CREMONA'S
TWO TREATISES ON GEAPHICAL STATICS
HUDSON SHARE
HENRY FROWDE
OXFORD UNIVERSITY PRESS WAREHOUSE
AMEN CORNER, E.G.
GRAPHICAL STATICS
TWO TREATISES
ON THE
GEAPHICAL CALCULUS
AND
BECIPBOCAL FIGURES IN GRAPHICAL STATICS
BY
LUIGI CEEMONA
LL.D. EDIN., FOE. MEMB. R.S. LOND., HON. F.B.S. EDIN.
HON. MEMB. CAMB. PHIL. SOC.
PEOFESSOE OF MATHEMATICS IN THE TJNIVEESITY OF SOME
TEANSLATED BY
THOMAS HUDSON BEARE
B.SC. LOND., ASSOC. M. INST. C.E., F.R.S. EDIN.
PROFESSOR OF ENGINEERING AND APPLIED MECHANICS, HERIOTWATT COLLEGE, EDINBURGH
AT THE CLARENDON PRESS
1890
[All rights reserved]
TEANSLATOE'S PEEFACE.
FOE some years I had used a rough English manuscript
summary of Professor CREMONA'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 chiefly to pressure of work both on the
part of myself and Professor CREMONA.
I feel sure that the translation will supply a longfelt 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.
HERIOTWATT COLLEGE, EDINBURGH.
CONTENTS.
Page
TRANSLATOR'S PREFACE v
ELEMENTS OF THE GRAPHICAL CALCULUS.
AUTHOR'S PREFACE TO THE ENGLISH EDITION xv
CHAPTER I.
THE USE OF SIGNS IN GEOMETRY.
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
89. 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
1314. Relation between the triangles determined by 4 points in
a plane 9
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 nonconcur
rent straight lines in its plane 10
Vlll CONTENTS.
Art. Page
17. Circuits, simple and selfcutting, Modes 12
1823. Areas of selfcutting circuits 12
24. Eeduction of selfcutting to simple circuits 18
25. Eelation between two polygons with equipollent sides .. .. 20
2630. Areal relation of a pole and system of segments equi
pollent to a closed or open circuit 21
CHAPTEE IT.
GRAPHICAL ADDITION.
3133. 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
3538. Cases where the sum is zero 27
39. Geometrical subtraction 28
4042. Projection of segments and circuits 28
4345. Theorems for 2 systems of points, when the resultants of
the segments joining each system to the same pole are
equal 29
4648. Extension of the word sum to include absolute position 31
4952. Constructions for completely determining the sum .. 32
53. Case of parallel segments 35
54. Case of 2 parallel segments 35
CHAPTEE III.
GRAPHICAL MULTIPLICATION.
5556. 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
5962. Multiplication of a number of segments of a straight line
by a constant ratio. Similar point rows 39
6364. Multiplication of a system of segments by a system of
ratios 42
6567. Case where these segments are parallel, and other cases 45
6869. Multiplication of a segment by a given series of ratios .. 49
7072. 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
CHAPTER V.
EXTRACTION OF ROOTS.
7677. Equiangular Spiral 59
7882. Properties and construction of the spiral 60
8384. 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.
9091. Lill's construction of a complete polynomial 70
9293. Reduction of the degree of an equation 73
94. Equations of the second degree 75
CHAPTER VII.
REDUCTION OF PLANE FIGURES.
9596. Reduction of a triangle to a given base 77
97100. Reduction of a quadrilateral 78
101103. Reduction of polygons .. 80
104105. Reduction of sectors, and segments of a circle .. .. 82
106107. Examples, figures bounded by circular arcs and
rectilinear segments 83
108110. 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
112114. Centroid of a system of n points 89
115. Centre of mean distances 90
116119. Ceutroid of a system of n points with integral coeffi
cients, or loads 90
120123. 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
128129. 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
133134. 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
138139. Centroid of a trapezium 104
140. Centroid of any rectilinear figure or polygon 106
141143. Examples (polygon, cross sections of Angle and Tee
irons) 106
144145. 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 77 117
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 RECIPROCAL
FIGURES 131
CHAPTER III.
APPLICATION OF RECIPROCAL DIAGRAMS TO FRAMEWORK .. 143
CHAPTER IV.
EXAMPLES OF FRAME AND STRESS DIAGRAMS 153
ELEMENTS OF THE GRAPHICAL CALCULUS
AUTHOE'S PEEFACE
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 MR. 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 Geometrla
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.
HOME, July 1888.
ELEMENTS OP THE GEAPHICAL CALCULUS,
CHAPTEK 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 J
as X remains between and Fig. i.
A y 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 2
tive. For example, if X
Jig. 2.
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 of 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.
2 THE USE OF SIGNS IN GEOMETRY. [2
number (x) of linear units it contains, its sense is shown by the
sign + or of the number #.
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
y , 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 CBAC + 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 BCAC + 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
always holds.
* That is to say, two magnitudes of equal arithmetical values, but with opposite
algebraical signs, such as + a and a.
f MOBIUS, BarycentriscTier Calcul (Leipzig, 1827), 1, 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 J, 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
therefore AB = OB OA,
or AB = AO+OB.
4. If A, B, C, ..., M, ^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
NA = NO+OA,
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 2 , ..., A 9 ,
segments A^A^ A^A 2 ', ..., A n A n ' all parallel to some given
direction in the plane until they intersect a fixed straight line
A^A^^.A^ then the sense of one segment determines that
B 2
4 THE USE OF SIGNS IN GEOMETRY. [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 Bellavitis.
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,
CA = CD AD,
now multiply these three equations by AD, BD, and CD re
spectively and add the results, the righthand 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
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, RP, and PQ
respectively, and the above equation may therefore be written
.qR + MB.RP + MC.PQ = 0.
7] THE USE OF SIGNS IN GEOMETRY. 5
If we now draw through any other point M f a new
transversal in the fixed direction PQJK, cutting p, q, r in
A', B', and C', we have similarly
M'A'.QR+M'B'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, , C
respectively, then the segments MA, MS, MC are connected by the
relation
where a, I), c are constants.
From the point M let fall perpendiculars MD, ME, MF
upon the three straight lines p, q, ;; and also from some
arbitrarily chosen point 8 of the line PQR perpendiculars SU,
SF, SW upon the same given straight lines. Then since the
triangles MAD, SPU are similar we have
Therefore MA = ~ MD,
and similarly MB = ME,
8r
The equation
MA. QR + M3.RP + MC.PQ = 0,
may therefore be written
that is to say :
If we 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 MD, 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 GEOMETKY. [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
Fig. 5. Fig. 6.
trorsum, in the sense of rotation of the hands of a watch], or
from right to left [sinis trorsum]. 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 R. 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 db 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),
ab3QOxn = ab.
If OA, OB are two positive segments
of the straight lines a, b, the angle ab
XX
can also be denoted by OA . OB, or
more briefly by AOB.
The sum of the angles ab, ba is equal
to any number of complete revolutions ;
we may therefore write Fig. 7.
ab + ba = 0,
or ba = ab, or ab = ba.
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.
But cb = be, ba = ab,
and therefore be + ca + ab = 0.
* BALTZEB, Analy. Geometric, 9.
8
THE USE OF SIGNS IN GEOMETRY.
[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\db =
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 0, 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, It, 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
Fig. 10.
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.
14] THE USE OF SIGNS IN GEOMETRY. 9
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,
ACS as equal but opposite : the first as positive, the second
negative. We may suppose the area ABC (or ACE) 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, Cmay lie in one straight line, is that the area ABC is zero.
13. PROPOSITION. If is any
point whatever , in the plane of the
triangle ABC (Fig. n),we always
have the identity
OBC + OCA + OAB = 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 + OAB = ABC.
Finally, if lies within the opposite vertex of 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
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 = ABCD,
For, using 8 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
Fig. 12.
OAB+ OCD = ABC = ABCD.
Q. E. D.
Since \AECD = DAB, the above equation may also be written
ODC= OABDAB.
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 , MFy 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.
\v
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
OCA
and similarly MCA = ME,
But from (Art. 13)
MBC + MCA + MAB = ABC,
therefore j . MD+.ME+.MF* ABC.
If we vary the position of the point M in the plane, whilst
keeping the directions OU, OF, OW 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 + (3.ME + y.MF=b,
the quantities a, (3, y, 5 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
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 GEOMETKY. [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 8 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 of
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+ OCD+ ... + OMN+ ONA,
is a constant quantity for any position whatever of the pole 0*.
Proof. Let Cf be another point in the plane ; then from the
proposition in Art. 13,
O'AB = OAB + OBO'+OO'A,
O'BC = OBC+ OCO' + 00' 'B,
0'CD = OCD+ODO'+OO'C,
&c. &c.
0'MN= OMN+ ONO' + OO'M,
O'NA = ONA + OAO' + OO'N.
* MOBIUS, Baryc. Calcul, 165, Ges. Werke, Bd. 1 ; Statik (Leipzig, 1837),
45, Ges. Werke, Bd. 3.
20]
THE USE OF SIGNS IN GEOMETRY.
13
Adding we have,
O'AB + 0'BC + O'CV + . . . + 0'MN+ O'NA
= OAB+ OBC + OCD + ...+ OMN + ONA = V,
since all the other terms cancel, because they occur in pairs of
equal and opposite terms, as, for example, 00' A and OAO f , OC/B
and OBO', and so on. We may consider the magnitude 2 as
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 T be 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 Y in passing through is in the
act of describing a positive or negative rotation.
Lemma. If a radius vector 07, 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 A, 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 pn 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.
14
THE USE OF SIGNS IN GEOMETKY.
[20
If a r and a r+l are of the same sign, the radius vector Y
will either describe the angle a r + will be equal, in sense and absolute value., to the number of
times 0r passes through 0. Therefore if (Jc h) 360 are the
number of complete rotations of the radius vector 7, th co
efficient of the elementary area w in the sum 2 will be k ^,
that is, is independent of the pole 0.
22. A given closed and selfcutting circuit (Fig. 15) divides
the plane into a definite number of finite spaces S 13 S< 2 ,..,
contiguous to one another. Each of these is bounded by a
circuit without nodes ; so that the whole plane consists of
Fig. 15.
these spaces and of the remaining (external) infinite region,
which latter we shall denote by S .
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 a/to. It is evident that the
radius vector OX cannot pass over a/ without at the same time
passing over co in the same sense ; co and co 7 will therefore enter
into 2 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 GEOMETKY. 17
o>' and o>", or between o>" 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 S will appear in the
sum 2 with the same coefficient c. That is to say, S appears
in the sum 2 with the coefficient c. If therefore e l9 c 2 ,... are
analogous coefficients for the spaces S 19 S. 2 , ..., we have
2 = ^ + ^ + ...,
if we understand that S 19 S 29 ... at the same time express
the areas of the spaces represented by these symbols.
Next let G), toj be two elements, between which the circuit
passes once; and let &> lie on the right and w l on the
left of the circuit which passes between co and co l5 in the
given sense. Take the pole upon the continuation of the
straight line b^o)...; now if X traverses that part of the
circuit which lies between &> and & 19 the radius vector OX
will describe &> once with a positive rotation, without describ
ing eoj, whilst for all other parts of the circuit the elements o>
and coj will be described simultaneously in the same sense.
