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You may copy it, give it away or re-use it under the terms of +% the Project Gutenberg License included with this eBook or online at % +% www.gutenberg.org. If you are not located in the United States, you'll have +% to check the laws of the country where you are located before using this ebook. +% % +% % +% % +% Title: Geometrical Solutions Derived from Mechanics % +% A Treatise of Archimedes % +% % +% Author: Archimedes % +% % +% Release Date: November 1, 2014 [EBook #7825] % +% % +% Language: English % +% % +% Character set encoding: ISO-8859-1 % +% % +% *** START OF THIS PROJECT GUTENBERG EBOOK GEOMETRICAL SOLUTIONS *** % +% % +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % + +\def\ebook{7825} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%% %% +%% Packages and substitutions: %% +%% %% +%% book: Required. %% +%% inputenc: Latin-1 text encoding. Required. %% +%% %% +%% ifthen: Logical conditionals. 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You may copy it, give it away or re-use it under the terms of +the Project Gutenberg License included with this eBook or online at +www.gutenberg.org. If you are not located in the United States, you'll have +to check the laws of the country where you are located before using this ebook. + + + +Title: Geometrical Solutions Derived from Mechanics + A Treatise of Archimedes + +Author: Archimedes + +Release Date: November 1, 2014 [EBook #7825] + +Language: English + +Character set encoding: ISO-8859-1 + +*** START OF THIS PROJECT GUTENBERG EBOOK GEOMETRICAL SOLUTIONS *** +\end{PGtext} +\end{minipage} +\end{center} +\newpage +%% Credits and transcriber's note %% +\begin{center} +\begin{minipage}{\textwidth} +\begin{PGtext} +Produced by Gordon Keener +\end{PGtext} +\end{minipage} +\vfill +\end{center} + +\begin{minipage}{0.85\textwidth} +\small +\BookMark{0}{Transcriber's Note.} +\subsection*{\centering\normalfont\scshape% +\normalsize\MakeLowercase{\TransNote}}% + +\raggedright +\TransNoteText +\end{minipage} +%%%%%%%%%%%%%%%%%%%%%%%%%%% FRONT MATTER %%%%%%%%%%%%%%%%%%%%%%%%%% +\PageSep{i} +\FrontMatter +\begin{center} +\LARGE GEOMETRICAL SOLUTIONS \\[12pt] +\normalsize DERIVED FROM \\ +\LARGE MECHANICS +\vfill + +\large A TREATISE OF ARCHIMEDES +\vfill + +\footnotesize RECENTLY DISCOVERED AND TRANSLATED~FROM~THE~GREEK~BY \\ +\normalsize DR.~J.~L. HEIBERG \\ +\scriptsize PROFESSOR OF CLASSICAL PHILOLOGY AT~THE~UNIVERSITY~OF~COPENHAGEN +\normalsize +\vfill + +\footnotesize WITH AN INTRODUCTION BY \\ +\small DAVID EUGENE SMITH \\ +\scriptsize PRESIDENT OF TEACHERS COLLEGE, COLUMBIA~UNIVERSITY,~NEW~YORK +\vfill + +\footnotesize ENGLISH VERSION TRANSLATED FROM~THE~GERMAN BY~LYDIA~G.~ROBINSON \\ +AND REPRINTED FROM ``THE MONIST,'' APRIL,~1909 +\vfill\vfill\vfill + +CHICAGO \\ +THE OPEN COURT PUBLISHING COMPANY \\ +LONDON AGENTS \\ +KEGAN PAUL, TRENCH, TRÜBNER \& CO., LTD. \\ +1909 +\end{center} +\PageSep{ii} +\newpage +\begin{center} +\null\vfill +\footnotesize\scshape +Copyright by \\ +The Open Court Publishing Co. \\ +1909 +\vfill +\end{center} +\PageSep{1} +\MainMatter + +\Section{Introduction.} + +\First{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, \textit{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 +\PageSep{2} +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 \textit{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 (\Title{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 +\PageSep{3} +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: ``I 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\dots. Indeed I assume that some one among the +\PageSep{4} +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 \Title{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 +\PageSep{5} +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\dots +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 $v = 2r^{2}h/3$, the volume of the prism being~$4r^{2}h$, +and requires a knowledge of the center of gravity of the +cylindric section in question. Archimedes is, so far as we +\PageSep{6} +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 +$\frac{1}{4}p : v = \frac{1}{2}b : (\frac{2}{3} · \frac{1}{2}b)$, +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 Archimedes 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. + +\Signature{David Eugene Smith.}{Teachers~College, Columbia~University.} +\PageSep{7} + +\Section{Geometrical Solutions Derived from Mechanics.} + +\noindent\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 parallelograms\SameMark\ +and its 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 +parallelograms\SameMark\ and whose surface touches the other four planes, +and if in the same cube a second cylinder is inscribed whose bases +lie in two other parallelograms\SameMark\ and whose 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 +\PageSep{8} +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 already known from Heron's recently discovered \Title{Metrica} that these + propositions were contained in this treatise, and Professor Heiberg made the + same comment in \Title{Hermes}.