CREMONA'S 
 
 TWO TREATISES ON GEAPHICAL STATICS 
 
 HUDSON BEARS 
 
HENRY FROWDE 
 
 OXFORD UNIVERSITY PRESS WAREHOUSE 
 AMEN CORNER, E.G. 
 
GRAPHICAL STATICS 
 
 TWO TREATISES 
 
 ON THE 
 
 GRAPHICAL CALCULUS 
 
 RECIPROCAL FIGURES IN GRAPHICAL STATICS 
 
 LUIGI CEEMONA 
 
 LL.D. EDO., FOR. MEMB. R.S. LOND., HON. F.ll.S. EDIN. 
 
 HON. MEMB. CAMB. PHIL. SOC. 
 PROFESSOR OF MATHEMATICS IN THE UNIVEBSITY OF ROME 
 
 TRANSLATED BY 
 
 THOMAS HUDSON BEARE 
 
 B.SC. LOND., ASSOC. M. INST. C.E., F.R.S. EDIN. 
 PROFESSOR OF ENGINEERING AND APPLIED MECHANICS, HERIOT-WATT COLLEGE, EDINBURGH 
 
 AT THE CLARENDON PRESS 
 
 1890 
 
 [All rights reserved'] 
 
60 
 
TBANSLATOE'S PKEFACE, 
 
 FOE some years I had used a rough English manuscript 
 summary of Professor CKEMONA'S works on the Graphical 
 Calculus and Reciprocal Figures, while reading with engineer- 
 ing students of University College, London. As English 
 versions were much wanted, I was advised by Professors 
 PEARSON and KENNEDY to ask the consent of Professor 
 CREMONA to my undertaking their translation, and at the 
 same time they supported my application to the Delegates of 
 the Clarendon Press that they should become the publishers. 
 To both applications a most cordial consent was given ; and 
 I take the opportunity of thanking both the Author and the 
 Delegates for the trust they have reposed in me. The trans- 
 lations have been revised by Professor CREMONA and certain 
 portions (in particular Chap. I. of Reciprocal Figures) have 
 been entirely written by him for the present English edition. 
 I regret that a long delay has occurred in the appearance 
 of this book, due chieny to pressure of work both on the 
 part of myself and Professor CREMONA. 
 
 I feel sure that the translation will supply a long-felt want, 
 and be found extremely useful by students of engineering and 
 the allied sciences, especially by those whose work compels 
 them to pay attention to graphical methods of solving pro- 
 blems connected with bridges, roofs, and structures presenting 
 similar conditions. 
 
 THE TRANSLATOR. 
 
 HERIOT-WATT COLLEGE, EDINBURGH. 
 
CONTENTS. 
 
 Page 
 
 TEANSLATOE'S PEEFACE v 
 
 ELEMENTS OF THE GRAPHICAL CALCULUS. 
 AUTHOB'S PEEFACE TO THE ENGLISH EDITION xv 
 
 CHAPTER I. 
 
 THE USE OF SIGNS IN GEOMETEY. 
 
 Art. 
 
 1. Rectilinear segments, negative and positive sense .. .. 1 
 
 2. Relation between the segments determined by 3 collinear 
 
 points 2 
 
 3. Distance between 2 points 3 
 
 4. Relation between the segments determined by n collinear 
 
 points 3 
 
 5. Positive and negative direction of a straight line 3 
 
 6. Relation between the segments determined by 4 points on a 
 
 straight line 4 
 
 7. Relation between the distances of any point from three con- 
 
 current straight lines in its plane 4 
 
 8-9. Angles, negative and positive sense 6 
 
 10. Relation between the angles formed by 3 straight lines in a 
 
 plane 7 
 
 11. Expression for the angle between two straight lines .. .. 8 
 
 12. Areas, negative and positive sense 8 
 
 13-14. Relation between the triangles determined by 4 points in 
 
 a plane 
 
 15. Relation between 5 points in a plane, 4 of which form a 
 
 parallelogram 10 
 
 16. Relation between the distances of a point and 3 non-concur- 
 
 rent straight lines in its plane 10 
 
viii CONTENTS. 
 
 Art. Page 
 
 17. Circuits, simple and self-cutting, Modes 12 
 
 18-23. Areas of self-cutting circuits 12 
 
 24. Reduction of self-cutting to simple circuits 18 
 
 25. Relation between two polygons with equipollent sides .. .. 20 
 26-30. Areal relation of a pole and system of segments 'equi- 
 pollent to a closed or open circuit 21 
 
 CHAPTER II. 
 
 GRAPHICAL ADDITION. 
 
 31-33. Geometrical sum of a series of segments given in magni- 
 tude and sense 24 
 
 34. The sum of segments independent of their order of construction 26 
 
 35-38. Cases where the sum is zero 27 
 
 39. Geometrical subtraction 28 
 
 40-42. Projection of segments and circuits 28 
 
 43-45. Theorems for 2 systems of points, when the resultants of 
 the segments joining each system to the same pole are 
 
 equal 29 
 
 46-48. Extension of the word sum to include absolute position 31 
 
 49-52. Constructions for completely determining the sum .. 32 
 
 53. Case of parallel segments 35 
 
 54. Case of 2 parallel segments 35 
 
 CHAPTER III. 
 
 GRAPHICAL MULTIPLICATION. 
 
 55-56. Multiplication of a straight line by a ratio 37 
 
 57. Division of a straight line into equal parts 38 
 
 58. Division of angles into equal parts. Spiral of Archimedes 39 
 59-62. Multiplication of a number of segments of a straight line 
 
 by a constant ratio. Similar point rows 39 
 
 63-64. Multiplication of a system of segments by a system of 
 
 ratios 42 
 
 65-67. Case where these segments are parallel, and other cases 45 
 
 68-69. Multiplication of a segment by a given series of ratios .. 49 
 
 70-72. Other constructions for same problem 51 
 
CONTENTS. IX 
 
 CHAPTER IV. 
 
 POWERS. 
 
 Art. Pag 
 
 73. Multiplication of a segment by the n^ power of a given ratio 54 
 7475. Other constructions for same problem 55 
 
 CHAPTEE V. 
 
 EXTRACTION OF ROOTS. 
 
 76-77. Equiangular Spiral .. ., 59 
 
 78-82. Properties and construction of the spiral .. .. .. 60 
 
 83-84. Application of it to the extraction of roots 63 
 
 85. Extraction of square roots 64 
 
 86. The Logarithmic Curve and its properties 64 
 
 87. Construction of the Curve 66 
 
 88. Construction of tangents to the curve 67 
 
 89. Applications of the curve 68 
 
 CHAPTER VI. 
 
 SOLUTION OF NUMERICAL EQUATIONS. 
 
 90-91. Lill's construction of a complete polynomial 70 
 
 92-93. Reduction of the degree of an equation 73 
 
 94. Equations of the second degree 75 
 
 CHAPTER VII. 
 
 REDUCTION OF PLANE FIGURES. 
 
 95-96. Reduction of a triangle to a given base 77 
 
 97-100. Reduction of a quadrilateral 78 
 
 101-103. Reduction of polygons 80 
 
 104-105. Reduction of sectors, and segments of a circle .. .. 82 
 106-107. Examples, figures bounded by circular arcs and 
 
 rectilinear segments 83 
 
 108-110. Reduction of curvilinear figures in general .. .. 86 
 111. Application of the reduction of areas to find the resultant 
 
 of a number of segments 87 
 
CONTENTS. 
 
 CHAPTER VIII. 
 
 CENTROIDS. 
 Art. Page 
 
 112-114. Centroid of a system of n points 89 
 
 115. Centre of mean distances .. .. 90 
 
 116-119. Centroid of a system of n points with integral coeffi- 
 cients, or loads 90 
 
 120-123. Centroid of a system of n points with any coefficients 
 
 or loads whatever 92 
 
 124. Construction of the centroid when all the points lie on one 
 
 straight line 94 
 
 125. Construction of the centroid for any 3 points 95 
 
 126. Construction of the centroid for the general case .. .. 96 
 
 127. Case where the sum of the coefficients or loads vanishes .. 97 
 128-129. Further properties and constructions of the centroid .. 98 
 130131. Centroids of homogeneous figures, linear, superficial, 
 
 and solid 99 
 
 132. Centroid of a system of rectilinear segments or triangular 
 
 areas 100 
 
 133-134. Centroid of a rectilinear circuit, regular polygon .. 100 
 
 135. Centroid of an arc of a circle 102 
 
 136. Centroid of the periphery of a triangle 102 
 
 137. Centroid of a quadrilateral 103 
 
 138-139. Centroid of a trapezium 104 
 
 140. Centroid of any rectilinear figure or polygon 106 
 
 141-143. Examples (polygon, cross sections of Angle and Tee- 
 irons) 106 
 
 144-145. Centroid of sectors and segments of circles .. .. 109 
 146. Centroid of a figure bounded by rectilinear segments and 
 
 circular arcs .. 110 
 
 CHAPTER IX. 
 
 RECTIFICATION OF CIRCULAR ARCS. 
 
 147. Rankine's constructions 113 
 
 148. Sayno's constructions, use of spiral of Archimedes, and 
 
 other curves 114 
 
 149. Kochansky's graphical construction of TT 117 
 
CONTENTS. XI 
 
 RECIPROCAL FIGURES IN GRAPHICAL STATICS. 
 
 Page 
 
 AUTHOR'S PREFACE TO THE ENGLISH EDITION 121 
 
 CHAPTER I. 
 POLE AND POLAR PLANE 123 
 
 CHAPTER IT. 
 
 POLYGON OF FORCES AND FUNICULAR POLYGON AS PIECIPROCAL 
 
 FIGURES 131 
 
 CHAPTER III. 
 APPLICATION OF PVECIPROCAL DIAGRAMS TO FRAMEWORK .. 143 
 
 CHAPTER IV. 
 EXAMPLES OF FRAME- AND STRESS- DIAGRAMS 153 
 
ELEMENTS OF THE GRAPHICAL CALCULUS 
 
AUTEOK'S PKEFACE 
 
 TO THE ENGLISH EDITION. 
 
 A GREAT many of the propositions, which form the Graphical 
 Calculus of the present day, have been known for a long 
 time ; but they were dispersed in various geometrical works. 
 We are indebted to CULMANN for collecting and placing 
 them at the head of his Graphical Statics ; a branch of science, 
 created by him, which is such a powerful help in engineering 
 problems. 
 
 The first chapter of this small work, which now appears in 
 English, treats of the use of signs in Geometry, as MOEBIUS 
 conceived them. The succeeding chapters, on Graphical 
 Addition and other arithmetical operations, contain chiefly 
 the graphical calculation of a system of forces in a plane 
 when they are represented by rectilinear segments. The 
 research on centroids, to which the reduction of plane figures 
 serves as an introduction, refers equally to the same subject, 
 being nothing else but the determining of the centres of 
 systems of parallel forces. A special chapter is dedicated to 
 LILL'S method of graphical resolution of numerical equations. 
 
 As MB. BEARE expressed a wish to translate my little 
 treatise II Calcolo Grafico, and also Le Figure Reciproche nella 
 Statica Grafica, for the use of English students, and as the 
 Clarendon Press authorities kindly agreed to publish them, I 
 have been happy to give my consent, as I gave it, some time 
 ago, to Mr. LEUDESDORF for the translation of my Geometria 
 projettiva. Whilst reading the translation I have profited by 
 the opportunity to revise the text, and to introduce some 
 improvements. 
 
 I take the opportunity of thanking both the Translator, 
 and the Delegates of the Clarendon Press. 
 
 THE AUTHOR. 
 
 ROME, July 1888. 
 
ELEMENTS OF THE GKAPHICAL CALCULUS, 
 
 CHAPTEE I. 
 
 THE USE OF SIGNS IN GEOMETKY. 
 
 1. LET 0, A, X be three points in a given straight line 
 (Fig. i), of which and A are fixed points whilst X 
 moves from in the direc- 
 tion OA. Further let the _o o A 
 
 segments (limited portions of 
 
 the straight line) OA, OX _2 
 
 contain a, x linear units 
 
 respectively^. Then as long ~xf~ 
 
 as X remains between and Fig. i. 
 
 A, we have x < a ; when 
 
 X coincides with A, x = a ; and as soon as X has passed 
 
 beyond A, we shall have x > a. 
 
 If the point X instead of moving from towards A, were 
 to travel in the opposite direction (Fig. 2), the number x of 
 linear units contained in 
 
 the segment OX would be o 
 
 considered negative, the 
 
 number a remaining posi- 
 
 tive. For example, if X 
 and A were equally dis- 
 tant from 0, we should have x a. 
 
 A straight line will always be considered to have been 
 described by a moving point. One of the two directions in 
 which the motion of the generating point can take place is 
 called positive, the other negative. Instead oi positive or negative 
 direction we may also speak of positive or negative sense. 
 
 When a segment of a straight line is designated by the 
 
 * The linear unit is supposed to be a segment of unit length measured in the 
 same direction as OA. 
 
 B 
 
2 THE USE OF SIGNS IN GEOMETKY. [2- 
 
 number (oo) of linear units it contains, its sense is shown by the 
 sign + or of the number x. 
 
 A segment may also be designated by means of the two 
 letters which stand at its ends ; for example AB (Fig. 3). In 
 
 this case we agree to write 
 
 A c B AB or BA, according as 
 
 the generating point is 
 conceived to move from A 
 
 c A B to B, or in the opposite 
 
 j>- 3 sense. In accordance with 
 
 this convention, the sym- 
 bols AB, BA denote two equal magnitudes of opposite* sense, 
 hence the identity 
 
 AB + BA = 0, 
 
 or AB = -BA, BA = -AB. 
 
 Of the two points A, B, the extremities of the segment AB, 
 the one A is called the initial, and the other B the final point of 
 the segment. On the other hand for the segment BA, B is the 
 initial point, and A the final point. 
 
 2. Let A, B, C be three points in a straight line. If C lies 
 between A and B (Fig. 3), then 
 
 AB = AC+CB, 
 
 and therefore CB-AC + AB = 0, 
 
 or, since [Art. 1] -CB - BC, and -AC = CA, 
 
 BC+CA + AB= 0. 
 If C lies on the prolongation of AB, then 
 
 AB + BC = AC, 
 
 hence BC-AC + AB=0, 
 
 and therefore BC+ CA + AB = 0. 
 
 And, finally, if C lies on the prolongation of BA, 
 
 CA + AB = CB, 
 
 hence -CB+CA + AB = 0, 
 
 or BC + CA + AB = 0. 
 
 We therefore conclude that f : 
 
 If A, B, C are three points (in any order whatever) in a 
 straight line, the identity 
 
 BC+CA + AB=0 
 
 always /tolas. 
 
 * That is to say, two magnitudes of equal arithmetical values, but with opposite 
 algebraical signs, such as + a and a. 
 
 ^ M.OBivs } J3arycentrisc?terCalcul (Leipzig, 1827), l.Gesammelte Werke,Bd. 1. 
 
-5] THE USE OF SIGNS IN GEOMETRY. 3 
 
 3. From this proposition we obtain an expression for the 
 distance between two points A, B in terms of the distances 
 of these points from a third point collinear with them 
 which we choose as the initial point of the segments. In fact, 
 since 0, A, B are three points in a straight line, we have 
 
 OA + AB + BO = 0, 
 therefore AB = OB- OA, 
 or AB = AO+OB. 
 
 4. If A, B, C, ..., M, J\ r are n points in a straight line, and if 
 the theorem expressed by the equation 
 
 AB + BC+ ... +MN+NA = 
 
 is true for them ; then the same theorem is true for n + 1 points. 
 For if is another point of the same straight line, then 
 since between the three points N, A, there exists the relation 
 
 the above assumed equation becomes 
 
 AB + BC+...+NO+OA = 0. Q. E. D. 
 
 Now it has already been proved (Article 2) that the theorem 
 is true for n = 3, therefore it is also true for n = 4, and 
 so on. 
 
 5. The sign of a segment AB is undetermined, unless a 
 positive segment of the same straight line has already been 
 given; the direction of this latter segment is called the 
 positive direction of the straight line. 
 
 For two different straight lines the positive direction of the 
 one is in general independent of that of the other. But if the 
 two straight lines are parallel, we can compare their directions 
 and say that they have the same positive direction when. 
 after having displaced the one line parallel to itself until it 
 coincides with the other, the two directions are found to be 
 identical. 
 
 Hence it follows, that two parallel segments AB, CD have 
 the same or opposite signs, according as the direction from 
 A to B coincides with the direction from C to D, or not. If. 
 for example, ABCD is a parallelogram, then 
 
 AB+CD = 0, and BC + DA = 0. 
 
 If we draw through n given points of a plane A lt A 29 . .., A n , 
 segments A 1 A l ', A^A^, ..., A H A n ' all parallel to some given 
 direction in the plane until they intersect a fixed straight line 
 AI A 2 ' . . . A n ' , then the sense of one segment determines that 
 
 B 2 
 
4 THE USE OF SIGNS IN GEOMETKY. [6- 
 
 of all the others. Two segments A r A r ', A S A' have the same 
 or opposite sense, according as the points A r , A s lie on the same 
 or opposite side of the given straight line A^A^... A n '. 
 
 Two equal parallel segments, with the same sign, are called 
 equipollent., after Bella vitis. 
 
 6. If A, B, C, D are four collinear points, we have the identity 
 
 AD.BC + BD.CA + CD.AB = 0. 
 
 For the segments BC> CA, AB can be expressed as follows, 
 BC=BD-CD, 
 CA = CD-AD, 
 AB = AD-BD; 
 
 now multiply these three equations by AD, BD, and CD re- 
 spectively and add the results, the right-hand side vanishes, 
 and we obtain the identity we wished to prove. 
 
 7. Let p, q, r be three straight lines intersecting in the 
 point (Fig. 4). Through any point M of the plane draw a 
 
 Fig. 4. 
 
 transversal, cutting p, q, <r in A, B, and C respectively ; then 
 from the proposition just proved, we have 
 
 MA.BC+MB.CA + MC.AB= 0. 
 
 Now draw, parallel to the transversal ABC, a straight line 
 cutting p, q, r in the points P, Q, R ; then the segments BC, 
 CA, AB are proportional to the segments QR, EP, and PQ 
 respectively, and the above equation may therefore be written 
 .QR + MB.RP + MC.PQ = 0. 
 
-7] THE USE OF SIGNS IN GEOMETKY. 5 
 
 If we now draw through any other point M f a new 
 transversal in the fixed direction PQR, cutting p, q, r in 
 A', B', and C', we have similarly 
 
 M'A'.Q + M''.RP + M'C'.PQ = 0; 
 
 that is to say : 
 
 If we draw, through any point M, in a given direction, a trans- 
 versal which cuts three given concurrent straight lines in A, B, C 
 respectively, then the segments MA, MB, MC are connected by the 
 relation 
 
 where a, I, c are constants. 
 
 From the point M let fall perpendiculars ML, ME, MF 
 upon the three straight lines p, q, r ; and also from some 
 arbitrarily chosen point S of the line PQR perpendiculars SV, 
 SF, 7Fupon the same given straight lines. Then since the 
 triangles MAD, SP U are similar we have 
 
 = SP:SU. 
 
 op 
 
 Therefore MA = ^- MD, 
 
 Su 
 
 and similarly MB = ^%- ME, 
 
 Sr 
 
 The equation 
 
 MA. QR + M3.RP + MC.PQ = 0, 
 
 may therefore be written 
 
 .. 
 
 that is to say : 
 
 If ive drop from any point M perpendiculars MD, ME, MFupon 
 three concurrent straight lines, the following relation holds 
 
 a.MD + p.ME+y.MF = 0, 
 where a, /3, y are constants. 
 
 The lines ML, ME, MF, instead of being perpendicular to 
 the given straight lines, may be inclined to them at any 
 the same arbitrarily chosen angle ; we should then obtain a 
 relation of similar form, by merely altering the values of the 
 constants a, /3, y ; the proof however remains the same. 
 
 The proof does not necessarily presuppose that the intersec- 
 tion of the three straight lines p, q, r lies at a finite distance ; 
 
6 THE USE OF SIGNS IN GEOMETRY. [8- 
 
 the proposition is therefore true even if the three given straight 
 lines are all parallel to one another. 
 
 8. A plane has two sides which face the two regions into 
 which it divides space. Let a perpendicular be drawn through 
 any point of the plane, and let the positive direction of this 
 perpendicular be fixed. If 01 be any positive segment of this 
 straight line, then the region in which I lies is called the 
 positive region, and the side which looks toward I is called 
 the positive face. 
 
 Now let an observer, standing with his feet at 0, and his 
 head at I observe a rotational motion in the plane (Fig. 5) ; 
 this can take place in two senses, either from left to right [dex- 
 
 trorsum, in the sense of rotation of the hands of a watch], or 
 from right to left [sinistrorsum]. The former sense is called 
 positive, the latter negative. 
 
 Let P, Q,R be three points on a circle in the plane (Fig. 6) ; the 
 points P, Q divide the circumference into two arcs PQ, one of 
 which contains 7^. If we take as positive the sense in which 
 one of the two arcs has been described, the other arc has 
 negative sense. If we fix the positive arc PQ, then the 
 sense of any arc, and of any rotational motion in the plane 
 will be fixed ; and thereby the positive face of the plane is 
 also fixed, as it is the one on which the observer must stand 
 in order that the positive arcs may seem to him to be 
 described in the sense of the motion of the hands of a, watch. 
 The positive sense of a plane is that of its positive arcs. 
 
 9. Let a, b be the positive directions of two straight lines 
 in a plane, intersecting in the point (Fig. 7), and let OP, 
 OQ be two positive segments of these straight lines, each of 
 length equal to unity. By the angle ab between these two 
 lines, we mean the circular arc PQ described in the positive 
 sense of the plane. In order that the angle may be fixed 
 
-10] THE USE OF SIGNS IN GEOMETRY. 7 
 
 it is necessary to fix both the positive directions of the two 
 straight lines and the positive sense of the plane; but we 
 may add to any angle any number of 
 complete rotations either positive or 
 negative, i. e. (if n is an integer), 
 
 ab360 xn = ab. 
 
 If OA, OS are two positive segments 
 of the straight lines a, 6, the angle ab 
 
 XX. 
 
 can also be denoted by OA . OB, or 
 more briefly by AOB. 
 
 The sum of the angles ab, la is equal 
 to any number of complete revolutions ; 
 we may therefore write Fig. 7. 
 
 ab + ba = 0, 
 
 or la = ab, or ab = da. 
 
 that is to say, ab and ba can be regarded as of equal magnitude 
 and opposite sense *. 
 
 This leads us to consider the positive rotation ab as equiva- 
 lent to the negative rotation ba; or in other words, the angle 
 ab is the circular arc PQ described in the positive sense of the 
 plane, or the circular arc QP described in the negative sense 
 and then taken with the sign : PQ = QP. 
 
 A negative angle is one described by a negative rotation, or 
 by negative arcs. 
 
 Analogously we have 
 
 AOB + BOA = 0; 
 
 that is, AOB, BOA are two angles of equal magnitude and 
 opposite sense. 
 
 10. Let the directions a, b, c of three straight lines in the 
 plane be given, and suppose them to be drawn from the same 
 point 0, and to be extended on only one side of it, for the 
 angle between two straight lines is independent of their 
 absolute position. Then if in turning round in the positive 
 sense of the plane, we meet with the three straight lines in 
 the order acb (Fig. 8), we have the identity 
 
 ca = cb + ba, 
 
 hence cb + ca ba = 0. 
 
 T But cb = be, ba = ab, 
 
 and therefore bc + ca + ab 0. 
 
 * BALTZEK, Analy. Geometric, 9. 
 
8 
 
 THE USE OF SIGNS IN GEOMETKY. 
 
 [11- 
 
 If the order of the succession is abc (Fig. 9), then 
 
 bc + ca = ba, 
 
 or bc + ca ba = 0, 
 
 and therefore be + ca + ab = 0. 
 Accordingly we have this proposition : 
 
 If a, b, c are three straight lines in the same plane, in 
 order whatever^ the identity 
 
 be -\-ca-\-ab = 
 is always true. 
 
 Fig. 8. 
 
 Fig. 9. 
 
 11. From this we obtain, by a procedure similar to that for 
 segments (Art. 3), an expression for the angle' between two 
 straight lines a, b, in terms of the angles, which they make 
 with a third straight line o, taken anywhere at pleasure in the 
 given plane. In fact if o, a, b are directions in one and the 
 same plane, we have 
 
 oa + ab + bo = 0, 
 therefore ab ob oa, 
 
 or ab = ao + ob. 
 
 12. Three points A, , C which do not lie in one straight line, 
 are the vertices of a triangle (Fig. 10). Let us consider that 
 
 we pass round its periphery con- 
 tinuously, that is, passing through 
 each point once and through no 
 point more than once : then each 
 vertex is the final point of one side 
 and the initial point of the follow- 
 ing side. This can be done in two 
 ways, that is to say in two opposite 
 
 directions ; namely in the sense ABC or in the sense ACB. 
 The sense BCA or CAB does not differ from ABC, and 
 
 similarly neither CBA nor BAG is different from ACB. 
 
 Fig. 10. 
 
-14] THE USE OF SIGNS IN GEOMETRY. 
 
 The area of the triangle lies to the right or to the left hand, 
 according as we go round the periphery in the positive or 
 negative sense ; for this reason we consider the areas ABC, 
 ACB as equal but opposite : the first as positive, the second 
 negative. We may suppose the area ABC (or ACB] to be de- 
 scribed by a revolving line of variable length, of which one end 
 is fixed at A, whilst the other describes the segment BC (or 
 CB). Now this rotation takes place in the positive (or nega- 
 tive) sense of the plane ; for this reason also we consider the 
 area as positive (or negative) *. 
 
 The necessary and sufficient condition that three points 
 A, B, may lie in one straight line, is that the area ABC is zero. 
 
 13. PROPOSITION. If is any 
 point whatever p , in the plane of the 
 triangle ABC (Fig. 1 1), we always 
 have the identity 
 
 OBC + OCA + AB = ABC f. 
 Proof. If lies within the 
 triangle ABC, then of course 
 the latter is the sum of the 
 triangles OBC, OCA, OAB. 
 
 If lies within the angle 
 BAG, but upon the other side 
 of BC, we have 
 
 OCA + OAB- OCB = ABC ; 
 but OCB = - OBC, 
 
 therefore OBC + OCA + AB = ABC. 
 Finally, if lies within the opposite vertex si BAG, we have 
 
 OBC- OAC- OB A = ABC, 
 
 and hence OBC + OCA + OAB = ABC. Q. E. D. 
 
 It follows from the remark at the end of Art. 12, that if 
 A, B, C are three points in a straight line, then wherever 
 may be we have 
 
 OBC+OCA+OAB=0. 
 
 14. It follows from this proposition, that the area of the 
 triangle ABC may be regarded as generated by the motion of 
 a revolving line of variable length (radius vector), of which one 
 end is fixed at (the pole], whilst the other describes the 
 
 * MO'BIUS, loc. cit. 17. t Ibid. 18. . 
 
10 
 
 THE USE OF SIGNS IN GEOMETRY. 
 
 [15- 
 
 Fig. 12. 
 
 periphery (outline) in the sense denoted by the given expres- 
 sion ABC. 
 
 This remark and the above proposition would remain un- 
 altered, even if BC were no longer a segment of a straight 
 line, but an arc of a curve *. 
 
 15. If is any point whatever in the plane of the parallelogram 
 ABCD (Fig. 12), we have 
 
 OAB + OCD = i ABCD, 
 
 For, using S to denote the point, in 
 which the side BC is cut by the 
 straight line, drawn through 
 parallel to AB, we have (Art. 13), 
 
 SAB + SBC + SCA = ABC. 
 But SBC = 0, SCA = SCD, 
 
 SAB = OAB, SCD = OCD, 
 therefore 
 
 OAB + OCD = ABC = \ ABCD. Q. E. D. 
 
 Since ^ABCD = DAB, the above equation may also be written 
 ODC= OAB -DAB. 
 
 16. Let (Fig. 13) p, q, r be three straight lines, which form a 
 triangle ABC ; and let and M be two points in its plane, of 
 
 which the first is considered as 
 fixed or given, and the other as 
 variable. Draw from the points 
 and M to the straight line p in 
 any direction the two parallels OU, 
 MD, and similarly to q the paral- 
 lels OF, ME, and to r the paral- 
 lels W, MF, also in any directions 
 whatever. 
 
 The areas of the triangles OBC, 
 MBC are proportional to the dis- 
 tances of their vertices 0, M from 
 the common base BC, and therefore 
 also to the segments U, MD ; 
 hence we have 
 OBC: MBC = OU-.MD, 
 
 Fig. 13- 
 
 * And therefore also, if BC, CA, AB were three arcs, which do not intersect, 
 see Art. 19. 
 
-16] THE USE OF SIGNS IN GEOMETRY. 11 
 
 or MBC=~.MD, 
 
 OCA 
 and similarly MCA = ME, 
 
 But from (Art. 13) 
 
 MBC + MCA + MAB = ABC, 
 
 OBC OCA OAB 
 
 therefore MD + -_ . ME+ -^ . MF = ABC. 
 
 If we vary the position of the point M in the plane, whilst 
 keeping the directions V, 07, W fixed, then in the above 
 equation only the lengths MD, ME, MF change ; we obtain 
 therefore this Theorem : 
 
 If we draw in given directions from any point M in the plane 
 of a given triangle, the straight lines MD, ME, MF meeting the sides 
 of this triangle, then these straight lines are connected by the relation 
 
 (f) a.MD + p.ME + y.MF=S, 
 the quantities a, (3, y, 8 being constants. 
 
 The proposition is still true if two of the three given 
 straight lines p, q, r are parallel to one another. For example, 
 let q, r be parallel, and let us draw a straight line s, which is 
 parallel neither to q, r, nor to p. If now we draw through 
 any point M, in directions chosen at pleasure, the straight 
 lines MD, ME, MF, MG to the straight lines p, q, r, s, then 
 from the proposition just proved, since p, q, s form a triangle, 
 MD, ME, MG are related by an equation of the form (f), 
 which may be written thus 
 
 a.MD + (3.ME + MG = 8 ; 
 
 and similarly since p, r, s form a triangle, we shall obtain 
 a relation of the same form 
 
 between MD, MF, MG. 
 
 Subtracting this equation from the foregoing one, we have 
 
 (a-a')MD + (3.ME-y.MF=b-b', 
 
 that is to say, MD, ME, MF are also connected by a relation of 
 the form (f). Q. E. D. 
 
 This proposition is a generalisation of the one (in Art. 7) 
 
12 THE USE OF SIGNS IN GEOMETRY. [17- 
 
 concerning three straight lines p, q, r which intersect in a 
 point situated at either a finite or infinite distance. In the 
 special case mentioned the constant 6 is zero. 
 
 17. We shall call that line a circuit which a point describes 
 whilst it moves in a plane from one position (the initial) to 
 another position (the final) continuously, that is without 
 ever leaving the plane. The circuit is closed if the final 
 position coincides with the initial position ; it is open if this 
 is not the case. If the circuit intersects itself, we call the 
 points of intersection nodes, and the circuit a self-cutting one. 
 
 If the circuit is formed of rectilinear segments, it is said to 
 be polygonal^ or simply a polygon. 
 
 Any circuit can be described, like the periphery of a triangle 
 (Art. 12), in two opposite senses. In order that the sense 
 of a circuit may be fixed, it is sufficient to know the order 
 of succession of two points of it, if the circuit is open, and oi 
 three, if it is closed. 
 
 18. A closed circuit without nodes encloses within itself an 
 internal finite region of the plane, and divides it from the rest 
 of the plane, which is external and infinite. The area bounded 
 by the circuit is the measure of the interior region, and it 
 is considered to be positive or negative, according as it lies to 
 the right or left of an observer on the plane, who passes along 
 the circuit in the given sense. 
 
 19. PROPOSITION. If ABCD...MNA (Fig. 14) is any closed 
 circuit., and a point in its plane, then the sum of all the triangles 
 (or sectors), 
 
 2 = OAB+OBC+ OC0+... + OMN+ ON A, 
 is a constant quantity for any position whatever of the pole 0*. 
 
 Proof. Let 0' be another point in the plane ; then from the 
 proposition in Art. 13, 
 
 O'AB = OAB + OBO'+OO'A, 
 O'BC = OJ3C + OCO' + 00' B, 
 0'CD = OCL+ODO'+OO'C, 
 
 &c. &c. 
 
 0'MN= OHN + ONO'+OO'M, 
 O'NA = ONA + OAO' + OO'N. 
 
 * MOBIUS, Baryc. Calcul, 165, Ges. Werke, Bd. 1 ; Staiik (Leipzig, 1837), 
 45, Ges. Werke, Bd. 3. 
 
-20] 
 
 THE USE OF SIGNS IN GEOMETRY. 
 
 13 
 
 Adding we have, 
 
 O'AB + 0'$C + O'CD + . . . + 0'MN+ O'NA 
 = OAB + OBC + OCD + ... -f OMN 4- ON A = 2, 
 since all the other terms cancel, because they occur in pairs of 
 equal and opposite terms, as, for example, 00' A and OAO*, OO'B 
 and OBO' } and so on. We may consider the magnitude 2 as 
 
 o f 
 
 Fig. 14. 
 
 generated by the motion of a revolving line OX (radius vector) 
 of variable length, which has one end fixed at the pole 0, 
 whilst the other describes the given circuit in the given sense. 
 
 20. If a radius vector Yloe rotated in a given plane about 
 a fixed point 0, and if it pass over any point of the given 
 plane, we shall call the passage, positive or negative, accord- 
 ing as the radius vector F in passing through is in the 
 act of describing a positive or negative rotation. 
 
 Lemma. If a radius vector Y, moveable in a plane about 
 a fixed point 0, starting from the original position @A, 
 describes successively the angles a l5 a 2 , &c., &c...., and if after 
 having passed p times positively, and n times negatively, 
 over a given point it returns to its original position QA, then 
 the difference p n is independent of the order of succession of 
 the angles a. 
 
 It will be sufficient to show, that if we interchange a r , a r+1 
 the difference p n is unaltered. We are at liberty to suppose 
 that the angles a are less than 180, because if a r were greater 
 than 180 we could divide it into parts each less than 180. 
 
THE USE OF SIGNS IN GEOMETRY. 
 
 [20- 
 
 If a r and a r+l are of the same sign, the radius vector &Y 
 will either describe the angle a r + a r+1 , or the angle a r+l + a r , 
 hence it will pass over the same positions and in the same 
 sense ; and therefore neither p nor n will be changed. 
 
 Now suppose that a r and a r+l are of opposite sign. Before 
 the interchange, let us suppose that at the completion of the 
 angles <z r _ l5 a r , a r+1 , the moving radius vector takes respec- 
 
 tively the positions Y r _ 
 
 Y r +i (Fig- I 4 a )> an d after 
 
 the interchange at the completion of the angle a r+1 let it take 
 up the position Y r '. Then, if the point lies in one of the 
 
 angles Y r Y r+l = F^j J/ = a r+l , the interchange will de- 
 crease or increase by unity each of the numbers ;;, n. If, on 
 the other hand, lies outside these angles, both these numbers 
 will be unaltered. In every case therefore the difference/? n 
 is unchanged. 
 
 COROLLARY. The difference p n is equal to the number 
 (positive or negative) of revolutions contained in the sum 
 aj + a 2 -f . . . In fact, let K . 360 + y be the sum of the positive 
 a's, and (h 360 + y') the sum of the negative a's. Now y and 
 y' are each less than 360, and as the final position of the 
 
-21] THE USE OF SIGNS IN GEOMETRY. 15 
 
 radius vector Zis supposed to coincide with its original posi- 
 tion A, we must necessarily have y = y '. But by virtue of the 
 preceding lemma the difference p n will remain unaltered 
 if, instead of describing the angles a 1? a 2 , a 3 , &c., in succession, 
 we describe the rotation y y +k 360 /$360, or the rotation 
 360 h 360 (as the equal and opposite angles y and y can 
 be neglected) ? since this leaves the numbers p, n unchanged, or 
 increases or diminishes each of them by unity. Now, in describ- 
 ing each of the Jc (or 7i) positive (negative) rotations, we make a 
 positive (negative) passage through the point ; therefore 
 p n = k h. 
 
