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﻿Title: Geometrical Solutions Derived from Mechanics; a Treatise of Archimedes
Author: Archimedes, 280? BC-211? BC
Language: English
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Gordon Keener

% gbn0305181551: Archimedes, Geometrical Solutions Derived from Mechanics.  Gordon Keener .  1909c.  5/19/2003.  ok.
\documentclass[12pt]{article}
\usepackage[greek,english]{babel}
\usepackage{wrapfig}
\usepackage{graphicx}

\title{Geometrical Solutions Derived from Mechanics}

\author{A Treatise of Archimedes}

\date{\vspace{\baselineskip}
{\small Recently discovered and translated from the Greek
by}\\ Dr. J. L. Heiberg\\ {\small Professor of Classical Philology at
the University of Copenhagen}\\
\vspace{\baselineskip} {\small with an introduction by}\\ David Eugene
Smith\\ {\small President of Teachers College, Columbia University, New York}\\
\vspace{\baselineskip}
{\small English version translated from the German by}\\ Lydia
G. Robinson\\ {\small and reprinted from The Monist,'' April,
1909}\\
\vspace{\baselineskip}
gbn0305181551: Archimedes, Geometrical Solutions Derived from Mechanics.
Gordon Keener $<$gkeener@nc.rr.com$>$.  1909c.  5/19/2003.  ok.}
}

\begin{document}

\maketitle

\vfill\pagebreak

\section*{Introduction}

If there ever was a case of appropriateness in discovery, the finding
of this manuscript in the summer of 1906 was one. In the first place
it was appropriate that the discovery should be made in
Constantinople, since it was here that the West received its first
manuscripts of the other extant works, nine in number, of the great
Syracusan. It was furthermore appropriate that the discovery should be
made by Professor Heiberg, \emph{facilis princeps} among all workers
in the field of editing the classics of Greek mathematics, and an
indefatigable searcher of the libraries of Europe for manuscripts to
aid him in perfecting his labors.  And finally it was most appropriate
that this work should appear at a time when the affiliation of pure
and applied mathematics is becoming so generally recognized all over
the world. We are sometimes led to feel, in considering isolated
cases, that the great contributors of the past have worked in the
field of pure mathematics alone, and the saying of Plutarch that
Archimedes felt that every kind of art connected with daily needs
was ignoble and vulgar''\footnote{Marcellus, 17.} may have
strengthened this feeling. It therefore assists us in properly
orientating ourselves to read another treatise from the greatest
mathematician of antiquity that sets clearly before us his
indebtedness to the mechanical applications of his subject.

Not the least interesting of the passages in the manuscript is the
first line, the greeting to Eratosthenes. It is well known, on the
testimony of Diodoros his countryman, that Archimedes studied in
Alexandria, and the latter frequently makes mention of Konon of Samos
whom he knew there, probably as a teacher, and to whom he was indebted
for the suggestion of the spiral that bears his name. It is also
related, this time by Proclos, that Eratosthenes was a contemporary of
Archimedes, and if the testimony of so late a writer as Tzetzes, who
lived in the twelfth century, may be taken as valid, the former was
eleven years the junior of the great Sicilian. Until now, however, we
have had nothing definite to show that the two were ever
acquainted. The great Alexandrian savant,---poet, geographer,
arithmetician,---affectionately called by the students Pentathlos, the
champion in five sports,\footnote{His nickname of \emph{Beta} is well
known, possibly because his lecture room was number 2.} selected by
Ptolemy Euergetes to succeed his master, Kallimachos the poet, as head
of the great Library,---this man, the most renowned of his time in
Alexandria, could hardly have been a teacher of Archimedes, nor yet
the fellow student of one who was so much his senior. It is more
probable that they were friends in the later days when Archimedes was
received as a savant rather than as a learner, and this is borne out
by the statement at the close of proposition I which refers to one of
his earlier works, showing that this particular treatise was a late
one. This reference being to one of the two works dedicated to
Dositheos of Kolonos,\footnote{We know little of his works, none of
which are extant. Geminos and Ptolemy refer to certain observations
made by him in 200 B. C., twelve years after the death of
Archimedes. Pliny also mentions him.} and one of these (\emph{De
lineis spiralibus}) referring to an earlier treatise sent to
Konon,\footnote{\selectlanguage{greek} T\~wn pot\i K\'onwna
\'apustal\'entwn jewrhm\'atwn.} we are led to believe that this was
one of the latest works of Archimedes and that Eratosthenes was a
friend of his mature years, although one of long standing. The
statement that the preliminary propositions were sent some time
ago'' bears out this idea of a considerable duration of friendship,
and the idea that more or less correspondence had resulted from this
communication may be inferred by the statement that he saw, as he had
previously said, that Eratosthenes was a capable scholar and a
prominent teacher of philosophy,'' and also that he understood how
to value a mathematical method of investigation when the opportunity
offered.'' We have, then, new light upon the relations between these
two men, the leaders among the learned of their day.

A second feature of much interest in the treatise is the intimate view
that we have into the workings of the mind of the author. It must
always be remembered that Archimedes was primarily a discoverer, and
not primarily a compiler as were Euclid, Apollonios, and Nicomachos.
Therefore to have him follow up his first communication of theorems to
Eratosthenes by a statement of his mental processes in reaching his
conclusions is not merely a contribution to mathematics but one to
education as well. Particularly is this true in the following
statement, which may well be kept in mind in the present day: l have
thought it well to analyse and lay down for you in this same book a
peculiar method by means of which it will be possible for you to
derive instruction as to how certain mathematical questions may be
investigated by means of mechanics. And I am convinced that this is
equally profitable in demonstrating a proposition itself; for much
that was made evident to me through the medium of mechanics was later
proved by means of geometry, because the treatment by the former
method had not yet been established by way of a demonstration. For of
course it is easier to establish a proof if one has in this way
previously obtained a conception of the questions, than for him to
seek it without such a preliminary notion. .  . . Indeed I assume that
some one among the investigators of to-day or in the future will
discover by the method here set forth still other propositions which
have not yet occurred to us.'' Perhaps in all the history of
mathematics no such prophetic truth was ever put into words.  It would
almost seem as if Archimedes must have seen as in a vision the methods
of Galileo, Cavalieri, Pascal, Newton, and many of the other great
makers of the mathematics of the Renaissance and the present time.

The first proposition concerns the quadrature of the parabola, a
subject treated at length in one of his earlier communications to
Dositheos.\footnote{\selectlanguage{greek} Tetragwnismds parabol\~hs.}
He gives a digest of the treatment, but with the warning that the
proof is not complete, as it is in his special work upon the
subject. He has, in fact, summarized propositions VII-XVII of his
communication to Dositheos, omitting the geometric treatment of
propositions XVIII-XXIV.  One thing that he does not state, here or in
any of his works, is where the idea of center of
gravity\footnote{\selectlanguage{greek} K\'entra bar\~wn,
\selectlanguage{english} for barycentric'' is a very old term.}
started. It was certainly a common notion in his day, for he often
uses it without defining it. It appears in Euclid's\footnote{At any
rate in the anonymous fragment \emph{De levi et ponderoso}, sometimes
attributed to him.} time, but how much earlier we cannot as yet say.

Proposition II states no new fact. Essentially it means that if a
sphere, cylinder, and cone (always circular) have the same radius,
$r$, and the altitude of the cone is $r$ and that of the cylinder
$2r$, then the volumes will be as $4 : 1 : 6$, which is true, since
they are respectively $\frac{4}{3}\pi r^3$, $\frac{1}{3}\pi r^3$, and
$2\pi r^3$.  The interesting thing, however, is the method pursued,
the derivation of geometric truths from principles of mechanics. There
is, too, in every sentence, a little suggestion of Cavalieri, an
anticipation by nearly two thousand years of the work of the greatest
immediate precursor of Newton. And the geometric imagination that
Archimedes shows in the last sentence is also noteworthy as one of the
interesting features of this work: After I had thus perceived that a
sphere is four times as large as the cone. . . it occurred to me that
the surface of a sphere is four times as great as its largest circle,
in which I proceeded from the idea that just as a circle is equal to a
triangle whose base is the periphery of the circle, and whose altitude
is equal to its radius, so a sphere is equal to a cone whose base is
the same as the surface of the sphere and whose altitude is equal to
the radius of the sphere.'' As a bit of generalization this throws a
good deal of light on the workings of Archimedes's mind.

In proposition III he considers the volume of a spheroid, which he had
already treated more fully in one of his letters to
Dositheos,\footnote{\selectlanguage{greek} Per\i kwnoeide\~wn kai
sfairoeide\~wn.} and which contains nothing new from a mathematical
standpoint. Indeed it is the method rather than the conclusion that is
interesting in such of the subsequent propositions as relate to
mensuration. Proposition V deals with the center of gravity of a
segment of a conoid, and proposition VI with the center of gravity of
a hemisphere, thus carrying into solid geometry the work of Archimedes
on the equilibrium of planes and on their centers of
gravity.\footnote{\selectlanguage{greek} 'Epip\'edwn \isorropi\~wn
\^h k\'entra bar\~wn \'epip\'edwn.} The general method is that already
known in the treatise mentioned, and this is followed through
proposition X.