The coefficient of o> will therefore exceed that of o>j 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 o> is
an element, which lies outside the spaces S l , S 2 , &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 S a
straight line to a point of $ , 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 S is equal to m n.
23. If the circuit has no nodes, we have a single finite space
S t and this has the coefficient +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 GEOMETEY.
[24
circuit has been described positively (Fig. 15 a) or negatively
(Fig. 156). In this case therefore we have
2= 8,
that is to say, If the circuit is not a selfcuttingone then the sum
2 is the area of the space enclosed ty the circuit.
Fig. 150.
Fig. 15 b.
This property naturally leads us to consider the sum 2 as
defining the area of any self cutting circuit*.
24. A selfcutting circuit can be decomposed into circuits
which are not selfcutting, by separating the (curvilinear)
i
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. 1 7 a.
Fig. 175.
of the branches themselves. Consider, for example, Figs. 16
and 17; in each a selfcutting 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. 18, where a selfcutting 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. iSa.
Fig. i8J.
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 ?*), S 2 is inside S l , S 3 inside S 2 ,..., S 2
inside #_ 15 .... 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 _ l , 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 _ 1} and so on.
Consider for example (Fig. 18) where the area is equal to
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'B ' ; on the other hand, the
* CULMANN, Graphische StatiJc, 2d ed. (Zurich, 1875), N ; . 26.
C 2
THE USE OF SIGNS IN GEOMETRY.
[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
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 ...M, C'D'W. . . 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 ED D'E'. Also
CC'D'E'EDEC = CC'D'E'ED'WC' = 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 A! 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' is another polygon, whose
sides are equipollent to those of the given one, we have
A 2 DE=A 2 D 'W + D 'EE',
A n MC =
summing up we have,
because, as we have shown above, the sum
2 \C'DD'+I)'EE'+...H'CC']
is equal to zero.
26. THEOREM. If the rectilinear segments A l B l , 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 selfcutting or not), then the sum of the triangles
OA^ + OA 2 2 + OA 3 3 + . . . + OA n B n ,
is constant wherever the pole 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, a*
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 l 1 , A 2 B 2 , A 3 J3 3 , ...A n B n ; so that the
figures A l 1 J)C, A 2 B 2 ED, ... , A n B n NM are parallelograms.
Then from (Art. 15),
OAiBi= OCDA^CD,
OA 2 2 = ODEA 2 DE,
&c. &c.
OA n n = OMNA n MN,
and also from the proposition in Art. 19
OCD+ODE+... OMN+ ONC = CDE ... MNC,
hence by addition we have
OA OAH+... +OA3= CDE...MNC+OCN
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^ +OA 2 2 +... + OA n B n
* APOLLONIUS, Loci Plani, lib. 1. L'HuiLiER, Polygonometrie, 1789, p. 92.
MOBius, Stattic, 46.
22 THE USE OF SIGNS IN GEOMETKY. [27
has a value independent of the position of 0. Hence it
follows that, in the special case where CN 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 (lying 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 + 0'A3+... + 0'A n B n = CDE...NC+ O'CN
Take the pole O f at such a distance from CN, that the
area of the triangle O'CN is equal to
then the sum O'A^ B l + 0'A 2 B z + . . . + O f 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
CLE ... , upon r, then the area of O'CN is zero, and therefore
the sum of the triangles
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 S 1 +OA 2 2 +... = OCN.
27. Conversely, if the sum OA l l , OA 2 J3 2 , &c. ... is the
same for every point in the plane, the segments A 1 l ,
A 2 H 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
AiCD + AtDE+...+A n MN, 4 , s 5t . . . t n . 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 , * 4f 3 , $5, . . ., n
But 3, 4 and s 4j3 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
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 closed 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 Fig. 31.
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
J3, and the segment A'B' the
projection of the segment AB.
The straight lines A A' BB'
are called the 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.
Fig. 34
42. Let ABC ... MNA be a closed circuit (Fig. 34), and
A',B', C", ... M', N' the projections of its vertices ; then since
A', B', &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 dosed circuit is zero.
Let A L B : , 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 1 B 1) A 2 B 2 , ...,A n B n \)e 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 Afl, OB 1J A 2 0, OB 2) ...,A n O, OB n will vanish
(Art. 38), i.e. the resultant of the segments OA^ OA 2 , ..., OA n
is equal to that of the segments OB l} OB 2 , ...,OB n .
Conversely. Given two groups of n points A 19 A 2 ,... t 4 nt
and BI , B 2 , . . . , B n ; if the resultant of the straight lines
OAL, 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 O# 15 OB 2 , ..., OB n , got by joining the same pole
to the points of the second group : then the resultant of the
segments A i 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 ^0, A 2 0, ...,A n 0,0 B l ,
O.Z? 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 1 B 1 ,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 / A l , 0'A 2 , ..., (/A n , is equal to the resultant of the
segments 'B^ , 0'B 2 , . . . , 0'B n .
Wherefore *
If, for two groups of n points A L ,A 2 , ...,A n ; _Z? l5 B 2 > ..., B n
and a fixed pole 0, the resultant of the segments OA l , OA 2 , ... , OA n
is equal to the resultant of the segments OB , 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 ',
Bi> B 2 ', . . . B^ 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
* GRASSMAN, Die A usdehnunyslehre (Leipzig, 1844), p. 41.
f See Fig. 35, and imagine the straight line r so displaced, that the points
and 0' coincide.
47] GRAPHICAL 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 n ',A l f A lr
A 2 A 2 , ...,A n 'A n is therefore equal to the resultant of the seg
ments OB^ OBJ, OBJ, . . . 05,', B^B^ , B 2 'B 2 , . . . , B n 'B n . But
(Art. 41) the resultant or sum of the segments OA^, OA 2 , ...
OA n 'ia equal to that of the segments OZ?/, OB 2 , ..., OB* 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 l ,A 2 , fyc. ; B lt B 2 , fyc. and a
fixed jjole 0, the resultant of the segments OA 19 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 1 B 1 +OA 2 B 2 +... + OA n B u (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 ny
then its closing side NC is equipollent to the resultant AB.
If the circuit is closed, i. e. if N coincides with C, but if the
sum of the areas OA l B l + OA 2 B 2 h . , . 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, ...,^, 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] GRAPHICAL 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
Z7, and draw from it the rays U? Q1 , U? 129 ..., UF n0 * to the
vertices of the circuit; where 7< >t  +1 means the vertex which
is the final point of the side i (equipollent to A i B i ) i and the
initial point of the side i+1 (equipollent to A i+1 B i+l ).
We next construct a second polygonal circuit with its
vertices 1,2, ..., n lying respectively on the lines to which
the segments A 1 S 19 A 2 B 2 , ..., A n B n belong, and with its
sides 01, 12, ..., nO respectively parallel to the rays UT 01 ,
UV 12 > ..., U7 no . 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 1 B 1 to be resolved into 2
others, situated in the sides 01, 12 of the second polygon,
and equipollent to the rays 7 ol U, ~UV Vi of the first ; in like
manner we suppose the segment A 2 B 2 resolved into two
others, situated in the sides 12, 23 of the second polygon,
and equipollent to the rays T' 12 U, UF 23 of the first ; and so on,
till finally A n B n is resolved into two segments situated on the
sides n 1 n, nO, and equipollent to the rays ^ B _j ttl U 9 U? n0 .
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 AjJB^ , 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 01, and equipollent to V G] U 9 and the last is situated
on wO, and equipollent to U? n0 , 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 12 and
fjff.
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.
f CULMANN, 1. c., Nos. 41 & 42.
D
GRAPHICAL ADDITION.
[50
than the resultant of the first and last component segments,
of which the first is situated in 01 and equipollent to P 01 E/,
and the other is situated in nO and equipollent to U?^.
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 , nQ of the second polygon.
50. If the pole U were taken in a straight line with the
two extreme points 7 ol , V nQ of the first polygon, then the
two extreme rays U? Q1 , Uf^o would coincide, and therefore
the two extreme sides 1 , nQ 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' not lying in the straight line F ol , V nQ and then
proceeding as above.
61. Even if that is not so, it may happen (Fig. 37) that the
points V nQ and T Ql 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 l 1 + OA 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 T ol U, U?^ 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]
GRAPHICAL ADDITION.
35
In this case the resultant is zero and situated at an infinite
distance ; and the sum OA l J3 l + 8tc. 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 +&c. 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 (n l) segments.
53. If we take the given segments A^B^A^B^ ... all parallel
to one another, then the first polygon T (
01
V
'23
(Fig. 39) reduces to a straight line, but the construction of the
\
2
^
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~J$^ A 2 B 2 ^ the con
struction may be simplified as follows (Figs. 40, 41). In the
unlimited straight line A l S 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 V B V Then the common point of the
straight lines CD', and CD lies on the required resultant. For
if we draw 1/E parallel to C'D and join to E, then , D, E
represent the vertices 7 ol , ? 12 , 7 2Q of the first polygon, and
takes the place of the point U ; the points L', 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.
Fig. 40.
Fig. 41.
From the similar triangles OCD, OD'C' we have
OC f : OD= C'tf.LC
= A^:B^
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.
GKAPHICAL MULTIPLICATION.
55. To multiply a straight line a by the ratio of two other
straight lines b : c, we must find a fourth straight line so such
that the geometrical proportion holds :
c : o = 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, b ; and in the
second the line homologous to c is a ; then x is the line in the
second triangle homologous to & ; or else
In the first there are two lines proportional or equal to o
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, ON, OL to represent the segments a, b, c, then
OQ = x. But if OL = b,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
rightangled 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 c.
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] GKAPHICAL 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 4 b, a + 2 b, &c., &c. ;
then their extremities M,M 19 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 OM 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 nl points of
division will be the lengths
of the nl 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 Fi s 4
equal parts *.
59. If several segments AB, AC,...,3C,...of& straight line
u have to be multiplied by a constant ratio b : 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
~AB = 1C~ ~C = '" = c h '
The straight lines u, uf are called similar pointrows, and the
points A and A', B and B',..., and also the segments AB and
A'B*..., are said to be corresponding.
60. If the straight lines u,
u f 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, A'P' = b, then AA' 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' are not parallel (Fig. 48), and if their common
point represents two coincident corresponding points A and
A', then the straight lines BB', CC'. &c. are all parallel to one
another. The common direction of these parallel lines may be
Fig. 4 8.
found, for example, by taking AB = c, A'B' b ; then every
straight line parallel to BB' cuts u and u' in two corresponding
points.
62. If, finally (Fig. 49), u and u' are not parallel, and their
common point represents two noncorresponding points P, Q',
then all the straight lines A A', 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]
GRAPHICAL MULTIPLICATION.
41
touch u in Q, and u' in P'. 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' 9 B and ', &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 .
and therefore
AW _ P'Q'
~AB = PQ
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" t in which two similiar pointrows (they
* COTJSINERY, Le calcul par le trait (Paris, 1840), p. 20. For another method
of solving this problem see SACHERI, Sul tracciamento delle punteggiate projettive
simili (Atti dell' Accademia di Torino, Novembre, 1873).