 21. THEOREM. Let any given closed circuit whatsoever, in 
 a plane, be described in a given sense by a point X, returning 
 to its original position, after having passed over all the points 
 of the circuit. Take a point in the plane, and let 2 be the 
 algebraic sum of the sectors described in succession by the 
 radius vector OX. Then the sum 2 remains constant 
 wherever may be taken *. 
 
 Let us imagine the plane divided by a close network of 
 lines into very small areas, which we shall call elementary areas, 
 so small that the circuit does not pass through the interior 
 of any one of them, with the exception of those that form part 
 of the contour. If while the point X describes the circuit it 
 happens that the radius vector OX in passing through certain 
 positions changes its sense of rotation, we shall suppose that 
 the straight lines forming these special positions of the radius 
 vector form part of the network. Then it is not possible for 
 any elementary area to be partially described by the radius 
 vector OX, but it will be either totally described or not at all. 
 
 Having premised this, then, during the whole movement of 
 the point X in the circuit, let any elementary area whatever co 
 be described by the radius vector OX,p times positively, n times 
 negatively. Then the area <o will be contained pn times in 
 2, or 2 will be the sum of (p n) <o extending over all the 
 elementary areas of the plane. It will be therefore sufficient 
 to show that the coefficient pn does not vary with the 
 pole 0. 
 
 * DE MORGAN, Extension of the word area, (Cambridge and Dublin Math. 
 Journal, vol. v. 1850). For the treatment of this argument the author is indebted 
 to the suggestions of Professor Gabriele Torelli, of Naples. 
 
16 
 
 THE USE OP SIGNS IN GEOMETKY. 
 
 [22- 
 
 If we join the point X to a point taken inside the area 
 co, and if we produce the straight line QX beyond 0, for 
 example, to meet the circuit in 7, then it is evident that every 
 time the radius vector OX describes the area co in one sense, the 
 straight line 07 passes through in the same sense, and con- 
 versely. Therefore the number of times OX passes through 
 co will be equal, in sense and absolute value, to the number of 
 times 07 passes through 0. Therefore if (k Ji) 360 are the 
 number of complete rotations of the radius vector 7, the co- 
 efficient of the elementary area co in the sum 2 will be k k, 
 that is, is independent of the pole 0. 
 
 22. A given closed and self-cutting circuit (Fig. 15) divides 
 the plane into a definite number of finite spaces S lt $,,,... 
 contiguous to one another. Each of these is bounded by a 
 circuit without nodes ; so that the whole plane consists of 
 
 (T 
 
 C -o 
 
 M 
 
 Fig. 15- 
 
 these spaces and of the remaining (external) infinite region. 
 which latter we shall denote by S Q . 
 
 Let co and a/ be two elementary areas or elements of the 
 plane, which can be joined by a straight line that does not 
 cross the circuit, and let us take the pole upon the con- 
 tinuation of the straight line coV It is evident that the 
 radius vector OX cannot pass over co' without at the same time 
 passing over co in the same sense ; co and co ' will therefore enter 
 into with the same coefficient. The elements co", a/"... have 
 also this same coefficient, if the circuit does not pass between 
 
-23] THE USE OF SIGNS IN GEOMETRY. 17 
 
 a/ and &>", or between &>" and a/", &c. Since we can thus 
 conjoin all the elements in succession of one and the same 
 space S, therefore all the elements of 8 will appear in the 
 sum 2 with the same coefficient c. That is to say, 8 appears 
 in the sum 2 with the coefficient c. If therefore c lt c 2i ... are 
 analogous coefficients for the spaces 15 S 2 , .,., we have 
 
 2 = ^ + ^+..., 
 
 if we understand that S lt S 2 , ... at the same time express 
 the areas of the spaces represented by these symbols. 
 
 Next let CD, coj be two elements, between which the circuit 
 passes once ; and let w lie on the right and u 1 on the 
 left of the circuit which passes between o> and <o l5 in the 
 given sense. Take the pole upon the continuation of the 
 straight line wjo)...; now if X traverses that part of the 
 circuit which lies between w and co l5 the radius vector OX 
 will describe o> once with a positive rotation, without describ- 
 ing o) 19 whilst for all other parts of the circuit the elements o> 
 and Wi will be described simultaneously in the same sense. 
 The coefficient of o> will therefore exceed that of o^ by 1 ; 
 that is to say, if in passing from one space to a neighbouring 
 one we cross the circuit once from right to left*, then 
 the coefficient of the first space exceeds that of the other by 
 unity. 
 
 The infinite region $ has the coefficient zero ; for if <D O is 
 an element, which lies outside the spaces S l , S 2t &c...., then it 
 is clear, that we can give the pole such a position, that 
 the (finite) radius OX never passes through co , wherever X 
 may lie on the circuit. 
 
 Any space from which we can get to S by crossing the 
 circuit only once, has the coefficient +1 or 1, according 
 as the crossing takes place from right to left, or from left to 
 right. In general if we draw from a point in any space 8 a 
 straight line to a point of S Q) and if this straight line crosses 
 the circuit m times from right to left and n times from left to 
 right, then the coefficient of 8 is equal to m n. 
 
 23. If the circuit has no nodes, we have a single finite space 
 8 t and this has the coefficient 4-1 or 1, according as the 
 
 * From right to left is always to be taken in the sense of a person describing 
 the circuit in the given sense ; the particular sense is indicated in the figure 
 by an arrow. 
 
 C 
 
18 
 
 THE USE OF SIGNS IN GEOMETRY. 
 
 [24- 
 
 circuit has been described positively (Fig. 15$) or negatively 
 (Fig. 15$). In this case therefore we have 
 
 * = 8 > 
 
 that is to say, If the circuit is not a self -cutting -one then the sum 
 2 is the area of the space enclosed by the circuit. 
 
 Fig. 15 1. 
 
 This property naturally leads us to consider the sum 2 as 
 defining the area of any self -cutting circuit*. 
 
 24. A self-cutting circuit can be decomposed into circuits 
 which are not self-cutting, by separating the (curvilinear) 
 
 Fig. 1 6 a. 
 
 Fig. 1 6 b. 
 
 angles, formed by the branches which intersect at each node, 
 without altering at all the sense (i. e. the direction of the arrows) 
 
 Fig. 170. 
 
 Fig. 1 7 ft. 
 
 of the branches themselves. Consider, for example, Figs. 16 
 and 17; in each a self-cutting circuit is resolved into two 
 
 * Besides the paper by DE MORGAN previously mentioned, see MOBIUS, Ueber 
 die Bestimmung des Inhalts eines Polyeders [Berichte der Konigl. Sachs Gesellsch. 
 der Wissenschaften zu Leipzig, 1865), 13 and following ; Ges. Werke, Bd. 2.] 
 
-24] 
 
 THE USE OF SIGNS IN GEOMETRY. 
 
 19 
 
 simple ones ; also Fig. 1 8, where a self-cutting circuit is 
 resolved into four simple ones. 
 
 The spaces with negative coefficients are in this way 
 separated from those with positive coefficients ; and of two 
 
 Fig. 18 a. 
 
 Fig. 18 5. 
 
 spaces whose coefficients have the same sign, the one whose 
 coefficient is greater in absolute value, lies wholly within 
 the other. Thus, for example (if we denote by S r the space 
 whose coefficient is r), S 2 is inside /S^; S B inside S 2 ,..., _ 2 
 inside _ 19 .... Hence it follows that the area 2 can be ex- 
 pressed as a sum of spaces, which all have positive or negative 
 unity for their coefficient. For this purpose it is sufficient to 
 take the area S r once for itself, and once more with the 
 area S r _ lt within which it lies; that is to say, we sum the 
 spaces S r and S r _ l + S r instead of 2S r and S r - ly and so on. 
 Consider for example (Fig. 18) where the area is equal to 
 
 8* + (S t + &) + (^ + S, + S 3 ) - S_*. 
 
 By the area of a system of closed circuits we understand the 
 algebraic sum of the areas of the single circuits. Thus, for 
 
 Fig. 19. 
 
 Fig. 20. 
 
 example, the ring inclosed between the two oval curves in Fig. 
 1 9 is the area of the circuits ABC, A'C'Jl'; on the other hand, the 
 
 * CULMANN, GrapMsche StatiJc, 2d ed. (Zurich, 1875), N r . 26. 
 C 2 
 
20 THE USE OF SIGNS IN GEOMETKY. [25- 
 
 area of the circuits ABC, A'B'C' (of Fig. 20) is equal to that ring 
 plus twice the internal area A'B'C'. In both cases we can sub- 
 stitute for the two circuits a single one AA'C'B'BCA (Fig. 19) or 
 AB'C'A'BCA (Fig. 20), where the points B, B' are considered 
 as infinitely near to A, A' respectively. In (Fig. 21) the two 
 
 Fig. 21. Fig. 22. 
 
 circuits intersect; the area of the circuits ABC, A'B'C' is equi- 
 valent to that of the circuits AA'B'C, ABB'C'. In Fig. 22 the 
 area of the circuits ABC, A'B'C' is equivalent to that of the 
 circuits ABA'B', AC'A'CA. The two circuits can, in each 
 case, be replaced by a single one. 
 
 25. If the two closed polygons CDE ... Hf, C f D'E'. . . M', in a 
 plane, have their sides CD, C'D', DE, D'E,' ... MC, M'C' re- 
 spectively equipollent, the sum of the parallelograms CDD'C' 
 DEE'D' ... MCC'M' is zero. It will be sufficient to prove 
 this for the case of the triangle CDE. 
 
 Taking D as the pole of the contour CC'D'E'E, we have 
 from the theorem of Art. 19, 
 
 DCC' + DC'D' + DD'E' + DE'E+DEC = CC'D'E'E. 
 
 But the two first triangles together form the parallelogram 
 DCC'D'\ similarly the third and fourth triangles form the 
 parallelogram EDD'E'. Also 
 
 CC'D'E'E-DEC = CC'D'E'E-D'E'C' = E'ECC', 
 which is a parallelogram. 
 
 Wherefore : 
 
 DCC'D' + EDD'E' + CEE'C' = 0. 
 
 From this it follows that if CDE . . . M is a closed polygon 
 whose n sides are the bases of n triangles whose vertices are the 
 points AI A 2 ... A n , respectively, (which are taken anywhere in 
 the plane of the polygon,) the sum of the triangles 
 
 does not change when the polygon is moved parallel to itself in 
 
-26] THE USE OF SIGNS IN GEOMETRY. 21 
 
 its plane. In fact, if C'D'W ... M f is another polygon, whose 
 sides are equipollent to those of the given one, we have 
 A 1 CD = A L C'D'+C'DD', 
 
 A 2 DE = 
 
 A n MC = 
 summing up we have, 
 
 because, as we have shown above, the sum 
 
 is equal to zero. 
 
 26. THEOREM. If the rectilinear segments A 1 B 1 , A 2 B 2 , 
 A 3 B 3 , ... A n B n of given magnitude and position in a plane are 
 equipollent to the sides of a polygon (i.e. of any rectilinear closed 
 circuit, whether self-cutting or not), then the sum of the triangles 
 
 OA^ + OA 2 B 2 + OA 3 S + . . . + OA n B n , 
 
 is constant wherever the jwle may be taken, at a finite distance. 
 But if the given segments are not equipollent to the sides of a closed 
 polygon, then this sum is not constant except for such points 0, as 
 are equidistant from a fixed straight line*. 
 
 Proof. Construct the crooked line CDE . . . MN, of which the 
 successive sides CD, DE, . . . MN are respectively equipollent to 
 the given segments A- [ B l , A 2 B 2 , A 3 B%, ... A n B n ; so that the 
 figures A 1 B 1 DC, A 2 B 2 ED, ... , A n B n NM are parallelograms. 
 Then from (Art. 15), 
 
 OA^= OCD-A^CD, 
 OA 2 B 2 = ODE-A 2 DE, 
 
 &c. &c. 
 
 OA n B n = OMN-A n MN, 
 and also from the proposition in Art. 19 
 
 OCD+ODE+... OMN+ ONC = CDE ... MNC, 
 hence by addition we have 
 
 OA l BI + OA 2 B 2 + . . . + OA n B n = CDE . . . MNC + OCN 
 
 -(A l CD^-A 2 DE+...+A n MN). 
 
 If the given equipollent segments form a closed polygon, 
 that is, if the point N coincides with C, then the area of OCN 
 is zero, provided that the point remains at a finite distance, 
 and therefore the sum OA 1 B 1 + OA 2 B 2 + . .. + OA n B n 
 
 * APOLLONIUS, Loci Plani, lib. 1. L'HUILIER, Polygonometrie, 1789, p. 92. 
 MOBIUS, Statik, 46. 
 
22 THE USE OF SIGNS IN GEOMETBY. [27- 
 
 has a value independent of the position of 0. Hence it 
 follows that, in the special case where ON is zero, the above- 
 mentioned sum either has the value zero for every point of 
 the plane, or else it vanishes for no single point (tying at a 
 finite distance). 
 
 If N does not coincide with C, the above sum will remain 
 unaltered, so long as the area of the triangle OCN does not 
 alter ; that is, so long as the point remains at the same 
 distance from the straight line CN. 
 
 If we change this distance, and take a new pole 0', we shall 
 have 
 
 0'A 1 B 1 + 0'A 2 B 2 +... + 0'A n B n = CDE...NC+ O'CN 
 
 Take the pole 0' at such a distance from CN, that the 
 area of the triangle O'CN is equal to 
 
 A 1 CD + A 2 DS+ ... + A n MN-CDE ... NC, 
 then the sum O'A^ 1 +0'A z 2 + ... + 0'A n B n = 0. 
 
 The straight line (parallel to CN), which is the locus of those 
 points 0' for which this sum is zero, we call r. If we take 
 the point C, i. e. the arbitrary initial point of the crooked line 
 CDE ... , upon r, then the area of O'CN is zero, and therefore 
 the sum of the triangles 
 
 A 1 CD + A 2 DE+ ... +A H MN 
 
 is equal to the area CDE . . . MNC. If we keep to this choice of 
 C, i.e. if we agree that C shall be a point in the line r, then 
 for any point whatever we shall have 
 
 OA 1 1 +OA 2 B 2 +... = OCN. 
 
 27. Conversely, if the sum OA 1 B 1 , OA 2 B 2 , &c. ... is the 
 same for every point in the plane, the segments A l B l . l 
 A 2 B 2 , &c. ... are equipollent to the sides of a closed polygon. 
 If there are two segments, they will therefore be parallel, 
 equal, and opposite in sense. If we take the point on one 
 of them, we see that the sum is half that of the parallelogram 
 formed by the two segments. 
 
 28. In the special case, where all the given segments meet 
 in a common point C, the sum of the triangles 
 
 A 1 CD + A 2 DE+ ... +A n MN, or else CLE+... + CMN 
 is identical with the area of the polygon CDE ... NC (Art. 23) ; 
 and therefore the common point C must also be a point in the 
 straight line r. This is tantamount to saying that in this case 
 
-30] THE USE OF SIGNS IN GEOMETRY. 23 
 
 / coincides with the straight line CN, which joins the ex- 
 tremities of the crooked line CDE . . . MN. 
 
 The same conclusion holds good if the given segments lie 
 upon straight lines, which all intersect in the same point C ; 
 since we can substitute for the triangle OA r B r the triangle 
 OCB ',., because the segments A r B r and CB' r lie on the same 
 straight line, and are equal to one another in magnitude and 
 similar in direction. 
 
 29. From this property of the straight line r, for the case 
 where all the segments lie upon straight lines which meet 
 in the same point, we obtain a construction for the straight 
 line ; in the general case, when the segments lie anywhere 
 upon the plane. 
 
 Let C be the point in which A 1 B 1 and A 2 B 2 intersect. With 
 
 C as the initial point construct the triangle OLE, whose sides 
 
 CD, DE &YQ equipollent to the straight lines A 1 B l , A 2 B 2 ; then 
 
 from what has just been proved for every position of the point 
 
 OCE= OA 1 J3 i +OA 2 S 2 . 
 
 Now let P be the point, in which CE cuts the straight line 
 A 3 B 3 ; with P as initial point construct the triangle PQlt, whose 
 sides PQ, QR are equipollent to the segments CE, A 3 J3 3 , then 
 OPE = OCE+ OA 3 3 = OA 1 B 1 +OA 2 B 2 +OA 3 3 . 
 
 And so we proceed continually until we ultimately reach a 
 segment AB such that 
 
 OAB = OA l B! + OA 2 B 2 + . . . + OA n B n . 
 
 This segment AB lies on the required straight line r, and is 
 equipollent to the straight line CN, which joins the extremities 
 of the crooked line CDE . , . MN, whose sides are respectively 
 equipollent to the given segments. 
 
 30. As in the general case, when CN is not zero, all the 
 points 0, for which the sum 
 
 OA 1 B 1 +OA 2 B 2 +... OA n B n 
 
 has the same value, lie upon a fixed straight line (par. 26), so 
 there is only one straight line r, the locus of the points 0, for 
 which the above sum is zero. Hence it follows, that whatever 
 be the order, in which we take the given segments in the 
 above construction, we shall always arrive at one and the 
 same straight line r. 
 
CHAPTER II. 
 
 GRAPHICAL ADDITION. 
 
 31. To geometrically add or combine a number of segments 
 1 , 2 , 3 , . . . , ft given in direction and magnitude, we must con- 
 struct a polygonal circuit, whose sides, taken in order, are 
 equipollent to the given segments (Fig. 23). 
 
 The straight line *!,..., n which joins the first and last 
 points of the circuit so constructed, is called the geometrical sum 
 
 Fig. 23. 
 
 1 
 
 Fig. 24. 
 
 Fig. 25. 
 
 or resultant of the given segments * ; and these are called its 
 components. If the given segments are all parallel to one 
 another, the polygonal circuit reduces to a straight line, 
 whose successive segments 01, 12, 23, ... (Fig. 24), or 11, 22, 
 
 * CHELINI, Saggio di Geometna Analitica, trattata con nuovo metodo (Koma, 
 1838), p. 35. 
 
GRAPHICAL ADDITION. 
 
 33,... (Fig. 25) are respectively equipollent to the given 
 segments. In this case the resultant of the given segments is 
 identical with their algebraical sum. The two figures show 
 two different methods of denoting a series of segments which 
 follow one another consecutively upon a straight line. 
 
 32. From the definition given above, it follows that the 
 resultant *!,..., of the n given segments 1, 2, 3, ..., n is iden- 
 tical with the resultant of the two segments s lt .,, ir * r+1 ,...,n> of 
 which s lt . . . f r is the resultant of the first r given segments, and 
 * r +i,...,n of the n r remaining segments. For since the 
 straight lines <? L ..., and * lt ...,r start from the same point as 
 the segment 1, and the straight lines 8 lt ..., n and * r+L ...,n end 
 in the same point as the segment n, therefore the straight line 
 *!,..., n begins at the same point 
 
 with #!,..., r and ends at the same 
 point with * r+1> ...,. 
 
 Fig. 26 corresponding to n = 8, 
 and r = 5 , shows that the result- 
 ant of the segments 1,2,3,4,5, 
 6, 7, 8 coincides with the geo- 
 metrical sum of two components, 
 one of them the resultant of the 
 segments 1 , . . . , 5, the other the 
 resultant of the segments 6, 7, 8. 
 
 From this we infer that, if we 
 divide the given segments [always 
 taken consecutively, i.e. in the 
 given order] into any number of groups, and if we sum the 
 segments of each group, the sum of the partial resultants 
 thus obtained will coincide 
 with the resultant of all the 
 given segments. 
 
 33. The resultant of a number ^\$ 
 of given segments is independent 
 
 of the position of the point as- 
 sumed as the initial point of 
 the circuit. Fig 27 
 
 In fact the circuits drawn 
 
 from two different initial points, and O lt are equal 
 similar and similarly situated (congruent) figures, and the 
 
 Fig. 26. 
 
GRAPHICAL ADDITION. 
 
 [34- 
 
 Fig. 28. 
 
 second may be found by moving the first parallel to itself, so 
 that each of its points describes a straight line equipollent to 
 the straight line 00 1 (Fig. 27). 
 
 34. THEOREM. The resultant * lf ... jW of several given segments 
 1, 2, 3,..., n is independent of the order in which they are 
 combined. 
 
 Proof. We begin by proving that two consecutive seg- 
 ments, for example 3, and 4 (Fig. 28), can be interchanged. 
 
 In the given order, the resultant 
 of all the segments is also the 
 resultant of the three partial 
 resultants # lf 2 , s 3> 4 , * 5> ...,. In 
 like manner, in the new order, 
 the resultant of all the segments 
 will be the resultant of the par- 
 tial resultants s lt 2 , s^ s , s 5t . . . t n . 
 But #3,4 and * 4>3 are the same 
 straight line, namely the dia- 
 gonal of the parallelogram, which we obtain by drawing first 
 two consecutive segments equipollent to the given ones 3, 
 and 4 , and then, starting from the same point, two other 
 consecutive segments equipollent to the same given seg- 
 ments with their order changed 4 ' 3 '. Thus the interchange 
 of the segments 3, and 4 has no influence on the required 
 resultant. 
 
 If we interchange first 3 and 4, then 3 with 5 , and finally 
 5 and 4 , the total effect is the same as if we had interchanged 
 
 3 and 5 (Fig. 29). In gen- 
 .4' 
 
 eral we interchange any 
 two non-consecutive seg- 
 ments we please by means 
 of interchanges of consecu- 
 tive segments. Therefore 
 the resultant of any num- 
 ber of segments is unal- 
 tered if we interchange any 
 two segments we please ; 
 or the resultant is inde- 
 pendent of the order in 
 
 Fig. 29. 
 
 which the segments are taken to form the figure. 
 
-37] 
 
 GRAPHICAL ADDITION. 
 
 27 
 
 Fig. 30 shows several circuits, constructed with the same 
 segments, taken in the different orders 12345, 13254, 
 15234. 
 
 Fig. 30. 
 
 35. If a closed circuit can be constructed with the given 
 segments, then from the proposition just proved it follows, that 
 all the circuits obtained by changing the order of the seg- 
 ments have this same property. In this case the resultant 
 of the given segments is zero, or 
 
 The resultant of any number of segments vanishes when they are 
 equipollent to the sides of a dosed polygon. 
 
 The simplest case in which the resultant vanishes is that of 
 only two segments, one of which is equipollent to the other 
 taken in the opposite sense. 
 
 36. If, out of some of the segments whose resultant is 
 required, a closed polygon can be 
 
 formed, then all these may be neg- 
 lected without affecting the required 
 resultant. 
 
 In Fig. 31 the resultant of the seg- 
 ments 1 ... 9 coincides with that of 
 1 , 2,8, 9 , because the resultant of 
 3 , 4 , 5 , 6 , 7 is zero. 
 
 If the component segments are in- 
 creased in any given ratio, then the 
 resultant is increased in the same 
 ratio, without changing its direction. 
 
 37. Two series of segments have equal (equipollent) result- 
 ants, if, after constructing the corresponding circuits starting 
 from the same point the final points of the two circuits coincide 
 (Fig. 32). If we combine the segments of the one series with 
 
28 
 
 GKAPHICAL ADDITION. 
 
 [38- 
 
 Fig. 32. 
 
 those of the other taken in the opposite sense, the total 
 resultant is zero. 
 
 38. Two series of segments have equal resultants, but of 
 opposite sense, when, the corresponding polygonal circuits being 
 
 so constructed that the initial 
 point of the second coincides 
 with the final point of the first, 
 the final point of the second 
 also falls on the initial point of 
 the first. If we combine the two 
 series of segments, their total 
 resultant is zero. Conversely, 
 if the resultant of several seg- 
 ments is zero, and if we split 
 them up into two distinct groups, the resultant of the one 
 group is equal, and of opposite sense, to that of the other 
 group. 
 
 39. Subtraction is not a distinct operation from addition. 
 To subtract a segment 1 from a segment 2 is the same as 
 adding to 2 a segment equipollent to the segment 1 taken in 
 the opposite sense. 
 
 40. If two series of segments have equal (equipollent) 
 resultants, by adding to, or taking away from, both the same 
 segment, we shall obtain two new series whose resultants will 
 also be equal (equipollent)*. 
 
 41. Given a segment AB (Fig. 33), and a straight line r ; then 
 if we draw through A and B in any arbitrarily chosen direction 
 
 two parallel straight lines to 
 meet r in the points A' and B', 
 the points A', B' are called the 
 projections of the points A and 
 B, and the segment A'B' the 
 projection of the segment AB. 
 The straight lines AA' BB' 
 are called ihe projecting rays. 
 The projections of two equipollent segments are themselves 
 
 equipollent (so long as we neither change the direction of r, 
 
 nor that of the projecting rays). 
 
 * The properties of Art. 32 and Art. 40 can be both deduced without further 
 proof from those of Art. 30. 
 
 Fig. 33- 
 
-43] GRAPHICAL ADDITION. 29 
 
 42. Let ABC . . . MNA be a closed circuit (Fig. 34), and 
 A',B f , C f , ... M' t N' the projections of its vertices ; then since 
 A', J5', &c. are points in a straight line, it follows, from (Art. 4), 
 that A'B' + B'C'+... M'N' + N'A'= ; i.e. the sum of the pro- 
 jections of the sides of a closed circuit is zero. 
 
 Let A l B l , A 2 B 2 , . . . , A n B n be n segments in a plane, whose 
 resultant is zero, that is to say, n segments which are equal in 
 magnitude and direction to the sides 
 of a closed polygon. Then since the 
 sum of the projections of the sides 
 of a closed polygon is zero, and since 
 the projections of two equipollent 
 segments are equal, therefore the 
 sum of the projections of the given 
 segments will vanish. 
 
 A number of given segments to- 
 gether with a segment equal, but 
 of opposite sense, to their resultant, 
 form a system of segments whose 
 resultant is zero. Hence the following proposition : 
 
 The projection of the resultant of a number of given segments is 
 equal to the sum of their projections. 
 
 From this we at once conclude that : 
 
 If two series of segments have equal resultants., the sum of the 
 projections of the segments of the one series is equal to the sum of 
 the projections of the segments of the other. 
 
 43. Let A^B-L, A 2 B 2 , ...,A n B n be n given segments in a 
 plane, whose resultant is zero (Fig. 35). If we take an 
 arbitrary point as pole, then we may suppose A r B r to be the 
 resultant of the segments A r O, OB r -, therefore the resultant 
 of the segments A,0, 03 lt A 2 0, OB 2 , ... , A n O, OB n will vanish 
 (Art. 38), i.e. the resultant of the segments OA lt OA 2 , ..., OA n 
 is equal to that of the segments OB-^ 2 , ...,OB n . 
 
 Conversely. Given two groups of n points A^ , A 2 , ...,A n , 
 and lt B 2 , ..., B n ; if the resultant of the straight lines 
 OA l , OA 2 , ..., OA n , obtained by joining any pole to 
 the points of the first group, is equal to the resultant of the 
 straight lines OB } , OB 2 , ... , OB n , got by joining the same pole 
 to the points of the second group : then the resultant of the 
 segments A l B l , A 2 B 2 , ,.., A n B n , which join the points of the 
 
30 
 
 GRAPHICAL ADDITION. 
 
 [44- 
 
 one group to those of the other, is zero. (It is here supposed 
 that the points of the one group can only be properly united 
 to those of the other, when no point is left out, or used more 
 than once.) In fact it follows from the proposition of Art. 37 
 
 Fig. 35- 
 
 that the resultant of the segments Afl , A 2 0, ...,A n 0,OB l , 
 OB 2 , ..., OB n is equal to zero ; but the resultant of A r O and 
 OB r is A r B r , therefore also the resultant of the segments 
 A l B l , A 2 B 2 , . . . , A n B n is zero. 
 
 44. Hence it follows from the first proposition (Art. 43), 
 that when a new pole 0' is assumed, the resultant of the 
 segments 0'A lt 0'A 2 , ..., 0'A n , is equal to the resultant of the 
 segments O'_# l5 0'B 2 , ..., 0'B n . 
 
 Wherefore * 
 
 If, for two groups of n points A i , A 2 , . . . , A n ; B , B 2 , . . . , B n 
 and a fixed pole 0, the resultant of the segments OA , OA 2 , ..., OA n 
 is equal to the resultant of the segments OB 1 , OB 2 , . . . , OB n ; then 
 the same equality holds for any other pole 0' '. Moreover the 
 resultant of the n segments, which join the points of the one group 
 with those of the other taken in any arbitrary order, is equal to zero. 
 
 45. Retaining the supposition just made as to the two 
 groups of n points, project them into the points J/, A 2 , ..., A n ', 
 -^i> -^2') -Bn on a straight line r by means of rays parallel 
 to any arbitrarily chosen direction. Now take the pole 
 on the straight line rf, then we may suppose the ray OA r to 
 
 * GKASSMAN, Die Ausdehnungslehre (Leipzig, 1844), p. 41. 
 f See Fig. 35, and imagine the straight line r so displaced, that the points 
 and 0' coincide. 
 
-47] GKAPHICAL ADDITION. 31 
 
 be formed by combining the two lines OA' r , A' r A r , and so on ; 
 the resultant of the segments OA{, OA 2 , ..., OA U ', A 1 f A 1 , 
 A 2 A 9 , ...,A n 'A n is therefore equal to the resultant of the seg- 
 ments OJ?/, OS,', OSJ, ... OS,,', S&, 2 '3 2 , ..., S,;S n . But 
 (Art. 41) the resultant or sum of the segments OA{, OA 2 ', ... 
 OA n ' is equal to that of the segments OB{, OB 2 , OBJ since 
 all these segments are the projections of two other series of 
 segments, whose resultants are equal ; 
 Therefore 
 
 If for two groups of n points A lt A 2 , fyc. ; _Z? l5 B 2 , fyc. and a 
 fixed pole 0, the resultant of the segments OA 1 , OA 2 , fyc. is equal to 
 the resultant of the segments OB l , OB 2 , fyc., and if we project all 
 the points by means of rays parallel to an arbitrarily chosen direction 
 on to the same straight line, the sum of the projecting rays of the points 
 of the one group is equal to the sum of the projecting rays of the 
 points of the other group. 
 
 46. So far, we have been speaking of the resultant of a 
 number of segments, considering only their magnitude, 
 direction, and sense, but not their absolute position. We shall 
 now give a more general definition, which includes the 
 one previously given (31), and takes account of all the 
 elements of the resultant straight line of a number of given 
 segments. 
 
 If n segments A l B l , A 2 B 2 , . . . , A n B n , are given (in sense, 
 position, and magnitude) their resultant will mean a segment 
 AB of such magnitude, position, and sense, that, for any 
 pole 0, the area of OAB is equal to the sum of the areas 
 OA^ + OA 2 B 2 + ... + OA n B H (26, 30). 
 
 47. For shortness we shall call the triangle OAB, the triangle 
 which joins the segment AB to 0. The sense AB of this seg- 
 ment shows the way in which the circuit OAB is traced out, 
 and therefore shows the sense of the area OAB. 
 
 This being premised, our definition may be expressed as 
 follows. By the resultant of a number of given segments, we mean 
 a segment such that the area of the triangle which joins it to an 
 arbitrary pole 0, is equal to the sum of the areas of the triangles 
 which join the given segments to the same pole. 
 
 Since the area of the triangle OAB does not change, if we 
 displace the segment AB along the straight line on which it 
 lies, therefore the resultant of a number of segments will not 
 
GRAPHICAL ADDITION. 
 
 [48- 
 
 change, if we displace each of them in an arbitrary manner 
 along the straight lines on which they respectively lie. 
 
 48. We know already from Art. 26 that if we construct a 
 polygonal circuit CDE . . . MN, the sides of which are respect- 
 ively equipollent to the given straight lines A L B L , ..., A n B n , 
 then its closing side NC is equipollent to the resultant AS. 
 If the circuit is closed, i. e. if N coincides with C, but if the 
 sum of the areas A l B + A 2 B 2 4- . . . OA n B n is not zero, then 
 the magnitude of the required resultant is zero and it is 
 situated at an infinite distance. If the circuit is closed, and 
 the above sum also zero, then the magnitude of the resultant 
 is still zero, and. its position is indeterminate. In this case 
 therefore it may be asserted that the given series of segments 
 has no resultant. 
 
 49. But if C does not coincide with N, then the problem 
 is uniquely solved by a segment AB of finite magnitude, 
 situated at a finite distance. As we already know its magni- 
 tude, its direction, and its sense, it will be sufficient, in order 
 to completely determine its position, to find one point in the 
 straight line of which it forms a part. For this purpose we 
 may use either the construction in Art. 29, or else the much 
 simpler one following (Fig. 36). 
 
 Fig. 36. 
 
 We begin by constructing a polygonal circuit, with its sides, 
 which we shall now denote by 1, 2, ...,n, respectively, equi- 
 pollent to the given segments ; then their resultant is equi- 
 pollent to the segment , which closes the circuit, taken in 
 the opposite sense, i.e. it is equal, but of opposite sense, to 
 
-49] GKAPHICAL ADDITION. 33 
 
 the segment which joins the final point of the side n to the 
 initial point of the side 1. We now choose at pleasure a pole 
 U, and draw from it the rays UT 01 , U7 12 , ..., UF n0 * to the 
 vertices of the circuit; where 77.i+i means the vertex which 
 is the final point of the side i (equipollent to A i B i \ and the 
 initial point of the side i+ 1 (equipollent to A i+l .Z? /+1 ). 
 
 We next construct a second polygonal circuit with its 
 vertices 1,2, ..., n lying respectively on the lines to which 
 the segments A-^B^ A 2 B 2 , ..., A n B n belong, and with its 
 sides 01, 12, ..., nO respectively parallel to the rays UT 01J 
 UT[ 2 , ..., U7 HQ . The extreme sides 01,wO of this polygon will, 
 if sufficiently produced, meet in a point which lies on the 
 required line of the resultant f . 
 
 Proof. We suppose the segment A l S l to be resolved into 2 
 others, situated in the sides 01, 12 of the second polygon, 
 and equipollent to the rays T 01 U, Uf^. 2 of the first; in like 
 manner we suppose the segment A^B 2 resolved into two 
 others, situated in the sides 12, 23 of the second polygon, 
 and equipollent to the rays T\ 2 U, UY^ of the first ; and so on, 
 till finally A n B n is resolved into two segments situated on the 
 sides n 1 -n, nQ, and equipollent to the rays 7 > n _ 1 >n 7, U"F H0 . 
 
 If we take any pole , then the area of the triangle, which 
 joins it to one of the given segments, is equal to the sum of 
 the two triangles which join its two component segments 
 to the same pole ; and consequently the resultant of the n 
 given segments A l B l , A 2 B 2 , . . . , A n B n coincides with the re- 
 sultant of the 2 n component segments into which the given 
 ones have been resolved. Now the first of these 2 n segments 
 is situated on 1 , and equipollent to 7 Ql U, and the last is situated 
 on n , and equipollent to UJ' n , whilst all the rest, 2 (n 1 ) 
 in number, are equal to one another in pairs, are of opposite 
 sense, and are situated on the same side of the second polygon. 
 For example, the second and third component segments lie 
 on the side 12, and are respectively equipollent to UT 19 and 
 
 f n u. 
 
