Parametric equation of a non-wrapping circle on the surface of a cylinder
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I have a cylinder of radius $R$, and I wish to draw a circle of radius $r$ on its surface that does not wrap around it. So, the centre will lie somewhere on the surface of the cylinder, and the whole of the circle will lie on the surface as well.
Can someone help me to find a parametric representation?
geometry parametric curves
asked Dec 2 '18 at 18:12
ap21ap21
385
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add a comment |
$begingroup$
I have a cylinder of radius $R$, and I wish to draw a circle of radius $r$ on its surface that does not wrap around it. So, the centre will lie somewhere on the surface of the cylinder, and the whole of the circle will lie on the surface as well.
Can someone help me to find a parametric representation?
geometry parametric curves
asked Dec 2 '18 at 18:12
ap21ap21
385
$endgroup$
1
$begingroup$
My suggestion: give each point a pair of coordinates describing where it is on the cylinder (as if you'd made it out of a tube of graph paper), then work out how to convert those into 3-d coordinates, and finally feed in a parametric representation of a circle.
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– timtfj
Dec 2 '18 at 18:52
1
$begingroup$
Assume cylinder has $z$-axis as symmetry axis. You can parametrize the surface as $$(u,v) mapsto (x,y,z) = left(Rcosleft(frac{u}{color{red}{R}}right), Rsinleft(frac{u}{color{red}{R}}right),vright)$$ To draw a circle on it, compose circle's parametrization $t mapsto (u,v) = (rcos t,rsin t )$ with that of cylinder. You get $$t mapsto (x,y,z) = left(Rcosleft(frac{r}{R}cos tright), Rsinleft(frac{r}{R}cos tright),r sin tright)$$ If you want circle looks like a circle, make sure you use the right scaling factor for $u, v$ (the two $color{red}{R}$ above)
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– achille hui
Dec 3 '18 at 0:37
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@achillehui Can you please explain why I need the scaling by $R$?
$endgroup$
– ap21
Dec 3 '18 at 1:58
1
$begingroup$
Imagine you don't include the scaling factor and just use the parametrization $(u,v) mapsto (Rcos u,Rsin u, v)$. When $R$ becomes large, the cylindrical surface becomes flat. The neighborhood of $(u,v) = (0,0)$ get mapped to something well approximated by the plane $x = R$ near $(x,y,z) = (R,0,0)$. A circle in $(u,v)$ plane, say $u^2+v^2 = epsilon^2$ get mapped to approximately an ellipse $x = 0, (y/R)^2 + z^2 = epsilon^2$ instead of something like a circle. The scaling factor is to compensate this distortion.
$endgroup$
– achille hui
Dec 3 '18 at 2:08
$begingroup$
@achillehui Your first comment looks to me like an answer. Maybe ought to be turned into one rather than hidden away in these comments?
$endgroup$
– timtfj
Dec 3 '18 at 4:01
add a comment |
$begingroup$
I have a cylinder of radius $R$, and I wish to draw a circle of radius $r$ on its surface that does not wrap around it. So, the centre will lie somewhere on the surface of the cylinder, and the whole of the circle will lie on the surface as well.
Can someone help me to find a parametric representation?
geometry parametric curves
asked Dec 2 '18 at 18:12
ap21ap21
385
$endgroup$
I have a cylinder of radius $R$, and I wish to draw a circle of radius $r$ on its surface that does not wrap around it. So, the centre will lie somewhere on the surface of the cylinder, and the whole of the circle will lie on the surface as well.
Can someone help me to find a parametric representation?
geometry parametric curves
geometry parametric curves
asked Dec 2 '18 at 18:12
ap21ap21
385
asked Dec 2 '18 at 18:12
ap21ap21
385
asked Dec 2 '18 at 18:12
ap21ap21
385
asked Dec 2 '18 at 18:12
ap21ap21
385
asked Dec 2 '18 at 18:12
ap21ap21
385
385
1
$begingroup$
My suggestion: give each point a pair of coordinates describing where it is on the cylinder (as if you'd made it out of a tube of graph paper), then work out how to convert those into 3-d coordinates, and finally feed in a parametric representation of a circle.
