1of the Circumſcribed and inſcribed Figures: and that it's poſſible, that

there be a Line in the middle betwixt thoſe Centers that is leſs than any

Line aſſigned; it followeth that the ſame given Line be much leſs that

lyeth betwixt one of the ſaid Centers and the ſaid point that divides

the Axis.

there be a Line in the middle betwixt thoſe Centers that is leſs than any

Line aſſigned; it followeth that the ſame given Line be much leſs that

lyeth betwixt one of the ſaid Centers and the ſaid point that divides

the Axis.

PROPOSITION.

The Center of Gravity divideth the Axis of any

Cone or Pyramid ſo, that the part next the

Vertex is triple to the remainder.

Cone or Pyramid ſo, that the part next the

Vertex is triple to the remainder.

Let there be a Cone whoſe Axis is A B. And in C let it be divided,

ſo that A C be triple to the remaining part C B. It is to be proved,

that C is the Center of Gravity of the Cone. For if it be not, the

Cone's Center ſhall be either above or below the point C. Let it be firſt

beneath, and let it be E. And draw the Line L P, by it ſelf, equal to

C E; which divided at pleaſure in N. And as both B E and P N to

gether are to P N, ſo let the Cone be to the Solid X: and inſcribe in the

Cone a Solid Figure of Cylinders that have equal Baſes, whoſe Center

of Gravity is leſs diſtant from the point C than is the Line L N, and

the exceſs of the Cone above it leſs than the Solid X. And that this

may be done is manifeſt from what hath been already demonſtrated.

Now let the inſcribed Figure be ſuch as

177[Figure 177]

was required, whoſe Center of Gravity

let be I. The Line I E therefore ſhall be

greater than N P together with L P. Let

C E and I C leſs L N be equal: And be

cauſe both together B E and N P is to N P

as the Cone to X: and the exceſs by which

the Cone exceeds the inſcribed Figure is

leſs than the Solid X: Therefore the Cone

ſhall have greater proportion to the ſaid

X S than both B E and N P to N P: and, by

Diviſion, the inſcribed Figure ſhall have

greater proportion to the exceſs by which

the Cone exceeds it, than B E to N P: But B E hath leſs proportion to

E I than to N P with I E. Let N P be greater. Then the inſcribed Fi

gure hath to the exceſs of the Cone above it much greater proportion

than B E to E I. Therefore as the inſcribed Figure is to the ſaid exceſs,

ſo ſhall a Line bigger than B E be to E I. Let that Line be M E. Becauſe,

therefore, M E is to E I as the inſcribed Figure is to the exceſs of the

Cone above the ſaid Figure, and D is the Center of Gravity of the

Cone, and I the Center of Gravity of the inſcribed Figure: Therefore

ſo that A C be triple to the remaining part C B. It is to be proved,

that C is the Center of Gravity of the Cone. For if it be not, the

Cone's Center ſhall be either above or below the point C. Let it be firſt

beneath, and let it be E. And draw the Line L P, by it ſelf, equal to

C E; which divided at pleaſure in N. And as both B E and P N to

gether are to P N, ſo let the Cone be to the Solid X: and inſcribe in the

Cone a Solid Figure of Cylinders that have equal Baſes, whoſe Center

of Gravity is leſs diſtant from the point C than is the Line L N, and

the exceſs of the Cone above it leſs than the Solid X. And that this

may be done is manifeſt from what hath been already demonſtrated.

Now let the inſcribed Figure be ſuch as

177[Figure 177]

was required, whoſe Center of Gravity

let be I. The Line I E therefore ſhall be

greater than N P together with L P. Let

C E and I C leſs L N be equal: And be

cauſe both together B E and N P is to N P

as the Cone to X: and the exceſs by which

the Cone exceeds the inſcribed Figure is

leſs than the Solid X: Therefore the Cone

ſhall have greater proportion to the ſaid

X S than both B E and N P to N P: and, by

Diviſion, the inſcribed Figure ſhall have

greater proportion to the exceſs by which

the Cone exceeds it, than B E to N P: But B E hath leſs proportion to

E I than to N P with I E. Let N P be greater. Then the inſcribed Fi

gure hath to the exceſs of the Cone above it much greater proportion

than B E to E I. Therefore as the inſcribed Figure is to the ſaid exceſs,

ſo ſhall a Line bigger than B E be to E I. Let that Line be M E. Becauſe,

therefore, M E is to E I as the inſcribed Figure is to the exceſs of the

Cone above the ſaid Figure, and D is the Center of Gravity of the

Cone, and I the Center of Gravity of the inſcribed Figure: Therefore