It's on an HP Designjet T2600, but this question applies to any wide format device printing onto roll paper.
This question was a dilemma for me, since the maximum roll length is only limited by the physical space in the spindle area ... in this specific case 5 7/8" (the diameter of the spindle end).
At first I went the intuitive route:
Paper_Roll_Size_Calculation.jpg
Let's say the outside of the 3" core is 3.3125" and the maximum roll diameter is 5.875". The
average diameter of every wrap of paper is 4.59375", or 14.43169" circumference. How many
average wraps of 20# bond (0.0038" caliper) are there?:
(5.875" - 3.3125")/0.0038"/2 = 337.17105 wraps
(14.43169" x 337.17105 wraps)/12" = 405.49'
Surprisingly accurate considering that I'm dealing in averages.
Then I found the formula on a sheet metal working website:
Paper_Roll_Dimensions.jpg
Length (feet) = (Π/4)*(ODē-IDē)/Caliper/12"
(3.14159265/4)*(5.875ē-3.3125ē)/0.0038/12 = 405.2901'
I'm a little rusty with my linear algebra, but I think that both of these formulas solve out to the same thing. The only difference is the rounding at various stages of the calculation. =^..^=
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