Here's an easy problem. If 500 sheets of 20 lb. paper make a stack 2 inches high, we can easily get a value for the average thickness (0.004 inches) of a piece of paper. (Here thickness is the average distance from the top of one piece of paper to the top of the next piece of paper.)
Here's a harder problem. In a similar manner we wish to find the thickness of a sheet of toilet paper. In general, we have a roll of toilet paper with n sheets, each of length l, an inner radius r where the sheets start, and an outer radius R where the roll ends.
1) Find the average thickness, d, of a sheet of the toilet paper in terms of R, r, n, and l.
Note that we can get the thickness d of any roll of paper in terms of R, r, and L, where L is the total length of the roll if it were unrolled. (Obviously L = n×l.)
Essentially we have a large number of concentric circles, until the length runs out. That is, at first the (tissue) paper goes around the cardboard tube with length r. So far, we say that 2p(r + d) has been wrapped around the roll. (For simplicity, we use the outer radius of the paper to compute the length of the first circle, and each circle hereafter. That is, the distance from the center of the roll to the outside edge of the paper.) This process continues until we run out of paper, i.e., until
2) Find the average thickness of a sheet from a roll with stated values: n = 425 sheets, l = 4 inches, and measured values R = 2.25 inches and r = 0.75 inches.
3) For the above roll in (2), find the value of n when R is 1.5 inches (the outer radius in now 0.75 inches larger than r, but it started out as 1.5 inches larger than r). That is, use the d value from (2) to find the new n for the new R value.
4) For the above roll in (2), find the value of R when n is 223 (about half what it started out as). That is, use the d value from (2) to find the new R for the new n value.
Many simplifications were made in the above work, but none will affect the answers to any real degree. (Famous last words!)
Also note that if the roll is pinched so that there is a new outer radius is R2, then since