42 GEAPHICAL MULTIPLICATION. [63
may be equal) ABODE, ..., A"B"C", &c. are so situated that
the common point of the two straight lines represents two
noncorresponding 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'\G'I) r il)'W:A!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 2/5 and lay off on it n + 1 equal segments,
A'B'=B'C'= C'D', &c.;
then having joined E to W we should, in like mariner, lay off
upon the joining line u", n+I segments each equal to EE', or
A"B"= B"C" = C"D"= & c ., &c. The rc+ 1 straight lines &c., &c. will meet BE in the required divisionpoints
C, D, &c.
Fig. 51.
63. Let (Fig. 51) # t , # 2 , . . . , a n be n segments given in magni
tude, direction, and sense, which have to be respectively
multiplied by the ratios
We construct a polygonal circuit P a , whose sides are re
spectively equipollent to the given segments a^ , 2 , &c., and
call its successive vertices 1, 12, 23, ... nl.n, n, beginning
63] GRAPHICAL MULTIPLICATION. 43
at the initial point of the first side a l 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 l ,
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 ,...,& respect
ively upon the sides of P c , and its sides 1 , 12, 23, ... , *~ 1 ., f
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 v , P b , P xb , having the following properties
(Fig. 51 a):
Fig. 51 a.
1. 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 .
2. 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 lt 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.
f 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 n. In order to construct this
polygon, we can take the side 1 at pleasure, provided it passes through the vertex
lofP a .
44 GKAPHICAL MULTIPLICATION. [64
similarly named side of P b , and each of its sides 1 , 12,23,
...,!, must pass through the similarly named vertex
of P..
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 ^ 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 l 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 a?
gives the result of the required multiplication. For the
triangle, which has os 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 b r , c r)
and therefore
x r :a r =b r :c r orx r = a r x~'
c r
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 d 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 x 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]
GRAPHICAL MULTIPLICATION.
45
metrical addition. Wherefore their resultant, i.e. the re
sultant of the segments a r  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 ^ to
the final point of x n ).
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 pointrow (Fig. 52)
Fig. 52.
Fig. 52 a.
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 Ql=a 1 ,12=a 2 , 23= 3 ,
&c., along a straight line a ; parallel to this line and at
distances l5 e 2 , x z ~ a z X 3 ~ a s < ^ c 5 respectively,
C l 2 C 3
and the segment, that lies between the side rlr and the
side s s f 1 , is equal to
i=s i=s j
2=2>4*
i=r i=r i
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 , x & have the same or opposite sense,
according as amongst the three pairs a r a s , b r b 8 , 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 c's become equal, in addition to all the as
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 1 .b l , a 2 .b 2 ,..., a n .b n to a common base c,
by determining the segments ac 19 # 2 , ..,, x n proportional to
them.
The solution is as follows (Fig. 53). Draw the rectilinear
pointrow a, whose consecutive segments are 01 = 0] , 12 = # 2 ,
* JAEGER, Das Graphische Eechnen (Speyer, 1867), p. 15.
67]
GRAPHICAL MULTIPLICATION.
47
&c., &c., and join each of the points 0, 1, 2, ..., n 1, n of
this pointrow to a point taken at a distance c from a ;
the distance being measured perpendicularly, or obliquely
Fig 53
at pleasure. Then construct a pencil of rays 1, 2, ...,#,
parallel to a, and at distances 6 19 # 2 ,..., #, respectively,
measured in the direction of c from a given line a?, also
parallel to a. Finally, draw a polygon whose vertices fall
respectively on the abovementioned parallel rays 1 , 2, . . . , n,
and whose sides 01, 12, 23, ..., nl.n, n are respectively
parallel to the rays 00, 01, 02, ..., On I, On of the pencil
0. The segments 01, 12, 23, ..., which the sides of this
polygon intercept upon the straight line as, will be the re
quired segments # 15 # 2 , ..., % n *.
67. If instead of all the e's, all the #'s are equal, and all the
's still parallel, then the problem may be stated thus :
Given the ratios
to determine segments a? ls a? 2 , a? 3 , ..., a? n , proportional to
them, so that the product of the multiplication of any x by its
Q
corresponding ratio  shall be the constant segment b.
* CULMANN, 1. C., No. 2.
48
GKAPHICAL MULTIPLICATION.
[67
After we have constructed (Fig. 54) the pointrow a, with
the segments 01 = a lt 12 = 2 , &c. 3 and the pencil of rays
1, 2, 3, ..., n parallel to the straight line a, their respective dis
tances from it being c l9 : 01 = b l , 2 = b 2 , On b n ; and on the second axis
cc: Ol = c lt 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 / 15 / 2 , / 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 th triangle of the second set, of which the
sides parallel to cc and lib are a r ^ lt a r , with the similar triangle
50 GRAPHICAL MULTIPLICATION. [69
of the first set, whose corresponding sides are c r and b r , we
X.
have : = ; and therefore
Multiply all these equations together, and we have
*i ^2 *.
n = fl.i .^...
Cj 2 C n Q. E. D.
69. We shall now prove that the result is not altered by the
interchange of two factors, for example, and . Taking
b b c i C 2
them in the order and , the construction is as follows
C l C 2
(Fig. 56) : on cc take OA a\ from A draw a parallel to bb,
cutting ^ in A l ; the segment AA = a l , carry this over on to
cc, i.e. set off OA 1 = a lt and from this new point A l draw a
parallel to bb, cutting 1 2 in A 2 ; the segment A l A 2 thus obtained
is a. 2 . Now take the factors in the other order and :
c 2 c l
and proceed as follows :
Make OA a as before, and
draw through A a parallel to
bb, let it cut / 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 l parallel to bb cutting ^
in A l  J the straight line A 2 A l
is a". The similar triangles
contained between ^ and cc, OA l A 2 , A l A give the following
relations, ^ QA, C 4> ^65 & c  5 an( i along the other axis (cc) (Fig. 58), b 2 , # 4 , &c.
and
2 = a,   = a ~ * . &c.
C 2 (?j C 2
Continuing this construction, we finally get, on the (rc+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 c's, 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),
\
' ~,
\
' """^
\
\

I
K
\
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 6 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, 0J , a 2 , . . . , a n _
..., a n , and therefore
whilst the sides parallel to b are a l , # 2 ,
f\
().* ()
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
c
parallel to b is equal to a, the side which lies upon Oc is
6 ( b \~ l
a i = a ' = a '() >
next, constructing a triangle with its side parallel to b = _ T ,
7 "" 2
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.
* EGGEBS, 1. c., p. 15. JAEGER, 1. c., pp. 1820.
56 POWERS. [75
The vertices of this crooked line determine on the first axis
segments, measured from 0, which have the values
f l \ (1>\
a, 2 = *(), a, = (),...,
and on the other axis
7 7i 3
l = fl (0)' *8 = () >&C*
Moreover the sides of the crooked line form a geometrical pro
gression ; for, if we call the first side a', the second is a'  ,
the third a'() , the fourth a' () , &c.
\C' ^G '
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 (n+l) ih side will then be
the result of the multiplication.
Continuing the crooked line in the opposite direction w r e
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 (a f 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 a", then
the sides of the new crooked line are respectively
75. Finally, if we employ the fourth method of construction
* COUSINEBY, 1. c., p. 24, 25.
f COUSINERY, 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 I 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 b ii
a, % = a  , a 2 = a (  J , &c.
c c
This progression may be continued in the opposite direction, so
as to give the products of a by the negative powers of  j :
Also the sides of the polygonal circuit form a geometrical
progression with the same common ratio  *.
* JAEGER, 1. c., p. 20.
58
POWEKS.
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.
(/?+ l) th radius coincides with the first, the (^ + 2) th 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
(Fig. 64).
* EEULEAUX, Der Constructeur, 3rd edition (Braunschweig, 1869), p. 84.
K. VON OTT, Orundzuge des grapUsclien Eechnens und der graphischen Statik,
(Prag. 1871), p. 10.
CHAPTER V.
EXTRACTION OF ROOTS.
76. CONSIDER the spiral polygon ABCDEFG . . . (Fig. 65),
whose radii vectores OA, OS, 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  = ^ , and
C \J /L
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 HOOTS.
[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
vectores) 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
Flg ' 66 ' curve completely deter
mine an equiangular spiral.
78. Any two points B } C (Fig. 66) of an equiangular spiral,
* COUSINERY, 1. C., p. 41, 42. CULMANN, 1. C., No. 5.
t WHITWOKTH, The equiangular spiral, its chief properties proved geome
trically (Oxford, Cambridge, and Dublin Messenger of Mathematics, vol. i. p. 5,
Cambridge, 1862).
80] EXTRACTION OF BOOTS. 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 .A 7 2'.
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 NBT 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 (7, its centre lies at N\ the tangents B T, 6Tare 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, will be the point of
intersection of the straight line CN~, 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 OCand 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 ),
we may suppose the constant angle between the tangent
and radius vector to be given. In this case, having drawn
US inclined to OB at the given angle, let 8 be the point
of intersection of this tangent BS with the internal bisector
/X>> :: f 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 8, 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
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 am also indebted to Prof. A. SAYNO.
84]
EXTRACTION OF ROOTS.
63
elementary triangle OA l B l (Fig. 68), of which the angle at is
very small, and between the sides OA l and 0^ construct the
crooked line A l B l C D 1 . . . , with its sides alternately parallel
and anti parallel to A 1 B r Then, upon the radii vectores
OA, OB, OC, &c., drawn at angular intervals each equal to
the constant angle A 1 B v take points A, B, C, &c., in such a
manner, that OA^ = 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
It i
segments a { , a. Write a i = a (), then the question becomes
that of finding the ratio  Take on the spiral (Fig. 69,
c
where i = 5) the radii vectores a and a it and divide the angle
included between them into i equal parts. The il dividing
radii vectores # j , 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 { . The ratio a : 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 bp 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 1 :b 1 ,
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 BOOTS.
[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 a?, given
by the equation
then x 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 a l , a z , ... 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, on = 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 0JT, drawn from to a circle described through A and
B ; or (Fig. 70 a) 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.
OB A
Fig. 70 a.
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 t and then x is the ordi
nate erected at the point 0.
86. The same ends for which the
equiangular spiral serves, are easily
attained by using another curve called
' The Logarithmic Curve!
Fig. 71.
Draw (Fig. 72) two axes Ox and Oy ; and on the first of
them, starting from the origin 0, take the segments 00, 01,
02 , 03 , &c., respectively equal to the terms
m
m
86] EXTRACTION OF ROOTS. 65
of a geometrical progression, of which the first term is # , and
rn.
the common ratio (where m is supposed greater than ri) ;
and on the second axis take the segments 00 , Ol , 02 , 03 ,
&c., also measured from 0,
and respectively equal to
the terms y Q = 0, y^ = I,
#2 = 2!, $3=31, &c., of
an arithmetical progres
sion, with its first term
equal to zero, and the
common difference* = I.
The terms of the two
progressions, which corre
spond to the index r, are
and therefore
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
a? r , and # r+1 , and the arithmetical between y r and y r+1 , we
obtain the two corresponding terms
* Or +JV+l) =
of the two new progressions.
In these progressions we can in like manner interpolate a
* In the succession of numbers on Cty, the zero coincides with the origin 0,
because y was taken = 0.
