Math CentralQuandaries & Queries


Question from Travis:

This question is probably close to the same question as "roll of paper"

We have Rolls of Window Film that we are trying to figure out an equation for a spreadsheet that we can use to "inventory" our window film.

We use a caliper tool to measure the thickness of the roll in millimeters.

the core thickness = 1.90mm
Full Roll thickness(including core) = 9.08mm to 9.12mm
Film thickness = 0.06

Full Roll of Film is supposed to average 1200" of film

What equation could we use to get the approximate inches left remaining on the roll if we measured the roll including the core with the Caliper tool in Millimeters?

Hi Travis,

You can use the expression from the "roll of paper" question which is that

\[t \times L = \pi\;\left(R^2 - r^2\right)\]

where $t$ is the thickness of the film, $L$ is the length of film on the roll, $R$ is the radius of the roll including the core and $r$ is the radius of the core. You didn't put units on the thickness but I assume it is 0.06 mm and hence for a full roll $(R = 9.12 \mbox{ mm})$ I get $R = \frac{9.12}{2} = 4.56$ and $r = \frac{1.90}{2} = 0.95$ both in mm so

\[L = \frac{\pi\;\left(R^2 - r^2\right)}{t} = \frac{\pi\;\left(4.56^2 - 0.95^2\right)}{0.06} = 1041 \mbox{ mm.}\]

You had 1200 inches. Did you mean 1200 mm?

This calculation is very sensitive to the thickness. If the thickness of the film is 0.05 mm the I get

\[L = \frac{\pi\;\left(R^2 - r^2\right)}{t} = \frac{\pi\;\left(4.56^2 - 0.95^2\right)}{0.05} = 1250 \mbox{ mm.}\]


About Math Central


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
Quandaries & Queries page Home page University of Regina PIMS