Chapter Xvii, May Be Found Out by the Numbers of the Second

row.]

11. The centre of equiponderation of the half of any of those crooked-lined figures, which are in the first row of the table of art. 3 of chap. XVII, is in that strait line which, being parallel to the axis, divides the base according to the numbers of the fraction next below it in the second row, so that the numerator be answerable to that part which is towards the axis.

For example, let the first figure of three means be taken, whose half is A B C D (in fig. 6), and let the rectangle A B E D be completed. The complement therefore will be B C D E. And seeing A B E D is to the figure A B C D (by the table) as 5 to 4, the same A B E D will be to the complement B C D E as 5 to 1. Wherefore, if F G be drawn parallel to the base D A, cutting the axis so that A G be to G B as 4 to 5, the centre of equiponderation of the figure A B C D will, by the precedent article, be somewhere in the same F G. Again, seeing, by the same article, the complete figure A B E D, is to the complement B C D E as 5 to 1, therefore if B E and A D be divided in I and H as 5 to 1 the centre of equiponderation of the complement B C D E will be somewhere in the strait line which connects H and I. Let now the strait line L K be drawn through M the centre of the complete figure, parallel to the base; and the strait line N O through the same centre M, perpendicular to it; and let the strait lines L K and F G cut the strait line H I in P and Q. Let P R be taken quadruple to P Q; and let R M be drawn and produced to F G in S. R M therefore will be to M S as 4 to 1, that is, as the figure A B C D to its complement B C D E. Wherefore, seeing M is the centre of the complete figure A B E D, and the distances of R and S from the centre M be in proportion reciprocal to that of the weight of the complement B C D E to the weight of the figure A B C D, R and S will either be the centres of equiponderation of their own figures, or those centres will be in some other points of the diameters of equiponderation H I and F G. But this last is impossible. For no other strait line can be drawn through the point M terminating in the strait lines H I and F G, and retaining the proportion of M R to M S, that is, of the figure A B C D to its complement B C D E. The centre, therefore, of equiponderation of the figure A B C D is in the point S. Now, seeing P M hath the same proportion to Q S which R P hath to R Q, Q S will be 5 of those parts of which P M is four, that is, of which I N is four. But I N or P M is 2 of those parts of which E B or F G is 6; and, therefore, if it be as 4 to 5, so 2 to a fourth, that fourth will be 2½. Wherefore Q S is 2½ of those parts of which F G is 6. But F Q is 1; and, therefore, F S is 3½. Wherefore the remaining part G S is 2½. So that F G is so divided in S, that the part towards the axis is in proportion to the other part, as 2½ to 3½, that is as 5 to 7; which answereth to the fraction 5⁄7 in the second row, next under the fraction 4⁄5 in the first row. Wherefore drawing S T parallel to the axis, the base will be divided in like manner.

By this method it is manifest, that the base of a semiparabola will be divided into 3 and 5; and the base of the first three-sided figure of two means, into 4 and 6; and of the first three-sided figure of four means, into 6 and 8. The fractions, therefore, of the second row denote the proportions, into which the bases of the figures of the first row are divided by the diameters of equiponderation. But the first row begins one place higher than the second row.

12. The centre of equiponderation of the half of any of the figures in the second row of the same table of art. 3, chap. XVII, is in a strait line parallel to the axis, and dividing the base according to the numbers of the fraction in the fourth row, two places lower, so as that the numerator be answerable to that part which is next the axis.

