent to the convexity of the surface, so that 
 the line vy may coincide with the bottom of the straight draught first 
 wrought. 
 
 In applying the mould y, the curved edge must be laid along the line 
 ua or firv ; and in directions parallel to these lines ; and several draughts 
 
56 
 
 A PRACTICAL TREATISE ON 
 
 [CH. III. 
 
 must be wrought on the spiral bed^ so as to coincide with the straight 
 edge and the angular with the line vw, or ux. 
 
 Having shown the developement of the intrados and its 
 projection, it will be proper to show how the cui-ves are 
 projected, and more particularly as it will not only show 
 the application of Problem 1, Ch. J, Section VI, but also 
 the positions of the sections of the cylinder, in order to find 
 the proper moulds. 
 
 Let the line AF, Plate XVIII, the edge of the triangle AEF, be the 
 developement of one of the longitudinal joints, and let HG at right an- 
 gles AF be the developement of one of the lines of direction of the head- 
 ing joints ; then, as the projection of all the longitudinal lines is equal 
 and similar, and the projection of the heading joints is equal and similar, 
 one curve of each being obtained, and a mould formed thereto, each series 
 of curves may be drawn by means of its proper mould. 
 
 Divide the arc ABC into any number of equal parts at the points 
 1, 2, 3, &c., and the straight line AF into the same number of equal 
 parts at the points 1, 2, 3, &c. ; but it will be most convenient to divide 
 each into as many equal parts as the ring-stones are in number, which in 
 this example are fifteen. From the points 1, 2- 3, <fec. of division in the 
 straight line AF, draw la, 2b, 3c, &c. perpendicular to AE, and through 
 the points 1, 2, 3, &c. in the arc CB, draw lines la, 2b, 3c, &c., parallel 
 to CD, and through the points a, b, c, &c. draw a curve, which is the pro- 
 jection of a cylindric spiral, and is the plan of one of the longitudinal joints 
 required. In the same manner, dividing HG into the same number of 
 equal parts as the arc ABC, and drawing lines as before from the divisions 
 of the arc, and from the divisions of the straight line HG, to intersect 
 each other respectively in the points a, b, c, &c. we shall have the curve 
 of direction of the heading joints. In order to find the direction of the 
 curve in the middle, it will be necessary to show the manner of finding a 
 tangent in the middle of the curve. For this purpose. 
 
 Make the angle EAA: equal to EAF, and let the point m be the mid- 
 dle of the curve Dot A. Through the point m draw pq parallel to kA, 
 and pq will be the tangent required. 
 
 In like manner, make the angle AH/ equal to AHG, and let g be the 
 middle point of the curve H^C; through g draw rs parallel to H/, and 
 rs will be a tangent to the curve HgC. 
 
 It is here evident from the tangents, that if these two curves had in- 
 
CH. III.] MASONRY AND STONE-CUTTING. 
 
 57 
 
 tersected each other in the middle, they would have been at right angles 
 to each other ; they are, however^ still the projections of two straight 
 lines bent upon the cylindric surface. 
 
 To draw a tangent to the point n. Draw w4 parallel to EA^ meeting 
 the curve AB in 4. Draw 4u perpendicular to the radical line^ and make 
 4ti equal to the developement of the arc 4A. Draw ut perpendicular to 
 AG, and join tn, which is the tangent required. 
 
 To find the curvature of a stone along the two edges of 
 the longitudinal joints, and along the heading joints of the 
 intrados. In ^g. 1, plate XIX, which is a developement of 
 the intrados, abed is the developement of the intrados of an 
 arch-stone, it is required to find the carvatur'e along be, and 
 ad, also in the direction ab, de at the ends. 
 
 In Jig. 2, make OA equal to the radius of the cylinder, and through 
 A draw BE perpendicular to AO. Make the angle BOA equal to the 
 complement of the angle which the joints in the developement of the 
 intrados make with the springing lines, that is equal to the angle DAE, 
 Jig, I. Make OC,^g. 2, equal to OB, and draw OD perpendicular to 
 BC. Make OD equal to OA. Then with the transverse axis BC, and 
 semi- conjugate OD, describe tlie semi-elliptic arc or curve CDB ; then 
 the portion of the elliptic arc on each side of the point D will be the 
 curvature in Jig, 1, along the longitudinal edge be or da of the soffit of a 
 stone. 
 
 Again, produce DO to E, and make OE equal to OD. In OB, take 
 OG, equal to OA, the radius of the circular end of the cylinder; then 
 with the transverse axis DE, and the semi-conjugate OG, describe the 
 semi-elliptic arc DGE, and the small portion of this arc on each side 
 of the point G has the same curvature as ab or dc, Jig. 1. There- 
 fore, the stone being wrought hollow, as directed in the description 
 of the preceding plate, then the mould shown at D is that for working 
 the longitudinal joints, or those which terminate on the soffit in the 
 lines ad and be. In like manner, the mould G is that for working the 
 heading joints which terminate upon the soffit in the lines ab, dc, &c. 
 It will hardly be necessary to remind the reader, that the convex edge 
 of the squares at D and G is to be applied upon the hollow soffit already 
 wrought. The curvature of these moulds may be shown by calculation 
 thus : let R be the radius of curvature, a ~ the semi-transverse axis, and 
 
 b ~ the semi-conjugate ; then b : a'.'.a : R — — , 
 
58 A PEACTICAL TREATISE ON [CH. III. 
 
 As for example to this formula, let the radius of the cylindric intrados, 
 or 6 =: 13 feet, and the semi-transverse axis, or a == 28 feet 
 28 
 28 
 
 224 
 56 
 
 13)784(60 feet 4 inches nearly 
 78 
 
 4 
 12 
 
 ~48( 
 
 To find the angle of the joints of the face of the arch, 
 and intrados of the oblique arch with spiral joints. 
 
 Let the semi-circular are ABC, Plate XX, be a section of the intrados at 
 right angles to the axis of the cylinder. Draw CD and AE perpendicular 
 to the diameter AC. Draw AD, making an angle with CD, equal to the 
 inclination which the plane of the face of the arch makes with the ver- 
 tical plane which is parallel to the axis of the cylinder, and which passes 
 through the springing line of the arch. 
 
 Find the edge D/G of the developement and face of the arch, or draw 
 the curve D/G with a mould made from the developement before 
 shown. Draw the face of the ring-stones AKD. Let it now be re- 
 quired to find the fourth joint from the point D. Make D/ equal to the 
 portion D4 of the intrados AKD. Draw // the developement of a part 
 of the longitudinal spiral joint corresponding to the point 4 of the elliptic 
 arc AKD. Draw the line st a tangent to the curve at /. To do this, 
 we shall again repeat the process of which the principle has already been 
 taught, viz. On CD, as a diameter, describe the semi-circle CqD, and 
 draw fq, intersecting CD perpendicularly. Draw qr a tangent to the 
 semi-circular arc at the point q, and make qr equal to the developement 
 of the portion qD of the semi-circular arc. Draw perpendicular to 
 CD, meeting CD, or CD produced in the point t. Through /draw the 
 straight line is, and ts will be a tangent to the curve at the point. By 
 this means we have the angles which the spiral joints in the intrados 
 make at the point 4 with the elliptic curve AKD. 
 
 To find the angle made by the normal and the curve, in 
 
 M 2. 
 
CH. III.] MASONRY AND STONE-CUTTING, 
 
 59 
 
 In Jig. 2 draw the straight line ab, and make a.b equal to the radius of 
 curvature of the elliptic arc AKD at the point 4. This radius would 
 be near enough to make it the half of the half sum of the semi-parame- 
 ters of the two axes. 
 
 But if greater nicety is required^ let the radius of curvature be de- 
 noted by r, the semi-transverse axis OD or OA be denoted by a, and 
 the semi-conjugate, which is the radius of the semi-circular arc ABC, be 
 denoted by h, and let the distance Op be denoted by x ; then will 
 
 _ fa*— (a8— 5 
 ^— \— ^ f which will be exact to the number of 6gTires 
 
 found in the operation here indicated. 
 
 Having thus found the radius of curvature, either mechanically or by 
 calculation, make ah, Jig. 2, equal to that radias. From the point a as a 
 centre, with the distance ab, describe the arc be ; and draw the straight 
 line bd a tangent to the curve. 
 
 To find the angle made by a tangent plane to the cy- 
 lindric surface at the point 4, /ig. 1, and the plane of 
 the face of the arch. 
 
 Draw the straight line 4m a tangent to the elliptic curve AKD at the 
 point 4, and draw 4v parallel to AD. Transfer the angle uiv to abc 
 
 In Jig. 3, at the point/*, in the straight line be, nnake the angle cbd equal 
 to the angle T>0¥,Jig. 1, which the axis makes with the plane of the face 
 of the arch. Again in Jig. 3, draw ef perpendicular to ab, intersecting 
 ab in the point a. Draw cd perpendicular to eb, and ce perpendicular to 
 cf. Make ce equal to cd, and join ea ; then will the angle eafhe the 
 inclination of the curved surface of the cylindric intrados, and the face of 
 the ring-stones. 
 
 We have now ascertained two sides, and the contained angle of a tre- 
 hedral; in order to find the remaining parts, the third side of this trehedral 
 is the angle of the joints of the intrados and face of the arch, by applying 
 the proper curved moulds to the angular point ; it is, however, rather 
 unfavourable to our purpose, that the angle abd^fig. 2, is a right angle, 
 and that the angles IJt and ljs,jig. J, differ but in a very small degree 
 from right angles. As from this circumstance the principle cannot be 
 made evident, we shall therefore suppose, that these angles have at 
 least a certain degree of obliquity. 
 
 In Jigs. 4 and 5, let ABC equal to angle lft,Jig. 1, and ABD,>-^*.4 and 
 6, equal to the angle abd,fig. 2 : thus, in Jigs. 4 and 5, draw De, intersect- 
 ing AB in/, or producing Be to meet AB in/. At the point /in the 
 
'60 
 
 A PRACTICAL TREATISE ON 
 
 [CH. III. 
 
 straight line efin Jig. 5, make the angle efg equal to the angle eac,Jig, S; 
 or in^^-. 4, make the angle efg equal to the supplement of the angle eaf. 
 In Jigs. 4 and 5, draw ek perpendicular to BC, meeting BC in i, or BC 
 produced in i. Draw eg perpendicular to ef, and eh to eK. Make eh 
 equal to eg, and join hi Make iK equal to ih, and join BK ; then will the 
 angle CBK be the angle of the joints of the intrados and face of the arch. 
 
 When each of the given sides is a right angle, then tlie remaining side 
 of the trehedral will be the same as the contained augle ; that is, the 
 angle of the joints of the intrados and face of the arch, will be the same 
 as the angle eaf. Jig. 3. In this case, no lines are necessary in order to 
 discover the angle of the joints. 
 
 In order to apply the angle CBK, one of the lines which applies to 
 the face must be straight, and the curved edge shown by the bevel at D 
 of the preceding plate must be so applied, that the other leg of the bevel 
 may be a tangent to the curve at the angular point B, and this will com- 
 plete what is necessary in the construction of an oblique arch with spiral 
 joints. 
 
 SECTION III. 
 A CIRCULAR ARCH IN A CIRCULAR WALL. 
 
 PROBLEM I. 
 
 To execute a semi-cylindric arch in a cylindric wall, 
 supposing the axes of the two cylinders to intersect each 
 other. Given the two diameters of the wall, and the diame- 
 ter of the cylindric arch, and the number of arch-stones. 
 
 Fig. \, plate XXI. From any point o with the ra(dius of the inner 
 circle of the wall describe the circle ABC, or as much of it as may be ne- 
 cessary ; and from the same point o, with the radius of the exterior face 
 of the wall describe the circle DEP, or as much of it as may be found 
 convenient. 
 
 Apply the chord AB equal to the width of the arcb, and draw DA 
 and EB perpendicular to AB or DE ; then ABED willl be the plan of 
 the cylindric arch. 
 
 Draw op perpendicular to AB, and draw tv perpendicuilar to op. From 
 the point p as a centre, with the radius of the intrados; of the arch (de- 
 scribe the semi-circular arc,^r* ; and from the same poinit p, with the la- 
 dius of the extrados, describe the semi-circular arc iuv. Divide the trc 
 qrs into as manv equal parts as the arch-stonos are intendled to be in nun- 
 
CH. III.] MASONRY AND STONE-CUTTING. 
 
 61 
 
 ber, that is, here into nine equal parts. From the centre p, draw lines 
 through the points of division to meet the curve tuv ; and these lines 
 will be the elevation of the joints ; and the joints, together with the in- 
 tradosal and extradosal arcs, will complete the elevation of the arch. 
 
 Find the developement, Jig. 2, as in fig. 3, plate IX, and the parallel 
 equi-distant lines to the same number as the joints in the elevation, will be 
 the joints of the soffits of the stones ; and the surfaces comprehended by 
 the parallel lines, and the edges of the developements, will be the 
 moulds for shaping the soffits of the stones. 
 
 In Jig. 3. Let AB be equal to the diameter of the external cylinder. 
 Draw AC and BD each perpendicular to AB. Bisect AB in p, from 
 which describe the intradosal and extradosal arcs, and draw the joints as 
 in fig. 1. Produce the joints to meet AC or BD, in the point e, /, g. 
 Sec. ; then it is evident that since every section of a cylinder is an 
 ellipse, the lines pA, pe, pfi, pg, &c. are the semi-transverse axis of the 
 curves, which form the joints in the face of the arch, and that these 
 curves have a common semi-conjugate axis equal to half the diameter of 
 the cylinder. 
 
 Therefore, upon any indefinite straight line pQ,fig. 4, set off the semi- 
 axis pA, pe, pfi, pg, &c. and draw pB perpendicular to pQ. From p, 
 with the radius pA, describe an arc AB. On the semi-axes pe and pB, 
 describe the quadrantal curve of an ellipse ; in the same manner describe 
 the quadrantal curves /B, gB, &c. Make pq equal to pg, fig. 3, and in 
 Jig.A^ draw qt parallel to pB, intersect the curves AB, eB,/B, &c. in the 
 points i, k, I, Sec. ; then him, hkn, hlo, &c. are the bevels to be 
 applied in forming the angles of the joints ; viz. the bevel hiin is that of 
 the impost, the straight side hi being applied upon the soffit or infirados ; 
 and the curved part im horizontally to the curve of the exterior side of 
 the wall : the point k, of the bevil hkn. Jig. 4, applies to the point k, 
 _Jig. 3, so that kh may coincide with the joint upon the intrados, and 
 the curved edge kn,Jig. 4, upon the face hi, fig. 3 ; and so on. 
 
