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Old 03-08-2019, 11:11 AM
RCBrust RCBrust is offline
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Question regarding how paper bends

I have a question about bending paper/card stock into cylinders, and dimensions. Say I want a 1/2" diameter cylinder, so I cut a piece of stock to length 0.5 * Pi and roll it into a cylinder with a butt joint. What ends up having a dimension of 1/2"? The cylinders ID, OD, or the center of one side to the center of the other side?

To get an accurate ID, do you typically have to add a paper thickness to the desired ID before multiplying by Pi?

Thanks,
Randy
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Old 03-08-2019, 12:17 PM
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Rubenandres77 Rubenandres77 is offline
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If it is a thin paper, the difference is negligible.
When using cardstock, then thickness starts to be an issue.
In your example, the end result will be: 1/2" -> ID.
So yes, card thickness needs to be considered if you need the 1/2" to be the OD.
But even using 0.22mm thick cardstock, the difference is almost unnoticeable for a 1/2" diameter cylinder.
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Old 03-08-2019, 12:22 PM
RCBrust RCBrust is offline
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Thanks for the reply. So basically the cardstock only stretches and doesn't compress.

Randy
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Old 03-08-2019, 12:27 PM
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The other thing that can be helpful in getting a smooth curve is to roll with the grain of the paper.
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Old 03-08-2019, 12:28 PM
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the only way cardstock compresses is into an Origami Boulder
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Old 03-08-2019, 12:40 PM
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Rick, think we've all made a few of those.
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Old 03-08-2019, 02:14 PM
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airdave airdave is offline
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You are using an incomplete forumla for your calcualtion.

Pi R squared is a formula for calculating the circumference of a circle...a 2D object.

Calculating the inside and outside diameters of a tube are different calculations based on 3D objects
and you must know the material/wall thickness of the "tube" to make this calculation.
And I.D and O.D calculations have nothing to do with the length of the sheet you used to make the tube.

And...paper/card thickness is not a negligible thing
...in fact its an important consideration when designing a paper model.

I use it all the time, on every part...factoring in an average tolerance of .25mm for material thickness
and .5mm as the inside and outside difference of tubes and circular fit.

See?...I have predetermined the wall thickness of my "tube"
and, I have predetermined a desired measurement for the inside or outside diameter of the "tube".
Two things you failed to do.
Once you do those, your question is uneccessary.
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Old 03-08-2019, 05:26 PM
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pi.r ˛ has long been, and currently remains, the formula for calculating the area of a 2D circular disc.
pi.d, or 2.pi.r, is the correct formula for calculating the circumference of a 2D circle or the minimum circumference of a right circular cylinder in 3D.
For a right circular cylinder with wall thickness of say 0.25mm the outside circumference will be 2.pi.R and the inside circumference will be 2.pi.r where r = (R - 0.25).
Both paper and card don't particularly like either tension or compression and it's difficult to know where the axis of bending will occur within the thickness of the material.
So theoretical calculations will only find the ball park and a bit of fumble will likely be needed with the actual material being used.
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Old 03-08-2019, 06:04 PM
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yes, I mistyped my first sentence...pi x radiusx2 or pi(r/2) is of course the circumference.
same as pi X diameter

I do the same thing when speaking...I always say "squared" when I know its "doubled".
my bad.
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Old 03-08-2019, 06:38 PM
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I can agree you mis-typed your first sentence but the second as well? Really?
See?...I predetermined the correct formula to use.
One of the things you failed to do.
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