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Read Ebook: Geometrical Solutions Derived from Mechanics; a Treatise of Archimedes by Archimedes BCE BCE Smith David Eugene Author Of Introduction Etc Heiberg J L Johan Ludvig Translator Robinson Lydia Gillingham Translator
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Ebook has 41 lines and 13304 words, and 1 pages
Edition: 10
Gordon Keener
% gbn0305181551: Archimedes, Geometrical Solutions Derived from Mechanics. Gordon Keener
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date \ Dr. J. L. Heiberg\ \ vspace \ David Eugene Smith\ \ vspace \ Lydia G. Robinson\ \ vspace }
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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 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 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.
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.
Proposition II states no new fact. Essentially it means that if a sphere, cylinder, and cone have the same radius, $r$, and the altitude of the cone is $r$ and that of the cylinder r$, then the volumes will be as : 1 : 6$, which is true, since they are respectively $fracpi r^3$, $fracpi r^3$, and 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 general, therefore, the greatest value of the work lies in the following:
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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:
These propositions differ essentially from those formerly discovered; for then we compared those bodies 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.
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 .
Let $alphabetagamma$ be the segment of a parabola bounded by the straight line $alphagamma$ and the parabola $alphabetagamma$. Let $alphagamma$ be bisected at $delta$, $deltabetaepsilon$ being parallel to the diameter, and draw $alphabeta$, and $betagamma$. Then the segrnent $alphabetagamma$ will be $frac$ as great as the triangle $alphabetagamma$.
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.
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:
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Draw the straight lines $phibetachi$ and $psideltaomega | alphagamma$ through $beta$ and $delta$ in the parallelogram $lambdazeta$ and imagine a cylinder whose bases are the circles on the diameters $phipsi$ and $chiomega$ and whose axis is $alphagamma$. Now since the cylinder whose axes form the parallelogram $phiomega$ is twice as large as the cylinder whose axes form the parallelogram $phidelta$ and the latter is three times as large as the cone the triangle of whose axes is $alphabetadelta$, as is shown in the Elements , the cylinder whose axes form the parallelogram $phiomega$ is six times as large as the cone whose axes form the triangle $alphabetadelta$. But it was shown that the sphere whose largest circle is $alphabetagammadelta$ 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.
In the parallelogram $lambdazeta$ draw the straight lines $phichi$ and $psiomega | alphagamma$ through the points $beta$ and $delta$ and imagine a cylinder whose bases are the circles on the diameters $phipsi$ and $chiomega$, and whose axis is $alphagamma$. Now since the cylinder whose axes form the parallelogram $phiomega$ is twice as great as the cylinder whose axes form the parallelogram $phidelta$ 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 $phidelta$ is three times as great as the cone whose vertex is at $alpha$ and whose base is the circle on the diameter $betadelta$ perpendicular to $alphagamma$, then the cylinder whose axes form the parallelogram $phiomega$ 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.
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:
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:
that the portion near the surface of the hemisphere is in the ratio of : 3$ to the remaining portion.
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
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.
It can also be seen by this method that 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 ; 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.
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 will be $frac$ 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.
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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 $alphabeta$ 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 $betagamma$; let the axis of the cylinder and the prism be $gammadelta$ which is bisected at right angles by $epsilonzeta$ and on $epsilonzeta$ let a plane be constructed perpendicular to $gammadelta$. This will intersect the prism in a square and the cylinder in a circle.
a perpendicular prism with square bases 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
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Gordon Keener
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