Ellipse on a Circular Cylinder in Cylindrical Coordinates
Is there an equation in cylindrical coordinates for an ellipse (tilted at some angle) on the surface of a right circular cylinder of radius r? For simplicity, I envision the cylinder to be coincident with the x-axis.
I am aware that the cylinder could be "unwrapped" into a plane, which would result in the ellipse becoming a sine curve. I am just not sure how that information ties into cylindrical coordinates.
EDIT: I have realized that I am looking for a parametric equation. For an ellipse on the surface of a cylinder of radius r which has a certain angle of inclination, is there a parameterization where I can calculate the axial coordinate seperately from the azimuth angle for a given t?
Thank you.
conic-sections
asked Nov 28 '18 at 17:36
kreeser1kreeser1
10110
add a comment |
Is there an equation in cylindrical coordinates for an ellipse (tilted at some angle) on the surface of a right circular cylinder of radius r? For simplicity, I envision the cylinder to be coincident with the x-axis.
I am aware that the cylinder could be "unwrapped" into a plane, which would result in the ellipse becoming a sine curve. I am just not sure how that information ties into cylindrical coordinates.
EDIT: I have realized that I am looking for a parametric equation. For an ellipse on the surface of a cylinder of radius r which has a certain angle of inclination, is there a parameterization where I can calculate the axial coordinate seperately from the azimuth angle for a given t?
Thank you.
conic-sections
asked Nov 28 '18 at 17:36
kreeser1kreeser1
10110
You need a pair of implicit Cartesian equations to describe a curve in 3D. If you want a single equation, it’s not too hard to come up with a parameterization of the curve.
– amd
Nov 28 '18 at 21:58
That is a very good point; In my mind I am looking for a parameterization of the curve that takes into account the angle of inclination, but I failed to put that in my original question. For a given cylinder radius, I believe I need two parameters: an expression for the axial coordinate and an expression for the azimuth. Could you help me with that parameterization of the ellipse?
– kreeser1
Nov 28 '18 at 23:35
You can choose a cylindrical coordinate system $(rho,phi,z)$ such that the ellipse is $rho = r, z = r tanalpha sinphi$, where $alpha$ is the angle between the plane normal and cylinder axis. You’ve already accepted an answer, but I can expand on this if you like.
– amd
Nov 30 '18 at 22:36
add a comment |
Is there an equation in cylindrical coordinates for an ellipse (tilted at some angle) on the surface of a right circular cylinder of radius r? For simplicity, I envision the cylinder to be coincident with the x-axis.
I am aware that the cylinder could be "unwrapped" into a plane, which would result in the ellipse becoming a sine curve. I am just not sure how that information ties into cylindrical coordinates.
EDIT: I have realized that I am looking for a parametric equation. For an ellipse on the surface of a cylinder of radius r which has a certain angle of inclination, is there a parameterization where I can calculate the axial coordinate seperately from the azimuth angle for a given t?
Thank you.
conic-sections
asked Nov 28 '18 at 17:36
kreeser1kreeser1
10110
Is there an equation in cylindrical coordinates for an ellipse (tilted at some angle) on the surface of a right circular cylinder of radius r? For simplicity, I envision the cylinder to be coincident with the x-axis.
I am aware that the cylinder could be "unwrapped" into a plane, which would result in the ellipse becoming a sine curve. I am just not sure how that information ties into cylindrical coordinates.
EDIT: I have realized that I am looking for a parametric equation. For an ellipse on the surface of a cylinder of radius r which has a certain angle of inclination, is there a parameterization where I can calculate the axial coordinate seperately from the azimuth angle for a given t?
Thank you.
conic-sections
conic-sections
asked Nov 28 '18 at 17:36
kreeser1kreeser1
10110
asked Nov 28 '18 at 17:36
kreeser1kreeser1
10110
edited Nov 28 '18 at 23:39
kreeser1
asked Nov 28 '18 at 17:36
kreeser1kreeser1
10110
asked Nov 28 '18 at 17:36
kreeser1kreeser1
10110
asked Nov 28 '18 at 17:36
kreeser1kreeser1
10110
10110
You need a pair of implicit Cartesian equations to describe a curve in 3D. If you want a single equation, it’s not too hard to come up with a parameterization of the curve.