$endgroup$
– timtfj
Dec 2 '18 at 18:52
1
$begingroup$
Assume cylinder has $z$-axis as symmetry axis. You can parametrize the surface as $$(u,v) mapsto (x,y,z) = left(Rcosleft(frac{u}{color{red}{R}}right), Rsinleft(frac{u}{color{red}{R}}right),vright)$$ To draw a circle on it, compose circle's parametrization $t mapsto (u,v) = (rcos t,rsin t )$ with that of cylinder. You get $$t mapsto (x,y,z) = left(Rcosleft(frac{r}{R}cos tright), Rsinleft(frac{r}{R}cos tright),r sin tright)$$ If you want circle looks like a circle, make sure you use the right scaling factor for $u, v$ (the two $color{red}{R}$ above)
$endgroup$
– achille hui
Dec 3 '18 at 0:37
$begingroup$
@achillehui Can you please explain why I need the scaling by $R$?
$endgroup$
– ap21
Dec 3 '18 at 1:58
1
$begingroup$
Imagine you don't include the scaling factor and just use the parametrization $(u,v) mapsto (Rcos u,Rsin u, v)$. When $R$ becomes large, the cylindrical surface becomes flat. The neighborhood of $(u,v) = (0,0)$ get mapped to something well approximated by the plane $x = R$ near $(x,y,z) = (R,0,0)$. A circle in $(u,v)$ plane, say $u^2+v^2 = epsilon^2$ get mapped to approximately an ellipse $x = 0, (y/R)^2 + z^2 = epsilon^2$ instead of something like a circle. The scaling factor is to compensate this distortion.
$endgroup$
– achille hui
Dec 3 '18 at 2:08
$begingroup$
@achillehui Your first comment looks to me like an answer. Maybe ought to be turned into one rather than hidden away in these comments?
$endgroup$
– timtfj
Dec 3 '18 at 4:01
add a comment |
1
$begingroup$
My suggestion: give each point a pair of coordinates describing where it is on the cylinder (as if you'd made it out of a tube of graph paper), then work out how to convert those into 3-d coordinates, and finally feed in a parametric representation of a circle.
$endgroup$
– timtfj
Dec 2 '18 at 18:52
1
$begingroup$
Assume cylinder has $z$-axis as symmetry axis. You can parametrize the surface as $$(u,v) mapsto (x,y,z) = left(Rcosleft(frac{u}{color{red}{R}}right), Rsinleft(frac{u}{color{red}{R}}right),vright)$$ To draw a circle on it, compose circle's parametrization $t mapsto (u,v) = (rcos t,rsin t )$ with that of cylinder. You get $$t mapsto (x,y,z) = left(Rcosleft(frac{r}{R}cos tright), Rsinleft(frac{r}{R}cos tright),r sin tright)$$ If you want circle looks like a circle, make sure you use the right scaling factor for $u, v$ (the two $color{red}{R}$ above)
$endgroup$
– achille hui
Dec 3 '18 at 0:37
$begingroup$
@achillehui Can you please explain why I need the scaling by $R$?
$endgroup$
– ap21
Dec 3 '18 at 1:58
1
$begingroup$
Imagine you don't include the scaling factor and just use the parametrization $(u,v) mapsto (Rcos u,Rsin u, v)$. When $R$ becomes large, the cylindrical surface becomes flat. The neighborhood of $(u,v) = (0,0)$ get mapped to something well approximated by the plane $x = R$ near $(x,y,z) = (R,0,0)$. A circle in $(u,v)$ plane, say $u^2+v^2 = epsilon^2$ get mapped to approximately an ellipse $x = 0, (y/R)^2 + z^2 = epsilon^2$ instead of something like a circle. The scaling factor is to compensate this distortion.
$endgroup$
– achille hui
Dec 3 '18 at 2:08
$begingroup$
@achillehui Your first comment looks to me like an answer. Maybe ought to be turned into one rather than hidden away in these comments?
$endgroup$
– timtfj
Dec 3 '18 at 4:01
1
1
$begingroup$
My suggestion: give each point a pair of coordinates describing where it is on the cylinder (as if you'd made it out of a tube of graph paper), then work out how to convert those into 3-d coordinates, and finally feed in a parametric representation of a circle.
