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I'm exploring the transformation of a 2D unit circle into a lemniscate (infinity symbol) by fixing two antipodal points and "twisting" the circle (in the 3rd dimension) such that the orthogonal pair of antipodal points meet at the origin. The total perimeter length $2\pi$ should be preserved during the transformation.

The unit circle in the $xy$-plane is described by:

$$ x = \cos(\theta), y = \sin(\theta), z = 0, $$

where $\theta \in [0, 2\pi)$.

I seek an exact mathematical formulation for this transformation. The primary challenge is in defining the function for the $z$-coordinate, $f(\theta)$, which should modify the curve's height without changing its length.

I would appreciate any advice, references, or ideas on how to approach this.

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  • $\begingroup$ The space curve has variable curvature and torsion with right angled self intersections at the origin, right? $\endgroup$
    – Narasimham
    Commented Dec 25, 2023 at 9:44
  • $\begingroup$ @Narasimham - The intersection point need not necessarily be at right angles $\endgroup$
    – swami
    Commented Dec 26, 2023 at 17:28

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Such a transformation can be written down fairly straightforwardly, except that the requirement to keep the length constant introduces a somewhat implicit element: $$ x=\cos \theta \ , \ y=\sin \theta \cos t \ , \ z=a(t) \sin \theta \cos \theta \sin t $$ where $\theta \in [0,2\pi ]$, and $t\in [0,\pi /2]$ is the deformation parameter; at $t=0$, one has the circle, at $t=\pi /2$, one has the lemniscate. The function $a(t)$ varies monotonically between $a(0)=1$ and $a(\pi/2)=1.061395$; it is defined by the requirement of constant length at any given $t$, $$ 2\pi =\int_{0}^{2\pi } d\theta \sqrt{1+(a^2 \cos^2 2\theta - \cos^2 \theta ) \sin^{2} t} $$ and can be easily determined numerically.

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    $\begingroup$ 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.) $\endgroup$
    – Steven Stadnicki
    Commented Dec 21, 2023 at 17:19
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    $\begingroup$ @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. $\endgroup$
    – Michael Engelhardt
    Commented Dec 21, 2023 at 18:20
  • $\begingroup$ 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? $\endgroup$
    – Narasimham
    Commented Dec 25, 2023 at 10:14
  • $\begingroup$ @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. $\endgroup$
    – Michael Engelhardt
    Commented Dec 25, 2023 at 16:08
  • $\begingroup$ @MichaelEngelhardt Is there a transformation that gives a Bernoulli Lemniscate? $\endgroup$
    – swami
    Commented Dec 26, 2023 at 10:09

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