 The areas of the two triangles, which join these pairs of 
 segments to 0, are equal to one another but of opposite 
 sense ; the resultant of the given segments is therefore no other 
 
 * In Fig. 36 all the letters V, A, B are left out, and n = 4. 
 t CULMANN, 1. c. ; Nos. 41 & 42. 
 D 
 
34 
 
 GRAPHICAL ADDITION. 
 
 [50- 
 
 than the resultant of the first and last component segments, 
 of which the first is situated in 01 and equipollent to T' 01 U, 
 and the other is situated in nO and equipollent to UP n0 . 
 But the resultant of two segments passes through the common 
 point of (Art. 28) the straight lines to which they belong, 
 therefore the required resultant passes through the common 
 point of the two extreme sides 01 , ?iO of the second polygon. 
 
 50. If the pole U were taken in a straight line with the 
 two extreme points J r olt 7 nQ of the first polygon, then the 
 two extreme rays VF ol , UT^ would coincide, and therefore 
 the two extreme sides 01, nO of the second polygon would be 
 parallel. In this case therefore the construction would not 
 give a point at a finite distance in the required resultant. 
 But this inconvenience could at once be remedied by choosing 
 a new pole U f not lying in the straight line ^ 13 ^ and then 
 proceeding as above. 
 
 61. Even if that is not so, it may happen (Fig. 37) that the 
 points V nQ and T' 01 coincide, and then, wherever U may be, 
 
 Fig- 37- 
 
 the extreme rays coincide, and therefore the sides 1 , #0 
 are either parallel or coincident. If they are parallel, the 
 sum OA 1 B 1 + CL4 2 7? 2 + &c. is equal to the sum of the two 
 triangles whose common vertex is 0, and whose bases are 
 equipollent to the equal and opposite rays F ol U, UT' Ql and 
 lie in the sides 01 , 0, or is equal to the half of the parallelo- 
 gram (Art. 15), of which those bases are the opposite sides. 
 
-54] 
 
 GEAPHICAL ADDITION. 
 
 35 
 
 In this case the resultant is zero and situated at an infinite 
 distance; and the sum OA l B 1 + &c. has a constant value not 
 zero, wherever (at a finite distance) the pole may lie. 
 
 52. If, on the contrary, the sides (Fig. 38) 01, nO coincide, 
 i.e. if the opposite sides of the parallelogram coincide, then the 
 sum OA 1 B 1 +OA 2 B 2 +&e. vanishes for every position of the 
 pole 0. In this case any one segment taken in the reverse 
 sense is the resultant of the remaining (nl) segments. 
 
 53. If we take the given segments A^B^ A 2 B 2 , ... all parallel 
 to one another, then the first polygon ? 01 f 12 T 23 ... ? n0 
 (Fig. 39) reduces to a straight line, but the construction of the 
 
 1 V 
 
 Fig. 39- 
 
 Fig. 38. 
 
 second polygon is just the same as in the general case. The 
 resultant is parallel to the components. 
 
 54. If there are only two segments A^B^, A 2 B 2 , the con- 
 struction may be simplified as follows (Figs. 40, 41). In the 
 unlimited straight line A 1 B l . . . take a segment CD equipollent 
 to A 2 B 2 , and in the unlimited straight line A 2 B 2 ... a segment 
 Cflf equipollent to A^B V Then the common point of the 
 straight lines CD', and CfD lies on the required resultant. For 
 if we draw If IE parallel to C'D and join to E, then C, D, E 
 represent the vertices F Ql , J 12 , 7 20 of the first polygon, and 
 takes the place of the point U ; the points D' , E are the vertices 
 1, 2 of the second polygon, which is here represented by the 
 
 D 2 
 
36 
 
 GEAPHICAL ADDITION. 
 
 triangle OD'E, and represents also the point of intersection 
 of the extreme sides of this second polygon. 
 
 -C' 
 
 XD' 
 
 Fig. 40. 
 
 
 Fig. 41. 
 
 From the similar triangles OCD, OD'C' we have 
 OC': OD= C f If -.DC 
 
 that is to say : 
 
 The ratio of the distances of the resultant of two parallel segments 
 from these segments is the negative reciprocal of the ratio of the 
 component segments. 
 
CHAPTER III. 
 
 GRAPHICAL MULTIPLICATION. 
 
 55. To multiply a straight line a by the ratio of two other 
 straight lines 6:c, we must find a fourth straight line so such 
 that the geometrical proportion holds : 
 
 c : I a : x. 
 
 For this purpose it is sufficient to construct two similar 
 triangles OLM and O'PQ with the following properties. 
 
 In the first there are two lines (two sides, or base and 
 altitude, and so on) equal or proportional to c, I ; and in the 
 second the line homologous to c is a ; then co is the line in the 
 second triangle homologous to b ; or else 
 
 In the first there are two lines proportional or equal to c 
 and a ; and in the second the line homologous to c is b ; then 
 x is the line of the second triangle homologous to a. 
 
 56. The relative position of the two triangles is purely 
 a matter of choice ; and the particular choice made gives rise 
 to different constructions. The choice will be chiefly deter- 
 mined by the position occupied by the given segments a, b, c, 
 or of that which we wish x to occupy. 
 
 (a) In (Fig. 42), for example, the two triangles have the angle 
 in common and the sides opposite to it parallel. If in them 
 
 Fig. 42, 
 
 Fig. 43- 
 
 we take OP, OM, OL to represent the segments , I, c, then 
 OQ = x. But if OL = c, OP = a, LM = I, then PQ = x. 
 
38 
 
 GKAPHICAL MULTIPLICATION. 
 
 [57- 
 
 (b) In (Fig. 43), on the contrary, the sides opposite to the 
 common angle are antiparallel, i.e. the angles OML and OQP 
 are equal (and therefore also the angles OLM and OPQ). 
 
 (c) (Fig. 44). We may take c and a to be the altitudes of the 
 two triangles ; and then, on the supposition that b is a side 
 ON or LM of the first triangle, OQ or PQ will be equal to x. 
 
 (d) Or again, let c and a be represented by OL, OP, or by 
 OM, OQ, and let b be the altitude of the triangle OLM, then x 
 is the altitude of the triangle OPQ. 
 
 M 
 
 M 
 
 Fig. 44. 
 
 (e) If (Fig. 45) the lines OM = b and O'P = a are drawn per- 
 pendicular to one another, supposing that <? >#, we may proceed 
 as follows. Construct the triangle OLM, so that the side LM 
 is parallel to O'P, whilst the hypothenuse OL = c. Then if 
 we drawPQ parallel to OL, and O'Q perpendicular to PQ, the 
 right-angled triangles OLM, O'PQ are similar because of the 
 equal angles L and P ; and therefore O'Q x. 
 
 The straight line O'Q, the orthogonal projection of O'P 
 upon a straight line at right angles to OL, is called the Anti- 
 projection of O'P on OL. If therefore a and b are perpen- 
 dicular to one another, then x is the antiprojection of a upon <?. 
 
 57. To divide a straight line a by the ratio of two other 
 straight lines b : c, is just the same as to multiply a by the 
 ratio c : b. 
 
 The division of a straight line a into n equal parts is the 
 same thing as multiplying a by the ratio c : b, where c is 
 an arbitrary segment, and b is equal to c repeated n times. 
 
 If a straight line b is to be divided into parts, which are 
 proportional to the given segments a lt a 2 , a 3) ...,a n , lying in 
 
-59] 
 
 GEAPHICAL MULTIPLICATION. 
 
 39 
 
 the same straight line, then we have only to multiply these 
 segments by the ratio b : c, where c = a l + a 2 + ... + a n 
 (Art. 59). 
 
 58. If we draw from a centre or pole radii vectores, each 
 consecutive pair of which contains the constant angle w, their 
 lengths forming the Arithmetical Progression, 
 
 a, a + b, a + 2d, &c., &c. ; 
 
 then their extremities 31, J/ x , J/ 2 , &c. are points on a curve, 
 called the Spiral of Archimedes ; which is the name given to the 
 curve described by a point M which moves uniformly along the 
 radius OM whilst the radius itself rotates about with constant 
 velocity, in such a manner that M describes the rectilinear seg- 
 ment b in the same time that the radius OJf describes the angle o>. 
 
 If we take the angle o> small enough, we shall obtain 
 points sufficiently close together to be able to draw the curve 
 with sufficient accuracy for all practical purposes. 
 
 After we have drawn the Spiral of Archimedes, we are able 
 to reduce the problem of dividing an angle to that of the 
 division of a straight line. For if two radii vectores are drawn, 
 which enclose the angle we wish to divide into n parts pro- 
 portional to n given straight lines, we need only divide the 
 difference of the radii vec- 
 tores into n parts propor- 
 tional to the same magni- 
 tudes ; and then the dis- 
 tances of the point 
 from the n\ points of 
 division will be the lengths 
 of the n\ radii vectores 
 to be inserted between 
 the two given ones, in 
 order to obtain the divi- 
 sion of the angle. Fig. 46 
 shows the division of the 
 angle MOM 5 into five 
 
 Fig. 46. 
 
 equal parts *. 
 
 59. If several segments AB, AC, ... , BC, ... of a straight line 
 u have to be multiplied by a constant ratio 6 : c, the problem 
 
 * PAPPUS, Collectiones Matliematicae, Lib. iv. Prop, xx, xxxv. 
 
40 
 
 GRAPHICAL MULTIPLICATION. 
 
 [61- 
 
 resolves itself into finding a series of points A', B', C', &c. on 
 another straight line u' ; such that the equations 
 A'B f A'C' B'C' b . 
 
 -Tn= ^777= ~T^ = ... = - hold. 
 
 AB AC BC c 
 
 The straight lines u, u' are called similar point-rows, and the 
 points A and A', B and B',..., and also the segments ^J5 and 
 A'B'..., are said to be corresponding. 
 
 60. If the straight lines n, 
 n are parallel (Fig. 47), then 
 all the joining lines A A', BB', 
 CC', &c., pass through a fixed 
 point (the centre of projec- 
 tion). If, for example, we make 
 AP = c, ^'P' = b, then ^^' and 
 PP' give by their intersection 
 the point 0, and every radius 
 vector drawn through the point 
 cuts u and u' in two corre- 
 sponding points. 
 
 61. If u and u f are not parallel (Fig. 48), and if their common 
 point represents two coincident corresponding points A and 
 A', then the straight lines BB*, CC f . &c. are all parallel to one 
 another. The common direction of these parallel lines may be 
 
 Fig. 48. 
 
 found, for example, by taking AB = c, A'B' = b ; then every 
 straight line parallel to BB' cuts u and u f in two corresponding 
 points. 
 
 62. If, finally (Fig. 49), u and u' are not parallel, and their 
 common point represents two non-corresponding points P, Q', 
 then all the straight lines AA\ BB', CC' , ... are tangents to the 
 same parabola. If, for example, we take PQ = c, P'Q' = b, 
 then the parabola is determined from the fact that, it must 
 
-62] 
 
 GEAPHICAL MULTIPLICATION. 
 
 41 
 
 touch u in Q, and u' in P f . Every tangent of this parabola 
 cuts u and u in two corresponding points. 
 
 In order to obtain pairs of corresponding points, such as A 
 and A', B and B f , &c., we need only draw from the different 
 
 Fig. 49. 
 
 points A", B", &c. of the straight line P'Q the straight lines A' A, 
 B"B, &c. parallel to u' , and the straight lines A" A', B"B', &c. 
 parallel to u. For then clearly we have 
 
 A'B' P'Q' AB _ PQ m 
 ' A"B" " P'Q ' 
 
 A f 7?' P' O f 
 and therefore ^=r = 
 
 Fig. 50. 
 
 If we wish to avoid drawing parallels * it is sufficient to 
 consider two tangents (Fig. 50) of the parabola as given, i. e. 
 two straight lines u, u", in which two similiar point-rows (they 
 
 * COUSINEKY, Le calculpar le trait (Paris, 1840), p. 20. For another method 
 of solving this problem see SACHERI, Sul tracciamento delle pur.tegyiate projettive 
 aimili (Atti dell' Accademia di Torino, Novembre, 1873). 
 
42 
 
 GRAPHICAL MULTIPLICATION. 
 
 [63 
 
 may be equal) ABODE, ..., A"B"C", &c. are so situated that 
 the common point of the two straight lines represents two 
 non-corresponding points E, A", and that the segment AE of u 
 [which is contained between the parabola and u"~\ is equal to 
 the denominator c of the given ratio. If we want now to 
 multiply the segments of u by the ratio I : c, we must place a 
 line A'E' of length b between u and u" in such a manner that 
 it joins two corresponding points BB". Then the straight 
 lines CC", DD", &c. which join corresponding points of u and 
 u", determine upon A'E' the required segments 
 B'C':C'D':D'E'iA'W = 
 BC : CD : DE : AE. 
 
 If, for example, it were required to divide a given length 
 BE into n equal parts, we should draw through B the straight 
 line u'> and lay off on it n + 1 equal segments, 
 
 A'B'=B'C'= C'D',&c.; 
 
 then having joined E to E' we should, in like manner, lay off 
 upon the joining line u", n+I segments each equal to EE f , or 
 A"B"= B"C" = C"D"= & c ., &c. The n + 1 straight lines C'C", 
 D'D", &c., &c. will meet BE in the required division-points 
 C, D, &c. 
 
 63. Let (Fig. 51) a l , a 2 , . . . , a n be n segments given in magni- 
 tude, direction, and sense, which have to be respectively 
 multiplied by the ratios 
 
 *i, i, &c., !. 
 
 C\ C 2 C n 
 
 We construct a polygonal circuit P at whose sides are re- 
 spectively equipollent to the given segments a lt a 2 , &c., and 
 call its successive vertices 1, 12, 23, ... nl.n, n, beginning 
 
63] 
 
 GKAPHICAL MULTIPLICATION. 
 
 43 
 
 at the initial point of the first side a 1 and ending with the 
 final point of the last side a n . 
 
 Then we construct two other circuits P c and P uc ; of which 
 the first is formed by the n straight lines 1, 2, ..., respectively 
 parallel to the sides of P a , and at the respective distances ^ , 
 c 2 , . . . , c n from them, each measured in a constant direction which 
 may be fixed arbitrarily, provided that c r be not parallel to a r * ; 
 and the second P ac must have its vertices 1 , 2 , . . . , n respect- 
 ively upon the sides of P ci and its sides 1, 12, 23, ...,nl.n,n-\ 
 must respectively pass through the similarly named vertices 
 of P a . The combination of these three circuits is called ' the 
 First Figure.' 
 
 Now construct a ' Second Figure,' which similarly consists 
 of three circuits P x , P b , P xb , having the following properties 
 (Fig. 51*): 
 
 Fig. 51 a. 
 
 i. The sides of P x are respectively parallel to the sides of 
 P a \ the sides of P b parallel to those of P c (and therefore to 
 those of P a and P x ) ; the sides of P xb to those of P ac . 
 
 1, The distances of the sides 1, 2, ... , n of P b from the 
 similarly named ones of P x are # 15 # 2 , ... , b n measured parallel 
 to the distances c l9 c 2 , ..., c n . 
 
 3. Each of the vertices 1, 2, ..., n of P xb lies on the 
 
 * According as c r is positive or negative, we draw the straight line r to the 
 right or left of a person who travels along a r in the sense belonging to that segment. 
 
 t The side 1 is that which goes through the vertex 1 ; the side 12 joins the 
 vertices 1, 2 ; . . . ; the side n passes through the vertex . In order to construct this 
 polygon, we can take the side 1 at pleasure, provided it passes through the vertex 
 lof P fL . 
 
44 GKAPHICAL MULTIPLICATION. [64- 
 
 similarly named side of P b , and each of its sides 1 , 12,23, 
 ..., nl n, n must pass through the similarly named vertex 
 of P x . 
 
 In order to construct the * Second Figure ' we may, for ex- 
 ample, proceed thus. The vertex 1 of P x is taken arbitrarily, 
 and through it two straight lines are drawn, respectively 
 parallel to the side a^ of P a , and the side 1 of P ac . These 
 determine the positions of the side 1 of P x and the side 1 of 
 P xb If now, at a distance l l from the side 1 of P x , a straight 
 line is drawn parallel to this side, then this line will be the 
 first side of P b , and the point where it meets the side 1 of P xb 
 will be the vertex 1 of P xb . 
 
 From this point draw (parallel to the side 1 2 of I ac ) the side 
 12 of P xb) then the intersection of this with the side 1 of 
 P x , will be the vertex 12 of P x . From this point we draw the 
 side 2 of the polygon P x in the direction of the segment a 2 , 
 and afterwards the side 2 of P b in the same direction, but at a 
 distance 2 from it, then the intersection of this with the side 
 12 of P xb , gives the vertex 2 of P xb , and so on. 
 
 The polygon P x , whose sides we shall call #!, # 2 , ...,#, 
 gives the result of the required multiplication. For the 
 triangle, which has x r for its base, and for its opposite angle 
 the vertex r of P xb) is, on account of the parallelism of the 
 sides, similar to the triangle of the ' First Figure/ which has 
 a r for base, and the vertex r of I ac for the opposite angle. The 
 dimensions of these triangles in the chosen directions are l r , c r) 
 and therefore 
 
 Q. E. D. 
 
 64. With regard to the sense of the segment x r we remark, 
 that the two triangles are similarly situated when c r and b r 
 have the same sense, i.e. when the vertices r lie both to the 
 right or both to the left of the bases (a r or a? r ) respectively 
 opposite to them ; if, on the contrary, c r and l r are of opposite 
 sense, then the two triangles have opposite positions. For this 
 reason in the first case the segments a r and x r are of the same 
 sense, in the second case of opposite sense. Hence it follows, 
 that the segments x are placed consecutively taking account 
 of their sense, i.e. in the way which is required by Geo- 
 
-65] 
 
 GKAPHICAL MULTIPLICATION. 
 
 45 
 
 metrical addition. Wherefore their resultant, i.e. the re- 
 sultant of the segments a r -, will be, in magnitude, sense, and 
 
 C r 
 
 direction, the straight line which closes the polygonal contour 
 P x (i.e. the straight line which joins the initial point of x to 
 the final point of #). 
 
 65. Special cases. 
 
 Let all the segments a become parallel ; then each of the two 
 circuits P a and P x reduces to a rectilinear point-row (Fig. 52) 
 
 Fig. 52. 
 
 Fig. 520?. 
 
 and each of the circuits P c and P b becomes a pencil of parallel 
 rays. That is to say, the construction reduces to the following. 
 We set off the consecutive segments 01= ls 12 = 2 , 23 = a z , 
 &c., along a straight line a ; parallel to this line and at 
 distances <? x , c 2 , c 3 , . . . , c n (measured in some constant direc- 
 tion, different from the direction of the as, but otherwise 
 arbitrary) we draw as many straight lines 1 , 2, . . . , ft, which we 
 may consider as rays of a pencil whose centre lies at infinity ; 
 
46 GRAPHICAL MULTIPLICATION. [66- 
 
 then draw a polygonal circuit, with its vertices 1, 2,...,n on 
 the similarly named parallel rays, and with its sides 01 , 12, 
 23, ..., nl n, n passing through the corresponding points 
 0, 1, 2, ..., n of the point-row a [i.e. through the extremities 
 of the segments 15 2 , ...,]. 
 
 Now construct the second figure, by drawing, first, a pencil 
 of rays 1, 2, ...,#, parallel to a, and at distances b lt b 2 , ..., b n 
 respectively from a straight line x (also parallel to a); and 
 then a circuit whose sides are respectively parallel to the 
 sides of the first polygon, and whose vertices fall on the rays 
 of the second pencil. The segments 01, 12, 23, &c. of x, which 
 are enclosed between the successive sides of this new polygon, 
 will be 
 
 a?! = a x . , # 2 = a 2 , # 3 = a B , &c., respectively, 
 c i J ^2 ^3 
 
 and the segment, that lies between the side r 1-r and the 
 side S'S+1, is equal to 
 
 i=r 
 
 In the case just considered it is an immediate deduction 
 from the remarks made about the sense of the segment a? r , 
 that two segments a? r , oc s have the same or opposite sense, 
 according as amongst the three pairs a r a s , l r b s , c r c s , an even 
 number (none or two) or an odd number (one or three) are 
 formed by segments of opposite sense. This agrees with the 
 rule of signs in algebraic multiplication. 
 
 66. If all the cs become equal, in addition to all the a's 
 being parallel, then the first pencil of rays reduces to a 
 single straight line, and therefore all the vertices of the first 
 polygon coincide in a single point of this straight line ; i.e. 
 the first polygon degenerates into a pencil of rays proceeding 
 from a point situated at a distance c from the straight line a. 
 
 In this case the problem may be stated thus. To reduce 
 the given products a^ . b lt a 2 . # 2 ,..., a n . l n to a common base c t 
 by determining the segments a? 15 a? 2 , ..,, x n proportional to 
 them. 
 
 The solution is as follows (Fig. 53). Draw the rectilinear 
 point-row a, whose consecutive segments are 01 = a 3 , 1 2 = a 2 , 
 * JAEGEK, Das Graphische Rechnen (Speyer, 1867), p. 15. 
 
-67] 
 
 GEAPHICAL MULTIPLICATION. 
 
 47 
 
 &c., &c., and join each of the points 0, 1, 2, ..., nl, n of 
 this point-row to a point taken at a distance c from a ; 
 the distance being measured perpendicularly, or obliquely 
 
 Fig. 53- 
 
 Fig. 53 a- 
 
 at pleasure. Then construct a pencil of rays 1, 2, ...,w, 
 parallel to a, and at distances b l9 2 ,..., #, respectively, 
 measured in the direction of c from a given line x, also 
 parallel to a. Finally, draw a polygon whose vertices fall 
 respectively on the above-mentioned parallel rays 1 , 2, . . . , n, 
 and whose sides 01, 12, 23, ..., nl.n, n are respectively 
 parallel to the rays 00, 01, 02, ..., OnI, On of the pencil 
 0. The segments 01, 12, 23, ..., which the sides of this 
 polygon intercept upon the straight line #, will be the re- 
 quired segments a? 1} a? 2 , .,., x n *. 
 
 67. If instead of all the c's, all the b's are equal, and all the 
 as still parallel, then the problem may be stated thus : 
 
 Given the ratios 
 
 a, a,, a,, 
 
 to determine segments x l , # 2 , a? 3 , ..., # M , proportional to 
 them, so that the product of the multiplication of any x by its 
 
 /? 
 
 corresponding ratio - shall be the constant segment b. 
 
 * CULMANN, 1. C., No. 2. 
 
48 
 
 GRAPHICAL MULTIPLICATION. 
 
 [67- 
 
 After we have constructed (Fig. 54) the point-row a, with 
 the segments 01 = 15 12 = a 2 , &c., and the pencil of rays 
 1, 2, 3, ..., n parallel to the straight line a t their respective dis- 
 tances from it being c 1 , c 2 , c 3 , &c. (all measured in a constant 
 direction), we draw a polygonal circuit, whose vertices 1, 2, ... , 
 n fall on these rays respectively, and whose sides 1, 12, 23, ..., 
 n I n, n pass through the similarly named points of the 
 point-row a. 
 
 We then construct a second pencil of rays, diverging from 
 a point 0, and respectively parallel to the sides of the poly- 
 
 Fig. 54 a. 
 
 54- 
 
 gonal contour ; finally, cut this second pencil by a straight line 
 a?, parallel to a, and at a distance b from measured in the 
 direction of the <?'s. The segments 01, 12, 23... which we thus 
 obtain on x are those required. 
 
 The first and last sides of the polygonal circuit intersect in 
 a point whose distance from a in the direction of the cs we 
 
 shall call d, and then we shall have -- = 2 - . For it is clear 
 
 d c 
 
 that 2 - = -7-, and from the two similar triangles, one of which 
 
 is bounded by a and the first and last sides of the polygonal 
 circuit, and the other by x and the first and last rays pro- 
 ceeding from 0, we have = -. 
 
 cl U 
 
 This problem is substantially the same as that of trans- 
 forming a number of given fractions, 
 a. 
 
 o a 
 
 &c.. 
 
-68] 
 
 GEAPHICAL MULTIPLICATION. 
 
 49 
 
 rp fy* 
 
 into equivalent ones =,-, &c., with a common denom- 
 inator b. 
 
 68. PROBLEM. To multiply a straight line a by the ratios 
 b* b n $,. 
 
 j. j 
 
 Draw (Fig. 55} two straight lines or axes bb, cc which cut 
 one another, at any angle whatever, in the point 0. 
 
 Fig. 55- 
 
 From set off along the first axis the segments b, and along 
 the second the segments c, so that we have on the first axis 
 
 bb: 0\ = b l , 2 b% , On = b n ; and on the second axis 
 
 cc: 01 q, 02 = c 2 , On = c n . Join the homonymous 
 
 points of the two axes, i.e. 1 and 1, 2 and 2, and so on, and 
 parallel to the joining lines draw through the same number 
 of straight lines 1 1 , 1 2 , 1 3 &c. (which are only denoted in the 
 figure by the numerical index). Two segments b r , c r with the 
 same index, and the line rr which joins their final points, form 
 a triangle. Construct a triangle similar to this, in which 
 the two sides corresponding to c r and rr are set off from 
 along cc and l r respectively ; the third side corresponding to 
 b r and parallel to bb, is called a r . In order to completely deter- 
 mine this triangle, we need only fix one side, that lying on 
 cc ; this is equal to a in the first triangle, a^ in the second, a 2 
 in the third, and a n-l in the last. Then a n , that is, that side 
 of the last triangle which is parallel to bb, is the result of the 
 multiplication we wished to perform. 
 
 For comparing the r ih triangle of the second set, of which the 
 sides parallel to cc and bb are a r - 19 a r , with the similar triangle 
 
 E 
 
50 GKAPHICAL MULTIPLICATION. [69- 
 
 of the first set, whose corresponding sides are c r and b r , we 
 have : = ; and therefore 
 
 a <?! ! 2 -i ,. 
 
 Multiply all these equations together, and we have 
 
 q. E. D. 
 
 69. We shall now prove that the result is not altered by the 
 interchange of two factors, for example, and . Taking 
 them in the order and , the construction is as follows 
 
 C l C 2 
 
 (Fig. 56) : on cc take A = a ; from A draw a parallel to bb, 
 cutting I, in A, ; the segment AA, = a, , carry this over on to 
 cc, i.e. set off OA 1 = a l , and from this new point A l draw a 
 parallel to 66, cutting 1 2 in A 2 \ the segment A, A 2 thus obtained 
 
 is a. 2 . Now take the factors in the other order and ; 
 
 C 2 C, 
 
 and proceed as follows : 
 
 Make OA a as before, and 
 draw through A a parallel to 
 66, let it cut 1 2 in the point A 2 , 
 and call the segment so ob- 
 tained a' '; then set off on cc 
 the segment OA 2 = a' ', and draw 
 A 2 A, parallel to 66 cutting I, 
 in A^\ the straight line A 2 A, 
 ri s- 56. is a ff . The similar triangles 
 
 contained between I, and cc, OA, A 2 , OA, A give the following 
 
 relations, A A OA //' // 
 
 ^2^*1 "-^2 ' 
 
 and the similar triangles OA,A 2 , OAA 2 lying between 1 2 and 
 cc, give in like manner 
 
 A,A 2 AA 2 a 2 a' 
 
 TJA^ := OA> cr ^ = 7' 
 
 And therefore a"= a 2 . Q. E. D.* 
 
 * EGGEES, Grundzilge einer grapMschen AritJimetik (Scliaffhausen, 1865), p. 12. 
 JAEGER, 1. c., p. 11. 
 
-71] 
 
 GRAPHICAL MULTIPLICATION. 
 
 51 
 
 70. In constructing the triangles of the first set, instead of 
 setting off the segments b on the straight line bb, we might, after 
 taking on cc the side 01 = c l , find on Ob a point 1 , such that the 
 joining line 11 would be equal in absolute length to b r Then 
 having drawn through 0, ^ parallel to 1 1 we might,, as above, 
 construct a triangle of the second set, similar to 01 1, setting off 
 
 on cc a side equal to a. Then the product (Fig. 57) a l = a . 
 
 Fig- 57- 
 
 is given not by the side parallel to ll t but by the side lying 
 on /j ; and similarly for the other triangles. 
 
 In this construction the signs of the segments b are not 
 taken account of, since they are all set off in different directions ; 
 it is therefore necessary in carrying over the segments, for 
 example 1} upon cc, in order to proceed with the construction 
 of the next triangle in the series, to give a the same sign as a, or 
 the opposite according as b^ and c l have the same or opposite 
 signs. 
 
 In this method the segments 15 a 2 , ..., , which we have 
 respectively obtained on l lt / 2 , ..., l n (the parallels to b l9 -b 2 , 
 ..., l n ), are carried over to cc by means of circular arcs de- 
 scribed around as centre. 
 
 71. A third method of performing the required multipli- 
 cation is as follows. Set off from the common point along- 
 one of the two axes (bb) the segments b l , # 3 , # 6 , &c., and 
 c 2 , c 4 , C G , &c. ; and along the other axis (cc) (Fig. 58), b 2 , 4 , &c. 
 and c 1 , <? 3; c 1 .-. &c., always joining the extremities 11, 22, 33, &c. 
 
 E 2 
 
GRAPHICAL MULTIPLICATION. 
 
 [72 
 
 of the segments b and c with the same index. Then it is only 
 necessary to inscribe between the two axes a crooked line 
 whose successive sides are respectively parallel to the 
 joining lines 11, 22, &c., and whose vertices lie alternately on 
 cc and bb. 
 
 Fig. 58. 
 
 If we take the first vertex, so that it is the final point of 
 that segment of cc which is equal to a, and has its initial point 
 at 0, then the second vertex, and the third, fourth, &c., are 
 likewise the final points of the segments 
 
 b-, b<y b 
 
 #, == a ) a = a-, > a* = a^ &c., 
 
 r 2 r 2 r 
 
 c 1 <? 2 c 3 
 
 whose common initial point is *. 
 
 This is evident, when we consider that in this construction 
 all the triangles of the second set have one side on cc and the 
 other on bb, whilst the third side, on the crooked line, is parallel 
 to the third side of the similar triangle of the first set. 
 
 72. When there is no need to take account of the signs of 
 the segments , d, c, i. e. when they may all be considered 
 positive, we can so order. the construction, that both the 
 triangles of the first and of the second sets are placed con- 
 secutively around a common vertex (Fig. 59) (just like a 
 
 * In Figs. 58, 59, and the following, each of the segments whose common 
 initial point is 0, is marked at its final point with the letter a, b, or c, which 
 indicates its measure. 
 
72] 
 
 GKAPHICAL MULTIPLICATION. 
 
 53 
 
 fan). Through draw n+1 straight lines, or radii vectores, 
 
 making arbitrary angles with one another. Between the first 
 
 and second radii construct the first triangle of the first set, 
 
 and the first of the second ; between the second and the 
 
 third radius vector the second triangles of both sets ; between 
 
 the third and the fourth radius vector the third triangles ; 
 
 and so on ; in such a manner that two consecutive triangles 
 
 of the second set always have 
 
 one side in common. That is to 
 
 say, starting from 0, take on 
 
 the first radius two segments 
 
 equal to a and ^ respectively; 
 
 on the second radius a segment, 
 
 with the same initial point, equal 
 
 to 1 1 ; join the final points of & 19 
 
 c\, and draw a parallel to the 
 
 joining line through the final 
 
 point of the segment a, this de- 
 
 termines a segment a l on the 
 
 second radius, such that 
 
 = a- 
 
 Fig. 59. 
 
 Now take, in the same way, the segment c 2 on the second 
 radius, and the segment 2 on the third radius, and we deter- 
 mine on the latter a segment 
 
 ^2 ^1 ^2 O 
 
 a 9 = a-, = a . &c. 
 c 2 q c 2 
 
 Continuing this construction, we finally get, on the (+ l) th 
 radius, a segment with its initial point at 0, whose value 
 will be 
 
CHAPTER IV. 
 
 POWERS. 
 
 73. IF, in the last problem, we make all the 's equal to one 
 another, as also all the cs, then the constructed segment a n is 
 
 the result of multiplying a by the n ih power of the ratio - 
 In this case, either in the first construction Art. 68 (Fig. 60), 
 
 Fig. 60. 
 
 or in the second, Art. 70 (Fig. 61), all the triangles of the 
 first set coincide, and form a single triangle, two of whose 
 
 Fig. 61. 
 
 sides are the given segments b and c. The n triangles of 
 the second set are all similar to one another and to the 
 
POWERS. 
 
 55 
 
 single triangle Obc. The sides lying on Oc are respectively 
 a, a l9 2 , ..., a n - v whilst the sides parallel to b are a lt a 2 , 
 ..., a n) and therefore 
 
 b ,b* f b* f b. 
 
 *! = -, * = *(-), %=(-),..., = (-) ' 
 
 This series of similar triangles can also be prolonged on the 
 opposite side, so as to give the product of a by the negative 
 
 powers of - In fact, constructing the triangle whose side 
 parallel to b is equal to a, the side which lies upon Oc is 
 
 6 b ~ l 
 
 next, constructing a triangle with its side parallel to b = _ x , 
 
 7 _ O 
 
 the side on OC = a_ 2 = # (-) > and so on*. 
 
 74. In the third method (Art. 71) the triangles of the first 
 set reduce to two equal, but differently situated, triangles Obc 
 (Fig. 62) ; the one has its side c on the first axis and its 
 
 Fig. 62 
 
 side b on the second ; whilst the other has its side b on the 
 first axis and c on the second. The directions of the third 
 sides are therefore antiparallel, and the sides of the crooked 
 line, inserted between the two axes, are parallel to them. 
 
 * EGGERS, 1. c., p. 15. JAEGER, 1. c., pp. 18-20. 
 
56 POWERS. [75 
 
 The vertices of this crooked line determine on the first axis 
 segments, measured from 0, which have the values 
 
 and on the other axis 
 
 Moreover the sides of the crooked line form a geometrical pro- 
 gression ; for, if we call the first side a 1 ', the second is a' - , 
 
 u 
 72 7 ^ 
 
 the third '(-), the fourth a' (-) , &c. 
 
 Hence we conclude that the given segment, which has to be 
 multiplied by (-\ , instead of being set off on the first axis, 
 
 may be placed in the angle between the axes so as to form the 
 first side of the crooked line; its (?-fl) th side will then be 
 the result of the multiplication. 
 
 Continuing the crooked line in the opposite direction we 
 obtain the products of the given segment (a or a'] by the 
 negative powers of the given ratio 
 
 If we wish to continue the progression between two suc- 
 cessive sides of the crooked line, for example, between the 
 
 two first (of and a' . - ) , then we need only draw between them 
 
 a new crooked line, whose sides are alternately parallel to 
 the axes ; and we obtain a figure analogous to the foregoing 
 one. 
 
 Let the segment of the first axis, which is intercepted by 
 the first two sides of the first crooked line, be called #", then 
 the sides of the new crooked line are respectively 
 
 *","*, "(^,&c. t 
 75. Finally, if we employ the fourth method of construction 
 
 * COTTSINERT, 1. c., p. 24, 25. 
 
 t COTTSINERY, 1. C., p. 24. CULMANN, 1. C., No. 3. 
 
75] 
 
 POWERS. 
 
 57 
 
 (Art. 72) and take the angle between consecutive radii vectores 
 constant (Fig. 63), all the triangles of the first series become 
 equal and their vertices (opposite 0) lie on two concentric 
 circles whose radii are respectively equal to b and c. The 
 triangles of the second series are all similar to one another, 
 because each is similar to the corresponding triangle of 
 
 Fig. 63. 
 
 the first series ; their vertices (opposite 0), and their sides 
 (lying opposite 0} are the vertices and sides of a polygonal 
 spiral circuit. 
 
 The radii vectores of this spiral, i.e. the straight lines drawn 
 from to the vertices, are the terms of a geometrical pro- 
 gression 2 
 
 a, % = a - , a 2 = a ( - J , &c. 
 
