“Addition” of two tilted ellipses
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I have two ellipses each centered at the origin and defined by:
- Semi major axis $a$
- Semi minor axis $b$
- Angle $psi$, which is the angle of $a$ counterclockwise from the positive $x$ axis
I add them together such that, if you parametrize an ellipse as $r(theta)$, then $r_{sum}(theta)=r_1(theta)+ r_2(theta)$.
Is it possible to express the new ellipse as a function of the parameters of the other two, so a function of $(a_1, a_2, b_1, b_2, psi_1, psi_2)$?
I hope this makes sense.
conic-sections
asked Nov 23 at 21:05
J. doe
112
|
show 2 more comments
up vote
2
down vote
favorite
I have two ellipses each centered at the origin and defined by:
- Semi major axis $a$
- Semi minor axis $b$
- Angle $psi$, which is the angle of $a$ counterclockwise from the positive $x$ axis
I add them together such that, if you parametrize an ellipse as $r(theta)$, then $r_{sum}(theta)=r_1(theta)+ r_2(theta)$.
Is it possible to express the new ellipse as a function of the parameters of the other two, so a function of $(a_1, a_2, b_1, b_2, psi_1, psi_2)$?
I hope this makes sense.
conic-sections
asked Nov 23 at 21:05
J. doe
112
I believe you are looking for the "implicit form" of the ellipse described above. Not sure how to get it (I'm sure it will depend on where they are with respect to each other etc), but that's the term you need
– Michael Stachowsky
Nov 23 at 21:27
2
Please include the specific parameterizations that you’re using. How do you know that the result is also an ellipse?
– amd
Nov 23 at 21:40
1
Which parametrization are you thinking of? We have e.g. $(x,y)=(acos u, bsin u)$ with the center of the ellipse in the origin. Or $r=frac{b^2}{a-costheta}$ in polar coordinates with a focal point in the origin.
– I like Serena
Nov 23 at 21:54
2
Addition of the radius $r(theta)$ in polar coordinates does not give an ellipse, as you can check by considering two ellipses $(a=1,b=epsilon,phi=0)$ and $(a=epsilon,b=1,phi=0)$ where $epsilonll1$.
– Rahul
Nov 23 at 22:17
1
The problem is that the angle $theta$ is a poor parameter for choosing corresponding points from the ellipses. It's worth noting that, because there's a linear transformation from any ellipse to another, you use that transformation to establish the correspondence between points. In that case, the "sum" of two ellipses is again an ellipse.
– Blue
Nov 23 at 23:16
|
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have two ellipses each centered at the origin and defined by:
- Semi major axis $a$
- Semi minor axis $b$
- Angle $psi$, which is the angle of $a$ counterclockwise from the positive $x$ axis
I add them together such that, if you parametrize an ellipse as $r(theta)$, then $r_{sum}(theta)=r_1(theta)+ r_2(theta)$.
Is it possible to express the new ellipse as a function of the parameters of the other two, so a function of $(a_1, a_2, b_1, b_2, psi_1, psi_2)$?
I hope this makes sense.
conic-sections
asked Nov 23 at 21:05
J. doe
112
I have two ellipses each centered at the origin and defined by:
- Semi major axis $a$
- Semi minor axis $b$
- Angle $psi$, which is the angle of $a$ counterclockwise from the positive $x$ axis
I add them together such that, if you parametrize an ellipse as $r(theta)$, then $r_{sum}(theta)=r_1(theta)+ r_2(theta)$.
Is it possible to express the new ellipse as a function of the parameters of the other two, so a function of $(a_1, a_2, b_1, b_2, psi_1, psi_2)$?
I hope this makes sense.
conic-sections
conic-sections
asked Nov 23 at 21:05
J. doe
112
asked Nov 23 at 21:05
J. doe
112
edited Nov 23 at 21:30
asked Nov 23 at 21:05
J. doe
112
asked Nov 23 at 21:05
J. doe
112
asked Nov 23 at 21:05
J. doe
112
112
I believe you are looking for the "implicit form" of the ellipse described above. Not sure how to get it (I'm sure it will depend on where they are with respect to each other etc), but that's the term you need
– Michael Stachowsky
Nov 23 at 21:27
2
Please include the specific parameterizations that you’re using. How do you know that the result is also an ellipse?
– amd
Nov 23 at 21:40
1
Which parametrization are you thinking of? We have e.g. $(x,y)=(acos u, bsin u)$ with the center of the ellipse in the origin. Or $r=frac{b^2}{a-costheta}$ in polar coordinates with a focal point in the origin.
– I like Serena
Nov 23 at 21:54
2
Addition of the radius $r(theta)$ in polar coordinates does not give an ellipse, as you can check by considering two ellipses $(a=1,b=epsilon,phi=0)$ and $(a=epsilon,b=1,phi=0)$ where $epsilonll1$.
– Rahul
Nov 23 at 22:17
1
The problem is that the angle $theta$ is a poor parameter for choosing corresponding points from the ellipses. It's worth noting that, because there's a linear transformation from any ellipse to another, you use that transformation to establish the correspondence between points. In that case, the "sum" of two ellipses is again an ellipse.