 As to the angles which the beds of the stones make with the intrados, 
 they are all equal, and may be found from the elevation svyx, fig. 1 ; 
 which is the same as a section of one of the arch-stones perpendicular to 
 any one of the joints on the soffits. • 
 
 The faces of the stones must be wrought by a straight edge, by per- 
 pendicular lines. The first thing to be done is to work one of the beds ; 
 secondly, work the intrados — at first as a plane surface at an angle sxtf, or 
 ^^"^tjig- 1 ; then gauge off the bed of the soffit, and work the other bed 
 of the stone by the angle vsx or i/xs ; then apply the proper soffit, 1, 2, 
 or 3, fig. 2 ; and lastly, the two moulds in fig. 4. 
 
62 A PEACTICAL TREATISE ON [CH. III. 
 
 SECTION IV. 
 A CONIC ARCH IN A CYLINDRIC WALL. 
 
 PROBLEM I. 
 
 To execute a semi-conic arch in a cylindric wall, sup- 
 posing the vertex of the cone to meet the axis of the 
 cylinder. Given the interior and exterior diameters of the 
 wall, the length of the axis of the cone, and the diameter 
 of its base. 
 
 Example I. 
 
 From the point o, plate XXII, with the radius of the interior sur- 
 face of the wall describe the arc ABC, and from the same point o, with 
 the radius of the exterior surface, describe the arc DEP, and the area 
 between the arcs ABC and DEF will contain the plan of the wall. 
 
 Draw any line op, and make op equal to the length of the axis of the cone. 
 Through p draw tv perpendicular to op. From p as a centre, with the 
 radius of the base of the cone, describe the semi-circle qrs meeting tv in 
 the points q and s. Divide the arc qrs into as many equal parts as the 
 arch-stones are to be in number, that is, in this example, into nine equal 
 parts. Through the points of division draw the joint lines, which will 
 of course radiate from the centre p. The extradosal line tuv is here de- 
 scribed, as we here suppose the cone to be of an equal thickness, and con- 
 sequently the axis of the exterior cone longer than that of the interior. 
 
 From the points 1, 2, 3, &c. where the lower ends of the joints of the 
 arch-stones meet the intradosal arc, draw lines perpendicular to tv, meet- 
 ing tv in the points i, k, I, m, &c. From these points draw lines to the 
 vertex of the cone at o, meeting the arc DE of plan of the wall under the 
 arch, in the points a, b, c, d, &c. Draw the lines ae, hf, eg, dh, &c., pa- 
 rallel to the chord DE, to meet op in the points e,f, g, k, &c. and let DE 
 meet op in rv. In Jig. 2, draw the straight line AB, in which take the 
 point p near the middle of it, and make pA, pB, each equal to the radius 
 of the exterior surface of the cyHndric wall. Through the points A 
 and B draw fg,fg, perpendicular to AB. 
 
 From the point p aa a centre, with any radius, describe a semi-circular 
 
CH. III.] MASONRY ANT) STONE-CUTTING. 
 
 63 
 
 arc, and divide it into nine equal parts as before. Through the points of 
 division draw the radiating lines to meet fg in the points e, /, g, &c. 
 From^g. 1 transfer the distances Ew, ae, bf, eg, &c. to pq, pr, ps, pt, 
 &c. on each side of the point p. Draw the perpendiculars rk, si, tm, &c. 
 to AB, vi^hich wiU intersect with the radials pe, pf, pg, See. in the points 
 k, I, m, &c. ; through the points k, I, m, &c. on each side draw a curve, 
 and this curve will be the elevation of the intrados of the arch. 
 
 Fig. 3 exhibits another method by which the heights of the points k, I, 
 m, might have been found. This method is as foUows : — Upon a straight 
 line ah, and from the point a make ah', ac, ad, ae, &c. and a/ respectively- 
 equal to 01, ok, ol, Sic.Jig. 1. In Jig. 3, draw the straight lines bgyck, 
 di, ek, fo, perpendicular to ab. Make bg, ch, di, ek, respectively equal to 
 the heights i\, k'i, Z3, m4. Draw the straight lines ag, ah, ai, ak, inter- 
 secting /o in the points I, m, n, o. 
 
 Injig. 2, make rk, si, tm, un, respectively equal to fl,fm,fn,fo,Jig. 2 ; 
 and thus the points k, I, m, &c. are found by a different method;, which is 
 more accurate for ascertaining the points near the top, as the radials and 
 the perpendiculars intersect more and more obliquely as they approach 
 the summit. 
 
 In some line pQ,Jig. 4, make pA, pe, pf, pg, &c. equal to pA, pe, pf, 
 pg, Sic.Jig. 2. Draw perpendicular topQ. From p with the radius 
 pk, describe the arc AB. With the several semi-axes pe, jt)B ; pf, pQ ; 
 pg, pB, &c. describe the quadrantal elliptic curves enB,foB, &c. Draw 
 Bm parallel to AQ. Make the angle Bpt equal to the angle PEop,Jig. 1 ; 
 and let i, k, I, &c. be the points where pt intersects the curves AB, eB, 
 fB, Sec. Then the bevels of the joints are him, hkn, hlo, &c. 
 
 Now, if EBCF, Jig. 1, be the developement of the intrados, with the 
 joints drawn on it, we shall have the soffits of the stones. 
 
 In^g. 5, draw ab and ac at a right angle Avith each other. Make ab 
 equal to the radius of the base of the cone, and ac equal to the length 
 of its axis. Join be. From a, with the radius ab, describe an arc, 
 dbe. Make bd, equal to the chord of the intrados of one of the arch- 
 stones. Produce be to any point f, and dxawfg perpendicular to ab, 
 meeting ab in g. Draw gi perpendicular to be, and gh parallel to be. 
 Make gh equal to gf, and join hi ; then hig is the angle which the 
 soffits of the stones, when wrought as planes, make with the beds. 
 
 Example II. 
 
 To construct an arch in a cylindric wall, of which arch 
 the intrados is a uniform conic surface, so that the axes of 
 
64. 
 
 A PRACTICAL TREATISE ON 
 
 [CH. III. 
 
 conic and cylindric surfaces may meet or intersect each 
 other. 
 
 In Jig. I, which is the plan and elevation of the arch, the elevation be- 
 ing above, and the plan below as usual, let AD be considered as the 
 ground-line, and ABD the elevation of the base of the cone, which base 
 is supposed to be a tangent plane to the surface of the wall ; let bd, pa- 
 rallel to the ground-line AD, be the half plan of the base of the cone ; 
 a'b'c'kgf the plan of the cylindric face of the wall ; and d'rmlk the plan of 
 the intersections of the conic and the intermediate cylindric surfaces 
 which terminate the interior of the aperture of the arch. 
 
 First, to find the elevation of the intersections of the cylindric face of 
 the wall and the conic surface of the intrados. Having divided the semi- 
 circular arc DBA, into the equal parts Dl', 2'3', &c. at the points 
 1', 2', 3', &c., draw the connecting lines Hd, I'l, 2'2, 3'3, &c. meeting 
 hd in the points in d,\,2, 3, &c. Draw be perpendicular to hd, and make 
 he equal to the axis of the conic surface. 
 
 Draw the straight lines de, Ic, 2c, &c. meeting the plan of the face of 
 the wall in the points/, g, h, and draw the connecting lines /F,gG, /«H, 
 &c. intersecting the lines FC, GC, HC, &c. in the points F, G, H, &c. 
 A sufficient number of points being found in the same manner, through 
 these points draw the curve EBF, and the curve EBF will be the eleva- 
 tion of the line of intersection of the conic and cylindrical surfaces re- 
 quired. 
 
 To find the elevation of the intersection of the conic surface with the 
 intermediate concentric cylindric surface. Let the arc d'rh be the plan of 
 this concentric cylindric surface, having the same centre as the arc d b' f, 
 which is the plan of the cylindric surface of the wall ; and let the straight 
 lines dc, Ic, 2c, 8zc. meet the arc drk in the points k, I, vi, &c. ; then, if 
 connectants be drawn from the points k, I, m, to the elevation to meet the 
 radial lines, we shall thus obtain the elevations K, L, M, of the corre- 
 sponding points. Let us now suppose that a suflicient number of points 
 are thus found, and the curve UK drawn through these points ; then 
 UK will be the elevation of the intersection of the conic and cylindric 
 surfaces required. 
 
 Let us now construct a mould for one of the joints, suppose for the 
 second joint UX, in the elevation. Draw the connectants Um, Vu, Ww, 
 X:r, meeting the line db prolonged in the points u,v,w, x', and prolong 
 the connectants Vu, Vd, Ww, &c. to meet the plan of the exterior cylindric 
 surface of the wall, in the points d, b', c ; and the connectant Xx to meet 
 
It' - 
 
 CH. III.] MASONRY AND STONE-CUTTING. 
 
 65 
 
 the plan of the intermediate cylindric surface in the point d' , and the 
 plan est of the inner cylindric surface on the point e. 
 
 Suppose No, 1, No. 2, No. 3, No. 4, to be the figures of the moulds 
 of the first, second, third, and fourth joints from the springing-line ; and 
 as it is proposed to find the figure of the joint. No. 2, draw the straight 
 line ux, No. 2, and in ux take vn, wx, respectively equal to UV, VW, 
 WX, in the elevation,/^. 1. Draw in No. 2, ua, vb, rvc, xe, perpendi- 
 cular to ux, and make ua, vb, wc, xd, xe, respectively equal to ua, vb', wc, 
 xd', xe, on the plan,/^. 1. Through the points a, b, c, No. 2, draw a por- 
 tion of an ellipse, and we shall have the edge of the joint that meets the 
 surface of the wall. Draw the straight line cd, No. 2, and this straight 
 line will be the intersection of the joint and the conic surface ; the por- 
 tion de. No. 2, will be the section of the inner cylindric surface. 
 
 The remaining lines of the figure of the mould will be found in the 
 same manner, and thus we shall have the complete figure, No. 2, of the 
 mould. 
 
 Fig. 2, exhibits the developement of the soffit of the horizontal cylin- 
 dritic surface next to the aperture, upon the supposition that the face 
 of the ring-stones are first wrought in horizontal lines from the curve 
 EBP, to meet the inner horizontal cylindritic surface, and afterwards 
 reduced to the conic form. The breadth of the stones in this develope- 
 ment are not equal, but increase from each extreme to the middle. 
 
 The mould for the springing-stone is the same as the plan of the 
 jamb. 
 
 It Avill be necessary to work the arch-stones into prisms, of which the 
 ends are the sections of the stones in the right section of the arch, viz. 
 the same as the compartments adjacent to the curve in the elevation. 
 The prisms being formed, draw the figure of the soffit of the stone upon 
 the surface intended for the same. Then apply the joint-mould upon 
 each face of the stone intended for the joint, and draw the figure of the 
 joints ; then reduce the end of the stone which is to form a part of the 
 face of the arch in such a manner that when the arch-stone is placed in 
 the position which it is to occupy, or in a similar situation, a straight 
 edge, applied in a horizontal position, may have all its points in contact 
 with the surface of the face of the stone now formed. The face being 
 thus formed, the conic surface must also be formed by means of a straight 
 edge, in such a manner that all points of the straight edge must coincide 
 with the surface when the straight edge is directed to the centre of the 
 cone. 
 
 F 
 
66 
 
 A PRACTICAL TREATISE ON 
 
 [ClI. IV. 
 
 CHAPTER IV. 
 
 CONSTRUCTION OF THE MOULDS FOR SPHERICAL 
 NICHES, BOTH WITH RADIATING AND HORIZONTAL 
 JOINTS, IN STRAIGHT WALLS. 
 
 When niches are small, the spherical heads are gene- 
 rally constructed with radiating joints meeting in a straight 
 line, which passes through the centre of the sphere perpen- 
 dicularly to the surface of the wall, when the wall is 
 straight; but when it is erected upon a circular plan, the 
 line of common intersection of all the planes of the joints is 
 a horizontal line tending to the axis of the cylindric wall. 
 
 Niches of large dimensions will be more conveniently con- 
 structed in horizontal courses, than with joints which meet 
 in the centre of the spheric head ; since in the latter, the 
 length and breadth of the stones are always proportional to 
 the diameter or radius of the sphere, and therefore when the 
 diameter is great, the stones would be difficult to procure. 
 
 The construction of niches depend also upon the nature 
 and position of the surface from which they are recessed ; 
 viz. a spherical niche may be made in a straight wall, either 
 vertical or inclined ; or it may be constructed in a circular 
 wall, or a spherical surface, such as a dome. 
 
 This subject, therefore, naturally divides itself under 
 several heads or branches; the principal are, a spherical 
 niche in a straight wall, with radiating joints ; a spherical 
 niche in a straight wall, in horizontal courses ; a spherical 
 niche in a circular wail, with radiating joints ; a spherical 
 niche in a circular wall, in horizontal courses ; and, a sphe- 
 j^cal iiiclie in a spherical surface or dome. 
 
CH. IV.] MASONRY AND STONE-CUTTING. 
 
 67 
 
 SECTION I. 
 
 EXAMPLES OF NICHES, WITH RADIATING JOINTS, IN 
 STRAIGHT WALLS, AS IN PLATE XXIV. 
 
 Niches of very small dimensions will be easily con- 
 structed in two equal cubical stones, hollowed out to the 
 spherical surface, with one vertical joint ; the portion of 
 the spherical surface, formed by both stones, being one 
 fourth of the entire surface of the sphere. 
 
 Fig. I, plate XXV., is the elevation, 2, the plan, and Jig. 3, the^^-. ' " 
 tical section perpendicular to the face of the straight wall of such a mche.^^^ r£-r^''£ 
 
 The first operation is to square the stone ; viz. to bring the heii«J-(if-^ ^ , 
 each stone to a plane surface, then the vertical joints and the upper as^ ^ ,y 
 lower beds to plane surfaces at right angles with the surface which forms ' . 
 the head. 
 
 The two stones as hollowed out are shown at Nos. 3 and 4. To show 
 how they are wrought, we will commence with one of the stones after 
 being brought to the cubical form. Let this stone be No. 3. In the 
 solid angle of the stone formed by the head, the vertical joint and the 
 lower bed meeting in the point p, apply the quadrantal mould. No. 2, 
 upon each side, so that the angular point of the two radiants may coin- 
 cide with the point />, and one of the radiants upon the arris of the stone 
 which joins the point p ; then if the face of the quadrantal mould coincide 
 with the surface of the stone, the other radiant line will also coincide, 
 because the angle of the mould, and all the angles of the faces of the 
 stone, are right angles. 
 
 By this means we obtain by drawing round the curved edge of the 
 mould, the three quadrantal arcs ahc, agli, and c'lh. The superfluous 
 stone being cut away, the spherical surface will be formed by trial of the 
 mould. No. 2. 
 