 This progression may be continued in the opposite direction, so 
 as to give the products of a by the negative powers of ( - ) : 
 
 Also the sides of the polygonal circuit form a geometrical 
 progression with the same common ratio - *. 
 
 JAEGER, 1. c., p. 20. 
 
58 
 
 POWERS. 
 
 If the constant angle between two consecutive radii vectores 
 is a commensurable fraction of four right angles, which has the 
 denominator p when reduced to its lowest terms, then the 
 
 Fig. 64. 
 
 (p + l) th radius coincides with the first, the (p + 2) ih with the 
 second, and so on. If, for example, the given constant angle 
 were a right angle * ; the angles between every pair of conse- 
 cutive sides of the spiral polygon would also be right angles 
 (Kg. 64). 
 
 * REULEAUX, Der Constructeur, 3rd edition (Braunschweig, 1869), p. 84. 
 K. VON OTT, Grundziige des grapMschen Rechnens und der yraphischen StatiJc, 
 (Prag.1871), p. 10. 
 
CHAPTER V. 
 
 EXTKACTION OF EOOTS. 
 
 76. CONSIDER the spiral polygon ABCDEFG ... (Fig. 65), 
 whose radii vectores OA, OB, OC, OD, &c. represent the pro- 
 ducts of a constant segment OA by the powers (corresponding 
 
 to the indices 0, 1,2, 3, &c.) of a given ratio - = jr- , and 
 
 c \J A 
 
 whose sides AB, BC, CD, &c. subtend a constant angle at the 
 
 Fig. 65. 
 
 pole (Art. 75). As already remarked, all the elementary 
 triangles, which have for a vertex and a side of the polygon 
 as base, are similar ; also all the figures, obtained by combining 
 2, 3 or 4, &c. consecutive triangles, are similar, because they 
 are made up of the same number of similar and similarly 
 situated triangles. Therefore all the angles ABO. BCO, CDO, 
 &c. are equal; also the angles AGO, BDO, CEO, &c. ; and so 
 on. In general all the triangles around the vertex 0, the 
 bases of which are chords, joining the extreme points of the 
 same number of consecutive sides of the polygon, are similar ; 
 these chords also subtend equal angles at the pole 0. 
 
 These properties are quite independent of the magnitude 
 
60 EXTRACTION OF ROOTS. [77- 
 
 of the angle AOB, which in the construction of the first 
 elementary triangle is chosen at pleasure. They would not 
 therefore cease to be true if this angle were made infinitely 
 small : in which case the polygonal circuit becomes a curve. 
 From the similarity of the elementary triangles we have 
 already deduced the equality of the angles at the bases OAB, 
 OBC, &c. ; but if the angles at the point become infinitely 
 small, the sides of the elementary triangles lying opposite 
 to will become tangents to the curve ; the curve obtained 
 has therefore the property, that its tangents (produced in 
 the same sense, for example, in that of the increasing radii 
 vector es) meet"*" the radii vectores, drawn from the pole to 
 the point of contact, at equal angles. From this property 
 this curve is called The Equiangular Spiral f . 
 
 77. Since the figures, which are made up of an equal 
 number of successive elementary triangles, are similar, so 
 also, if we draw in the equiangular spiral the radii vectores 
 OA, OB, OC, &c., at equal angular intervals, the triangles 
 OAB, OBC, OCD, &c., will be similar to one another. There- 
 fore the radii vectores in question form a geometrical pro- 
 gression, i.e. the polygonal circuit ABCD... inscribed in 
 the spiral is exactly the same as the one constructed by 
 the rule of Art. 75, starting from the elementary triangle 
 
 AOB. If therefore we 
 take the triangle A OB at 
 pleasure, and construct 
 the polygonal circuit 
 ABCD..., all its vertices 
 lie on the same equi- 
 angular spiral with its 
 pole at 0. Hence it 
 follows, that the pole 
 and two points of the 
 curve completely deter- 
 mine an equiangular spiral. 
 
 78. Any two points B, C (Fig. 66) of an equiangular spiral, 
 
 * COUSINEKY, 1. C., p. 41, 42. CULMANN, 1. C., No. 5. 
 
 t WHITWORTH, The equiangular spiral, its chief properties proved geome- 
 trically (Oxford, Cambridge, and Dublin Messenger of Mathematics, vol. i. p. 5, 
 Cambridge, 1862). 
 
-80] EXTKACTION OF ROOTS. 61 
 
 the pole 0, the point of intersection T of the tangents at those 
 points, and the point of intersection N of the corresponding 
 normals, are five points on the circle whose diameter is NT. 
 Of the truth of this we are easily convinced if we consider, 
 (i) that the circle drawn on NT as a diameter will pass 
 through the points B and C, since the angles NET and 
 NCT are right angles ; (2) that the angles OBT and OCT being 
 supplementary (since the angle made by a tangent with the 
 radius vector drawn to its point of contact is constant), the 
 four points OTBC belong to the same circle. Hence it 
 follows that NOT is a right angle. 
 
 79. Now take the points B and C so close together, that 
 the spiral arc between them can be replaced by a circular 
 arc. Since this arc must touch BT and CT in the points 
 B and C, its centre lies at N ; the tangents BT, CT are equal, 
 and therefore the chord BC is bisected at right angles by the 
 straight line NT; hence also, N and T are the points of 
 bisection of the arcs BC of the circle OBC, i.e. OT is the 
 internal, and ON the external bisector of the angle BOC. 
 The point N, which will serve as a centre from which to 
 describe the arc BC substituted for the spiral arc, can there- 
 fore be constructed as the extremity of that diameter of the 
 circle OBC, which is perpendicular to the chord BC. The 
 centre P of the next arc CD, which must be the point of 
 intersection of the normals at C and D t will be the point of 
 intersection of the straight line CN 9 with the straight line 
 which bisects the chord CD at right angles, or with the 
 external bisector of the angle COD. And so on. 
 
 80. From this we obtain a construction for the equiangular 
 spiral by means of circular arcs. We divide (Fig. 67) the 
 angular space (four right angles) round the pole into a 
 certain number of equal parts, so small that the spiral arc 
 corresponding to each part can be replaced by a circular arc. 
 On two consecutive radii vectores points A and B are taken, 
 through which the spiral must pass. The centre M for the 
 arc AB is then the end of that diameter of the circle OAB, 
 which is at right angles to the chord AB. Let N be the point 
 where BM cuts the external bisector of the angle between OB 
 and the next radius vector. With the centre N describe the 
 arc BC. Similarly let P be the point, in which CN cuts the 
 
62 
 
 EXTRACTION OF EOOTS. 
 
 [81- 
 
 external bisector of the angle between 0(7 and the radius vector 
 immediately following it ; then with P as centre we describe 
 the arc CD ; and so on*. 
 
 Fig. 67. 
 
 81. Instead of assuming the point A (as well as and B), 
 we may suppose the constant angle between the tangent 
 and radius vector to be given. In this case, having drawn 
 BS inclined to OB at the given angle, let 8 be the point 
 of intersection of this tangent B8 with the internal bisector 
 
 of the angle between OB and its pre- 
 ceding radius vector ; then the point 
 A is given by the intersection of that 
 radius vector with the circle OBS. 
 After we have found the point M 
 of this circle, which is diametrically 
 opposite to S, we proceed with the 
 construction in the manner explained 
 above f. 
 
 82. We are often able to avoid 
 drawing these circular arcs, and to 
 restrict ourselves to finding a series 
 of points on the curve sufficiently 
 near together to be united to one 
 
 Fig. 68. 
 
 another by a continuous line. For this purpose we take the 
 
 * For this construction I am indebted to Prof. A. SAYNO, of Milan, 
 f For this construction I ana also indebted to Prof. A. SAYNO. 
 
-84] 
 
 EXTRACTION OF ROOTS. 
 
 63 
 
 elementary triangle OA 1 B 1 (Fig. 68), of which the angle at is 
 very small, and between the sides OA t and OB l construct the 
 crooked line A l B l 6^ D 1 . . . , with its sides alternately parallel 
 and anti parallel to A l B r Then, upon the radii vectores 
 OA, OB, OC, &c., drawn at angular intervals each equal to 
 the constant angle A^ B v take points A, B, C, &c., in such a 
 manner, that OA l = OA, OB l = OB, &c. 
 
 83. This spiral when drawn serves for the solution of 
 problems involving the extraction of roots. 
 
 We want, say, the i ih root of the ratio between two given 
 
 segments a it a. Write a { = a (-) , then the question becomes 
 
 7 ^ 
 
 that of finding the ratio - Take on the spiral (Fig. 69, 
 
 where i = 5) the radii vectores a and a { , and divide the angle 
 included between them into i equal parts. The i 1 dividing- 
 radii vectores a^ , a 2 , &c,, will be the intermediate terms of a 
 geometrical progression of i + 1 terms, the first of which is 
 a and the last a. t . The ratio a i :a of the two first terms is 
 
 therefore the required ratio (-) . 
 
 84. Two radii vectores containing a constant angle have 
 a constant ratio. From this 
 
 it follows, that, if we take 
 the sum or difference of the 
 angles contained by two pairs 
 of radii vectores a 1 ^ and 
 a 2 b 2 , the resulting angle is 
 contained by two radii vec- 
 tores, whose ratio is in the 
 first case equal to the pro- 
 duct, in the second to the 
 quotient, of the ratios a : : ^ , 
 and a 2 : b 2 . That is to say, the 
 equiangular spiral renders 
 the same service in graphical 
 
 Fig. 69. 
 
 calculations which a table of logarithms does in numerical 
 methods. The ratios of the radii vectores correspond to 
 the numbers, the angles to their logarithms. 
 
 On account of this property the curve we are speaking of is 
 also called the Logarithmic Spiral. If we take a radius vector 
 
64 EXTRACTION OF ROOTS. [85- 
 
 equal to the linear unit as the common denominator of these 
 ratios, it is obvious that the radii vectores themselves may be 
 considered instead of their ratios to unity. 
 
 If, for example, we wish to construct the segment x, given 
 by the equation 
 
 x=^/a l .a 2 a n , 
 
 then SG is the radius vector of the spiral, which makes with the 
 radius 1 an angle equal to the arithmetic mean of the angles, 
 which the radii # ! , # 2 , ... a n make with the same radius 1 . 
 
 85. But when the extraction of a square root only is 
 wanted, instead of employing the spiral, it is much easier to 
 use the known constructions of elementary geometry. If, 
 for example, x = Vab, we construct x as the geometric mean of 
 the segments a and b. 
 
 If the segments OA = a, OB = b are set off on a straight 
 line in the same sense, then (Fig. 70) as is the length of the 
 tangent OX, drawn from to a circle described through A and 
 B ; or (Fig. 700) a circle may be drawn with diameter = OA 
 (the greater segment), and then x is the chord OX, whose pro- 
 jection on the diameter is the other segment b. 
 
 Again, if the segments OA a, OB = b lie in a straight 
 line, but have opposite sense (Fig. 71), we describe a semi- 
 circle on AB, and then x is the ordi- 
 nate erected at the point . 
 
 86. The same ends for which the 
 equiangular spiral serves, are easily 
 A attained by using another curve called 
 7 1 - < The Logarithmic Curve! 
 
 Draw (Fig. 72) two axes Ox and Oy ; and on the first of 
 them, starting from the origin 0, take the segments 00, #1, 
 02 , 03 , &c., respectively equal to the terms 
 
 m f m \ 2 f m \ 3 
 
 -, * 2 = *> (- ) , # 3 = *o (-.) > 
 
-86] EXTKACTION OF BOOTS. 65 
 
 of a geometrical progression, of which the first term is # , and 
 the common ratio (where m is supposed greater than n) ; 
 
 and on the second axis take the segments 00,01, 02, 03, 
 &c., also measured from 0, 
 and respectively equal to 
 the terms y = , y^ = I, 
 
 O 7 o 7 
 
 2/2 = yz~^'i &c.j 01 
 
 an arithmetical progres- 
 sion, with its first term 
 equal to zero, and the 
 common difference* = /. 
 The terms of the two 
 progressions, which corre- 
 spond to the index r, are 
 
 Fig. 72. 
 
 and therefore 
 
 ,m 
 
 "A. &{\ I "~~* 
 
 Between each pair of consecutive terms in each of the 
 two progressions we can interpolate a new term, so as to 
 obtain two new progressions, of which the first has a common 
 
 ratio ( ) or - > and the other a common difference 
 v n ' n 2 
 
 This follows, from the fact that in every geometrical (arith- 
 metical) progression any term whatever is the geometrical 
 (arithmetical) mean between the terms preceding and follow- 
 ing it. 
 
 If we construct, for example, the geometrical mean between 
 #V, and # r+1 , an d the arithmetical between y r and ^ v+1 , we 
 obtain the two corresponding terms 
 
 of the two new progressions. 
 
 In these progressions we can in like manner interpolate a 
 
 * In the succession of numbers on Oy, the zero coincides with the origin 0, 
 because y ti was taken = 0. 
 
66 EXTRACTION OF ROOTS. [87- 
 
 term between each pair of consecutive terms, and so on, until we 
 
 i 
 
 arrive at two progressions, for which the ratio ( ) and the 
 difference . are as small, as we please *. If we use x and y 
 to denote two corresponding terms, we have always 
 
 ff\ T r (\ , 
 (I) X XQ I I , 
 
 or (a) ,- 
 
 the logarithms being taken in any system whatever. We 
 shall call those points of the axes Ox, Oy corresponding points, 
 in which corresponding segments x and y terminate. We 
 draw parallels to the axes through these corresponding points, 
 i.e. through the final point of x a parallel to Oy, and through the 
 final point of y a parallel to Ox. The straight lines so drawn 
 will intersect in a point M\ x and y are then called the co- 
 ordinates of the point M, and in particular as is called the 
 'abscissa' and y the ' ordinateJ The equation (i) or (2) ex- 
 pressing the relation between the co-ordinates of the point M, 
 is called the equation to the curve which is the locus of all 
 points analogous to M. We call this curve the ' Logarithmic 
 Curve' because the ordinate is proportional to the logarithm 
 of a number which is proportional to the abscissa. 
 
 87. We construct this curve 'by points' in the following 
 manner. After drawing the two axes Ox, Oy (Fig. 73) (usually 
 at right angles) we take on Oy a segment OB = (2 *) = I, where 
 I may be considered as the unit of the scale of lengths on Oy ; 
 
 tYl 
 
 and, upon Ox we take a' segment OA = (2*) = OC Q ~, where 
 
 n 
 
 00 = x Q is the unit of length of the scale for 0#f, and - 
 
 % 
 
 the base of the logarithmic system (the number 10). 
 
 * i is the number of interpolations. 
 
 t Since x increases much faster than y, it is convenient, in order to keep 
 the construction within reasonable limits, to take the unit x t much smaller than 
 I, for ex., Xo =l. 
 
-88] 
 
 EXTKACTION OF ROOTS. 
 
 67 
 
 Let OB be divided into 2* equal parts, and let 1,2, 3, 
 2*" 1 , ... , 2* ( = B) be the points of division. 
 
 01 2 3 4 5 6 
 
 Fig. 73- 
 
 In order to find the corresponding points of Ox, take the 
 
 M9 
 
 geometrical mean between # and # , i.e. describe a semicircle 
 
 on OA as diameter, and set off along OA, starting from 0, the 
 length of the chord of this semicircle, which has for a pro- 
 jection ; we shall thus obtain the point 2 i ~ 1 of Ox, which 
 corresponds to the similarly named point of Oy (i.e. to the middle 
 point of 0). Similarly by taking the geometrical mean of 
 and 2 *~\ and the geometrical mean between 2 *~ 1 and OA , 
 we shall obtain the points on Ose, corresponding to the middle 
 points of the segments 02*" 1 , and 2 i ~ 1 of Oy ; and so on. 
 
 Now draw parallels to Oy through the points of division of 
 Ox, and parallels to Ox through the points of division of Oy ; 
 the points, in which the lines drawn through similarly named 
 points intersect, lie on the logarithmic curve we are con- 
 structing. Since to the value y = y = 
 corresponds the value x = OC Q = , the 
 curve passes through the point marked 
 on Ox. 
 
 88. It is also very easy to construct 
 a tangent to the curve at any one of 
 its points (Fig. 74). Let M and N be Fi s- 74- 
 
 any two points on the curve, at a small distance from one 
 
 F 2 
 
68 EXTRACTION OF ROOTS. [89 
 
 another ; J/P, NQ parallels to Ox , MR a parallel to Oy , and T 
 the point in which Oy is cut by the chord MN. The similar 
 triangles TPM and MRN give 
 
 or TP : MP = OQ- OP : NQ-MP. 
 
 Let OP = y, PQ = h, then MP and NQ are the abscissae 
 corresponding to the ordinates y , ^ + /, and therefore 
 
 .v y+fe 
 
 Jff -=' *-='. 
 
 whence TP = = I 
 
 Now let the point N approach continually nearer and nearer 
 to the point M, i. e. until li approximates to the value zero, 
 then J\ J MTwi\l also continually approach towards the position 
 of the tangent at M, and the segment TP , the projection of 
 TM upon Oy, has for its limiting value what is usually called 
 the ' subtangentJ But the limit * which the fraction 
 
 h 
 1 
 approaches, when h tends towards zero, is the natural 
 
 /W) AM 
 
 logarithm of , which we shall denote by \ ; therefore in the 
 n J n 
 
 limit we have / 
 
 TP = -~ 
 
 n 
 i.e. the subtangent is constant for all the points on the curve f. 
 
 Hence it follows, that a single construction suffices for 
 drawing tangents at all the different points on the curve. 
 
 89. Having thus constructed the logarithmic curve, we 
 can solve by its aid all those problems for which the 
 ordinary logarithmic tables are used. We want for instan ce 
 
 * BALTZER, Elements der MathematiJc, i. 4ed. (Leipzig, 1872), p. 200. 
 f SALMON, Higher plane curi-en, 2nd ed. (Dublin, 1873), p. 314. 
 
89] EXTRACTION OF BOOTS. 69 
 
 to construct the f th root of the ratio between two straight lines 
 p,q. Take upon Ox the abscissae of = p, x" = q, and find by 
 means of the curve the corresponding ordinates y' and y". 
 
 The abscissa corresponding to the ordinate -{tf y"} has the 
 
 value r /In 
 
 *oA/ - 
 
 V q 
 
 Secondly, say we want the ? >th root of the product of the r 
 straight lines p l , p 2 , . . . , p r . Take on Ox the abscissae x^ =p 1 , 
 x 2 =p 2 , &c., and find the corresponding ordinates y 15 y 2 , 
 yz y r 5 t nen the abscissa a? corresponding to the ordinate 
 1 
 
 T 
 
 is equal to the required quantity 
 
CHAPTEE VI. 
 
 SOLUTION OF NUMEEICAL EQUATIONS*. 
 
 90. LET , fl x , 2 , . . . , a n be n + 1 numbers given in magnitude 
 and sign, and let (Fig. 75) a polygonal right-angled circuit 
 be constructed, the lengths of whose successive sides 01, 12, 
 
 23, ... are proportional to the given 
 numbers. The sense of each side 
 is determined by the following law : 
 the r ih and the (r + 2) ih sides, which 
 are parallel to one another, have the 
 same or opposite sense, according as 
 the signs of the numbers a r _ lt a r+l , 
 which are proportional to these sides, 
 are unlike or alike f. 
 
 Fig- 75- 
 
 Assume a point A 1 in the straight line 12, and take 
 OA 1 as the first side of a second right-angled circuit of n sides, 
 whose respective vertices A lt A 2 , A 3 , ... lie on the sides 12, 
 23, 34, ... of the first circuit. 
 
 * LILL, Resolution grapMque des equations mtmerique <Xun degre quelconque a 
 line inconnue (Nouvelles Annales de Mathe'matiques, 2 e serie, t. 6, Paris, 1867), 
 p. 359. 
 
 t In order to fix with the greatest possible precision the sense of each side of 
 the crooked line, the following convention is useful. We take two rectangular 
 axes XOX, YO Y } and determine for each of them the positive sense ; and we 
 agree to give the number, which expresses the length of a segment, the coefficient 
 + 1, or 1, according as it is in the positive or negative direction of XOX, and 
 the coefficient +i, or i (where i */ 1, i. e., % 2 1), according as it is in 
 the positive or negative direction of YO Y. Now let a circuit be formed whose 
 successive sides, 
 
 01, 12, 23, 34, 45, 56, ... . 
 a 0) ia, i?a 2 i 3 a 3 , * 4 4 , i 5 a 5t ... , 
 a , ia l} a 2 , ia 3 , a t ,ia it ... , 
 sides will be parallel to XOX, and the others to YOY-, 
 
 are equal to 
 i.e. equal to 
 then the l^ } 3 rd , 5 th , 
 
 moreover two parallel sides, separated by a single side at right angles to both, 
 will have the same, or opposite sense according as the corresponding numbers a 
 have opposite signs or the same sign. 
 
SOLUTION OF NUMEKICAL EQUATIONS. 
 
 71 
 
 The triangles QlA l ,A l 2A 2 ,A 2 3A 3 ,A 3 4A 4: ,...a l i'Q all similar 
 to one another, and therefore give 
 01 A 
 
 A 2 3 __ A 3 4 
 
 whence, remembering the identities, 
 
 , 
 
 01 = 
 12 = !, 
 23 =a 
 
 A. 
 
 A^ = A l l+a lJ 
 A3 = 
 
 n 1, ft = fl n _ 
 n.n+ 1 = , 
 
 and putting 
 we obtain : 
 
 A n .n+l = 
 
 y- = a?, or J x 1 = o?, 
 
 A 2 2 =a ( 
 A 2 3 = # 
 
 A n .n = a SB n + x x n ~ l + . . . + _!?, 
 A n .n+l = a () x n + a l x n - l +...+a n 
 
 Thus the segment ^ n .+l, included between the final 
 points of the first and second polygonal circuits, represents 
 the value which the polynomial 
 
 takes, when we substitute for x the ratio of the segment A^ 1 to 
 the segment 1 or a . Keeping a Q positive, the signs of x and 
 a l will be equal or opposite, according as A l l, 12 have the 
 same or opposite sense. 
 
 If the final points of the two circuits coincide, we have the 
 identity F(x} = ; and then x is called a root of the equation 
 F(z) = 0. The real roots of the equation F(z) = are there- 
 fore the ratios A l 1 : 1 , which correspond to those right-angled 
 inscribed circuits whose final points coincide with the point 
 
 On account of this property we say, that the circuit 
 0123 ... w-f 1, represents the whole polynomial F(z). 
 
 91. If in 01 23 ... n+ 1 we inscribe a new right-angled circuit, 
 
 4 
 
 UK. I r\ 
 
72 
 
 SOLUTION OF NUMERICAL EQUATIONS. 
 
 [01- 
 
 OS 1 S 2 ...S n , and if we denote the ratio of B L l :01 by^, we 
 shall have in like manner 
 
 Bn.n+l = F(y) = a y n + a l ?/ n -i + ...+a n . 
 For the coefficients a we substitute their values 
 a = 01, 
 
 ^ = 12 = A 1 2-A 1 1 = y42-01 .a?, 
 a 2 23 = J 2 3 J 2 2 = J 2 3 ^2.^, 
 # = 34 = A4 A3 = A4A3.x 
 
 -i = n-l.n = A n _ l .n-A n _ l .n~l = A n _ l ,n~A n __ 2 .n- 
 = A n n+ \A n n = A n n+ I A n _^ .n.se; 
 
 a n = n.n + 
 and thus obtain 
 
 t n .n+l = 
 
 + (A-i-^~A-2^-" 
 
 + (A n n + 1 A n _^n.x) 
 
 01 
 
 But B n n+l A n n + 
 and y x 
 
 therefore 
 
 01. -~-f 01 
 "i ^i 
 
 01 
 
 Fig. 76. 
 
 This result may be expressed as follows (Fig. 76, where 
 n = 6): In a rectangular circuit of n+1 sides 0123...W + 1, 
 
-92] 
 
 SOLUTION OF NUMEKICAL EQUATIONS. 
 
 73 
 
 two other rectangular circuits of n sides OA^A^... A n , OB^B^, 
 ...B n are inscribed; we t hen form a new rectangular circuit 
 01 %'3 f ...n' of n sides, which are respectively parallel to the sides of 
 the frst circuit and equal to 01, A 1 2, A 2 3,... , A n _ l n. In this 
 inscribe the rectangular circuit of n 1 sides OJ3 l B 2 '. . . J3' n ~i 5 
 having the side OB l in common with the circuit already described 
 
 Then we have 
 
 
 That is to say, the segment S / n _ l .n / is the result of the division of 
 ~F(y) F(x) by yx: where F signifies the Polynomial represented 
 by the first circuit 0123. . .n + 1, and x and y are the ratios ^1:01; 
 J&jliOl. 
 
 Or, in other words, 
 
 The circuit 01 2 / 3 / ...^ / represents the polynomial 
 
 or, in the case where x is a root of the equation F(z) = 0, the 
 polynomial F(z) : (z sc). 
 
 92. The similar triangles considered above give 
 
 so that our equation may also be written, 
 
 This result may be interpreted as follows (Fig. 77) : 
 
 A. 
 
 If we inscribe in the rectangular circuit 0123...W+1 of n-\-\ 
 
74 SOLUTION OP NUMERICAL EQUATIONS. [93- 
 
 (Fig. 77, where n = 6) two new rectangular circuits of n sides 
 OA l A 2 ...A n , OjBjj^g... .B tt , and then inscribe in the Jirst of these 
 a rectangular circuit ofnl sides OC^ C 2 . . . C n _ l , 
 
 , <v*i .V 
 
 -oA^ == -oT' 
 
 then 
 
 B, A, QA, 
 
 X JL 1 
 
 that is to say, C n _ l A n is also equal to the quotient 
 
 F(y}-F(x) 
 y-x ' 
 
 A 
 
 multiplied however by - 
 
 In other words : 
 
 The circuit OA^ A 2 ... A n represents the polynomial 
 F(z)-F(x), 
 
 z x 
 
 or, in the case where x is a root of the equation F (z) = 0, the poly- 
 nomial F(z):z x, provided the lengths be reduced in the ratio 
 OA 1 :Ol. 
 
 Every rectangular circuit of n sides inscribed in the given 
 circuit, and having the same extremities, is therefore a re- 
 solvent circuit in regard to the given one, because it represents 
 the quotient, obtained by the division of the Polynomial 
 represented by the given circuit, by one of its linear 
 factors. 
 
 93. Again, let the entire polynomial of the th degree F(z] be 
 represented by the circuit 0123...0+1. In it let the two 
 circuits OA^...A n , OB^B 2 ...B n (Fig. 78) be inscribed. We 
 assume that both the points A n , B n coincide with the extremity 
 n+1 of the given circuit; i.e. let OA 1 A 2 ...A n) OB^B 2 ...B n 
 be two resolvent circuits of the given circuit. Moreover let 
 L 19 L 2 ,... 9 L n _ 2 be the points of intersection of the pairs of sides 
 A 1 A 2 , H l 2 ; A 2 A 3 , B 2 B 3 ; ... ; A n _ 2 A n _ l9 B n _ 2 B n _-^. Then the 
 triangles QA^B^ L 1 A 2 B 2 are similar, since their corresponding 
 sides are perpendicular to one another ; for the same reason the 
 triangles A^B^L^ A 2 B 2 L 2 are similar, and therefore also the 
 quadrilaterals 0^ 1 ^ 1 Z 1 , L 1 A 2 B 2 L 2 are similar, whence it 
 follows that the sides OZ 15 L^L 2 are perpendicular to one 
 
-94] 
 
 SOLUTION OF NUMEKICAL EQUATIONS. 
 
 75 
 
 another. In the same way we show that the angles L 1 L 2 L 3 , 
 L 2 L.^L^ ...,L n ^L n _ 2 n+l are right angles. 
 
 Fig. 78. 
 
 Hence it follows that the points OL 1 L 2 ...L n _ 2 n + I are the 
 vertices of a circuit of n I sides, which is right-angled, and 
 is inscribed both in the circuit QA 1 A 2t ... 9 and in Ol^-Z^... ; 
 that is to say, OL 1 L 2 ... is a resolvent circuit in regard to each 
 of the circuits OA^..., OjB 1 2 .... In other words, if we 
 
 reduce the lengths in the ratio ~ , the circuit 
 represents the polynomial of the (n 2) th degree 
 
 L ... -Z/_ 
 
 n _ 2 
 
 where oc = OA : : 1 , y = J5 X : 1 . 
 
 94. Let an equation of the second degree 
 be given 
 
 After constructing the circuit 0123 (Fig*. 
 79), whose sides 01, 12, 23 represent the 
 
 coefficients a , # 
 
 2 , 
 
 it is sufficient, in 
 
 Fig. 79. 
 
 order to find a root, to construct a right 
 angle, with its vertex A upon 12, and its 
 legs passing through 03. We describe therefore a semicircle 
 
76 SOLUTION OF NUMERICAL EQUATIONS. 
 
 on 03 as diameter ; if this cuts 12 in the points A 19 A^ the roots 
 of the given equation will be 
 
 A 2 l 
 
 From known properties of the circle we have : 
 
 A 2 l = 2A^ 
 and therefore 
 
 = 2A 1 + A 1 1 = 21, 
 
 or 
 
 that is to say, the sum of the roots is 
 
 a o 
 
 Further the similar triangles 01^ 15 ^23 give 
 
 01:^1=^2:32, 
 or a :A 1 l=A 2 l:a 2 , 
 
 and therefore '-- = , 
 
 o 2 
 
 i.e. the product of the roots is equal to 
 
 a o 
 
 A simple apparatus depending on the foregoing theorem 
 (Art. 91) has been designed by Lill, with the object of 
 determining the roots of a given numerical equation. The 
 apparatus consists of a perfectly plane circular disc, which 
 may be made of wood ; upon it is pasted a piece of paper 
 ruled in squares. In the centre of the disc, which should 
 remain fixed, stands a pin, around which as a spindle another 
 disc of ground glass of equal diameter can turn. Since the 
 glass is transparent, we can, with the help of the ruled paper 
 underneath, immediately draw upon it the circuit corresponding 
 to the given equation. If we now turn the glass plate, the 
 ruled paper assists the eye in finding the circuit which deter- 
 mines a root. A division upon the circumference of the ruled 
 disc enables us, by means of the deviation of the first side of 
 the first circuit from the first side of the second, to immediately 
 determine the magnitude of the root. For this purpose the 
 first side of the circuit corresponding to the equation must be 
 directed to the zero point of the graduation. 
 
CHAPTER VII. 
 
 REDUCTION OF PLANE FIGURES*. 
 
 95. To reduce a given figure to a given base I, we must 
 transform the figure into a rectangle whose base is b, or deter- 
 mine a straight line /, which when multiplied by b gives the 
 area of the given figure. Instead of constructing a rectangle 
 on the base b we may construct a triangle on the base 2 b ; 
 the height of this triangle is the straight line /. The segment 
 b is called the base of reduction. 
 
 When several figures are reduced to the same base b, their 
 areas are proportional to the corresponding straight lines 
 /i /2 5 /a > &c - 5 whence it follows, that the reduction of a 
 figure to a given base is the same thing as finding its area. 
 
 Let the given figure be the triangle OAB (Fig. 80), whose base 
 OA is denoted by a, and its height by li. Then, since the area 
 must remain unaltered by the transformation, fb = \ah, 
 
 and therefore f = a . j = Ji . r , 
 
 that is to say, we have either to multiply a by the ratio Ji : 2 b, 
 or k by the ratio a : 2 b. 
 
 We therefore take OC = 2 b, B 
 
 join C to B, and draw AD 
 parallel to CB. 
 
 Or else, take on OB the 
 point D, whose distance from 
 OA = 2 b, join DA, and draw 
 BC parallel to DA. Fi s- 8o - 
 
 If we join CD, the triangles OAB, and OCD are equivalent, 
 because we obtain them, if we subtract the equal triangles 
 ADB, ADC from, or add them to the same triangle OAD 
 (according as OC is smaller than, or greater than OA). The 
 required segment / is therefore in the first construction the 
 
 * CULMANN, 1. c., No. 15 ct scqq. 
 
78 
 
 REDUCTION OF PLANE FIGURES. 
 
 [96- 
 
 height of the point D above OC, and in the second is the 
 length OC. 
 
 96. It is not necessary that one of the dimensions / or 2 #, 
 should fall along a side of the given triangle. We may take 
 as the doubled base 2b a straight line BC (Fig. 81) drawn 
 from the vertex B to the opposite side OA, provided 2 b is not 
 less than the distance of B from OA ; the corresponding 
 height f will then be OD, the antiprojection of OA on BC*. 
 Or if 2 b is not greater than OA, we can take as the doubled 
 base 2 b a chord OD of the semicircle drawn on OA as 
 diameter In this case the parallel BC to the supplementary 
 chord DA is the required height/. 
 
 p 
 
 Fig. 81. 
 
 97. Let it be required to reduce the quadrilateral ABCD to 
 the base I (Fig. 82). Draw CO parallel to the diagonal BD ; 
 then the quadrilateral reduces to the triangle OAB, and we 
 proceed as above, viz. we make BC' = 2b, and the antipro- 
 jection OB' of OA upon BC' is the required length/. 
 
 98. The reduction may also be performed without first 
 reducing the given quadrilateral ABCO to a triangle. Take 
 the diagonal OB (Figs. 83, 84, 85, 86), which must not be 
 less than 2 b, as hypothenuse, and construct the right-angled 
 triangle ODE of which the side BD = 2 b. Let the points 
 A and C be projected, by means of rays parallel to OB, 
 into A', Cf on OD, the other side of the triangle OBD ; the 
 
 * The triangles SPC, DO A are similar, hence SP : C = OD : OA, or 
 
 h : 2 b =/ : a ; therefore/ = a . -7 . 
 
 Q. E. D. 
 
-100] 
 
 EEDUCTION OF PLANE FIGUBES. 
 
 79 
 
 triangles OCB, OBA are equivalent to the two triangles OC B, 
 and OB A' ; but in each of these the distance of the base OC', or 
 
 OA', from the opposite vertex =. 2 #, and therefore the required 
 height /for the quadrilateral is equal to OC' + A'O = A'C'. 
 
 Fig. 85. 
 
 Fig. 86. 
 
 In the crossed quadrilateral (Fig. 87) if AC is parallel to 
 BO, the points A and C' coincide, and therefore /=. 0. In 
 fact, in this case the area ABCO is equal 
 to the sum of the two triangles UAB, 
 and UCO, which are of equal area but 
 of opposite sign. 
 
 99. The length / is also equal to that 
 segment of the straight line drawn 
 through A or C parallel to A C' which is 
 intercepted by the straight lines AA f , CO '. 
 
 100. The foregoing construction assumes that 2 1 is not 
 greater than the greatest diagonal OB of the quadrilateral. 
 If 2 b is > OB, the lengths 2 b and / can be interchanged. 
 We draw, namely, AE parallel to OB and make CE = 2 I ; 
 
 Fig. 87. 
 
80 REDUCTION OF PLANE FIGURES. [101- 
 
 then, construct on the hypothenuse OB a right-angled triangle 
 ODB, of which the side 01) is parallel to CE ; and then the 
 other side BD = f. 
 
 101. In order to reduce a polygon to a given base, whether 
 its periphery is self- cutting or not, we begin by reducing it 
 to an equivalent quadrilateral. We then apply the above 
 construction to the quadrilateral and thus obtain the segment 
 f, which multiplied by the base b gives the area of the 
 polygon proposed. 
 
 Fig. 88. 
 