Proposition XI is the interesting case of a segment of a right
cylinder cut off by a plane through the center of the lower base and
tangent to the upper one. He shows this to equal one-sixth of the
square prism that circumscribes the cylinder. This is well known to us
through the formula $v = 2r^2h/3$, the volume of the prism being
$4r^2h$, and requires a knowledge of the center of gravity of the
cylindric section in question.  Archimedes is, so far as we know, the
first to state this result, and he obtains it by his usual method of
the skilful balancing of sections.  There are several lacunae in the
demonstration, but enough of it remains to show the ingenuity of the
general plan. The culminating interest from the mathematical
standpoint lies in proposition XIII, where Archimedes reduces the
whole question to that of the quadrature of the parabola. He shows
that a fourth of the circumscribed prism is to the segment of the
cylinder as the semi-base of the prism is to the parabola inscribed in
the semi-base; that is, that $\frac{1}{4}p : v = \frac{1}{2}b : (\frac{2}{3} \cdot \frac{1}{2}b)$, whence $v = \frac{1}{6}p$.
Proposition XIV is incomplete, but it is the conclusion of the two
preceding propositions.

In general, therefore, the greatest value of the work lies in the
following:

1. It throws light upon the hitherto only suspected relations of
Archi\-medes and Eratosthenes.

2. It shows the working of the mind of Archimedes in the discovery of
mathematical truths, showing that he often obtained his results by
intuition or even by measurement, rather than by an analytic form of
reasoning, verifying these results later by strict analysis.

3. It expresses definitely the fact that Archimedes was the discoverer
of those properties relating to the sphere and cylinder that have been
attributed to him and that are given in his other works without a
definite statement of their authorship.

4. It shows that Archimedes was the first to state the volume of the
cylinder segment mentioned, and it gives an interesting description of
the mechanical method by which he arrived at his result.

{\hspace*{\fill}\textsc{David Eugene Smith.}}\linebreak
{\hspace*{\fill}\textsc{Teachers~College,~Columbia~University.}}\linebreak

\vfill\pagebreak

\section*{Geometrical Solutions Derived from Mechanics.}

\textsc{Archimedes to Eratosthenes, Greeting:}

Some time ago I sent you some theorems I had discovered, writing down
only the propositions because I wished you to find their
demonstrations which had not been given. The propositions of the
theorems which I sent you were the following:

1. If in a perpendicular prism with a parallelogram\footnote{This must
mean a square.} for base a cylinder is inscribed which has its bases
in the opposite
surface touching the other planes of the prism, and if a plane is
passed through the center of the circle that is the base of the
cylinder and one side of the square lying in the opposite plane, then
that plane will cut off from the cylinder a section which is bounded
by two planes, the intersecting plane and the one in which the base of
the cylinder lies, and also by as much of the surface of the cylinder
as lies between these same planes; and the detached section of the
cylinder is $\frac{1}{6}$ of the whole prism.

2. If in a cube a cylinder is inscribed whose bases lie in opposite
surface touches the other four planes, and if in the same cube a
second cylinder is inscribed whose bases lie in two other
surface touches the four other planes, then the body enclosed by the
surface of the cylinder and comprehended within both cylinders will be
equal to $\frac{2}{3}$ of the whole cube.

These propositions differ essentially from those formerly discovered;
for then we compared those bodies (conoids, spheroids and their
segments) with the volume of cones and cylinders but none of them was
found to be equal to a body enclosed by planes. Each of these bodies,
on the other hand, which are enclosed by two planes and cylindrical
surfaces is found to be equal to a body enclosed by planes. The
demonstration of these propositions I am accordingly sending to you in
this book.

Since I see, however, as I have previously said, that you are a
capable scholar and a prominent teacher of philosophy, and also that
you understand how to value a mathematical method of investigation
when the opportunity is offered, I have thought it well to analyze and
lay down for you in this same book a peculiar method by means of which
it will be possible for you to derive instruction as to how certain
mathematical questions may be investigated by means of mechanics. And
I am convinced that this is equally profitable in demonstrating a
proposition itself; for much that was made evident to me through the
medium of mechanics was later proved by means of geometry because the
treatment by the former method had not yet been established by way of
a demonstration. For of course it is easier to establish a proof if
one has in this way previously obtained a conception of the questions,
than for him to seek it without such a preliminary notion. Thus in the
familiar propositions the demonstrations of which Eudoxos was the
first to discover, namely that a cone and a pyramid are one third the
size of that cylinder and prism respectively that have the same base
and altitude, no little credit is due to Democritos who was the first
to make that statement about these bodies without any
demonstration. But we are in a position to have found the present
proposition in the same way as the earlier one; and I have decided to
write down and make known the method partly because we have already
talked about it heretofore and so no one would think that we were
spreading abroad idle talk, and partly in the conviction that by this
means we are obtaining no slight advantage for mathematics, for indeed
I assume that some one among the investigators of to-day or in the
future will discover by the method here set forth still other
propositions which have not yet occurred to us.

In the first place we will now explain what was also first made clear
to us through mechanics, namely that a segment of a parabola is
$\frac{4}{3}$ of the triangle possessing the same base and equal
altitude; following which we will explain in order the particular
propositions discovered by the above mentioned method; and in the last
part of the book we will present the geometrical demonstrations of the
propositions.\footnote{In his Commentar,'' Professor Zeuthen calls
attention to the fact that it was aiready known from Heron's recently
discovered \emph{Metrica} that these propositions were contained in
this treatise, and Professor Heiberg made the same comment in
\emph{Hermes}.---Tr.}

1. If one magnitude is taken away from another magnitude and the same
point is the center of gravity both of the whole and of the part
removed, then the same point is the center of gravity of the remaining
portion.

2. If one magnitude is taken away from another magnitude and the
center of gravity of the whole and of the part removed is not the same
point, the center of gravity of the remaining portion may be found by
prolonging the straight line which connects the centers of gravity of
the whole and of the part removed, and setting off upon it another
straight line which bears the same ratio to the straight line between
the aforesaid centers of gravity, as the weight of the magnitude which
has been taken away bears to the weight of the one remaining [\emph{De
plan. aequil.} I, 8].

3. If the centers of gravity of any number of magnitudes lie upon the
same straight line, then will the center of gravity of all the
magnitudes combined lie also upon the same straight line [Cf.
\emph{ibid.} I, 5].

4. The center of gravity of a straight line is the center of that line
[Cf. \emph{ibid.} I, 4].

5. The center of gravity of a triangle is the point in which the
straight lines drawn from the angles of a triangle to the centers of
the opposite sides intersect [\emph{Ibid.} I, 14].

6. The center of gravity of a parallelogram is the point where its
diagonals meet [\emph{Ibid.} I, 10].

7. The center of gravity [of a circle] is the center [of that circle].

8. The center of gravity of a cylinder [is the center of its axis].

9. The center of gravity of a prism is the center of its axis.

10. The center of gravity of a cone so divides its axis that the
section at the vertex is three times as great as the remainder.

11. Moreover together with the exercise here laid down I will make use
of the following proposition:

If any number of magnitudes stand in the same ratio to the same number
of other magnitudes which correspond pair by pair, and if either all
or some of the former magnitudes stand in any ratio whatever to other
magnitudes, and the latter in the same ratio to the corresponding
ones, then the sum of the magnitudes of the first series will bear the
same ratio to the sum of those taken from the third series as the sum
of those of the second series bears to the sum of those taken from the
fourth series [\emph{De Conoid.} I].

\section*{Proposition I}

Let $\alpha\beta\gamma$ [Fig.~1] be the segment of a parabola bounded
by the straight line $\alpha\gamma$ and the parabola
$\alpha\beta\gamma$.  Let $\alpha\gamma$ be bisected at $\delta$,
$\delta\beta\epsilon$ being parallel to the diameter, and draw
$\alpha\beta$, and $\beta\gamma$.  Then the segrnent
$\alpha\beta\gamma$ will be $\frac{4}{3}$ as great as the triangle
$\alpha\beta\gamma$.

