mlawrence2102 said:
Fascinating read... although totally bewildering to someone of my mathematical ability. Would it be possible to input this formula into an excel spreadsheet as a quick calculation tool?
EDIT: went through and converted things that were supposed to be diameters into their corresponding radii.
Well, I didn't quite finish the problem, so I don't think that that formula is particularly useful to you. I worked on it some more, and I figured out that you can eliminate both the paper thickness and the number of windings, and get a result that depends only upon the ratio of the radii of a partial reel to a full reel. Here's how you do it (I've used the subscript '0' or 'naught' to represent quantites in a full reel):
Let R_s = (3.5/2)" be the spool radius, n_0 be the number of labels in a full roll (which is given by the manufacturer to within some guaranteed error...we hope). Let R_0 = (8/2)" be the radius of a full reel, so that a full "ring" has a width of (\Delta R)_0 = R_0 - R_s = (4.5/2)" in this case. Let N_0 be the number of windings in a full roll. Using the formula I derived earlier, we can actually figure out both the paper thickness and the number of windings in full roll. First, the paper thickness:
t = \frac{(\Delta R)_0}{N_0} = \frac{2.25^{\prime\prime}}{N_0}
Using the formula I derived earlier, and applying it specifically to a full reel:
n_0 \ell = \sum_{i=0}^{N_0-1}2\pi(R_s + it) = 2\pi R_s \sum_{i=0}^{N_0-1}1 + 2\pi t \sum_{i=0}^{N_0-1} i
The nice thing is that both of these sums can be evaluated without actually manually adding up each of the 'N' terms. The answers are: \sum_{i=0}^{N_0-1}1 = N_0 and \sum_{i=0}^{N_0-1}i = (1/2)N_0(N_0 -1). Substituting in those answers, as well as the expression for t, the result is:
n_0 \ell = N_02\pi R_s + \pi \frac{(\Delta R)_0}{N_0} N_0(N_0 -1)
This equation can be rearranged to solve for N_0, the number of windings in a full roll, with the result that:
N_0 = \frac{n_0 \ell + \pi (\Delta R)_0}{2\pi R_s + \pi (\Delta R)_0}
So, now we know both the number of windings in a full reel, AND the paper thickness. A neat trick you can use is that it must always be true that the difference between the inner and outer radii of a label reel is equal to the number of windings times the paper thickness. This is true for both a full roll and a partial roll of radius R and number of windings N. Therefore:
t = \frac{(\Delta R)_0}{N_0} = \frac{\Delta R}{N}
which leads to the result that:
\frac{\Delta R}{(\Delta R)_0} = \frac{N}{N_0}
What this formula says is that if the radial width of the reel decreaes by a certain factor (compared to a full reel), then the number of windings will decrease by that same factor. For example, if the partial reel has decreased in width by a factor of 2 (so that is it now 1.125" instead of 2.25"), then the number of windings will also be half as many. Practically speaking, what this means is that you don't have to worry about the number of windings in a partial reel at all. All you have to do is measure the width. You'll see that below. We finish the problem by going back to the equation I derived for the length of a reel, only this time we apply it generally to some partial reel of radius R and number of windings N (so that the subscript 'naughts' are gone):
n \ell = N2\pi R_s + \pi \frac{(\Delta R)_0}{N_0} N(N -1)
Using the relationship between the number of windings in a partial reel vs. a full reel, we can rewrite this as:
n \ell = \frac{\Delta R}{(\Delta R)_0}N_0 2\pi R_s + \pi \Delta R \left(\frac{\Delta R}{(\Delta R)_0}N_0 -1\right)
n = \frac{1}{\ell} \left[\frac{\Delta R}{(\Delta R)_0}N_0 2\pi R_s + \pi \Delta R \left(\frac{\Delta R}{(\Delta R)_0}N_0 -1\right)\right]
This is the formula you would want to put into Excel.
(just teasing). Note also, that, you made a minor mistake. The diameter of the spool is 3.5", not 4.5". Oh well. It doesn't matter -- I used your value to check.
