1to the Cylinder Y Z as S Y to R Z, that is, as Y A to A Z: and, by the
ſame reaſon, the Cylinder whoſe Axis is Z Y is to that whoſe Axis is
Z V, as Z A is to A V. The ſaid Cylinders, therefore, are to one ano
ther as the Lines D A, A Y; Z A, A V: But theſe are equally exceed
ing to one another, and the exceſs is equal to the leaſt, ſo that A Z is
double to A V; and A Y is triple the
167[Figure 167]
ſame; and D A Quadruple. Thoſe
Cylinders, therefore, are certain Mag
nitudes in order equally exceeding one
another, whoſe exceſs is equal to the
leaſt of them, and is the Line X M,
in which they are ſuſpended at equal
diſtances (for that each of the Cy
linders hath its Center of Gravity in
the miaſt of the Axis.) Wherefore,
by what hath been above demonſtra
ted, the Center of Gravity of the Mag
nitude compounded of them all divi
deth the Line X M ſo, that the part
towards X is double to the reſt. Divide it, therefore, and, let X α be
double α M: therefore is α the Center of Gravity of the inſcribed Fi
gure. Divide A V in two equal parts in ε: ε X ſhall be double to
M E: But X α is double to α M: Wherefore ε E ſhall be triple E α. But
α E is triple E N: It is manifeſt, therefore, that E N is greater than
E X; and for that cauſe α, which is the Center of Gravity of the in
ſcribed Figure, cometh nearer to the Baſe of the Conoid than N. And
becauſe that as A E is to E N, ſo is the part taken away ε E to the part
taken away E α: and the remaining part ſhall be to the remaming part,
that is, A ε to N α, as A E to E N. Therefore α N is the third part of
A ε, and the ſixt part of A V. And in the ſame manner the Cylinders of
the circumſcribed Figure may be demonſtrated to be equally exceeding
one another, and the exceſs to me equal to the least; and that they have
their Centers of Gravity at equal diſtances in the Line ε M. If therefore
ε M be divided in π, ſo as that ε π be double to the remaining part π M;
π ſhall be the Center of Gravity of the whole circumſcribed Magnitude.
And ſince ε π is double to π M; and A ε leſs than double EM: (for
that they are equal:) the whole A E ſhall be leſs than triple E π: Where
fore E π ſhall be greater than E N. And, ſince ε M is triple to M π,
and M E with twice ε A is likewiſe triple to M E: the whole A E with
A ε ſhall be triple to E π: But A E is triple to E N: Wherefore the
remaining part A ε ſhall be triple to the remaining part π N. Therefore
N π is the ſixth part of A V. And theſe are the things that were to be
demonſtrated.
ſame reaſon, the Cylinder whoſe Axis is Z Y is to that whoſe Axis is
Z V, as Z A is to A V. The ſaid Cylinders, therefore, are to one ano
ther as the Lines D A, A Y; Z A, A V: But theſe are equally exceed
ing to one another, and the exceſs is equal to the leaſt, ſo that A Z is
double to A V; and A Y is triple the
167[Figure 167]ſame; and D A Quadruple. Thoſe
Cylinders, therefore, are certain Mag
nitudes in order equally exceeding one
another, whoſe exceſs is equal to the
leaſt of them, and is the Line X M,
in which they are ſuſpended at equal
diſtances (for that each of the Cy
linders hath its Center of Gravity in
the miaſt of the Axis.) Wherefore,
by what hath been above demonſtra
ted, the Center of Gravity of the Mag
nitude compounded of them all divi
deth the Line X M ſo, that the part
towards X is double to the reſt. Divide it, therefore, and, let X α be
double α M: therefore is α the Center of Gravity of the inſcribed Fi
gure. Divide A V in two equal parts in ε: ε X ſhall be double to
M E: But X α is double to α M: Wherefore ε E ſhall be triple E α. But
α E is triple E N: It is manifeſt, therefore, that E N is greater than
E X; and for that cauſe α, which is the Center of Gravity of the in
ſcribed Figure, cometh nearer to the Baſe of the Conoid than N. And
becauſe that as A E is to E N, ſo is the part taken away ε E to the part
taken away E α: and the remaining part ſhall be to the remaming part,
that is, A ε to N α, as A E to E N. Therefore α N is the third part of
A ε, and the ſixt part of A V. And in the ſame manner the Cylinders of
the circumſcribed Figure may be demonſtrated to be equally exceeding
one another, and the exceſs to me equal to the least; and that they have
their Centers of Gravity at equal diſtances in the Line ε M. If therefore
ε M be divided in π, ſo as that ε π be double to the remaining part π M;
π ſhall be the Center of Gravity of the whole circumſcribed Magnitude.
And ſince ε π is double to π M; and A ε leſs than double EM: (for
that they are equal:) the whole A E ſhall be leſs than triple E π: Where
fore E π ſhall be greater than E N. And, ſince ε M is triple to M π,
and M E with twice ε A is likewiſe triple to M E: the whole A E with
A ε ſhall be triple to E π: But A E is triple to E N: Wherefore the
remaining part A ε ſhall be triple to the remaining part π N. Therefore
N π is the ſixth part of A V. And theſe are the things that were to be
demonſtrated.
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