From the points $\alpha$ and $\gamma$ draw $\alpha\zeta \| \delta\beta\epsilon$, and the tangent $\gamma\zeta$; produce
[$\gamma\beta$ to $\kappa$, and make $\kappa\theta = \gamma\kappa$].
Think of $\gamma\theta$ as a scale-beam with the center at $\kappa$
and let $\mu\xi$ be any straight line whatever $\| \epsilon\delta$. Now since $\gamma\beta\alpha$ is a parabola,
$\gamma\zeta$ a tangent and $\gamma\delta$ an ordinate, then
$\epsilon\beta = \beta\delta$; for this indeed has been proved in the
Elements [i.e., of conic sections, cf.  \emph{Quadr. parab.} 2].  For
this reason and because $\zeta\alpha$ and $\mu\xi \| \epsilon\delta$,
$\mu\nu = \nu\xi$, and $\zeta\kappa = \kappa\alpha$.  And because
$\gamma\alpha : \alpha\xi = \mu\xi : \xi o$ (for this is shown in a
corollary, [cf. \emph{Quadr. parab.} 5]), $\gamma\alpha : \alpha\xi = \gamma\kappa : \kappa\nu$; and $\gamma\kappa = \kappa\theta$,
therefore $\theta\kappa : \kappa\nu = \mu\xi : \xi o$.  And because
$\nu$ is the center of gravity of the straight line $\mu\xi$, since
$\mu\nu = \nu\xi$, then if we make $\tau\eta = \xi o$ and $\theta$ as
its center of gravity so that $\tau\theta = \theta\eta$, the straight
line $\tau\theta\eta$ will be in equilibrium with $\mu\xi$ in its
present position because $\theta\nu$ is divided in inverse proportion
to the weights $\tau\eta$ and $\mu\xi$, and $\theta\kappa : \kappa\nu = \mu\xi : \eta\tau$; therefore $\kappa$ is the center of gravity of
the combined weight of the two.  In the
%
\begin{wrapfigure}{r}{0.5\textwidth}
\includegraphics[width=0.5\textwidth]{fig01.png}
\begin{center}
{\small Fig.~1.}
\end{center}
\end{wrapfigure}
%
same way all straight lines drawn in the triangle $\zeta\alpha\gamma \| \epsilon\delta$ are in their present positions in equilibrium with
their parts cut off by the parabola, when these are transferred to
$\theta$, so that $\kappa$ is the center of gravity of the combined
weight of the two. And because the triangle $\gamma\zeta\alpha$
consists of the straight lines in the triangle $\gamma\zeta\alpha$ and
the segment $\alpha\beta\gamma$ consists of those straight lines
within the segment of the parabola corresponding to the straight line
$\xi o$, therefore the triangle $\zeta\alpha\gamma$ in its present
position will be in equilibrium at the point $\kappa$ with the
parabola-segment when this is transferred to $\theta$ as its center of
gravity, so that $\kappa$ is the center of gravity of the combined
weights of the two. Now let $\gamma\kappa$ be so divided at $\chi$
that $\gamma\kappa = 3\kappa\chi$; then $\chi$ will be the center of
gravity of the triangle $\alpha\zeta\gamma$, for this has been shown
in the Statics [cf. \emph{De plan. aequil.} I, 15, p. 186, 3 with
Eutokios, S. 320, 5ff.]. Now the triangle $\zeta\alpha\gamma$ in its
present position is in equilibrium at the point $\kappa$ with the
segment $\beta\alpha\gamma$ when this is transferred to $\theta$ as
its center of gravity, and the center of gravity of the triangle
$\zeta\alpha\gamma$ is $\chi$; hence triangle $\alpha\zeta\gamma :$
segm. $\alpha\beta\gamma$ when transferred to $\theta$ as its center
of gravity $= \theta\kappa : \kappa\chi$. But $\theta\kappa = 3\kappa\chi$; hence also triangle $\alpha\zeta\gamma = 3$
segm. $\alpha\beta\gamma$.  But it is also true that triangle
$\zeta\alpha\gamma = 4\Delta\alpha\beta\gamma$ because $\zeta\kappa = \kappa\alpha$ and $\alpha\delta = \delta\gamma$; hence
segm. $\alpha\beta\gamma = \frac{4}{3}$ the triangle
$\alpha\beta\gamma$. This is of course clear.

It is true that this is not proved by what we have said here; but it
indicates that the result is correct. And so, as we have just seen
that it has not been proved but rather conjectured that the result is
correct we have devised a geometrical demonstration which we made
known some time ago and will again bring forward farther on.

\section*{Proposition II}

That a sphere is four times as large as a cone whose base is equal to
the largest circle of the sphere and whose altitude is equal to the
radius of the sphere, and that a cylinder whose base is equal to the
largest circle of the sphere and whose altitude is equal to the
diameter of the circle is one and a half times as large as the sphere,
may be seen by the present method in the following way:

\begin{wrapfigure}{r}{0.5\textwidth}
\includegraphics[width=0.5\textwidth]{fig02.png}
\begin{center}
{\small Fig.~2.}
\end{center}
\end{wrapfigure}

Let $\alpha\beta\gamma\delta$ [Fig.~2] be the largest circle of a
sphere and $\alpha\gamma$ and $\beta\delta$ its diameters
perpendicular to each other; let there be in the sphere a circle on
the diameter $\beta\delta$ perpendicular to the circle
$\alpha\beta\gamma\delta$, and on this perpendicular circle let there
be a cone erected with its vertex at $\alpha$; producing the convex
surface of the cone, let it be cut through $\gamma$ by a plane
parallel to its base; the result will be the circle perpendicular to
$\alpha\gamma$ whose diameter will be $\epsilon\zeta$. On this circle
erect a cylinder whose axis $= \alpha\gamma$ and whose vertical
boundaries are $\epsilon\lambda$ and $\zeta\eta$. Produce
$\gamma\alpha$ making $\alpha\theta = \gamma\alpha$ and think of
$\gamma\theta$ as a scale-beam with its center at $\alpha$. Then let
$\mu\nu$ be any straight line whatever drawn $\| \beta\delta$
intersecting the circle $\alpha\beta\gamma\delta$ in $\xi$ and $o$,
the diameter $\alpha\gamma$ in $\sigma$, the straight line
$\alpha\epsilon$ in $\pi$ and $\alpha\zeta$ in $\rho$, and on the
straight line $\mu\nu$ construct a plane perpendicular to
$\alpha\gamma$; it will intersect the cylinder in a circle on the
diameter $\mu\nu$; the sphere $\alpha\beta\gamma\delta$, in a circle
on the diameter $\xi o$; the cone $\alpha\epsilon\zeta$ in a circle on
the diameter $\pi\rho$. Now because $\gamma\alpha \times \alpha\sigma = \mu\sigma \times \sigma\pi$ ( for $\alpha\gamma = \sigma\mu$,
$\alpha\sigma = \pi\sigma$), and $\gamma\alpha \times \alpha\sigma = \alpha\xi^2 = \xi\sigma^2 + \alpha\pi^2$ then $\mu\sigma \times \sigma\pi = \xi\sigma^2 + \sigma\pi^2$. Moreover, because
$\gamma\alpha : \alpha\sigma = \mu\sigma : \sigma\pi$ and
$\gamma\alpha = \alpha\theta$, therefore $\theta\alpha : \alpha\sigma = \mu\sigma : \sigma\pi = \mu\sigma^2 : \mu\sigma \times \sigma\pi$. But it has been proved that $\xi\sigma^2 + \sigma\pi^2 = \mu\sigma \times \sigma\pi$; hence $\alpha\theta : \alpha\sigma = \mu\sigma^2 : \xi\sigma^2 + \sigma\pi^2$. But it is true that
$\mu\sigma^2 : \xi\sigma^2 + \sigma\pi^2 = \mu\nu^2 : \xi\alpha^2 + \pi\rho^2 =$ the circle in the cylinder whose diameter is $\mu\nu :$
the circle in the cone whose diameter is $\pi\rho$ + the circle in the
sphere whose diameter is $\xi o$; hence $\theta\alpha : \alpha\sigma =$ the circle in the cylinder $:$ the circle in the sphere $+$ the
circle in the cone. Therefore the circle in the cylinder in its
present position will be in equilibrium at the point $\alpha$ with the
two circles whose diameters are $\xi o$ and $\pi\rho$, if they are so
transferred to $\theta$ that $\theta$ is the center of gravity of
both. In the same way it can be shown that when another straight line
is drawn in the parallelogram $\xi\lambda \| \epsilon\zeta$, and upon
it a plane is erected perpendicular to $\alpha\gamma$, the circle
produced in the cylinder in its present position will be in
equilibrium at the point $\alpha$ with the two circles produced in the
sphere and the cone when they are transferred and so arranged on the
scale-beam at the point $\theta$ that $\theta$ is the center of
gravity of both. Therefore if cylinder, sphere and cone are filled up
with such circles then the cylinder in its present position will be in
equilibrium at the point $\alpha$ with the sphere and the cone
together, if they are transferred and so arranged on the scale-beam at
the point $\theta$ that $\theta$ is the center of gravity of both. Now
since the bodies we have mentioned are in equilibrium, the cylinder
with $\kappa$ as its center of gravity, the sphere and the cone
transferred as we have said so that they have $\theta$ as center of
gravity, then $\theta\alpha : \alpha\kappa =$ cylinder $:$ sphere $+$
cone. But $\theta\alpha = 2\alpha\kappa$, and hence also the cylinder
$= 2 \times$ (sphere $+$ cone). But it is also true that the cylinder
= 3 cones [Euclid, \emph{Elem.} XII, 10], hence 3 cones = 2 cones + 2
spheres. If 2 cones be subtracted from both sides, then the cone whose
axes form the triangle $\alpha\epsilon\zeta =$ 2 spheres. But the cone
whose axes form the triangle $\alpha\epsilon\zeta =$ 8 cones whose
axes form the triangle $\alpha\beta\delta$ because $\epsilon\zeta = 2\beta\delta$, hence the aforesaid 8 cones = 2 spheres. Consequently
the sphere whose greatest circle is $\alpha\beta\gamma\delta$ is four
times as large as the cone with its vertex at $\alpha$, and whose base
is the circle on the diatneter $\beta\delta$ perpendicular to
$\alpha\gamma$.

