Reply #124: Here's what I settled on as a solution to (3) [View All]

[BrewerJohn]

The first figure shows how I divided up
half the total area into three regions: region 1 is just a quarter-circle of radius 100 ft,
region 2 is the area over which I integrate the involute, and region 3 is the little sliver
left over when I stop the integration.


The second figure shows how to do the integral, labeling the radius of the silo r and the
length of the rope L. I use the formula that the area of a wedge
of a circle is (1/2)r^2 times the included angle. The infinitesimal area is not quite a
circular wedge, but it only differs from one by a factor of d(theta). Since the area is already
proportional to d(theta), and second-order infinitesimals always vanish in the limit, this
should be OK. I integrate this dA from theta = 0 to theta = theta_m, where theta_m is such
that the rope left over just touches the dividing line between north and south.


In the third figure I show a closeup of region 3. This area can be calculated by taking
the difference between the area of the triangle (known from simple geometry) and the
area of the circular wedge.

The only really sticky part is that theta_m can't be solved for analytically. In equation
form, (L/r) - theta_m = tan(theta_m). This has to be solved either numerically or graphically.