 Let the given polygon be 0123456780 (Fig. 88). Draw 
 the straight line 8 7' parallel to the diagonal 07 , 
 
 33 3 
 
 7'6' 06, 
 
 33 33 6'5' 3J 35 05 3 
 
 54 04 
 
 33 33 33 33 U ^ 3 
 
 33 33 43 ,, ,, 03 , 
 
 and the polygon is successively transformed into the equiva- 
 lent polygons 01234567', 0123456', 012345', 01234', 0123', 
 each of them having a side less than the preceding one"*. We 
 finally arrive at the quadrilateral 0123'. 
 
 102. In this construction the new sides 07', 06', 05',... of the 
 reduced polygons are rays proceeding from the fixed vertex 
 0. But we can also proceed in such a manner, that all the 
 new vertices 7', 6', 5', &c. lie on a specified side. If we have, 
 for example, the polygon Aabcde (7012345, and if we draw 
 
 11' parallel to 20 
 
 22' 31' 
 
 33' 42' 
 
 44' 53' 
 
 52) A4' 
 
 >. till each intersects the side OC, 
 
 * The triangles 087, 077' are equivalent, because the straight lines 07, 87' are 
 parallel ; if we take the first triangle away from the given polygon, and add the 
 second one to it, we obtain the new polygon 012345 67'0. And so on. 
 
-103] 
 
 SEDUCTION OF PLANE FIGURES. 
 
 81 
 
 we determine a straight line AD which, can be substituted for 
 the crooked line A 5 43210. 
 
 For, since 11', 20 are parallel, the triangles 120 1'20 are 
 equivalent, and if we subtract the former from the given 
 polygon, and add the latter to it, the polygon reduces to 
 Aabcde Cl'2345. Similarly, from the equivalence of the tri- 
 angles 1'23, 1'2'3 the latter polygon reduces to Aabcde C2'345, 
 and proceeding in this manner we finally arrive at the poly- 
 gon Aabcde CD. 
 
 D 
 
 In order to effect a like transformation for the crooked line 
 AabcdeC, we draw 
 IV parallel to ca \ 
 
 cc' dV\ 
 
 ,7,7' O j \ till each intersects the side Aa, 
 
 35 VL 
 
 eB Cdf) 
 
 and now the whole polygon Aabcde (7012345 is reduced to the 
 equivalent quadrilateral ABCD. 
 
 103. This is the easiest and most convenient way of 
 finding the areas of figures, the perimeters of which take the 
 most different forms. With a little practice we learn to 
 perform the reduction quite mechanically, and without paying 
 any attention to the actual form of the proposed circuit. This 
 construction moreover permits us to take account of signs, 
 so that in dealing with areas of different signs, the result 
 gives the actual sign belonging to their sum without further 
 
 G 
 
82 
 
 REDUCTION OF PLANE FIGURES. 
 
 [104- 
 
 trouble*. Take, for example, the self-cutting circuit (Fig. 90) 
 ABC 01234, which represents the cross section of an embank- 
 ment and excavation in earthwork. 
 
 Draw 
 
 Fig. 90. 
 
 ' parallel to 20 , 
 
 22' 
 33' 
 4D 
 
 31', 
 42' 
 
 until they meet the side CO, then the given polygon is re- 
 duced to the equivalent quadrilateral ABCD, which there- 
 fore represents the difference between the areas ABC '2 4 of the 
 embankment and 70123 of the excavation, which have neces- 
 sarily different signs. The circuit ABCD has the same sign 
 as the circuit ABC 1 4, or as the circuit 70123, according as the 
 embankment or the excavation is the larger. 
 
 104. Circular Figures. A sector of a circle (Fig. 91) OAB is 
 equivalent to a triangle OAC, with its vertex at the centre 
 and its base a portion AC of the tangent equal to the arc AB. 
 In order to obtain approximately the length of the arc AB 
 measured along the tangent, we take an 
 arc a, so small that it can without any 
 sensible error be replaced by its chord 
 a ; we then apply the chord a to the given 
 arc AB starting from its extremity 7?, 
 and continue doing so as long as necessary 
 till we reach A or a point A' very near A. 
 Then starting from A or A' we set off the chord a the same 
 number of times along the tangent AC-\. The sector OAB is 
 now replaced by the rectilinear triangle OAC. 
 
 * CULMANN, 1. C., No. 17. 
 
 f CULMANN, 1. c., No. 21. In Chapter IX is given a method of Rankine for the 
 approximate rectification of circular arcs, and also some methods of Professor 
 Sayno. 
 
 Fig. 91. 
 
-106] 
 
 KEDUCTION OF PLANE FIGUKES. 
 
 83 
 
 The segment AB (i.e. the area between the arc AB and its 
 chord) is the difference of the two triangles OA C, OAB, and is 
 equivalent therefore to the crossed quadrilateral OB AC. 
 
 105. It is not necessary that the tangent upon which we 
 set off the arc should pass through an extremity of the arc ; 
 instead of doing so it may (Fig. 92) touch the arc at any other 
 
 point T. In such case we set off the arc AT on CT, and BT 
 on DT. The sector OAB is transformed into the triangle OCD, 
 and the segment AB is the difference between OCD and OAB, 
 i.e. is equal to the doubly crossed figure OCDOBAO, which 
 may be considered as a hexagon (Fig. 93) with two coinci- 
 dent vertices at 0. If we draw OB / and OA' respectively 
 parallel to AC emd BD, the triangles OAC, OBD are transformed 
 into the two other triangles B'AC, A'BD, and therefore the 
 segment is equal to the quadrilateral A'B'CD. 
 
 106. Example 
 
 Let the figure to be reduced be the four-sided figure ACD3, 
 contained between two 
 non-concentric circular 
 arcs, AC and 3 D, and the 
 straight lines CD, A 3 
 
 (Fig- 94). 
 
 Let 0, 1 be the centres of 
 the two circles ; the given 
 figure is then equal to the 
 sector AC the sector 
 13 D the quadrilateral 
 QAlC, Change the sectors 
 into the triangles OAB, Fir 
 
 132, by setting off the two 
 
 arcs along their respective tangents AB and 32, starting from 
 corresponding extremities A, 3 ; and now the given figure is 
 
 G 2 
 
84 
 
 REDUCTION OF PLANE FIGURES. 
 
 [107 
 
 equal to the triangle OAB the area 0.4321 Co, i.e. is equal to 
 the self-cutting polygon ABOC123A. 
 
 Draw 11' parallel to 2C\ 
 
 22' 31' ( till they cut the fixed side CO, 
 
 3C' A2'l 
 
 and the polygon reduces to the crossod quadrilateral AB C'. 
 The area If of this quadrilateral is found in the usual manner ; 
 i.e. on the diagonal AO as hypothenuse a right-angled triangle 
 is constructed, of which one side AE = 2 b ; the length f is 
 then the distance, measured parallel to the other side, of the 
 point B from a straight line parallel to AO and passing 
 through C'. 
 
 107. As another example suppose we wish to determine the 
 area of Fig. 95, which represents the cross section of a so-called 
 C/-iron. 
 
 It consists; (i) of a lime-shaped area AEA'F, bounded by two 
 circular arcs, one having ?7as centre, the other 0\ (2) of a 
 crown-shaped piece CBFIfC' bounded by two concentric cir- 
 cular arcs Btf, CCf drawn with as centre ; (3) of two equal 
 rectilinear pieces * BCJIH and B'C'JTH', symmetrically 
 situated with regard to the straight line OUFE, which is an 
 axis of symmetry for the whole figure. 
 
 r 
 
 Fig. 95- 
 
 The lune is equal to the sector UAEA' plus the quadrilateral 
 OAUA' minus the sector OAF A', i.e. it equals the sum 
 UAEA' + OA UA' + A OA 'F. After transforming the two sectors 
 spoken of into the triangles UAD, OAG (where AD, AG are 
 the arcs AEA' and AFA f set off along their respective initial 
 
 * We say rectilinear, because we suppose the small arcs CJ, C'J' to be 
 replaced by their chords. 
 
107] 
 
 REDUCTION OF PLANE FIGURES. 
 
 85 
 
 tangents), the lime becomes equal to the sum UAD -f OA UA' 
 + AOG, or finally, if we merge these three circuits in one, it is 
 equal to the area of the circuit ADUA'OAOGA, Here we can 
 neglect the part OA, which is twice passed over in opposite 
 senses ; and consequently (Art. 23) the lune is equal to the 
 self-cutting hexagon ADUA'OGA. 
 
 The crown-piece we consider as the difference of the sectors 
 0ff, OCC f . After setting off the arcs along their middle 
 tangents PP / , QQ', since PQ, P'Q' both pass through 0, the 
 crown becomes equal to the trapezium PP'Q'Q, which is the 
 difference between the two triangles OPP', OQQ', equivalent 
 to the two sectors in question. 
 
 If we now reduce the hexagon ADUA'OG, the trapezium 
 PP'Q'Q, and the two pentagons BCJIH to a common base > 
 and find their corresponding segments to be/ , / 15 2f 2 , then 
 (/o+/i + 2/2) will be the required area of the given figure*. 
 
 Or we may consider the given figure as an aggregate of 
 triangles and trapeziums made up in the following way 
 
 UAJ)+2 OAV- OAG+OPP'- OQQ' + 2BCKH+ 2CJ1K, 
 where CK is drawn parallel to BH t and JL We consider the 
 areas of these triangles and trapeziums as the products of two 
 
 Fig-. 95 a. 
 
 Fig. 95 a. 
 
 factors, and reduce these products to a base I by means of the 
 multiplication polygon (Fig. 95 a). It is, of course, understood 
 that for each area to be subtracted, one of the two factors 
 must be taken negatively. 
 
 * In Fig. 95 we find 2/ 2 directly, if in the reduction of the figure B'C'J'TH' 
 we substitute b for 2 b. 
 
86 REDUCTION OF PLANE FIGURES. [108- 
 
 108. Curvilinear figures in general*. It is a well-known 
 property of the parabola, that a parabolic segment (Fig. 96) is 
 
 equivalent to -J of the triangle, whose base 
 is that chord of the parabola which forms 
 the base line of the segment, and the 
 
 vertex of which is that point of the arc 
 Fig. 96. 
 
 where the tangent is parallel to the base ; 
 
 that is to say, the segment of a parabola is equal to a triangle 
 whose base is the chord, and whose altitude is f the Sagitta : 
 where we understand by Sagitta the perpendicular distance 
 between the chord and that tangent of the arc which is 
 parallel to the chord. 
 
 109. One method then of reducing curvilinear figures, con- 
 sists in considering each small portion of the curved periphery 
 to be a parabolic arc. 
 
 If a curved line (Fig. 97) is divided into small arcs each of 
 which may be approximately regarded as a parabolic arc, and 
 if the parabolic segments between these arcs and their respec- 
 tive chords are reduced to triangles on these chords as bases ; 
 then the vertices of all these triangles can be taken anywhere 
 at pleasure on the straight lines drawn parallel to the chords 
 at distances from them equal to their respective Sagittae. 
 
 r^ Let these vertices be taken so that 
 
 the vertex of each new triangle lies 
 on the prolongation of one side 
 of the preceding triangle, i.e. so 
 that the vertices of two successive 
 triangles and the point of intersec- 
 lg ' 97 ' tion of their bases always lie in the 
 
 same straight line. Then the curvilinear circuit is reduced to 
 an equivalent rectilinear circuit formed of sides whose number 
 is equal to that of the parabolic segments into which the given 
 circuit was divided. The rectilinear circuit or polygon is next 
 reduced to its equivalent quadrilateral, and finally this is 
 reduced to the given base in the manner previously explained. 
 
 110. Suppose, for example, we wish to replace the irregular 
 boundary line AB between two fields by another consisting 
 of two rectilinear segments which form an angle with its 
 extremities at A and B (Fig. 98). We consider the curve AB 
 
 * CULMANN, 1. C., No. 23. 
 
-Ill] 
 
 REDUCTION OF PLANE FIGURES. 
 
 87 
 
 and the straight line BA as a circuit, and reduce it to a triangle 
 on the base BA. 
 
 For this purpose we divide the curve into small arcs ; draw 
 their chords and for the segments thus formed substitute 
 triangles, by the method we have just given above. In this 
 way we transform the given circuit into the rectilinear polygon 
 AQ12345B. Then we draw 
 
 11' parallel to 20 
 
 22' 31' 
 
 33' 42' I till each cuts the fixed line Ao ; 
 
 44' 53'' 
 
 5C 34') 
 
 and thus transform the polygon into the triangle ACB. We 
 have therefore substituted the two rectilinear segments AC, CB 
 for the given irregular line. The point C can be displaced at 
 pleasure along a line parallel to AB, since by doing so we 
 do not alter the area ABC. 
 
 111. The reduction of areas to a given base furnishes another 
 construction for the resultant of a number of segments A l B l , 
 A 2 B 2 , &c., &c. given in magnitude, sense, and position (Fig. 99). 
 Take a point as the initial point of a polygonal circuit whose 
 sides are respectively equipollent to the given segments ; let N 
 be its final point. Now transform the triangles OA 1 B 1 , OA. 2 B^ 
 &c., by reducing them to a common base ON, and let them be 
 so transformed that they have a vertex at 0, and the side 
 opposite to it equipollent to ON. Then the sum, 
 
 OA^ + OA 2 R 2 + &c., &c., 
 
 will also have been transformed into a triangle OAB, where AB 
 is equipollent to ON. The segment AB is the required re- 
 sultant (Art. 46). 
 
88 
 
 REDUCTION OF PLANE FIGUKES. 
 
 In order to effect the above-mentioned transformation, it 
 will be convenient to take the initial points A ly A 2 , &c., &c. of 
 
 Fig. 99. 
 
 the segments in a line with 0. Project the points I? 15 _Z? 2 , 
 &c., &c. into the points J9/, _Z?/, &c., &c. upon ON, by rays 
 parallel to OA 1 A 2 ... , and then the triangles OA 1 B 1 , OA 2 B 2 , &c., 
 are transformed into the triangles OA l J3 l ', OA 2 B 9 ', &c. Then 
 draw the straight lines B^ C^ BJ C.^ &c., parallel to -A^ 15 ^V"^, 
 &c., respectively, and let the points in which they cut the 
 straight line OA l A 2 ... be C lt C 2 , &c. 
 
 We thus obtain the triangles OC-^N, OC. 2 N, &c., respectively 
 equivalent to OA 1 1 f , OA 9 B 2 , &c., &c. Therefore, if the seg- 
 
 ment A OC l +OC 2 
 
 is taken on OA 1 A 2 ... J and if 
 
 through A the straight line AB is drawn equipollent to ON, 
 then OAB is equal to OA 1 B 1 + OA 2 B 2 
 
CHAPTER VIII 
 
 CENTEOIDS. 
 
 112. LET us suppose that, in the theorems of Articles 43 and 
 44, all the points 1 , H 2 , ..., B n coincide in a single point G ; 
 these theorems may then be stated as follows : 
 
 If A 1 G, A 9 G, A 3 G,..., A n G are n segments, whose resultant 
 vanishes, and is any arbitrarily assumed point in the plane, the 
 resultant of tlie segments OA^ OA 2 ,..., OA n is equal (equipollent) 
 to n times the segment OG (Fig. 100). 
 
 Fig. 100. 
 
 Fig. 101. 
 
 Conversely : 
 
 Let there be given n points A l , A 2 , ... , A n) and let the resultant 
 of the straight lines OA^ , OA 2 , . . . , A n , which join the pole to 
 the given points, be equal to n times the straight line OG drawn 
 from to G, then the same equality holds for any other pole 0' ; 
 that is to say, the resultant of the straight lines 0'A l , 0'A 2 , ... , 
 0'A n is equal to n times the segment O f G; and the resultant of the 
 straight lines GA^ GA 2 , ... , GA n is equal to zero*. 
 
 113. The point G is called the Ceniroid of the points A v 
 A 2 , ...,A n . Let the (Fig. 101, where n = 4) points Aj^,A 2 , ..., 
 A n be given, to construct their centroid G we proceed as 
 follows. An arbitrary pole is taken, and a circuit 
 
 * GRASSMANN, 1. c., p. 141. CHELINI, Sui centri de* sistemi geometrici 
 Raccolta scientifica; Roma, marzo 1849), 1. 
 
90 CENTROIDS. [114- 
 
 OA : 23 ... n constructed, whose initial point is and whose 
 successive sides are equipollent to the segments OA^ OA 2 , ... , 
 OA n . The straight line On, which closes the circuit, passes 
 
 On 
 
 through the point G } and OG "-. Instead of dividing On 
 
 into n equal parts in order to obtain G, we may construct a 
 second circuit starting from another initial point 0'; the 
 straight line which closes this new circuit will cut On in the 
 
 O 
 
 required point G. 
 
 114. The system of n given points cannot have another 
 centroid G'. For if both the resultant of GA 19 GA 2) ... , GA nt 
 and the resultant of G / A 1 , G'A 2 , . . . , G'A n should vanish, then the 
 general resultant of all the segments GA 1 , A 1 G', GA 2 , A 2 G', . . . , 
 GA n , A n G' would also vanish. But if we combine the two 
 segments GA rt A r G', we obtain the segment GG / ; and therefore 
 GG' must vanish, that is to say, G' must coincide with G. 
 
 115. Again, if in the proposition of Article 45, all the points 
 B 1 , B 2 , . . . , B n are supposed to coincide with a single point G, 
 the theorem may be stated as follows : 
 
 If G is the centroid of the points 1, 2, 3, ..., n, and if all these 
 points are projected by means of parallel rays into the points G' , 
 l', 2', 3', ..., n' upon one straight line, then the sum of the straight 
 lines 11', 22', 33', ..., nn' is equal to n times the straight line GG' 
 (Fig. 102). 
 
 As a result of this proposition, since rr' is 
 the (oblique) distance of the point r from the 
 straight line upon which we project, the point 
 G is also called the centre of mean distances * of 
 the given points 1, 2. 3, . .., n. 
 
 116. Instead of supposing in the pro- 
 positions of Articles 43 and 44, that all the 
 jfl* 102 points B lt B 2 , ..., B n coincide with a single 
 
 point G, we now imagine some of them B l , _Z? 2 , 
 ..., B i to remain distinct, and the rest to coincide with a 
 single point G\ so that the resultant of the segments A^B^, 
 A 2 B 2 , . . . , A i B t , A i+1 G, . , . , A n G vanishes ; and, whatever the 
 position of may be, the resultant of OA 1 , OA 2 , ..., OA n is 
 equal to the resultant of OB lt OR 2 , ... , OB it . (n-i) . OG. The 
 
 * CARNOT, Correlation de* figures de geomttrie (Paris, 1801), No. 209. 
 
-119] CENTROIDS. 91 
 
 first of these equalities does not change, if we substitute for the 
 segment A r B r the two others A r G, GB r or A r G,-B r G\ the 
 second equality will also continue to subsist if we add to both 
 resultants the segments B^O, B 2 0, ..., B { 0, so that it becomes 
 an equality between the resultant of OA i} OA 2 , ..., OA n , B 1 0, 
 B 2 0, ..., .0 and the resultant of OB^ OB 2 , ... , OE i9 B^O, 
 B 2 0, ..., B i O t (n i).OG; that is, between the resultant of 
 OA 19 OA 2 , ..., OA n , OB 19 -OB 2 , ..., -O^and (n-i) . OG. 
 
 Hence : 
 
 If the resultant of the segments A L G, A 2 G, ... , A n G, B 1 G, 
 
 B 2 G,..., B { G vanishes; then , for any point whatever , the 
 resultant of the segments OA^, OA 2 , . .., OA n , OB l , OB 2 , ... , 
 
 OB i is equal to (n i) . OG: ami conversely if this equality sub- 
 sists for any pole 0, it will also hold for any other pole 0', and the 
 resultant of the segments A l G, A 2 G, . . . , A n G, - B^ G, - B 2 G, . . . , 
 
 Bfi will vanish. 
 
 117. Now let us assume that of the n points A I} A 2 , ..., A H 
 some coincide with one point, others (in like manner) with a 
 second point, and so on; and that the points B lt B 2 , ..., B L 
 also unite in groups and coincide. Then if we use a l9 a 2 , u s , . . . 
 to denote positive or negative integral numbers whose sum 
 is m, the foregoing proposition may be stated as follows : 
 
 Jf the points A lt A 2 , A 3 , ... and the point G are so situated, that 
 the resultant of the segments a l .A : G, a. 2 ,A 2 G, a z . A^G,... vanishes, 
 then, wherever the pole way be, m . OG ivill be equal to the resultant 
 of the segments a 1 . OA 1} o 2 . OA 2 , 3 . OA 3 , ... &c. 
 And conversely : 
 
 Jf this property holds for any pole 0, viz. that m .OG is equal to the 
 resultant of a l . OA l , o 2 . OA 2 , 3 . OA 3 , ..., then the same property 
 holds for every other pole O ' , that is to say, the resultant of 04 . 0' A l 
 o 2 . 0'A 2 , a 3 . O f A z , ... is equal to m ,0'G\ and the resultant of 
 the segments a 1 .GA l , o 2 . GA 2 , o 3 . GA 3 , ... vanishes. 
 
 118. The point G is called the centroid of the points A 1 , A 2 , A 3 , 
 ... weighted with the coefficients a lt a 2 , o 3 , .... For shortness 
 however we say that G is the centroid of the points a x . A l , 
 a 2 .A 2 , a 3 .A 3 , . . . , writing before each point the coefficient 
 which belongs to it. 
 
 119. Furthermore, from the proposition of Article 45 we 
 obtain the following theorem : 
 
 If G is the centroid of the points a l . A 1 , a 2 .A 2 , a^.A 3 , ... and if, 
 
92 CENTEOIDS. [120- 
 
 by means of parallel rays, the points G, A l , A 2 , A 3) ... are projected 
 into the points G', A\, A' 2 , A' 3 , ... which lie on a straight line, then 
 the sum of the straight lines a x . J^/, o 2 . A 2 4 2 ' t a 3 . A 3 A 3 , ... is 
 equal to m . GG\ where m = a^ + a^ + o^ + .... 
 
 On account of this property G is also called the centre of 
 mean distances of the points a l .A 1 , a 2 . A 2 , a 3 . A 3 , . . . *. 
 
 120. Hitherto the coefficients a 15 a 2 , a 3 , ... have been positive 
 or negative integral numbers; we shall now extend the idea of a 
 Centroid to the case where a 1 , o 2 ,a 3 , ... are any numbers whatever ', 
 or rather parallel segments proportional to any given homo- 
 geneous magnitudes 
 
 Let then the points A l , A 2 , A 3 , . . . be given, weighted with 
 the numbers or parallel segments a 1? a 2 , 3 , .... Project the 
 given points on to a straight line p', by means of rays parallel 
 to some arbitrarily chosen direction, into A\, A' 2 , ... ; and by 
 means of rays parallel to another direction chosen at pleasure, 
 project the same points into A'\, A" 2 , A"*, ... on a second 
 straight line p", not parallel to j/. Now determine a straight 
 line / parallel to p' t such that the distance from / to // 
 measured parallel to the rays A 1 A\, A 2 A f 2 , A 3 AS, . . ., is equal to 
 
 similarly determine a straight line /" parallel to //', such 
 that the distance from /' top", measured parallel to the rays 
 A 1 A f \, A 2 A 2 ", ..., is equal to 
 
 Let G denote the point of intersection of the straight lines 
 /, /', and G f , G" the projections of G upon the straight lines 
 p', p" (by means of rays parallel to AA', AA" respectively), 
 then we shall have : 
 
 a l .A 1 A l ' + a 2 .A 2 A' 2 + a 3 .A 3 A' 3 +... =(a 1 + a 2 + o 3 + ...). GG' 
 a, . A^A'\ + a 2 . A 2 A" 2 + o 3 . A.^\ + ... =(% + o 2 + a s + ...) . GG". 
 
 Next, let //" be any third given line, let us project upon it 
 the given points and the point G, into the points A"\, A'" 2 , 
 A'" 3,..., G'"> by rays parallel to a new direction. Between the 
 three rays which project the same point A l or J 2 or A. 6 , there 
 
 * L'HUILIEE, EUmens cVanalyse gtomttrique et d'analyse algclrique etc. 
 (Paris, 1809), 2. 
 
-122] CENTROIDS. 93 
 
 exists (Article 16) a linear relation with constant coefficients, 
 i. e. we have : 
 
 k'. A^A\ + k". A 1 A'\ + k'". A^A"\ = /;, 
 
 K An A 2 ~}~ fC -"2 2 * * >' 2 ^ = ' 
 
 ft -/I o J--L o -J- A/ . ^l o J-L o ~P A> -^J. o -^L o A 1 
 
 k'. G G' + /&". G G". + k'". G G"' = k. 
 
 Multiply these equations by 04 , a 9 , a 3 , . . ., (a x + a 2 + a 3 + . . .) 
 respectively, and add the products ; then we obtain, taking 
 the equations already established into account, 
 tf". {a l .A l A"\ + a 2 .A 2 A m . 2 + a 3 . ^ 3 A "' ^ + ... 
 
 -(04 + 02 + 03 + . ..).'"} = 0, 
 or, 
 
 QI . A.A'^ + a^ A 2 A m ,4 a 3 .A 3 A'" 3 +... = ( ai + a 2 + a 3 +...).6^ w . 
 That is to say : 
 
 If we project the points A 19 A 2 , A 3) ..., G upon any straight line 
 whatever ly means of rays, which are parallel to an arbitrarily 
 chosen direction, then the product of the ray which projects G 
 bt/ ( 1 + 2 + 3 + ...) is equal to the sum of the products formed Ijy 
 multiplying each of the rays which project A lt A 2t A B . tt ly a lt a.,, 
 3 , ... respectively. 
 
 We call the point G, so defined, the centroid of the points A 1 , 
 A. 2 , A 3 , . . . loaded with the numbers or segments 04 , a. 2 , a 3 , .... 
 
 The centroid does not change if we substitute for the coeffi- 
 cients 04, a.,, 0.3, ... others proportional to them, for by so doing 
 we do not change the ratios of 04 , a 2 , a 3 . . . , to 04 + a. 2 + a 3 + . . . . 
 
 121. If the points A 19 J 2 , A 3 , . . . , and G are projected, by means 
 of rays parallel to a straight linep" ', on to another straight line p' , and 
 if we use 0' to denote any point whatever of 'p l \ we have identically : 
 
 a, . C/A\ + a, . 0'A' 2 + a 3 . V A' ^ + . . . = (a x + a 2 + o 3 + . ..)VG'. 
 
 If we draw through 0' a straight line parallel to p", and pro- 
 ject on to it, by rays parallel toX, the points A lt A. 2 , A 3 , ... , G 
 into the points A'\, A".,, A" 3 , ... , ", we have the identities 
 A^A'\ = A\a, A. 2 A" 2 = A\V> A 3 A" 3 = A' 3 0' ..., GG" = G'0'>, 
 but from the foregoing theorem we have 
 
 and therefore the above proposition is true. 
 
 122. If is an arbitrary point , the resultant of the segments 
 a, . OA 19 a 2 . OA 2 , 0-3. OJ 3 , ..., is ( ai + a 2 + az + ...).OG. 
 
 By the segment a . OA we understand a segment parallel to 
 
94 CENTEOIDS. [123- 
 
 OA, drawn either in the sense of OA or in the opposite sense, 
 according as a is positive or negative, and whose magnitude is 
 equal to that of OA increased in the ratio of a : 1 . Draw 
 through the point a straight line p', and project upon it, by 
 means of parallel rays, the points A lt A 2 , A 3 , ..., G into the 
 points A\ , A\ ,A' 3 ,..,, G'. Then the segment OA is the resultant 
 of the segments OA', A' A, and therefore if we increase these 
 segments in the ratio a : 1, the resultant of a . OA', a . A' A will 
 be a . OA. It follows, that the resultant of c^ . OA 1 , a. 2 . OA 2 , 
 a 3 . OA 3 , ..., may be obtained by combining all the segments 
 a x . OA\ , a 2 . OA'... , a 3 . OA' B , ... with the segments a l . A^A^ , 
 a 2 .A' 2 A 2 ,a 3 .A' s A 3) .... But the resultant (i.e. the sum) of 
 a x . OA\, a. 2 . OA'. 2 , a 3 . OA' 3 ... is (a 1 + a 2 + a 3 + ...) . OG', and the 
 resultant (or sum) of c^ . A\ A l} a 2 . A' 2 A 2 , a 3 . A' 3 A 3 , ... is 
 (aj + a 2 + a 3 +...). G r G\ therefore the resultant of the segments 
 ctj . OA 1 , a. 2 . OA 2 , a 3 . 0^ 3 , . . . , can be obtained by combining 
 the two segments (%-f o 2 -\-a 3 ...) . 0r', (a 1 + a 2 + a 3 ...) . G 1 '^, and 
 consequently, it coincides with the segment (a l + a 2 + a 3 + ...).OG. 
 
 123. j?jf -H" is the centroid of the points a 1 . A 13 a. 2 . A 2 , a 3 . A 3 , ..., 
 and K the centroid of the points ft l .S l ^ (3 2 . J3 2 , . . . , then the centroid 
 of all tJie given points a 1 .A 1 , ct 2 . A 2 , . . . , f3 lt j$ lt /3 2 . J3 2 ,. . . coincides 
 with the centroid of the two points m . H, n . K, where 
 
 w = (o 1 + o 2 + ...), = (ft + /3 2 + ...). 
 
 For, taking an arbitrary pole 0, if we combine the straight 
 line m. OH, the resultant of the segments a l OA 1 , a,OA. 2 , ..., 
 with the straight line n . OK, the resultant of the segments 
 ft 1 . l , p. 2 . B 2 , ..., we find that (m + n).OG, the resultant of 
 m . Oil, and n . OJT, is also the resultant of all the segments 
 a^OA^a^OA,, ...,ft.0^ 15 /3 2 .0^, .... 
 
 124. If all the points A^, A. 2 , A 3 , ... lie on a straight line, their 
 centroid G lies in the same straight line. 
 
 This is clear, if we take the pole upon the straight line 
 A^A^A^ . . . ; for then all the segments a^ . OA l , a. ; . OA 2 , a 3 . OA 3 ,., . 
 lie in this straight line, and therefore also their resultant 
 m . OG lies in the same straight line. 
 
 From this it follows : 
 
 If we project A l , A 2 , A 3 , ..., A n , G upon an arbitrarily chosen 
 straight line into the points A\ ,A' 2 ,A' 3 ..., A' n , G', then the point 
 G' is the centroid of the points a l . A\ , o 2 . A' 2 , a 3 . A' 3 ..., a H . A' n . 
 
 Let there be only two points A 1 , A 2 (Fig. 1033 where the 
 
-125] CENTEOIDS. 95 
 
 segments a l5 a 2 are simply denoted by the numbers 1, 2) with 
 coefficients a l9 a 2 , then their centroid G is a point of the 
 straight line A 1 A 2 . Since the resultant 
 of the straight lines 04 . G A l , a 2 . GA 2 
 is equal to zero, we have 
 
 or A : G: GA 2 = a 2 : a 19 
 
 and therefore 
 
 AI G '. GA 2 i AI A 2 = a 2 i cij i c? 2 -f- QJ j 
 
 that is to say, the point G divides the segment A 1 A 2 into 
 two parts, which are inversely proportional to the numbers 
 a x , o 2 , and it lies inside or outside the given segment, according 
 as a lt a 2 have the same or opposite signs. 
 
 If aj = a 2 , then A 1 G = GA 2 , i. e. G is the middle point of 
 A^Ay If a l + a 2 = Q, we obtain from the proportion A l G : 
 A^A^ = a 2 :a 1 + a 2 the value A 1 G = cc, i.e. G is the point at 
 infinity of the straight line A l A 2 . 
 
 125. Let there be three given points A 19 A 29 A 3 not in one 
 straight line (Fig. 104); and let a l9 a 2 , a 3 , 
 whose sum is not zero, be their coefficients. 
 The centroid of the points c? 2 . A 2 , a 3 . A 3 
 is a point l on the straight line A 2 A 3 , 
 and the centroid of the given points 04 . A lt 
 a 2 . A 2 , a 3 . ^4 3 is therefore the centroid of 
 the points a x . A 1 , (a 2 + a 3 ) . -Sj , that is, it 
 is the point G on the straight line A^B^ 
 which is determined by the relation 
 
 But the triangles A 1 A 2 A 3 , G A 2 A 3 are proportional to 
 their altitudes, therefore also to the oblique distances A^B^ 
 GB l of their vertices from the common base A 2 A 3 ; therefore 
 
 Similarly we prove that 
 
 GA 1 A 2 :A 3 A 1 A 2 = a 3 : a 1 -\-a 2 -\- 3 , 
 and therefore 
 
 GA 2 A S : GA 3 A l : GA 1 A 2 = a-^: a 2 : o 3 . 
 
 That is to say ; the centroid G of the three points o l . A l , a 2 . 
 o 3 .^ 3 divides the area A^ 2 A 3 into three triangles GA^ 
 
CENTEOIDS. 
 
 [126- 
 
 GA%A ly GA 1 A 2 , which are proportional to the coefficients a 19 
 cr 2 , a 3 . 
 
 Given the points A l , A 2 , J 3 , every system of values for the 
 loads a ls a 2 , a 3 , determines a point G on the plane A 1 A 2 A 3 , and 
 conversely to every point G of the plane there corresponds 
 a fixed system of values equivalent to the above. This is the 
 principle of the calculus of the centroid of Mobius. 
 
 126. It follows (from the foregoing articles) that if we wish 
 to find the centroid G of the given points A l ,A 2 ,A^... (Fig. 1 05) 
 
 weighted with the coefficients 
 (numbers or segments) a x , a 2 , 
 a 3 , . . . , we must construct two 
 circuits starting from two 
 different initial points 0, 0' ; 
 the sides of the first being 
 equipollent to c^ . OA l , a 2 . A 2 , 
 a 3 . A 3 , . . . , and those of the 
 second to a x . O'A^ , a 2 . 0'A 2 , 
 Fig. 105. a 3 . 0'A 3 , .... The straight 
 
 lines OR, O'R' which respect- 
 ively close the two circuits, intersect in the required point 
 G, and we have 
 
 If the coefficients a l5 a 2 , &c., &c. are proportional to given 
 segments %, a 2 , &c., &c. they will also be proportional to 
 
 loads -r>-f' &c., &c. where h is any arbitrary segment; we 
 
 Us fls 
 
 can therefore make the sides of the first circuit equal to 
 the lengths T T OA L ,^OA 2 , &c., &c., and if OR is the closing 
 
 / / 
 
 line then h. OR (a l + a 2 + a 3 +...). OG. Hence it follows, 
 that G is found, without constructing a second circuit, by 
 determining on the closing line the segment 
 
 OG=- h - R '-. 
 
 If the coefficients a l9 a 2 , &c., ... are proportional to the areas 
 9 lt * 2 , &c., &c., which when reduced to a common arbitrary base 
 k are equivalent to the rectangles ka l , ka 2 , ..., &c. ; they will 
 
 * GRASSMANN, 1. c., p. 142. 
 
-127] CENTEOIDS. 97 
 
 also be proportional to the segments a x , a 2 , . . . , or to the loads 
 Y , ~r> ' ' anc ^ ^ the circuit be constructed with the sides 
 
 ib Ib 
 
 C OA OA 2 , &c.,...,&c., then OG = - ^? - . 
 