Draw the straight lines $\phi\beta\chi$ and $\psi\delta\omega \| \alpha\gamma$ through $\beta$ and $\delta$ in the parallelogram
$\lambda\zeta$ and imagine a cylinder whose bases are the circles on
the diameters $\phi\psi$ and $\chi\omega$ and whose axis is
$\alpha\gamma$.  Now since the cylinder whose axes form the
parallelogram $\phi\omega$ is twice as large as the cylinder whose
axes form the parallelogram $\phi\delta$ and the latter is three times
as large as the cone the triangle of whose axes is
$\alpha\beta\delta$, as is shown in the Elements [Euclid, \emph{Elem.}
XII, 10], the cylinder whose axes form the parallelogram $\phi\omega$
is six times as large as the cone whose axes form the triangle
$\alpha\beta\delta$. But it was shown that the sphere whose largest
circle is $\alpha\beta\gamma\delta$ is four times as large as the same
cone, consequently the cylinder is one and one half times as large as
the sphere, Q. E. D.

After I had thus perceived that a sphere is four times as large as the
cone whose base is the largest circle of the sphere and whose altitude
is equal to its radius, it occurred to me that the surface of a sphere
is four times as great as its largest circle, in which I proceeded
from the idea that just as a circle is equal to a triangle whose base
is the periphery of the circle and whose altitude is equal to its
radius, so a sphere is equal to a cone whose base is the same as the
surface of the sphere and whose altitude is equal to the radius of the
sphere.

\section*{Proposition III}

By this method it may also be seen that a cylinder whose base is equal
to the largest circle of a spheroid and whose altitude is equal to the
axis of the spheroid, is one and one half times as large as the
spheroid, and when this is recognized it becomes clear that if a
spheroid is cut through its center by a plane perpendicular to its
axis, one-half of the spheroid is twice as great as the cone whose
base is that of the segment and its axis the same.

For let a spheroid be cut by a plane through its axis and let there be
in its surface an ellipse $\alpha\beta\gamma\delta$ [Fig.~3] whose
diameters are $\alpha\gamma$ and $\beta\delta$ and whose center is
$\kappa$ and let there be a circle in the spheroid on the diameter
$\beta\delta$ perpendicular to $\alpha\gamma$; then imagine a cone
whose base is the same circle but whose vertex is at $\alpha$, and
producing its surface, let the cone be cut by a plane through $\gamma$
parallel to the base; the intersection will be a circle perpendicular
to $\alpha\gamma$ with $\epsilon\zeta$ as its diameter. Now imagine a
cylinder whose base is the same circle with the diameter
$\epsilon\zeta$ and whose axis is $\alpha\gamma$; let $\gamma\alpha$
be produced so that $\alpha\theta = \gamma\alpha$; think of
$\theta\gamma$ as a scale-beam with its center at $\alpha$ and in the
parallelogram $\lambda\theta$ draw a straight line $\mu\nu \| \epsilon\zeta$, and on $\mu\nu$ construct a plane perpendicular to
$\alpha\gamma$; this will intersect the cylinder in a circle whose
diameter is $\mu\nu$, the spheroid in a circle whose diameter is $\xi o$ and the cone in a circle whose diameter is $\pi\rho$. Because
$\gamma\alpha : \alpha\sigma = \epsilon\alpha : \alpha\pi = \mu\sigma : \sigma\pi$, and $\gamma\alpha = \alpha\theta$, therefore
$\theta\alpha : \alpha\sigma = \mu\sigma : \sigma\pi$. But $\mu\sigma : \sigma\pi = \mu\sigma^2 : \mu\sigma \times \sigma\pi$ and $\mu\sigma \times \sigma\pi = \pi\sigma^2 + \sigma\xi^2$, for $\alpha\sigma \times \sigma\gamma : \sigma\xi^2 = \alpha\kappa \times \kappa\gamma : \kappa\beta^2 = \alpha\kappa^2 : \kappa\beta^2$ (for both ratios are
equal to the ratio between the diameter and the parameter [Apollonius,
\emph{Con.} I, 21]) $= \alpha\sigma^2 : \sigma\pi^2$ therefore
$\alpha\sigma^2 : \alpha\sigma \times \sigma\gamma = \pi\sigma^2 : \sigma\xi^2 = \sigma\pi^2 : \sigma\pi \times \pi\mu$, consequently
$\mu\pi \times \pi\sigma = \sigma\xi^2$. If $\pi\sigma^2$ is added to
both sides then $\mu\sigma \times \sigma\pi = \pi\sigma^2 + \sigma\xi^2$. Therefore $\theta\alpha : \alpha\sigma = \mu\sigma^2 : \pi\sigma^2 + \sigma\xi^2$. But $\mu\sigma^2 : \sigma\xi^2 + \sigma\pi^2 =$ the circle in the cylinder whose diameter is $\mu\nu :$
the circle with the diameter $\xi o$ + the circle with the diameter
$\pi\rho$; hence the circle whose diameter is $\mu\nu$ will in its
present position be in equilibrium at the point $\alpha$ with the two
circles whose diameters are $\xi o$ and $\pi\rho$ when they are
transferred and so arranged on the scale-beam at the point $\alpha$
that $\theta$ is the center of gravity of both; and $\theta$ is the
center of gravity of the two circles combined whose diameters are $\xi o$ and $\pi\rho$ when their position is changed,
%
\begin{wrapfigure}{r}{0.5\textwidth}
\includegraphics[width=0.5\textwidth]{fig03.png}
\begin{center}
{\small Fig.~3.}
\end{center}
\end{wrapfigure}
%
hence $\theta\alpha : \alpha\sigma =$ the circle with the diameter
$\mu\nu :$ the two circles whose diameters are $\xi o$ and
$\pi\rho$. In the same way it can be shown that if another straight
line is drawn in the parallelogram $\lambda\zeta \| \epsilon\zeta$ and
on this line last drawn a plane is constructed perpendicular to
$\alpha\gamma$, then likewise the circle produced in the cylinder will
in its present position be in equilibrium at the point $\alpha$ with
the two circles combined which have been produced in the spheroid and
in the cone respectively when they are so transferred to the point
$\theta$ on the scale-beam that $\theta$ is the center of gravity of
both. Then if cylinder, spheroid and cone are filled with such
circles, the cylinder in its present position will be in equilibrium
at the point $\alpha$ with the spheroid $+$ the cone if they are
transferred and so arranged on the scale-beam at the point $\alpha$
that $\theta$ is the center of gravity of both. Now $\kappa$ is the
center of gravity of the cylinder, but $\theta$, as has been said, is
the center of gravity of the spheroid and cone together. Therefore
$\theta\alpha : \alpha\kappa =$ cylinder $:$ spheroid $+$ cone. But
$\alpha\theta = 2\alpha\kappa$, hence also the cylinder = 2 $\times$
(spheroid $+$ cone) = 2 $\times$ spheroid + 2 $\times$ cone. But the
cylinder = 3 $\times$ cone, hence 3 $\times$ cone = 2 $\times$ cone +
2 $\times$ spheroid. Subtract 2 $\times$ cone from both sides; then a
cone whose axes form the triangle $\alpha\epsilon\zeta$ = 2 $\times$
spheroid. But the same cone = 8 cones whose axes form the
$\Delta\alpha\beta\delta$; hence 8 such cones = 2 $\times$ spheroid, 4
$\times$ cone = spheroid; whence it follows that a spheroid is four
times as great as a cone whose vertex is at $\alpha$, and whose base
is the circle on the diameter $\beta\delta$ perpendicular to
$\lambda\epsilon$, and one-half the spheroid is twice as great as the
same cone.

In the parallelogram $\lambda\zeta$ draw the straight lines $\phi\chi$
and $\psi\omega \| \alpha\gamma$ through the points $\beta$ and
$\delta$ and imagine a cylinder whose bases are the circles on the
diameters $\phi\psi$ and $\chi\omega$, and whose axis is
$\alpha\gamma$. Now since the cylinder whose axes form the
parallelogram $\phi\omega$ is twice as great as the cylinder whose
axes form the parallelogram $\phi\delta$ because their bases are equal
but the axis of the first is twice as great as the axis of the second,
and since the cylinder whose axes form the parallelogram $\phi\delta$
is three times as great as the cone whose vertex is at $\alpha$ and
whose base is the circle on the diameter $\beta\delta$ perpendicular
to $\alpha\gamma$, then the cylinder whose axes form the parallelogram
$\phi\omega$ is six times as great as the aforesaid cone. But it has
been shown that the spheroid is four times as great as the same cone,
hence the cylinder is one and one half times as great as the
spheroid. Q. E. D.

