Circumference of a Tube
Date: 10/11/2001 at 14:29:20 From: Ernie Subject: Circumference How can I predict the circumference from a tubed length of stock? Maybe a different way to ask the same question would be, What effect does the thickness of a tube have on the circumference when it's formed into a circle?
Date: 10/11/2001 at 19:32:48 From: Doctor Douglas Subject: Re: Circumference Hi Ernie, and thanks for writing. If we bend a thin wire of negligible thickness with length L into a circle, the circumference of the formed circle is L and its radius (the distance between the center of the circle and the wire) is L/(2*pi). Say you cut a length L of tube from stock. Let's say that the stock has diameter D. What I mean by this is that the overall diameter of the tube (including the interior and the wall thickness) is D. Or to put it another way, its cross section has diameter D. Now if we bend this tube around into a circle, we expect that to a first approximation, the circumference of the formed circle is approximately equal to L. Now, as you bend the tube around into a circle, the material has to deform somewhat in order to make the ends join up. In particular, the part of the cross-section that is close to the center of the circle [i.e. at radius r < L/(2*pi)] is compressed, and the remainder of the material [r > L/(2*pi)] is stretched. What happens to the actual circumference depends on the shape of the cross section. In the case of a circular cross section, the compression and the stretching exactly cancel (to a first approximation), and the result is that along the axis of the tube, there is no change in length: in other words, the measured distance along the tube axis is still L, just as in the case of the wire. Now we can also see that the distance from the center of the object to its _outermost_ edge is L/(2*pi) + D/2. The circumference measured at its largest point is (2*pi) times this, or L + D*pi. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/
Date: 10/12/2001 at 13:21:03 From: Ernie Subject: Re: Circumference I think I understand most of your answer. For example, if tubed stock (rubber) = 36" in length (L) and 3" in width (D)... L/(2*pi)+D/2 = 7.2297" center to outside edge (r) Cir = (2*pi)(7.2297) Cir = 6.2832*7.2297 Cir = 45.43"
Date: 10/12/2001 at 14:23:51 From: Doctor Douglas Subject: Re: Circumference Yes, your formula is absolutely correct. C = 2*pi*[L/(2*pi) + D/2] = L + D*pi For the values in your problem, L = 36.0, D = 3.0, so C = 36.0 + 3.0*pi = 36 + 3*3.1416 = 45.4 inches, which agrees with what you calculated above. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/
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