$endgroup$
– timtfj
Dec 2 '18 at 18:52
$begingroup$
My suggestion: give each point a pair of coordinates describing where it is on the cylinder (as if you'd made it out of a tube of graph paper), then work out how to convert those into 3-d coordinates, and finally feed in a parametric representation of a circle.
$endgroup$
– timtfj
Dec 2 '18 at 18:52
1
1
$begingroup$
Assume cylinder has $z$-axis as symmetry axis. You can parametrize the surface as $$(u,v) mapsto (x,y,z) = left(Rcosleft(frac{u}{color{red}{R}}right), Rsinleft(frac{u}{color{red}{R}}right),vright)$$ To draw a circle on it, compose circle's parametrization $t mapsto (u,v) = (rcos t,rsin t )$ with that of cylinder. You get $$t mapsto (x,y,z) = left(Rcosleft(frac{r}{R}cos tright), Rsinleft(frac{r}{R}cos tright),r sin tright)$$ If you want circle looks like a circle, make sure you use the right scaling factor for $u, v$ (the two $color{red}{R}$ above)
$endgroup$
– achille hui
Dec 3 '18 at 0:37
$begingroup$
Assume cylinder has $z$-axis as symmetry axis. You can parametrize the surface as $$(u,v) mapsto (x,y,z) = left(Rcosleft(frac{u}{color{red}{R}}right), Rsinleft(frac{u}{color{red}{R}}right),vright)$$ To draw a circle on it, compose circle's parametrization $t mapsto (u,v) = (rcos t,rsin t )$ with that of cylinder. You get $$t mapsto (x,y,z) = left(Rcosleft(frac{r}{R}cos tright), Rsinleft(frac{r}{R}cos tright),r sin tright)$$ If you want circle looks like a circle, make sure you use the right scaling factor for $u, v$ (the two $color{red}{R}$ above)
$endgroup$
– achille hui
Dec 3 '18 at 0:37
$begingroup$
@achillehui Can you please explain why I need the scaling by $R$?
$endgroup$
– ap21
Dec 3 '18 at 1:58
$begingroup$
@achillehui Can you please explain why I need the scaling by $R$?
$endgroup$
– ap21
Dec 3 '18 at 1:58
1
1
$begingroup$
Imagine you don't include the scaling factor and just use the parametrization $(u,v) mapsto (Rcos u,Rsin u, v)$. When $R$ becomes large, the cylindrical surface becomes flat. The neighborhood of $(u,v) = (0,0)$ get mapped to something well approximated by the plane $x = R$ near $(x,y,z) = (R,0,0)$. A circle in $(u,v)$ plane, say $u^2+v^2 = epsilon^2$ get mapped to approximately an ellipse $x = 0, (y/R)^2 + z^2 = epsilon^2$ instead of something like a circle. The scaling factor is to compensate this distortion.
$endgroup$
– achille hui
Dec 3 '18 at 2:08
$begingroup$
Imagine you don't include the scaling factor and just use the parametrization $(u,v) mapsto (Rcos u,Rsin u, v)$. When $R$ becomes large, the cylindrical surface becomes flat. The neighborhood of $(u,v) = (0,0)$ get mapped to something well approximated by the plane $x = R$ near $(x,y,z) = (R,0,0)$. A circle in $(u,v)$ plane, say $u^2+v^2 = epsilon^2$ get mapped to approximately an ellipse $x = 0, (y/R)^2 + z^2 = epsilon^2$ instead of something like a circle. The scaling factor is to compensate this distortion.
$endgroup$
– achille hui
Dec 3 '18 at 2:08
$begingroup$
@achillehui Your first comment looks to me like an answer. Maybe ought to be turned into one rather than hidden away in these comments?
$endgroup$
– timtfj
Dec 3 '18 at 4:01
$begingroup$
@achillehui Your first comment looks to me like an answer. Maybe ought to be turned into one rather than hidden away in these comments?
$endgroup$
– timtfj
Dec 3 '18 at 4:01
add a comment |
1 Answer
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Credit goes to @achillehui for giving me the answer. I am just elaborating on what he said, and clearing up why the scaling by $R$ is needed.