\section*{Proposition IV}

That a segment of a right conoid cut by a plane perpendicular to its
axis is one and one half times as great as the cone having the same
base and axis as the segment, can be proved by the same method in the
following way:

Let a right conoid be cut through its axis by a plane intersecting the
surface in a parabola $\alpha\beta\gamma$ [Fig.~4]; let it be also cut
by another plane perpendicular to the axis, and let their common line
of intersection be $\beta\gamma$. Let the axis of the segment be
$\delta\alpha$ and let it be produced to $\theta$ so that
$\theta\alpha = \alpha\delta$. Now imagine $\delta\theta$ to be a
scale-beam with its center at $\alpha$; let the base of the segment be
the circle on the diameter $\beta\gamma$ perpendicular to
$\alpha\delta$; imagine a cone whose base is the circle on the
diameter $\beta\gamma$, and whose vertex is at $\alpha$. Imagine also
a cylinder whose base is the circle on the diameter $\beta\gamma$ and
its axis $\alpha\delta$, and in the parallelogram let a straight line
$\mu\nu$ be drawn $\| \beta\gamma$ and on $\mu\nu$ construct a plane
perpendicular to $\alpha\delta$; it will intersect the cylinder in a
circle whose diameter is $\mu\nu$, and the segment of the right conoid
in a circle whose diameter is $\xi o$. Now since $\beta\alpha\gamma$
is a parabola, $\alpha\delta$ its diameter and $\xi\sigma$ and
$\beta\delta$ its ordinates, then [\emph{Quadr. parab.} 3]
$\delta\alpha : \alpha\sigma = \beta\delta^2 : \xi\sigma^2$.  But
$\delta\alpha = \alpha\theta$, therefore $\theta\alpha : \alpha\sigma = \mu\sigma^2 : \sigma\xi^2$. But $\mu\sigma^2 : \sigma\xi^2$ = the
circle in the cylinder whose diameter is $\mu\nu$ : the circle in the
segment of the right conoid whose diameter is $\xi o$, hence
$\theta\alpha : \alpha\sigma$ = the circle with the diameter $\mu\nu$
: the circle with the diameter $\xi o$; therefore the circle in the
cylinder whose
%
\begin{wrapfigure}{r}{0.5\textwidth}
\includegraphics[width=0.5\textwidth]{fig04.png}
\begin{center}
{\small Fig.~4.}
\end{center}
\end{wrapfigure}
%
diameter is $\mu\nu$ is in its present position, in equilibrium at the
point $\alpha$ with the circle whose diameter is $\xi o$ if this be
transferred and so arranged on the scale-beam at $\theta$ that
$\theta$ is its center of gravity. And the center of gravity of the
circle whose diameter is $\mu\nu$ is at $\sigma$, that of the circle
whose diameter is $\xi o$ when its position is changed, is $\theta$,
and we have the inverse proportion, $\theta\alpha : \alpha\sigma$ =
the circle with the diameter $\mu\nu$ : the circle with the diameter
$\xi o$. In the same way it can be shown that if another straight line
be drawn in the parallelogram $\epsilon\gamma \| \beta\gamma$ the
circle formed in the cylinder, will in its present position be in
equilibrium at the point $\alpha$ with that formed in the segment of
the right conoid if the latter is so transferred to $\theta$ on the
scale-beam that $\theta$ is its center of gravity. Therefore if the
cylinder and the segment of the right conoid are filled up then the
cylinder in its present position will be in equilibrium at the point
$\alpha$ with the segment of the right conoid if the latter is
transferred and so arranged on the scale-beam at $\theta$ that
$\theta$ is its center of gravity. And since these magnitudes are in
equilibrium at $\alpha$, and $\kappa$ is the center of gravity of the
cylinder, if $\alpha\delta$ is bisected at $\kappa$ and $\theta$ is
the center of gravity of the segment transferred to that point, then
we have the inverse proportion $\theta\alpha : \alpha\kappa$ =
cylinder : segment. But $\theta\alpha = 2\alpha\kappa$ and also the
cylinder = 2 $\times$ segment. But the same cylinder is 3 times as
great as the cone whose base is the circle on the diameter
$\beta\gamma$ and whose vertex is at $\alpha$; therefore it is clear
that the segment is one and one half times as great as the same cone.

\section*{Proposition V}

That the center of gravity of a segment of a right conoid which is cut
off by a plane perpendicular to the axis, lies on the straight line
which is the axis of the segment divided in such a way that the
portion at the vertex is twice as great as the remainder, may be
perceived by our method in the following way:

Let a segment of a right conoid cut off by a plane perpendicular to
the axis be cut by another plane through the axis, and let the
intersection in its surface be the parabola $\alpha\beta\gamma$
[Fig.~5] and let the common line of intersection of the plane which
cut off the segment and of the intersecting plane be $\beta\gamma$;
let the axis of the segment and the diameter of the parabola
$\alpha\beta\gamma$ be $\alpha\delta$; produce $\delta\alpha$ so that
$\alpha\theta = \alpha\delta$ and imagine $\delta\theta$ to be a
scale-beam with its center at $\alpha$;
%
\begin{wrapfigure}{r}{0.5\textwidth}
\includegraphics[width=0.5\textwidth]{fig05.png}
\begin{center}
{\small Fig.~5.}
\end{center}
\end{wrapfigure}
%
then inscribe a cone in the segment with the lateral boundaries
$\beta\alpha$ and $\alpha\gamma$ and in the parabola draw a straight
line $\xi o \| \beta\gamma$ and let it cut the parabola in $\xi$ and
$o$ and the lateral boundaries of the cone in $\pi$ and $\rho$.  Now
because $\xi\sigma$ and $\beta\delta$ are drawn perpendicular to the
diameter of the parabola, $\delta\alpha : \alpha\sigma = \beta\delta^2 : \xi\sigma^2$ [\emph{Quadr. parab.} 3]. But $\delta\alpha : \alpha\sigma = \beta\delta : \pi\sigma = \beta\delta^2 : \beta\delta \times \pi\sigma$, therefore also $\beta\delta^2 : \xi\sigma^2 = \beta\delta^2 : \beta\delta \times \pi\sigma$.  Consequently
$\xi\sigma^2 = \beta\delta \times \pi\sigma$ and $\beta\delta : \xi\sigma = \xi\sigma : \pi\sigma$, therefore $\beta\delta : \pi\sigma = \xi\sigma^2 : \sigma\pi^2$.  But $\beta\delta : \pi\sigma = \delta\alpha : \alpha\sigma = \theta\alpha : \alpha\sigma$, therefore
also $\theta\alpha : \alpha\sigma = \xi\sigma^2 : \sigma\pi^2$.  On
$\xi o$ construct a plane perpendicular to $\alpha\delta$; this will
intersect the segment of the right conoid in a circle whose diameter
is $\xi o$ and the cone in a circle whose diameter is $\pi\rho$. Now
because $\theta\alpha : \alpha\sigma = \xi\sigma^2 : \sigma\pi^2$ and
$\xi\sigma^2 : \sigma\pi^2$ = the circle with the diameter $\xi o$ :
the circle with the diameter $\pi\rho$, therefore $\theta\alpha : \alpha\sigma$ = the circle whose diameter is $\xi o$ : the circle
whose diameter is $\pi\rho$. Therefore the circle whose diameter is
$\xi o$ will in its present position be in equilibrium at the point
$\alpha$ with the circle whose diameter is $\pi\rho$ when this is so
transferred to $\theta$ on the scale-beam that $\theta$ is its center
of gravity.  Now since $\sigma$ is the center of gravity of the circle
whose diameter is $\xi o$ in its present position, and $\theta$ is the
center of gravity of the circle whose diameter is $\pi\rho$ if its
position is changed as we have said, and inversely $\theta\alpha : \alpha\sigma$ = the circle with the diameter $\xi o$ : the circle with
the diameter $\pi\rho$, then the circles are in equilibrium at the
point $\alpha$.  In the same way it can be shown that if another
straight line is drawn in the parabola $\| \beta\gamma$ and on this
line last drawn a plane is constructed perpendicular to
$\alpha\delta$, the circle formed in the segment of the right conoid
will in its present position be in equilibrium at the point $\alpha$
with the circle formed in the cone, if the latter is transferred and
so arranged on the scale-beam at $\theta$ that $\theta$ is its center
of gravity.  Therefore if the segment and the cone are filled up with
circles, all circles in the segment will be in their present positions
in equilibrium at the point $\alpha$ with all circles of the cone if
the latter are transferred and so arranged on the scale-beam at the
point $\theta$ that $\theta$ is their center of gravity. Therefore
also the segment of the right conoid in its present position will be
in equilibrium at the point $\alpha$ with the cone if it is
transferred and so arranged on the scale-beam at $\theta$ that
$\theta$ is its center of gravity. Now because the center of gravity
of both magnitudes taken together is $\alpha$, but that of the cone
alone when its position is changed is $\theta$, then the center of
gravity of the remaining magnitude lies on $\alpha\theta$ extended
towards $\alpha$ if $\alpha\kappa$ is cut off in such a way that
$\alpha\theta : \alpha\kappa$ = segment : cone. But the segment is one
and one half the size of the cone, consequently $\alpha\theta = \frac{3}{2}\alpha\kappa$ and $\kappa$, the center of gravity of the
right conoid, so divides $\alpha\delta$ that the portion at the vertex
of the segment is twice as large as the remainder.

\section*{Proposition VI}

[The center of gravity of a hemisphere is so divided on its axis] that
the portion near the surface of the hemisphere is in the ratio of $5 : 3$ to the remaining portion.