– amd
Nov 28 '18 at 21:58
That is a very good point; In my mind I am looking for a parameterization of the curve that takes into account the angle of inclination, but I failed to put that in my original question. For a given cylinder radius, I believe I need two parameters: an expression for the axial coordinate and an expression for the azimuth. Could you help me with that parameterization of the ellipse?
– kreeser1
Nov 28 '18 at 23:35
You can choose a cylindrical coordinate system $(rho,phi,z)$ such that the ellipse is $rho = r, z = r tanalpha sinphi$, where $alpha$ is the angle between the plane normal and cylinder axis. You’ve already accepted an answer, but I can expand on this if you like.
– amd
Nov 30 '18 at 22:36
add a comment |
You need a pair of implicit Cartesian equations to describe a curve in 3D. If you want a single equation, it’s not too hard to come up with a parameterization of the curve.
– amd
Nov 28 '18 at 21:58
That is a very good point; In my mind I am looking for a parameterization of the curve that takes into account the angle of inclination, but I failed to put that in my original question. For a given cylinder radius, I believe I need two parameters: an expression for the axial coordinate and an expression for the azimuth. Could you help me with that parameterization of the ellipse?
– kreeser1
Nov 28 '18 at 23:35
You can choose a cylindrical coordinate system $(rho,phi,z)$ such that the ellipse is $rho = r, z = r tanalpha sinphi$, where $alpha$ is the angle between the plane normal and cylinder axis. You’ve already accepted an answer, but I can expand on this if you like.
– amd
Nov 30 '18 at 22:36
You need a pair of implicit Cartesian equations to describe a curve in 3D. If you want a single equation, it’s not too hard to come up with a parameterization of the curve.
– amd
Nov 28 '18 at 21:58
You need a pair of implicit Cartesian equations to describe a curve in 3D. If you want a single equation, it’s not too hard to come up with a parameterization of the curve.
– amd
Nov 28 '18 at 21:58
That is a very good point; In my mind I am looking for a parameterization of the curve that takes into account the angle of inclination, but I failed to put that in my original question. For a given cylinder radius, I believe I need two parameters: an expression for the axial coordinate and an expression for the azimuth. Could you help me with that parameterization of the ellipse?
– kreeser1
Nov 28 '18 at 23:35
That is a very good point; In my mind I am looking for a parameterization of the curve that takes into account the angle of inclination, but I failed to put that in my original question. For a given cylinder radius, I believe I need two parameters: an expression for the axial coordinate and an expression for the azimuth. Could you help me with that parameterization of the ellipse?
– kreeser1
Nov 28 '18 at 23:35
You can choose a cylindrical coordinate system $(rho,phi,z)$ such that the ellipse is $rho = r, z = r tanalpha sinphi$, where $alpha$ is the angle between the plane normal and cylinder axis. You’ve already accepted an answer, but I can expand on this if you like.
– amd
Nov 30 '18 at 22:36
You can choose a cylindrical coordinate system $(rho,phi,z)$ such that the ellipse is $rho = r, z = r tanalpha sinphi$, where $alpha$ is the angle between the plane normal and cylinder axis. You’ve already accepted an answer, but I can expand on this if you like.
– amd
Nov 30 '18 at 22:36
add a comment |
2 Answers
2
active
oldest
votes
For greater simplicity, is better that cylinder axis $=$ $z$-axis. Let be $R$ the cylinder radius, $ax + by + cz + d = 0$ the plane containing the ellipse. Then, $r = R$ and
$$a Rcostheta + b Rsintheta + cz = d.$$
answered Nov 28 '18 at 17:46
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.1k42771
I think that it’s worth emphasizing that this is a pair of implicit Cartesian equations. Without the constraint $r=R$, you get an intersecting-looking surface, not an ellipse.
