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But, to find the ratio of A to B, it is not requisite to know the values of the remainders C, D, E, &c. Suppose the subdivision to terminate at B; then A=mB, and consequently A : B, as mB : B, or m : 1. If the subdivision extend to C, then A=m'C, and B-nc.; whence A : B, as m': n. In general, therefore, the second term, in the expressions for A and B, may be rejected, and the letter which precedes it considered as the ultimate measure, and corresponding to the arithmetical unit. Hence, resuming the substitutions, and combining the whole in one view, it follows, that the ratio of A to B may thus be successively represented:
The formation of these numbers will evidently stop, when the corresponding subdivision terminates. But even though the successive decomposition should never terminate, as in the case of incommensurable quantities, yet the expression thus obtained must constantly approach to the ratio of A : B, since they suppose only the omission of the remainder of the last division, and which is perpetually diminishing.
pansion of the same residue woas a repeated integral quotient. Hence m being 1 and n, p, q, r, &c. all equal to 2, the successive approximations are, by the last note, 1 : 1, 2:3, 5:7, 12: 17, 29 :41, 70 : 99, &c. The ratios of the squares of these numbers are 4: 9, 25:49, 144 : 289, 841 : 1681,4900 : 9801, thus approaching rapidly to the ratio of one to two, but alternately in excess and defect.
, which therefore gives 2
1. Proposition first. The consideration of diverging lines furnishes the simplest and readiest means, for transferring the doctrine of proportion to geometrical figures. The order which Euclid has followed, beginning with parallelograms, and thence passing from surfaces to lines, appears to be less natural.
2. Proposition fourth. It will be proper here to notice the several methods adopted in practice, for the minute subdivision of lines. The earliest of these—the diagonal scale—depending immediately on the proposition in the text, is of the most extensive use, and constituted the first improvement on astronomical instruments. Thus, in the figure annexed, the extreme portion of the horizontal line is divided into ten equal parts, each of which again is virtually subdivided into ten secondary parts. The subdivision is effected by means of diagonal lines, which decline from the perpendicular by intervals equal to the primary divisions, and which are cut transversely into ten equal segments by equidistant parallels. Suppose, for example, it were required to find the length of 2 and 38–100 parts of a division; place one foot of the compasses in the second vertical at the eight interval which is marked with a dot, and extend the other foot, along the parallel, to the dot on the third diagonal. The distance between these dots may, however, express ifidifferently 2.38, 23.8, or 238, according to the assumed magnitude of the primary unit.
Nunez, or Nonius, in a Treatise De Crepusculis, printed at Eisbon in 1542, proposed one more complicated. He placed a number of parallel scales, or concentric circles, differently divided, and forming a regular ascending gradation of 89, 88, 87, &c. equal parts, from 90 to 46 inclusive. An index laid any where across these scales might, therefore, be presumed to cut at least one of them at some of the divisions, and hence the intercepted space would be expressed by a corresponding fraction.
But the method of subdivision which was afterwards introduced by Peter Wernier, a gentleman of Franche Comté, and published by him in a small tract printed at Brussels in 1631, being itself an improvement on the method used in the construction of Tycho Brahe’s astronomical instruments, is much simpler and far more ingenious. It is founded on the difference of two approximating scales, one of which is moveable. Thus, if a space equal to n—I parts on the limb of the instrument be divided into n parts, these evidently will each of them be smaller than the former, by the nth part of a division. Wherefore, on shifting forward this parasite or Wernier scale, the quantity of aberration will diminish at each successive division, till a new coincidence obtains, and then the number of those divisions on that scale will mark the fractional value of the displacement.
Thus in the annexed figure, nine divisions of the primary scale, forming ten equal parts in the attached or sliding scale, the moveable zero stands 9 8 7 6 5 4 3 24 beyond the ===HH=H. first interval between the third and fourth division. To find this minute difference, observe where the opposite sections of the scales come to coincide, which occurs under the fourth division of the sliding scale, and therefore indicates the quantity 1.34.
3. Proposition fifth. This problem could be otherwise solved. Through B draw the inclined straight line CBG extended both ways, in this take any point C, and make BD, DE, EF, FG, &c. each equal to BC, complete the parallelogram ABCI, and join ID, IE, IF, IG, &c. cutting AB in the point K, L, M, N, &c.; then is the segment AK the half of AB, AL the third, AM the fourth, and AN the fifth part of the same given line.
For the segments of the straight line AB must be proportional to the segments of the parallels AI and BG, intercepted by the diverging lines ID, IE, IF, IG, &c. Thus, AK : KB ; : Al : BD; but, by con- struction, BC or AI-BD, whence (V. 4.) AK=KB, and therefore AK is the half of AB. Again, AL : LB : : AI : BE; and since BE=2AI, it follows that LB=2AL, or AL is the third part of AB. In the same manner, AM : MB : : AI : BF; but BF=3AI, whence MB-3AM, or AM is the fourth part of AB. And, by a like process, it may be shown that AN is the fifth part of AB.
4. Proposition seventeenth. The solution of this important problem now inserted in the text, was suggested to me by Mr Thomas Carlyle, an ingenious young mathematician, formerly my pupil. But I here subjoin likewise the original construction given by Pappus, which, though rather more complex, has yet some peculiar advantages. Let AB be a straight line, which it is required to cut, so that the rectangle under its segments shall be equivalent to a given rectangle.
On AB describe the semicircle AFB, at A and B apply tangents AD and BE equal to the sides of the given rectangle, and both in the same or in opposite directions, according as the line is to be cut internally or externally; join DE, and from the point F where it meets the circumference, draw the perpendicular FC; this will divide the given line AB into AC and BC, the segments required. For the right angle DFC is equal (III. 19.) to the angle AFB contained in the semi- SJF circle, and consequently their I. difference from AFC or the /N angles DFA and CFB are JA C B equal. For the same rea- TE son, the angle AFB being likewise equal to CFE, add or take away CFB, and the angle BFE will be equal to AFC. But AD being a tangent, and AF a straight line inflected to the circumference, the exterior angle DAF is equal (III. 21.) to D the angle in the alternate segment AF or the angle CBF (III. 17. cor. 2.). Again, BE being a tangent and BF an inflected line, the exterior angle EBF is equal to BAF. Wherefore the triangles DAF and AFC are similar to BFC and BFE ; and hence AD : AF : : CB: BF, and AF : AC : : BF : BE; consequently (V. 16.) AD : AC : : CB : BE, and (V. 6.) AD.BE=AC.CB. Cor. If the sides of the given rectangle be equal, the construction of the problem will become materially simplified. First, in the case of internal section: The tangents AD, BE being equal, it is evident that DE must be parallel to AB and the per- T E pendicular FC parallel to EB. Whence, >employing this construction, or erect- N \ ing the perpendicular BE equal to the v &# sides of the given square, and drawing the parallel EF to meet the circumference F, from which is let fall on AB the perpendicular FC, the rectangle under the segments AC and CB is equivalent to the square of BE; which also follows from Prop. 26, cor. 1. Book III,