 127. If G is the centroid of the points a 1 .A 1 , a. 2 .A 2 , a 3 .A 3 , 
 , . . , and any point whatever, we have seen that the resultant 
 OR of the segments o x . OJ 15 a 2 . OA 2) a 3 . OA 3 , .,., is given by 
 the equation 
 
 whence OR 
 
 If a l + a 2 +a 5 + ...= 0, while OR is not zero, then 
 OG = oo, or the centroid is at an infinite distance. To 
 find in what direction G lies, let B be the centroid of the points 
 a 2 A 2 , a 3 A 3 , ..., a n A n . Then B l is at a finite distance, because 
 a 2 -|- # 3 4. . . . , a n is not equal to zero, but is equal to a r Let 
 0.Z2 be the resultant of a x 0-Sj and c^ 0^ x , this resultant will 
 be equipollent to a^^, that is it will be independent of 
 the point 0. Consequently the resultant of a l OA l , a. 2 OA 2 , ... , 
 a n OA n , where a l + a 2 , ..., a n = 0, is constant in direction and 
 in magnitude wherever may be, and is equipollent to the. 
 segments a^A, = a 2 R 2 A 2 = ... a n B n A n ; 
 
 where B r is the centroid of the points a^A^ a 2 A 2 , ... , a r _ 1 ^ r _ 1 , 
 a r+1 A r+lt ..., a n A n . The point at infinity common to the 
 segments BA^ B 2 A 2 , ..., is the centroid G of the given 
 points. 
 
 Let parallel straight lines be drawn through each of the 
 points A lt A 2 , ..., -Z? 1; B 2 , ... in any arbitrary direction, and 
 let them be cut by a transversal in J/, A 2 , ..., B^ B 2 ..., the 
 theorem of Art. 121 applied to the points a 2 A 2 , a 3 ,4 3 ...a n 
 to their centroid B l gives 
 a 2 A 2 A 2 +a 3 A 3 A 3 ', ...,a n A n A n ' [a 2 + o 3 , ...,oj B 1 B l ' 
 
 Therefore a^A^A{ + a 2 A 2 AJ + . . . = Oj [ J^/- ^ 1 ^ 1 / ], 
 consequently a^j J/ + a 2 ^ 2 ^ 2 7 + ... a n A n A n ' = Q 
 if the transversal is parallel to B^A^ i.e. is drawn towards the 
 centroid G at infinity. 
 
 In the particular case when OR = , or when B 1 coincides 
 
 H 
 
98 
 
 CENTEOIDS. 
 
 [128- 
 
 with A l , B 2 will also coincide with A 2 , &c., &c. The centroid 
 G is then quite indeterminate ; or, in other words, the system 
 of points a 1 A l , a 2 A 2 , ... has no centroid. The sum 
 
 c^ A 1 A 1 ' + a 2 A 2 A 2 ' + . 3 <* n d-n A^ 
 
 is then zero, whatever the direction of the parallel lines 
 A 1 A l / ) A 2 A 2 ..., and of the transversal A^A 2 ...*. 
 
 128. Through the points A 1 , A 2 ,'A 3 , ..., and through their 
 centroid G segments A l B lt A 2 B 2 , A 3 JB 3 , ..., Gil are drawn in 
 an arbitrary direction parallel to one another, and proportional 
 to the co-efficients a l3 a 2 , a 3 , . . . , m = a + a 2 + . . . , taking account 
 of signs ; that is, having chosen the positive direction of the 
 segments, let the segments proportional to the positive co- 
 efficients be drawn in that direction, and those proportional to 
 the negative coefficients in the opposite direction. Let be 
 an arbitrary point, and through it draw a straight line parallel 
 to the segments AS, and upon this line project the points 
 A 19 A 2 , A 3 , ..., G by parallel rays into A^ t A 2i A 3t .... G' ; 
 then by the theorem of Art. 1 1 9 we have 
 
 a^.A^ + a 2 .A 2 A 2 +a 3 .A 3 A 3 + ... = m.GG'. 
 But the numbers a : , a 2 , a 3 , ...,m are proportional to the bases 
 of the triangles OA^, OA 2 B 2 , OA 3 R 3 , ..., OGIT, and the 
 segments A l A l / , A 2 A 2 , A 3 A 3 ', ..., GG f to the heights of the 
 same triangles, hence the following theorem ; 
 
 The sum of the triangles which join the segments A 1 B l , A 2 B 2 , 
 A 3 J3 3 ,..., to 0, is equal to the triangle which joins the straight 
 line GH to the same pole 0. Whence it follows that GH is the 
 resultant of the segments A l B l > A 2 B 2 , A 3 B 3 , ..., (Art. 47). 
 
 129. This furnishes another construction for the centroid G. 
 After drawing through A 19 A 2 ,A 3 , ... (Fig. 106) the segments 
 
 20- 
 
 .*& 
 
 Fig. 106. 
 
 a 15 a 2 ,a 3 , ..., in an arbitrarily chosen direction, we combine 
 them in the manner of Article 53. 
 
 * MOBIUS, Bary. Calcul, 9, 10. BALTZER, Stereom., 11. 
 
-131] CENTKOIDS. 99 
 
 We shall thus obtain a straight line r, in which the result- 
 ant segment lies, and which must therefore pass through G. 
 We now repeat this combination, only changing the common 
 direction of the segments a 15 o 2 , o 3 , ..., and obtain another 
 straight line r' ; the lines r and / intersect in the required 
 centroid. 
 
 130. A figure (linear, superficial, or solid) is called homogeneous 
 if all its points are weighted with equal coefficients. Geo- 
 metrical figures are understood to be homogeneous, unless the 
 contrary is stated. 
 
 If the points in a figure are collinear two and two with a 
 fixed point, and situated at equal opposite distances from it ; 
 the fixed point is evidently the centroid of the figure. For 
 instance, the centroid of a rectilinear segment is its middle 
 point ; the centroid of a parallelogram is the point of inter- 
 section of its diagonals ; the centroid of a circle, of a circum- 
 ference, and of a regular polygon, is the geometrical centre of 
 the figure (Figs. 107 and 108). 
 
 Fig. 108. 
 
 If the figure has an axis of symmetry, that is, if its points 
 are two and two on chords bisected normally by an axis, this 
 axis will also contain the centroid. 
 
 131. Let the figure be the triangle ABC (Fig. 109). If D is the 
 middle point of BC, the straight line AD divides the area ABC 
 into two equal parts. To every point X in one half there cor- 
 responds a point X' in the other half, such that the segment 
 XX' is parallel to BC, and bisected by AD. 
 The centroid of every couple XX' is there- 
 fore on AD, hence the centroid G of the 
 area ABC lies on AD. Therefore G is the f 
 point of concourse of the three median 
 lines AD, BE, CF. It divides each of the 
 three median lines into two segments which are in the pro- 
 portion of 2:1. For, since the triangle ABD is cut by the 
 transversal FGC, we have 
 
 H 2 
 
100 
 
 CENTROIDS. 
 
 AF BC DG_ 
 
 [132- 
 
 But AF = FB, BC = 2 DC, therefore 
 
 or GD = i AD, and similarly GE = i BE, GF = CF. The 
 point G is also the centroid of the three points A, J3,C. 
 
 132. If a (linear or areal) figure is made up of a system of 
 rectilinear segments, or triangular areas, then its centroid 
 is that of the points a 1 .A lJ a 2 .A 2 , a 3 .A 3 , ..., where A^A^A^ ... 
 are the centroids of the segments or triangles, of which the 
 figure is made up, and the (numerical or segmental) co- 
 efficients a ls o 2 , a 3 , ... are proportional to the segments or 
 triangles themselves. 
 
 133. Let the figure be a circuit with rectilinear sides. Let 
 A 19 A 2 , A 3 , ... be the middle points of the sides, and a lt a 
 
 2 , 
 
 a 3 , ... segments proportional to the sides. Then, if we find 
 by one of the methods already described (Articles 126, 129) 
 the centroid G of the points A 1 , A 2 , A. 3 , ..., weighted with 
 the segments o^ , a 2 , a 3 , . . . ; G is the centroid of the given 
 circuit. 
 
 134. If the circuit is part of the perimeter of a regular 
 polygon (Fig. 1 10), its centroid can be found in a much simpler 
 way. Draw a diameter of the inscribed circle, and let the 
 sides of the circuit be projected orthogonally upon it. Let <r 
 be a side, A its projection, r the radius of the circle which is 
 drawn through the middle point of o-, and p the perpendicular 
 let fall from the latter point on to the diameter ; then the 
 right-angled triangle of which a- is the hypothenuse and A one 
 
 Fig. no. 
 
 of the other sides, is similar to the triangle, whose hypo- 
 thenuse is r and one of its other sides p. 
 
-134] CENTROIDS. 101 
 
 Therefore we have 
 
 A cr 
 
 - - or Xr = p(r. 
 
 p r 
 
 Write down equations corresponding to these for all the sides 
 of the circuit, and by addition we get 
 
 rl=p l o- l +p 2 o- 2 + ..., 
 where I is the projection of the whole circuit. 
 
 Let G be the centroid, y the perpendicular let fall from 
 G upon the diameter. Since G is the centroid of the middle 
 points of the sides, supposed to be loaded with the co- 
 efficients <r l , o- 2 , &c. respectively, we have (120, 133) 
 
 where * means the length of the whole circuit. Therefore 
 
 rl 
 
 rl = ys, i.e. y = T - 
 s 
 
 This equation gives the distance of the point G from the 
 diameter; the point G must also lie on that radius (OC) of 
 the circle, which bisects the circuit, since this radius is an axis 
 of symmetry of the circuit. Draw a straight line EF = s, of 
 which one extremity E lies on the diameter, and the other 
 extremity F upon the tangent of the circle, which is parallel 
 to this same diameter ; and then take upon EF a segment 
 EH I, and through H draw a parallel to 1)F, cutting the 
 axis of symmetry OC in G ; then, since the straight line EF is 
 cut by the parallels EO, HG, DF, we obtain : 
 EF_ _ distance of DF&om EO 
 1S~ distance of HG from EO ' 
 s r r 
 
 I distance of EO from G y 
 
 and therefore G is the required centroid. We notice in the 
 formula obtained above, that I is the projection of the (unclosed) 
 circuit upon any diameter chosen at pleasure, and y is the per- 
 pendicular distance of the point G from the same diameter. 
 
 Another construction. Upon the tangent CM, drawn at right 
 angles to the axis of symmetry OC, set off a segment 
 CM \s, join OM, and draw from that extremity (A) of the 
 given circuit, which lies on the same side of the axis of 
 symmetry as M, a parallel to OC, to cut OM in N; through N 
 draw a parallel to CM, till it cuts OC in G. In the similar 
 
102 
 
 CENTEOIDS. 
 
 [135- 
 
 s I 
 triangles OCM, OGN the bases are respectively -> -, where 
 
 2- 
 
 by I we understand the projection of the circuit upon the 
 diameter perpendicular to OC. The altitude of the first 
 triangle is r, and therefore that of the second is equal to the 
 distance of G from the centre 0*. 
 
 135. This construction is applicable even when the regular 
 polygon, of whose perimeter the given circuit is a part, has 
 an infinite number of sides, that is, when it becomes a circle. 
 Hence let the given line be an arc AB of a circle whose centre 
 is (Fig. in); let s be the length of the arc, the half of 
 which CM is set off along the tangent at 
 its middle point. Project the extremity 
 A into N upon OM by means of a parallel 
 to the axis of symmetry OC, and through 
 N draw a parallel to MC cutting OC in G, 
 then G is the centroid of the arc AB. 
 For we have 
 CM:CO = GN: GO, 
 
 A 
 
 j? 
 
 Fig. in. 
 
 therefore GO = y. 
 
 136. If the given circuit is the perimeter of a triangle 
 ABC (Fig. 112), its centroid G is the centre of the circle 
 inscribed in the triangle DEF, whose vertices are the middle 
 
 points of the sides of the given 
 triangle. For, D, E, F are the cen- 
 troids of the rectilinear segments 
 BC, CA, AB ; and therefore G is the 
 centroid of the points a.D,{$ .E,y . F, 
 where 
 
 a:(3:y = BC:CA:AB. 
 The centroid A f of the points 
 (3.E, y.F divides the segment EF 
 into two segments EA', A'F, such that 
 
 EA':A'F= y:(3 = AB:CA = AB:^CA = ED : DF. 
 Therefore DA' is the bisector of the angle EDF, and conse- 
 quently G, which is the centroid of the points a . D, (fi + y). A', 
 lies on the (internal) bisector of the angle D of the triangle 
 
 Fig. 112. 
 
 * CULMANN, 1. c., No. 94. 
 
-137] 
 
 CENTEOIDS. 
 
 103 
 
 DEF. Similarly G must also lie upon the bisectors EB', FC' 
 of the other two angles, and therefore G is the centre of the 
 circle inscribed in the triangle DEF. Q. E. D. 
 
 137. Let the given figure be the quadrilateral ABCD (Figs. 
 113, 114, 115), which may be regarded as the algebraical sum 
 of the two triangles ABD, CDB into which it is divided by the 
 diagonal BD. Let E be the middle point of BD. The centroids 
 G! , G 2 of the two triangles are respectively so situated on the 
 
 Fig. 114. 
 
 straight lines AE, CE, that G 1 E = \AE and G 2 E= CE. There- 
 fore the centroid G of the quadrilateral is the centroid of the 
 two points a l .G i ,o 2 .G 2 , where a- L : a 2 = ABD : CBD = AF: FC, 
 where F is the point of intersection of the two diagonals BD, 
 AC. Since G l G 2 divides two sides of the triangle AEG into 
 proportional parts, it is* parallel to the third side AC ; whence 
 it follows, that the straight line EG divides G^ G 2 , and AC in 
 the same ratio, namely GG-^: GG 2 a 2 : a l = FC:AF. In 
 order to divide AC in the ratio FCiAF, it is sufficient to 
 
104 
 
 CENTKOIDS. 
 
 [138 
 
 interchange the segments AF, FC, that is, to make AH = FC, 
 and IIC = AF. The line joining E to H divides G 1 G 2 in the 
 required point G. 
 
 The parallels G l G 2 and AC divide FA, FC, FH in the same 
 ratio; and therefore GF=%HF, since G 1 F = ^ AF, and 
 G 2 F= CF. 
 
 If instead of BD we employ the diagonal AC, whose middle 
 point is K, and if we interchange the segments BF, FD of BD 
 (i. e. if we take BL = FD, and LD = BF) ; then the point G 
 is so situated on LK, that GK = \LK. 
 
 But F, the middle point of BD, is also the middle point of 
 FL, and similarly K is the middle point of FH\ hence is 
 the centroid of the triangle FLU, that is to say : 
 
 The centroid of a quadrilateral coincides with that of the triangle, 
 whose vertices are the point of intersection of the diagonals, and 
 the two points obtained by interchanging the segments on each of the 
 two diagonals. 
 
 Hence it follows that the straight line FG passes through 
 the middle point / of HL *. 
 
 138. If AD, BC are parallel (Figs. 116, 117), and if we draw 
 through the centroids of the triangles BCD, ABD parallels to 
 
 Fig. 1 1 6. 
 
 AD, these parallels divide tne straight line MN which joins 
 the middle points of AD, BC into three equal parts. Since 
 the straight line MN contains the middle points of all 
 
 * CULMANN, 1. c., No. 95. Cfr. Quarterly Journal of Mathematics, vol. 6 
 (London 1864), p. 127. 
 
138] CENTROIDS. 105 
 
 chords parallel to AD, it is a diameter of the figure, 
 and therefore the point G lies in it, and divides its central 
 segment into two parts proportional to the areas of the 
 triangles in question, i. e. proportional to BC, AD. The parts 
 of this central segment (since their sum is %MN, and their 
 ratio AD : BC) are respectively equal to 
 
 MN.AD MN.BC 
 3(AD + BC)' 3(AD + BC)' 
 and consequently 
 
 MN x AD MN (BC + 2 AD) 
 
 3(AD + BC)~ 3(AD + BC) 
 MNx BC MN(AD + 2BC], 
 
 whence MG : GN = BC+ 2 AD : AD + 2 BC. 
 
 Every straight line therefore which passes through G, and is 
 contained between the parallels AD, BC, will be divided by G 
 into two parts proportional to BC+2AD and AD+2BC re- 
 spectively. If now on BC we take CP = AD, and if on AD 
 we take AQ = CB, it follows that the straight line PQ will be 
 divided by MNinto two segments proportional to MP, QN; but 
 
 MP = ^BC+AD, QN = BC+^AD, 
 or MP : QN = BC+2AD :AD + 2 BC. 
 
 Hence PQ passes through G. Since BP, QD are equal and 
 parallel, PQ and BD bisect one another; therefore PQ passes 
 through E the middle point of BD, that is, PQ coincides 
 with HE. 
 
 If moreover we take on AD 
 
 D8 = CB, AA' = $ AS, 
 and if on BC we take CC' AA' ; then, because 
 
 A'N=AN-AA r = \AD-l(AD-BC] = l(AD + 2BC) 
 and MC = MC+ CC' = \BC + % (AD-BC) = l(BC+2AD) 
 
 therefore A'N: Mtf= JD + 2BC: BC+ 2 AD, 
 that is A'C' passes through G. 
 
 Hence we obtain two simple constructions for the centroid 
 of a quadrilateral with two parallel sides (i.e. a trapezium), 
 either as the intersection of MN with PQ, or as the intersection 
 
 * CULMANN, ibid. WALKER, On an easy construction of the centre of 
 gravity of a trapezium. (Quarterly Journal of Mathematics, vol. 9, London, 
 1868, p. 339.) 
 
D 
 
 106 CENTROIDS. [139- 
 
 139. The construction given above for the centroid of a 
 quadrilateral fails in the case where the diagonals AC,BD are 
 
 parallel (Fig. 118). But in this case the 
 triangles ABD, CD are equivalent, and of 
 opposite sign, so that a x -f o 2 0. It fol- 
 lows that the area of the figure is zero, and 
 the centroid lies at infinity in the direction 
 common to AC and BD. 
 
 140. Now let it be required to find the centroid of any recti- 
 linear figure whatever. We may consider the area of the figure 
 to be the algebraic sum of the triangles, formed by joining the 
 sides of the circuit to an arbitrary point 0. Having found 
 the centroids J 15 A 2 , A 3 , ... of these triangles, and reduced their 
 areas to a common base so that they are proportional to the 
 segments a 13 a.,, a 3 , ..., the centroid in question is the centroid 
 of the points a 1 . /^ , o 2 . A 2 , a 3 . A 3 , ... which may be constructed 
 by one or other of the methods already explained. 
 
 If the pole is taken quite arbitrarily, then the number 
 of triangles is equal to the number of sides of the circuit ; 
 but if we take upon one of the sides, or at the point of 
 intersection of two of them, then the number of triangles is 
 reduced by one or two units respectively. 
 
 Instead of regarding the proposed figure as the sum of 
 triangles, we may also consider it as the aggregate of the 
 quadrilaterals and triangles, into which it can be decom- 
 posed by means of straight lines conveniently drawn. 
 
 141. Example. Let the given figure be the self-cutting 
 hexagon ABCDEF(Y\g. 119), which is the sum of the triangles 
 OBC, OCX), ODE, OFA, being the point of intersection of 
 the sides A, EF. Of these four triangles, the first and last 
 are positive, the other two negative. Let their centroids 
 6r 1? G 2 , 6r 3 , 6r 4 be found, and let the areas of the triangles, 
 reduced to a common base, be proportional to the segments 
 a 19 a 2 ,0 3 , a 4 . These segments a have the same signs as the 
 triangles, the first and last of them are positive, the second 
 and third negative. If now we wish to employ the method of 
 Art. 126, we must first reduce the four products a r . OG r to a 
 common base fi. In the figure, an arbitrary straight line as is 
 drawn through 0, its positive direction is fixed, and upon it 
 the segments //, a l5 o 2 , a 3 , a 4 are set off from their common initial 
 
-142] CENTROIDS. 107 
 
 point (/i, a 15 a 4 in one sense; o 2 ,a 3 in the opposite sense*). 
 Then the final point of h is joined to G r) and through the 
 
 D 
 
 Fig. 119. 
 
 final point of r a parallel is drawn to this joining line cutting 
 OG r in //,.. Thus we obtain OG r \h = OH r :a r , and therefore 
 a r . OG r = h . OH r . Now construct a circuit starting from 
 with its sides equipollent to 0// 15 0// 2 , 0// 3 , O7/ 4 ; the closing 
 line is OR. Finally to construct the point G, given by the 
 relation /, . R 
 
 U(jr = j 
 
 a 1 + a 2 + a 3 + a 4 
 
 we set off along Ox from its initial point the segment 
 
 08 = a 1 + a 2 + a 3 + a 4 , 
 
 join its final point to R, and draw through the final point 
 of k a parallel to this joining line cutting OR in G. 
 
 142. Again, let the figure be the cross-section of a so-called 
 Angle-iron (Fig. 120). Divide it into six parts, four trapeziums, 
 one triangle, and one parallelogram, denoted in the figure by 
 the numbers 1, 2, 3, 4, 5, 6. Construct the centroids of these six 
 parts, and reduce the areas to a common base, determining the 
 proportional segments 1, 2, 3, 4,5, 6 ; and set off these six segments 
 
 * In Fig. 119 the final points of the segments h, a are denoted by these letters 
 themselves. Some of the straight lines mentioned in the text are not drawn in 
 the figure. 
 
108 
 
 CENTEOIDS. 
 
 [142- 
 
 in succession along a straight line zz. Then through an arbi- 
 trarily chosen pole U draw rays to the points of zz, which 
 
 Fig. 1 20. 
 
 bound the segments ; next draw through the centroids of the six 
 component figures parallels to zz t and construct a polygon, with 
 its vertices lying on these parallels, and its sides respectively 
 parallel to the rays emanating from U. The two extreme sides 
 of this polygon will intersect in a point; through which if 
 a parallel to zz is drawn, then this straight line must 
 contain the required centroid. In order to obtain a second 
 straight line, possessing the same property, we either repeat the 
 above detailed operations for another direction different to zz ; 
 or else construct, as shown in the figure, a new polygon, whose 
 vertices lie upon straight lines drawn through the centroids 
 1, 2, 3, 4, 5, 6 perpendicular to zz, and whose sides are respect- 
 ively perpendicular to the corresponding rays of U. It is quite 
 clear that this is just the same thing, as if we drew a new 
 straight line zz perpendicular to the first, and then dealt 
 with it just as we formerly dealt with the first zz. It should 
 
-145] 
 
 CENTKOIDS. 
 
 109 
 
 1 2 
 
 12I 
 
 be remembered, that in setting off the segments 1, 2, ... 
 along zz, attention must be paid to their signs if the partial 
 areas into which the figure is divided are not all of the same 
 sign *. 
 
 143. In the foregoing construction two polygons were used 
 for the purpose of finding two straight lines, passing through 
 the centroid we were in search of. 
 
 But whenever we know a priori one 
 straight line in which the centroid 
 must lie, one polygon is sufficient, 
 for example, when the figure has a 
 diameter. This case is illustrated 
 by the example (Fig. 121), where 
 the figure possesses an axis of 
 symmetry. 
 
 The figure represents the cross- 
 section of a double Tee-iron. 
 
 144. We proceed now to the case of centroids of curvilinear 
 figures, and first we examine that of a circular sector OAB 
 (Fig. 122). We consider it to 
 
 be divided into an indefinitely 
 
 large number of concentric ele- 
 
 mentary sectors. The centroid 
 
 of each of these, regarded as a 
 
 triangle, lies upon a circle 
 
 drawn with radius OA' = %OA. 
 
 The required centroid is there- 
 
 fore the centroid G of the arc A'E' . In order to find that 
 
 point (Art. 135), set off the semi- 
 
 arc CA along the tangent CM, join 
 
 OM, and draw A'N parallel to 
 
 OC until it intersects OM in N. 
 
 Then G is the foot of the perpen- 
 
 dicular let fall from N upon the 
 
 mean radius OC f. 
 
 145. Next, let the circular seg- 
 ment ABC (Fig. 123) be given. 
 This is the difference between the 
 
 sector OAB and the triangle OAB, or the sum of the sector OAB 
 
 Fig. 123. 
 
 * CULMANN, 1. c., Nos. 96 & 116. 
 
 t CCLMANX, 1. c., No. 96. 
 
110 CENTROIDS. [146- 
 
 and the triangle OB A. Therefore the centroid G of the segment 
 lies on the straight line (the mean radius OC) joining the 
 centroids G l , G 2 of the sector and triangle, and divides the 
 segment G r 1 G 2 into two parts inversely proportional to the 
 areas of these figures. If we take OA' = ^OA, and find the 
 point N as just shown (Art. 144), then G l , G 2 are the feet 
 of the perpendiculars let fall from 'N and A' upon the mean 
 radius OC. Let F be the point of intersection of AB and OC, 
 and II the foot of the perpendicular let fall from F upon OA. 
 Then the areas of the sector and triangle are respectively 
 equal to CM. OA, and FH . OA, that is to say, they are pro- 
 portional to the lengths CM smd FH; therefore, if through G l 
 and G 2 two parallel segments G 1 I and G 2 K are drawn in the 
 same sense, equal or proportional to FH, and CM respectively, 
 KI and OC will intersect in G, the required centroid. In 
 fact from the similar triangles GG 1 I, GG 2 K we have 
 G l G : G 2 G =GJ: G 2 K = FH: CM*. 
 
 146. If the perimeter of the figure, whose centroid we are 
 finding, consists of rectilinear segments and circular arcs, we 
 decompose the figure by drawing the chords of these arcs 
 or radii to their extremities ; then we know how to find the 
 centroid and area of each part, and are able to apply the 
 process of Art. 142. 
 
 Example. Let us find the centrcid of the figure already 
 dealt with in Art. 107 (Fig. 124). For this purpose we first 
 consider it to be broken up into three parts, the lune, the 
 crown-piece, and the sum of the rectilinear parts ; then, 
 regarding the lune as the algebraic sum of two sectors and 
 one quadrilateral, the crown-piece as the algebraic sum of 
 two sectors, and having divided the rectilinear parts by means 
 of the straight line KCC'K', we finally have the given figure 
 equal to the sum of the following parts : 
 
 1 ...... Sector UAEA', 
 
 2 ...... Quadrilateral OAUA', 
 
 3 ...... Sector AOA'F, 
 
 4 ...... Sector OB'B, 
 
 5 ...... Sector 
 
 CULMANN, Hid. 
 
146] 
 
 CENTHOIDS. 
 
 Ill 
 
 . . . Trapeziums BCKH+ H'K'C'ff, 
 ...Trapeziums CJIK+K'I'J'C'. 
 
 Fig. 124. 
 
 We know how to determine the areas of all these, and by 
 reducing them to a common base we are also able to construct 
 their centroids. In order to find the centroid of the sum of 
 ECKR and R'K'C'B, it is sufficient (Art. 138), to find the 
 centroid of the trapezium BCKH, and then to draw through 
 it a parallel to KG until it intersects the axis of symmetry 
 EO-, the point of intersection is the centroid required. 
 Now to apply the process of Art. 142, we draw, in a direction 
 different to EO, say in that of KCC'K', a straight line zz, on 
 which we set off in succession the segments 1, 2, 3, 4, 5, 6, 7 
 respectively proportional to the areas of the seven partial 
 figures, noticing that the segments 3 and 5 must be set off in 
 the opposite direction to the others, because they represent 
 negative areas. Through any point whatever V lying outside 
 zz, draw rays to the limiting points of the above segments ; 
 then draw lines parallel to zz through the centroids of the 
 partial figures, and construct a polygon whose vertices lie on 
 these parallels, and whose successive sides are parallel respec- 
 tively to the rays emanating from V. Now draw through the 
 point of intersection of the first and last sides of this polygon a 
 parallel to zz ; this line cuts the axis OU in the required 
 centroid of the given figure. This point G falls in our figure 
 
112 CENTROIDS. 
 
 very near to the point 2, the centroid of the quadrilateral 
 OAUA'. If we produce the sides of the polygon sufficiently, in 
 order to find the point in which the first side cuts the fourth, 
 and also that in which the fourth and sixth intersect, and if 
 through these points we draw parallels to zz till they intersect 
 the axis of symmetry, these latter points of intersection will 
 be the centroids of the lune and the crown-piece. 
 
CHAPTEE IX. 
 
 RECTIFICATION OF CIRCULAR ARCS. 
 
 147. IN order to develope a circular arc AB along its 
 tangent (Fig. 125) we may proceed in the following way. 
 On BA produced mark off a part 
 AC J BA, and with C as centre and 
 CB as radius, describe an arc cutting 
 the tangent AD in D. Then AD is the 
 length of the given arc, with a negative 
 error, whose ratio to the whole arc is 
 
 Fig. 125. 
 
 1080 54432 
 
 being the ratio of the arc to the 
 radius *. 
 
 Otherwise (Fig. 126) : let J9 be the middle point of the arc 
 AB, and E the middle point of the arc AD ; let the radius 
 OE intersect the tangent at A in 
 C, and join CB ; then AC+CB 
 is the length of the given arc with 
 a positive error, whose ratio to 
 the whole arc is 
 
 o 
 
 ---=0 
 
 Fig. 126. 
 
 4320 3484648" 
 Since 4320=4x1080, if we 
 add to | of the length found by 
 
 the second construction i of that found by the first, the sum 
 obtained will be very approximately equal to the length 
 of the arc, with a positive error, whose ratio to the whole 
 length of the arc is 
 
 lie* 
 
 870912 " 
 
 * RANKINE, On the approximate drawing of circular arcs of given length 
 (Philosophical Magazine, October, 1867), p. 286. 
 
 f RANKINE, On the approximate rectification of circular arcs (Philoso- 
 phical Magazine, November, 1867), p. 381. 
 
 I 
 
114 RECTIFICATION OF CIRCULAR ARCS. [148 
 
 For the proof of these rules we refer the reader to the 
 original memoirs of Professor Rankine, cited in the foot- 
 notes. 
 
 148. In regard to this question, it will be convenient to 
 mention at this point some methods suggested by Professor 
 A. Sayno, of Milan. 
 
 The method given by Culmann for developing a circular 
 arc AB along the tangent at one of its points is much too long. 
 The length of a circular arc may be found graphically in 
 a much simpler fashion, by having recourse to auxiliary 
 curves, which drawn once for all can be employed in every 
 example. 
 
 Consider a convolution OMRS of the Spiral of Archimedes, 
 which when referred to its polar axis OX and its pole 0, has 
 
 Fig. 127. 
 
 the equation p = a o> *, and the circle drawn with centre and 
 radius OA' = a. Let OM be any radius vector of the spiral, 
 which cuts the circle in M f ; then the arc A'M'= ON. If now 
 we wish to find the length of an arc A"M" of any radius what- 
 ever OA", it is sufficient to place the spiral (supposed moved 
 from its previous position) so that its polar axis coincides with 
 the radius OA" of the given arc, to mark on OA" the point A', 
 and on the other radius OM" the point M in which it cuts 
 the curve. Now take the spiral away and draw through A" 
 a parallel to A'M, cutting OM" in M'", then OM'" is the 
 required length of the arc. We can construct this spiral 
 upon a thin plate of brass, horn, or ivory; it is sufficient 
 
 * p is the radius vector OM, and ca the corresponding vectorial angle A'OM. 
 
148] 
 
 RECTIFICATION OF CIRCULAR ARCS. 
 
 115 
 
 to mark upon it the pole and the point A '. This would be a 
 new instrument, which might be added as a ' Graphometer ' to 
 the case of drawing implements of an Engineer. 
 
 The Spiral of Archimedes p = ao> (Fig. 128) enables us also 
 to develope the arc along the tangent. Having drawn the 
 
 Fig. 128. 
 
 circle whose radius OA = a, and the circle whose diameter OC 
 OA, if i>, II are the points in which these circles are 
 cut by any radius vector OM, then OM= the &TcAH = 
 arc Off. Therefore, if we wish to set off the arc 0V along 
 the tangent OX, we need only place the spiral in such a 
 manner that the pole and the polar axis coincide respectively 
 with the point of contact and the tangent OX of the given 
 arc, and then mark the points ff, M in which the chord 07 
 cuts the circle on OC as diameter, and the spiral. We then take 
 away the spiral, and mark off on OX the segment OM' = OM\ 
 draw through V a parallel W to ffM', and OF' is the 
 required length of the arc. 
 
 In order .to increase the stiffness of the plate which 
 forms the instrument, it is best to use the circle of radius 
 
 I 2 
 
116 
 
 RECTIFICATION OF CIRCULAR ARCS. 
 
 [148- 
 
 OC' = OC, and then, supposing the chord FO to be produced, 
 we obtain OH' = HO. 
 
 Another curve, which serves the same purpose, is the hyper- 
 bolic spiral, whose equation in polar coordinates is a = p o>. 
 Draw (Fig. 129) a convolution of this curve NMDCBA, and 
 
 M'" 
 
 Fig. 129. 
 
 mark off a point A' on the polar axis, such that OA' = a. Then 
 the length of the circular arc MM', of radius OM, is OA' ; hence 
 the length of any circular arc whatever M" M"', drawn with its 
 centre at 0, is OA", where A" is got by drawing M" A" parallel 
 to MA'. This curve however is of no use in determining the 
 lengths of small arcs, so that for practical purposes the first 
 curve is to be preferred. 
 
 The hyperbolic spiral enables us also to divide angles in a 
 
 very elegant manner. Thus, to find the arc M'N' = - M'M 
 
 (Fig. 129), we need only produce the radius vector OM, take 
 OM" = n - OM, and draw an arc of radius OM" to cut the spiral 
 
 in N\ the radius ON meets 
 the arc M'M in the required 
 point N'. 
 
 In order to set off the 
 arc along the tangent we 
 can also employ another 
 auxiliary curve, namely the 
 involute of the circle. Take 
 (Fig. 130) a circle of radius 
 OA', and let A'M'ff(fff 
 be its involute. From the 
 figure we have at once 
 the arc MA'=MM', where 
 ' is the tangent of the circle at M. If now it is required 
 
 X 3<>. 
 
-149] 
 
 RECTIFICATION OF CIRCULAR ARCS. 
 
 117 
 
 to set off the arc N" M'" (whose centre is 0), along its 
 tangent from M", we need only draw OH ', which if sufficiently 
 produced cuts the tangent in question in M fy , and M" H'\ 
 is the required length of the arc. 
 
 149. By far the simplest method of rectifying the semi- 
 circumference is that of a Polish Jesuit, Kochansky, which was 
 published in the Acta Eruditorum Lipsise, year 1685, page 
 397, according to Dr. BoUcher*. Let be the centre and 
 
 Fig. 131. 
 
 AB a diameter of the circle of radius = 1 , the angle CO A = 30 
 Then if we take CD = three times the radius, we have 
 
 i.e. 3D 3-14153, 
 
 a value of the semi-circumference true to four places of decimals. 
 
 By means of this method, the rectification of an arc greater 
 
 than 90 can be reduced to the rectification of its supplementary 
 
 arc. 
 
 * [In the XVI vol. (Leipsic, ] 883) of Hoffmann's Zeitschrift fur math, und 
 naturl. Unterricht.] 
 
RECIPROCAL FIGURES IN GRAPHICAL STATICS. 
 
AUIHOK'S PEEFACE 
 
 TO THE ENGLISH EDITION. 
 
 AT a time when it was the general opinion that problems 
 in engineering could be solved by mathematical analysis only, 
 Culmann's genius suddenly created Graphical Statics, and 
 revealed how many applications graphical methods and the 
 theories of modern (projective) geometry possessed. 
 
 No section of Graphical Statics is more brilliant or shows 
 more effectually the services that geometry is able to render 
 to mechanics, than the one dealing with reciprocal figures and 
 framed structures with constant load. 
 
 It is to this circumstance that I owe the favourable 
 reception my little work (Le figure reciproche nella statica 
 grajica, Milano, 1872) met with everywhere; and not the 
 least from Culmann himself. It has already had the honour 
 of being translated into German and French. Having been 
 requested to allow an English version of it, to be published 
 by the Delegates of the Clarendon Press, I consented with 
 pleasure to Professor Beare undertaking the translation. 
 
 I have profited by this occasion to introduce some improve- 
 ments, which I hope will commend themselves to students of 
 the subject. 
 