 Fig. 1, plate XXVI., is the elevation, and Jig. 2, the plan of a niche 
 in a straight wall. 
 
 The elevation, 1, not only shows the number of stoj^^s which must 
 be odd, and the number of radiating joints which must in consequence 
 be one less than the number of stones, but also the thic^^ness «f these 
 stones, and the moulds for forming the heads and opposite sides. 
 
 F 2 
 
68 
 
 A PRACTICAL TREATISE ON 
 
 [CH. IV. 
 
 The head of the niche being spherical, makes it a surface of revolution. 
 It follows therefore, that the sections through the joints are equal and 
 similar figures ; hence, if all the joints were of one length, one mould 
 would be sufficient for the whole ; but since in this example, they are of 
 diflferent lengths, every two joint moulds will have a common part ; and 
 thus if the mould for the longest joint be found, each of the other moulds 
 will only be a part of the mould thus found. 
 
 In order to ascertain the mould for each joint, the longest being AD, 
 %. 1, extending from the centre to the extremity of the stone upon one 
 side of the plan, the next longest is AF, extending from the centre to 
 the extremity of the keystone, and the shortest AG. 
 
 Upon PQ, Jig. 1, make AF equal to AF", and AG equal to AG'. 
 Perpendicular to PQ draw T)d, ¥f, Gg, meeting the front line RS of the 
 plan Jig. 2, in the points d,J, g, intersecting the back line of the stone 
 in the points m, n, o : then will hikedm be the mould for the first stone 
 raised upon the plan, hikefn the mould for the joint on each side of the 
 keystone, kikego the mould for the first stone above the springing line. 
 These moulds are shown separately at I, II, III, and identified by similar 
 letters. 
 
 Nos. 1, 2, 3, exhibit the first, second, and third stones of the niche as 
 if wrought to the form of the spherical surface ; No. 3 being the key- 
 stone ; therefore the two remaining stones are wrought in a reverse order 
 to the stones exhibited at No. 1, and No. II. 
 
 The first part of the operation is to work the stones into a wedge-like 
 form, so that the right sections of these stones may correspond to the 
 figures formed by the radiations of the joints to the centre A, fig. \, 
 and by the horizontal and vertical joints of the stones adjacent to those 
 which form the niche ; for this purpose, two moulds for each head will 
 be necessary, viz. one whole mould must be made for each stone, and 
 one mould for the part within the circle, which will apply to every 
 stone, in order to form the extent of the part within the recess : thus 
 a mould formed to the sectoral frustrum EE'K'K in the elevation, 1, 
 will apply alike to all stones as will be shown presently. 
 
 The next thing is to form the moulds K'KDSG', K"K'G'TF" anc 
 K"K"F"F" of the heads, the application of these moulds is as follows : — 
 
 Having wrought the under bed, the head and back of each stone, anc 
 having formed a draught next to the edge of the bed, upon the side 
 which is to lie ipon the cylindric part in the centre, at a right angle 
 with the head, apply the mould K'KDSG', Jig. 1, upon the head of the 
 stone, I'fc. i, so that the straight edge KD may be close upon the bed o: 
 the stone, and draw by the other edges of the mould ; thus applied the 
 
CH. IV.] MASONRY AND STONE-CUTTING. 
 
 69 
 
 figure r'rdsg ; and, in the same line rd, close to the bed, apply the 
 mould K'KEE', Jig. 1 , and by the other edges of this mould draw the 
 figure r'ree. Apply the mould K'KDSG', to the opposite, or parallel side 
 of the stone, close to the bed, and draw a similar and equal figure as was 
 done by the same mould when it was applied to the head ; this done, 
 work the upper bed of the stone. 
 
 Proceed in like manner with the stones exhibited at No. 2 and No. 3, 
 and similarly with the stones on the left-hand side of the arch ; the 
 stones No. 1 and No. 2, answering to those on the right hand of the key- 
 stone. 
 
 In order to show the application of the moulds marked I. II. III. at 
 the bottom of the plate, taken from the plan. Jig. 2 ; the mould I. ap- 
 plies to the under bed of the stone. No. 1 ; the next mould II. applies 
 upon the upper bed of No. 1, and upon the under bed of No. 2 ; and 
 the mould III. applies upon the upper bed of No. 2, and upon each side 
 of the keystone. No. 3. 
 
 As every arch has both a right and left-hand side, and as every joint is 
 formed by the surfaces of two stones, every mould has four applications, 
 one on each of the four stones. 
 
 In order to render these applications of the moulds I. II. III. as 
 clear as possible, the corresponding situations of the points marked upon 
 each stone by each respective mould, are marked by similar letters to 
 those on the moulds I. II. III. or their correspondents on the plan, fig. 2, 
 viz. on the under bed of the stone. No. 1, will be found the letters 
 h, i, k, e, d, m, as in the mould I. ; upon the under bed of No. 2, will be 
 found h', i', k', e , g, o ; as also upon the upper bed, of No. 1, i', k', e, g, 
 and upon the right hand side of the keystone, No. 3, will be found the 
 letters k", i", k", e",f", «", as also similar letters upon the upper bed. 
 No. 2, to those of the mould III. 
 
 SECTION II. 
 
 E.vamples of Niches in straight Walls with horizontal 
 Courses, as in Plate XXVII. 
 
 Plate XXVIII. Let Jig. I represent a niche with horiz ontal courses. 
 No. I being the elevation, exhibiting three arch-stones on each side of 
 the key-stone, and No. 2 the plan, consisting of two stones, making to- 
 gether a semicircle, each being one quadrant. 
 
70 
 
 A PRACTICAL TREATISE ON 
 
 [CH. IV. 
 
 The heads of the stones in the wall, on the right-hand side of the arch, 
 which also form a portion of the concave surface, are ABCDE, FDCGHI, 
 HGKLM, and the Icey-stone JkKKL. Round each of these figures cir- 
 cumscribe a rectangle, so that two sides may be parallel and two per- 
 'pendicular to the horizon : thus round the head of the stone ABODE 
 circumscribe the rectangle ANOE', round the figure FDCGHI, the head 
 of the second stone, circumscribe the rectangle PQRI, &c. 
 
 Draw the straight lines am, and ai, jig. 2, No. 1 , forming a right angle 
 with each other ; from the point a as a centre, with the radius TC,fig. J, 
 describe the arc cc, meeting the lines am and ai in the points c,c'. 
 
 Let the quadrangular figure hgfe, No. 1, be considered as the upper 
 bed of a stone, which, as well as the lower bed, is wrought smooth, tliese 
 two surfaces being parallel planes at a distance from each other equal 
 to the line AE or NO, Jig. 1. Moreover, let mcc'ikl be considered as 
 a mould made to the figure before described and laid flat on the upper 
 bed of the stone in its true position, the points c,c' of the mould being 
 brought as near to the side he as will just leave a sufficient quantity of 
 stone, in order to work it complete. By the edges of the mould thus 
 placed draw the curve cc , the straight lines cm and c'i, and the rough 
 edges ik and ml. 
 
 Perpendicular to the upper bed, and along the arc cc', cut the stone so 
 as to form a surface perpendicular to the upper bed, and the surface thus 
 formed will necessarily be cylindric ; through each of the straight lines 
 cm and c'i, cut a surface perpendicular to the said upper bed, and these 
 surfaces will be the planes of the vertical joints, and will be at a right 
 angle with each other ; then with a guage, of which the head is made to 
 the cylindric surface, and which is set to the distance OD, Jig. 1, No. 1, 
 draw the curve line dd on the upper bed of the stone. Upon the lower 
 bed of the stone, with the guage set to the distance NB, draw the arc 
 bb'. 
 
 The thickness of the stone is exhibited at No. 2, Jig. 2, the upper bed 
 being represented by the line Jir, and the lower bed by the line qti, so that 
 7ir and qu are parallel lines, the distance between them being equal to 
 the thickness of the stone, viz. equal to AFj, Jig. 1, No. J. Lastly, with 
 a plane or common guage set to the distance NC, Jig. 1, No. 1, draw the 
 line cc on the cylindric surface, .y?^. 2, No. 1. 
 
 Now, in Jig. 2, the line dd', No. 2, represents the arc dd', No. 1 ; cc'. 
 No. 2, represents the arc cc, No. 1 ; and bb'. No. 2, represents the arc 
 bb. No. 1 : so that the stone must be cut away between the line dd' on 
 the upper bed, and cc on the cylindric surface, by means of a straight 
 edge, so as to form a conic surface ; this may be done by setting a bevel 
 
CH. IV.] MASONRY AND STONE-CUTTING. 
 
 71 
 
 to the angle EBC,^g. 1, No. 1. The conic surface thus formed will be 
 one side of the joint within tlie spheric surface. 
 
 Agaiuj cut away the stone between the line cc on the cylindric sur- 
 face, and the arc bb' before drawn on the lower bed by means of the cur- 
 ved bevel shown at A, Jig. 1, No. 2, so as to form a spherical surface. 
 This may be done in the most complete manner, by applying the straight 
 side of the curved bevel B,Jig. 1, No. 2, to the under bed of the stone, so 
 as to be perpendicular to the curve ; then, if the curved edge coincide at 
 all points, the surface between these lines will be spherical, and will 
 form that portion of the head of the Niche which is contained on the 
 stone. 
 
 In the same manner all the other stones may be cut to the form re- 
 quired. 
 
 Fig. 3 exhibits the stone in the middle of the second course, anijg. 4, 
 the stone on the left of the same course in the angle, which last stone is 
 only half of the stone represented hy Jig. 3. 
 
 Fig. 5 exhibits the left-hand stone of the third course, and^g. 6, the 
 keystone, which is wrought into the frustrum of a cone to a given height 
 in order to agree with the circular courses ; and to prevent any tendency 
 of the keystone from coming out of its place, the upper part is cut into 
 the frustrum of a pyramid. 
 
 Plate XXIX, Jig. 1, represents a spheric headed niche in a straight 
 wall with four arch-stones on each side of the keystone, and therefore, 
 also, with four horizontal courses ; and as the joints are broken, if we begin 
 the first course with four whole stones, as exhibited on the plan. No. 2, 
 the next course will consist of three whole stones and two half stones in 
 one in each angle. As the stones are here in this example projected on 
 the plan as well as on the eltvation, the elevation. No. l,not only exhibits 
 the number of courses, but the number of stones also in each course. 
 
 Fig. 1 represents a spheric headed niche in four courses besides the 
 keystone. 
 
 It may be observed once for all, that the greater the dimensions of a 
 niche, the greater must also be the number of courses in the height. 
 
 The principles for cutting the stones of these niches, is the same as has 
 already been explained for Plate XXVIII. 
 
72 
 
 A PRACTICAL TREATISE ON 
 
 [CH. V. 
 
 CHAPTER V. 
 
 CONSTRUCTION OF THE MOULDS, AND FORMATION OF 
 THE STONES, FOR DOMES UPON CIRCULAR PLANES. 
 AS IN PLATE XXX. 
 
 ON THE CONSTRUCTION OF SPHERICAL DOMES. 
 
 Since walls and vaults are generally built in horizontal 
 courses, the sides of the coursing joints in spherical domes 
 are the surfaces of right cones, having one common vortex 
 m the centre of the spheric surface, and one common axis ; 
 hence the conic surfaces will terminate upon the spheric 
 surface in horizontal circles : again, because the joints be- 
 tween any two stones of any course are in vertical planes 
 passing through the centre of the spheric surface, the planes 
 passing through all the joints between every two stones 
 of every course will intersect each other in one common 
 vertical straight line, passing through the centre of the 
 spheric surface. ^ 
 
 The line in which all the planes which pass through the 
 vertical joints intersect, is called the axis of the dome. 
 
 Because a straight line drawn through the centre of a 
 spheric surface, perpendicular to any plane cutting the 
 spheric surface, will intersect the cutting plane in the centre 
 of the circle of which the circumference is the common 
 section of the plane and spheric surface, the axis of the 
 dome will intersect all the circles parallel to the horizon in 
 their centre. 
 
 The circumference of the horizontal circle, which passes 
 through the centre of the spheric surface, is called the equa- 
 
CH. v.] MASONRY AND STONE-CUTTING. 7B 
 
 torial circumference, and any portion of this circumference 
 is called an equatorial arc. 
 
 The circumferences of circles, which are parallel to the 
 equatorial circle, are called parallels of altitude, and any 
 portions of these circumferences are called arcs of the paral- 
 lels of altitude. 
 
 The intersection of the axis, and the spheric surface, is 
 called the pole of the dome^ 
 
 The arcs between the pole and the base of the dome, of 
 the circles formed on the spheric surface by the planes 
 which pass along the axis, are called meridians, and any 
 portions of these meridians are called meridianal arcs. 
 
 The conical surfaces of the coursing-joints terminate 
 upon the spheric surface of the dome in the parallels of 
 altitude, and the surfaces of the vertical joints terminate in 
 the meridional arcs. 
 
 Hence in domes, where the extrados and intrados are 
 concentric spheric surfaces, two apparent sides of each 
 stone contained by two meridional arcs, and the < arcs of 
 two parallel circles are spheric rectangles, the two sides 
 which form the vertical joints are equal and similar frus- 
 trums of circular sectors, and the other two sides forming 
 the beds are frustrums of sectors of conic surfaces. 
 
 In the execution of domes, since the courses are placed 
 upon conical beds which terminate upon the curved sur- 
 faces in the circumferences of horizontal circles, they are 
 comprised between horizontal planes, and therefore may 
 be said to be horizontal. Hence the general principle of 
 forming the stones of a niche constructed in horizontal 
 courses may likewise be applied in the construction of 
 domes. 
 
 Each of the stones of a course is first formed into six 
 such faces as will be most convenient for drawing the lines 
 which form the arrises between the real faces. Two of 
 
74 
 
 A PRACTICAL TREATISE ON 
 
 [CII. V. 
 
 these preparatory faces are formed into uniform concentric 
 cylindric surfaces, passing through the most extreme points 
 of the axal section of the course in which the stone is in- 
 tended to be placed, the axis of the dome being the common 
 axis of the two cylindric surfaces of every course. 
 
 Two of the other surfaces are so formed as to be in plans 
 perpendicular to the axis of the dome, and to pass through 
 the most extreme points of the axal or right section of the 
 course, as was the case with the two cylindric surfaces. 
 
 The extreme distance of the two remaining surfaces 
 depends upon the number of stones in the course. These 
 surfaces are in planes passing through the axis, and are 
 therefore perpendicular to the other two planes. As these 
 planes, which pass through the axis, form the vertical 
 joints, they remain permanent, and undergo no alteration 
 except in the boundary, which is reduced to the figure of 
 the axal section of the course. 
 
