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Torus and inversion yielding a given cyclide

torus intersectionTorus, maps and metricsAlgebraic $n$-torus and topological $n$-torusRectilinear polygons winding around a torusA set in the plane mapping elements with an inversionInversion problem: Given 2 intersecting circles, orthogonal to a third, prove that the points of intersection and center of the third are collinear.What's the name of the curves obtained by intersecting a n-holed torus and a plane?Equation of torus: stucked in removing one variableEquation of an egg figure created by intersecting a torus with a planeQuestion about inversion images

0

$begingroup$

Let $mathcalC$ be a ring cyclide with symmetry plane $z=0$ and parameters $a$, $c$, $mu$ as in this picture:

How to find a torus and an inversion such that $mathcalC$ is the image of this torus by this inversion?

geometry

share|cite|improve this question

edited Jun 14 '18 at 19:36

paf

4,0281824

asked Jun 14 '18 at 19:23

Stéphane LaurentStéphane Laurent

770317

$endgroup$

add a comment | 

0

$begingroup$

Let $mathcalC$ be a ring cyclide with symmetry plane $z=0$ and parameters $a$, $c$, $mu$ as in this picture:

How to find a torus and an inversion such that $mathcalC$ is the image of this torus by this inversion?

geometry

share|cite|improve this question

edited Jun 14 '18 at 19:36

paf

4,0281824

asked Jun 14 '18 at 19:23

Stéphane LaurentStéphane Laurent

770317

$endgroup$

add a comment | 

$begingroup$

Let $mathcalC$ be a ring cyclide with symmetry plane $z=0$ and parameters $a$, $c$, $mu$ as in this picture:

How to find a torus and an inversion such that $mathcalC$ is the image of this torus by this inversion?

geometry

share|cite|improve this question

edited Jun 14 '18 at 19:36

paf

4,0281824

asked Jun 14 '18 at 19:23

Stéphane LaurentStéphane Laurent

770317

$endgroup$

share|cite|improve this question

edited Jun 14 '18 at 19:36

paf

4,0281824

asked Jun 14 '18 at 19:23

Stéphane LaurentStéphane Laurent

770317

share|cite|improve this question

edited Jun 14 '18 at 19:36

paf

4,0281824

asked Jun 14 '18 at 19:23

Stéphane LaurentStéphane Laurent

770317

edited Jun 14 '18 at 19:36

paf

4,0281824

edited Jun 14 '18 at 19:36

paf

4,0281824

asked Jun 14 '18 at 19:23

Stéphane LaurentStéphane Laurent

770317

asked Jun 14 '18 at 19:23

Stéphane LaurentStéphane Laurent

770317

1 Answer
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0

$begingroup$

An answer is given in Garnier & al's paper Computation of Yvon-Villarceau circles on Dupin cyclides and construction of circular edge right triangles on tori and Dupin cyclides.

Let $k>0$ and define
$$
b = sqrta^2-c^2,
$$
$$
r = frackc^2(mu-c)bigl((a+c)(mu-c)+bsqrtmu^2-c^2bigr)bigl((a-c)(mu-c)+bsqrtmu^2-c^2bigr),
$$
$$
R = frackc^2(a-c)bigl((a-c)(mu+c)+bsqrtmu^2-c^2bigr)bigl((a-c)(mu-c)+bsqrtmu^2-c^2bigr),
$$
$$
omega_2 = fracamu+bsqrtmu^2-c^2c,
$$
$$
omega_mathcalT = omega_2 - frackb^2(omega_2-c)bigl((a-c)(mu+omega_2)-b^2bigr)bigl((a+c)(omega_2-c)+b^2bigr).
$$
Then the inversion of center $(omega_2,0,0)$ and ratio $k$ maps the torus centered at $(omega_mathcalT,0,0)$ with major radius $R$, minor radius $r$, and symmetry plane $z=0$ to the cyclide $mathcalC$ of implicit equation
$$
(x^2+y^2+z^2-mu^2-b^2)^2 - 4(cx-amu)^2 -4b^2z^2 = 0.
$$

EDIT

The above formula for $omega_mathcalT$ is wrong. With the notations of the paper, the formula $omega_mathcalT = fracb'_1+b'_22$ is correct.

share|cite|improve this answer

edited Mar 16 at 19:50

answered Jun 29 '18 at 9:36

Stéphane LaurentStéphane Laurent

770317

$endgroup$

add a comment | 

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0

$begingroup$

An answer is given in Garnier & al's paper Computation of Yvon-Villarceau circles on Dupin cyclides and construction of circular edge right triangles on tori and Dupin cyclides.