F
66 EXTEACTTON OF ROOTS. [87
term between each pair of consecutive terms, and so on, until we
:L
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
y_
(!) . = *,()',
x
log
n>
or (2) y=l
m
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 O?/, and through the
final point of y a parallel to Ox. The straight lines so drawn
will intersect in a point M; IK and y are then called the co
ordinates of the point M , and in particular x is called the
' abscissa' and y the ' ordinate.' The equation (i) or (2) ex
pressing the relation between the coordinates 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 ;
4MM
and, upon Ox we take a segment OA = (2*) = # , where
?l
/wi
OQ = X Q is the unit of length of the scale for 0#f, and 
the base of the logarithmic system (the number 1 0).
* i is the number of interpolations.
f Since x increases much faster than y, it is convenient, in order to keep
the construction within reasonable limits, to take the unit # much smaller than
Z, for ex.,
88]
EXTRACTION OF ROOTS.
67
Let OB be divided into 2* equal parts, and let 1,2, 3, ...,
'" 1 , ... , 2* (= ff) be the points of division.
y
IB
01 2 3 4 5
Fig. 73
In order to find the corresponding points of Ox, take the
geometrical mean between # and X Q , i.e. describe a semicircle
on (9^4 as diameter, and set oft' 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*" 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 02 i ~ 1 , and the geometrical mean between 02 1  1 and OA,
we shall obtain the points on Ox, corresponding to the middle
points of the segments 02 i ~ l , and 2 i ~ 1 of Oy ; and so on.
Now draw parallels to Of/ 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^ = y
corresponds the value x = X 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  74
any two points on the curve, at a small distance from one
F 2
68 EXTEACTION OF BOOTS. [89
another; MP, NQ parallels to Oca, MR a parallel to Oy, and T
the point in which Oy is cut by the chord MN. The similar
triangles TPM and MEN give
TP:MP=MR:NR,
or TP : HP = OQ OP : NQMP.
Let OP = y, PQ = li, then HP and NQ are the abscissae
corresponding to the ordinates y , ^ + /, and therefore
y y+h
whence 27> =
D 7  1
Now let the point ^V approach continually nearer and nearer
to the point M, i. e. until Ji approximates to the value zero,
then NMTwill also continually approach towards the position
of the tangent at M, and the segment TP, the projection of
TM upon 0y, has for its limiting value what is usually called
the ' sublanqent.' But the limit * which the fraction
1
approaches, when h tends towards zero, is the natural
logarithm of , which we shall denote by A ; therefore in the
TV %
limit we have /
TP =
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, Elemente der Mathematik, i. 4ed. (Leipzig, 1872), p. 200.
t SALMON, Higher plane curves, 2nd ed. (Dublin, 1873), p. 314.
80] EXTRACTION OF ROOTS. 69
to construct the r th root of the ratio between two straight lines
p, q. Take upon Ox the abscissae af = p, x" ' = q, and find by
means of the curve the corresponding ordinates i/ and y" .
The abscissa corresponding to the ordinate (//') has the
value r /~n
Q \/ '
V q
Secondly, say we want the r ih root of the product of the r
straight lines p lt p 2 , ...,p r . Take on Ox the abscissae x^ =j 1?
x 2 =j) 2 , &c., and find the corresponding ordinates y 15 y 2 ,
y$ t "*y r \ ^ nen the abscissa x corresponding to the ordinate
&i+y 2 +"+yr)
is equal to the required quantity
CHAPTEK VI.
SOLUTION OF NUMERICAL EQUATIONS*.
90. LET 05 tf 15 2 , ... , a n be n+ I numbers given in magnitude
and sign, and let (Fig. 75) a polygonal rightangled 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) th sides, which
are parallel to one another, have the
same or opposite sense, according as
the signs of the numbers r _ 15 # r+ i,
which are proportional to these sides,
/5 ' are unlike or alike f.
Assume a point A 1 in the straight line 12, and take
OA l as the first side of a second rightangled circuit of n sides,
whose respective vertices A 19 A 2 , A^ ... lie on the sides 12,
23, 34, ... of the first circuit.
* LILL, Resolution grapMque des equations numerique (Pun degre quelconque a
une inconnue (Nouvelles Annales de Mathe'matiques, 2 e se*rie, t. 6, Paris, 1867),
p. 359.
f 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 +', or i (where i V 1, i e., ^ 2 = 1), according as it is in
the positive or .negative direction of YOY. Now let a circuit be formed whose
successive sides,
01, 12, 23, 34, 45, 56, ... .
are equal to a t) ia lt fia^ * 3 ^3> * 4 ^ 4 , i 5 a 5t ... ,
i.e. equal to a 0) ia i} a 2 , ia 3) a if ia 5 , ... ,
then the 1 st , 3 rd , 5 th , ... sides will be parallel to XOX, and the others to YOY;
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 NUMERICAL EQUATIONS. 71
The triangles OU 1} A^A^ A 2 3A 3 , A 3 4 A, ... are all similar
to one another, and therefore give
01 _A 1 2_A 2 3_A^_ _ A n _ lt n
1^1 ~ 1^2 ~~ A^S ~ 2^4 ~ A n .n '
whence, remembering the identities,
01 = a , ^j2 = Aj^l+a^
12 = !, A 2 3 = A 2 2 + a 2 ,
23 =tf 4 =
.+ 1 = w ,
^ 1
and putting y = x, or ^ 1
we obtain :
Thus the segment A n .n+l, included between the final
points of the first and second polygonal circuits, represents
the value which the polynomial
F(x) = a () x n + a 1 x n  1 +...+a n
takes, when we substitute for as the ratio of the segment A^ 1 to
the segment 1 or a . Keeping a positive, the signs of x and
a l will be equal or opposite, according as ^1, 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 , which correspond to those rightangled
inscribed circuits whose final points coincide with the point
+!.
On account of this property we say, that the circuit
0123 ... + 1, represents the whole polynomial F(z).
91. If in 1 23 ... n + ' 1 we inscribe a new rightangled circuit,
SOLUTION OF NUMERICAL EQUATIONS.
[01
Qfi l JB 2 ...JB n , and if we denote the ratio of ^1 : 01 by^ 3 we
shall have in like manner
For the coefficients a we substitute their values
= 01,
! = 12 = ^2^1 = ^201 .a?,
fl 2 = 23 = ^ 2 3 J 2 2 =^ 2 3^2. a?,
= 34 = A4A3 A4A3.x
and thus obtain
But
and
therefore
A n n + 1 = n A
01
01 '
This result may be expressed as follows (Fig. 76, where
= 6): In a rectangular circuit of n+1 sicks 0123...^+!,
92]
SOLUTION OF NUMERICAL EQUATIONS.
73
two other rectangular circuits of n sides QA^A^... A nt OB 1 B 2
...B n are inscribed; we then form a new rectangular circuit
01 2 / 3 / . ..#' of n sides, which are respectively parallel to the sides of
the Jirst circuit and equal to 01, A 1 2 ) A 2 3, ..., A n _^n. In this
inscribe the rectangular circuit of n 1 sides OB^ B%. . . J5' n _i ,
having the side OB^ in common with the circuit already described
OB.B,. ...
!'
01 B^AI
That is to say, the segment B f n _ l .n / is the result of the division of
F(y} F(x) by yx: where F signifies the Polynomial represented
ly the first circuit 0123...^ + 1, and x and y are the ratios ^1:01;
1^1:01.
Or, in other words,
The circuit 01 2 / 3 / ...# / represents the polynomial
z x
or, in the case where x is a root of the equation F(z) = 0, the
polynomial F(z) : (z x).
92. The similar triangles considered above give
so that our equation may also be written,
>* + A 2 A B .y n * + ...+A n _ 1 A n .
This result may be interpreted as follows (Fig. 77) :
If we i\
Fig. 77
the rectangular circuit 0123.,.^
of n+l
74 SOLUTION OF NUMERICAL EQUATIONS. [93
sides (Fig. 77, where n = 6) two new rectangular circuits of n sides
QA l A 2 ...A n , QB 1 I! 2 ...13 n , and then inscribe in the Jirst of these
a rectangular circuit of n 1 sides QC l C 2 ...C n _ l ^
or * 2 = ?1 ,
o ft o
that is to say, the sum of the roots is
Further the similar triangles 01^ 15 J X 23 give
01 \A,\ = A 1 2:32,
or a :A 1 l=A 2 l:a 2 ,
and therefore ' = ,
o ^o
i.e. the product of the roots is equal to
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.
CHAPTEE VII.
REDUCTION OF PLANE FIGURES*.
95. To reduce a given figure to a given base b, 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
A > /2 > /s ' &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 L Then, since the area
must remain unaltered by the transformation, fb = \ah,
and therefore / = a . 7 = h . 7 ,
that is to say, we have either to multiply a by the ratio Ji : 2 b,
or Ji by the ratio a : 2 b.
We therefore take OC = 2b,
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.
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
* CTJLMANN, 1. c., No. 15 et seqq.
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 b,
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/ 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.
Fig. 82.
97. Let it be required to reduce the quadrilateral ABCD to
the base l> (Fig. 82). Draw CO parallel to the diagonal BD ;
then the quadrilateral reduces to the triangle OAJ3, and we
proceed as above, viz. we make BC' = 2b, and the antipro
jection OI/ 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 rightangled
triangle ODB of which the side BD = 2 b. Let the points
A and C be projected, by means of rays parallel to OB,
into A\ C' on OD, the other side of the triangle OBD ; the
* The triangles BPC, DO A are similar, hence P : BC = OD : OA, 01
h : 2 b =f : a ; therefore/ =a .. Q. E. D.
100]
REDUCTION OP PLANE FIGURES.
79
triangles OCB, OBA are equivalent to the two triangles OCf B,
and OBA' ; but in each of these the distance of the base 06", or
Fig. 83.
OA', from the opposite vertex 2 b, and therefore the required
height /for the quadrilateral is equal to OC' + A'O = A'C'.
.85.
Fig. 86.
In the crossed quadrilateral (Fig. 87) if AC is parallel to
BO, the points A' and Cf 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' Cf which is
intercepted by the straight lines AA\ CCf.
100. The foregoing construction assumes that 2 b is not
greater than the greatest diagonal OB of the quadrilateral.
If 2 b is > OB, the lengths 2 I 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 rightangled triangle
ODB, of which the side OD is parallel to CE\ and then the
other side BD = /.
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 I gives the area of the
polygon proposed.
Fig. 88.
Let the given polygon be 0123456780 (Fig. 88).
the straight line 8 7' parallel to the diagonal 07 ,
5 , 76 06 ,
Draw
6 ' 5 '
05 3
04.
U * 5
4'3' 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'
33'
44'
5D
31'
42'
53'
>. till each intersects the side C,
* 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 01234567'0. And so on.
103] KEDUCTION OF PLANE FIGUBES. 81
we determine a straight line AD which can be substituted for
the crooked line J543210.
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
AabcdeC l / 2345. Similarly, from the equivalence of the tri
angles 1'23, 1'2'3 the latter polygon reduces to Aabcde C'2'345,
and proceeding in this manner we finally arrive at the poly
gon Aabcde CD.
In order to effect a like transformation for the crooked line
AabcdeC) we draw
lib' parallel to ca \
cc' db'\
ftff e( / r till each intersects the side Aa,
eB " Of]
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
a
REDUCTION OF PLANE FIGURES.