- The general way to go about finding the parameterisation of a circle on a cylinder (or to a sphere or a cone or any mappable surface) is to find a mapping from the plane to the cylinder, and then apply this mapping to a circle on the plane.
Quoting @achillehui from above:
Assume cylinder has z-axis as symmetry axis. You can parametrize the
surface as $$(u,v)↦(x,y,z)=(Rcos(frac{u}{R}),Rsin(frac{u}{R}),v)$$. To draw a circle on
it, compose circle's parametrization $$t↦(u,v)=(rcos t,rsin t)$$ with that
of cylinder. You get $$t↦(x,y,z)=(Rcos(frac{r cos t}{R}),Rsin(frac{r cos t}{R}),r sin t)$$. If
you want circle looks like a circle, make sure you use the right
scaling factor for $u,v$ (the two $R$ above).
- The composition of mappings is what I said at the beginning. What really piques my interest is the re-scaling by $R$. @achillehui explains this using a Taylor expansion around $(u,v) = (0,0)$ for the large $R$ limit. Thus, the unscaled parametrisation $(u,v)↦(x,y,z)=(Rcos(u),Rsin(u),v)$ gives $(u,v)↦(x,y,z)=(R,Ru,v)$, so that we have a mapping from circle to ellipse:
$$u^2+v^2=1 : (rm{circle}) mapsto (y/R)^2 + z^2 =1 :rm{(ellipse)}.$$
Scaling this by $R$ gives $(u,v)↦(x,y,z)=(R,u,v)$ so that we have:
$$u^2+v^2=1 :(rm{circle}) mapsto y^2 + z^2 =1 :rm{(circle)}.$$
- I have a different argument, more intuitive for me. It is that the first unscaled mapping is not an isometry, whereas the second is. Only isometries preserve the geodesic curvature $kappa_g$ (see: Why the geodesic curvature is invariant under isometric transformations?), so that circles ($kappa_g =1$) remain circles under the mapping.
This can be seen clearly by looking at the action of the mapping on the line $u=rm{const.}$, which maps onto a circle concentric with the cylinder's axis. For the unscaled map, the interval of length $2pi$ is mapped onto a circle of length $2pi R$, which clearly involves a scaling by factor $R$. Dividing by $R$ thus ensures isometry.
------------------------------------------- IMAGES ------------------------------------------------
- Unscaled parametrisation (with $R=1$):
$$t↦(x,y,z)=(cos(r cos t),sin(r cos t),r sin t), :t in [0, 2pi]$$
- Scaled parametrisation: $$t↦(x,y,z)=(Rcos(frac{r cos t}{R}),Rsin(frac{r cos t}{R}),r sin t), :t in [0, 2pi]$$
answered Dec 3 '18 at 19:18
ap21ap21
385
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add a comment |
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$begingroup$
Credit goes to @achillehui for giving me the answer. I am just elaborating on what he said, and clearing up why the scaling by $R$ is needed.
- The general way to go about finding the parameterisation of a circle on a cylinder (or to a sphere or a cone or any mappable surface) is to find a mapping from the plane to the cylinder, and then apply this mapping to a circle on the plane.
Quoting @achillehui from above:
Assume cylinder has z-axis as symmetry axis. You can parametrize the
surface as $$(u,v)↦(x,y,z)=(Rcos(frac{u}{R}),Rsin(frac{u}{R}),v)$$. To draw a circle on
it, compose circle's parametrization $$t↦(u,v)=(rcos t,rsin t)$$ with that
of cylinder. You get $$t↦(x,y,z)=(Rcos(frac{r cos t}{R}),Rsin(frac{r cos t}{R}),r sin t)$$. If
you want circle looks like a circle, make sure you use the right
scaling factor for $u,v$ (the two $R$ above).
