



 
Hi, A while ago we got a similar question from Tuomas and in my response to him I used the expression \[L \times t = \pi \left(R^2  r^2\right)\] where $R$ is the outside radius of the roll, $r$ is the radius of the core, $t$ is the thickness of the material being rolled and $L$ is the length of material on the roll. In the example you gave $R = 20$ inches, $r = 1.875$ inches, and $T = 0.002$ inches. Using these dimensions and the expression above I get \[L = \frac{\pi \left(20^2  1.875^2\right)}{0.002} = 622,796 \mbox{ inches.}\] If each impression is the same length then the length of an impression is \[\frac{622796}{2700} = 231 \mbox{ inches}\] which is about 19 feet. Does this seem correct? If so then you can use the same expression with $R = 3.25$ inches to calculate the length of material remaining on the roll and then the number of impressions. Let me know if you need any further assistance,  


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