Chapter XXII: Of Other Variety of Motion (2)

First, that the centre of equiponderation of the figure C I A P E is somewhere in the axis A B is manifest of itself; and therefore A B is a diameter of equiponderation. Let A E be drawn, and let B E be put for the moment of the strait line C E; the triangle A B E will therefore (by the third article) be the moment of the complete figure C D F E. Let the axis A B be equally divided in L, and let G L H be drawn parallel and equal to the strait line C E, cutting the crooked line C I A P E in I and P, and the strait lines A C and A E in K and M. Moreover, let Z O be drawn parallel to the same C E; and let it be, as L G to L I, so L M to another, L N; and let the same be done in all the rest of the strait lines possible, parallel to the base; and through all the points N, let the line A N E be drawn; the three-sided figure A N E B will therefore be the moment of the figure C I A P E. Now the triangle A B E is (by the 9th article of chapter XVII) to the three-sided figure A N E B, as A B C D + A I C B is to A I C B twice taken, that is, as C D F E + C I A P E is to C I A P E twice taken. But as C I A P E is to C D F E, that is, as the weight of the deficient figure is to the weight of the complete figure, so is C I A P E twice taken to C D F E twice taken. Wherefore, setting in order C D F E + C I A P E. 2 C I A P E. 2 C D F E; the proportion of C D F E + C I A P E to C D F E twice taken will be compounded of the proportion of C D F E + C I A P E to C I A P E twice taken, that is, of the proportion of the triangle A B E to the three-sided figure A N E B, that is, of the moment of the complete figure to the moment of the deficient figure, and of the proportion of C I A P E twice taken to C D F E twice taken, that is, to the proportion reciprocally taken of the weight of the deficient figure to the weight of the complete figure.

Again, seeing by supposition A Z. Z B:: C D F E. C I A P E are proportionals; A B. A Z:: C D F E + C I A P E. C D F E will also, by compounding, be proportionals. And seeing A L is the half of A B, A L. A Z:: C D F E + C I A P E. 2 C D F E will also be proportionals. But the proportion of C D F E + C I A P E to 2 C D F E is compounded, as was but now shown, of the proportions of moment to moment, &c., and therefore the proportion of A L to A Z is compounded of the proportion of the moment of the complete figure C D F E to the moment of the deficient figure C I A P E, and of the proportion of the weight of the deficient figure C I A P E to the weight of the complete figure C D F E; but the proportion of A L to A Z is compounded of the proportions of A L to B Z and of B Z to A Z. Now the proportion of B Z to A Z is the proportion of the weights reciprocally taken, that is to say, of the weight C I A P E to the weight C D F E. Therefore the remaining proportion of A L to B Z, that is, of L B to B Z, is the proportion of the moment of the weight C D F E to the moment of the weight C I A P E. But the proportion of A L to B Z is compounded of the proportions of A L to A Z and of A Z to Z B; of which proportions that of A Z to Z B is the proportion of the weight C D F E to the weight C I A P E. Wherefore (by art. 5 of this chapter) the remaining proportion of A L to A Z is the proportion of the distances of the points Z and L from the centre of the scale, which is A. And, therefore, (by art. 6) the weight C I A P E shall hang from O in the strait line O Z. So that O Z is one diameter of equiponderation of the weight C I A P E. But the strait line A B is the other diameter of equiponderation of the same weight C I A P E. Wherefore (by the 7th definition) the point Z is the centre of the same equiponderation; which point, by construction, divides the axis so, that the part A Z, which is the part next the vertex, is to the other part Z B, as the complete figure C D F E is to the deficient figure C I A P E; which is that which was to be demonstrated.

Coroll. I. The centre of equiponderation of any of those plane three-sided figures, which are compared with their complete figures in the table of art. 3, chap. XVII, is to be found in the same table, by taking the denominator of the fraction for the part of the axis cut off next the vertex, and the numerator for the other part next the base. For example, if it be required to find the centre of equiponderation of the second three-sided figure of four means, there is in the concourse of the second column with the row of three-sided figures of four means this fraction 5⁄7, which signifies that that figure is to its parallelogram or complete figure as 5⁄7 to unity, that is, as 5⁄7 to 7⁄7, or as 5 to 7; and, therefore the centre of equiponderation of that figure divides the axis, so that the part next the vertex is to the other part as 7 to 5.

