



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


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