- The composition of mappings is what I said at the beginning. What really piques my interest is the re-scaling by $R$. @achillehui explains this using a Taylor expansion around $(u,v) = (0,0)$ for the large $R$ limit. Thus, the unscaled parametrisation $(u,v)↦(x,y,z)=(Rcos(u),Rsin(u),v)$ gives $(u,v)↦(x,y,z)=(R,Ru,v)$, so that we have a mapping from circle to ellipse:
$$u^2+v^2=1 : (rm{circle}) mapsto (y/R)^2 + z^2 =1 :rm{(ellipse)}.$$
Scaling this by $R$ gives $(u,v)↦(x,y,z)=(R,u,v)$ so that we have:
$$u^2+v^2=1 :(rm{circle}) mapsto y^2 + z^2 =1 :rm{(circle)}.$$
- I have a different argument, more intuitive for me. It is that the first unscaled mapping is not an isometry, whereas the second is. Only isometries preserve the geodesic curvature $kappa_g$ (see: Why the geodesic curvature is invariant under isometric transformations?), so that circles ($kappa_g =1$) remain circles under the mapping.
This can be seen clearly by looking at the action of the mapping on the line $u=rm{const.}$, which maps onto a circle concentric with the cylinder's axis. For the unscaled map, the interval of length $2pi$ is mapped onto a circle of length $2pi R$, which clearly involves a scaling by factor $R$. Dividing by $R$ thus ensures isometry.
------------------------------------------- IMAGES ------------------------------------------------
- Unscaled parametrisation (with $R=1$):
$$t↦(x,y,z)=(cos(r cos t),sin(r cos t),r sin t), :t in [0, 2pi]$$
- Scaled parametrisation: $$t↦(x,y,z)=(Rcos(frac{r cos t}{R}),Rsin(frac{r cos t}{R}),r sin t), :t in [0, 2pi]$$
answered Dec 3 '18 at 19:18
ap21ap21
385
$endgroup$
add a comment |
$begingroup$
Credit goes to @achillehui for giving me the answer. I am just elaborating on what he said, and clearing up why the scaling by $R$ is needed.
- The general way to go about finding the parameterisation of a circle on a cylinder (or to a sphere or a cone or any mappable surface) is to find a mapping from the plane to the cylinder, and then apply this mapping to a circle on the plane.
Quoting @achillehui from above:
Assume cylinder has z-axis as symmetry axis. You can parametrize the
surface as $$(u,v)↦(x,y,z)=(Rcos(frac{u}{R}),Rsin(frac{u}{R}),v)$$. To draw a circle on
it, compose circle's parametrization $$t↦(u,v)=(rcos t,rsin t)$$ with that
of cylinder. You get $$t↦(x,y,z)=(Rcos(frac{r cos t}{R}),Rsin(frac{r cos t}{R}),r sin t)$$. If
you want circle looks like a circle, make sure you use the right
scaling factor for $u,v$ (the two $R$ above).
- The composition of mappings is what I said at the beginning. What really piques my interest is the re-scaling by $R$. @achillehui explains this using a Taylor expansion around $(u,v) = (0,0)$ for the large $R$ limit. Thus, the unscaled parametrisation $(u,v)↦(x,y,z)=(Rcos(u),Rsin(u),v)$ gives $(u,v)↦(x,y,z)=(R,Ru,v)$, so that we have a mapping from circle to ellipse:
$$u^2+v^2=1 : (rm{circle}) mapsto (y/R)^2 + z^2 =1 :rm{(ellipse)}.$$
Scaling this by $R$ gives $(u,v)↦(x,y,z)=(R,u,v)$ so that we have:
$$u^2+v^2=1 :(rm{circle}) mapsto y^2 + z^2 =1 :rm{(circle)}.$$
- I have a different argument, more intuitive for me. It is that the first unscaled mapping is not an isometry, whereas the second is. Only isometries preserve the geodesic curvature $kappa_g$ (see: Why the geodesic curvature is invariant under isometric transformations?), so that circles ($kappa_g =1$) remain circles under the mapping.
This can be seen clearly by looking at the action of the mapping on the line $u=rm{const.}$, which maps onto a circle concentric with the cylinder's axis. For the unscaled map, the interval of length $2pi$ is mapped onto a circle of length $2pi R$, which clearly involves a scaling by factor $R$. Dividing by $R$ thus ensures isometry.