– amd
Nov 28 '18 at 21:57
@amd, correct. The ellipse is the intersection of the cylinder $x^2 + y^2 = R^2$ and the plane $ax+by+cz+d=0$.
– Martín-Blas Pérez Pinilla
Nov 29 '18 at 15:39
That was meant to be “interesting,” not “intersecting.” My fingers have minds of their own at times.
– amd
Nov 29 '18 at 19:27
add a comment |
As stated before assume the (normalized) equation of the plane is
$$ frac{a x + b y + c z}{sqrt{a^2+b^2+c^2}} = d $$
and the parametric equation of the cylinder
$$ pmatrix{x & y & z}= f(varphi,z) = pmatrix{ R cosvarphi, Rsinvarphi, z} $$
Where the two intersect you have your ellipse in cartesian coordinates
$$ z(varphi) = frac{d sqrt{a^2+b^2+c^2}-R (a cosvarphi+b sinvarphi) }{c} $$
or
$$vec{r}_{rm curve}(varphi) = pmatrix{x\y\z} = pmatrix{ R cosvarphi \ R sinvarphi \ frac{d sqrt{a^2+b^2+c^2}-R (a cosvarphi+b sinvarphi) }{c} } $$
Now let's find the properties of this ellipse.
The center of the ellipse is at $vec{r}_{rm cen} = pmatrix{0 & 0 & frac{d sqrt{a^2+b^2+c^2}}{c}} $
The ellipse in polar coordinates is
$$ r(varphi) = | vec{r}_{rm curve}-vec{r}_{rm cen} | = sqrt{R^2 + frac{R^2}{c^2} left( frac{a^2+b^2}{2} + a b sin(2 varphi) + frac{a^2-b^2}{2} cos(2varphi) right)} $$
This allows us to find the major and minor radii
$$ begin{aligned}
r_{rm major} & = R frac{ sqrt{a^2+b^2+c^2}}{c} \
r_{rm minor} & = R
end{aligned} $$
The principal axes of the ellipse are on
$$ varphi = frac{1}{2} {rm atan}left( frac{2 a b}{a^2-b^2} right) + n frac{pi}{2} $$
answered Dec 1 '18 at 0:22
ja72ja72
7,42212044
add a comment |
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2 Answers
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2 Answers
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For greater simplicity, is better that cylinder axis $=$ $z$-axis. Let be $R$ the cylinder radius, $ax + by + cz + d = 0$ the plane containing the ellipse. Then, $r = R$ and
$$a Rcostheta + b Rsintheta + cz = d.$$
answered Nov 28 '18 at 17:46
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.1k42771
I think that it’s worth emphasizing that this is a pair of implicit Cartesian equations. Without the constraint $r=R$, you get an intersecting-looking surface, not an ellipse.
– amd
Nov 28 '18 at 21:57
@amd, correct. The ellipse is the intersection of the cylinder $x^2 + y^2 = R^2$ and the plane $ax+by+cz+d=0$.
– Martín-Blas Pérez Pinilla
Nov 29 '18 at 15:39
That was meant to be “interesting,” not “intersecting.” My fingers have minds of their own at times.
– amd
Nov 29 '18 at 19:27
add a comment |
For greater simplicity, is better that cylinder axis $=$ $z$-axis. Let be $R$ the cylinder radius, $ax + by + cz + d = 0$ the plane containing the ellipse. Then, $r = R$ and
$$a Rcostheta + b Rsintheta + cz = d.$$
answered Nov 28 '18 at 17:46
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.1k42771
I think that it’s worth emphasizing that this is a pair of implicit Cartesian equations. Without the constraint $r=R$, you get an intersecting-looking surface, not an ellipse.
– amd
Nov 28 '18 at 21:57
@amd, correct. The ellipse is the intersection of the cylinder $x^2 + y^2 = R^2$ and the plane $ax+by+cz+d=0$.