[104
trouble*. Take, for example, the selfcutting circuit (Fig. 90)
ABC 01234, which represents the cross section of an embank
ment and excavation in earthwork.
Fig. 90.
Draw
11' parallel to 20 ,
22' 31',
33' 42',
4D A3',
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 14; of the
embankment and 701 23 of the excavation, which have neces
sarily different signs. The circuit ABCD has the same sign
as the circuit ABCI, 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.
Fig. 91.
Then starting from A or A' we set off the chord a the same
number of times along the tangent AC"f. 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 Bankine for the
approximate rectification of circular arcs, and also some methods of Professor
Sayno.
106]
REDUCTION OF PLANE FIGURES.
83
The segment AB (i.e. the area between the arc AB and its
chord) is the difference of the two triangles OAC, 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 f respectively
parallel to AC and 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 foursided 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
0^1 C. Change the sectors
into the triangles OAB, Firf
1 32, 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 QAB the area 0^321 C 0, i.e. is equal to
the selfcutting polygon AJBOC123 A.
Draw 11' parallel to 2(7
22' 31' till they cut the fixed side CO,
3C' A2'
and the polygon reduces to the crossed quadrilateral AB C'.
The area bfof this quadrilateral is found in the usual manner ;
i.e. on the diagonal Ao as hypothenuse a rightangled triangle
is constructed, of which one side AE 2 b ; the length / is
then the distance, measured parallel to the other side, of the
point 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 socalled
?7iron.
It consists; (i) of a luneshaped area AEA'F, bounded by two
circular arcs, one having 27 as centre, the other 0; (2) of a
crownshaped piece CBFB'C' bounded by two concentric cir
cular arcs BI?, CC' drawn with as centre ; (3) of two equal
rectilinear pieces * BCJIH and B'C'J'I'H', symmetrically
situated with regard to the straight line OUFE, which is an
axis of symmetry for the whole figure.
Fig. 95
The lune is equal to the sector UAEA' plus the quadrilateral
OAUA' minus the sector OAFA' t i.e. it equals the sum
UAEA' + OA UA' + AOA 'F. After transforming the two sectors
spoken of into the triangles UAD, OAG (where AD, AG are
the arcs ASA' and AFA' 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 lune becomes equal to the sum UAD + 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
selfcutting hexagon ADUA'OGA.
The crownpiece we consider as the difference of the sectors
OBB f , OCC'. 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 It
and find their corresponding segments to be./o, f 19 2f 2 , then
^(/o +/i + 2 ./2) w iH b e ^ ne 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
UAD +20AUOAG+ OPP' OQQ' + 2 BCKH+ 2 CJ1K,
where CK is drawn parallel to BH, and JL We consider the
areas of these triangles and trapeziums as the products of two
673 524<
Fig. 95 a.
Fig. 95 a.
factors, and reducs these products to a base b 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 S'C'J'I'H'
we substitute b for 2 J.
86 REDUCTION OF PLANE FIGURES. [108
108. Curvilinear figures in 'general*. It is a wellknown
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 f their respective Sagittae.
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.
Fig. 98.
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'
44' 53'
5C B4'
till each cuts the fixed line A ;
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 2 ,
&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 B 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 FIGURES.
In order to effect the abovementioned 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 .Z? 1} i? 2 ,
&c., &c. into the points B^ 2 , &c., &c. upon ON, by rays
parallel to OA^A^ . . . , and then the triangles OA 1 JB 1 , OA 2 B 2 , &c.,
are transformed into the triangles OA l B l / ) OA 2 B 2 , &c. Then
draw the straight lines B C^ B 2 'C. 2 , &c., parallel to NA^NA 2 ,
&c., respectively, and let the points in which they cut the
straight line A 1 A 2 . . . be C^ , C 2 , &c.
We thus obtain the triangles OQN, OC 2 N, &c., respectively
equivalent to OA 1 J3 1 / , OA 2 JB 2 ', &c., &c. Therefore, if the seg
ment OA = OC^ + OC 2 + &c., is taken on OA 1 A 2 ... ) and if
through A the straight line AB is drawn equipollent to ON,
then OAB is equal to OA 1 B 1 + OA 2 2
CHAPTER VIII.
CENTROIDS.
112. LET us suppose that, in the theorems of Articles 43 and
44, all the points J? 19 J3 2 , .. , B n coincide in a single point G ;
these theorems may then be stated as follows :
If A 1 G, A 2 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 the segments OA l , OA 2 , . .
to n times the segment OG (Fig. 100).
OA n is equal (equipollent)
o'
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 l , OA 2 , ..., OA 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 (/ ;
that is to say, the resultant of the straight lines (/A 1 , 0'A 2 , . . . ,
0'A n is equal to n times the segment O'G ; and the resultant of the
straight lines GA^ GA 2 , ..., GA n is equal to zero*.
113. The point G is called the Centroid of the points A 19
A 2 , ...,A n . Let the (Fig. 101, where n 4) points A 1 ,A 2 , ...,
A n be given, to construct their centroid G we proceed as
follows. An arbitrary pole is taken, and a circuit
* GKASSMANN/ 1. c.,*. p. 141. CHELINI, '~Sui centri de 1 sistemi geometrici
Eaccolta scientifica; Roma, marzo 1849), 1.
90 CENTKOIDS. [114
0^2 3... ft constructed, whose initial point is and whose
successive sides are equipollent to the segments OA lt OA 2 , ... ,
OA n . The straight line On, which closes the circuit, passes
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
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 n ,
and the resultant of G / A 1 , G'A 2 , . . . , G'A n should vanish, then the
general resultant of all the segments GA l , A l G', GA 2 ,A 2 G',...,
GA n , A n G f would also vanish. But if we combine the two
segments GA r , 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 , _Z? 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 ll/, 22', 33', ..., nn f 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
Fi I02 points 19 B%, ..., B n coincide with a single
point G, we now imagine some of them B , B 2 ,
..., JBi 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 i , A i+1 G, . . . , A n G vanishes ; and, whatever the
position of may be, the resultant of OA^ OA 2 , ..., OA n is
equal to the resultant of OB^ OB 2 , ... , OB L , . (ni) . OG. The
* CARNOT, Correlation dei 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 B0, B \0, ..., B { 0, so that it becomes
an equality between the resultant of OA L , OA 2 , ..., OA n , B 1 0,
B 2 0, ..., J^O and the resultant of OB 1 , 2 , ..., OB { , B^,
B 2 0, ..., BiO, (n i).OG; that is, between the resultant of
OA^ OA 2) ...,OA n , OB^ OB 2 ,..., Off, and (*). OG.
Hence :
If the resultant of the segments A L G, A 2 G, ... , A n G, B l G }
B 2 G,..., B i G vanishes; then , for any point whatever, the
resultant of the segments OA 19 OA 2) ... , OA n , OB l , OB 2 , ... ,
OB i is equal to (n i) . OG: and 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 1 G,A 2 G ) . . . , A n G, B 1 G, B 2 G, . . . ,
QG will vanish.
117. Now let us assume that of the n points A 1 , A 2 , ..., A n
some coincide with one point, others (in like manner) with a
second point, and so on; and that the points H lf B 2 , ..., B (
also unite in groups and coincide. Then if we use a l5 a 2 , a 3 , . . .
to denote positive or negative integral numbers whose sum
is m, the foregoing proposition may be stated as follows :
If the points A ) A 2 , A 3 , ... and the point G are so situated, that
the resultant of the segments a l .A l G, a 2 . A 2 G, a 3 . A 3 G y . . . vanishes,
then, wherever the pole may be, m.OG will be equal to the resultant
of the segments a x . OA l , o 2 . OA 2 , a 3 . OA 3 , . . . &c.
And conversely :
If this property holds for any pole 0, viz. that m .OG is equal to the
resultant of a x . A l , a 2 . OA 2 , a 3 . OA 3 , ..., then the same property
holds for every other pole 0, that is to say, the resultant of 04 . O f A l
a 2 . 0*A 2 , a 3 . O^gj * equal to m . C/G\ and the resultant of
ike segments c^ .GA l , o 2 . GA 2 , o s . GA 3) ... vanishes.
118. The point G is called the centroid of the points A^ , A 2 , A 3 ,
... weighted with the coefficients a lf a 2 , o 3 , .... For shortness
however we say that G is the centroid of the points a l .A 1 ,
o 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 1 .A 1 , a 2 .A 2) a 3 ,^ 3 , ... and if,
92 CENTKOIDS. [120
by means of parallel rays, the points G, A 1} A 2 , A 3 , ... are projected
into the points G', A\, A' 2 , A\, ... which lie on a straight line, then
the sum of the straight lines a^.A^A^ a 2 .A 2 A 2 , a 3 .A 3 A 3) ... is
equal to m . GG', where m = ajfo^ + ctg + ....
On account of this property G is also called the centre of
mean distances of the points c^ . A 19 a 2 . A 2 , a 3 . A 2 , . . . *.
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 c^ , a 2 , a g , . . . are any mimbers 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 , a 3 , .... Project the
given points on to a straight line p', by means of rays parallel
to some arbitrarily chosen direction, into A\ , A\ , . . . ; and by
means of rays parallel to another direction chosen at pleasure,
project the same points into A" lt A" 2 , A" 3 , ... on a second
straight line p", not parallel to p' '. Now determine a straight
line / parallel to j/ 9 such that the distance from / to p f
measured parallel to the rays A^A\, A 2 A' 2 ,A 3 A 3 ', ..., is equal to
similarly determine a straight line r" parallel to //', such
that the distance from /' to _//', measured parallel to the rays
A l A" l , A 2 A 2 ", ..., is equal to
a . AA' + a.A A" + a . AA" + ...
Let G denote the point of intersection of the straight lines
/, /', and ', G" the projections of G upon the straight lines
p', p" (by means of rays parallel to AA f , AA" respectively),
then we shall have :
a 1 .^ 1 ^ 1 / + a 2 .A 2 A' 2 + a 3 .A 3 A' 3 +... =( ai + a 2 + a 3 +...). GG'
a i A\ A "\ + a z  A i A "i + 3 A *A\ + . =(! + a 2 + a 3 + ...) . GG".
Next, let p f " be any third given line, let us project upon it
the given points and the point G, into the points A'" 19 A'" 2 ,
A"' 3 ,..., G" f , by rays parallel to a new direction. Between the
three rays which project the same point A^ or A 2 or A 3 , there
* L'HUILIER, EUmens d'analyse geomttrique et d'analyse algtbrique etc,
(Paris, 1809), 2.
122] CENTROIDS. 93
exists (Article 1 6) a linear relation with constant coefficients,
i. e. we have :
k'. A t A\ + k". A,A\ + k'". A l A ff \ = k,
k'. A 2 A\ + k". A 2 A" 2 + k m . A., A'" 2 = k,
k'. A 3 A' 3 + k". A Z A" Z 4 V". A 3 A"\ = k,
k'. G G' +k".G G". + k f ". G G m = k.
Multiply these equations by a x , a 2 , a 3 , . . ., (04 + a 2 + 3 + . . .)
respectively, and add the products ; then we obtain, taking
the equations already established into account,
k'". {a, . A^
or,
a l .A 1 A f \ + a 2 .A 2 A^^a 3 .A 3 A^ 3 +...=(a 1 + a 2 + a 3 +...).GG^.
