Timeline for Formulating a 3D "twist" transformation for a unit circle into a lemniscate while preserving arc length
Current License: CC BY-SA 4.0
9 events
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| Dec 26, 2023 at 14:38 | comment | added | Michael Engelhardt | @swami - sure, you can do the same thing with a Bernoulli lemniscate, just all the expressions become longer, because you have an additional $\theta $-dependent denominator in the parametrization. | |
| Dec 26, 2023 at 10:10 | comment | added | Narasimham | Ah, I see, ok. The problem is in mechanics of materials I suppose in shape finding of an Elastic ring ( maybe of rectangular section, moments of inertia ratio of $I_x/I_y$ ) when antipodal points are squeezed together. Static equilibrium, moment/curvature & torque/torsion relations, material elastic constants formulate an ode, which when integrated define its bending/twisting elastic deformations. | |
| Dec 26, 2023 at 10:09 | comment | added | swami | @MichaelEngelhardt Is there a transformation that gives a Bernoulli Lemniscate? | |
| Dec 25, 2023 at 16:08 | comment | added | Michael Engelhardt | @Narasimham - well, at the intersection point, the curvature vanishes and the direction of the binormal vector flips, for both of the line segments present there. Together, these are equal in magnitude, opposite in sign effects. In the rest of the figure, indeed, there trivially can be no torsion. The particular lemniscate I parametrized here is just a Lissajous figure - apparently also known as the lemniscate of Gerono. | |
| Dec 25, 2023 at 10:14 | comment | added | Narasimham | If the space curve is entirely in the $xy$ plane then it has no torsion anywhere , has curvature only, so is like a Bernoulli Lemniscate... ( length to be adjusted $2\pi$ ) right? | |
| Dec 22, 2023 at 19:41 | history | edited | Michael Engelhardt | CC BY-SA 4.0 |
added 3 characters in body
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| Dec 21, 2023 at 18:20 | comment | added | Michael Engelhardt | @StevenStadnicki - I chose the most symmetric arrangement, which indeed places the final curve into the $xz$ plane. From what I can see, the OP expressed no preference in this respect. Certainly, one can straightforwardly compose this with a rotation about the $x$ axis such as to place the final curve back in the $xy$ plane. | |
| Dec 21, 2023 at 17:19 | comment | added | Steven Stadnicki | This puts the lemniscate orthogonal to the circle — in the $xz$ plane, as opposed to the $xy$ plane the original circle is in. I think OP is looking for a $180^\circ$ twist. (Though I suppose composing this with a rotation of $t$ about the x axis will do it.) | |
| Dec 21, 2023 at 15:47 | history | answered | Michael Engelhardt | CC BY-SA 4.0 |