– Martín-Blas Pérez Pinilla
Nov 29 '18 at 15:39
That was meant to be “interesting,” not “intersecting.” My fingers have minds of their own at times.
– amd
Nov 29 '18 at 19:27
add a comment |
For greater simplicity, is better that cylinder axis $=$ $z$-axis. Let be $R$ the cylinder radius, $ax + by + cz + d = 0$ the plane containing the ellipse. Then, $r = R$ and
$$a Rcostheta + b Rsintheta + cz = d.$$
answered Nov 28 '18 at 17:46
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.1k42771
For greater simplicity, is better that cylinder axis $=$ $z$-axis. Let be $R$ the cylinder radius, $ax + by + cz + d = 0$ the plane containing the ellipse. Then, $r = R$ and
$$a Rcostheta + b Rsintheta + cz = d.$$
answered Nov 28 '18 at 17:46
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.1k42771
answered Nov 28 '18 at 17:46
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.1k42771
answered Nov 28 '18 at 17:46
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.1k42771
answered Nov 28 '18 at 17:46
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.1k42771
34.1k42771
I think that it’s worth emphasizing that this is a pair of implicit Cartesian equations. Without the constraint $r=R$, you get an intersecting-looking surface, not an ellipse.
– amd
Nov 28 '18 at 21:57
@amd, correct. The ellipse is the intersection of the cylinder $x^2 + y^2 = R^2$ and the plane $ax+by+cz+d=0$.
– Martín-Blas Pérez Pinilla
Nov 29 '18 at 15:39
That was meant to be “interesting,” not “intersecting.” My fingers have minds of their own at times.
– amd
Nov 29 '18 at 19:27
add a comment |
I think that it’s worth emphasizing that this is a pair of implicit Cartesian equations. Without the constraint $r=R$, you get an intersecting-looking surface, not an ellipse.
– amd
Nov 28 '18 at 21:57
@amd, correct. The ellipse is the intersection of the cylinder $x^2 + y^2 = R^2$ and the plane $ax+by+cz+d=0$.
– Martín-Blas Pérez Pinilla
Nov 29 '18 at 15:39
That was meant to be “interesting,” not “intersecting.” My fingers have minds of their own at times.
– amd
Nov 29 '18 at 19:27
I think that it’s worth emphasizing that this is a pair of implicit Cartesian equations. Without the constraint $r=R$, you get an intersecting-looking surface, not an ellipse.
– amd
Nov 28 '18 at 21:57
I think that it’s worth emphasizing that this is a pair of implicit Cartesian equations. Without the constraint $r=R$, you get an intersecting-looking surface, not an ellipse.
– amd
Nov 28 '18 at 21:57
@amd, correct. The ellipse is the intersection of the cylinder $x^2 + y^2 = R^2$ and the plane $ax+by+cz+d=0$.
– Martín-Blas Pérez Pinilla
Nov 29 '18 at 15:39
@amd, correct. The ellipse is the intersection of the cylinder $x^2 + y^2 = R^2$ and the plane $ax+by+cz+d=0$.
– Martín-Blas Pérez Pinilla
Nov 29 '18 at 15:39
That was meant to be “interesting,” not “intersecting.” My fingers have minds of their own at times.
– amd
Nov 29 '18 at 19:27
That was meant to be “interesting,” not “intersecting.” My fingers have minds of their own at times.
– amd
Nov 29 '18 at 19:27
add a comment |
As stated before assume the (normalized) equation of the plane is
$$ frac{a x + b y + c z}{sqrt{a^2+b^2+c^2}} = d $$
and the parametric equation of the cylinder
$$ pmatrix{x & y & z}= f(varphi,z) = pmatrix{ R cosvarphi, Rsinvarphi, z} $$
Where the two intersect you have your ellipse in cartesian coordinates
$$ z(varphi) = frac{d sqrt{a^2+b^2+c^2}-R (a cosvarphi+b sinvarphi) }{c} $$
or
$$vec{r}_{rm curve}(varphi) = pmatrix{x\y\z} = pmatrix{ R cosvarphi \ R sinvarphi \ frac{d sqrt{a^2+b^2+c^2}-R (a cosvarphi+b sinvarphi) }{c} } $$
Now let's find the properties of this ellipse.