That is to say :
If we project the points A 19 A 2 , A 3) ..., G upon any straight line
whatever by means of rays, which are parallel to an arbitrarily
chosen direction, then the product of the ray which projects G
by (a l + a 2 + a 3 + ...) is equal to the sum of the products formed by
multiplying each of the rays which project A l) A. 2 ,A 3t .. by a lt # 9 ,
# 3 , ... respectively.
We call the point (?, so defined, the centroid of the points A t ,
A 29 A 3 ,... loaded with the numbers or segments a 1} a. 2 , a 3 , ....
The centroid does not change if we substitute for the coeffi
cients a x , a 2 , a 3 , . . . others proportional to them, for by so doing
we do not change the ratios of c^ , a 2 , a 3 . , . , to a t + a 2 + a 3 + . . . .
121. If the points A 19 A 2 , A 3 , . . . , and G are projected, by means
of rays parallel to a straight linep"> on to another straight linep', and
if we use 0' to denote any point whatever off/, we have identically.
If we draw through 0' a straight line parallel to j>/', and pro
ject on to it, by rays parallel toy, the points A l9 A. 2i A 3 , ... , G
into the points A" 19 A". n A" 3 , ..., G'\ we have the identities
A^A'\ = A\C/, A 2 A" 2 = A\0\ AzA"s=A\0' ..., GQ"
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
fll . OA 19 a 2 . OA 2 , a 8 . OA 3 , .,., is (a L + a 2 + a 3 + ...). OG.
By the segment a . OA we understand a segment parallel to
94 CENTROIDS. [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: I. Draw
through the point a straight line p', and project upon it, by
means of parallel rays, the points A l , A 2 , A 3 , ..., G into the
points A\ , A' 2 ,A' 3 ,..,,G'. Then the segment A 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 a a . OA,, a. 2 . OA 2 ,
a 3 . OA 3 , ..., may be obtained by combining all the segments
a, . OA\ , a 2 . OA' , a 3 . OA' Z , ... with the segments a, . A 1 f A 1 ,
a. 2 .A' 2 A 2 ,a 3 .A'zA 3 , But the resultant (i.e. the sum) of
a, . OA\, a. 2 . OA'. 2 , a 3 . OA' 3 ... is (a, + a 2 + a 3 +...). OG', and the
resultant (or sum) of a l . A\ A l} a 2 . A' 2 A 2 , a 3 . A' 3 A 3 , ... is
(a x + 2 + a 3 +...). G'G\ therefore the resultant of the segments
a, . A l , a 2 . A 2 , a 3 . OA 3 . . . . , can be obtained by combining
the two segments (a : j a 2 + a 3 . . .) . OG', (a x + a 2 + a 3 . . .) . G'G, and
consequently, it coincides with the segment (a x + a 2 + a 3 +...). OG.
123. If H is the centroid of the points a x . A l , a 2 . A 2 , a 3 . A 3 , . . . ,
and K the centroid of the points fa . B^ , (3 2 . B 2 , . . . , then the centroid
of all the given points a 1 .A 1 , a 2 .A 2 , ..., /3 X . JB 1 , f3 2 . H 2 ,. . . coincides
with the centroid of the two points m . H, n . K, where
m = (^ + 02+...), n = (ft+/3, + ...).
For, taking an arbitrary pole 0, if we combine the straight
line m . OH, the resultant of the segments a x OA l , a . OA. 2 , . . . ,
with the straight line n . OK, the resultant of the segments
/V^, Pt'Bji we fi n ^ tlmt (m + n).OG, the resultant of
m . Oil, and n . OK, is also the resultant of all the segments
a, . OA,, a 2 .OA 2 , ..., ft . OS 19 p. 2 .OR 2 , ....
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^AZ . . . ; for then all the segments 04 . OA, , a. 2 . 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, , A 2 , A 3 , . . . , A n , G upon an arbitrarily chosen
straight line into the points A\, A' 2 , A' 3 ..., A f n , G', then the point
G' is the centroid of the points a, . A\ , o 2 . A' 2 , o 3 . A'% ..., a n . A' n .
Let there be only two points A 19 A 2 (Fig. 103, where the
125]
CENTROIDS.
95
Fig. 103.
segments a 1? o 2 are simply denoted by the numbers 1, 2) with
coefficients a l9 o 2 , then their centroid G is a point of the
straight line A^A 2 . Since the resultant
of the straight lines 04 . GA l , o 2 . G A 2
is equal to zero, we have
a x . GA l \a 2 . GA 2 = 0,
or A^G : GA 2 = a 2 : 04,
and therefore
^f, : GA 9 : A, A 9 = a : a, : a 9 + a, :
1. L JL A L i \. '
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 1} o 2 , and it lies inside or outside the given segment, according
as a 19 a 2 have the same or opposite signs.
If a : = a 2 , then A l G = GA 2 , i. e. G is the middle point of
A 1 A 2 . If 0^ + 02=0, we obtain from the proportion A 1 G:
A^A 2 = a^.aL + o^ the value A^ G = oo, i.e. G is the point at
infinity of the straight line A 1 A 2 .
125. Let there be three given points A^
straight line (Fig. 104); and let a 15 a 2 , a 3 ,
whose sum is not zero, be their coefficients.
The centroid of the points o 2 . A 2 , a 3 . A 3
is a point B l on the straight line A 2 A 3 ,
and the centroid of the given points a l .A 1 ,
L I> A 2)
A not in one
.A
2 .
B . A 3
is therefore the centroid of
the points a t . A L , (a 2 + a 3 ) . B l , that is, it
is the point G on the straight line A^B^
which is determined by the relation
Fig. 104.
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 l B l ,
GB l of their vertices from the common base A A.^ ; therefore
GA 2 A 3 :A 1 A 2 A 3 =a l :a l + <
Similarly we prove that
3
'2'
= o
and therefore
That is to say ; the centroid G of the three points c^ . A l , a 2 . A 2 ,
a 3 .^ 3 divides the area A^A 2 A 3 into three triangles GA 2 A 3 ,
96
CENTROIDS.
[126
GA 1 A 2 , which are proportional to the coefficients a
1?
2 , 3
Given the points A l , A 2 , A 3 , every system of values for the
loads a l5 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 3 ,... (Fig. 105)
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, (7 ;
the sides of the first being
equipollent to c^ . A l , a 2 . OA 2 ,
a 3 . A B , . . . , 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
(7, and we have
If the coefficients a l5 a 2 , &c., &c. are proportional to given
segments a lt a 2 , &c., &c. they will also be proportional to
loads r^r' ^ c ' 5 ^ c * wnere ^ ^ s anv arbitrary segment; we
it iL
can therefore make the sides of the first circuit equal to
the lengths j OA L , j OA 2 , &c., &c., and if OR is the closing
it it/
line then k . OR = (a L f a 2 + %+...). OG. Hence it follows,
that G is found, without constructing a second circuit, by
determining on the closing line the segment
If the coefficients a ls a 2 , &c., ... are proportional to the areas
* 1} s 2 , &c., &c., which when reduced to a common arbitrary base
k are equivalent to the rectangles Jca^ 7ca 2 , ..., &c. ; they will
* GBASSMANN, 1. c., p. 142.
127] CENTROIDS. 97
also be proportional to the segments a ls a 2 , ..., or to the loads
T^ , T^J 5 and if the circuit be constructed with the sides
It It
^ OA, , ^ OA &c., , . , &c. 5 then OG =  h ' OE
h
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 a x . OJ 15 a 2 . OA 2) a 3 . OA 3 , ..., is given by
the equation
(~) 7? ( n _L n J _L \ DC 1
whence Q~R
OG=
If a 1 + a 2 ia 3 + ...= 0, while OR is not zero, then
OG = oc, or the centroid G is at an infinite distance. To
find in what direction G lies, let B l 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
2 + # 3 + . . . , a n is not equal to zero, but is equal to a r Let
OR be the resultant of a x l and a l OA l) this resultant will
be equipollent to a^Z?^, that is it will be independent of
the point 0. Consequently the resultant of a^OA^ a 2 OA 2 , ... ,
a n OA n) where o 1 + a 2 , ,.., a n = 0, is constant in direction and
in magnitude wherever may be, and is equipollent to the
segments a^B^ = a 2 2 A 2 = ... a n n A n ;
where B r is the centroid of the points a^A^ a 2 A 2 , ... , a r _ 1 ^4 r _ 1 ,
a r+1 A r+l , ..., a n ^ n . The point at infinity common to the
segments R^, B^A^ ..., is the centroid G of the given
points.
Let parallel straight lines be drawn through each of the
points A 1 ^A 2) ..., 1 ,R 2)t .. in any arbitrary direction, and
let them be cut by a transversal in A^. A 2 , . . . , B, 2 ..., the
theorem of Art. 121 applied to the points a 2 ^ 2 , a 3 ^ 3 .
to their centroid 1 gives
a 2 A 2 A 2 +a 3 A 3 As, ,.., a n A n A n ' = [a 2 + o 3 , ...,a n ] B^
Therefore o^A^A{ + a 2 A 2 A 2 ' + . . . = a x \A^B^B{\,
consequently a^A^ A^ + a 2 A 2 A 2 + ... 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 coincides
H
98
CENTROIDS.
[128
with A ly 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 l A
1)
a 2 A 2 ,
has no centroid. The sum
2 ^ 2 ^2 / + ...,a n A n A n '
is then zero, whatever the direction of the parallel lines
A^AI, A 2 A 2 ... , and of the transversal A{ A ...*.
128. Through the points A 19 A 2 , A 3 , ... , and through their
centroid G segments A 1 B l , A 2 B 2 , A 3 B 3 , ..., (7//are drawn in
an arbitrary direction parallel to one another, and proportional
to the coefficients a x , a 2 , a 3 , . . . , m = a x + 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^ A 2 , A 3 , ..., G by parallel rays into A L ', J 2 , A 3 , .... G' ;
then by the theorem of Art. 1 1 9 we have
a l .A 1 A l ' + a 2 .A 2 A 2 \a 3 .A 3 A 3 + ... = rn.GG'.
But the numbers a : , a 2 , a 3 , . . . , m are proportional to the bases
of the triangles OA 1 B 1 , OA 2 B 2 , OA 3 B 3 , ..., OGH, and the
segments A l A l f , 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 1 , A 2 B 2 ,
A 3 S 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 1 , A 2 B 2 , A 3 J3 3 , ..., (Art. 47).
129. This furnishes another construction for the centroid G.
After drawing through A 1 ,A 2 ,A 3 , ... (Fig. 106) the segments
7
^
G
2 C
\ \ A
\ x ^
,,,H /"
r*^O
''x
/ T4 /
t' /
/
V
**
4 2
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] CENTROIDS. 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 lt a 2 , o 3 , . . . , and obtain another
straight line / ; 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. 107.
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. 1 09). 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 (7 of the
area ABC lies on AD. Therefore G is the
point of concourse of the three median
Jf IOT. 1OQ.
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.
[132
BC DG
FB' CD' GA
But AF = FB, BC = 2 DC, therefore
~GA
i
"25
or GD=%AD, and similarly GE = i BE, GF = F. The
point G is also the centroid of the three points A, B, 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 l ,a 2 . A 2 , a 3 .^ 3 , ..., where A lt A 2 ,A 39 ...
are the centro'ids of the segments or triangles, of which the
figure is made up, and the (numerical or segmental) co
efficients a l9 a 2 , a 3 , ... are proportional to the segments or
triangles themselves.
