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
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