 In order to find the terminating lines of the last and per- 
 manent faces, draw the figure of the section of the course 
 upon one of the two vertical joints in its proper position, 
 then two of the corners of the mould will be in the two 
 cylindric surfaces, one point in the one, and the other in 
 the other, and the two remaining corners of the mould will 
 be in the two surfaces which are perpendicular to the axis, 
 one point pf the mould being in the one plane surface, and 
 the other point in the other plane surface. 
 
 Draw a line on each of the cylindric surfaces through the 
 point where the axal section meets the surface parallel to 
 one of the circular edges, and the line thus drawn on each 
 of the cylindric surfaces will be the arc of a circle in a plane 
 perpendicular to the axis of the two cylindric surfaces, and 
 will be equal and similar to each of the edges of the cylin- 
 dric surface to which it is parallel ; but in the first course of 
 a hemispheric dome, there will be no intermediate line on 
 
CII. v.] MASONRY AND STONP]-CUTTING. 75 
 
 the convex side, since the circular arc terminating the lower 
 edge, will also be the arris line of the convex spheric surface 
 and the lower bed of the stone, which, in this course, is a 
 plane surface. 
 
 In all the intermediate courses of the dome between the 
 summit and the first course, the line drawn on the convex 
 cylindric surface will be the arris line between the convex 
 spheric surface, and the convex conic surface which forms 
 the lower bed of the stone ; and in ail the courses from the 
 base to the summit, the line drawn on concave cylindric 
 surface will be the arris line between the concave conic sur- 
 face forming the upper bed, and the concave spheric surface 
 of the stone, which concave surface will form a portion of 
 the interior surface of the dome. 
 
 On the upper plane surface of each stone to be wrought 
 for the first course, draw a line parallel to one of the cir- 
 cular edges ; but in each of the stones for the interme- 
 diate courses between the first course and the key-stone at 
 the summit, draw a line on each of the planes which are 
 perpendicular to the axis parallel to either of the edges of 
 the face upon which the line is made through the com- 
 mon point in the vertical plane of the joint and the hori- 
 zontal plane, then the line drawn on the top of every 
 stone will be the arris line between the convex spheric, 
 and the concave conic surfaces to be formed, and the line 
 drawn on the under side of any stone in each of the inter- 
 mediate courses will be the arris between the convex conic 
 and the concave spheric surfaces to be formed ; that is be- 
 tween the surfaces which will form the lower bed and a 
 portion of the interior surface of the dome. 
 
 Draw the form of the section of the course upon the 
 plane of the other joint, so that the corners of the quadrila- 
 teral figure thus drawn, may agree with the four lines drawn 
 on the two cylindric, and on the two parallel plane surfaces. 
 
76 
 
 A PRACTICAL TREATISE ON 
 
 [CH. Y. 
 
 Lastly, reduce the stone to its ultimate figure by cutting 
 away the parts between every two adjacent lines which are 
 to form the arrises between every two adjacent surfaces, 
 until each surface acquire its desired form. 
 
 Each of the spherical surfaces must be tried with a 
 circular edged rule, in such a manner that the plane of 
 curve must in every application be perpendicular to each 
 of the arris lines, the mould for the convex spheric surface 
 being concave on the trying edge which must be a por- 
 tion of the convex side of the section, jig. 1, and the mould 
 for the concave side convex on the trying edge, and a 
 portion of the concave arc forming the inside of the sec- 
 tion. 
 
 The two conical surfaces of the beds, and the two plane 
 surfaces of the vertical joints, must be each tried with a 
 straight edge, in such a manner that the trying edge must 
 always be so placed as to be in a plane perpendicular to 
 each of the circular terminating arcs ; so that the surfaces 
 between these arcs must always be prominent until the 
 trying edge coincide with the two circular edges, and 
 every intermediate point of the trying edge with the 
 surface. 
 
 Vlate XXX.lfJig. 1. Let Abcdef . ... Y, be the exterior curve of the 
 section divided into the equal parts Xb, be, cd, &c. at the points b, c, &c. 
 so that each of the chords Ab, he, cd, &c. may be equal to the breadth of 
 the stones in each of the circular courses ; also let ghij'kl .... X, be the 
 inner curve of the section, divided likewise into the equal arcs gh, hi, ij, 
 &c.. by the radiating lines bi, ci. Sec. ; hence Abkg is a right section 
 of the first course ; and, therefore, the figure of the joint at each end of 
 every stone in the first course ; likewise bcih is the right section of the 
 second course ; and, therefore, the figure of the joint at each end of 
 every stone in the second course. 
 
 Since the entire exterior curve of the axal section of the dome is 
 divided into equal parts alike from the base on each side of the section ; 
 and since the exterior and interior sides of the section are each a semi- 
 circular arc, and described from the same centre ; and since the dividing 
 
CH. v.] MASONRY AND STONE-CUTTING. 
 
 77 
 
 lines bh, ci, &c. radiate to this centre^ all the sections of the courses, 
 and the boundaries of the vertical joints will be equal and similar 
 figures ; and, therefore, a mould made to the figure of the section of any 
 course will serve for the vertical joints of all the stones. 
 
 Fig. 2 exhibits one-fourth part of the plan of the convex side of the 
 dome, showing the number of courses, and the number of stones in each 
 quarter-course, there being three stones of equal length in each quarter- 
 course. 
 
 In the first or bottom course, mnop is the plan of the convex side of 
 one of the stones, and mndf the plan of the concave side of the same 
 stone ; and, in the second course, qrst is the plan of the convex side of 
 one of the stones, and qrsi is the plan of the concave side of the same 
 stone ; so that in the first course mndp is the figure of the top and bot- 
 tom of one of the ring-stones, /;o is the intermediate line on the top, and 
 mn that on the bottom, and so on for the remaining stones. 
 
 All the stones of any course being equal and similar solids, and alike 
 situated, the same mould which serves to execute any stone of any one 
 course will serve to execute every stone of that course ; but every course 
 must have a different set of moulds from those of another, except the 
 figures of the vertical joints, which will be all found by one mould, as has 
 been already observed. 
 
 The reader, who has a competent knowledge of the construction of 
 niches in horizontal courses, will not be at any great loss to understand 
 the construction of domes ; or if the construction of domes is well under- 
 stood, he cannot be at any loss to comprehend the construction of niches ; 
 however, as there are many observations respecting the construction of 
 domes that do not apply to niches, particularly as the dome in the pre- 
 sent article has two apparent sides, in order to prevent the reader from 
 wasting his time in referring to both articles, we shall here conduct him 
 through the formation of one of the stones in the first two courses, the 
 figure of the stones in the remaining courses being found in a similar 
 manner. 
 
 \r\Jig. 1 draw AD perpendicular to the ground-line AY, and through 
 h draw BC also perpendicular to the ground-line AY. Now AB as well 
 as being upon the ground-line, therefore to complete the rectangle 
 ABCD, so as to circumscribe the section A6%, and to have two vertical 
 and two horizontal sides, draw through the point h the remaining side 
 DC parallel to AY. 
 
 The rectangle ABCD is the section of a circular course of stone, or 
 that of a ring contained by two vertical concentric uniform cylindric sur- 
 faces and by two horizontal plane rings, the radius of the concave cylin- 
 
78 
 
 A PRACTICAL TREATISE ON 
 
 [CH. V. 
 
 dric surface being aB, and the radius of the convex cylindric surface 
 being oA, and the height of the ring being AD or BC. 
 
 Make a mould to the plan of one of the stones in the first course, that 
 is, to 7nnop,Jig. 2. 
 
 From any point ij, Jig. 3, with a radius zm, jig, 2, or the radSus 
 rtA, fg. 1, describe the arc mn. Make the arc mn,fig. 3, equal to the 
 arc 7nn,Jig. 2, and draw the lines mu and nv radiating to the point ij. 
 Again, from the centre y, and with the radius aB, Jig. 1, describe the 
 arc vu. 
 
 Make a face-mould to mnvu, and this mould will serve for dra-wimg 
 the figure of the two horizontal surfaces of each stone in the first or 
 bottom. 
 
 To cut one of the stones in the first course to the required form : — 
 Reduce the stone from one of the sides till the surface becomes a 
 plane. Apply the mould made to the figure mnvu on this surface, whi ch 
 is one of the two horizontal faces, and having drawn the figure of tlie 
 mould, reduce the stone so as to form three of the arris lines of the faces, 
 which are to be vertical, and these arrises will be square to the face ail- 
 ready wrought. On each of the three arrises thus formed, set the heiglit 
 of the stone from the plane surface already made ; reduce the substan ce 
 till the surface becomes a plane parallel to that first formed. 
 
 Apply then the face mould mnvu, upon the plane surface last wrought, 
 so that three points of the mould may join the corresponding points in 
 the meeting of the three arrises, and having drawn the figure of the 
 mould upon the second formed face, run a draught on the outside of 
 each line upon each of the intermediate surfaces from each of the parallel 
 faces. So that there will be four draughts receding from the face first 
 formed, and four receding from the face last formed, and that upon the 
 whole, including the two draughts upon each side of each of the four 
 perpendicular arrises, there will be sixteen in all. 
 
 The two draughts along the edges of the convex cylindric surface to 
 be formed, must be tried with a concave circular rule, made to the form 
 of the arc mn,Jig. 2, and the two draughts along the edges of the con- 
 cave cvlindric surface, must be tried with a convex circular rule made 
 to the form of the arc po,fig. 2. Moreover, the two draughts which are 
 made along each of the edges of each opposite intermediate plane surface, 
 must be tried with a straight edge. 
 
 Having regularly formed the draughts, so that the circular and straight 
 edges of each of the three rules may coincide in all points with the bot- 
 tom surface of each respective draught, and with the arris line at each 
 
CH. v.] MASONRY AND STONE-CUTTING. 
 
 79 
 
 extremity, the workman may then cut away the superfluous parts of the 
 stone, as far as he can discern to be just prominent, or something raised 
 above the four draughts, bordering the four edges of each of these sur- 
 faces. 
 
 The rough part of the operation being done, each of the four interme- 
 diate faces maybe brought to a smooth surface and to the required form, by 
 means of a common square ; the face of coincidence of the stock, or thick 
 leg, being applied upoii one of the two parallel faces, and the thin leg, 
 called the blade, to the surface of the stone, in the act of reducing, until 
 it has acquired the figure desired, or the two cylindric surfaces may also 
 be tried, by means of circular edged rules, the edge of each rule being 
 placed so as to be parallel to one of the parallel faces ; a concave circular 
 edge being applied upon the convex side, and a convex circular edge 
 upon the concave side. 
 
 The six faces which contain the solid being thus formed, we shall now 
 proceed to find the upper arris : — for this purpose apply the mould made 
 to the form mnop, fig. 2, upon the top of the stone drawn by the means 
 of the mould mnvu^fig. 3. 
 
 Suppose mnvu, fig, 3, to be the figure drawn on the top of the stone 
 itself, by means of the mould made to 7nnvu ; and mnop, fig. 3, to 
 be the mould made from mnop, fig. 2. Lay the edge mn,fig. 2, upon the 
 edge mn^ fig. 3, on the top of the stone, so that the equal circular arcs 
 may coincide in all their points ; and draw the line op along the concave 
 edge of the mould, and op will be the arris line of the spherical and 
 conical surfaces which are yet to be formed. 
 
 Let the rectangle mnnm ,fig. 4, be the elevation of the convex cylin- 
 dric surface of the same stone, projected on a plane parallel to each of 
 the chords of the circular arcs, and to one of the straight arrises of this 
 surface ; the straight line vin representing the upper circular edge, 
 mm , nn the two vertical arrises ; so that the convex spherical surface 
 is terminated at the top by the arc op, and at the bottom by the 
 arc n'm'. 
 
 Let the rectangle nmmrt',fig. 5, be the elevation of the concave cylin- 
 dric face projected on a plane, parallel to one of the chords of one of the 
 circular boundaries, and to one of the straight-lined boundaries of this 
 face ; then the upper and lower planes will be projected into the parallel 
 lines nm, n'm'. Therefore all the lines of each of these three planes will 
 be projected upon the lines nm, n'm', and as the rectilineal figure formed 
 by the two chords and the two straight lines is parallel to the plane of 
 projection, it will be projected into an equal and similar figure ; there- 
 
80 
 
 A PRACTICAL TREATISE ON 
 
 [CH. V. 
 
 fore the projected figure is a rectangle, and the sides nm, nun are equal 
 to each other, and to the chords of the two circular arcs ; and the lin^es 
 mm, nn are each equal to the height of the hollow cylinder, or equal to 
 the distance between the parallel planes. 
 
 Hence the concave surface will be projected also into a rectangle, 
 and the middle of the chords of the arcs terminating the parallel edgies 
 of the concave surface upon the middle of the chords of the arcs, ter- 
 minating two of the opposite edges of the convex surface, as also the 
 two opposite parallel straight-lined sides in the height of the solid, will 
 be projected into straight lines equi-distant from the projections of tbe 
 corresponding lines in the height of the solid on the convex side. 
 
 Therefore, the straight lines nri, mm, vv', uii, are all equal to the 
 height of the hollow cylindric solid, or equal to the distance between the 
 parallel planes and the distance between the lines nn', vv, equal to the 
 distance between the lines mm', uu'. 
 
 To form the common termination between the upper conical and 
 the lower spherical surfaces, let vv', uu, represent the concave cylindric 
 surface ; and, therefore vv, uu', will represent the opposite circular 
 arcs, which terminate two of the sides of this concavity. Upon this 
 surface draw the line v" u, parallel to the circular edge vu, on the 
 top at the distance hC, fig. 1, and the line v", u will be the arris now 
 required between the concave conic surface at the top, and the concave 
 spheric surface. These two surfaces being as yet to be formed. 
 
 To form the remaining and common termination of the concave 
 spherical surface, and the lower or level bed of the stone : — Draw a cir- 
 cular arc on the level surface, underneath parallel to the circular, to the 
 circular edge on the lower edge of the concave cylindric surface, and 
 this line will be the remaining arris required. 
 
 The two cylindric surfaces, and the upper plane surface, are entirely 
 cut away ; but the intermediate line drawn on the top, and that drawn 
 on each cylindric surface, remain, as well as the outer edge of the lower- 
 bed. 
 
 To form the intermediate faces of the stone, into the two upper and 
 lower conical beds, aud into the two apparent concave and convex 
 spherical surfaces -.—Reduce each side of the solid as near to the required 
 surface as possible, so that all the intermediate parts between the ar- 
 rises or lines drawn on the former faces, may be prominent. 
 