Coroll. II. The centre of equiponderation of any of the solids of those figures, which are contained in the table of art. 7 of the same chap. XVII, is exhibited in the same table. For example, if the centre of equiponderation of a cone be sought for, the cone will be found to be 1⁄3 of its cylinder; and, therefore, the centre of its equiponderation will so divide the axis, that the part next the vertex to the other part will be as 3 to 1. Also the solid of a three-sided figure of one mean, that is, a parabolical solid, seeing it is 2⁄4, that is ½ of its cylinder, will have its centre of equiponderation in that point, which divides the axis, so that the part towards the vertex be double to the part towards the base.

[Sidenote: The diameter of equiponderation of the complement of the half
of any of the said deficient figures, divides that line which
is drawn through the vertex parallel to the base, so that the
part next the vertex is to the other part as the complete
figure to the complement.]

10. The diameter of equiponderation of the complement of the half of any of those figures which are contained in the table of art. 3, chap. XVII, divides that line which is drawn through the vertex parallel and equal to the base, so that the part next the vertex will be to the other part, as the complete figure to the complement.

For let A I C B (in the same fig. 5) be the half of a parabola, or of any other of those three-sided figures which are in the table of art. 3, chap. XVII, whose axis is A B, and base B C, having A D drawn from the vertex, equal and parallel to the base B C, and whose complete figure is the parallelogram A B C D. Let I Q be drawn at any distance from the side C D, but parallel to it; and let A D be the altitude of the complement A I C D, and Q I a line ordinately applied in it. Wherefore the altitude A L in the deficient figure A I C B is equal to Q I the line ordinately applied in its complement; and contrarily, L I the line ordinately applied in the figure A I C B is equal to the altitude A Q in its complement; and so in all the rest of the ordinate lines and altitudes the mutation is such, that that line, which is ordinately applied in the figure, is the altitude of its complement. And, therefore, the proportion of the altitudes decreasing to that of the ordinate lines decreasing, being multiplicate according to any number in the deficient figure, is submultiplicate according to the same number in its complement. For example, if A I C B be a parabola, seeing the proportion of A B to A L is duplicate to that of B C to L I, the proportion of AD to A Q in the complement A I C D, which is the same with that of B C to L I, will be subduplicate to that of C D to Q I, which is the same with that of A B to A L; and consequently, in a parabola, the complement will be to the parallelogram as 1 to 3; in a three-sided figure of two means, as 1 to 4; in a three-sided figure of three means, as 1 to 5, &c. But all the ordinate lines together in A I C D are its moment; and all the ordinate lines in A I C B are its moment. Wherefore the moments of the complements of the halves of deficient figures in the table of art. 3 of chap. XVII, being compared, are as the deficient figures themselves; and, therefore, the diameter of equiponderation will divide the strait line A D in such proportion, that the part next the vertex be to the other part, as the complete figure A B C D is to the complement A I C D.

Coroll. The diameter of equiponderation of these halves may be found by the table of art. 3 of chap. XVII, in this manner. Let there be propounded any deficient figure, namely, the second three-sided figure of two means. This figure is to the complete figure as 3⁄5 to 1, that is 3 to 5. Wherefore the complement to the same complete figure is as 2 to 5; and, therefore, the diameter of equiponderation of this complement will cut the strait line drawn from the vertex parallel to the base, so that the part next the vertex will be to the other part as 5 to 2. And, in like manner, any other of the said three-sided figures being propounded, if the numerator of its fraction found out in the table be taken from the denominator, the strait line drawn from the vertex is to be divided, so that the part next the vertex be to the other part, as the denominator is to the remainder which that subtraction leaves.

[Sidenote: The centre of equiponderation of the half of any of the
deficient figures in the first row of the table of art. 3,
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.

[Sidenote: 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.

[Sidenote: 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.

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.

[Sidenote: The centre of equiponderation of a solid sector is in the
axis so divided, that the part next the vertex be to the
whole axis, wanting half the axis of the portion of the
sphere, as 3 to 4.]

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.

------------------------------------------------------------------------

_Vol. 1. Lat. & Eng._
C.XXIII.
_Fig. 1-9_
]

------------------------------------------------------------------------