en it will
describe an arc of a circle, and the rod will generate an area 1/2l squared[theta],
where [theta] is the angle AQB through which the rod has turned. The wheel
will roll over an arc c[theta], where c is the distance of the wheel from
Q. The "roll" is now w = c[theta]; hence the area generated is

P = 1/2 l squared/c w,

and is again determined by w.

[Illustration: FIG. 8.]

Next let the rod be moved parallel to itself, but in a direction not
perpendicular to itself (fig. 8). The wheel will now not simply roll.
Consider a _small_ motion of the rod from QT to Q'T'. This may be resolved
into the motion to RR' perpendicular to the rod, whereby the rectangle
QTR'R is generated, and the sliding of the rod along itself from RR' to
Q'T'. During this second step no area will be generated. During the first
step the roll of the wheel will be QR, whilst during the second step there
will be no roll at all. The roll of the wheel will therefore measure the
area of the rectangle which equals the parallelogram QTT'Q'. If the whole
motion of the rod be considered as made up of a very great number of small
steps, each resolved as stated, it will be seen that the roll again
measures the area generated. But it has to be noticed that now the wheel
does not only roll, but also slips, over the paper. This, as will be
pointed out later, may introduce an error in the reading.

[Illustration: FIG. 9.]

We can now investigate the most general motion of the rod. We again resolve
the motion into a number of small steps. Let (fig. 9) AB be one position,
CD the next after a step so small that the arcs AC and BD over which the
ends have passed may be considered as straight lines. The area generated is
ABDC. This motion we resolve into a step from AB to CB', parallel to AB and
a turning about C from CB' to CD, steps such as have been investigated.
During the first, the "roll" will be p the altitude of the parallelogram;
during the second will be c[theta]. Therefore

w = p + c[theta].

[Illustration: FIG. 10.]

The area generated is lp + 1/2 l^2[theta], or, expressing p in terms of w, lw
+ (1/2l^2 - lc)[theta]. For a finite motion we get the area equal to the sum
of the areas generated during the different steps. But the wheel will
continue rolling, and give the whole roll as the sum of the rolls for the
successive steps. Let then w denote the whole roll (in fig. 10), and let
[alpha] denote the sum of all the small turnings [thet