The Essential Galileo (55 page)

Proposition 3: Consider now the case of a prism or cylinder lying horizontal and growing longer in a horizontal direction.We must find out in what ratio the moment of its own weight increases in comparison
[159]
with its resistance to fracture.This moment I find increases in proportion to the square of the length.

In order to prove this, let
AD
be a prism or cylinder lying horizontal with its end
A
firmly fixed in a wall. Let the length of the prism be increased by the addition of the portion
BE.
It is clear that merely changing the length of the lever from
AB
to
AC
will, if we disregard its weight, increase the moment of the force tending to produce fracture at
A
in the ratio of
CA
to
BA.
But, besides this, the weight of the solid portion
BE
, added to the weight of the solid
AB
, increases the moment of the total weight in the ratio of the weight of the prism
AE
to that of the prism
AB
, which is the same as the ratio of the length
AC
to
AB.
It follows, therefore, that, when the length and weight are simultaneously increased in any given proportion, the moment, which is the product of these two, is increased in a ratio that is the square of the preceding proportion. The conclusion is then that the bending moments due to the weight of prisms and cylinders that have the same thickness but different lengths bear to each other a ratio that is the square of the ratio of their lengths, or, what is the same thing, the ratio of the squares of their lengths.

We shall next show in what ratio the resistance to fracture in prisms and cylinders increases with increasing thickness while [160] the length remains unchanged. Here I say that (
Proposition 4
):
In prisms and cylinders of equal length but of unequal thicknesses, the resistance to fracture increases in the same ratio as the cube of the diameter of the thickness, i.e., of the base.

Let
A
and
B
be two cylinders of equal lengths
DG, FH;
let their bases be unequal, namely, the circles with the diameters
CD, EF.
Then I say that the resistance to fracture offered by the cylinder
B
is to that offered by
A
as the cube of the diameter
EF
is to the cube of the diameter
CD.
For if we consider the resistance to fracture by longitudinal pull as dependent upon the bases, i.e., upon the circles
EF
and
CD
, no one can doubt that the resistance of the cylinder
B
is greater than that of
A
in the same proportion in which the area of the circle
EF
exceeds that of
CD;
this is so because it is precisely in this ratio that the number of fibers binding the parts of the solid together in the one cylinder exceeds that in the other cylinder. But in the case of a force acting transversely, it must be remembered that we are employing two levers in which the forces are applied at distances
DG
and
FH
, and the fulcrums are located at the points
D
and
F;
and the resistances act at distances that are equal to the radii of the circles
CD
and
EF
, since the fibers distributed over these entire cross sections act as if concentrated at the centers. Remembering this and remembering also that the arms,
DG
and
FH
, through which the forces
G
and
H
act are equal, we can understand that the resistance located at the center of the base
EF
and acting against the force H is greater than the resistance at the center of the base
CD
opposing the force
G
in the ratio of the radius
EF
to the radius
CD
. Accordingly, the resistance to fracture offered by the cylinder
B
is greater than that of the cylinder
A
in a ratio which is compounded of that of the area of the circles
EF
and
CD
and that of their radii, or of their diameters. But the areas of circles are as the squares of their diameters. Therefore, the ratio of the resistances, being the product of the two preceding ratios, is the same as that of the cubes [161] of the diameters. This is what I set out to prove.

Moreover, since the volume of a cube varies as the third power of its edge, we may say that the resistance of a cylinder whose length remains constant varies as the third power of its diameter. So from the preceding we are also able to derive a
Corollary: The resistance of a prism or cylinder of constant length varies as the three-halves power of its volume or weight
. This is evident as follows. The volume of a prism or cylinder of constant altitude varies directly as the area of its base, i.e., as the square of a side or diameter of this base. But as just demonstrated, the resistance varies as the cube of this same side or diameter. Hence, the resistance varies as the three-halves power of the volume—and consequently also of the weight—of the solid itself.

