



 
Hi Stephen, The inside diameter of the cardboard core is 76 mm so the inside radius of the cardboard core is $\large \frac{76}{2} \normalsize = 38$ mm. If I understand what you mean by "the thickness of the wall of the cardboard core" then the outside radius of the cardboard core is 38 + 5 = 43 mm. Now look at our response to a previous question from Tuomas. Using the notation in that response you have $r = 43 \mbox{ mm }, R = \large \frac{320}{2} \normalsize = 160 \mbox{ mm },$ and since there are 1000 microns in a millimeter, $t = 0.138$ mm. Now you can use the expression \[t \times L = \pi \; \left( R^2  r^2\right)\] to find $L.$ Write back if you need more assistance. Harley 



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