Let $k>0$ and define
$$
b = sqrta^2-c^2,
$$
$$
r = frackc^2(mu-c)bigl((a+c)(mu-c)+bsqrtmu^2-c^2bigr)bigl((a-c)(mu-c)+bsqrtmu^2-c^2bigr),
$$
$$
R = frackc^2(a-c)bigl((a-c)(mu+c)+bsqrtmu^2-c^2bigr)bigl((a-c)(mu-c)+bsqrtmu^2-c^2bigr),
$$
$$
omega_2 = fracamu+bsqrtmu^2-c^2c,
$$
$$
omega_mathcalT = omega_2 - frackb^2(omega_2-c)bigl((a-c)(mu+omega_2)-b^2bigr)bigl((a+c)(omega_2-c)+b^2bigr).
$$
Then the inversion of center $(omega_2,0,0)$ and ratio $k$ maps the torus centered at $(omega_mathcalT,0,0)$ with major radius $R$, minor radius $r$, and symmetry plane $z=0$ to the cyclide $mathcalC$ of implicit equation
$$
(x^2+y^2+z^2-mu^2-b^2)^2 - 4(cx-amu)^2 -4b^2z^2 = 0.
$$

EDIT

The above formula for $omega_mathcalT$ is wrong. With the notations of the paper, the formula $omega_mathcalT = fracb'_1+b'_22$ is correct.

share|cite|improve this answer

edited Mar 16 at 19:50

answered Jun 29 '18 at 9:36

Stéphane LaurentStéphane Laurent

770317

$endgroup$

add a comment | 

0

$begingroup$

An answer is given in Garnier & al's paper Computation of Yvon-Villarceau circles on Dupin cyclides and construction of circular edge right triangles on tori and Dupin cyclides.

Let $k>0$ and define
$$
b = sqrta^2-c^2,
$$
$$
r = frackc^2(mu-c)bigl((a+c)(mu-c)+bsqrtmu^2-c^2bigr)bigl((a-c)(mu-c)+bsqrtmu^2-c^2bigr),
$$
$$
R = frackc^2(a-c)bigl((a-c)(mu+c)+bsqrtmu^2-c^2bigr)bigl((a-c)(mu-c)+bsqrtmu^2-c^2bigr),
$$
$$
omega_2 = fracamu+bsqrtmu^2-c^2c,
$$
$$
omega_mathcalT = omega_2 - frackb^2(omega_2-c)bigl((a-c)(mu+omega_2)-b^2bigr)bigl((a+c)(omega_2-c)+b^2bigr).
$$
Then the inversion of center $(omega_2,0,0)$ and ratio $k$ maps the torus centered at $(omega_mathcalT,0,0)$ with major radius $R$, minor radius $r$, and symmetry plane $z=0$ to the cyclide $mathcalC$ of implicit equation
$$
(x^2+y^2+z^2-mu^2-b^2)^2 - 4(cx-amu)^2 -4b^2z^2 = 0.
$$

EDIT

The above formula for $omega_mathcalT$ is wrong. With the notations of the paper, the formula $omega_mathcalT = fracb'_1+b'_22$ is correct.

share|cite|improve this answer

edited Mar 16 at 19:50

answered Jun 29 '18 at 9:36

Stéphane LaurentStéphane Laurent

770317

$endgroup$

add a comment | 

$begingroup$

An answer is given in Garnier & al's paper Computation of Yvon-Villarceau circles on Dupin cyclides and construction of circular edge right triangles on tori and Dupin cyclides.

Let $k>0$ and define
$$
b = sqrta^2-c^2,
$$
$$
r = frackc^2(mu-c)bigl((a+c)(mu-c)+bsqrtmu^2-c^2bigr)bigl((a-c)(mu-c)+bsqrtmu^2-c^2bigr),
$$
$$
R = frackc^2(a-c)bigl((a-c)(mu+c)+bsqrtmu^2-c^2bigr)bigl((a-c)(mu-c)+bsqrtmu^2-c^2bigr),
$$
$$
omega_2 = fracamu+bsqrtmu^2-c^2c,
$$
$$
omega_mathcalT = omega_2 - frackb^2(omega_2-c)bigl((a-c)(mu+omega_2)-b^2bigr)bigl((a+c)(omega_2-c)+b^2bigr).
$$
Then the inversion of center $(omega_2,0,0)$ and ratio $k$ maps the torus centered at $(omega_mathcalT,0,0)$ with major radius $R$, minor radius $r$, and symmetry plane $z=0$ to the cyclide $mathcalC$ of implicit equation
$$
(x^2+y^2+z^2-mu^2-b^2)^2 - 4(cx-amu)^2 -4b^2z^2 = 0.
$$

EDIT

The above formula for $omega_mathcalT$ is wrong. With the notations of the paper, the formula $omega_mathcalT = fracb'_1+b'_22$ is correct.

share|cite|improve this answer

edited Mar 16 at 19:50

answered Jun 29 '18 at 9:36

Stéphane LaurentStéphane Laurent

770317

$endgroup$

share|cite|improve this answer

edited Mar 16 at 19:50

answered Jun 29 '18 at 9:36

Stéphane LaurentStéphane Laurent

770317

answered Jun 29 '18 at 9:36

Stéphane LaurentStéphane Laurent

770317

answered Jun 29 '18 at 9:36

Stéphane LaurentStéphane Laurent

770317