How long of a paper roll can I load?

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  • blackcat4866
    Master Of The Obvious

    Site Contributor
    10,000+ Posts
    • Jul 2007
    • 22702
    #1

    How long of a paper roll can I load?

    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
    Last edited by blackcat4866; 05-09-2021, 06:10 PM.
    If you'd like a serious answer to your request:
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    blackcat: Master Of The Obvious =^..^=
    Tags: diameter, length, paper, roll, spindle
    • MoochPage
      Technician

      Site Contributor
      50+ Posts
      • Jul 2019
      • 57
      #2
      Re: How long of a paper roll can I load?

      Based on the specs for that machine that HP has put out it is 300 ft., or this to be more specific. I am personally not going to go through all the math I will go off of what they have put out in the specs for the machine.

        Comment

        • DRichard
          Trusted Tech

          250+ Posts
          • May 2008
          • 482
          #3
          Re: How long of a paper roll can I load?

          Originally posted by blackcat4866
          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:
          [ATTACH=CONFIG]49127[/ATTACH]
          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:
          [ATTACH=CONFIG]49128[/ATTACH]
          This is the best I can do...
          "Enjoy every sandwich."

          -- Warren Zevon

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