133. Let the figure be a circuit with rectilinear sides. Let
A^A^A Z ^... be the middle points of the sides, and a lf o 2 ,
a ;} , ... 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 a 15 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 a
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
rightangled triangle of which o is the hypothenuse and A one
A
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 ^" opposite sign, so that a^ + o 2 = 0. It fol
^ ^ lows that the area of the figure is zero, and
B D
T,. the centroid lies at infinity in the direction
rig. II o. '
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 A lt A 2 , A 3 , ... of these triangles, and reduced their
areas to a common base so that they are proportional to the
segments a 15 o 2 , a 3 , ... , the centroid in question is the centroid
of the points c^ . ^ , a 2 . ^ 2 , a 3 . .4 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 selfcutting
hexagon AJBCDjEF(Fig. 119), which is the sum of the triangles
OUC, OCD, ODE, OFA, being the point of intersection of
the sides A~B, EF. Of these four triangles, the first and last
are positive, the other two negative. Let their centroids
G lt 2 , (r 3 , G be found, and let the areas of the triangles,
reduced to a common base, be proportional to the segments
a u a 2> a 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 sc is
drawn through 0, its positive direction is fixed, and upon it
the segments /i, a lt o 2t a 3 , o 4 are set off from their common initial
142] CENTROIDS. 1C7
point (h, a 1? a 4 in one sense ; a 2 ,a 3 in the opposite sense *).
Then the final point of h is joined to G r) and through the
Fig. 119.
final point of r a parallel is drawn to this joining line cutting
OG r in H r . Thus we obtain OG r :h = Off r :a r , and therefore
o r . OG r = k. OH r . Now construct a circuit starting from
with its sides equipollent to OH^ OH 2 , Off B , OH 4 , the closing
line is OH. Finally to construct the point G, given by the
relation
we set off along Ox from its initial point the segment
OS = a 1 + a 2 + a 3 + a 4:J
join its final point to R, and draw through the final point
of h a parallel to this joining line cutting OR in G.
142. Again, let the figure be the crosssection of a socalled
Angleiron (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 Ti, a are denoted by these letters
themselves. Some of the straight lines mentioned in the text are not drawn in
the figure.
108
CENTKOIDS.
[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, 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
Fig. 121.
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 Teeiron.
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'B'. In order to find that
point (Art. 135), set off the semi
arc CA along the tangent CM, join
OM 9 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
Fig. 123.
sector OAB and the triangle OAB, or the sum of the sector OAB
* CULMANN, I.e., Nos. 96 & 116.
CULMANN, 1. C., No. 96.
110 CENTROIDS. [146
and the triangle OS 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 l 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 lt 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 If 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 and FH\ therefore, if through G l
and G 2 two parallel segments 6^7 and G 2 K BXQ 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^I, GG 2 K we have
G 1 G:G 2 G= G 1 I: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 centroid 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
crownpiece, and the sum of the rectilinear parts ; then,
regarding the lune as the algebraic sum of two sectors and
one quadrilateral, the crownpiece 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 OC f C }
* CULMANN, ibid.
146]
CENTROIDS.
6 Trapeziums BCKH+ Il'K'C'B',
7 Trapeziums CJIK+KTJ'C'.
Ill
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
ECKH and H'K'C'tf, 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 U in the required
centroid of the given figure. This point G falls in our figure
112 CENTKOIDS.
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 crownpiece.
CHAPTER 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  J3A, and with C as centre and ^^ i\
CB as radius, describe an arc cutting / I \
the tangent AD in D. Then AD is the /^ I \
length of the given arc, with a negative
error, whose ratio to the whole arc is
1080 54432
6 being the ratio of the arc to the
radius*. Fig  125 '
Otherwise (Fig. 126) : leiD 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
*, 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! ; then the arc A'M'= OM. 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'" 9 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 (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 J3, H are the points in which these circles are
cut by any radius vector OM, then OM= the arc AB =
arc OH. Therefore, if we wish to set off the arc 07 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 H, M in which the chord 07
cuts the circle on OC&s diameter, and the spiral. We then take
away the spiral, and mark off on OX the segment OM r OM\
draw through 7 a parallel 77' to HM', 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
KECTIFICATION OF CIECULAR AllCS.
[148
OC' = OC, and then, supposing the chord 70 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 = pco.
Draw (Fig. 129) a convolution of this curve NMDCBA, and
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 f ; hence
the length of any circular arc whatever M " M'" 9 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
n
(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 JV; 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'B'C'D'
be its involute. From the
figure we have at once
the arc MA'=MM', where
Fig. 130.
MM' is the tangent of the circle at M. If now it is required
149]
RECTIFICATION OF CIRCULAR ARCS.
117
to set off the arc M" ' M'" (whose centre is 0), along its
tangent from M ft ', we need only draw OM', which if sufficiently
produced cuts the tangent in question in 3/' v , and M" M" 1
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. Bottcher*. 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. BD = 344153,
a value of the semicircumference 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, 1 883) of Hoffmann's Zeitschrift fur math, und
naturl. TJnterricht.]
RECIPROCAL FIGURES IN GRAPHICAL STATICS.
AUTHOK'S PKEFACE
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 redproche nella statica
grafica, 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 M6bius*,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 Nullsystem by German
mathematicians.
Such a correspondence is obtained in the following manner.
Let there be a plane 8, 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,
yo, a/3 in P, Q, R respectively, then the planes PBC, QCA,
EAB will all intersect in the same point 8 of the plane y ~~^'
The straight line
ax + ly + c
=
is reciprocal to the straight line
ax + ly + c' =
px + qy + / =
where rc f = r'c = k (aq bp).
11. Hence :
(#) 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.
(#) 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, /3, 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 wellknown theorem of Euler. Hence, if
the first orthographic figure possesses p nodes, p' polygons, and
* sides, we have p + y _ s + 2 .
The second figure will have p' 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 I at infinity on the central axis is a vertex
common to n faces of the first polyhedron, then the other
* TODHDNTER, Spherical Trigonometry, Chapter xiii, Polyhedra.
K
130 POLE AND POLAE PLANE.
polyhedron has in the plane at infinity a polygonal face of
n sides. In this case, the first orthographic figure has p I
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 p , p, s are still
connected by the relation
CHAPTEK II.
POLYGON OP 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 be
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 framework
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 wellknown 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 Rankine, 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 to the lines of
resistance of the bars of a polygonal frame, then the sides of any
* Philosophical Magazine, April 1864, p. 258.
f VAEIGNON, Nouvelle Mtcanique ou Statique, dont le projet fut donnt en
1687 : Paris, 1725.
K 2
132 POLYGON OF FORCES AND FUNICULAR POLYGON [16
polygo* whose angles He in these radiating lines will represent a
system of forces ', which, being applied to the joints of the frame,
will 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
due 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 jigures
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 If. In that memoir, after quoting
* Page 142 of the sixth edition (1872).
f Philosophical Magazine, Feb. 1864, p. 92.
J On reciprocal figures and diagrams offerees (Philosophical Magazine, April
1864, p. 250).
On reciprocal figures, frames and diagrams of forces (Transactions of the
Royal Society of Edinburgh, vol. xxvi. p. 1). See also a letter of Professor
Rankine in the ' Engineer? Feb. 1872.
 Or, rather, that of MONGE. (See CHASLES, Apercu, historique, p. 378.)
*![ 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 paragraph*
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 ly the circumstance that it icas
independently discovered by a practical draughtsman, Mr. Taylor,
working in the Office of the wellknown contractor Mr.. J. S. 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
^vhich 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 Bridges
and Roofs, London, 1869 ; Bow, Economies of construction in relation to framed
structures, London, 1873 ; CLARKE, Graphic Statics, London, 1880 ; EDDY, New
constructions in Graphical Statics, in Van Nostrand's Engineering Magazine, New
York, 18778, and American Journal of Mathematics, vol. i., Baltimore, 1878,
&c., &c.
134 POLYGON OF FORCES AND FUNICULAR POLYGON [19
and uniform method, which reduces itself in substance to the
construction of two figures, which he calls Krdflepolygon, 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
Culmann 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 Krdftepolygon 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 JJ, ^, ..., P n in
equilibrium, then by the polygon of forces we understand a
polygon, whose sides 1, 2, ..., & 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^ I, ^, ..., P n , and whose
sides are respectively parallel to the rays proceeding from
0J, in such a manner that the side comprised between
the lines of action of P r and P 8 , 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.
The direction only of the first side of that polygon is determinate.
20]
AS RECIPROCAL 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 a, a, in which the
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, 1,2,3,7... ,*...; 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 0', 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 polygonf 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 ' coincide, the corre
sponding sides of the two funicular polygons are parallel
(Fig. 4 a). 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.
f 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 FORCES 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),
(s , s + 1 ) of the polygon of forces, that is to say, a straight
line equipollent to the resultant of the forces P r+ i, P r+2 P s ;
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, P r+2 ,..., P K
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+l), (s , *+l) will then coincide in one point A (see
Fig. 5, 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+l , P r+2 , ...,P S
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,
>^> tne one which acts along the side (r, r + l) is
Fig. 5 a.
directed from A towards ; and the one which acts along the
side (s , s + 1) is directed from towards A.
25. Given the forces P l5 ^ 2 , P 3 ,..., P n _ (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 nl vertices 1, 2,3, ..., (n i), lie in the lines
of action of the given forces .ZJ, P 2 , P 3 , ..., P n _ 1 ; 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 potygon, (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 quotcumque rectae lineae sese mutuo secent, non plures qnam
duae per idem punctum, omnia autem in una ipsarum data sint, et
reliqiiorum multitudinem Jiabentium triangulum numerum, Jiujus latus
singula Jiabet puncta tangentia rectam lineam positione datam^ quorum
trium non ad angulum exist ens trianguli spatii unumquodque reli
quum punctum 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 1 ) ,
n ; and, in the same plane, n 1 straight lines 1 , 2 , 3 , . . . ,
(n l) , respectively parallel to the first n l 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 nl
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),
* [Mathematicae 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 Traitd des proprUUs projectives (Paris,
1822)].
28] AS RECIPROCAL FIGURES. 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),
($, # + 1) 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
selfevident, 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 0' \ 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 O f 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 FT be the poly
hedron enclosed by the surface 2, and the pyramidal sur
face whose vertex is a point 12 , taken arbitrarily in space, and
whose baseline 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 n will have n+p faces, and 2#f m edges,
and therefore m + n p + 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 .4) acting on it at the joint A. can be reduced to a single resultant : so
may those at the joint S; 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 (CBOFTON, Lectures on Applied Mechanics, at the Royal Military
Academy, London, 1877).
f A surface with an edge is simply connected, if its edge is a single closed
continuous line which does not intersect itself.
t 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.
}\ Therefore m can never be less than p 1.
144 APPLICATION OF RECIPKOOAL [30
Reciprocally, U' has m + n~j) + 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 pf 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 o> , will be the funicular
polygon of the external forces, corresponding to the pole 0,
the projection of 12 ; 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 H 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] DIAGKAMS 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 two 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 ; winch 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 EECIPKOCAL [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 contradiagonals. 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 i 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.