Let a sphere be cut by a plane through its center intersecting the
surface in the circle $\alpha\beta\gamma\delta$ [Fig.6],
$\alpha\gamma$ and $\beta\delta$ being two diameters of the circle
perpendicular to each other. Let a plane be constructed on
$\beta\delta$ perpendicular to $\alpha\gamma$.  Then imagine a cone
whose base is the circle with the diameter $\beta\delta$, whose vertex
is at $\alpha$ and its lateral boundaries are $\beta\alpha$ and
$\alpha\delta$; let $\gamma\alpha$ be produced so that $\alpha\theta = \gamma\alpha$, imagine the straight line $\theta\gamma$ to be a
scale-beam with its center at $\alpha$ and in the semi-circle
$\beta\alpha\delta$ draw a straight line $\xi o \| \beta\delta$; let
it cut the circumference of the semicircle in $\xi$ and $o$, the
lateral boundaries of the cone in $\pi$ and $\rho$, and $\alpha\gamma$
in $\epsilon$.  On $\xi o$ construct a plane perpendicular to
$\alpha\epsilon$; it will intersect the hemisphere in a circle with
the diameter $\xi o$, and the cone in a circle with the diameter
$\pi\rho$.  Now because $\alpha\gamma : \alpha\epsilon = \xi\alpha^2 : \alpha\epsilon^2$ and $\xi\alpha^2 = \alpha\epsilon^2 + \epsilon\xi^2$
and $\alpha\epsilon = \epsilon\pi$, therefore $\alpha\gamma : \alpha\epsilon = \xi\epsilon^2 + \epsilon\pi^2 : \epsilon\pi^2$.  But
$\xi\epsilon^2 + \epsilon\pi^2 : \epsilon\pi^2$ = the circle with the
diameter $\xi o$ + the circle with the diameter $\pi\rho$ : the circle
with the diameter $\pi\rho$, and $\gamma\alpha = \alpha\theta$, hence
$\theta\alpha : \alpha\epsilon$ = the circle with the diameter $\xi o$
+ the circle with
%
\begin{wrapfigure}{r}{0.33\textwidth}
\includegraphics[width=0.33\textwidth]{fig06.png}
\begin{center}
{\small Fig.~6.}
\end{center}
\end{wrapfigure}
%
the diameter $\pi\rho$ : circle with the diameter $\pi\rho$. Therefore
the two circles whose diameters are $\xi o$ and $\pi\rho$ in their
present position are in equilibrium at the point $\alpha$ with the
circle whose diameter is $\pi\rho$ if it is transferred and so
arranged at $\theta$ that $\theta$ is its center of gravity.  Now
since the center of gravity of the two circles whose diameters are
$\xi o$ and $\pi\rho$ in their present position [is the point
$\epsilon$, but of the circle whose diameter is $\pi\rho$ when its
position is changed is the point $\theta$, then $\theta\alpha : \alpha\epsilon$ = the circles whose diameters are] $\xi o$ [,$\pi\rho$
: the circle whose diameter is $\pi\rho$.  In the same way if another
straight line in the] hemisphere $\beta\alpha\delta$ [is drawn $\| \beta\delta$ and a plane is constructed] perpendicular to
[$\alpha\gamma$ the] two [circles produced in the cone and in the
hemisphere are in their position] in equilibrium at $\alpha$ [with the
circle which is produced in the cone] if it is transferred and
arranged on the scale at $\theta$.  [Now if] the hemisphere and the
cone [are filled up with circles then all circles in the] hemisphere
and those [in the cone] will in their present position be in
equilibrium [with all circles] in the cone, if these are transferred
and so arranged on the scale-beam at $\theta$ that $\theta$ is their
center of gravity; [therefore the hemisphere and cone also] are in
their position [in equilibrium at the point $\alpha$] with the cone if
it is transferred and so arranged [on the scale-beam at $\theta$] that
$\theta$ is its center of gravity.

\section*{Proposition VII}

By [this method] it may also be perceived that [any segment whatever]
of a sphere bears the same ratio to a cone having the same [base] and
axis [that the radius of the sphere + the axis of the opposite segment
: the axis of the opposite segment] \dotfill and [Fig.~7] on $\mu\nu$
construct a plane perpendicular to $\alpha\gamma$; it will intersect
the cylinder in a circle whose diameter is $\mu\nu$, the segment of
the sphere in a circle whose diameter
%
\begin{wrapfigure}{r}{0.6\textwidth}
\includegraphics[width=0.6\textwidth]{fig07.png}
\begin{center}
{\small Fig.~7.}
\end{center}
\end{wrapfigure}
%
is $\xi o$ and the cone whose base is the circle on the diameter
$\epsilon\zeta$ and whose vertex is at $\alpha$ in a circle whose
diameter is $\pi\rho$. In the same way as before it may be shown that
a circle whose diameter is $\mu\nu$ is in its present position in
equilibrium at $\alpha$ with the two circles [whose diameters are $\xi o$ and $\pi\rho$ if they are so arranged on the scale-beam that
$\theta$ is their center of gravity. [And the same can be proved of
all corresponding circles.] Now since cylinder, cone, and spherical
segment are filled up with such circles, the cylinder in its present
position [will be in equilibrium at $\alpha$] with the cone + the
spherical segment if they are transferred and attached to the
scale-beam at $\theta$. Divide $\alpha\eta$ at $\phi$ and $\chi$ so
that $\alpha\chi = \chi\eta$ and $\eta\phi = \frac{1}{3}\alpha\phi$;
then $\chi$ will be the center of gravity of the cylinder because it
is the center of the axis $\alpha\eta$. Now because the above
mentioned bodies are in equilibrium at $\alpha$, cylinder : cone with
the diameter of its base $\epsilon\zeta$ + the spherical segment
$\beta\alpha\delta = \theta\alpha : \alpha\chi$. And because
$\eta\alpha = 3\eta\phi$ then [$\gamma\eta \times \eta\phi$] =
$\frac{1}{3}\alpha\eta \times \eta\gamma$. Therefore also $\gamma\eta \times \eta\phi = \linebreak\frac{1}{3}\beta\eta^2$. \dotfill

\section*{Proposition VIIa}

In the same way it may be perceived that any segment of an ellipsoid
cut off by a perpendicular plane, bears the same ratio to a cone
having the same base and the same axis, as half of the axis of the
ellipsoid + the axis of the opposite segment bears to the axis of the
opposite segment. \dotfill

