Dropping a shoe...

[alexpgp]

From time to time, someone will come into the store and ask about buying bubble wrap in lengths longer than the six-foot sections we have out on sale (we create such sections by cutting appropriate lengths from what starts out as a 125-foot bulk roll). Often, said someone will spy some part such a roll and ask: "Say, how much do you want for that?"

It's hard to tell just by looking how much bubble wrap is left, and if you assume that a roll that is half the diameter of a full roll contains half the amount of bubble wrap, well... you're going to be dramatically wrong.

So the question is: how can you get a good estimate of how much of a product there is in a package where the goods are wrapped around a core? Would you believe the little math exercise illustrated in the previous post helps? Follow:

First, we must recall the formula relating the diameter of a circle (d) to its circumference (c): c = πd. Since the diameter is just twice the radius, the formula can be rewritten c = π2r (or, more conventionally, c = 2πr) and that's where we'll start.

To make things simple, imagine that, instead of forming a spiral around the core, the product forms a set of concentric circles. Now what remains is to add up the individual circumferences of all these circles.

The innermost circle of product has a length 2πr, where r is the radius of that innermost circle. The next circle out has a length 2π(r+t), where r+t is the radius of that next circle, and is equal to the radius of the first plus the thickness of the product. The third circle from the core has a length 2π(r+2t), where r+2t is the radius of the third circle (which again, is the radius of the previous circle, r+t, plus another thickness of product, t).

The length of the fourth circle?  2π(r+3t)
Fifth? 2π(r+4t)
Sixth? 2π(r+5t)

See the pattern?

For the n-th circle, the length is going to be 2π(r+(n-1)t).

So, to figure out the overall length of n concentric circles whose radius increases by thickness t in each ring, we add up an equation that looks like:

2π(r)+2π(r+t)+2π(r+2t)+...+2π(r+(n-1)t)

rearranging, we get:

n x 2π + 2π x (t+2t+...+(n-1)t)

after one more simplification, we have:

n x 2π + 2π x (1+2+...+(n-1))t

Does that series look familiar? (The sum of digits between 1 and n-1? It's (n-1)(n)/2.) So we can rewrite the equation as:

n x 2π + 2π x ((n-1)(n)/2)t

So if the bubble wrap is 1/2-inch thick, if the radius of the cardboard core is 2 inches, and there are 14 layers of product, how much bubble wrap is on the roll? Substituting, we get

14x2x3.14 + 2x3.14x((13)(14)/2)x0.5 = 87.92 + 45.5 = 133.42 inches or a little over 11 feet.

Remember what I said at the beginning about guessing how much might be in a roll having twice the diameter?

Let's calculate how much product is in a roll with 28 layers of bubble wrap (14 as in the first example + 14 to double that number, which will come to a little less than double the diameter, since we're not accounting for the core):

28x2x3.14 + 2x3.14x((27)(28)/2)x0.5 = 175.84 + 189.00 = 364.84 inches or a little over 30 feet.

Surprise! A roll twice the diameter contains about three times as much product!

Simplifying the above formula gives us the formula:

   Length on roll = 6.28 x n + 3.14 x (n-1)(n)/2

So now it's easy to determine the length of bubble wrap on the roll: count the number of layers, plug that number in instead of n, and do the calculation.

The result is, BTW, very close to the real length.

Cheers...

Comments

[bandicoot.livejournal.com]

Given that the result is close enough to the actual length, wouldn't it be faster and easier to modify the equation so that you simply measured the OD of the roll and plugged that in?

[alexpgp.livejournal.com]

Yes, it would, although if we want to shoot for excruciating exactness, there will be two formulas: one for the large-bubble product (~0.5-in thick) and one for the small-bubble product (~0.25-in thick).

I left the example where it was to keep the exposition (relatively) simple. Thanks for your note.

Cheers...