Wikipedia:Reference desk/Archives/Mathematics/2006 August 25

3D Geometry problem
I have a thick tube of radius R with an initially straight line drawn on its surface parallel to the axis. Now I bend, twist, extend and shear the tube arbitrarily. Assume that while bending plane sections remain plane, etc. How can I find the length of the line after the deformation? Is the following in the right direction?

$$ \mathbf {r_2}(s) = \mathbf {r_1}(s) + R \mathbf {n}(s) $$

Where r2 defines the line, r1 defines the tube centerline and n is the normal vector, Therefore,

$$ d\mathbf {r_2}(s) = d\mathbf {r_1}(s) + R d\mathbf {n}(s) $$

Now,

$$ \frac {ds'} {ds} = \sqrt { \frac {d\mathbf {r_2}(s)} {ds}. \frac {d\mathbf {r_2}(s)} {ds} } $$

$$ = \sqrt { \frac {d\mathbf {r_1}(s)} {ds}. \frac {d\mathbf {r_1}(s)} {ds} + 2R \frac {d\mathbf {r_1}(s)} {ds}. \frac {d\mathbf {n}(s)} {ds}+R^2\frac {d\mathbf {n}(s)} {ds}.\frac {d\mathbf {n}(s)} {ds}}$$

$$ = \sqrt { \frac {d\mathbf {r_1}(s)} {ds}. \frac {d\mathbf {r_1}(s)} {ds} +R^2\frac {d\mathbf {n}(s)} {ds}.\frac {d\mathbf {n}(s)} {ds}}$$

$$ = \sqrt { \left(\frac {dx(s)} {ds}\right)^2+\left(\frac {dy(s)} {ds}\right)^2+\left(\frac {dz(s)} {ds}\right)^2 +R^2\left(\left(\frac {dn_1(s)} {ds}\right)^2+\left(\frac {dn_2(s)} {ds}\right)^2+\left(\frac {dn_3(s)} {ds}\right)^2\right)}$$

Where x(s), y(s) and z(s) describe the deformed centerline and s is the (unstretched) arclength. Now I can integrate this expression wrt ds. Is this correct ? deeptrivia (talk) 19:38, 25 August 2006 (UTC)


 * I think the shearing makes your first equation, $$ \mathbf {r_2}(s) = \mathbf {r_1}(s) + R \mathbf {n}(s) $$, incorrect (if we wish n(s) to be a normal to r1(s)). I'm also not sure about cancelling $$2R \frac {d\mathbf {r_1}(s)} {ds} . \frac {d\mathbf {n}(s)} {ds}$$ - even if n was the normal, it would be iself orthogonal to the derivative of r1, not n's derivative. -- Meni Rosenfeld (talk) 19:55, 25 August 2006 (UTC)

Oh yes, that's clearly a blunder. I'm also concerned with the fact that the information regarding the location of the line on the tube surface is nowhere in the equations -- it matters whether the line is getting stretched or compressed (or sometimes compressed and sometimes stretched.) I'm talking about the relative orientation of the plane containing the line and axis with respected to the plane containing the curvature. Let's make it simpler: let's have no shear and torsion. Could you suggest a solution now? Thanks, deeptrivia (talk) 21:10, 25 August 2006 (UTC)


 * The problem as posed is underdefined. To each point on the central line there are many "normals" (vectors at a right angle). What happens under deformation depends on which one we have. Given the new curve for the original centreline, there are several non-equivalent ways of deforming the tube that result in that curve, even with the (hard to formalize) additional constraint of "no shear and torsion". In the general case, without that constraint, you must specify, next to how the directional vector of the centreline twists, also how the space around it "torques". --Lambiam Talk 01:49, 26 August 2006 (UTC)


 * I kind of get what you're saying. To get over this for the time being, let's assume the tube is just bent in a(n arbitrary) plane. Torsion of the curve defining the centerline will be zero in this case. Another case useful to me is the one with a constant torque (leading to a constant torsion.) The third is a general case with a varying torsion M3(s)/EI. Thanks a ton for your help! deeptrivia (talk) 02:29, 26 August 2006 (UTC)

This sounds like an Arc length problem to me. When you say "normal", in 3D, there is a plane that is perpendicular to a line... if you pick one vector on that plane as the normal, then there is a corresponding second vector also on that plane and perpendicular to them both called the "binormal", but I think you want the "tangent". If you have f(x), the continuous set of points beloning to the curve, f'(x) is its tangent (a vector in the direction). If you integrate the magnitude of this tangent, you then get the length of the curve, or Arc length. The only problem is, arc length tends to not have closed form solutions (i.e. Elliptic integral), so taking that integral will probably force you into an infinite series. - Rainwarrior 04:10, 26 August 2006 (UTC)


 * It's not that problem. Here, we have a straight tube deformed so that its centerline now becomes a known curve given by {x(s), y(s), z(s)}, where s is the arclength. We furthermore know the torsion as a function of s exactly. The question is to find the length of an initially straight axial line marked on the tube surface, which is at a distance R from the axis. deeptrivia (talk) 13:03, 26 August 2006 (UTC)


 * Ahh, okay, I see now that you are already doing the arc length (I had difficulty reading your notation at first). Actually, what you have looks fine, everything except for the sudden removal of the 2R term... where did it go? But, if you have a curve, and you have a function for a normal to that curve, you can certainly take the arc length of your new curve on the tube's surface. How well is it working out? - Rainwarrior 01:01, 28 August 2006 (UTC)


 * Assume your centreline curve is sufficiently smooth so that we can talk about its curvature, and that after torsion-less deformation it wiggles in one plane. Let parameter s be equal to the path length. Take a jointly deformed line, drawn on the tube and lying in that plane, with path-length parameter s'. If the curvature at s equals κ(s), we have ds ' /ds = 1 ± κ(s)R, where the sign depends whether you're on the inside or the outside of the bend. So "all" you have to do is integrate that with respect to s. For a line drawn elsewhere, with the vector pointing to it from the centreline at an angle φ with the third dimension, you have to integrate 1 + κ(s)Rsin φ. --Lambiam Talk 06:13, 28 August 2006 (UTC)