Let the half of the second three-sided figure of two means be taken; and let it be A B C D (in fig. 7); whose complement is B C D E, and the rectangle completed A B E D. Let this rectangle be divided by the two strait lines L K and N O, cutting one another in the centre M at right angles; and because A B E D is to A B C D as 5 to 3, let A B be divided in G, so that A G to B G be as 3 to 5; and let F G be drawn parallel to the base. Also because A B E D is (by art. 9) to B C D E as 5 to 2, let B E be divided in the point I, so that B I be to I E as 5 to 2; and let I H be drawn parallel to the axis, cutting L K and F G in P and Q. Let now P R be so taken, that it be to P Q as 3 to 2, and let R M be drawn and produced to F G in S. Seeing, therefore, R P is to P Q, that is, R M to M S, as A B C D is to its complement B C D E, and the centres of equiponderation of A B C D and B C D E are in the strait lines F G and H I, and the centre of equiponderation of them both together in the point M; R will be the centre of the complement B C D E, and S the centre of the figure A B C D. And seeing P M, that is I N, is to Q S, as R P is to R Q; and I N or P M is 3 of those parts, of which B E, that is F G, is 14; therefore Q S is 5 of the same parts; and E I, that is F G, 4; and F S, 9; and G S, 5. Wherefore the strait line S T being drawn parallel to the axis, will divide the base A D into 5 and 9. But the fraction 5⁄9 is found in the fourth row of the table, two places below the fraction ⅗ in the second row.

By the same method, if in the same second row there be taken the second three-sided figure of three means, the centre of equiponderation of the half of it will be found to be in a strait line parallel to the axis, dividing the base according to the numbers of the fraction 6⁄10, two places below in the fourth row. And the same way serves for all the rest of the figures in the second row. In like manner, the centre of equiponderation of the third three-sided figure of three means will be found to be in a strait line parallel to the axis, dividing the base, so that the part next the axis be to the other part as 7 to 13, &c.

Coroll. The centres of equiponderation of the halves of the said figures are known, seeing they are in the intersection of the strait lines S T and F G, which are both known.

13. The centre of equiponderation of the half of any of the figures, which (in the table of art. 3, chap. XVII) are compared with their parallelograms, being known; the centre of equiponderation of the excess of the same figure above its triangle is also known.

For example, let the semiparabola A B C D (in fig. 8) be taken, whose axis is A B; whose complete figure is A B E D; and whose excess above its triangle is B C D B. Its centre of equiponderation may be found out in this manner. Let F G be drawn parallel to the base, so that A F be a third part of the axis; and let H I be drawn parallel to the axis, so that A H be a third part of the base. This being done, the centre of equiponderation of the triangle A B D will be I. Again, let K L be drawn parallel to the base, so that A K be to A B as 2 to 5; and M N parallel to the axis, so that A M be to A D as 3 to 8; and let M N terminate in the strait line K L. The centre, therefore, of equiponderation of the parabola A B C D is N; and therefore we have the centres of equiponderation of the semiparabola A B C D, and of its part the triangle A B D. That we may now find the centre of equiponderation of the remaining part B C D B, let I N be drawn and produced to O, so that N O be triple to I N; and O will be the centre sought for. For seeing the weight of A B D to the weight of B C D B is in proportion reciprocal to that of the strait line N O to the strait line I N; and N is the centre of the whole, and I the centre of the triangle A B D; O will be the centre of the remaining part, namely, of the figure B D C B; which was to be found.

Coroll. The centre of equiponderation of the figure B D C B is in the concourse of two strait lines, whereof one is parallel to the base, and divides the axis, so that the part next the base be ⅗ or 9⁄15 of the whole axis; the other is parallel to the axis, and so divides the base, that the part towards the axis be ½, or 12⁄24 of the whole base. For drawing O P parallel to the base, it will be as I N to N O, so F K to K P, that is, so 1 to 3, or 5 to 15. But A F is 5⁄15, or ⅓ of the whole A B; and A K is 6⁄15, or ⅖; and F K 1⁄15; and KP 3⁄15; and therefore A P is 9⁄15 of the axis A B. Also A H is ⅓, or 8⁄24; and A M ⅜, or 9⁄24 of the whole base; and therefore O Q being drawn parallel to the axis, M Q, which is triple to H M, will be 3⁄24. Wherefore A Q is 12⁄24, or ½ of the base A D.

The excesses of the rest of the three-sided figures in the first row of the table of art. 3, chap. XVII, have their centres of equiponderation in two strait lines, which divide the axis and base according to those fractions, which add 4 to the numerators of the fractions of a parabola 9⁄15 and 12⁄24; and 6 to the denominators, in this manner:—

In a parabola, the axis 9⁄15, the base 12⁄24. In the first three-sided figure, the axis 13⁄21, the base 16⁄30. In the second three-sided figure, the axis 17⁄27, the base 20⁄36, &c.