\section*{Proposition VIII}

\dotfill\linebreak produce $\alpha\gamma$ [Fig.~8] making
$\alpha\theta = \alpha\gamma$ and $\gamma\xi$ = the radius of the
sphere; imagine $\gamma\theta$ to be a scale-beam with a center at
$\alpha$, and in the plane cutting
%
\begin{wrapfigure}{r}{0.4\textwidth}
\includegraphics[width=0.4\textwidth]{fig08.png}
\begin{center}
{\small Fig.~8.}
\end{center}
\end{wrapfigure}
%
off the segment inscribe a circle with its center at $\eta$ and its
radius = $\alpha\eta$; on this circle construct a cone with its vertex
at $\alpha$ and its lateral boundaries $\alpha\epsilon$ and
$\alpha\zeta$.  Then draw a straight line $\kappa\lambda \| \epsilon\zeta$; let it cut the circumference of the segment at
$\kappa$ and $\lambda$, the lateral boundaries of the cone
$\alpha\epsilon\zeta$ at $\rho$ and $o$ and $\alpha\gamma$ at $\pi$.
Now because $\alpha\gamma : \alpha\pi = \alpha\kappa^2 : \alpha\pi^2$
and $\kappa\alpha^2 = \alpha\pi^2 + \pi\kappa^2$ and $\alpha\pi^2 = \pi o^2$ (since also $\alpha\eta^2 = \epsilon\eta^2$), then
$\gamma\alpha : \alpha\pi = \kappa\pi^2 + \pi o^2 : o\pi^2$. But
$\kappa\pi^2 + \pi o^2 : \pi o^2$ = the circle with the diameter
$\kappa\lambda$ + the circle with the diameter $o\rho$ : the circle
with the diameter $o\rho$ and $\gamma\alpha = \alpha\theta$; therefore
$\theta\alpha : \alpha\pi$ = the circle with the diameter
$\kappa\lambda$ + the circle with the diameter $o\rho$ : the circle
with the diameter $o\rho$. Now since the circle with the diameter
$\kappa\lambda$ + the circle with the diameter $o\rho$ : the circle
with the diameter $o\rho$ = $\alpha\theta : \pi\alpha$, let the circle
with the diameter $o\rho$ be transferred and so arranged on the
scale-beam at $\theta$ that $\theta$ is its center of gravity; then
$\theta\alpha : \alpha\pi$ = the circle with the diameter
$\kappa\lambda$ + the circle with the diameter $o\rho$ in their
present positions : the circle with the diameter $o\rho$ if it is
transferred and so arranged on the scale-beam at $\theta$ that
$\theta$ is its center of gravity. Therefore the circles in the
segment $\beta\alpha\delta$ and in the cone $\alpha\epsilon\zeta$ are
in equilibrium at $\alpha$ with that in the cone
$\alpha\epsilon\zeta$. And in the same way all circles in the segment
$\beta\alpha\delta$ and in the cone $\alpha\epsilon\zeta$ in their
present positions are in equilibrium at the point $\alpha$ with all
circles in the cone $\alpha\epsilon\zeta$ if they are transferred and
so arranged on the scate-beam at $\theta$ that $\theta$ is their
center of gravity; then also the spherical segment $\alpha\beta\delta$
and the cone $\alpha\epsilon\zeta$ in their present positions are in
equilibrium at the point $\alpha$ with the cone $\epsilon\alpha\zeta$
if it is transferred and so arranged on the scale-beam at $\theta$
that $\theta$ is its center of gravity. Let the cyIinder $\mu\nu$
equal the cone whose base is the circle with the diameter
$\epsilon\zeta$ and whose vertex is at $\alpha$ and let $\alpha\eta$
be so divided at $\phi$ that $\alpha\eta = 4\phi\eta$; then $\phi$ is
the center of gravity of the cone $\epsilon\alpha\zeta$ as has been
previously proved.  Moreover let the cylinder $\mu\nu$ be so cut by a
perpendicularly intersecting plane that the cylinder $\mu$ is in
equilibrium with the cone $\epsilon\alpha\zeta$. Now since the segment
$\alpha\beta\delta$ + the cone $\epsilon\alpha\zeta$ in their present
positions are in equilibrium at $\alpha$ with the cone
$\epsilon\alpha\zeta$ if it is transferred and so arranged on the
scale-beam at $\theta$ that $\theta$ is its center of gravity, and
cylinder $\mu\nu$ = cone $\epsilon\alpha\zeta$ and the two cylinders
$\mu + \nu$ are moved to $\theta$ and $\mu\nu$ is in equilibrium with
both bodies, then will also the cylinder $\nu$ be in equilibrium with
the segment of the sphere at the point $\alpha$.  And since the
spherical segment $\beta\alpha\delta$ : the cone whose base is the
circle with the diameter $\beta\delta$, and whose vertex is at $\alpha = \xi\eta : \eta\gamma$ (for this has previously been proved [\emph{De
sph. et cyl.} II, 2 Coroll.])  and cone $\beta\alpha\delta$ : cone
$\epsilon\alpha\zeta$ = the circle with the diameter $\beta\delta$ :
the circle with the diameter $\epsilon\zeta = \beta\eta^2 : \eta\epsilon^2$, and $\beta\eta^2 = \gamma\eta \times \eta\alpha$,
$\eta\epsilon^2 = \eta\alpha^2$, and $\gamma\eta \times \eta\alpha : \eta\alpha^2 = \gamma\eta : \eta\alpha$, therefore cone
$\beta\alpha\delta$ : cone $\epsilon\alpha\zeta = \gamma\eta : \eta\alpha$.  But we have shown that cone $\beta\alpha\delta$ :
segment $\beta\alpha\delta$ = $\gamma\eta : \eta\xi$, hence
{\selectlanguage{greek}di' \~isou} segment $\beta\alpha\delta$ : cone
$\epsilon\alpha\zeta$ = $\xi\eta : \eta\alpha$.  And because
$\alpha\chi : \chi\eta = \eta\alpha + 4\eta\gamma : \alpha\eta + 2\eta\gamma$ so inversely $\eta\chi : \chi\alpha = 2\gamma\eta + \eta\alpha : 4\gamma\eta + \eta\alpha$ and by addition $\eta\alpha : \alpha\chi = 6\gamma\eta + 2\eta\alpha : \eta\alpha + 4\eta\gamma$. But $\eta\xi = \frac{1}{4} (6\eta\gamma + 2\eta\alpha)$
and $\gamma\phi = \frac{1}{4} (4\eta\gamma + \eta\alpha)$; for that is
evident.  Hence $\eta\alpha : \alpha\chi = \xi\eta : \gamma\phi$,
consequently also $\xi\eta : \eta\alpha = \gamma\phi : \chi\alpha$.
But it was also demonstrated that $\xi\eta : \eta\alpha$ = the segment
whose vertex is at $\alpha$ and whose base is the circle with the
diameter $\beta\delta$ : the cone whose vertex is at $\alpha$ and
whose base is the circle with the diameter $\epsilon\zeta$; hence
segment $\beta\alpha\delta$ : cone $\epsilon\alpha\zeta$ = $\gamma\phi : \chi\alpha$. And since the cylinder $\mu$ is in equilibrium with the
cone $\epsilon\alpha\zeta$ at $\alpha$, and $\theta$ is the center of
gravity of the cylinder while $\phi$ is that of the cone
$\epsilon\alpha\zeta$, then cone $\epsilon\alpha\zeta$ : cylinder
$\mu$ = $\theta\alpha : \alpha\phi = \gamma\alpha : \alpha\phi$. But
cylinder $\mu\nu$ = cone $\epsilon\alpha\zeta$; hence by subtraction,
cylinder $\mu$ : cylinder $\nu$ = $\alpha\phi : \gamma\phi$. And
cylinder $\mu\nu$ = cone $\epsilon\alpha\zeta$; hence cone
$\epsilon\alpha\zeta$ : cylinder $\nu$ = $\gamma\alpha : \gamma\phi = \theta\alpha : \gamma\phi$.  But it was also demonstrated that segment
$\beta\alpha\delta$ : cone $\epsilon\alpha\zeta$ = $\gamma\phi : \chi\alpha$; hence {\selectlanguage{greek}di' \~isou} segment
$\beta\alpha\delta$ : cylinder $\nu$ = $\zeta\alpha : \alpha\chi$.
And it was demonstrated that segment $\beta\alpha\delta$ is in
equilibrium at $\alpha$ with the cylinder $\nu$ and $\theta$ is the
center of gravity of the cylinder $\nu$, consequently the point $\chi$
is also the center of gravity of the segment $\beta\alpha\delta$.

\section*{Proposition IX}

In a similar way it can also be perceived that the center of gravity
of any segment of an ellipsoid lies on the straight line which is the
axis of the segment so divided that the portion at the vertex of the
segment bears the same ratio to the remaining portion as the axis of
the segment + 4 times the axis of the opposite segment bears to the
axis of the segment + twice the axis of the opposite segment.

\section*{Proposition X}

It can also be seen by this method that [a segment of a hyperboloid]
bears the same ratio to a cone having the same base and axis as the
segment, that the axis of the segment + 3 times the addition to the
axis bears to the axis of the segment of the hyperboloid + twice its
addition [\emph{De Conoid.} 25]; and that the center of gravity of the
hyperboloid so divides the axis that the part at the vertex bears the
same ratio to the rest that three times the axis + eight times the
addition to the axis bears to the axis of the hyperboloid + 4 times
the addition to the axis, and many other points which I will leave
aside since the method has been made clear by the examples already
given and only the demonstrations of the above given theorems remain
to be stated.

\section*{Proposition XI}

When in a perpendicular prism with square bases a cylinder is
inscribed whose bases lie in opposite squares and whose curved surface
touches the four other parallelograms, and when a plane is passed
through the center of the circle which is the base of the cylinder and
one side of the opposite square, then the body which is cut off by
this plane [from the cylinder] will be $\frac{1}{6}$ of the entire
prism. This can be perceived through the present method and when it is
so warranted we will pass over to the geometrical proof of it.

\begin{figure}[t]
\begin{minipage}[5]{0.45\textwidth}
\includegraphics[width=\textwidth]{fig09.png}
\begin{center}
{\small Fig.~9.}
\end{center}
\end{minipage}
\hfill
\begin{minipage}[5]{0.45\textwidth}
\includegraphics[width=\textwidth]{fig10.png}
\begin{center}
{\small Fig.~10.}
\end{center}
\end{minipage}
\end{figure}

Imagine a perpendicular prism with square bases and a cylinder
inscribed in the prism in the way we have described. Let the prism be
cut through the axis by a plane perpendicular to the plane which cuts
off the section of the cylinder; this will intersect the prism
containing the cylinder in the parallelogram $\alpha\beta$ [Fig.~9]
and the common intersecting line of the plane which cuts off the
section of the cylinder and the plane lying through the axis
perpendicular to the one cutting off the section of the cylinder will
be $\beta\gamma$; let the axis of the cylinder and the prism be
$\gamma\delta$ which is bisected at right angles by $\epsilon\zeta$
and on $\epsilon\zeta$ let a plane be constructed perpendicular to
$\gamma\delta$. This will intersect the prism in a square and the
cylinder in a circle.

Now let the intersection of the prism be the square $\mu\nu$
[Fig.~10], that of the cylinder, the circle $\xi o\pi\rho$ and let the
circle touch the sides of the square at the points $\xi$, $o$, $\pi$
and $\rho$; let the common line of intersection of the plane cutting
off the cylinder-section and that passing through $\epsilon\zeta$
perpendicular to the axis of the cylinder, be $\kappa\lambda$; this
line is bisected by $\pi\theta\xi$.  In the semicircle $o\pi\rho$ draw
a straight line $\sigma\tau$ perpendicular to $\pi\chi$, on
$\sigma\tau$ construct a plane perpendicular to $\xi\pi$ and produce
it to both sides of the plane enclosing the circle $\xi o\pi\rho$;
this will intersect the half-cylinder whose base is the semicircle
$o\pi\rho$ and whose altitude is the axis of the prism, in a
parallelogram one side of which = $\sigma\tau$ and the other = the
vertical boundary of the cylinder, and it will intersect the
cylinder-section likewise in a parallelogram of which one side is
$\sigma\tau$ and the other $\mu\nu$ [Fig.~9]; and accordingly $\mu\nu$
will be drawn in the parallelogram $\delta\epsilon \| \beta\omega$ and
will cut off $\epsilon\iota = \pi\chi$.  Now because $\epsilon\gamma$
is a parallelogram and $\nu\iota \| \theta\gamma$, and
$\epsilon\theta$ and $\beta\gamma$ cut the parallels, therefore
$\epsilon\theta : \theta\iota = \omega\gamma : \gamma\nu = \beta\omega : \upsilon\nu$. But $\beta\omega : \upsilon\nu$ = parallelogram in the
half-cylinder : parallelogram in the cylinder-section, therefore both
parallelograms have the same side $\sigma\tau$; and $\epsilon\theta = \theta\pi$, $\iota\theta = \chi\theta$; and since $\pi\theta = \theta\xi$ therefore $\theta\xi : \theta\chi$ = parallelogram in
half-cylinder : parallelogram in the cylinder-section. lmagine the
parallelogram in the cylinder-section transferred and so brought to
$\xi$ that $\xi$ is its center of gravity, and further imagine
$\pi\xi$ to be a scale-beam with its center at $\theta$; then the
parallelogram in the half-cylinder in its present position is in
equilibrium at the point $\theta$ with the parallelogram in the
cylinder-section when it is transferred and so arranged on the
scale-beam at $\xi$ that $\xi$ is its center of gravity. And since
$\chi$ is the center of gravity in the parallelogram in the
half-cylinder, and $\xi$ that of the parallelogram in the
cylinder-section when its position is changed, and $\xi\theta : \theta\chi$ = the parallelogram whose center of gravity is $\chi$ :
the parallelogram whose center of gravity is $\xi$, then the
parallelogram whose center of gravity is $\chi$ will be in equilibrium
at $\theta$ with the parallelogram whose center of gravity is
$\xi$. In this way it can be proved that if another straight line is
drawn in the semicircle $o\pi\rho$ perpendicular to $\pi\theta$ and on
this straight line a plane is constructed perpendicular to $\pi\theta$
and is produced towards both sides of the plane in which the circle
$\xi o\pi\rho$ lies, then the parallelogram formed in the
half-cylinder in its present position will be in equilibrium at the
point $\theta$ with the parallelogram formed in the cylinder-section
if this is transferred and so arranged on the scale-beam at $\xi$ that
$\xi$ is its center of-gravity; therefore also all parallelograms in
the half-cylinder in their present positions will be in equilibrium at
the point $\theta$ with all parallelograms of the cylinder-section if
they are transferred and attached to the scale-beam at the point
$\xi$; consequently also the half-cylinder in its present position
will be in equilibrium at the point $\theta$ with the cylinder-section
if it is transferred and so arranged on the scale-beam at $\xi$ that
$\xi$ is its center of gravity.

