PROP. XIII. PROBL. IV.
There being given the greateſt Weight that can be ſup
ported at the middle of a Cylinder or Priſme, where
the Reſiſtance is leafl; and there being given a
Weight greater than that, to find in the ſaid Cylin
der, the point at which the given greater Weight may
be ſupporited as the greateſt Weight.
ported at the middle of a Cylinder or Priſme, where
the Reſiſtance is leafl; and there being given a
Weight greater than that, to find in the ſaid Cylin
der, the point at which the given greater Weight may
be ſupporited as the greateſt Weight.
Let the given weight greater than the greateſt weight that
can be ſupported at the middle of the Cylinder A B, have
unto the ſaid greateſt weight, the proportion of the line E
to F: it is required to find the point in the Cylinder at which the
ſaid given weight commeth to be ſupported as the biggeſt. Be
tween E and F let G be a Mean-Proportional; and as E is to G,
ſo let A D be to S, S ſhall be leſſer than A D. Let A D be the
Diameter of the Semicircle A H D: in which ſuppoſe A H equal
to S; and joyn together H and D, and take D R equal to it.
I ſay that R is the point ſought, at which the given weight,
greater than the greateſt that can be ſupported at the middle of the
Cylinder D, would become as the greateſt weight. On the length
BA deſcribe the Semicircle A N B, and raiſe the Perpendicular
RN, and conjoyn N and
D: And becauſe the two
67[Figure 67]
Squares N R and R D are
equal to the Square N D;
that is, to the Square A D;
that is, to the two A H and
and H D; and H D is equal
to the Square D R: There
fore the Square N R, that
is, the Rectangle A R B
ſhall be equal to the Square A H; that is, to the Square S: But
the Square S is to the Square A D, as F to E; that is, as the
greateſt ſupportable Weight at D to the given greater Weight:
Therefore this greater ſhall be ſupported at R, as the greateſt
that can be there ſuſtained. Which is that that we ſought.
can be ſupported at the middle of the Cylinder A B, have
unto the ſaid greateſt weight, the proportion of the line E
to F: it is required to find the point in the Cylinder at which the
ſaid given weight commeth to be ſupported as the biggeſt. Be
tween E and F let G be a Mean-Proportional; and as E is to G,
ſo let A D be to S, S ſhall be leſſer than A D. Let A D be the
Diameter of the Semicircle A H D: in which ſuppoſe A H equal
to S; and joyn together H and D, and take D R equal to it.
I ſay that R is the point ſought, at which the given weight,
greater than the greateſt that can be ſupported at the middle of the
Cylinder D, would become as the greateſt weight. On the length
BA deſcribe the Semicircle A N B, and raiſe the Perpendicular
RN, and conjoyn N and
D: And becauſe the two
67[Figure 67]
Squares N R and R D are
equal to the Square N D;
that is, to the Square A D;
that is, to the two A H and
and H D; and H D is equal
to the Square D R: There
fore the Square N R, that
is, the Rectangle A R B
ſhall be equal to the Square A H; that is, to the Square S: But
the Square S is to the Square A D, as F to E; that is, as the
greateſt ſupportable Weight at D to the given greater Weight:
Therefore this greater ſhall be ſupported at R, as the greateſt
that can be there ſuſtained. Which is that that we ſought.
SAGR. I underſtand you very well, and am conſidering that
the Priſme A B having alwayes more ſtrength and reſiſtance a
gainſt Preſſion in the parts that more and more recede from the
middle, whether in very great and heavy Beams one may take
the Priſme A B having alwayes more ſtrength and reſiſtance a
gainſt Preſſion in the parts that more and more recede from the
middle, whether in very great and heavy Beams one may take