 L. CREMONA. 
 
 ROME, October, 1888. 
 
CHAPTEE I. 
 
 POLE AND POLAR PLANE. 
 
 1. THAT dual and reciprocal correspondence between figures 
 in space, discovered by Mobius *, in which, to any plane what- 
 soever, corresponds a pole situated in the same plane, and all 
 planes passing through any one point have their poles on the 
 polar plane of that point is called a Null-system by German 
 mathematicians-. 
 
 Such a correspondence is obtained in the following manner. 
 Let there be a plane S, and four points in it A, B, C, D , no three 
 of which are in one straight line ; and let there be three 
 other planes a, /3, y passing through AD,BD, CD, respectively. 
 These will be the fixed elements in the construction. 
 
 Draw any plane whatever a- cutting the straight lines /3y, 
 ya, a/3 in P, Q, R respectively, then the planes PBC, QCA, 
 EAB will all intersect in the same point S of the plane a-. 
 
 Demonstration. Let X, Y, Z, X lt T 15 Z l be the points in 
 which the straight line 0-6 intersects the sides DC, CA, AB, 
 AD, BD, CD of the complete quadrilateral ABCD; these 
 points form three pairs of conjugate points of an involution f, 
 by Desargue's Theorem. Since the planes 8, <r, a, meet in 
 Xj the straight line QR common to the planes <r, a passes 
 through that point; similarly RP passes through Y ly and 
 PQ through Z l . Of the six points in involution, taken now 
 in the plane o-, three, X l , Y l , Z l , belong to the sides QR, RP, 
 PQ of a triangle PQR', therefore { the straight lines XP, 
 YQ, ZR meet in one point S, which with PQR forms a com- 
 plete quadrilateral. 
 
 * MOBIUS, Ueber eine besondere Art dualer Verhaltnisse zivisclien Figuren in 
 Raume, in vol. x. of Crelle's Journal, Berlin, 1833, or in vol. i. p. 489, G-esam- 
 melte Werke, Leipzig, 1883. 
 
 In reality this system of reciprocal figures in space had been already discovered 
 by GIORGINI (1827), (Meinorie della Societa Italiane delle Scienze Modena, vol. xx). 
 
 t CREMONA, Protective Geometry (Oxford, 1885), Art. 131. 
 
 J CREMONA, Protective Geometry (Oxford, 1885), Art. 135. 
 
124 POLE AND POLAR PLANE. [2- 
 
 Therefore the planes BCPX, CAQT, ABRZ meet in a point 
 8 of the plane PQR. 
 
 This theorem may be expressed as follows. 
 
 If the faces of a tetrahedron ABC 8 pass respectively through 
 the vertices of another tetrahedron PQRD, and if three faces 
 of the latter pass through three vertices of the former, then 
 the fourth face of the second tetrahedron will pass through 
 the fourth vertex of the first (Theorem of Mobius*). 
 
 2. Starting from the fixed elements A, B, C, a, /3, y, let any 
 plane a- whatever be given, and let it be required to determine 
 by means of this theorem the point S lying in it. 
 
 The plane a- meets the straight lines /3y, ya, a/3 in three 
 points P, Q, R, and the three planes PBC, QCA, RAB inter- 
 sect in the required point S. 
 
 Conversely, given any point S whatever, to determine the 
 corresponding plane <r, which passes through S. 
 
 The planes SBC, SCA, SAB intersect /3y, ya, afi in three 
 points P, Q, R, the plane of these points is the required 
 plane. 
 
 The point S is called ihepole of the plane o-, and the latter 
 is termed the polar plane of S. 
 
 3. If the plane o- change its position, the points Q, R in it 
 remaining fixed, the planes QCA, RAB will remain fixed, and 
 therefore the point S will move on the straight line (which 
 passes through A) common to these two planes. When 
 the point P falls on D, that is, when o- coincides with 
 QRZ) (i.e. a), the plane PBC coincides with ABCD, and 8 
 falls on A. Then A is the pole of the plane a, and similarly 
 B and C are the poles of /3, y. 
 
 If the arbitrary plane o- passes through BC, the traces of the 
 planes QCA , RAB on it, will be the straight lines QC, RB 
 which are the traces of the planes y, (S ; therefore the pole 
 falls in the straight line /3y, i.e. on P. The points P, Q, R 
 are consequently the poles of the planes PBC, QCA, RAB. 
 
 If the arbitrary plane coincides with ABC, the point P falls 
 on D, i.e. D is the pole of the plane ABC. 
 
 * MOBIUS, Kann von zwei dreiseitigen Pyramiden eine jede in Bezug auf die 
 andere um und ein-geschreiben zugleich heissen! vol. iii. of Crelle's Journal 
 (Berlin, 1828), or Gesammelte Werke, vol. i. p. 439. 
 
-6] POLE AND POLAR PLANE. 125 
 
 4. The pole S of the arbitrary plane o- (or conversely the polar 
 plane o- of the arbitrary point S) has been determined starting 
 from the system, supposed given, of three planes a, (3, y 
 (having no straight line in common) and their poles A , , O. 
 But in the tetrahedron ABCS the relations between the vertices 
 (or the faces) are perfectly reciprocal, that is, are interchange- 
 able ; so that just as 8 has been deduced from ABCa(3y , so A 
 may be determined from SBC<j(3y ; and so on. From this it 
 follows that if S lt S 2 , $ 3 are the poles of any three arbitrary 
 planes o- l9 cr 2 , <r 3 (not passing through the same straight line), 
 deduced in the manner above described from the system 
 ABCapy, the pole S of the plane a, determined from this 
 same system, coincides with that which would be determined 
 by a similar construction starting from S l S 2 S 3 a- 1 a- 2 cr 3 as the 
 given system. 
 
 5. From the theorem of Mobius it follows that if the plane a- 
 be drawn through the pole P of a plane TT = PBC, the pole S 
 of the plane a- falls in TT ; therefore : 
 
 If a plane passes through the pole of another plane, con- 
 versely the latter contains the pole of the former, that is 
 to say : 
 
 If a point lies in the polar plane of a second point, the latter 
 lies in the polar plane of the former. 
 
 From this it follows that the poles of all the planes passing 
 through a point S lie in a single plane <r , the polar plane of S ; 
 and the polars of all the points of a plane a pass through one 
 and the same point S, the pole of o-. 
 
 6. Let a, /3 be two planes, and A, B their poles. Any plane 
 whatever through AB will have its pole in a and in /3, that 
 is, in the straight line a/3 ; conversely, the polar plane of any 
 point whatever of a/3 will pass through A and B, i.e. through 
 the straight line AB. And any plane whatever through the 
 straight line a/3 , which contains the poles of the planes through 
 AB, will have its pole on the straight line AB ; and conversely, 
 any point whatever of AB, being on the polar plane of the 
 points of a/3 , will be the pole of a plane through a/3 . 
 
 Two straight lines, such as aft and AB, each of which is the 
 locus of the poles of planes passing through the other, are 
 called reciprocal straight lines. 
 
 \ UNIVERSITY 
 
126 POLE AND POLAR PLANE. [7- 
 
 Hence it follows that if a straight line r passes through a 
 point A, its reciprocal r' lies in a, the polar plane of A ; and 
 conversely. 
 
 7. A straight line r, which lies in a plane a and passes 
 through A, the pole of a, coincides with its reciprocal, that is 
 to say, it is reciprocal to itself. In fact, if M is any other 
 point whatever of r , since M lies in a, the polar plane of A , 
 then //, the polar plane of M, passes through A. And since /u 
 must also pass through M, the polar plane of it or of any 
 point whatever of the given straight line, r passes through 
 the straight line r. 
 
 From this it follows, that two reciprocal straight lines 
 r and r', which are non-coincident, cannot lie in one plane. If 
 a plane a passes through both r and /', the pole A of the plane 
 will be on both r and r', and r would lie in a plane and 
 contain its pole, therefore r would be reciprocal to itself. 
 
 All straight lines reciprocal to themselves and passing 
 through a given point A lie in a, the polar-plane of A. All 
 straight lines reciprocal to themselves and lying in a given 
 plane a pass through A, the pole of a. 
 
 A system of straight lines reciprocal to themselves is 
 called a linear complex, and the straight lines are called rays oi 
 the complex. 
 
 Each ray of the complex which meets a given straight line 
 r (not itself a ray) meets also its reciprocal straight line /. 
 In fact, if A is the point common to the ray and to r, the 
 plane a, the polar of A, must pass through the ray and 
 the straight line /. 
 
 Conversely, if a straight line t meets two reciprocals r and 
 /, the straight line t is necessarily a ray. For, the point tr 
 is the pole of a plane which passes through this point, and 
 through / ; therefore the plane also passes through t. Hence 
 t lies in a plane polar to one of its own points, or t is a ray. 
 
 From this it follows that all the straight lines (necessarily 
 rays) cutting two reciprocal straight lines r and /, and 
 another line <?, also meet the straight line / reciprocal to s. 
 Two pairs r/, $s' of reciprocal straight lines are therefore 
 situated on the same hyperboloid, the generators of which are 
 all rays of the complex of another system. 
 
-9] POLE AND POLAR PLANE. 127 
 
 8. All planes parallel to the same plane may be considered * 
 as having in common a line / situated at infinity, therefore 
 their poles all lie on a straight line r, the reciprocal of /. 
 Changing the bundle of parallel planes, the straight line r re- 
 mains parallel to itself, because it passes through a fixed point 
 / lying at infinity, that is, through the pole of the plane i at 
 infinity, in which the straight line r' is always situated. 
 
 Such lines r, whose reciprocals lie at infinity, are called 
 diameters of the complex. 
 
 Planes perpendicular to the common direction of the dia- 
 meters are parallel to each other, therefore their poles are on 
 a diameter. This diameter a, which is distinguished from the 
 other diameters by being perpendicular to the planes whose 
 poles it contains, is called the central axis of the complex. 
 
 Straight lines parallel to the central axis are reciprocals 
 to straight lines in the plane at infinity t ; and in particular 
 the central axis is reciprocal to the line at infinity common to 
 all planes perpendicular to the central axis itself. The point 
 /, at infinity on the central axis, is the pole of the plane 
 at infinity. 
 
 9. If r and / are any two reciprocal straight lines whatever, 
 the straight line which passes through their points at infinity 
 will be a ray of the complex, and will therefore pass through 
 the pole / of the plane at infinity ; that is, the points at 
 infinity of two reciprocal straight lines and of the central axis 
 are all three in one straight line. Hence it appears that 
 two reciprocal straight lines and the central axis are parallel 
 to the same plane. 
 
 Therefore, planes parallel to the central axis and passing 
 through two reciprocal straight lines are parallel to each 
 other. 
 
 From this it follows that : 
 
 If two reciprocal straight lines are projected parallel to 
 the central axis, on a plane, not containing the direction of 
 the central axis, their projections will be two parallel straight 
 lines. 
 
 We shall suppose that the projection is made on a plane 
 perpendicular to the central axis. 
 
 * CEEMONA, Protective Geometry (Oxford, 1885), Art. 26. 
 
128 POLE AND POLAR PLANE. [10- 
 
 Moreover, it follows that the straight line meeting two 
 reciprocal straight lines and perpendicular to them cuts the 
 central axis orthogonally. 
 
 10. Suppose the central axis horizontal, and let us call that 
 plane of projection which intersects the central axis in its 
 own pole the orthographic plane. 
 
 Take that point as the origin of a system of rectangular 
 coordinates x, y , z, and let the axis of z coincide with the 
 central axis : then the preceding theorems and laws of re- 
 ciprocity will be expressed by the following equations. 
 The point (x,y l , z-^) is the pole of the plane 
 
 xy l yx l -\-k(z~z l ] = 
 where k is some constant. 
 Conversely the plane 
 
 ax + by + cz -f d = 
 corresponds to the pole 
 
 kb ka d 
 
 c c c 
 
 The straight line 
 
 px + qy + rz = 
 is reciprocal to the straight line 
 
 ax + ly + c' =0 
 
 px + gy + r'z 
 where re' r'c = k(aq bp). 
 
 11. Hence : 
 
 (a) To any number of straight lines r in space, the projections of 
 which coincide in a single straight line, straight lines / correspond, 
 whose projections are coincident or parallel, according as the straight 
 lines r (necessarily lying in a plane parallel to the central axis) are 
 parallel or not. 
 
 (b) To any number of straight lines r in space, the projections of 
 which are parallel, straight lines r correspond, whose projections are co- 
 incident or parallel, according as the straight lines r (necessarily parallel 
 to a plane passing through the central axis) are parallel or not. 
 
 12. If the points A, B, C, D . . .in space are considered as 
 vertices of a polyhedron, the polar planes a, ft, y, 8, ... are 
 the faces of a second polyhedron, whose vertices a/3y,... 
 are the poles of the faces ABC,... of the first. The two 
 
-13] POLE AND POLAR PLANE. 129 
 
 polyhedra are called reciprocal ' to the vertices of each corre- 
 spond the faces of the other, to the edges the edges. Each 
 polyhedron is simultaneously inscribed and circumscribed to 
 the other (Art. 1). Two corresponding edges are reciprocal 
 straight lines (Art. 6). 
 
 Let the two polyhedra be projected on the orthographic 
 plane ; the projections will be two figures possessing reciprocal 
 properties. To each side of the first figure there will cor- 
 respond a parallel side of the other, since two corresponding- 
 sides are the projections of two reciprocal edges of the two 
 polyhedra. If one of the polyhedra has a solid angle, at 
 which m edges meet, the other will have a polygonal face of 
 m sides; and therefore, if in one of the orthographic figures 
 there are m sides diverging from a point or node, the m 
 corresponding sides of the other orthographic figure will be 
 the sides of a closed polygon. 
 
 In a polyhedron, each edge is common to two faces, and 
 joins two vertices ; each face has at least three edges, and 
 in each vertex at least three edges meet ; hence in both 
 orthographic figures, each side is common to two polygons, 
 and joins two nodes, three sides at least meet in every 
 node, and each polygon has at least three sides. 
 
 Suppose that one of the polyhedra, and consequently the 
 other, belongs to the class of Eulerian polyhedra"*; then the sum 
 of the numbers of vertices and faces exceeds by two the number 
 of edges, from the well-known theorem of Euler. Hence, if 
 the first orthographic figure possesses p nodes, p r polygons, and 
 * sides, we have p + y _ s + 2 . 
 
 The second figure will have yf nodes, p polygons, and s 
 sides. 
 
 13. If one polyhedron has a vertex at infinity, the other 
 has a face perpendicular to the orthographic plane, and con- 
 versely ; consequently, if one of the orthographic figures has 
 a vertex at infinity, the other contains a polygon whose sides 
 all lie in the same straight line, and conversely. 
 
 If the point / at infinity on the central axis is a vertex 
 common to n faces of the first polyhedron, then the other 
 
 * TODHUNTEB, Spherical Trigonometry, Chapter xiii, Polyhedra. 
 K 
 
130 POLE AND POLAR PLANE. 
 
 polyhedron has in the plane at infinity a polygonal face of 
 n sides. In this case, the first orthographic figure has p 1 
 nodes, p' n polygons, and s n sides; and the second (not 
 reckoning the straight line at infinity) possesses p 1 polygons, 
 p' n nodes, and s n sides: where the numbers jo, j' s are 
 connected by the relation 
 
CHAPTER II. 
 
 POLYGON OF FORCES AND FUNICULAR POLYGON AS 
 RECIPROCAL FIGURES. 
 
 14. THOSE reciprocal diagrams, which are obtained as the 
 orthographic projections of two reciprocal polyhedra, present 
 themselves directly in the study of graphical statics. The 
 mechanical property of reciprocal diagrams is expressed in the 
 following theorem due to the late Professor Clerk Maxwell * : 
 
 ' If forces represented in magnitude by the lines of a figure lie 
 made to act between the extremities of the corresponding lines of the 
 reciprocal figure, then the points of the reciprocal figure will all be in 
 equilibrium under the action of these forces' 
 
 The truth of the theorem is at once apparent, if we observe 
 that the forces applied at any node whatever of the second 
 diagram are parallel and proportional to the sides of the corre- 
 sponding closed polygon of the first diagram. 
 
 The theorem is particularly useful, in the graphical deter- 
 mination of the stresses, which are developed in frame- work 
 structures. 
 
 15. The first germs of the theory are met with in the 
 properties of the polygon of forces, whose sides represent in 
 magnitude and direction a system of forces in equilibrium 
 applied at any point ; and also in the well-known geometrical 
 constructions which enable us to determine the tensions of the 
 sides of a plane funicular polygon f . But the first to apply 
 the theory to frame- work structures was the late Professor 
 Macquorn Eankine, who, in Art. 150 of his excellent work 
 Manual of Applied Mechanics (1857), proved the following 
 theorem : 
 
 ' If lines radiating from a point be drawn parallel fo the lines of 
 resistance of the bars of a polygonal frame, then the sides of any 
 
 * Philosophical Magazine, April 1864, p. 258. 
 
 f VARIGNON, Nouvelle Mecanique on Statique, dont le projet fat donne en 
 1687: Paris, 1725. 
 
 K 2 
 
132 POLYGON OF FORCES AND FUNICULAK POLYGON [16- 
 
 polygon wJiose angles lie in these radiating lines will represent a 
 system of forces, which, Icing applied to the joints of the frame, 
 f mll balance each other ; each such force being applied to the joint 
 between the bars whose lines of resistance are parallel to the pair of 
 radiating lines that enclose the side of the polygon of forces, repre- 
 senting the force in question. Also, the lengths of the radiating lines 
 will represent the stresses along the bars to whose lines of resistance 
 they are respectively parallel* '.' 
 
 Rankine afterwards published an analogous theorem for a 
 system of polyhedral frames f. 
 
 16. The geometrical theory of reciprocal diagrams is specially 
 clue to the late Professor Clerk Maxwell, who first in 1864J, 
 and again in 1870, defined them generally, and obtained 
 them from the projections of two reciprocal polyhedra. 
 
 But his polyhedra are reciprocal in respect to a certain paraboloid 
 of revolution, in the sense of the theory of reciprocal polar fgures 
 of Poncelet || ; so that, projecting orthogonally and parallel to 
 the axis, the corresponding sides of their projections are not 
 parallel, but perpendicular to one another. Hence we must 
 rotate one of the diagrams through 90 in its own plane, in 
 order that it may assume that position which it ought to take 
 in statical problems. 
 
 On the contrary, by the more general process, explained in 
 this treatise, the orthographic projections of two reciprocal 
 polyhedra give precisely those diagrams which occur in 
 graphical statics. 
 
 17. The practical application of the method of reciprocal 
 figures was made the subject of a memoir by the late Professor 
 Fleeming Jenkin, communicated in March 1869 to the Royal 
 Society of Edinburgh^. In that memoir, after quoting 
 
 * Page 142 of the sixth edition (1872). 
 
 t Philosophical Magazine, Feb. 1864, p. 92. 
 
 * On reciprocal figures and diagrams offerees (Philosophical Magazine, April 
 1864, p. 250). 
 
 On reciprocal figures, frames and diagrams of forces (Transactions of the 
 
 Hoyal Society of Edinburgh, vol. xxvi. p. 1). See -also a letter of Professor 
 
 Kankine in the Engineer' Feb. 1872. 
 
 || Or, rather, that of MONGE. (See CHASLES, Aperqu historique, p. 378.) 
 
 H On the practical application of reciprocal figures to the calculation of 
 
 strains on framework (Transactions of the Royal Society of Edinburgh, vol. xxv. 
 
 p. 441). See also, by the same author: On braced arches and suspension bridges, 
 
-18] AS RECIPROCAL FIGURES. 133 
 
 the definition of reciprocal figures, and their statical pro- 
 perty, as enunciated by Maxwell in his memoir of 1864, he 
 adds : 
 
 ' Few engineers would, however ^ suspect that the two paragraphs 
 quoted put at their disposal a remarkably simple and accurate method 
 of calculating the stresses in framework ; and the author s attention 
 was drawn to the method chiefly by the circumstance that it was 
 independently discovered by a practical draughtsman, Mr. Taylor, 
 working in the Office of the well-known contractor Mr.. J. H. Cochrane.' 
 
 He also presents several examples, accompanied by figures, 
 and finishes with this observation : 
 
 ' When compared with algebraic methods, the simplicity and 
 rapidity of execution of the graphical method is very striking ; and 
 algebraic methods applied to frames, such as the Warren girders, in 
 which there are numerous similar pieces, are found to result in 
 frequent clerical errors, owing to the cumbrous notation which is 
 necessary, and especially owing to the necessary distinction between 
 odd and even diagonals' 
 
 18. But, whilst speaking of the geometrical solution of 
 problems relating to the science of construction, it is impos- 
 sible to pass over in silence the name of Professor Culmann, 
 the ingenious and esteemed creator of graphical statics*, for 
 to him are due the elegant methods of that science, which, 
 issuing from the Polytechnic School at Zurich, are now taught 
 in technical schools throughout the world. 
 
 Numerous questions of theoretical statics, as well as many 
 others which relate more particularly to certain branches of 
 practical science, are solved by Professor Culmann by a simple 
 
 read before the Royal Scottish Society of Arts (Edinburgh, 1870) ; and the memoir 
 On the application of graphic methods to the determination of the efficiencies of 
 machinery (Transactions of the Royal Society of Edinburgh, vol. xxviii. p. 1, 
 1877). 
 
 * Die graphische StatiJc, Zurich, 1866. In 1875 appeared the second edition of 
 vol. i. with rich additions. The reader is advised to read the Preface to that second 
 German edition, also Nos. 81 and 82. Graphical Statics have been treated since 
 in a whole series of elementary works. See UNWIN, Wrought Iron Bridyes 
 and Roofs, London, 1869 ; Bow, Economies of construction in relation to framed 
 structures, London, 1873 ; CLAKKE, Graphic Statics, London, 1880 ; EDDY, New 
 constructions in Graphical Statics, in Van Nostrand's Engineering Magazine, New 
 York, 1877-8, and American Journal of Mathematics, vol. i., Baltimore, 1878, 
 &c., &c. 
 
134 POLYGON OF FORCES AND FUNICULAK POLYGON [19- 
 
 and uniform method, which reduces itself in substance to the 
 construction of two figures, which he calls Kraftepolygon, and 
 Seilpolygon. And although he has not considered these figures 
 as reciprocal, in Maxwell's sense, still they are so sub- 
 stantially ; in particular the geometrical constructions which 
 Culinann gives in Chapter V of his work, devoted to systems 
 of framework (Das Fachwerk), almost always coincide with 
 those derived from Maxwell's own methods. 
 
 Moreover Culmann's constructions include certain cases, 
 (which are not treated by the English geometer,) in which it is 
 impossible to construct the reciprocal diagrams. 
 
 19. First of all, I wish to show that the Kraftepolygon and 
 the Seilpolygon (polygon of forces and funicular polygon) of 
 Culmann can be reduced to reciprocal diagrams. 
 
 Let there be given in a plane (which suppose always 
 to be the orthographic plane) n forces P l , P 2 , . . . , P n in 
 equilibrium, then by the polygon of forces we understand a 
 polygon, whose sides 1, 2, . .., n are equipollent* to the straight 
 lines which represent the forces f. 
 
 Take in the same plane a point 0, which will be called the 
 pole of the polygon of forces, and join the vertices of the 
 above polygon to that pole ; denote by (rs) the ray connecting 
 with the vertex common to the two sides r and s. The 
 funicular polygon corresponding to the pole is a polygon, 
 whose vertices lie in the lines of action (which we shall 
 call 1,2, ,.., n) of the forces P^ P 2 , P^, ..., P n , and whose 
 sides are respectively parallel to the rays proceeding from 
 OJ, in such a manner that the side comprised between 
 the lines of action of P r and P s , is parallel to the ray (rs) , 
 this side will be denoted by the symbol (rs). 
 
 The funicular polygon will be a closed one, like the 
 polygon of forces. 
 
 20. If the lines of action of the given forces meet in the 
 same point (Fig. la), we have two reciprocal diagrams, since 
 evidently the two polygons will be the orthographic pro- 
 
 * Equipollent, that is, equal in magnitude, direction and sense, a term due to 
 Professor Bellavitis. 
 
 t The position of the first side of that polygon is a matter of choice. 
 J The direction only of the first side of that polygon is determinate. 
 
-20] 
 
 AS RECIPEOCAL FIGURES. 
 
 135 
 
 jections of two pyramids, having each a polyhedral angle 
 of n faces. 
 
 If the forces are parallel, the polygon of forces is reduced to a 
 straight line, which corresponds to the case where the base of 
 
 Fig. i a. 
 
 the first pyramid is perpendicular to the orthographic plane, 
 and the vertex of the second is at infinity, that is to say, the 
 second polyhedron is a prism having only one base at a finite 
 distance. This case is illustrated in Figure 2 a, in which the 
 
 Fig. 2 a. 
 
 sides of the polygon of forces are not designated by one 
 number only, but by two numbers, placed at the ends of 
 each segment; so that the segments 01, 12, 23, 34, ..., cor- 
 respond to the straight lines 1, 2,3, 4 of the second diagram. 
 
136 
 
 POLYGON OF FOKCES AND FUNICULAR POLYGON 
 
 [21- 
 
 Here, as in all which follows, we adopt in the text two 
 series of numbers, 1 , 2 , 3 , r . . . , s . . . ; 1 , 2 , 3 , r , s , to distinguish 
 the lines of the one diagram from the corresponding lines 
 of the other. 
 
 21. Let us consider now the general case, in which the forces 
 do not all meet in the same point. 
 
 Take a second pole 0'; join it by straight lines to the 
 vertices of the polygon of forces, and construct a second 
 
 Fig. 3 a. 
 
 funicular polygon corresponding to the new pole 0', that 
 is to say, a polygon with its sides parallel to the rays 
 proceeding from 0' ', and its vertices situated in the lines of 
 action of the forces. See Figs. 3 and 5, in which the rays 
 proceeding from the second pole O f , and the corresponding 
 sides of the funicular polygon, are denoted by dotted lines. 
 
-23] AS RECIPROCAL FIGURES. 137 
 
 By operating in this way, it is plain that the two diagrams, 
 the one formed by the polygon of forces and the rays issuing 
 from the poles and 0' ', and the other formed by the 
 two funicular polygons and the lines of action of the forces, 
 are two reciprocal figures. The first is the projection of a 
 polyhedron*, formed by two solid angles of n faces, whose 
 corresponding faces form by their respective intersections 
 a twisted polygon f of n sides; the second is the projection 
 of a polyhedron comprised within two plane polygons of 
 n sides, in such a way that the sides of the one meet 
 the corresponding sides of the other. The straight line, in 
 space, which joins the vertices of the two solid angles of 
 n faces of the first polyhedron is conjugate to the straight 
 line, which the two planes of the bases of the second 
 polyhedron have in common. As a result of this, and 
 of the property that two conjugate straight lines are or- 
 thographically projected into two parallel straight lines, 
 it follows, that any two corresponding sides whatever (rs), 
 (rs)' of the two funicular polygons, intersect in a fixed 
 straight line, parallel to that which joins the two poles 
 and 0'. 
 
 This theorem is fundamental in Culmann's methods. 
 
 22. If we make the two poles and 0' coincide, the corre- 
 sponding sides of the two funicular polygons are parallel 
 (Fig. 40). In this case the straight line which joins the 
 vertices of the solid angles of the first polyhedron is per- 
 pendicular to the orthographic plane, whilst the bases of 
 the second polyhedron are parallel. 
 
 23. The diagonal which joins the vertices of two tetra- 
 hedral angles of the first polyhedron (Art. 21), or what is the 
 same thing, the diagonal between two vertices of the twisted 
 polygon, is conjugate to the line of intersection of the corre- 
 sponding quadrilateral faces of the second polyhedron, which 
 
 * This polyhedron has 3 n edges, 2 n triangular faces, 2 polyhedral angles 
 of n faces, and n of 4 faces ; the other polyhedron has 3 n edges, 2 n trihedral 
 angles, 2 bases which are polygons of n sides, and n quadrilateral faces. 
 
 t If this polygon degenerates into a continuous curve, the polygon of forces, 
 and the funicular polygon become respectively the curve of forces, and the 
 funicular curve (catenary) of a plane continuous system of forces. 
 
138 
 
 POLYGON OF FOEGES AND FUNICULAR POLYGON 
 
 [24- 
 
 line unites the point common to two sides of the one of the 
 bases to the point common to the corresponding two sides 
 of the other base. In an orthographic projection, the first 
 straight line is a diagonal joining the two vertices (r, r+ 1), 
 (5,5+1) of the polygon of forces, that is to say, a straight 
 line equipollent to the resultant of the forces P r+1 , P r+2 ..., P g ; 
 the second straight line is the line of action of the same 
 resultant. Hence the line of action of the resultant of any 
 
 Fig. 4 a. 
 
 number whatever of consecutive forces P r+ i, I r+2 ,..., P s 
 passes through the point common to the sides (r , r + 1) (s , s + 1) 
 of the funicular polygon ; another fundamental theorem of 
 graphical statics. (See, for example, Fig. 30, the resultant of 
 the forces 6,1,2.) 
 
 24. If the diagonal in question of the first polyhedron is 
 perpendicular to the orthographic plane, the conjugate straight 
 line is at infinity. Two vertices of the polygon of forces 
 (r, r+1), (s , <?+!) will then coincide in one point A (see 
 Fig. 5 a, where r = 1 , s = 4), and the sides (r, r + l), (s, s + l) 
 of each of the funicular polygons are parallel. 
 
 The magnitude of the resultant of the forces P r+ ^ , P r+2 , . . . , P 8 
 will be infinitely small, and its line of action the straight 
 line at infinity of the orthographic plane ; it is consequently 
 
-25] 
 
 AS RECIPROCAL FIGURES. 
 
 139 
 
 an infinitely small force, acting at infinity, equivalent to a 
 couple acting in the aforesaid parallel sides of the funicular 
 polygon, and represented in magnitude by the straight line 
 which joins the corresponding pole to the point A. Since 
 these two forces are equivalent to the system of forces P r+ i, 
 P r+2 ,...,P s , the one which acts along the side (r, r + l) is 
 
 n* 
 
 Fig. 5 a. 
 
 directed from A towards ; and the one which acts along the 
 side (s, s + l) is directed from towards A. 
 
 25. Given the forces P l9 ^ 2 , P 3 ,..., P tt _^ (Art. 19), the 
 two polygons (that is the force and funicular polygons) serve 
 to determine the force P n , equal and opposite to the resultant 
 of the given forces (see Fig. 3, in which n = 6). In fact, if we 
 construct a crooked line 1 , 2 , 3 , . . . , (n 1 ), whose sides are 
 equipollent to the given forces : it is clear that the straight 
 line n which joins the extremities of the crooked line (when 
 its direction is from the final to the initial point) is equipollent 
 to P n . Next take a pole 0, and construct a funicular polygon, 
 
140 POLYGON OF FORCES AND FUNICULAR POLYGON [26- 
 
 whose first n 1 vertices 1 , 2 , 3 , , . . , (n l) , lie in the lines 
 of action of the given forces P^ P 2 , P 3 , ..., J^ ; and whose 
 sides (n , l) (l , 2) (2 , 3) . . . (n 1 , n) are respectively parallel to 
 the rays connecting with the similarly named vertices of 
 the first polygon. Then the straight line drawn through the 
 last vertex n of the funicular polygon, (that is to say through 
 the point where the first side (n.l) meets the last (n 1, n),) 
 parallel to the last side n of the polygon of forces, is the line 
 of action of P n . 
 
 If the first side of the funicular polygon passes through a 
 fixed point, and the pole moves in a straight line, then all 
 the sides pass through fixed points situated on a straight line 
 parallel to the one described by the pole (Art 21). This 
 is contained in the celebrated porism of Pappus : 
 
 'Si quotcnmque rectae lineae sese mutuo secent, non plures quam 
 duae per idem pimctum, omnia autem in una ipsarum data sint, et 
 reliqiiorum multitudinem liabentium triangulum numerum, kujus lotus 
 singida habet puncta tangentia rectam lin earn positions dafam, quorum 
 trium non ad ongulum existens trianguli spatii unumquodqiie reli- 
 quum punctnm rectam lineam positione datam tanget*! 
 
 26. If we consider the point to be capable of occupying 
 any position whatever in the plane, the properties of the two 
 polygons (that is the polygon of forces and the funicular 
 polygon) may be compendiously stated in the following geo- 
 metrical enunciation: 
 
 Let a plane polygon be given of n sides 1 , 2 , 3 , ...,(!), 
 n\ and, in the same plane, n\ straight lines 1, 2, 3,..., 
 (n l) , respectively parallel to the first n\ sides of the poly- 
 gon. Join the point (i. e. a pole, moveable in any manner 
 whatever in the plane) to the vertices of the given polygon. 
 Imagine further a variable polygon of n sides, the first, n\ 
 vertices of which 1 , 2 , 3 , . . . , (n l) , lie in the corresponding 
 similarly named straight lines, whilst its n sides (n.l), (1.2), 
 (2.3) ...,(n 1, n) are parallel to the rays which join the simi- 
 larly named vertices of the given polygon to the pole 0. 
 Then the intersection of any two sides whatever (r, r + l), 
 
 * \_Mathematlcae Collectiones, preface to Book VII. p. 162, of the edition of 
 COMMANDING (Venice, 1689). See also the translation or paraphrase of the porism, 
 given by PONCELET in No. 498 of his Traitt des propridtes projectives (Paris, 
 
 1822)]. 
 
-28] AS BECIPKOCAL FIGUKES. 141 
 
 (s . s + 1) , of the variable polygon lies on a determinate straight 
 line, parallel to the diagonal which joins the vertices (r.r+ 1), 
 (,?,s+l) of the given polygon. 
 
 This theorem, which is not very readily proved by means 
 of the resources of Plane Geometry alone, is on the contrary 
 self-evident, if we consider the two plane figures as ortho- 
 graphic projections of two reciprocal polyhedra. 
 
 27. The polygon of forces is the projection of a plane poly- 
 gon, or twisted polygon, according as the directions of the 
 forces P do or do not meet in the same point. As we have 
 seen in Art. 20, one of the two reciprocal diagrams in the first 
 case is formed by the polygon of forces and the pole 0, the 
 other, by the lines of action of the forces, and the funicular 
 polygon corresponding to the pole 0. In the second case, 
 on the contrary, another pole 0' must be added to the first 
 diagram, and to the second a funicular polygon corresponding 
 to this pole O f ; we have further seen from Art. 22 that the 
 two poles may be made to coincide, and that then the first 
 diagram becomes as simple as possible. But, if we wish on 
 the other hand to simplify the second, it is best to remove the 
 pole 0' to infinity in an arbitrary direction ; and then the 
 polyhedron, of which the first diagram is the orthographic 
 projection, has the vertex of one of its polyhedral angles at 
 infinity ; and since the polar plane of a point at infinity is 
 parallel to the central axis, the new funicular polygon cor- 
 responding to O f has all its sides on the same straight line 
 (whose point at infinity is 0'\ The absolute position of this 
 straight line in the orthographic plane is still arbitrary, and 
 therefore it may be removed to infinity. 
 
 Very simple results are also obtained by the following 
 method : 
 
 Suppose that the previously mentioned polyhedral angle of 
 the first solid coincides with the infinite point of the central 
 axis ; in the first diagram the pole alone appears, since the 
 edges corresponding to the other polyhedral angle are pro- 
 jected orthographically into the vertices of the polygon of 
 forces. The polar plane of the vertex ' is now at infinity ; 
 hence the whole of the second funicular polygon is at an infi- 
 nite distance (see Art. 13). 
 