The center of the ellipse is at $vec{r}_{rm cen} = pmatrix{0 & 0 & frac{d sqrt{a^2+b^2+c^2}}{c}} $
The ellipse in polar coordinates is
$$ r(varphi) = | vec{r}_{rm curve}-vec{r}_{rm cen} | = sqrt{R^2 + frac{R^2}{c^2} left( frac{a^2+b^2}{2} + a b sin(2 varphi) + frac{a^2-b^2}{2} cos(2varphi) right)} $$
This allows us to find the major and minor radii
$$ begin{aligned}
r_{rm major} & = R frac{ sqrt{a^2+b^2+c^2}}{c} \
r_{rm minor} & = R
end{aligned} $$
The principal axes of the ellipse are on
$$ varphi = frac{1}{2} {rm atan}left( frac{2 a b}{a^2-b^2} right) + n frac{pi}{2} $$
answered Dec 1 '18 at 0:22
ja72ja72
7,42212044
add a comment |
As stated before assume the (normalized) equation of the plane is
$$ frac{a x + b y + c z}{sqrt{a^2+b^2+c^2}} = d $$
and the parametric equation of the cylinder
$$ pmatrix{x & y & z}= f(varphi,z) = pmatrix{ R cosvarphi, Rsinvarphi, z} $$
Where the two intersect you have your ellipse in cartesian coordinates
$$ z(varphi) = frac{d sqrt{a^2+b^2+c^2}-R (a cosvarphi+b sinvarphi) }{c} $$
or
$$vec{r}_{rm curve}(varphi) = pmatrix{x\y\z} = pmatrix{ R cosvarphi \ R sinvarphi \ frac{d sqrt{a^2+b^2+c^2}-R (a cosvarphi+b sinvarphi) }{c} } $$
Now let's find the properties of this ellipse.
The center of the ellipse is at $vec{r}_{rm cen} = pmatrix{0 & 0 & frac{d sqrt{a^2+b^2+c^2}}{c}} $
The ellipse in polar coordinates is
$$ r(varphi) = | vec{r}_{rm curve}-vec{r}_{rm cen} | = sqrt{R^2 + frac{R^2}{c^2} left( frac{a^2+b^2}{2} + a b sin(2 varphi) + frac{a^2-b^2}{2} cos(2varphi) right)} $$
This allows us to find the major and minor radii
$$ begin{aligned}
r_{rm major} & = R frac{ sqrt{a^2+b^2+c^2}}{c} \
r_{rm minor} & = R
end{aligned} $$
The principal axes of the ellipse are on
$$ varphi = frac{1}{2} {rm atan}left( frac{2 a b}{a^2-b^2} right) + n frac{pi}{2} $$
answered Dec 1 '18 at 0:22
ja72ja72
7,42212044
add a comment |
As stated before assume the (normalized) equation of the plane is
$$ frac{a x + b y + c z}{sqrt{a^2+b^2+c^2}} = d $$
and the parametric equation of the cylinder
$$ pmatrix{x & y & z}= f(varphi,z) = pmatrix{ R cosvarphi, Rsinvarphi, z} $$
Where the two intersect you have your ellipse in cartesian coordinates
$$ z(varphi) = frac{d sqrt{a^2+b^2+c^2}-R (a cosvarphi+b sinvarphi) }{c} $$
or
$$vec{r}_{rm curve}(varphi) = pmatrix{x\y\z} = pmatrix{ R cosvarphi \ R sinvarphi \ frac{d sqrt{a^2+b^2+c^2}-R (a cosvarphi+b sinvarphi) }{c} } $$
Now let's find the properties of this ellipse.