– Blue
Nov 23 at 23:16
|
show 2 more comments
I believe you are looking for the "implicit form" of the ellipse described above. Not sure how to get it (I'm sure it will depend on where they are with respect to each other etc), but that's the term you need
– Michael Stachowsky
Nov 23 at 21:27
2
Please include the specific parameterizations that you’re using. How do you know that the result is also an ellipse?
– amd
Nov 23 at 21:40
1
Which parametrization are you thinking of? We have e.g. $(x,y)=(acos u, bsin u)$ with the center of the ellipse in the origin. Or $r=frac{b^2}{a-costheta}$ in polar coordinates with a focal point in the origin.
– I like Serena
Nov 23 at 21:54
2
Addition of the radius $r(theta)$ in polar coordinates does not give an ellipse, as you can check by considering two ellipses $(a=1,b=epsilon,phi=0)$ and $(a=epsilon,b=1,phi=0)$ where $epsilonll1$.
– Rahul
Nov 23 at 22:17
1
The problem is that the angle $theta$ is a poor parameter for choosing corresponding points from the ellipses. It's worth noting that, because there's a linear transformation from any ellipse to another, you use that transformation to establish the correspondence between points. In that case, the "sum" of two ellipses is again an ellipse.
– Blue
Nov 23 at 23:16
I believe you are looking for the "implicit form" of the ellipse described above. Not sure how to get it (I'm sure it will depend on where they are with respect to each other etc), but that's the term you need
– Michael Stachowsky
Nov 23 at 21:27
I believe you are looking for the "implicit form" of the ellipse described above. Not sure how to get it (I'm sure it will depend on where they are with respect to each other etc), but that's the term you need
– Michael Stachowsky
Nov 23 at 21:27
2
2
Please include the specific parameterizations that you’re using. How do you know that the result is also an ellipse?
– amd
Nov 23 at 21:40
Please include the specific parameterizations that you’re using. How do you know that the result is also an ellipse?
– amd
Nov 23 at 21:40
1
1
Which parametrization are you thinking of? We have e.g. $(x,y)=(acos u, bsin u)$ with the center of the ellipse in the origin. Or $r=frac{b^2}{a-costheta}$ in polar coordinates with a focal point in the origin.
– I like Serena
Nov 23 at 21:54
Which parametrization are you thinking of? We have e.g. $(x,y)=(acos u, bsin u)$ with the center of the ellipse in the origin. Or $r=frac{b^2}{a-costheta}$ in polar coordinates with a focal point in the origin.
– I like Serena
Nov 23 at 21:54
2
2
Addition of the radius $r(theta)$ in polar coordinates does not give an ellipse, as you can check by considering two ellipses $(a=1,b=epsilon,phi=0)$ and $(a=epsilon,b=1,phi=0)$ where $epsilonll1$.
– Rahul
Nov 23 at 22:17
Addition of the radius $r(theta)$ in polar coordinates does not give an ellipse, as you can check by considering two ellipses $(a=1,b=epsilon,phi=0)$ and $(a=epsilon,b=1,phi=0)$ where $epsilonll1$.
– Rahul
Nov 23 at 22:17
1
1
The problem is that the angle $theta$ is a poor parameter for choosing corresponding points from the ellipses. It's worth noting that, because there's a linear transformation from any ellipse to another, you use that transformation to establish the correspondence between points. In that case, the "sum" of two ellipses is again an ellipse.
– Blue
Nov 23 at 23:16
The problem is that the angle $theta$ is a poor parameter for choosing corresponding points from the ellipses. It's worth noting that, because there's a linear transformation from any ellipse to another, you use that transformation to establish the correspondence between points. In that case, the "sum" of two ellipses is again an ellipse.
– Blue
Nov 23 at 23:16
|
show 2 more comments
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I believe you are looking for the "implicit form" of the ellipse described above. Not sure how to get it (I'm sure it will depend on where they are with respect to each other etc), but that's the term you need
– Michael Stachowsky
Nov 23 at 21:27
2
Please include the specific parameterizations that you’re using. How do you know that the result is also an ellipse?
– amd
Nov 23 at 21:40
1
Which parametrization are you thinking of? We have e.g. $(x,y)=(acos u, bsin u)$ with the center of the ellipse in the origin. Or $r=frac{b^2}{a-costheta}$ in polar coordinates with a focal point in the origin.
– I like Serena
Nov 23 at 21:54
2
Addition of the radius $r(theta)$ in polar coordinates does not give an ellipse, as you can check by considering two ellipses $(a=1,b=epsilon,phi=0)$ and $(a=epsilon,b=1,phi=0)$ where $epsilonll1$.
– Rahul
Nov 23 at 22:17
1
The problem is that the angle $theta$ is a poor parameter for choosing corresponding points from the ellipses. It's worth noting that, because there's a linear transformation from any ellipse to another, you use that transformation to establish the correspondence between points. In that case, the "sum" of two ellipses is again an ellipse.
– Blue
Nov 23 at 23:16