 Suppose then, that we proceed to finish the stone required to be 
 formed, in the following order : first, by proceeding with the convex 
 spherical surface ; secondly, the upper concave conical surface ; thirdly 
 and lastly, the concave spherical surface. Having approached as nearly 
 
CII. v.] MASONUY AND STONE-CUTTING. 
 
 81 
 
 to the required surfaces as can be done with safety, the upper conical 
 concave surface will be reduced to its ultimate form by cutting away the 
 substance carefully, so that the surface between the two arris lines may 
 at last coincide with all the points of a straight edge applied perpendicu- 
 larly to the two arrises. 
 
 The convex spherical face will be formed ultimately by cutting the 
 substance of the stone carefully, so that the surface between the arris- 
 line on the top, and the circular convex arris-line on the outside of the 
 lower bed, may at last agree with all the points of the circular concave 
 edge of the rule made to a portion of the aic . Abed, Jig. 1, of the section 
 of the dome. This circular-edged rule must be frequently applied ; and 
 in each application the plane of the arc must be perpendicular to the 
 surface, gradually approaching to its required sphericity. 
 
 To form the concave surface of the upper bed of the stone, reduce 
 the solid by carefully cutting parts away, so as at length the surface 
 between the upper arris and the intermediate line drawn on the inside 
 formerly concave, may coincide with all the points of a straight edge 
 applied perpendicularly to the upper ^rris-line from any point of this 
 arris. 
 
 The concave spherical surface will be formed in the same manner as 
 the convex spherical surface already supposed to be formed, with this dif- 
 ference, that the circular edge which proves the sphericity, by trial must 
 be convex instead of being concave. This convex surface lies between 
 the lower arris, terminating the upper conic bed, and the inner arris of 
 the lower bed. 
 
 As to the lower bed it is already formed, being part of the plane sur- 
 face, formerly one of the ends of the hollow cylinder, in a plane perpen- 
 dicular to the common axis ; and as to the ends forming the vertical 
 joints, they were at first formed in making the hollow cylindric solid ; so 
 that one of the stones in the lower course is now finished. 
 
 One of the stones in the second course being first formed into the 
 frustrum of a cylindric wedge, as was done with the stone formed for the 
 first course, the several faces which contain this solid are as follow : — 
 qrxtv. Jig. 3, represents the plane truncated sector forming the top, st 
 being the arris-line between the spheric surface on the convex side of si, 
 and the conic surface in the concave side of ; qrr'q', fig. 6, the convex 
 cylindric surface, q"r" the arris between the convex spheric and the con- 
 vex conic surfaces, and rq<ir',Jig. 7, the concave cylindric surface; x"'w'' , 
 the arris between the concave spheric surface underneath and the con- 
 cave conic surface above, the arris-line being drawn upon the lower plane 
 
 G 
 
82 
 
 A PRACTICAL TREATISE ON 
 
 [CH. V. 
 
 surface, we shall thus have the arris-lines between the spheric and conic 
 surfaces. 
 
 The solid being cut as before directed between the arris-lines until the 
 surfaces are duly formed, we shall have also one of the stones in the se- 
 cond course completely prepared for setting. 
 
 Perhaps for preparing the stones for the first and second courses, as 
 also the stones near the summit, no better method can be followed than 
 that which we have employed in preparing a stone in each of the two 
 lower courses, yet as the saving of an expensive material and labour is a 
 desirable object, we shall here show how the waste of stone and the 
 labour of the workman may in a considerable degree be prevented. 
 
 Plate, XXXII. Another Method. 
 
 Let fig. 1 be a section of the dome, and fig. 2, a plan of the same, 
 showing the convex side. Now as the saving of material will be prin- 
 cipally in the stones which constitute the intermediate courses, we shall 
 select, for an example, the fifth stone from the bottom and from the 
 summit. The section of this stone is ahcd,fig. 1. 
 
 Draw de parallel, and ae perpendicular to the base of the dome. 
 Then instead of first working the sides of the stone, so that the section 
 may be a rectangle, of which two sides are parallel and two perpendicu- 
 lar to the liorizon ; let it be wrought into the form abode, so that the 
 part de may be parallel to the horizon. 
 
 Let the section abi de be transferred to No. 1, at abode, and let fghi. 
 No. 1, be the section of the rough stone, out of which the coursing- stone of 
 the dome is to be wrought ; the sides of the section of the rough stone 
 having two parallel and two perpendicular faces to the lower bed of the 
 stone. The wrought stone must be selected sufficiently large, so that, 
 when it is reduced to the intended form, all the spherical and conical 
 surfaces must be entire, and thus the arrises will also be entire. 
 
 The first operation is to reduce the stone by taking away a triangular 
 prism from the top ; the sectioii of which prism is represented by kit, 
 No. 1, so that the surface, of which the section is dc, may be a plane 
 surface. 
 
 No. 2 is an orthographical projection of the stone, of which the section 
 is Ifghkl, after being thus reduced, qrst representing the plane surface, 
 of which the section is Id, No. 1, is parallel to the plane of projection. On 
 the plane surface qrst, No. 2, apply a mould xuvrv, so that the radius of the 
 curved edge uv, may be equal to tlie line dx,Jig. 1, dx being parallel to the 
 
CH. v.] MASONRY AND STONE-CUTTING. 
 
 83 
 
 base^ mieeting the axis in x, and that vu and ?vx may be straight lines 
 tending to the centre of the arc ux ; and tliat the chord of the arc iix 
 may be equal to the length of the chord of the upper arris of the stone. 
 Draw lines along vrv, ut, and rvx, of the mould, and let vw be the line 
 drawn by the curved edge vw of the mould, uv the line drawn by the 
 straight edge uv of the mould, and xw the line drawn by the straight 
 edge xxo of the mould. 
 
 Take the mould away, and there will remain the three lines, viz. the 
 arc vw, and the straight lines vu and wx, which radiate to the centre. 
 Then vw is the upper arris of the stone, and the straight lines vu and lux, 
 as in the planes of the meeting joints of the two adjacent' stones in the 
 same course to that which is now in the act of working. 
 
 The second operation is to work the spherical surface by means of the 
 bevel edc, fig. 1, in such a manner, that while the point d is upon any 
 point of the arc vw. No. 2, the straight edge de may coincide with the 
 plane surface xuviv, No. 2, and the curved edge dc may coincide with the 
 spherical surface required to be formed, and lastly, that the plane of the 
 bevel cde may be perpendicular to the arris line vvj. 
 
 The third operation is to find the vertical joints of the stone : these will 
 be formed by means of a common square, of which the right angle is con- 
 tained by two straight lines, so that when the vortex of the angle of the 
 square is upon any point of the line v-w or ux, No. 2, the inner face of 
 application of the third part must be upon the plane surface tuvrv, and 
 the edge of application of the thin part upon the vertical joint, and that 
 both edges of application may be perpendicular to the line vxo or ux. 
 
 The fourth operation is to form the conical upper bed of the stone by 
 means of the \ie\Q\fgh,Jig. 1, so that when this conic surface is wrought 
 to the required form, and the vortex g of the angle is applied upon any 
 point of the curve uv, No. 2, the curved edge gh may then coincide with 
 the spherical surface, and the straight edge gf with the conical bed 
 thus formed, the edges gf and gh being perpendicular to the arris ux. 
 
 Thus four sides of the stone are now formed, viz. the convex spherical 
 surface, the concave conical surface, and the two vertical joints of the 
 stone. By gauging the spherical surface to its breadth, the under or 
 convex conical surface may be formed by means of the sain,e bevel 
 fig. \, and gauging the sides of the stone which form the joints, viz. the 
 concave and convex conic surfaces which form the upper and lower beds, 
 and the two vertical joints from the spherical convex surface, we shall 
 now be enabled to form the concave spherical surface by means of a slip 
 of wood, of which one edge is formed to the curve of the inside of the 
 
 G 2 
 
84 
 
 A PRACTICAL TREATISE ON 
 
 [CH. V 
 
 section. No. 1, and thus we have formed a stone of the fifth course as re- 
 quired to be done. In the same manner the stones of every course may 
 be formed. 
 
 This method will neither require so much stone as the former or first 
 methodj nor yet the quantity of workmanship ; but it requires greater 
 care in the execution. This last method is that which the author used 
 in the construction of the dome of the Hunterian Museum at Glasgow. 
 
CH. VI.] MASONRY AND STONE-CUTTING. 
 
 85 
 
 CHAPTER VI. 
 
 CONSTRUCTION OF CIRCULAR ROOFS, OF WHICH THE 
 EXTERIOR AND INTERIOR SURFACES ARE CONICAL 
 AND CONCENTRIC WITH EACH OTHER. 
 
 PROBLEM. 
 
 To execute a vault, of which both the extrados and 
 intrados are conic surfaces, having a common vertical axis, 
 the solid being equally thick between the conic surfaces, 
 so that in the joint lines those of beds may be horizontal, 
 and those of the headings in vertical planes passing along 
 the axis. 
 
 The easiest method of executing this, is to form the beds so that when 
 built they will unite in horizontal planes, and the headings in vertical 
 planes. 
 
 Let ABC, fig. 1, be a section of the exterior surface, and EFG a sec- 
 tion of the interior surface ; the lines AB and EF being parallel, as also 
 the lines CB and GF. 
 
 In order for the easy application of the bevels, it will be convenient to 
 work the exterior faces of the stones first as plane surfaces ; then form the 
 joints by means of a face mouldy and the angles which the joints make 
 with the planes of the faces by means of the bevels, and lastly, run a 
 draught upon each end of the face first wrought according to the proper 
 curve of the cone. 
 
 Let dsv be the exterior line of the plan, D being the centre of all the 
 circles which form the seats of the joint lines in the plan. Divide the 
 semi-circular arc dsv into as many equal parts as the number of vertical 
 joints in the semi-circumference. 
 
 Let there be five stones, for instance, in each quadrant ; therefore, if 
 and src be quadrants, divide ds into five equal parts, and let de be the 
 first part. Through the point e, draw the radius fJ). Bisect the arc de 
 in Q, and draw C/ a tangent to the semi-circular arc dsv at the point 0. 
 
86 
 
 A PRACTICAL TREATISE ON 
 
 [CH. VI. 
 
 Bisect each of the arcs between the points of division in the quadrantal 
 arc ds, and the tangents being drawn at each point of bisection, will form 
 the polygonal base Cfmnop. 
 
 To form the angle of tlie mitre at the meeting of two heading joints. 
 In Cf, or C/ produced, take any point g, and draw gh perpendicular to 
 the diameter AC;, meeting AC in the point h. Draw hi perpendicular 
 to CB, meeting CB in the point i. In DC make hk equal to hi and join 
 kg ; then will the angle Yikg be the bevel of the mitre. 
 
 The sections of each of the stones as they rise, being de'b'G', e'i'f'b', 
 tjk'f, the dimensions of the stones will be found as follows. Through 
 the points e, i',f, draw the straight lines d'c. Kg, ^7', intersecting the 
 inner line GF in the points b', f, k'. Through b', /', k' , draw the lines 
 ab', d'f, Kk't perpendicular to AC. Also through the points e, i' , f, 
 draw eg, i'l', as also Cc, which will complete the sections of the stones. 
 The other side, AEFB of the section, exhibits the sections of the stones 
 perpendicular to the intrados and extrados of the lines ; the sections of 
 the stones being AEr, E/3/r, (iytd, and the sections of the joints Er, j3t, 
 yu. To find the curve of the stone at any section as Er at the point r. 
 With the horizontal radius 5r,Jig. I, and from the centre 5, describe an 
 arc 7-3. From the point 3, draw 32 perpendicular to 5r, meeting 5r in 2. 
 In 2r make 21 equal to the nearest distance between the point 2 and the 
 line AB. From some point found in the line 5r, describe an arc 13, and 
 the arc 13 will be the curvature of the top of the stone at the joint. This 
 is shown at fig. 2. 
 
 Figs. 3 and 4 exhibit another method of finding the curve at the joint, 
 by means of the radius of curvature. 
 
 PLATE XXXIV. 
 The construction exhibited in this plate being similar to 
 the foregoing, and may therefore be understood from the 
 description now given. Only one particular which we shall 
 observe is, that in the curve of the joints in the former 
 case they are portions of the hyperbolic arcs, and in the 
 latter case they are portions of elliptic arcs. 
 
CH. VII.] MASONRY AND STONE-CUTTING. 
 
 87 
 
 CHAPTER VIL 
 
 CONSTRUCTION OF THE MOULDS, AND FORMATION OF 
 THE STONES, FOR RECTANGULAR GROUND VAULTS. 
 
 CONSTRUCTION OF GROINED VAULTS, WITH CYLIN- 
 DRETIC SUBFACES. 
 
 A CYLiNDRETic surface is every surface which may be 
 generated by a straight line moving parallel to itself, and 
 intersecting a given curve line. 
 
 Since, in good masonry, the sides of the joints of any 
 course of a vault are made to terminate upon the intrados, 
 in a horizontal plane perpendicularly to the intrados, if the 
 intrados be a cylindric surface, of which the sides are 
 straight lines parallel to the horizon, the sides of the 
 coursing joints will be in planes intersecting the intrados, 
 perpendicularly in straight lines, and the course will form 
 one prismatic solid; hence all the right sections will be 
 equal and similar figures, and will be in vertical planes. 
 
 The stones of a groin, which have any difficulty in their 
 construction, are those at the meeting of two adjacent 
 sides, and it is only the formation cf these which we 
 shall describe. 
 
 In order to form the stone of any course, circumscribe 
 a rectangle round each corresponding right section of the 
 course, so that the sides of the rectangle may each pass 
 through the point of meeting of every two sides of the 
 section of the course, and that two of these sides may be 
 parallel, and two perpendicular to the horizon, as was 
 
88 A PRACTICAL TREATISE ON [CH. VII. 
 
 done in respect of the execution of niches in horizontal 
 courses, and in the formation of the stones of a dome. 
 
 In the first place, the stone must be squared in such 
 a manner, that every two faces which meet each other may 
 form a right angle, and that two of the faces may be pa- 
 rallel and six perpendicular to the horizon, and that only 
 two of the six faces which are perpendicular to the horizon 
 may form a receding angle ; and, moreover, that the figure 
 of the two faces which are parallel to the horizon may be 
 formed to the plan of the stone, as formed by the rectangular 
 planes. 
 
 The two vertical faces which form a right angle with 
 each other, but which do not join in consequence of the 
 two vertical faces which form the receding angle coming 
 between them, are those two faces in the plane of the 
 vertical joints. 
 
 The figures of these faces must be made to the rectangle, 
 circumscribing each respective section. 
 