The center of the ellipse is at $vec{r}_{rm cen} = pmatrix{0 & 0 & frac{d sqrt{a^2+b^2+c^2}}{c}} $
The ellipse in polar coordinates is
$$ r(varphi) = | vec{r}_{rm curve}-vec{r}_{rm cen} | = sqrt{R^2 + frac{R^2}{c^2} left( frac{a^2+b^2}{2} + a b sin(2 varphi) + frac{a^2-b^2}{2} cos(2varphi) right)} $$
This allows us to find the major and minor radii
$$ begin{aligned}
r_{rm major} & = R frac{ sqrt{a^2+b^2+c^2}}{c} \
r_{rm minor} & = R
end{aligned} $$
The principal axes of the ellipse are on
$$ varphi = frac{1}{2} {rm atan}left( frac{2 a b}{a^2-b^2} right) + n frac{pi}{2} $$
answered Dec 1 '18 at 0:22
ja72ja72
7,42212044
As stated before assume the (normalized) equation of the plane is
$$ frac{a x + b y + c z}{sqrt{a^2+b^2+c^2}} = d $$
and the parametric equation of the cylinder
$$ pmatrix{x & y & z}= f(varphi,z) = pmatrix{ R cosvarphi, Rsinvarphi, z} $$
Where the two intersect you have your ellipse in cartesian coordinates
$$ z(varphi) = frac{d sqrt{a^2+b^2+c^2}-R (a cosvarphi+b sinvarphi) }{c} $$
or
$$vec{r}_{rm curve}(varphi) = pmatrix{x\y\z} = pmatrix{ R cosvarphi \ R sinvarphi \ frac{d sqrt{a^2+b^2+c^2}-R (a cosvarphi+b sinvarphi) }{c} } $$
Now let's find the properties of this ellipse.
The center of the ellipse is at $vec{r}_{rm cen} = pmatrix{0 & 0 & frac{d sqrt{a^2+b^2+c^2}}{c}} $
The ellipse in polar coordinates is
$$ r(varphi) = | vec{r}_{rm curve}-vec{r}_{rm cen} | = sqrt{R^2 + frac{R^2}{c^2} left( frac{a^2+b^2}{2} + a b sin(2 varphi) + frac{a^2-b^2}{2} cos(2varphi) right)} $$
This allows us to find the major and minor radii
$$ begin{aligned}
r_{rm major} & = R frac{ sqrt{a^2+b^2+c^2}}{c} \
r_{rm minor} & = R
end{aligned} $$
The principal axes of the ellipse are on
$$ varphi = frac{1}{2} {rm atan}left( frac{2 a b}{a^2-b^2} right) + n frac{pi}{2} $$
answered Dec 1 '18 at 0:22
ja72ja72
7,42212044
answered Dec 1 '18 at 0:22
ja72ja72
7,42212044
answered Dec 1 '18 at 0:22
ja72ja72
7,42212044
answered Dec 1 '18 at 0:22
ja72ja72
7,42212044
7,42212044
add a comment |
add a comment |
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You need a pair of implicit Cartesian equations to describe a curve in 3D. If you want a single equation, it’s not too hard to come up with a parameterization of the curve.
– amd
Nov 28 '18 at 21:58
That is a very good point; In my mind I am looking for a parameterization of the curve that takes into account the angle of inclination, but I failed to put that in my original question. For a given cylinder radius, I believe I need two parameters: an expression for the axial coordinate and an expression for the azimuth. Could you help me with that parameterization of the ellipse?
– kreeser1
Nov 28 '18 at 23:35
You can choose a cylindrical coordinate system $(rho,phi,z)$ such that the ellipse is $rho = r, z = r tanalpha sinphi$, where $alpha$ is the angle between the plane normal and cylinder axis. You’ve already accepted an answer, but I can expand on this if you like.
– amd
Nov 30 '18 at 22:36