------------------------------------------- IMAGES ------------------------------------------------
- Unscaled parametrisation (with $R=1$):
$$t↦(x,y,z)=(cos(r cos t),sin(r cos t),r sin t), :t in [0, 2pi]$$
- Scaled parametrisation: $$t↦(x,y,z)=(Rcos(frac{r cos t}{R}),Rsin(frac{r cos t}{R}),r sin t), :t in [0, 2pi]$$
answered Dec 3 '18 at 19:18
ap21ap21
385
$endgroup$
add a comment |
$begingroup$
Credit goes to @achillehui for giving me the answer. I am just elaborating on what he said, and clearing up why the scaling by $R$ is needed.
- The general way to go about finding the parameterisation of a circle on a cylinder (or to a sphere or a cone or any mappable surface) is to find a mapping from the plane to the cylinder, and then apply this mapping to a circle on the plane.
Quoting @achillehui from above:
Assume cylinder has z-axis as symmetry axis. You can parametrize the
surface as $$(u,v)↦(x,y,z)=(Rcos(frac{u}{R}),Rsin(frac{u}{R}),v)$$. To draw a circle on
it, compose circle's parametrization $$t↦(u,v)=(rcos t,rsin t)$$ with that
of cylinder. You get $$t↦(x,y,z)=(Rcos(frac{r cos t}{R}),Rsin(frac{r cos t}{R}),r sin t)$$. If
you want circle looks like a circle, make sure you use the right
scaling factor for $u,v$ (the two $R$ above).
- The composition of mappings is what I said at the beginning. What really piques my interest is the re-scaling by $R$. @achillehui explains this using a Taylor expansion around $(u,v) = (0,0)$ for the large $R$ limit. Thus, the unscaled parametrisation $(u,v)↦(x,y,z)=(Rcos(u),Rsin(u),v)$ gives $(u,v)↦(x,y,z)=(R,Ru,v)$, so that we have a mapping from circle to ellipse:
$$u^2+v^2=1 : (rm{circle}) mapsto (y/R)^2 + z^2 =1 :rm{(ellipse)}.$$
Scaling this by $R$ gives $(u,v)↦(x,y,z)=(R,u,v)$ so that we have:
$$u^2+v^2=1 :(rm{circle}) mapsto y^2 + z^2 =1 :rm{(circle)}.$$
- I have a different argument, more intuitive for me. It is that the first unscaled mapping is not an isometry, whereas the second is. Only isometries preserve the geodesic curvature $kappa_g$ (see: Why the geodesic curvature is invariant under isometric transformations?), so that circles ($kappa_g =1$) remain circles under the mapping.
This can be seen clearly by looking at the action of the mapping on the line $u=rm{const.}$, which maps onto a circle concentric with the cylinder's axis. For the unscaled map, the interval of length $2pi$ is mapped onto a circle of length $2pi R$, which clearly involves a scaling by factor $R$. Dividing by $R$ thus ensures isometry.
------------------------------------------- IMAGES ------------------------------------------------
- Unscaled parametrisation (with $R=1$):
$$t↦(x,y,z)=(cos(r cos t),sin(r cos t),r sin t), :t in [0, 2pi]$$
- Scaled parametrisation: $$t↦(x,y,z)=(Rcos(frac{r cos t}{R}),Rsin(frac{r cos t}{R}),r sin t), :t in [0, 2pi]$$
answered Dec 3 '18 at 19:18
ap21ap21
385
$endgroup$
Credit goes to @achillehui for giving me the answer. I am just elaborating on what he said, and clearing up why the scaling by $R$ is needed.
- The general way to go about finding the parameterisation of a circle on a cylinder (or to a sphere or a cone or any mappable surface) is to find a mapping from the plane to the cylinder, and then apply this mapping to a circle on the plane.
Quoting @achillehui from above:
Assume cylinder has z-axis as symmetry axis. You can parametrize the
surface as $$(u,v)↦(x,y,z)=(Rcos(frac{u}{R}),Rsin(frac{u}{R}),v)$$. To draw a circle on
it, compose circle's parametrization $$t↦(u,v)=(rcos t,rsin t)$$ with that
of cylinder. You get $$t↦(x,y,z)=(Rcos(frac{r cos t}{R}),Rsin(frac{r cos t}{R}),r sin t)$$. If
you want circle looks like a circle, make sure you use the right
scaling factor for $u,v$ (the two $R$ above).
