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