' The figure can be drawn in a few 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.
f The frame or truss is always supposed to be formed by triangles only.
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 KECIPKOCAL [30
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, 1O, 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] DIAGRAMS 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 by 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 b 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 n (Art. 29), remembering that the
projection of the surface 2 forms the polygons of the diagram
b, 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 socalled Law of Edges (KANTEN
GESETZ) of polyhedra possessing one internal surface and one external.
See MOBIUS, Ueber die Bestimmung des Inhalts eines Potyeders (Leipziges
Berichte, 1865, vol. 17, p. 33 and following), or Gesammelte Werke, 2nd Band,
p. 473 ; also BALTZEB, Stereometric, 8, Art. 16.
40]
DIAGRAMS 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 be made in the structure, then in each of the
parts so obtained^ the external forces are in equilibrium with the
reactions of the bars cut across by 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
struct ing the triangle of forces
, 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 1670, p. 30, already cited on p. 132.
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. BITTER, Elementare Theorie und Berechnung eiserner Dach und
UruckenConstructionem, 2nd edition (Hannover, 1870).
CHAPTEK IV.
EXAMPLES OF FRAME AND STRESS 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 betiveen 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&.
parallel to 11 and 12; we notice that the straight line 11
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 13
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, 10, ..., 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 b 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 OP FRAME 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. 8.
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.
so
?.+
42
46
Fig. 8fc.
Construct the polygon corresponding to the joint (16, 18,
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 centradiagonals are all in compression ;
finally two of the verticals 23 , 39 are in tension, and all the
rest in compression.
44. Figure 9 a* represents one half of a locomotive shed.
The external forces are the weights 1,2,3,4,5 applied at
Fig. 9 a.
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 Grapkische
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 , 1 7) , 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. 10 a.
Fig. io&.
may be considered as the resultants of the deadloads and
wind pressure ; the forces 8,9 represent the reactions at
the supports.
The polygon of external forces is drawn in diagram b 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. n 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 8TKES8 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
Fig. ii 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 6.
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 centradiagonals 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 STRESS 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. 13 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, ...,0; 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, 1), (1, 2), (2, 3),
... (8, 9), (9, 0)*, and construct the corresponding funicular
polygon. The vertical through the point where the extreme
sides (O,l), (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 oi the segment 1, and (9,0) the
final point of segment 9. In the figure the rays, from the pole O and the sides of
the corresponding funicular polygon, are shown by dotted lines
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In Parchment, 45. 6d.
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of DRYDEN and POPE only, stiff covers, 2s, 6d.
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IV. Wit and Wisdom of Samuel Johnson. Edited by
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Cowper. Edited, with Life, Introductions, and Notes, by
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I. The Didactic Poems of 1782, with Selections from
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Keats. Hyperion, Book I. With Notes by W. T. ARNOLD,
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Scott. Lay of the Last Minstrel. Edited by W. MINTO, M.A.
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Marmion. Edited, with Introduction and Notes, by
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Campbell. Gertrude of Wyoming. Edited, with Introduction
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Shairp. Aspects of Poetry ; being Lectures delivered at
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Palgrave. The Treasury of Sacred Song. With Notes Ex
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SECTION III.
EUROPEAN LANGUAGES. MEDIAEVAL AND
MODERN.
(l) FRENCH AND ITALIAN.
Brachet's Etymological Dictionary of the French Language.
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Historical Grammar of the French Language. Trans
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Saintsbury. Primer of French Literature. By GEORGE
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Short History of French Literature. Crown 8vo. 105. 6d.
Specimens of French Literature, from Villon to Hugo.
Crown 8vo. 9*.
Beaumarchais' Le Barbier de Seville. Edited, with Intro
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Oxford: Clarendon Press.
French and Italian. n
Corneille's Horace. Edited, with Introduction and Notes,
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Moliere's Les Precieuses Ridicules. Edited, with Introduction
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Musset's On ne badine pas avec 1' Am our, andFantasio. Edited,
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Racine's Esther. Edited, with Introduction and Notes, by
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Voltaire's Merope. Edited, with Introduction and Notes,
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MASSON'S FRENCH CLASSICS.
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Louis XIV and his Contemporaries ; as described in Extracts
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SainteBeuve. Selections from the Causeries du Lundi.
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Tasso. La Gerusalemme Liberata. Cantos i, ii. With In
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(2) GERMAN AND GOTHIC.
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' History of German Literature,' by F. LICHTENSTEIN. 2 vols. Crown
8VO. 215.
Scherer. A History of German Literature by WILHELM
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C. CONYBEARE. Edited by F. MAX MULLER. 2 vols. 8vo. 2 is.
Skeat. The Gospel of St. Mark in Gothic. By W. W.
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Wright. An Old High German Primer. With Grammar,
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LANGE'S GERMAN COURSE.
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I. Germans #t Home ; a Practical Introduction to German
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German and Gothic. 13
III. Grammar of the German Language. 8vo. 3*. 6d.
IV. German Composition ; A Theoretical and Practical Guide
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German Spelling ; A Synopsis of the Changes which it
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BUCHHEIM'S GERMAN CLASSICS.
Edited, with Biographical, Historical, and Critical Introductions, Arguments
(to the Dramas}, and Complete Commentaries, by C. A. BUCHHEIM, Phil.
Doc., Professor in King's College, London.
Becker (the Historian). Friedrich der Grosse. Edited, with
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Goethe :
(a) Egmont. A Tragedy. ' 35.
(6) Iphigenie auf Tauris. A Drama. 30.
Heine:
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(b) Harzreise. Cloth, 25. 6d. ; paper covers, is. 6d.
Lessing :
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(b) Minna von Barnhelm. A Comedy. 35. 6d.
Schiller :
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(b) Wilhelm Tell. School Edition. With Map. 2s.
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Modern German Reader. A Graduated Collection of Ex
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Part II. Extracts in Prose and Poetry. With English Notes
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German Poetry for Beginners. Edited with English Notes
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Chamisso. Peter Schlemihl's Wundersame Geschichte.
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Lessing. The Laokoon, with Introduction, English Notes,
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Niebuhr: Griechische HeroenGeschichten (Tales of Greek
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Second, Revised Edition. Extra fcap. 8vo. cloth, 2s., stiff covers, I*. 6d.
Edition A. Text in German Type.
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Schiller's Wilhelm Tell. Translated into English Verse by
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(3) SCANDINAVIAN.
Cleasby and Vigfdsson. An IcelandicEnglish Dictionary,
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Sweet. Icelandic Primer, with Grammar, Notes, and
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Vigfdsson. Sturlunga Saga, including the Islendinga Saga
of Lawman STURLA THORDSSON and other works. Edited by Dr.
GUDBRAND VIGFUSSON. In 2 vols. 8vo. 2l. 2s.
Vigfdsson and Powell. Icelandic Prose Reader, with Notes,
Grammar, and Glossary. By G. VIGFUSSON, M.A., and F. YORK
POWELL, M.A. Extra fcap. 8vo. 10*. 6d.
Corpvs Poeticvm Boreale. The Poetry of the Old
Northern Tongue, from the Earliest Times to the Thirteenth Century.
Edited, classified, and translated, with Introduction, Excursus, and Notes,
by GUDBKAND VIGFUSSON, M.A., and F. YORK POWELL, M.A. 2 vols.
8vo. 2l. 2s.
The LandnamaB6k. Edited and translated by the
same. In the Press.
SECTION IV.
CLASSICAL LANGUAGES.
(1) LATIN.
STANDARD WORKS AND EDITIONS.
Ellis. Harleian MS. 2610 ; Ovid's Metamorphoses I, II, III,
1622; XXIV Latin Epigrams from Bodleian or other MSS. ; Latin
Glosses on Apollinaris Sidonius from MS. Digby 172. Collated and
Edited by EOBINSON ELLIS, M.A., LL.D. (Anecdota Oxon.) 45.
Oxford : Clarendon Press.
Latin: Standard Works. 15
King and Cookson. The Principles of Sound and Inflexion,
as illustrated in the Greek and Latin Languages. By J.E. KING, M.A.,
and CHRISTOPHER COOKSON, M.A. 8vo. i8s.
Lewis and Short. A Latin Dictionary, founded on Andrews'
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part rewritten by CHARLTON T. LEWIS, Ph.D., and CHARLES SHORT,
LL.D. 4to. il. 5*.
Nettleship. Contributions to Latin Lexicography. By
HENRY NETTLESHIP, M.A. 8vo. 2 is.
Lectures and Essays on Subjects connected with
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Crown 8vo. 7*. 6d.
, The Roman Satura. 8vo. sewed, is.
Ancient Lives of Vergil. 8vo. sewed, 2$.
Papillon. Manual of Comparative Philology. By T. L.
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Finder. Selections from the less known Latin Poets. By
NORTH FINDER, M.A. 8vo. 15*.
Sellar. Roman Poets of the Augustan Age. VIRGIL. By
W. Y. SELLAR, M.A. New JEdition. Crown 8vo. 95.
Roman Poets of the Republic. Third Edition. Crown
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Wordsworth. Fragments and Specimens of Early Latin.
With Introductions and Notes. By J. WORDSWORTH, D.D. 8vo. i8s.
Avianus. The Fables. Edited, with Prolegomena, Critical
Apparatus, Commentary, etc., by KOBINSON ELLIS, M.A., LL.D. 8vo.
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Catulli Veronensis Liber. Iterum recognovit, apparatum
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Catullus, a Commentary on. By ROBINSON ELLIS, M.A.
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Cicero. De Oratore. With Introduction and Notes. By A.
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Select Letters. With English Introductions, Notes, and
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Horace. With a Commentary. Vol. I. The Odes, Carmen
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Nonius Marcellus, de Compendiosa Doctrina (Harleian MS.
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Ovid. P. Ovidii Nasonis Ibis. Ex Novis Codicibus edidit,
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P. Ovidi Nasonis Tristium Libri V. Recensuit S. G.
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Persius. The Satires. With a Translation and Commen
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Plautus. Bentley's Plautine Emendations. From his copy
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Scriptores Latini rei Metricae. Edidit T. GAISFORD, S.T.P.
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LATIN EDUCATIONAL WORKS.
GKAMMAKS, EXERCISE BOOKS, &c.
ALLEN.
B/udimenta Latina. Comprising Accidence, and Exercises
of a very Elementary Character, for the use of Beginners. By JOHN
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Latin: Educational Works. 17
Passages for Translation into Latin. For the use of Passmen
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Catulli Veronensis Carmina Selecta, secundum recognitionem
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CICERO. Selection of Interesting and Descriptive Passages.
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Select Orations (for Schools). In Verrem Actio Prima.
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Selected Letters (for Schools). With Notes. By the
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Select Letters. Text. By ALBERT WATSON, M.A.
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Cornelius Nepos. With Notes. By OSCAR BROWNING, M.A.
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Juvenal. Thirteen Satires. Edited, with Introduction and
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Latin : Educational Works. 19
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Extra fcap. 8vo. 5*. 6d.
Tristia. Book I. The Text revised, with an Intro
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Pliny. Selected Letters (for Schools). With Notes. By
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Virgil. With Introduction and Notes. By T. L. PAPILLON,
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