\section*{Proposition XII}

Let the parallelogram $\mu\nu$ be perpendicular to the axis [of the
circle] $\xi o$ [$\pi\rho$] [Fig.~11]. Draw $\theta\mu$ and
$\theta\eta$ and erect upon them two planes perpendicular to the plane
in which the semicircle $o\pi\rho$ lies and extend these planes on
both sides. The result is a prism whose base is a triangle similar to
$\theta\mu\eta$ and whose altitude is equal to the axis of the
cylinder, and this prism is $\frac{1}{4}$ of the entire prism which
contains the cylinder. In the semicircle $o\pi\rho$ and in the square
$\mu\nu$ draw two straight lines $\kappa\lambda$ and $\tau\upsilon$ at
equal distances from $\pi\xi$; these will cut
%
\begin{wrapfigure}{r}{0.5\textwidth}
\includegraphics[width=0.5\textwidth]{fig11.png}
\begin{center}
{\small Fig.~11.}
\end{center}
\end{wrapfigure}
%
the circumference of the semicircle $o\pi\rho$ at the points $\kappa$
and $\tau$, the diameter $o\rho$ at $\sigma$ and $\zeta$ and the
straight lines $\theta\eta$ and $\theta\mu$ at $\phi$ and $\chi$. Upon
$\kappa\lambda$ and $\tau\upsilon$ construct two planes perpendicular
to $o\rho$ and extend them towards both sides of the plane in which
lies the circle $\xi o\pi\rho$; they will intersect the half-cylinder
whose base is the semicircle $o\pi\rho$ and whose altitude is that of
the cylinder, in a parallelogram one side of which = $\kappa\sigma$
and the other = the axis of the cylinder; and they will intersect the
prism $\theta\eta\mu$ likewise in a parallelogram one side of which is
equal to $\lambda\chi$ and the other equal to the axis, and in the
same way the half-cylinder in a parallelogram one side of which =
$\tau\zeta$ and the other = the axis of the cylinder, and the prism in
a parallelogram one side of which = $\nu\phi$ and the other = the axis
of the cylinder.\dotfill

\section*{Proposition XIII}

Let the square $\alpha\beta\gamma\delta$ [Fig.~12] be the base of a
perpendicular prism with square bases and let a cylinder be inscribed
in the prism whose base is the circle $\epsilon\zeta\eta\theta$ which
touches the sides of the parallelogram $\alpha\beta\gamma\delta$ at
$\epsilon$, $\zeta$, $\eta$, and $\theta$. Pass a plane through its
center and the side in the square opposite the square
$\alpha\beta\gamma\delta$ corresponding to the side $\gamma\delta$;
this will cut off from the whole prism a second prism which is
$\frac{1}{4}$ the size of the whole prism and which will be bounded by
three parallelograms and two opposite triangles. In the semicircle
$\epsilon\zeta\eta$ describe a parabola whose origin is $\eta\epsilon$
and whose axis is $\zeta\kappa$, and in the parallelogram $\delta\eta$
draw $\mu\nu \| \kappa\zeta$; this will cut the circumference of the
semicircle at $\xi$, the parabola at $\lambda$, and $\mu\nu \times \nu\lambda = \nu\zeta^2$ (for this is evident [Apollonios, \emph{Con.}
I, 11]). Therefore $\mu\nu : \nu\lambda = \kappa\eta^2 : \lambda\sigma^2$.  Upon $\mu\nu$ construct a plane parallel to
$\epsilon\eta$; this will intersect the prism cut off from the whole
prism in a right-angled triangle one side of which is $\mu\nu$ and the
other a straight line in the plane upon $\gamma\delta$ perpendicular
to $\gamma\delta$ at $\nu$ and equal to the axis of the cylinder, but
whose hypotenuse is in the intersecting plane. It will intersect the
portion which is cut off from the cylinder by the
%
\begin{wrapfigure}{r}{0.5\textwidth}
\includegraphics[width=0.5\textwidth]{fig12.png}
\begin{center}
{\small Fig.~12.}
\end{center}
\end{wrapfigure}
%
plane passed through $\epsilon\eta$ and the side of the square
opposite the side $\gamma\delta$ in a right-angled triangle one side
of which is $\mu\xi$ and the other a straight line drawn in the
surface of the cylinder perpendicular to the plane $\kappa\nu$,
\linebreak and the hypotenuse\dotfill\linebreak and all the triangles
in the prism : all the triangles in the cylinder-section = all the
straight lines in the parallelogram $\delta\eta$ : all the straight
lines between the parabola and the straight line $\epsilon\eta$. And
the prism consists of the triangles in the prism, the cylinder-section
of those in the cylinder-section, the parallelogram $\delta\eta$ of
the straight lines in the parallelogram $\delta\eta \| \kappa\zeta$
and the segment of the parabola of the straight lines cut off by the
parabola and the straight line $\epsilon\eta$; hence prism :
cylinder-section = parallelogram $\eta\delta$ : segment
$\epsilon\zeta\eta$ that is bounded by the parabola and the straight
line $\epsilon\eta$.  But the parallelogram $\delta\eta = \frac{3}{2}$
the segment bounded by the parabola and the straight line
$\epsilon\eta$ as indeed has been shown in the previously published
work, hence also the prism is equal to one and one half times the
cylinder-section. Therefore when the cylinder-section = 2, the prism =
3 and the whole prism containing the cylinder equals 12, because it is
four times the size of the other prism; hence the cylinder-section is
equal to $\frac{1}{6}$ of the prism, Q. E. D.

\section*{Proposition XIV}

[Inscribe a cylinder in] a perpendicular prism with square bases [and
let it be cut by a plane passed through the center of the base of the
cylinder and one side of the opposite square.] Then this plane will
cut off a prism from the whole prism and a portion of the cylinder
from the cylinder. It may be proved that the portion cut off from the
cylinder by the plane is one-sixth of the whole prism. But first we
will prove that it is possible to inscribe a solid figure in the
cylinder-section and to circumscribe another composed of prisms of
equal altitude and with similar triangles as bases, so that the
circumscribed figure exceeds the inscribed less than any given
magni-\-\linebreak tude.\dotfill

But it has been shown that the prism cut off by the inclined plane
$<\frac{3}{2}$ the body inscribed in the cylinder-section. Now the
prism cut off by the inclined plane : the body inscribed in the
cylinder-section = parallelogram $\delta\eta$ : the parallelograms
which are inscribed in the segment bounded by the parabola and the
straight line $\epsilon\eta$. Hence the parallelogram $\delta\eta <\frac{3}{2}$ the parallelograms in the segment bounded by the
parabola and the straight line $\epsilon\eta$. But this is impossible
because we have shown elsewhere that the parallelogram $\delta\eta$ is
one and one half times the segment bounded by the parabola and the
straight line $\epsilon\eta$, consequently is \dotfill not greater
\dotfill

And all prisms in the prism cut off by the inclined plane : all prisms
in the figure described around the cylinder-section = all
parallelograms in the parallelogram $\delta\eta$ : all parallelograms
in the figure which is described around the segment bounded by the
parabola and the straight line $\epsilon\eta$, i. e., the prism cut
off by the inclined plane : the figure described around the
cylinder-section = parallelogram $\delta\eta$ : the figure bounded by
the parabola and the straight line $\epsilon\eta$. But the prism cut
off by the inclined plane is greater than one and one half times the
solid figure circumscribed around the
cylinder-section\dotfill\linebreak.\dotfill\linebreak

\vfill

\end{document}

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