And by the same method, any man, if it be worth the pains, may find out the centres of equiponderation of the excesses above their triangles of the rest of the figures in the second and third row, &c.

14. The centre of equiponderation of the sector of a sphere, that is, of a figure compounded of a right cone, whose vertex is the centre of the sphere, and the portion of the sphere whose base is the same with that of the cone, divides the strait line which is made of the axis of the cone and half the axis of the portion together taken, so that the part next the vertex be triple to the other part, or to the whole strait line as 3 to 4.

For let A B C (in fig. 9) be the sector of a sphere, whose vertex is the centre of the sphere A; whose axis is A D; and the circle upon B C is the common base of the portion of the sphere and of the cone whose vertex is A; the axis of which portion is E D, and the half thereof F D; and the axis of the cone, A E. Lastly, let A G be 3⁄4 of the strait line A F. I say, G is the centre of equiponderation of the sector A B C.

Let the strait line F H be drawn of any length, making right angles with A F at F; and drawing the strait line A H, let the triangle A F H be made. Then upon the same centre A let any arch I K be drawn, cutting A D in L; and its chord, cutting A D in M; and dividing M L equally in N, let N O be drawn parallel to the strait line F H, and meeting with the strait line A H in O.

Seeing now B D C is the spherical superficies of the portion cut off with a plane passing through B C, and cutting the axis at right angles; and seeing F H divides E D, the axis of the portion, into two equal parts in F; the centre of equiponderation of the superficies B D C will be in F (by art. 8); and for the same reason the centre of equiponderation of the superficies I L K, K being in the strait line A C, will be in N. And in like manner, if there were drawn, between the centre of the sphere A and the outermost spherical superficies of the sector, arches infinite in number, the centres of equiponderation of the spherical superficies, in which those arches are, would be found to be in that part of the axis, which is intercepted between the superficies itself and a plane passing along by the chord of the arch, and cutting the axis in the middle at right angles.

Let it now be supposed that the moment of the outermost spherical superficies B D C is F H. Seeing therefore the superficies B D C is to the superficies I L K in proportion duplicate to that of the arch B D C to the arch I L K, that is, of B E to I M, that is, of F H to N O; let it be as F H to N O, so N O to another N P; and again, as N O to N P, so N P to another N Q; and let this be done in all the strait lines parallel to the base F H that can possibly be drawn between the base and the vertex of the triangle A F H. If then through all the points Q there be drawn the crooked line A Q H, the figure A F H Q A will be the complement of the first three-sided figure of two means; and the same will also be the moment of all the spherical superficies, of which the solid sector A B C D is compounded; and by consequent, the moment of the sector itself. Let now F H be understood to be the semidiameter of the base of a right cone, whose side is A H, and axis A F Wherefore, seeing the bases of the cones, which pass through F and N and the rest of the points of the axis, are in proportion duplicate to that of the strait lines F H and N O, &c., the moment of all the bases together, that is, of the whole cone, will be the figure itself A F H Q A; and therefore the centre of equiponderation of the cone A F H is the same with that of the solid sector. Wherefore, seeing A G is ¾ of the axis A F, the centre of equiponderation of the cone A F H is in G; and therefore the centre of the solid sector is in G also, and divides the part A F of the axis so that A G is triple to G F; that is, A G is to A F as 3 to 4; which was to be demonstrated.

Note, that when the sector is a hemisphere, the axis of the cone vanisheth into that point which is the centre of the sphere; and therefore it addeth nothing to half the axis of the portion. Wherefore, if in the axis of the hemisphere there be taken from the centre ¾ of half the axis, that is, ⅜ of the semidiameter of the sphere, there will be the centre of equiponderation of the hemisphere.

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Vol. 1. Lat. & Eng. C.XXIII. Fig. 1-9 ]

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