 The next operation is to gauge two lines on the upper 
 level surface, so as to form the return arris between the 
 upper bed and the convex cylindretic surface on each side 
 of the groin ; this operation being done, gauge two lines on 
 the lower level surface, so as to form the return arris be- 
 tween the lower bed and the concave cylindretic surface on 
 each side of the groin. These two lines will thus form a 
 right angle, which being drawn, gauge a line upon each of the 
 vertical sides which form the internal right angle, and these 
 lines will be the arris of the stone on each side of the groin 
 between the upper bed of the stone and the concave cylin- 
 dretic surface ; and, lastly, gauge a line upon each of the 
 vertical surfaces which are opposite to those forming the 
 internal angle, so that each of the two lines thus drawn 
 may form the arrises between the convex surface and the 
 lower bed. 
 
CH. VII.] MASONRY AND STONE-CUTTING. 
 
 89 
 
 The arrises of the stone being thus drawn, it must be 
 reduced to such surfaces, that each of the lines may be the 
 arris of every two adjacent surfaces. 
 
 The two beds of the stone are plane surfaces, and are 
 therefore formed by means of a straight edge. The other 
 cylindretic surfaces are brought to form by means of a 
 curved edge made to the place where the stone is to be set. 
 It is evident that when the curve varies, a mould must be 
 made to every stone. 
 
 Fig. 1, pZa^eXXV, is the plan of a ground vault with its vertical right 
 sections upon each side of it. In the plan A, B, C, D, exhibit the spring- 
 ing points of the groins, AC and BD are the plans of the groins, or inter- 
 sections of the cylindretic surfaces. These plans of the surfaces of the 
 stones in the intrados, which form the ground angles, are exhibited along 
 the lines AC, BD. 
 
 IKL is a section of the intrados, and pqr a section of the extrados, the 
 intrados and extrados being concentric semi-circular arcs ; EFG and mno 
 are sections of the intrados and extrados of the other vault, being each a 
 surbased semi-elliptic arc, equal in height respectively to the semi-circu- 
 lar arcs of the other vault. 
 
 These two sections of each vault exhibit the section of each course of 
 stone, with the circumscribing rectangle. These stones are exhibited 
 separately at No. 1, No. 2, No. 3, &c. 
 
 No. 1 is that over the centre of the section of the semi-circular vault ; 
 No. 2, that next to the stone over the centre ; No. 3, the second stone 
 from that over the centre, and so on. 
 
 No. 1, A, is a section of the course, or of a stone over the centre of one 
 of the semi-circular branches of the groined vault, showing the circum- 
 scribing rectangle ; and No. 1, B, is the underside of the same stone, 
 forming a part of the intrados of the vault. This exhibits the stone as if 
 squared with the portions of the plans of the groins, which are to be 
 wrought on this stone, as also the plans of the intersections of the joints 
 with the upper surface or intrados. 
 
 Having wrought a concave draught along the lines ab, cd to the middle 
 of the intrados EFG of the section of the elliptic vault, the intermediate 
 surface between ab and cd may be formed by means of a straight edge 
 applied parallel to ac or bd, and having wrought the concave draught 
 along the lines ef, gh, so that the points e, /, g, h may remain, while the 
 
90 
 
 A PRACTICAL TREATISE ON 
 
 [CH. VII. 
 
 intermediate is sunk, and so that the draught thus sunk, may have the 
 same curve as the intrados line IKL of the ssmi-oircular branch. The 
 intermediate part may be formed by means of a straight edge applied 
 parallel to eg or fh, and thus the two cylindretic surfaces crossing each 
 other will form the groins ik, Im, which belong to the central stone, and 
 whicli are a portion of the whole groins resting on the springing points. 
 
 These arrises being formed by the intersection of the cylindretic sur- 
 faces, which meet each other at very obtuse angles, ought to be done with 
 care, otherwise the beauty of the intersections would be destroyed. 
 
 No. 2, 3, &c. require a similar description to that of No. 1, and there- 
 fore will be sufficiently understood from that now given. 
 
 P and Q exhibit the manner of forming one of the stones agreeable to 
 the section of one of the elliptic branches of the groined vault after hav- 
 ing squared the stone, this stone being supposed to be the second from 
 that over the centre. 
 
 It is worthy of notice, that except the stone in the summit of the 
 groined vault, any four stones equally distant from the centre of the 
 ground ceiling, though reduced by the same moulds to the same number 
 of similar surfaces, and though every two corresponding similar surfaces 
 meet each other ; yet nevertheless any one of the four stones can only fit 
 one of the four situations ; so that the same moulds will serve for the 
 formation of four stones equally distant from the summit. 
 
CH. Vlll.] MASONRY AND STONE-CUTTING 
 
 91 
 
 CHAPTER VIII. 
 
 CONSTRUCTION OF THE STONES FOR GOTHIC VAULTS, 
 IN RECTANGULAR COMPARTMENTS UPON THE PLAN. 
 
 GllOlNED ARCHES SPRINGING FROM POLYGONAL 
 PILLARS. 
 
 To execute a ribbed-groined ceiling in severies, upon a 
 rectangular plan, so that the ribs may spring from points in 
 the quadrantal arc of a circle, of which the centres are in the 
 angular points of the plan, and to terminate in a horizontal 
 ridge parallel to the sides of the severies, and in a vertical 
 plane, bisecting each side of the plan. 
 
 Let STVW, Plate XXXVL/^. 1, be a portion of the plan consisting 
 of two severies STUX, XUVW, the points S, U, V, W, X, being the 
 points into which the axes of the pillars are projected. 
 
 Bisect VW by the perpendicular rh, and bisect VU by the perpendi- 
 cular ph. Draw the straight lines uq, vh, wm, xn, yo, radiating from V 
 to meet the ridge lines rh and hp in the points r, q, h, m, n, o, and the 
 arc tz described from v in the points u, v, tv, x, y, and these lines will be 
 the plans of the ribs for one quarter of a severy. 
 
 Suppose now the rib over tr to be given, and let this rib he fig. 2, which 
 is here made double. The half abc is the rib which stands upon rt, the 
 curve he, fig. 2, and the plans tr, uq, vh, win, xn, yo, zp,fig. I, of the ribs 
 are given by the architect in the plan and sections of the work : it is the 
 workman's province to find the curvature of the ribs, and tlie formation 
 of the stones for the ceiling. 
 
 For this purpose we shall suppose that the chords which are formed by 
 the joints in the intrados upon the meeting of the rib over tr to be equal; 
 therefore divide the curve be, fig. 2., into equal partvS, so as to admit of 
 vault stones of a convenient size. 
 
93 
 
 A PRACTICAL TREATISE ON [CH. VIII, 
 
 From the points ], 2, 3, &cc.fig. % in the arc he, draw lines perpendi- 
 cular to ah the base of the rib. Transfer the parts of the line ah to rt, 
 fig. 1, and let A be one of the points representing e,fig. 2. In fig. I, draw 
 ut and produce ut, and Lr to meet each other in the point 2. Draw the 
 straight line AB radiating to the point 2, to meet the plan uq in B. Join 
 uv and produce vu, and Jjv to meet in 3, and draw the straight line BC 
 radiating from 3, to meet the plan tL in C. Join vrv, and produce vw 
 and Lp to meet each other in H. Draw CD radiating to the point H, to 
 meet 7Vin in D. Join wx and produce wx, and Lp to meet each other in I, 
 and draw DE radiating to the point I, to meet xn in E. Find the points 
 F and G in the same manner as each of the points B, C, D, E, have been 
 found, and the compound line ABCDEFG will be the line of joints cor- 
 responding to the point 5, Jig' 2. Find the lines corresponding to the 
 other joints in the same manner. Transfer the divisions in the line uq 
 to the base line oifig. 3, and draw lines perpendicular to the base as or- 
 dinates. Transfer the ordinates oifig. 2 to their corresponding ordinate 
 \nfi,g. 3, and draw the curves which Avill complete the inner edge of the 
 x\h,Jig. 3. In the same manner find the curve of the ribs,^o-^, 4, 5, 6, 
 &c. which stand over the lines rL, rvm, xn, &c. 
 
 Fig. 7 exhibits a part of the plan of a groin-ceiling, consisting of two 
 severies when the plans of the piers are squares, of which the angular 
 points terminate in the sides of the plan of each severy, and then we 
 have only to find the diagonal ribs and those upon the narrow side of the 
 severy. It must, however, be observed, both in Jigs. 1 and J, that only 
 one of the curves which belong to arches of the two sides of a severy can 
 be given, the other must be found in the same manner as the curves of 
 the intermediate ribs. In Jig. 7 the plan of the joints has only two 
 points of convergence, which are found by producing the side of the 
 square which forms the plan of the pillars, and the plan of the ridge- 
 lines, till they meet each other. 
 
 We shall now proceed towards the formation of the 
 stones of the vaulting. 
 
 Let ABCD be the plan of one quarter of a severy, and let hC and ij 
 be the seats of two adjacent ribs, and let hyJzC be the rib which stands 
 upon hC, and let klmn be the plan of the soffit of a stone. Perpendicular 
 to hC draw % and Ij, and draw i/g parallel to kC. Produce nk to * and 
 7im to 0. Draw lo and Is respectively parallel to S7i and nm. Draw lr 
 perpendicular to Is ; make lr equal to gi, and join sr. Draw lu perpen- 
 dicular to sn ; and from s, with the radius sr, describe an arc meeting 
 lu in the point n. Draw uv and nv respectively parallel to sn and su. 
 
CH. VIII.] MASONRY AND STONE-CUTTING. 
 
 93 
 
 Perpendicular to no draw oq and mp. Make oq and mp each equal to gi, 
 and join np and nq. Draw pt perpendicular to nq, meeting nq in the 
 point t. To form the winding surface of the intrados, first work the 
 soffit as a plane surface ; on the plane surface describe the Jigs. usnv. 
 Make jiiv equal to nt. 
 
 In Jig. 2 make the angle ale equal to sno,Jig. \, and make the angle cbe, 
 Jig. 2, equal to onq. By problem 3, page 17, having the two legs cba, cbe of 
 a right-angled trehedral, find the angle ghi, which the hypotenuse makes 
 with the leg cbe. Secondly, form the bed of the stone to make an angle 
 at the arris-line nv with the surface usnv, equal to the angle ghi. Jig. 2. 
 Draw rvx upon the end of the stone thus formed perpendicular to mw, and 
 make mx equal to tp, and on the end of the stone draw nx. Join ku ; 
 then the four points n, k, u, x, are the four angular points of the soffit of 
 the stone. The other end of the stone will be formed in a similar 
 manner. 
 
 On the nature and construction of Gothic ceilings. 
 
 Let A, B, C, D, Plate XXXVIII, be the springing points, AC and BD 
 the plans of the groins disposed in the vertices of the angle of a rect- 
 angle, their plans bisecting each other in the point e ; also let QU and 
 SX, passing through the point e, and bisecting the angles AeB, BeC, CeD, 
 DeA, be the plans of the ridges of the gothic arches, and let AE, AH, BJ, 
 BK, CM, CN, DP, DG, be the springing lines of the gothic ceiling. 
 
 Moreover, let the four straight lines EG, HJ, KM, NP, at right an- 
 gles to QU and SX, be the plans of four right sections to each wing of 
 the groined vault ; and let the springing-lines AE, DG, AH, JB, &c. 
 be such as to meet respectively in the points Q, S, &c. 
 
 To construct the ribs which are at right angles to the ridge-lines, 
 and of which their plans are EG, HJ, &c. Let us suppose that the 
 given rib is EFG, standing upon EG as its plan. Prolong AE and DG 
 to meet each other in the point Q. Divide the half curve EF of the 
 arch into as many equal parts as the number of courses is intended to 
 be in the ceiling on each side of the ridge-line of the intrados of the 
 arch ; let us suppose that this number is six, and that h is the first point 
 of division from the bottom point E of the rib, the succession of parts 
 being EA, hi, &c. From the points h, i, &c. draw the straight lines hp, 
 iq, &c. perpendicular to EG, meeting EG in the points p, q, Sec Through 
 the joints p, q, &c. draw from the point Q the lines Qr, Qs, &c., meet- 
 ing AC, the plan of the groin in the points r, s. Sec, and perpendi- 
 cularly to AC draw the straight lines rj, sk, &c. Make rj, sk, Sec, each 
 respectively equal to ph, qi. Sec. through the points A, j, k, &c., draw 
 
94 
 
 A PRACTICAL TREATISE ON [CH. VIII. 
 
 the curve A/A:V for one-half of the curve of the groin rib, the other half 
 is symmetrical, and therefore the same curve in a reversed order. 
 
 To find the rib HIJ. Prolong AH and BJ to meet each other in the 
 point S, and draw the lines rS, sS, &c. intersecting HJ in the points t, u, 
 &c. Draw tn, vo, &c. perpendicular to HJ, and make tn, uo, &c. respec- 
 tively equal to ph, qi, &c. Through the points H, n, o, &c. draw the 
 curve HI, and HI will be the curve of one-half of the arch over the line 
 HJ for the plan. 
 
 Hence we see that the lines Jh, ki. Sec. prolonged will meet the line 
 QR perpendicular to the plane ABCD in the points /, g, &c. at the 
 same heights Qf, Qg, &c. as ph, pi, &c. of the heights of the ordinates of 
 the given rib. Since both sides are symmetrical, one description will 
 serve each of them. 
 
CH. IX.] MASONRY AND STONE-CUTTING-. 
 
 95 
 
 CHAPTER IX. 
 
 THE MANNER OF FINDING THE SECTIONS OF RAKING 
 
 MOULDINGS. 
 
 To find the raking mouldings of a canted bow-window, 
 with munions and transoms. 
 
 Let the plan of the window be^^. 1, Plate XXXIXj consisting of three 
 sides^ the middle one being parallel to the walls, and the other two at an 
 angle of 135 degrees each, with the middle face of the window. 
 
 Also, let UflQ, Jig. 2, be a horizontal section of one of the angles, 
 No. 1 being a right section of one of the munions, the same as the right 
 section of the transom sill or lintel, and let ar, No. 2, be the line of mitre 
 corresponding to AR, No. 1, AR being perpendicular to aQ. 
 
 In order to find the right section, No. 2, of the angular munion. In 
 the curves of the given section. No. 1, draw lines through a sufficient 
 number of points perpendicular to aQ, and draw ac perpendicular to 
 ar ; transfer the points B C from A, No. 1, made by the perpendiculars 
 to No. 2 ; from a to c upon ac, and from a to h through the points in ac 
 draw lines parallel to ar, to intersect the corresponding lines parallel 
 to Qa from the assumed points K, L, M, N, in the curves. No. 1, and 
 through these points trace the curves which will form one side of the 
 section. No. 2 ; repeat the same operation on the other side, and we shall 
 have the complete section required. 
 