 28. We conclude from these very simple cases, that it is 
 
142 POLYGON OF FORCES AND FUNIC. POLYGON AS RECIP. FIGURES. 
 
 possible to consider the polygon of forces, and the funicular 
 polygon, of a system of forces in equilibrium, situated in a 
 plane (the orthographic plane), but not meeting in the same 
 point, as reciprocal diagrams. The one diagram is formed by 
 the polygon of forces and the rays joining its vertices to a 
 pole 0, and the other by the lines of action of the forces, 
 the funicular polygon relative to the pole 0, and the straight 
 line at infinity; the first diagram is simply the projection 
 of a polyhedron, whose faces are obtained by projecting the 
 n sides of a twisted polygon perpendicularly to the orthogra- 
 phic plane, from a point in space at infinity. The reciprocal 
 polyhedron, which has for its projection the second diagram, 
 is the infinite portion of space, limited by a plane polygon 
 and the n planes passing through the sides of that polygon 
 and prolonged everywhere to the plane at infinity. 
 
CHAPTER III. 
 
 APPLICATION OF RECIPROCAL DIAGRAMS TO FRAMEWORK. 
 
 29. LET us pass on now to the study of the more com- 
 plicated diagrams, which present themselves in the theory of 
 frames'*. Consider two polyhedral reciprocal surfaces 2 and 
 2', which possess an 'edge,' are simply connected f, and whose 
 edges are two closed twisted polygons J; let IT be the poly- 
 hedron enclosed by the surface 2, and the pyramidal sur- 
 face whose vertex is a point 1 , taken arbitrarily in space, and 
 whose base-line is the polygonal edge of 2 ; let FI' be the 
 polyhedron reciprocal to n , i. e. the polyhedron enclosed by 
 the surface 2', the polar plane o> of H, and the planes of the 
 angles of the polygonal edge of 2'. Project orthographically 
 the two polyhedra, and we obtain two reciprocal diagrams, 
 which we will now proceed to study. 
 
 Suppose that the polygonal edge of 2 has n sides, and that 
 the surface has besides these m ordinary edges , and p faces. 
 The polyhedron IT will have n+p faces, and 2n + m edges, 
 and therefore m + np + 2 vertices. Hence 2 has, besides 
 those on its polygonal edge, m p + I vertices ||. 
 
 * A Frame is a structure composed of bars or rods attached together by 
 joints, which are considered merely as hinges or pivots. Let AB be any one 
 bar (whose weight is neglected) of such a frame ; and assume that no force 
 acts upon it, except at the joints A , B. Then the whole of the forces (some 
 external, some consisting of pressures from the bar or the bars which meet it at 
 the joint A] acting on it at the joint A can be reduced to a single resultant : so 
 may those at the joint IB ; and these resultants being necessarily equal and 
 opposite, must act along the bar AS. Hence the bar is in a simple state of 
 tension, when these resultants act outwards ; or of compression, or thrust, when 
 they act inwards. A bar is called a tie when in tension ; a strut when in com- 
 pression (CEOPTON, Lectures on Applied Mechanics, at the Royal Military 
 Academy, London, 1877). 
 
 \" A surface with an edge is simply connected, if its edge is a single closed 
 continuous line which does not intersect itself. 
 
 J If the edge of 2 is a plane polygon of n sides, that of 2' will be a point, 
 the vertex of a polyhedral angle of n faces. 
 
 We have evidently m ~~! n. 
 
 J| Therefore m can never be less than p 1. 
 
144 APPLICATION OF RECIPROCAL [30- 
 
 Reciprocally, FT has m + np + 2 faces, n+p vertices, and 
 2 n + m edges. 
 
 30. Suppose now that the projection of 2 ' is the skeleton of 
 a frame with p joints, and m rectilinear bars, and that the 
 external forces which are applied to it have for their lines of 
 action the projections of the sides of the polygonal edge of 2', 
 and are represented in magnitude by the n sides of the pro- 
 jection of the polygonal edge of 2*. Then the projection of 
 the face of n ', which lies in the plane co , will be the funicular 
 polygon of the external forces, corresponding to the pole 0, 
 the projection of 2, ; and the projections of the m edges of 
 2, not pertaining to its polygonal edge, represent the values 
 of the internal forces or stresses to which the corresponding 
 bars of the structure are subjected, in consequence of the 
 given system of external forces. 
 
 31. If the point il is removed to infinity in a direction per- 
 pendicular to the orthographic plane, the plane o> will coincide 
 with the plane at infinity. Then the first diagram reduces to 
 the projection of 2, i.e. to the entire system of the straight 
 lines which represent the magnitudes of the external and 
 internal forces ; and the second diagram, from which the 
 funicular polygon has completely disappeared, merely contains 
 the skeleton of the structure (i.e. the lines of actions of the 
 internal forces), and the lines of action of the external forces. 
 In the figures which accompany the text, the first diagram is 
 indicated by the letter b , and the second by the letter a . 
 
 32. If the external forces are all parallel to one another, as 
 very frequently happens in practice, the edge of 2 will be 
 a polygon situated entirely within a plane perpendicular to 
 the orthographic plane ; and therefore the sides of the polygon 
 of external forces will all fall on one and the same straight 
 line. 
 
 33. The diagrams may be formed by other degenerate poly- 
 gonal figures arising from analogous degenerations of the 
 figures in space. 
 
 Suppose, for example, that we have in space a solid tetra- 
 hedral angle, corresponding to a quadrilateral face in the 
 
 * This is only possible when 2 has no vertex at infinity ; i.e. when 2' has no 
 face perpendicular to the orthographic plane. 
 
-34] DIAGRAMS TO FRAMEWORK. 145 
 
 reciprocal figure; and let two edges (not opposite) of the 
 solid angle approach one another indefinitely, in their plane, 
 and ultimately coincide. The solid tetrahedral angle will 
 be replaced by a system composed of a trihedral angle and a 
 plane passing through one of its edges. Consequently the 
 quadrilateral face of the reciprocal figure will have tw r o sides 
 which, without ceasing to have a common vertex, will be super- 
 posed and may have either the same or the contrary direction. 
 Passing from the figures in space to their orthographic 
 projections, we shall have in one of the diagrams a point from 
 which four straight lines diverge, two of which will be super- 
 posed ; and in the other diagram a quadrilateral with three 
 collinear vertices'*. 
 
 34. Given the skeleton of a framework and the system of 
 external forces, it is necessary first of all to construct the 
 polygon of these forces, i.e. a polygon whose sides are equi- 
 pollent to them. In the figures contained in this work both 
 the external forces and the sides of their polygon are denoted 
 by the numbers 1 , 2 , 3 , . . . , so disposed that, if we go round 
 the contour of the polygon in the increasing order of the 
 numbers, each side is passed over in the sense of the force 
 which it represents. This way of going round the polygon is 
 called the cyclical order of its contour. 
 
 When we wish to construct the diagram reciprocal to the 
 one formed by the bars of the frame and by the lines of 
 action of the external forces, the order in which the forces 
 are made to follow one another when their polygon is con- 
 structed is not arbitrary; this order is determined by the 
 following considerations : 
 
 In the polygon of external forces > which forms part of the diagram 
 b, the sides equipollent to two forces will be adjacent, when the lines 
 of action of those forces belong to the contour of the same polygon in 
 diagram a, because that polygon corresponds to the vertex which is 
 common to those two sides. 
 
 Let us then give the index 1 to any one whatever of the 
 external forces ; the line of action of the selected force is a 
 side common to two polygons of diagram a ; the contour of 
 
 * Examples of these degenerate forms are to be found at p. 444 and in the first 
 two tables of the memoir of Professor Fleeming Jenkin, already cited on p. 132, 
 1869, and in Fig. 9 of our examples. 
 
 L 
 
146 APPLICATION OF RECIPEOCAL [35- 
 
 each of these contains the line of action of another external 
 force ; thus there are two external forces which may be re- 
 garded as contiguous to the force 1 , and the index 2 may 
 be attributed to either of them indifferently, and the index 
 n to the other, where n is the number of external forces. 
 After this, the order of the other sides of the polygon of 
 external forces is completely determined. Suppose that 
 the joints, to which the external forces are applied, all lie 
 on the contour* of the skeleton of the framework, then 
 the forces must be taken in the order in which we meet 
 them in passing round the contour. When we do not follow 
 these rules, as well as those previously laid down, we are 
 still able to determine the internal forces graphically, but 
 we no longer have two reciprocal diagrams, and the figures 
 will be very complicated ; since any segment which does not 
 lie in its proper place will have to be repeated or removed 
 to another place in view of further f constructions ; just 
 what happens in the old method, which consists in con- 
 structing a polygon of forces for each separate joint of 
 the framework. 
 
 35. The polygon of external forces being thus constructed, 
 we complete the diagram b, by constructing successively the 
 polygons which correspond to the different joints of the 
 framework. 
 
 The problem, of constructing a polygon all of whose sides 
 have given directions, is soluble when only two of the sides 
 are unknown. For this reason we ought to commence at 
 a joint through which only three straight lines pass ; the 
 lines of resistance of two bars, and the line of action of an ex- 
 ternal force. The segment equipollent to the external force will 
 be a side of the triangle corresponding to the joint in question, 
 and consequently we are able to construct the triangle. 
 
 * The contour of certain structures (trusses) is composed of two systems of 
 bars, an upper and lower. The bars which unite the joints of one of these systems 
 to those of the other (we consider them as going from the upper to the lower) are 
 diagonals or braces, if they are inclined from left to right, and if inclined in 
 the opposite sense contra- diagonals. We call the upright bars verticals. 
 
 "t* For this reason Figs. I and 3 of PI. xvi. in the atlas of Culmann's Graphical 
 Statics are not reciprocal, and similarly Figs. 7 and 7, of pi. xix., &c. ; on 
 the other hand, diagrams 168 and 169 of p. 422 (1st edition) are perfectly 
 reciprocal. 
 
-36] DIAGRAMS TO FRAMEWORK. 147 
 
 The construction presents no ambiguity, if we remember 
 that to a bar of the framework belonging to the contour of a 
 polygon of the diagram a, to which the lines of action of two 
 external forces also belong, corresponds in the diagram b a 
 straight line passing through the vertex common to the sides 
 equipollent to those two forces. 
 
 Then we pass on successively to the other joints, taking 
 them in such an order that in each new polygon to be 
 constructed only two unknown sides remain. 
 
 In the figures given, all the lines of each of the diagrams 
 have numbers attached to them indicating in what order the 
 operations are to be performed. 
 
 1 The figure can be drawn in a feio minutes, whereas the algebraic 
 computation of the stresses, though offering no mathematical diffi- 
 culty, is singularly apt, from mere complexity of notation, to result 
 in error*' 
 
 36. A superficial consideration might lead us to conclude 
 that the solution of the above problem is possible and deter- 
 minate, even in the case where the frame has no joint at which 
 three straight lines only intersect f. 
 
 Suppose, for example, that the skeleton of the structure is 
 formed by the sides 5,6,7,8 of a quadrilateral and the 
 straight lines 9,10,11,12 which join its vertices to a fifth 
 point; and let the external forces 1,2,3,4 be applied at 
 the vertices (8, 5, 9), (5, 6,10), (6, 7, ll), (7, 8, 12) of the 
 quadrilateral J. Construct the polygon 1,2,3,4 of external 
 forces and through the points (1 , 2), (2, 3), (3, 4), (4, 1) re- 
 spectively, draw the indefinite straight lines 5,6, 7 , 8 . 
 
 Then our problem is, to construct a quadrilateral, whose 
 sides 9,10,11,12 are respectively parallel to the lines denoted 
 by these numbers in the given diagram, and whose vertices 
 (9, 10), (10, 11), (11 , 12), (12,9) lie respectively on the straight 
 lines 5,6,7,8. Since the problem of constructing a quadrila- 
 teral whose sides have given directions (or pass through given 
 
 * Professor Fleeming Jenkin, p. 443 of the volume of the Transactions of Edin- 
 burgh already cited on p. 132. 
 
 t The frame or truss is always supposed to be formed by triangles only. 
 
 J Exactly the same reasoning applies to the structure formed by any polygon 
 whatever, and the straight lines joining its vertices to a fixed point. 
 
 We have given no figures for this article, but the reader can easily supply 
 them for himself. 
 
 L 2 
 
148 APPLICATION OF EECIPKOCAL [36- 
 
 points on the same straight line), and whose vertices lie on 
 four fixed straight lines admits in general of one, and only 
 one, solution ; we might at first sight suppose that the diagram 
 of forces is completely determined. 
 
 But this illusion vanishes when we remember that the geo- 
 metrical problem presents certain cases which are impossible 
 and indeterminate. In a word, suppress one of the con- 
 ditions, that is, assume that the quadrilateral has its sides 
 parallel to the given directions, and that its first three vertices 
 only lie on given straight lines 5 , 6 , 7 ; then we know that 
 the fourth vertex describes a straight line r* whose point of 
 intersection with the given straight line 8 , will determine 
 the fourth vertex, and give the required solution. Now if the 
 data are such that r is parallel to 8 , we arrive at an impossi- 
 bility. Again, making a still more special hypothesis, if the 
 straight line r coincides with 8, the problem is indeterminate, 
 and an infinite number of quadrilaterals will satisfy the con- 
 ditions of the problem. 
 
 In order to show that the construction of the diagram reci- 
 procal to the given diagram is indeterminate or impossible, it 
 is enough to reflect, that if we consider the given diagram as 
 the polygon of forces whose magnitudes are represented by 
 the segments 5, 6, 7, 8, the pole of the polygon being the 
 point (9, 10, 11, 12), then the reciprocal diagram (9, 10, 11, 
 12) is simply the corresponding funicular polygon. But, 
 in order that the construction of the funicular polygon 
 may be possible, it is necessary that the forces should be in 
 equilibrium : if then we suppose the magnitude of the forces 
 5, 6, 7, 8 given, and also the lines of action 5,6,7 of three of 
 them, the line of action of the fourth is perfectly determined, 
 and is the straight line r of which we have just spoken. 
 Hence if r and 8 do not coincide, the forces in question 5,6,7, 8 
 are not in equilibrium, but are equivalent to an infinitely 
 small force at an infinite distance, and the problem is impos- 
 sible ; if, however, r and 8 do coincide, that is to say, if 
 the forces in question are in equilibrium, the problem is inde- 
 terminate, since for a given pole and system of forces we are 
 able to construct an infinite number of funicular polygons. 
 
 * This is the Porism of PAPPUS, for which see CEEMONA'S Protective Geometry, 
 Art. 114. 
 
-37] DIAGBAMS TO FRAMEWORK. 149 
 
 In the first of these two cases, equilibrium might be 
 obtained by combining the forces 5, 6, 7, 8 with a force 
 equal and opposite to their infinitely small resultant, situated 
 at infinity, i.e. by considering the polygon 5, 6,7,8 as the 
 projection not of a quadrangle but of a pentagon, two succes- 
 sive vertices of which project into one and the same point 
 (7, 8, 12). The straight line 12 would then be the projection 
 of two distinct straight lines in space, and consequently in 
 the reciprocal diagram, to the point (9 , 10, 11, 12) there would 
 correspond an open pentagon 9, 10, 11, 12,12', having its 
 vertices (9, 10), (10, 11), (11 , 12), (12'- 9) situated respectively 
 on the straight lines 5,6,7,8, and its vertex 1 2, 1 2' at infinity. 
 
 37. Each rectilinear bar of a framework is the line of 
 action of two equal and opposite forces, applied respectively 
 at the two joints connected by the bar. The common magni- 
 tude of these two forces, that is to say, the measure of the 
 stress which they exert on the bar, is given by the length of 
 the corresponding straight line of diagram b. These two 
 forces may either be considered as actions or as reactions : 
 to pass from one case to the other, it is only necessary to 
 reverse their directions'*. 
 
 38. Each joint of the framework is the point of application 
 of a system of at least three forces, in equilibrium ; one of them 
 may be an external force, the others are the reactions which 
 are called into play in the bars which meet at the joint in 
 question. It is sufficient to know the sense of one of these 
 forces in order to obtain that of all the others. Two cases are 
 possible. 
 
 First, if an external force be applied at the joint con- 
 sidered ; then if we pass along the corresponding side of the 
 polygon of forces in the sense of that force, each of the other 
 sides of the polygon will be passed over in the sense which 
 belongs to its corresponding internal force, considered as a 
 reaction applied at the joint in question. If, on the con- 
 trary, we wish to find the sense in which the internal forces 
 would act when considered as actions, it is sufficient to reverse 
 the direction of the external force. 
 
 * In the figures of this work the ties are shown by finer lines than the struts. 
 In the figures of Culmann and Reuleaux the struts are shown in double lines, the 
 ties fey single ones. See the first note on p. 143. 
 
150 APPLICATION OF RECIPROCAL [39- 
 
 If the only forces which act at the joint in question are 
 internal forces, it is likewise sufficient to know the sense of 
 one of them in order to find by the process just explained the 
 sense of all the others. 
 
 We shall call that order which corresponds to the internal 
 forces considered as actions the cyclical order of the contour of 
 a polygon of the diagram b. We see then, that by commenc- 
 ing at any joint at which an external force is applied, we are 
 able to determine in succession the magnitude and sense of all 
 the internal forces. By considering one of the internal forces 
 as an action applied at one of the two joints between which it 
 acts, we are able to recognise at once whether the bar connect- 
 ing the same joints is in compression or tension. 
 
 Every straight line in diagram I is a side common to two 
 polygons : in going round the contour of each of them in their 
 respective cyclical order, the sides will be described once in 
 the one sense, and once more in the contrary sense *. 
 
 This corresponds to the fact that the straight line in ques- 
 tion represents two equal and opposite forces acting along 
 the corresponding bar of the framework. 
 
 39. We know that the algebraic sum of the projections of 
 the faces of a polyhedron is equal to zero. By applying this 
 theorem to the polyhedron IT (Art. 29), remembering that the 
 projection of the surface 2 forms the polygons of the diagram 
 #, corresponding to the joints of the structure, whilst the pro- 
 jection of the rest of the polyhedron IT is simply the polygon 
 of external forces, we arrive at the following theorem : 
 
 Regarding the area of a polygon as positive or negative according 
 as that area lies to the right or to the left of an observer passing 
 round its contour in the cyclical order which belongs to it, then the 
 sum of the areas of the polygons of diagram b which correspond 
 to the joints of the framework is equal and opposite to the area of 
 the polygon of external forces. 
 
 Clerk Maxwell has arrived at this theorem in another 
 
 * This property is in accordance with the so-called Law of Edges (KANTEN- 
 GESETZ) of polyhedra possessing one internal surface and one external. 
 
 See M&BIUS, Ueber die Bestimmiing des InJialts eines Polyeders (Leipziges 
 Berichte, 1865, vol. 17, p. 33 and following), or Gesammelte Werke, 2nd Baud, 
 p. 473 ; also BALTZEE, Stereometric, 8, Art. 16. 
 
-40] 
 
 DIAGKAMS TO FRAMEWORK. 
 
 151 
 
 way by investigating whether it is possible or not to con- 
 struct the diagrams of forces * for any plane frame whatever. 
 
 40. The method of sections generally employed in the study 
 of variable systems furnishes a valuable means of verification. 
 
 If an ideal section le made in the structure, then in each of the 
 parts so obtained, the external forces are in equilibrium with the 
 reactions of the Lars cut across liij the section. 
 
 If only three of the reactions are unknown, we can deduce 
 them from the conditions of equilibrium, since the problem 
 of decomposing a force P into three components, whose lines 
 of action 1,2,3 are given and form with 0, the line of action of 
 P, a complete plane quadri- 
 lateral, is a determinate pro- 
 blem and admits of only one 
 solution. 
 
 In fact (Fig. 6) it is only 
 necessary to draw one of the 
 diagonals of the quadrilate- 
 ral, for example, the straight 
 line 4 which joins the points 
 (0,1), (2,3); to decompose 
 the given force into two 
 components along the straight 
 lines 1, 4 (we do this by con- 
 structing the triangle offerees 
 , 4 , 1 of which the side o is 
 given in magnitude and direction) ; and finally to decompose 
 the force 4 along the straight lines 2 and 3 (by constructing 
 in like manner the triangle of forces 4,3,2). 
 
 This method, which may be called the static method, is all- 
 sufficient for the graphical determination of the internal forces, 
 equally with the geometrical method, previously explained, 
 which is deduced from the theory of reciprocal figures, consists 
 in the successive construction of the polygons corresponding 
 to the different joints of the structure. The static method is 
 at least as simple, it can be rendered very useful in combina- 
 tion with the latter method, and it permits the rapid verifica- 
 tion of the constructions. The external forces applied to a 
 portion of the structure, obtained by means of any section 
 
 * Memoir of 1870, p. 30, already cited on p. 132. 
 
 Fig. 6. 
 
152 APPLICATION OF RECIPROCAL DIAGRAMS TO FRAMEWORK. 
 
 whatever, and the reactions of the bars that are cut, must have 
 the property that the corresponding lines of diagram b form a 
 closed polygon. This polygon must be the projection of a 
 closed twisted polygon, and not merely of an open crooked 
 line whose extremities are situated in a straight line perpen- 
 dicular to the orthographic plane ; this condition requires 
 that the twisted reciprocal polygon shall also be closed, i.e. we 
 are able to unite the corresponding lines of the diagram a by 
 a closed funicular polygon. 
 
 The method of sections may also be presented in another 
 form. Denoting again by the resultant of all the known 
 forces applied to the portion of the structure considered, and 
 by 1, 2, 3 the three unknown reactions, the sum of the moments 
 of these four forces in regard to any point whatever is zero. 
 Now, by taking as the centre of moments the point where 
 two lines of resistance meet, for example, the point (2 , 3), the 
 moment of the third reaction 1 will be equal and opposite 
 to that of the force 0. We thus obtain a proportion between 
 four magnitudes (the two forces and their moments) among 
 which the only unknown quantity is the magnitude of the 
 force 1. This is the method of statical moments, by which 
 the internal forces developed in the different bars of a frame- 
 work can be calculated numerically, instead of being con- 
 structed graphically *. 
 
 * See A. KITTEK, Elementdre Theorie und BerecJinung eiserner Dach- und 
 BrucJcen-Constructionem, 2nd edition (Hannover, 1870). 
 
CHAPTEE IV. 
 
 EXAMPLES OF FRAME- AND STEESS- DIAGRAMS. 
 
 41. WE will now pass on to the study of some suitable 
 examples to show the simplicity and elegance of the graphic 
 method. We do not always adhere to regularity and sym- 
 metry of form in the structures which we are about to study, 
 although in practice engineers hardly ever depart from these 
 conditions. But the symmetrical forms of practice are only par- 
 ticular cases of the irregular ones of abstract geometry : and there- 
 fore the forms which we shall treat include all the cases 
 which are possible in practice. In what follows, the expres- 
 sion 'framed structure' will be used in the general and 
 theoretical sense which Maxwell attributed to the word 
 frame. 
 
 * A frame is a system of lines connecting a number of points. A 
 stiff frame is one in which the distance between any two points 
 cannot be altered without altering the length of one or more of the 
 connecting lines of the frame. A frame of s points in a plane 
 requires in general 2s 3 connecting lines to render it stiff*' 
 
 We confine ourselves to the study of plane figures formed 
 by triangular parts. 
 
 42. As a first (general and theoretical) example, let 1,2, 
 3 , . . . , 10 (Fig. 7 a) be a system of ten external forces in equi- 
 librium ; construct the corresponding polygon of forces, and 
 join its vertices to an arbitrary pole (Fig. 7 &, in which the 
 polygon of forces is represented by double lines). Draw next 
 a funicular polygon, having its sides respectively parallel to 
 the rays from (Fig. I), and its vertices lying in the lines of 
 action of the forces 1,2,3, . . . , 10 . The forces in question are 
 applied at the different joints of a framed structure, the bars 
 of which are numbered from 11 up to 27 (Fig. 7 a). 
 
 * Page 294, Phil. Mag., April 1864. 
 
154 
 
 EXAMPLES OF FRAME- AND STRESS- DIAGRAMS. [42- 
 
 We commence by constructing the triangle corresponding 
 to the joint (10,11, 12), drawing through the extremities 
 of the straight line 1 two straight lines 11, 12, respectively 
 
 Fig. 7 a. 
 
 Fig. 76. 
 
 parallel to 11 and 12 ; we notice that the straight line 1 1 
 must pass through the point (1, 10), because in the dia- 
 gram , the lines 1, 10, 11 belong to the contour of the same 
 polygon* ; for the same reason 12 must pass through the 
 point (9 , 10) . Passing round the contour of the triangle just 
 
 * This polygon is a quadrilateral, whose fourth side is the side of the funi- 
 cular polygon comprised between the forces 1 and 10. As previously stated 
 (Arts. 27, 31) the whole of the funicular polygon might have been removed to 
 infinity. 
 
-43] EXAMPLES OF FRAME- AND STRESS- DIAGRAMS. 155 
 
 obtained, in a sense contrary to that of the force 10, we 
 obtain the sense of the actions called into play at the joint we 
 are dealing with, along the lines 11 and 12 ; and it is thus seen 
 that the bar 11 is in a state of compression and the bar 12 of 
 tension. 
 
 Now construct the quadrilateral corresponding to the joint 
 at which the force 9 is applied, and for this purpose, draw 1 3 
 through the point (11, 12) and 14 through the point (8,9). 
 The bar 13 is in compression, 14 in tension. 
 
 Next construct the pentagon corresponding to the joint at 
 which the force 1 is applied, by drawing 1 5 through the point 
 (13, 14), and 16 through the point (1,2). The pentagon thus 
 obtained is a crossed one. The bar 15 is in tension, 16 in 
 compression. 
 
 Then construct the pentagon corresponding to the joint at 
 which the external force 8 is applied ; by drawing the line 
 17 through the point (15, 16), and the line 18 through the 
 point (7, 8). The bar 17 is in compression, 18 in tension. 
 
 Continuing in this manner we find all the other internal 
 forces. The last partial construction gives the triangle which 
 corresponds to the point of application of the force 5 . The 
 bars 20, 21, 24, 25, 27 are in compression; 19, 22, 23, 26 
 are in tension. 
 
 43. Figure 8 a represents a bridge girder, at the joints of 
 which are applied the forces 1 , 2 , 3 , . . . , 8 , 9 , 1 , . . . , 16 all 
 vertical; the forces 1 and 9 acting upwards represent the 
 reactions of the supports ; the forces 2,3, ... , 8 are the weights 
 applied at the joints on the upper platform; and 10,11,..., 
 16 the weights applied at the joints of the lower platform. 
 
 These forces are taken in the order in which they are met 
 with in going round the contour of the structure ; and in 
 diagram b the sides of the polygon of external forces are 
 disposed in the same order. This polygon has all its sides 
 lying in the same vertical straight line ; the sum of the 
 segments 1 , 9 is equal and opposite to that of the segments 
 2, 3,.. .8. 10, 11, ...,16, because the system of external forces 
 is necessarily in equilibrium. 
 
 The diagram I is completed by following precisely the same 
 rules as those just laid down. Commence at the joint (l, 17, 
 18) ; draw the straight line 17 through the point (1,2), where 
 
156 EXAMPLES OF FEAME- AND STRESS- DIAGRAMS. [43- 
 
 the upper extremity of the segment 1 meets the upper 
 extremity of the segment 2 ; and the straight line 1 8 through 
 the point (16, 1), which is both the lower extremity of the 
 segment 1 6 and that of the segment 1 . 
 
 Fig. So. 
 
 Pass on to the joint (2, 17, 19, 20). Draw 19 through the 
 point (17,18), and 20 through the point (2, 3), the lower end 
 of 2 and upper extremity of 3 ; and we obtain the polygon 
 2,17,19,20, which is a rectangle. 
 
 Fig. SI. 
 
 Construct the polygon corresponding to the joint (16, IS, 
 19, 21, 22). For this purpose draw 21 through the point 
 (19, 20), and 22 through the point (15, 16) ; we thus obtain 
 a crossed pentagon. Continue to deal in the same manner 
 with each of the points of application of the forces 3 , 15 , 4 , 
 14 , 13 , 5 , 12 , 6 , 11 , 7 . 10 , 9 , taken in succession. 
 
 Since the diagram a, which represents the skeleton of the 
 structure and all the external forces, has for its axis of sym- 
 metry the vertical which passes through the centre of the 
 figure, the diagram b has for its axis of symmetry the median 
 horizontal line. For example, the triangle 9, 45, 44 is sym- 
 
-44] 
 
 EXAMPLES OF FRAME- AND STRESS- DIAGRAMS. 
 
 157 
 
 metrical to the triangle 1, 17, 18 ; the rectangle 8, 45, 43, 42 
 to the rectangle 2, 17, 19, 20: and so on. 
 
 All the upper bars are in compression, and all the lower 
 ones are in tension. 
 
 The diagonals and contra-diagonals are all in compression ; 
 finally two of the verticals 23, 39 are in tension, and all the 
 rest in compression. 
 
 44. Figure qa* represents one half of a locomotive shed. 
 The external forces are the weights 1,2,3,4,5 applied at 
 
 the upper joints of the frame, and the reactions 6 and 7 of the 
 wall and column. Again, all the external forces are parallel, 
 and consequently the polygon of forces reduces, in diagram 
 #, to one straight line. The force 6 (taken in the opposite 
 sense to that in which it really acts) is equal to a certain part 
 of the weight 5 ; by adding the difference 
 to the other weights we get the magni- 
 tude of the force 7 . 
 
 In the diagram b the direction of the 
 lines 8 and 1 3 coincide ; the first is a 
 part of the second. Here then we have 
 for the polygon corresponding to the joint 
 (8, 10, 12, 13) one of those degenerate 
 forms about which we spoke in Art. 33 ; the 
 polygon is in fact a quadrilateral 8, 10, 
 12, 13, having three of its vertices (13,8), 
 (8, 10), (12, 13) in one straight line. 
 
 The polygon 5 , 17, 18, 6 , corresponding to the point where 
 
 * This example is taken from PI. xix. of the atlas of Graphische StatiJc of CULMANN, 
 1st edition. As previously stated, the two diagrams are not rigorously reciprocal. 
 
158 EXAMPLES OF FRAME- AND STRESS- DIAGRAMS. [45- 
 
 the roof is supported by the wall, presents an analogous 
 degenerate form, since the vertices (6 , 5), (upper point of the 
 segment (6) ,(5,17), and (1 8 , 6) all lie in the same straight line. 
 The lower bars 8, 13, 18 are in compression, as well as 
 the diagonals 10 , 14 , 16 , the column 7 and the wall 6 ; while 
 the upper pieces 9 , 11 , 15 , 17 and the diagonal 12 are in tension. 
 
 45. Diagram a of Fig. 10 represents a truss at the upper 
 joints of which are applied the oblique forces 1,2, . . . , 7 , which 
 
 Fig. ioa. 
 
 Fig. lob. 
 
 may be considered as the resultants of the dead-loads and 
 wind pressure ; the forces 8 , 9 represent the reactions at 
 the supports. 
 
 The polygon of external forces is drawn in diagram l> with 
 double lines. 
 
 We construct successively the triangle 1, 10, 11 , the quad- 
 rilateral 9,10,12,13, the pentagon 2,11,12,14,15, the 
 quadrilateral 13, 14, 16, 17, the crossed pentagon 3, 15,16, 
 18, 19 ; the crossed quadrilateral 4, 19, 20, 21, the pentagon 
 17, 18, 20, 22, 23, and so on. 
 
 The upper bars 15, 19, 21, 25 are in compression, as well as 
 the lower bars 1O, 13, 30, and the verticals 12, 16, 24, 28 ; 
 whilst all the remaining bars of the structure are in tension. 
 
 46. The diagram a of Fig. 1 1 represents a suspension 
 bridge, loaded at each of its upper joints with weights 1,2, 
 ... , 8, and at each of its lower joints with weights 10, 11, 21, 
 ...,16; the weights are kept in equilibrium by the two 
 
-46] EXAMPLES OF FRAME- AND STRESS- DIAGRAMS. 159 
 
 oblique reactions 9, 17 at the two extreme points of the 
 structure *. 
 
 The polygon of external forces has its first eight sides in 
 succession along the same vertical straight line, and its seven 
 
 3') 
 
 / w\wv / W V 2 
 
 '24 \/2S \/32 \/j6 \/4 
 
 ^ 37 \ 
 
 I r I 
 
 JJ 12 11 
 
 Fig. 1 1 a. 
 
 last sides situated in another vertical straight line. The 
 oblique sides 9 and 17 intersect, so that the polygon is a 
 crossed one. We construct successively the polygons 1,17, 
 19, 18; 16, 19,20, 21 ; 2, 18,20, 22, 23; 15, 21, 22, 24, 25; 
 3, 23, 24, 26, 27 ; and so on ; most of which are crossed. 
 
 Diagram b shows that the upper bars are all in tension, 
 and that the tension decreases from the ends towards the 
 
 Fig. 1 1 1. 
 
 middle of the structure ; the bars of the lower boom are also 
 all in tension, but in them the tension decreases from the 
 middle towards the ends. 
 
 The extreme diagonals and contra-diagonals are in tension ; 
 in the portion situated to the left of the axis of symmetry, the 
 diagonals or braces are alternately in tension and compression ; 
 similarly they are on the right but in the reverse order. Con- 
 sidering separately the ties and struts, we see that the internal 
 
 * This example is analogous to one of those studied by Maxwell in his memoir 
 of 1870. 
 
160 EXAMPLES OF FEAME- AND STEESS- DIAGRAMS. [47- 
 
 forces decrease from the ends towards the middle of the 
 structure. 
 
 In this example again the diagrams have axes of symmetry. 
 
 4 
 
 Fig. 12. 
 
 47. Diagram a of Fig. 1 2 represents a framed crane- 
 post; the weight of the machine is distributed over the 
 
-48] EXAMPLES OF FRAME- AND STRESS- DIAGRAMS. 161 
 
 different joints, and is represented by the sum of the forces 
 1. 2, 3, ..., 9 ; the force 5 also includes the load the crane is 
 required to lift. 
 
 All these forces are kept in equilibrium by the reactions at 
 the supports, the magnitudes of which are obtained by re- 
 solving the resultant weight into three forces acting along the 
 lines 10 , 11 , 12 . These forces, taken in the contrary sense, 
 furnish the pressures which are supported by the strut 10, 
 and the column 11 , and the tension in the tie 12. 
 
 48. These external forces may be determined as follows : 
 
 We take on the same vertical line segments representing 
 the magnitudes of the forces 1 , 2,3 ... 9 , and choose a pole 
 arbitrarily; join the pole to the points (0, l), (1, 2), (2, 3), 
 ... (8, 9), (9, 0)*, and construct the corresponding funicular 
 polygon. The vertical through the point where the extreme 
 sides (0,1), (9, 0) meet will be the line of action of the total 
 weight of the crane and load, a weight represented in magni- 
 tude by a segment, which has the same initial point as the 
 segment 1, and the same final point as the segment 9. If now 
 we decompose the resultant weight, which is now known, 
 into three components, whose lines of action are the straight 
 lines 10, 11, 12, employing the construction of Art. 40 (Fig. 6), 
 we obtain the three forces 10, 11, 12. That is to say, these 
 taken in the opposite sense and the given weights complete 
 the system of external forces. 
 
 In order to obtain the diagram b, we construct first the 
 polygon of the external forces, taking these forces in the order 
 in which we encounter them in going round the contour of the 
 structure. Then construct in succession the polygons corre- 
 sponding to the joints : (5, 13, 14), (4, 13, 15, 16), (6, 14, 15, 17, 
 18), and so on in the manner just described. 
 
 The diagram thus obtained enables us to see that the bars 
 of the upper part are in tension, and those of the lower 
 part in compression ; while the diagonals are alternately in 
 tension and compression. 
 
 * Here (0, 1) represents the initial point ot the segment 1, and (9,0) the 
 final point of segment 9. In the figure the rays, from the pole and the sides of 
 the corresponding funicular polygon, are shown by dotted lines 
 
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