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.
The centre of equiponderation of the half of any of the figures of the second row of the same table may be found out by the numbers of the fourth 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.
The centre of equiponderation of the half of any of the figures in the same table being known, the centre of the excess of the same figure above a triangle of the same altitude and base is also 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.