S
IMP.
    Before proceeding further I should like to have one of my difficulties removed. Up to this point you have not taken into consideration a certain other kind of resistance that, it appears to me, diminishes as the solid grows longer, and this is quite as true in the case of pulling as of bending. For example, in the case of a rope we observe that a very long one is less able to support a large weight than a short one. Thus, I believe, a short rod of wood or iron will support a greater weight than if it were long, provided that the force be always applied longitudinally and not transversely, and provided also that we take into account its own weight, which increases with its length.

S
ALV.
    I fear, Simplicio, that in this particular you are making the same mistake as many others, if I correctly catch your meaning; that is, if you mean to say that a long rope, one of perhaps forty cubits, cannot hold up so great a weight as a shorter length, say one or two cubits, of the same rope.

S
IMP.
    That is what I meant, and as far as I see the proposition is highly probable.

S
ALV.
    On the contrary, I consider it not merely improbable but false; and I think I can easily convince you of your error. Let
AB
represent the rope, fastened at the upper end
A.
At the lower end, attach a weight
C
whose force is just sufficient to break the rope. Now, Simplicio, point out the exact place where you think the break ought to occur.

[162] S
IMP.
    Let us say
D
.

S
ALV.
    And why at
D?

S
IMP.
    Because at this point the rope is not strong enough to support, say, one hundred pounds, made up of the portion of the rope
DB
and the stone
C.

S
ALV.
    Accordingly, whenever the rope is stretched with the weight of one hundred pounds at
D
, it will break there.

S
IMP.
    I think so.

S
ALV.
    But tell me, if instead of attaching the weight at the end of the rope,
B
, one fastens it at a point nearer
D
, say, at
E;
or if instead of fixing the upper end of the rope at
A
, one fastens it at some point
F
, just above
D;
will not the rope, at the point
D
, be subject to the same pull of one hundred pounds?

S
IMP.
    It would, provided you include with the stone
C
the portion of rope
EB
.

S
ALV.
    Let us therefore suppose that the rope is stretched at the point
D
with a weight of one hundred pounds. Then, according to your own admission, it will break. But
FE
is only a small portion of
AB
. How can you therefore maintain that the long rope is weaker than the short one? Give up then this erroneous view which you share with many very intelligent people, and let us proceed.

We have already demonstrated that in the case of prisms and cylinders of constant thickness, the moment of force tending to produce fracture varies as the square of the length; and likewise we have shown that when the length is constant and the thickness varies, the resistance to fracture varies as the cube of the side or diameter of the base; so let us go on to the investigation of the case of solids that simultaneously vary in both length and thickness. Here I formulate
Proposition 5: Prisms and cylinders that differ in both length and thickness offer resistances to fracture that are directly proportional
[163]
to the cubes of the diameters of their bases and inversely proportional to their lengths.

Let
ABC
and
DEF
be two such cylinders; then the resistance of the cylinder
AC
bears to the resistance of the cylinder
DF
a ratio that is the product of the cube of the diameter
AB
divided by the cube of the diameter
DE
, and the length
EF
divided by the length
BC
. Make
EG
equal to
BC;
let
H
be the third proportional to the lines
AB
and
DE;
let
I
be the fourth proportional; and let
I
be to
S
as
EF
is to
BC
.

Now, since the resistance of the cylinder
AC
is to that of the cylinder
DG
as the cube of
AB
is to the cube of
DE
, that is, as the length
AB
is to the length
I;
and since the resistance of the cylinder
DG
is to that of the cylinder
DF
as the length
FE
is to
EG
, that is, as
I
is to
S;
it follows, by equidistance of ratios, that the resistance of the cylinder
AC
is to that of the cylinder
DF
as the length
AB
is to
S.
But the line
AB
bears to
S
a ratio that is the product of
AB/I
and
I/S
. Hence the resistance of the cylinder
AC
bears to the resistance of the cylinder
DF
a ratio that is the product of
AB/I
(that is, the cube of
AB
to the cube of
DE
) and
I/S
(that is, the length
EF
to the length
BC
). This is what I meant to prove.