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,
Penny