 Fig. 3 exhibits the same species of lines applied to an architrave, or 
 mouldings run round tlie two jambs and soffit of a window or door parallel 
 to the face of the wall, the sofFet being level, and the jambs splayed ; the 
 middle section, No. 1, is that which is given, being a right section of the 
 moulding on the soffit. 
 
 Fig. 4, No. 1, is the right section of the raking moulding on a pedi- 
 ment, which if supposed to be given, the section No. 2 may be found as 
 that at No. 2, from No. 1, Jig. 2 ; but in this case No. 2 is generally 
 that which is given, and the section No. 1 is traced therefrom. 
 
96 
 
 A PRACTICAL TREATISE ON 
 
 [CH. IX. 
 
 In all these cases of raking mouldings, draw ac perpendicular to ar 
 the line of mitre. To find any point m, take the point M in the sec- 
 tion No. 1, and draw MB perpendicular to AC, Jigs. 2, 3, 4, meeting AC 
 in and draw Mm parallel to Rr. Make ab equal to AB, and draw 
 bm parallel to ar, and m will be a point in the curve. In the same man- 
 ner will be found the points J, k, I, n, No. 2, from the points J, K, h, N, 
 No. 1 ; and hence the section No. 2 may be traced from No. 1. 
 
CH. X.] MASONRY AND STONE-CUTTi:s'G. 
 
 97 
 
 CHAPTER X. 
 
 CONSTRUCTION OF A LINTEL, OR AN ARCHITRAVE, IN 
 THREE OR MORE PARTS, OVER AN OPENING, AND 
 THE STEPS OF A STAIR OVER AN AREA. 
 
 On the method of building a lintel, or architrave, with 
 several stones, so that the soffit and top of the lintel, or 
 architrave, may be level ; and that the connecting joints of 
 the course may appear to be vertical in the front and rear 
 of the lintel, or architrave. 
 
 A lintel, or architrave, is frequently formed in several 
 stones, from the difficulty of procuring one of sufficient 
 length. The method of doing this is founded upon the 
 principle of arching, the arch being concealed within the 
 thickness of the stones. 
 
 Fig. 1, plate XL, represents the upper part of an aperture, linteled 
 as specified in the contents of this chapter ; the centre of the radiating 
 joints being the vertex of an equilateral triangle. 
 
 Fig. 2 represents the top of the lintel, exhibiting the thickness of the 
 radiating joints, and the thickness of the square joints on each side of 
 the concealed arch. 
 
 Fig, 3 represents the soffit of the lintel, exhibiting the joint lines 
 perpendicular to the two edgeSj as the radiating as well as the vertical 
 joints, all terminate in these lines. 
 
 No. I. exhibits the first abutment-stone over the pier ; No. 2, the first 
 stone of the lintel ; No. 3, the second stone, which forms the key ; the 
 two remaining stones are the same as the first stone of the lintel, and 
 the abutment-stone being placed in reverse order. 
 
 The three stones here exhibited, show the manner of indenting the 
 stones so as to form a series of wedges ; and in order to regulate the stint, 
 the radiations are stopped at half their height. 
 
 H 
 
98 
 
 A PEACTICAL TREATISE ON 
 
 [CH. X. 
 
 No. 1, plate XLI, exhibits the method of constructing an architrave 
 over columns when the stone is not of sufficient length to reach thfe two 
 columns. No. 2, Plan of the upper horizontal side of the architrave 
 exhibiting a chain-bar of wrought -iron;, with collars let in flush with tbe 
 top bed, the sockets being filled Avith melted lead round the collars. 
 
 In the plan and elevation, the same letters express different sides of 
 the same parts ; thus in the elevation. No, 1, the letter A is written upon 
 the part expressing the vertical face of the stone, over the angular c<»- 
 lumn; and A on the plan, No. 2, expresses the horizontal side or bed of 
 the same stone. The letter B, on the elevation No. 1 , represents the 
 vertical face of the middle stone of the architrave ; and B, on the ])laii, 
 represents the bed of the middle stone. The letter C, on the elevation^ 
 represents the vertical face of the stone over the second column ; and C 
 represents the upper horizontal surface or bed. The stones A and C 
 serve as abutments to the middle stone B, which is let in in the manner 
 of a keystone, and therefore acts as a wedge. In order to lessen the 
 effect of the pressure of the inclined sides from forcing the columns to a 
 greater distance, the joint Imnop has two. horizontal ledges, ««, op, which 
 will prevent the middle part from descending. 
 
 D exhibits a stone in the act of setting, and is let down by means of a 
 lewis; e/g- represents a brick arch over the architrave, in order to-^dis-- 
 charge the weight from above, and is resisted by the abutments x, x. 
 The lateral pressure of the brick arch, and of the stone B, is entirely 
 counteracted by means of the chain-bar, of which the top is repre- 
 sented in No. 2- 
 
 No. 4 exhibits a section of the work, z being a section of the arch 
 in the middle, and ^ shows the void between. The right section through 
 the middle of the arch at /, between the columns, is the same as shown 
 at yz. 
 
 No. 3 exhibits the manner of cutting the joints of the stones ov«r 
 the column v and tv, being the steps of the socket and uim the square 
 part of the joint. 
 
 On the construction of stairs over an area to an entrance 
 door. 
 
 Stairs of this description, which consist of one flight, must 
 either be supported upon a solid foundation raised from the 
 ground ; or, if over a hollow, the steps must be supported 
 upon a brick arch, or otherwise, by working the soffits in 
 the form of a concave curve. 
 
CH. XI.] MASONRY AND STONE-CUTTIXG. 99 
 
 Since the joints should always be perpendicular to the 
 curve, they must all tend to the centre of the circle which 
 forms the soffit ; and since the steps should rest firmly upon 
 one another, they ought to rest upon a horizontal surface. 
 To accomplish these ends, every joint between two steps 
 ought to consist of two surfaces, one horizontal, and the 
 other part a plane, radiating to the axis of the cylinder, of 
 which the soffit of the steps is the curved surface. 
 
 Fig. 2, plate XLII, is the plan ; ^fig. 1, the elevation of a door-way with 
 steps, and^^. 3, a section of the same ; ab is the curve-line, represent- 
 ing a section of the soffits. The joints are here drawn to the centre c 
 of the arc ah. 
 
 In this case, where there are no brick arches below, the joints should 
 be plugged. Fig. 4 exhibits a section of the steps, showing the plugs, 
 one in each end perpendicular to the surface of the joint. 
 
 CHAPTER XI. 
 
 OF WATERLOO BRIDGE. 
 
 Plate XLIII. exhibits a longitudinal section of one of the arches, the ad- 
 jacent piers, and part of the next adjacent arches, withtlie elevation of one 
 of the trusses forming the centre. This elegant structure was built under 
 the direction of Mr. John Rennie, civil engineer. The cu^'ve of equili- 
 brium passes through the middle of the length of the arch stones, or very 
 nearly so. The hollows over the piers are raised to the level of the sum- 
 mits of the arches by parallel brick walls, and connected with blocks of 
 stone from wall to wall, for supporting the road-way. 
 
 The centering was composed of eight trusses. 
 
 This plate is given with a view of showing the construction of masonry 
 as generally applied to bridge building. The geometrical principle of 
 constructing arches, and drawing the joint lines so as to be perpendicular 
 to the curve, is sufficiently explained in the second section of this work. 
 
 H 2 
 
ERRATA. 
 
 Insert Plate viii. in page 22, line 6 from the top after fig. 1 ; Plate ix. pa^e 32, 
 in line 27, after fig. 1 ; Plate xxiii. page 64, in line 3, after fig. 1 ; Plate 
 xxxiii.page 85, in line 15, after ABC; xxxv. page 89, line 10, instead of xxv, 
 and Plate xxxvii. page 92, in line 33, after A B C D. 
 
 In page 15, line 8, Instead oft in gt, write f ; in page 22, line 19, insert a 
 between the words in plane ; in page 25, line 22, for a in ^ a, write a ; in page 32, 
 line 28, instead of P, at the end of the line, write F ; in page 40, line 6, in DC 
 write G, in place of Z> ; in page 41, line 13, after x tu, insert fig. 2 ; and in line 36, 
 in ^ B i, instead of i, write I ; in page 49, line 33, for 1, 2, 3, write c ; and for c 
 before the last word, write 1, 2, 3 ; in page 50, line 31, instead of a, after the 
 vrordfiTid, write u ; in page 51, line 1 6, after plan, insert fig. 2 ; in page 63, line 
 24, forPEo p, write PoE; and in line 32, after bd, write b e; in page 76, hne 
 28, in bi write h in place of i ; in page 87, hne 3, instead of ground, write 
 groined j in page 92, line 34, inhyz C, for x write N ; and inline 38, in gi, write 
 j instead ofi; in page 93, line 1, in gi, write j for i ; and in page 96, Une 5, in 
 a rywritejfor r. 
 
GLOSSARY 
 
 OF 
 
 TECHNICAL TERMS USED IN THIS TREATISE. 
 
 A. 
 
 Abscissa, a right line which bisects all the ordinates of a curve. 
 Acute, any sharp edge. 
 
 Acute angle, an angle less than a right angle. 
 
 Angle, the space contained between two right lines which meet each 
 other in a point, but which are not both in the same right line. 
 
 Angle of the joint lines, the angle made on either of the beds of a stone 
 comprised between the face of an arch and the intrados. 
 
 Arc, any small portion of a curve ; but in a circle, an arc is any portion 
 of the circumference. 
 
 Arch, in masonry, a mass of wedge-formed stones, supported at the 
 extremities by abutments, and supporting each other by their mutual 
 pressure. 
 
 Arch-stone, one of the stones of an arch. 
 
 Architrave, the lowest part of an entablature, and that which rests 
 upon the columns. 
 
 Arris, the line in Avhich two surfaces meet. 
 
 Axal plane, a plane passing along the axis. In domes, all the axal planes 
 are perpendicular to the horizon. 
 
 Axal section, the section of a body through its axis. 
 
 Axis of a curve, a right line which bisects all the ordinates. 
 
 Axis of a cone, a right line passing from the vertex to the centre of the 
 base. 
 
 Axis of a cylinder, a right line passing through the solid from the 
 centre of one of the circular ends to the centre of the other. 
 
 Axis of a dome, a right line perpendicular to the horizon, passing 
 through the centre of its base. 
 
 H 3 
 
102 
 
 GLOSSARY. 
 
 B. 
 
 Base, the lower line of a figure, or the lowest face of a solid. 
 Base line, the line upon which a figure is su])posed to stand. 
 Base of a cone, the circular end opposite the vertex. 
 Base of a cylinder, either of the two circular ends. 
 Base of a prism, either of the parallel ends. 
 
 Base of a pyramid, the figure Av^hich is joined at the vertices of its 
 angles to the summit by straight lines separating every two of the sides. 
 
 Battering wall, a wall of which the upper part of the surface falls 
 within the base. 
 
 Beds of a stone in walling, those horizontal faces v/hich form the sides of 
 the joints. 
 
 Bevel bridge, a bridge in which the axes of the cylindretic surface is 
 not at right angles to the face. 
 
 Bisect, to divide any thing into two equal parts. 
 Bridge upon an oblique plan, see bevel bridge. 
 
 C. 
 
 Canted, a prismatic body. 
 
 Canted bow windoiv, a window which has three or more upright faces. 
 
 Catenarian curve, the form of the iron chain which supports the road- 
 way in suspension bridges. 
 
 Centre, a mould for supporting an arch in its progress of building. 
 
 Centre of a figure, the point, through which a straight line may be 
 drawn in any direction which will divide tlie figure into two equal parts. 
 
 Centre of a circle, the point from which all right lines being drawn, to 
 points in the line surrounding the figure, are equal. 
 
 Centre of an ellipse, the point through which any diameter must pass. 
 
 Chord, a straight line drawn from any point of an arc to any other 
 point of that arc. 
 
 Circle, a plane figure of which its boundary is every where at an equal 
 distance from a point witliin its surface, called its centre. 
 
 Circular arc, any portion of the circumference of a circle- 
 
 Circular arch, an arch of which the profile is a portion of the circum- 
 ference of a circle, not exceeding the half. 
 
 Circular roofs, all roofs upon a circular plan are so called. 
 
 Circular course, a course or row of stones in a circular wall. 
 
 Cii cular edges, those edges of a stone where two surfaces meet in the 
 arc of a circle. 
 
 Circular pilan, a plan of which the exterior or interior edges are the 
 circumferences of circles of any portions of them. 
 
GLOSSARY. 
 
 108 
 
 Circular wall, a wall built upon a circular plan. 
 Circumference, the curve line which bounds the area of a circle. 
 Close curve, that which encloses a space. 
 
 Common axis, an axis which equally belongs to two or several things, 
 as the axis of a spherical dome belongs to all the vertical great circles of 
 the sphere. 
 
 Common section, when two or more lines or surfaces all meet in the 
 same line or point, the line or point is called the common section. 
 
 Common vertex, is the point in which two or more plain angles or sur- 
 faces meet each other. 
 
 Concave surface, that in which, if any two points whatever be taken, 
 and if a straight line stretched out between them cannot meet the sur- 
 face in any intermediate point, the side of the surface on which the line 
 is extended is called the concave surface, and the other is the convex sur- 
 face. 
 
 Concentric circles, are those that have the same centre. 
 
 Concentric conical surfaces, are those which hav^e the same axis. 
 
 Concentric cylindric surfaces, are those that have the same axis. 
 
 Concentric spherical surfaces, are those that have the same centre. 
 
 Conic arch, the arch of a circular headed aperture, applied over splayed 
 jambs as in doors and windows. 
 
 Conic parabola, that parabola which is one of tiie three conic sections. 
 
 Cotiic sections, are the plane figures made by cutting a cone, which 
 do not include the triangle nor the circle. These three sections are the 
 ellipse, parabola, and hyperbola. 
 
 Conic surface, is the surface of a piece of masonry presenting the whole 
 or a portion of the surface of a cone. In the construction of domes and 
 tapering buildings upon a circular plan, the beds of every joint are fre- 
 quently conic surfaces. 
 
 Conic wall, a battering wall built upon a circular plan, of which wall 
 the line of batter is a straight line. 
 
 Conjugate diameter, is the term applied to the least axis of an ellipse, 
 being the shortest of all the diameters of this curve. 
 
 Construction, a drawing or building performed by certain rules, and is 
 the result of the operations by which it Avas made to exist. 
 
 Convex surface, see concave surface. 
 
 Convex co7iical face, the convex surface of a cone. 
 
 Convex cylindrical face, the convex surface of a cylinder. 
 