---Tr.} +\PageSep{9} + +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 [\Title{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.\ \ibid\ +I,~5]. + +4. The center of gravity of a straight line is the center of that +line [Cf.\ \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 [\Ibid\ I,~14]. + +6. The center of gravity of a parallelogram is the point where +its diagonals meet [\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 [\Title{De Conoid.}~I]. +\PageSep{10} + + +\Subsection{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 +segment~$\alpha\beta\gamma$ 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 [\ie, of +conic sections, cf.\ \Title{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.\ \Title{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. +\Figure{1}{fig01} +In the 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 +\PageSep{11} +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.\ \Title{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 $\Tri \alpha\zeta\gamma : \segm \alpha\beta\gamma$ when +transferred to~$\theta$ as its center of gravity $= \theta\kappa : \kappa\chi$. But $\theta\kappa = 3\kappa\chi$; +hence also $\Tri \alpha\zeta\gamma = 3 \segm \alpha\beta\gamma$. But it is also true that $\Tri \zeta\alpha\gamma = 4\triangle\alpha\beta\gamma$ +because $\zeta\kappa = \kappa\alpha$ and $\alpha\delta = \delta\gamma$; hence $\segm \alpha\beta\gamma = \frac{4}{3}$ the +$\Tri \alpha\beta\gamma$. 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. + + +\Subsection{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: + +\Figure{2}{fig02} + +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~$\pi\rho$. +\PageSep{12} +Now because $\gamma\alpha × \alpha\sigma = \mu\sigma × \sigma\pi$ (for $\alpha\gamma = \sigma\mu$, $\alpha\sigma = \pi\sigma$), and +$\gamma\alpha × \alpha\sigma = \alpha\xi^{2} = \xi\sigma^{2} + \alpha\pi^{2}$ then $\mu\sigma × \sigma\pi = \xi\sigma^{2} + \sigma\pi^{2}$. Moreover, because +$\gamma\alpha : \alpha\sigma = \mu\sigma : \sigma\pi$ and $\gamma\alpha = \alpha\theta$, therefore +$\theta\alpha : \alpha\sigma = \mu\sigma : \sigma\pi = \mu\sigma^{2} : \mu\sigma × \sigma\pi$. +But it has been +proved that $\xi\sigma^{2} + \sigma\pi^{2} = \mu\sigma × \sigma\pi$; +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 $\cylinder = 2 × (\sphere + \cone)$. But it is also +true that the $\cylinder = 3 \cones$ [Euclid, \Title{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 +\PageSep{13} +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 diameter~$\beta\delta$ 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, \Title{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,~\QED + +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. + + +\Subsection{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~$\gamma$ +\PageSep{14} +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 $\mu\sigma : \sigma\pi = \mu\sigma^{2} : \mu\sigma × \sigma\pi$ and $\mu\sigma × \sigma\pi = \pi\sigma^{2} + \sigma\xi^{2}$, for +$ \alpha\sigma × \sigma\gamma : \sigma\xi^{2} = \alpha\kappa × \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, \Title{Con.}\ I,~21]) $= \alpha\sigma^{2} : \sigma\pi^{2}$ therefore +$\alpha\sigma^{2} : \alpha\sigma × \sigma\gamma = \pi\sigma^{2} : \sigma\xi^{2} = \sigma\pi^{2} : \sigma\pi × \pi\mu$, +consequently $\mu\pi × \pi\sigma = \sigma\xi^{2}$. +If $\pi\sigma^{2}$~is added to both +sides then $\mu\sigma × \sigma\pi = \pi\sigma^{2} + \sigma\xi^{2}$. +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, +\Figure{3}{fig03} +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 +\PageSep{15} +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 $\cylinder = 2 × (\spheroid + \cone) = 2 × \spheroid + 2 × \cone$. But the +$\cylinder = 3 × \cone$, hence $3 × \cone = 2 × \cone + 2 × \spheroid$. Subtract +$2 × \cone$ from both sides; then a cone whose axes form the triangle +$\alpha\epsilon\zeta = 2 × \spheroid$. But the same $\cone = 8$~cones whose axes form +the~$\triangle\alpha\beta\delta$; hence $8$~such cones $= 2 × \spheroid$, $4 × \cone = \spheroid$; +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.~\QED + + +\Subsection{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 +\PageSep{16} +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 [\Title{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 +\Figure{4}{fig04} +whose 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 +\PageSep{17} +to that point, then we have the inverse proportion $\theta\alpha : \alpha\kappa = \cylinder : \segment$. +But $\theta\alpha = 2\alpha\kappa$ and also the $\cylinder = 2 × \segment$. +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. + + +\Subsection{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 +\Figure{5}{fig05} +at~$\alpha$; 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, $\delta\alpha : \alpha\sigma = \beta\delta^{2} : \xi\sigma^{2}$ [\Title{Quadr.\ parab.