Fig. 151.
Let us now return to our quadrant cell (OAPB), which we have found to be divided into a triangular and a quadrilateral portion, as in Fig. [147] or Fig. [151]; and let us now suppose the whole system to grow, in a uniform fashion, as a prelude to further subdivision. The whole quadrant, growing uniformly (or with equal radial increments), will still remain a quadrant, and it is obvious, therefore, that for every new increment of size, more will be added to the margin of its triangular portion than to the {368} narrower margin of its quadrilateral portion; and these increments will be in proportion to the angles of arc, viz. 55° 22′ : 34° 38′, or as ·96 : ·60, i.e. as 8 : 5. And accordingly, if we may assume (and the assumption is a very plausible one), that, just as the quadrant itself divided into two halves after it got to a certain size, so each of its two halves will reach the same size before again dividing, it is obvious that the triangular portion will be doubled in size, and therefore ready to divide, a considerable time before the quadrilateral part. To work out the problem in detail would lead us into troublesome mathematics; but if we simply assume that the increments are proportional to the increasing radii of the circle, we have the following equations:—
Let us call the triangular cell T, and the quadrilateral, Q (Fig. [151]); let the radius, OA, of the original quadrantal cell = a = 1; and let the increment which is required to add on a portion equal to T (such as PP′A′A) be called x, and let that required, similarly, for the doubling of Q be called x′.
Then we see that the area of the original quadrant
= T + Q = ¼ π a2 = ·7854a2 ,
while the area of T
= Q = ·3927a2 .
The area of the enlarged sector, p′OA′,
= (a + x)2 × (55° 22′) ÷ 2 = ·4831(a + x)2 ,
and the area OPA