 Course of stones, a row of stones generally placed on a level bed. Tha 
 stones round the face and intrados of au arch, are also called a course of 
 stones. 
 
104 
 
 GLOSSARY. 
 
 Coursing johit, the joint between two courses of stones. 
 Coursing joint lines, the edges of the coursing joints in the face of the 
 work. 
 
 Curved edge, a mould with one of its edges curved, in order to draw a 
 curve line on the surface of a stone, or to ascertain its concavity or con- 
 vexity. 
 
 Curve line, a concave or convex line. 
 
 Curve lined joints, those joints which meet curved surfaces. 
 
 Curved surface, is that which is concave or convex, see concave. 
 
 Cylindretic oblique arch, an arch of which the axis of the surface is not 
 perpendicular to the face. 
 
 Cylindrical inlrados, is the intrados of an arch, of which the surface is 
 that of a cylinder, or a portion of a cylindrical surface. 
 
 Cylindrical spiral, a spiral on the surface of a cylinder. 
 
 Cylindrical surface, is the whole or a portion of the surface of a cylinder. 
 
 Cylindrical wall, a vertical wall on a circular plan. 
 
 Cylindroidic wall, a wall of which the surface is the whole or a portion 
 of the surface of a cylindroid. 
 
 D. 
 
 Design, a scheme or drawing, of something intended to be constructed 
 of stone or other material, as, the design of a house, the design of a 
 bridge. Sec. 
 
 Developable surface, such as can be extended upon a plane, the sur- 
 faces of prisms, cylinders, and cones are developable surfaces. 
 
 Developement, the extension of a surface upon a plane, so that every 
 point of the surface may coincide with the plane. 
 
 Diagram, any scheme or geometrical construction of a proposition. 
 
 Dimensions, such measures of extension as will be sufficient to ascertain 
 the superfices or solidity of a body, or to construct a surface or solid. 
 
 Double curvature, a curve of which its parts cannot be brought into one 
 plane. 
 
 Double ordinate, two equal ordinates of a curve in a right line, separated 
 by another right line called the abscissa. 
 
 Douelle, the surface of a stone, intended to be that which is to form a 
 portion of the intrados of an arch. 
 
 Draught, a grove, or rebat, sunk in a stone for the purpose of directing 
 its reduction to the required surface. 
 
 E. 
 
 Ellipse, a close curve -which may be divided into two equal and similar 
 
GLOSSARY. 
 
 105 
 
 parts by a diameter drawn in any direction : moreover the semi-ellipse, 
 terminated by either axis, may be divided into two synrmetrical parts. 
 
 Elliptic arc, any small portion of the curve of an ellipse. 
 
 Equatorial circumference of a dome, the circumference at the base of a 
 hemispheric dome. 
 
 Equilateral triangle, one having three equal sides. 
 
 Exterior cylindric surface, the curved surface of a cylinder, whether 
 solid or hollow. 
 
 Extrados, the outer surface of an arch. 
 
 Extradosal arc, the outer curve of the section of an arch. 
 
 F. 
 
 Focus is one of the two points to which a string may be fixed, so as to 
 describe the curve of an ellipse. 
 
 Foot, an extension containing twelve inches. 
 
 Figtire, any area enclosed on all sides. l " ' 
 
 Figures of the faces of a stone, the two beds, the face or faces, and the 
 vertical joint or joints. 
 
 G. 
 
 Geometry, the science which explains, and the art which shows, the 
 construction of lines, angles, plane figures, and solids. 
 
 Geometrical elevation, an orthographical projection of an object of which 
 the surfaces are plane figures, either parallel or perpendicular to the plane 
 of projection. 
 
 Gothic arch, an arch of which the two sides of the intrados meet in a 
 point or line at the summit. 
 
 Gothic isosceles arch, a pointed symmetrical arch, of which the spring- 
 ing lines are in the same level. 
 
 Ground line, the straight line upon which the vertical plane of projection 
 is placed. 
 
 H. 
 
 Heading, the vertical side of a stone perpendicular to the face. 
 Heading Joint, the thin stratum of mortar comprised between the ver- 
 tical surfaces of two adjacent stones. 
 
 Helix, a spiral winding round the surface of a cylinder. 
 Horizontal Joint, the same as the bed. 
 
 Horizontal plane of projection, the plane which contains the plan of an 
 object, or its horizontal projection. 
 
106 
 
 GLOSSARY. 
 
 Horizontal projection of an object, the same as the plan of the ob- 
 ject. 
 
 Horizontal trace, the intersection of any plane^ and the horizontal plane 
 of projection. 
 
 Hyi^erhola, an open curve being one of the three conic sections of 
 ^hicli the curve will ever meet a certain right line. 
 
 Hypotenuse, the longest side of a right angled triangle. 
 
 I. 
 
 Inclination, the angle contained between a line and a plane^ or between 
 two planes. W 
 
 Irregular, a term expressing the inequality of the sides and angles of a 
 body. 
 
 Intersection, the point on which two lines meet or cut each other ; the 
 line in which two surfaces cut or meet each other. 
 Intrados, the inner curve of an arch. 
 
 Intradosal curve, the inner cu^Je of the profile of an arch. 
 
 Intradosal Joints, those joints which are seen in the intrados of an arch. 
 
 J. 
 
 Joiyits of a stohe, the mortar comprehended between the adjacent sides 
 of two stones and the face of the work. 
 
 K. 
 
 Key course, the horizontal range of stones in the summit of a vaults in 
 which the course is placed. 
 
 Key-stone, the stone which appears in the front and in the summit of 
 an arch. 
 
 L. 
 
 Leg of a right-angled triangle, one of the three sides which contain 
 the right angle. 
 
 Leg of a trehedral angle, is either of the two planes of a right trehedral 
 angle, which contains the right angle. 
 
 Line in space, any line of which the projectioji is required, but not in 
 the plane of projection. 
 
 Litie of hatter, the line of section made b^ plane and the surface of 
 battering wall, the plane being perpendicular both to the surface of the 
 wall and to the horizon. 
 
 Lintel, the stone which extends over the aperture of a door or window. 
 
GLOSSAUr. 
 
 107 
 
 M. 
 
 Masonry, the art of constructing buildings of stone. 
 Meridians, are the curves on the surface ^of a dome made in vertical 
 planes. 
 
 Meridional arc, a portion of the meridional curve of a dome. 
 Meridional joint, the vertical joint of a vault of which the horizont^||p| 
 sections are all circles. 
 
 N. 
 
 Normal, a right line perpendicular to a curve. 
 
 O. 
 
 \^^^ ( 
 
 Oblique angles, adjacent angles of which their line of separation is not 
 perpendicular to the base. 
 
 Oblique arch, a cylindretic arch of which the axis is not perpendicular 
 to the face. 
 
 Oblique bridge, a bridge which crossWra river, and of ^jj|iici|^the faces 
 of the arch are not perpendicular to the direction of the stream. 
 
 Oblique cylindfier, a cylinder of which the axis is not perpendicular to 
 the circular ends. 
 
 Oblique cylindroid, a cylindroid in which the axis is not perpendicular 
 to the two bases. 
 
 Oblique plan, a parallelogramatic plan of which the sides are not at 
 right angles. fj^ 
 
 Obi ique trehedrd^f a trehedral of which the angles contained by any 
 two of its faces are not a right angle. 
 
 Open curve, that which does not enclose an area. 
 
 Ordinate, a right line comprised between a curve and its abscessa, and 
 is parallel to a tangent at the extremity of this abscessa. 
 
 P. 
 
 Parabola, an open curve, being one of the three conic sections, of which 
 both of its branches may be extended infinitely without ever meet- 
 ing. 
 
 " Parallel right li7ies, are those which can never meet. 
 
 Parallelogram, a quadrilateral figure of which the opposite sides are 
 equal and parallel. 
 
 Parallels, the same as parallel right 'lines. 
 
 Parameter, the ordinate of a conic section, which passes through the 
 focus perpendicular to the axis. . 
 
108 
 
 GLOSSARY. 
 
 Perpendicular, a right line perpendicular to another right line or to a 
 surface. 
 
 Perpendicular surface, a surface perpendicular to a right line, or to 
 a plane. 
 
 Plane, a surface in which all the points of a right line will coincide. 
 Plane angles, those drawn upon a plane surface. 
 Plane curve, that which has all its parts in one plane. 
 Plane wall, a wall of which the surface is a plane. 
 Plane of projection, the plane on which an object is to be represented. 
 Pole of a dome, the summit or upper extremity of the axis. 
 Position, the situation in which one thing is placed in respect of 
 another. 
 
 Prism, a solid bounded on the sides by parallelograms, and on the two 
 remaining faces, by polygonal figures in parallel planes. 
 
 Problem, a proposition which proposes something to be done. 
 
 Projectant, the distance of a point from its projection. 
 
 Projection^ the art of finding the representation of a point, line, surface, 
 or solid. 
 
 Proportion, the parts of two things so that the whole of the one may 
 be to any one of its parts as the whole of the other is to its corresponding 
 part. 
 
 Q. 
 
 Quadrant, the fourth part of a circle. 
 
 Quantity of batter, the angular distance between the plumb-line and 
 the line of batter. 
 
 R. 
 
 Radius, a right line, of which if one end be fixed in a certain point, the 
 other, if moved round, may be made to coincide with all the points of 
 another line, or with the points of a surface. 
 
 Radius of a circle, is any right line drawn from the centre to the cir- 
 cumference. 
 
 Radius of a cylinder, the radius of the circle which is the profile of 
 the cylinder. 
 
 Radius of a sphere, the right line extending from the centre to the 
 surface. 
 
 Radius of curvature, the radius of a circle which has the same curvature 
 as the curve at the point to which this radius belongs. 
 Radiating joints, those joints which tend to a centre. 
 Raking mouldings, moulding which run in an inclined position. 
 Rear line, a line on the back part of any thing. 
 
GLOSSAHY. 
 
 109 
 
 Regulating line, a line which fixes the position of other lines. 
 Retreating sides of the Joints, those which recede from the surface*. 
 Right angle, an angle of ninety degrees. 
 
 Right arch, an arch of which the intrados is perpendicular to the face. 
 Right cone, that of which its axis is perpendicular to the base. 
 Right section, the section of a body at right angles to the axis. 
 Right trehedral, a trehedral having one of its angles a right angle. 
 Ring-stones, the stones which appear in the face and intrados of an 
 arch. 
 
 Ruler surface, a curved surface on which two parallel straight lines may 
 be drawn through any two given points. 
 
 S. 
 
 Section, the figure, formed by cutting a solid by a plane. 
 Segment, the part of a surface or solid containing the upper extremity 
 or summit. 
 
 Segment of a circle, a portion of the circle contained by an arc and its 
 chord. 
 
 Segment of an ellipse, a portion of an ellipse contained by a part of the 
 curve and its chord. 
 
 Segment of a cylinder, a portion cut off by a plane parallel to the axis. 
 
 Semi-axis major, the longest diameter of an ellipse. 
 
 Semi-axis minor, the shortest diameter of an ellipse. 
 
 Semi-parameter, half the parameter, or the focal ordinate. 
 
 Semi-cylinder, the half of a cylinder contained by the curved surface 
 and a plane passing along the axis. 
 
 Semi-cylindric surface, the whole or a portion of the surface of a cylin^ 
 der. 
 
 Severies, the compartments of grained ceilings. 
 
 Similar Jigures or bodies, those which are of the same shape. 
 
 Soffit, the under surface of any part of a ceiling. 
 
 Soffit joints f those joints which appear on the under surface. 
 
 Solid, any body whatever. 
 
 Solid angles, angles in which three or more surfaces all meet in one 
 point. 
 
 Solid geometry, tlie consideration of the properties and construction of 
 solids. 
 
 Solid of revohition, that which may be generated round an axis. 
 Spherical dome, a dome having a spherical surface. 
 Spherical niche, a niche of which the surface of the head is spherical. 
 Spherical triangle, a triangle of which the surface is spherical. 
 
110 
 
 GLOSSARY. 
 
 Spiral curves, those consisting of one or more revolutions. 
 Spiral joints, those joints Avhich run in a spiral line. 
 Spiral surface, a ruler surface, of which the direction of the straight 
 lines tend to an axis, such are the soffits of winding stairs. 
 Splayed, a bevelled jamb. 
 
 Springing plane of an arch or vault, the plane from which the first 
 arch-stones rise. 
 
 Squaring a stone, is the form to which it is made in order to apply the 
 moulds, so as to obtain its ultimate form as inserted in the building. 
 Stone-cutting, the art of reducing stones to their intended form. 
 Straight edge, a rule with a straight edge for trying a surface. 
 Straight vaults, those which have their axis straight. 
 Straight walls, those which have plane surfaces. 
 Surface of revolution, such as may be generated round an axis. 
 Surmounted arch, an arch or vault of greater height thaa half its width. 
 
 T. 
 
 Tallus line, the battering line of a wall. 
 
 Tangeiit, a straight line which touches a curve without being able to 
 cut it. 
 
 Tangent plane, a plane which touches a curved surface without being 
 able to cut it. 
 
 Third proportional, the fourth term of four proportionals, when the two 
 middle terms are equal ; it is called a third proportional. 
 
 Traces of a plane, the intersections of an oblique plane with the hori- 
 zontal and vertical planes of projection. 
 • Transverse axis, the longest diameter of an ellipse. 
 
 Trehedral, a solid angle consisting of three plane angles. 
 
 Triangle, a figure consisting of three equal sides. 
 
 Trying edge, the edge of a rule for trying a given surface. 
 
 U. 
 
 Uniform conic surface, the surface of a right cone- 
 Uniform cylindric surface, the surface of a right cylinder. 
 Upper bed of a stone, the side of the stone which comes in contact v/ith 
 that above it. 
 
 V. 
 
 Vault, a concave ceiling. 
 
 Vertex, the summit of any thing. 
 
 Vertex of a cone, the point in which the surface ends. 
 
GLOSSARY. 
 
 Ill 
 
 Vertical angle, the opposite angle. 
 Vertical of a hatier, the perpendicular distance. 
 Vertical plane, a plane perpendicular to the horizon. 
 Vertical plane of projection, the plane on which the elevation is made. 
 Vertical projectant, the distance of a point from its projection, in the 
 horizontal plane of projection. 
 
 Vertical projection, the elevation of an object. 
 
 Vertical trace, the trace of a plane in the vertical plane of projection. 
 Vertical ivall, an upright wall. 
 
 W. 
 
 Wall in talbis, a battering wall. 
 Winding joints, spiral joints. 
 
 THE END. 
 
 LONDON 
 
 rUlNTED »Y S. AND U. BENTLEY. DORSET STREET. 
 
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