}~3]. But +$\delta\alpha : \alpha\sigma = \beta\delta : \pi\sigma = \beta\delta^{2} : \beta\delta × \pi\sigma$, therefore also $\beta\delta^{2} : \xi\sigma^{2} = \beta\delta^{2} : \beta\delta × \pi\sigma$. +Consequently $\xi\sigma^{2} = \beta\delta × \pi\sigma$ 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 +\PageSep{18} +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. + + +\Subsection{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 +\PageSep{19} +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 \Chg{semi-circle}{semicircle}~$\beta\alpha\delta$ 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 +\Figure[0.5\textwidth]{6}{fig06} +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. +\PageSep{20} + + +\Subsection{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 +\Figure{7}{fig07} +of the sphere in a circle whose diameter 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 +%[** TN: Opening bracket has no matching close bracket in the original] +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 [$\gamma\eta × \eta\phi$] $= \frac{1}{3}\alpha\eta × \eta\gamma$. +Therefore also $\gamma\eta × \eta\phi = \frac{1}{3}\beta\eta^{2}$.~\dotfill + + +\Subsection{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 +\PageSep{21} +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 + +\Subsection{VIII.} + +\noindent\dotfill\\ +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 +\Figure[0.5\textwidth]{8}{fig08} +cutting 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 scale-beam +at~$\theta$ that $\theta$~is their center of gravity; then also the spherical $\segment \alpha\beta\delta$ +\PageSep{22} +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 $\cylinder \mu\nu$ +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 [\Title{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 $\beta\eta^{2} = \gamma\eta × \eta\alpha$, +$\eta\epsilon^{2} = \eta\alpha^{2}$, and $\gamma\eta × \eta\alpha : \eta\alpha^{2} = \gamma\eta : \eta\alpha$, 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 $\segment \beta\alpha\delta$. + + +\Subsection{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. + + +\Subsection{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 [\Title{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. + + +\Subsection{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. + +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 +\PageSep{24} +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. + +\Figures{9}{fig09}{10}{fig10} + +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. +Imagine the parallelogram in the cylinder-section transferred and +so brought to~$\xi$ that $\xi$~is its center of gravity, and further imagine +\PageSep{25} +$\pi\xi$~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. + + +\Subsection{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~$\pi\xi$; +\Figure{11}{fig11} +these will cut the circumference of the semicircle~$o\pi\rho$ at the points +\PageSep{26} +$\kappa$~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 + + +\Subsection{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 +$\mu\nu × \nu\lambda = \nu\zeta^{2}$ (for this is evident [Apollonios, \Title{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 +\Figure{12}{fig12} +intersect the portion which is cut off from the cylinder by the 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 +\PageSep{27} +to the plane~$\kappa\nu$, and the hypotenuse \dotfill \\ +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,~\QED + + +\Subsection{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 magnitude.~\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\null + +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~$\epsilon\eta$, \ie, 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 + +\vfill + +%%%%%%%%%%%%%%%%%%%%%%%%% GUTENBERG LICENSE %%%%%%%%%%%%%%%%%%%%%%%%%% +\BackMatter +\PGLicense +\begin{PGtext} +End of the Project Gutenberg EBook of Geometrical Solutions Derived from +Mechanics, by Archimedes + +*** END OF THIS PROJECT GUTENBERG EBOOK GEOMETRICAL SOLUTIONS *** + +***** This file should be named 7825-pdf.pdf or 7825-pdf.zip ***** +This and all associated files of various formats will be found in: + http://www.gutenberg.org/7/8/2/7825/ + +Produced by Gordon Keener + +Updated editions will replace the previous one--the old editions +will be renamed. + +Creating the works from public domain print editions means that no +one owns a United States copyright in these works, so the Foundation +(and you!) can copy and distribute it in the United States without +permission and without paying copyright royalties. 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+jurisdictions other than the United States. Anyone seeking to utilize +this eBook outside of the United States should confirm copyright +status under the laws that apply to them. diff --git a/README.md b/README.md new file mode 100644 index 0000000..36f57b1 --- /dev/null +++ b/README.md @@ -0,0 +1,2 @@ +Project Gutenberg (https://www.gutenberg.org) public repository for +eBook #7825 (https://www.gutenberg.org/ebooks/7825) |
