How to conserve 2D lattice distances when layed onto a surface?
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Given a surface $varphi(x,y) = begin{pmatrix}x \ y \ f(x, y)end{pmatrix} in mathbb{R}^3$, where $f$ is continous (and differentiable), $x, y in mathbb{R}$, I am trying to find a map $begin{pmatrix} hat{x} \ hat{y}end{pmatrix} = Phibegin{pmatrix}x\yend{pmatrix} $ such that the distances of given points in the $x$-$y$-plane (e.g. square lattice, hexagonal lattice) are conserved when "layed" on the surface under the transformation $Phi$. By intuition, such a transformation should contract points if the surface is steep and just adopt the points from the $x$-$y$-plane where it is flat.
Does such a transformation $Phi$ exist? And can it be derived or approximated for a given function $f$?
I tried to approach this problem in 1D: for a curve $begin{pmatrix}x \ f(x)end{pmatrix}$ this can be solved by reparameterizing the curve such that it has unit speed. One can think of this as laying a chain of equidistant points onto the curve.
For the 1D case I created a Jupyter Notebook but now I am stuck on how to expand this to two dimensions.
multivariable-calculus differential-geometry
asked Nov 28 '18 at 16:39
tam724tam724
12
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add a comment |
$begingroup$
Given a surface $varphi(x,y) = begin{pmatrix}x \ y \ f(x, y)end{pmatrix} in mathbb{R}^3$, where $f$ is continous (and differentiable), $x, y in mathbb{R}$, I am trying to find a map $begin{pmatrix} hat{x} \ hat{y}end{pmatrix} = Phibegin{pmatrix}x\yend{pmatrix} $ such that the distances of given points in the $x$-$y$-plane (e.g. square lattice, hexagonal lattice) are conserved when "layed" on the surface under the transformation $Phi$. By intuition, such a transformation should contract points if the surface is steep and just adopt the points from the $x$-$y$-plane where it is flat.
Does such a transformation $Phi$ exist? And can it be derived or approximated for a given function $f$?
I tried to approach this problem in 1D: for a curve $begin{pmatrix}x \ f(x)end{pmatrix}$ this can be solved by reparameterizing the curve such that it has unit speed. One can think of this as laying a chain of equidistant points onto the curve.
For the 1D case I created a Jupyter Notebook but now I am stuck on how to expand this to two dimensions.
multivariable-calculus differential-geometry
asked Nov 28 '18 at 16:39
tam724tam724
12
$endgroup$
add a comment |
$begingroup$
Given a surface $varphi(x,y) = begin{pmatrix}x \ y \ f(x, y)end{pmatrix} in mathbb{R}^3$, where $f$ is continous (and differentiable), $x, y in mathbb{R}$, I am trying to find a map $begin{pmatrix} hat{x} \ hat{y}end{pmatrix} = Phibegin{pmatrix}x\yend{pmatrix} $ such that the distances of given points in the $x$-$y$-plane (e.g. square lattice, hexagonal lattice) are conserved when "layed" on the surface under the transformation $Phi$. By intuition, such a transformation should contract points if the surface is steep and just adopt the points from the $x$-$y$-plane where it is flat.
Does such a transformation $Phi$ exist? And can it be derived or approximated for a given function $f$?
I tried to approach this problem in 1D: for a curve $begin{pmatrix}x \ f(x)end{pmatrix}$ this can be solved by reparameterizing the curve such that it has unit speed. One can think of this as laying a chain of equidistant points onto the curve.
For the 1D case I created a Jupyter Notebook but now I am stuck on how to expand this to two dimensions.
multivariable-calculus differential-geometry
asked Nov 28 '18 at 16:39
tam724tam724
12
$endgroup$
Given a surface $varphi(x,y) = begin{pmatrix}x \ y \ f(x, y)end{pmatrix} in mathbb{R}^3$, where $f$ is continous (and differentiable), $x, y in mathbb{R}$, I am trying to find a map $begin{pmatrix} hat{x} \ hat{y}end{pmatrix} = Phibegin{pmatrix}x\yend{pmatrix} $ such that the distances of given points in the $x$-$y$-plane (e.g. square lattice, hexagonal lattice) are conserved when "layed" on the surface under the transformation $Phi$. By intuition, such a transformation should contract points if the surface is steep and just adopt the points from the $x$-$y$-plane where it is flat.
Does such a transformation $Phi$ exist? And can it be derived or approximated for a given function $f$?
I tried to approach this problem in 1D: for a curve $begin{pmatrix}x \ f(x)end{pmatrix}$ this can be solved by reparameterizing the curve such that it has unit speed. One can think of this as laying a chain of equidistant points onto the curve.
For the 1D case I created a Jupyter Notebook but now I am stuck on how to expand this to two dimensions.
multivariable-calculus differential-geometry
multivariable-calculus differential-geometry
asked Nov 28 '18 at 16:39
tam724tam724
12
asked Nov 28 '18 at 16:39
tam724tam724
12
asked Nov 28 '18 at 16:39
tam724tam724
12
asked Nov 28 '18 at 16:39
tam724tam724
12
asked Nov 28 '18 at 16:39
tam724tam724
12
12
add a comment |
add a comment |
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