- The composition of mappings is what I said at the beginning. What really piques my interest is the re-scaling by $R$. @achillehui explains this using a Taylor expansion around $(u,v) = (0,0)$ for the large $R$ limit. Thus, the unscaled parametrisation $(u,v)↦(x,y,z)=(Rcos(u),Rsin(u),v)$ gives $(u,v)↦(x,y,z)=(R,Ru,v)$, so that we have a mapping from circle to ellipse:
$$u^2+v^2=1 : (rm{circle}) mapsto (y/R)^2 + z^2 =1 :rm{(ellipse)}.$$
Scaling this by $R$ gives $(u,v)↦(x,y,z)=(R,u,v)$ so that we have:
$$u^2+v^2=1 :(rm{circle}) mapsto y^2 + z^2 =1 :rm{(circle)}.$$
- I have a different argument, more intuitive for me. It is that the first unscaled mapping is not an isometry, whereas the second is. Only isometries preserve the geodesic curvature $kappa_g$ (see: Why the geodesic curvature is invariant under isometric transformations?), so that circles ($kappa_g =1$) remain circles under the mapping.
This can be seen clearly by looking at the action of the mapping on the line $u=rm{const.}$, which maps onto a circle concentric with the cylinder's axis. For the unscaled map, the interval of length $2pi$ is mapped onto a circle of length $2pi R$, which clearly involves a scaling by factor $R$. Dividing by $R$ thus ensures isometry.
------------------------------------------- IMAGES ------------------------------------------------
- Unscaled parametrisation (with $R=1$):
$$t↦(x,y,z)=(cos(r cos t),sin(r cos t),r sin t), :t in [0, 2pi]$$
- Scaled parametrisation: $$t↦(x,y,z)=(Rcos(frac{r cos t}{R}),Rsin(frac{r cos t}{R}),r sin t), :t in [0, 2pi]$$
answered Dec 3 '18 at 19:18
ap21ap21
385
answered Dec 3 '18 at 19:18
ap21ap21
385
answered Dec 3 '18 at 19:18
ap21ap21
385
answered Dec 3 '18 at 19:18
ap21ap21
385
385
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1
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My suggestion: give each point a pair of coordinates describing where it is on the cylinder (as if you'd made it out of a tube of graph paper), then work out how to convert those into 3-d coordinates, and finally feed in a parametric representation of a circle.
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– timtfj
Dec 2 '18 at 18:52
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Assume cylinder has $z$-axis as symmetry axis. You can parametrize the surface as $$(u,v) mapsto (x,y,z) = left(Rcosleft(frac{u}{color{red}{R}}right), Rsinleft(frac{u}{color{red}{R}}right),vright)$$ To draw a circle on it, compose circle's parametrization $t mapsto (u,v) = (rcos t,rsin t )$ with that of cylinder. You get $$t mapsto (x,y,z) = left(Rcosleft(frac{r}{R}cos tright), Rsinleft(frac{r}{R}cos tright),r sin tright)$$ If you want circle looks like a circle, make sure you use the right scaling factor for $u, v$ (the two $color{red}{R}$ above)
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– achille hui
Dec 3 '18 at 0:37
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@achillehui Can you please explain why I need the scaling by $R$?
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– ap21
Dec 3 '18 at 1:58
1
$begingroup$
Imagine you don't include the scaling factor and just use the parametrization $(u,v) mapsto (Rcos u,Rsin u, v)$. When $R$ becomes large, the cylindrical surface becomes flat. The neighborhood of $(u,v) = (0,0)$ get mapped to something well approximated by the plane $x = R$ near $(x,y,z) = (R,0,0)$. A circle in $(u,v)$ plane, say $u^2+v^2 = epsilon^2$ get mapped to approximately an ellipse $x = 0, (y/R)^2 + z^2 = epsilon^2$ instead of something like a circle. The scaling factor is to compensate this distortion.
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– achille hui
Dec 3 '18 at 2:08
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@achillehui Your first comment looks to me like an answer. Maybe ought to be turned into one rather than hidden away in these comments?
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– timtfj
Dec 3 '18 at 4:01