From a Cone or Pyramid whoſe Axis is A D, and equidiſtant to

the Plane of the Baſe, let a Fruſtum be cut whoſe Axis is V D.

And as the triple of the greateſt Baſe with the double of the

mean and leaſt is to the triple of the leaſt and double of the mean and

greateſt, ſo is \ O to O D. It is to be proved that the Center of Gra

vity of the Fruſtum is in O. Let V M be the fourth part of V D.

Set the Line H X by the by, equal to A D: and let K X be equal to A V:

and unto H X K let X L be a third proportional, and X S a fourth.

And as H S is to S X, ſo let M D be to the Line taken from O towards

A: which let be O N. And becauſe the greater Baſe is in proportion

to that which is mean betwixt the

greater and leſſer as D A to A V; that

179[Figure 179]

is, as H X, to X K, but the ſaid

mean is to the leaſt as K X to X L;

the greater, mean, and leſſer Baſes

ſhall be in the ſame proportion as

H X, X K, and X L. Wherefore as

triple the greater Baſe, with double

the mean and leſſer, is to triple the

leaſt with double the mean and grea

teſt; that is, as V O is to O D; ſo is

triple H X with double X K and X L

to triple X L, with double X K and

X H: And by Compoſition and Converting the proportion, O D ſhall

be to V D, as H X, with double X K and triple X L, to quadruple H X,

X K, and X L. There are, therefore, four proportional Lines, H X,

X K, X L, and X S: And as X S is to S H, ſo is the Line taken N O

to 3/4 of D V, to wit, to D M; that is, to 3/4 of H K: And as H X

with double X K and triple X L is to quadruple H X, X K and X L;

ſo is another Line taken O D to D V; that is, to H K. Therefore, by

the things demonſtrated, D N ſhall be the fourth part of H X; that

is, of A D. Wherefore the point N ſhall be the Center of Gravity

of the Cone or Pyramid whoſe Axis is A D. Let the Center of Gra

vity of the Pyramid or Cone whoſe Axis is A V be I. It is therefore

manifeſt that the Center of Gravity of the Fruſtum is in the Line

I N inclining towards the part N, and in that point of it which with

the point N include a Line to which I M hath the ſame proportion that

the Fruſtum cut hath to the Pyramid or Cone whoſe Axis is A V.

It remaineth therefore to prove that I N hath the ſame proportion

to N O, that the Fruſtum hath to the Cone whoſe Axis is A V. But

as the Cone whoſe Axis is D A is to the Cone whoſe Axis is A V, ſo

is the Cube D A to the Cube D V; that is, the Cube H X to the

Cube X K: But this is the ſame proportion that H X hath to X S.

Wherefore, by Diviſion, as H S is to S X, ſo ſhall the Fruſtum whoſe

the Plane of the Baſe, let a Fruſtum be cut whoſe Axis is V D.

And as the triple of the greateſt Baſe with the double of the

mean and leaſt is to the triple of the leaſt and double of the mean and

greateſt, ſo is \ O to O D. It is to be proved that the Center of Gra

vity of the Fruſtum is in O. Let V M be the fourth part of V D.

Set the Line H X by the by, equal to A D: and let K X be equal to A V:

and unto H X K let X L be a third proportional, and X S a fourth.

And as H S is to S X, ſo let M D be to the Line taken from O towards

A: which let be O N. And becauſe the greater Baſe is in proportion

to that which is mean betwixt the

greater and leſſer as D A to A V; that

179[Figure 179]

is, as H X, to X K, but the ſaid

mean is to the leaſt as K X to X L;

the greater, mean, and leſſer Baſes

ſhall be in the ſame proportion as

H X, X K, and X L. Wherefore as

triple the greater Baſe, with double

the mean and leſſer, is to triple the

leaſt with double the mean and grea

teſt; that is, as V O is to O D; ſo is

triple H X with double X K and X L

to triple X L, with double X K and

X H: And by Compoſition and Converting the proportion, O D ſhall

be to V D, as H X, with double X K and triple X L, to quadruple H X,

X K, and X L. There are, therefore, four proportional Lines, H X,

X K, X L, and X S: And as X S is to S H, ſo is the Line taken N O

to 3/4 of D V, to wit, to D M; that is, to 3/4 of H K: And as H X

with double X K and triple X L is to quadruple H X, X K and X L;

ſo is another Line taken O D to D V; that is, to H K. Therefore, by

the things demonſtrated, D N ſhall be the fourth part of H X; that

is, of A D. Wherefore the point N ſhall be the Center of Gravity

of the Cone or Pyramid whoſe Axis is A D. Let the Center of Gra

vity of the Pyramid or Cone whoſe Axis is A V be I. It is therefore

manifeſt that the Center of Gravity of the Fruſtum is in the Line

I N inclining towards the part N, and in that point of it which with

the point N include a Line to which I M hath the ſame proportion that

the Fruſtum cut hath to the Pyramid or Cone whoſe Axis is A V.

It remaineth therefore to prove that I N hath the ſame proportion

to N O, that the Fruſtum hath to the Cone whoſe Axis is A V. But

as the Cone whoſe Axis is D A is to the Cone whoſe Axis is A V, ſo

is the Cube D A to the Cube D V; that is, the Cube H X to the

Cube X K: But this is the ſame proportion that H X hath to X S.

Wherefore, by Diviſion, as H S is to S X, ſo ſhall the Fruſtum whoſe