Solution to The October 16, 2005 PoW

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

2p(r + d) + 2p(r + 2d) + 2p(r + 3d) + ... + 2p(r + md) = n×l
where m is the number of circles (the number of times the paper goes around the roll). Since there are m terms on the left hand side of the above equation, and the sum of the first m integers is m(m + 1)/2 we have
2prm + 2pdm(m + 1)/2 = n×l
We also know that
md = R - r
since the thickness of the paper times the number of circles around equals the total thickness of the roll. So d = (R - r)/m. Putting this into the previous equation, we get
2prm + p(R-r)(m + 1)= n×l
=> .... 2prm + pRm - prm+ p(R-r) = n×l
=> .... p(R+r)m + p(R-r)= n×l
Putting m = (R - r)/d into the above we get
nl = p(R + r)(R - r )/d + p(R-r)
nl = p(R.² - r.²)/d + p(R - r)
And solving for d we get
d = p(R.² - r.²)/[nl - p(R - r)]

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.

d = p(2.25.² - 0.75.²)/[425×4 -p(2.25 - 0.75)] = 5.625p/[1700 - 1.5p] ~ 0.0104 inches
(It was two ply tissue.)

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.

4n = p(1.5.² - 0.75.²)/0.0104 + p(1.5 - 0.75) ~ 512.11
So n is about 128 sheets.

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.

4×223 = p(R.² - 0.75.²)/0.0104 + p(R - 0.75)
Using a numeric solver, we find R is about 1.872 inches. This leaves 1.122 inches from the inner radius (the initial amount was 1.5 inches from the inner radius).

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

md = R - r => m = (R - r)/d
we will have a new thickness level d2 given by
d2 = (R2 - r)/m = (R2 - r)/(R2 - rd
This is not too surprising. The new thickness (or average distance between sheets) has been reduced by the same proportion that the roll has been pinched. This would probably be a better estimate for the thickness. That is, first find m for the roll (or just d), then see how much you can pinch the roll to get a better determination of d2, the thickness of each sheet. So for example, if we pinch the above roll to a new radius R2 = 1.40625 inches (1 and 13/32 inches), the thickness of each sheet is given by
d2 = (R2 - r)/(R2 - rd = (1.40625 - 0.75)/(2.25 - 0.75)× 0.0104 = 0.00455