What algorithm can I use to find this ellipse inscribed in a quadrilateral?
$begingroup$
There's a certain drawing exercise designed to improve a drawing student's understanding of perspective and ability to draw shapes freehand. (The actual exercise is described by Irshad Karim on Drawabox.com, on the pages "Ghosted Planes" and "Ellipses in Planes".)
The exercise consists of drawing a convex quadrilateral, then drawing a window-like structure inside the quadrilateral, and finally drawing the ellipse implied by this structure.
Ideally, the result will look similar to this diagram:
The exact steps of the exercise are:
- Draw any convex quadrilateral $ABCD$.
- Draw the diagonals of the quadrilateral; call their intersection point $E$.
- Draw a line segment passing through $E$, which is concurrent to the edges $AD$ and $BC$. (Line segments are concurrent if they are all parallel, or if, when they are extended to lines, the resulting lines all intersect at a single point.) One endpoint should lie on $AB$, and be labeled $F$; the other endpoint should lie on $CD$, and be labeled $G$.
- Likewise, draw a line segment passing through $E$, which is concurrent to the edges $AB$ and $CD$. One endpoint lies on $AD$ and is labeled $H$, the other lies on $BC$ and is labeled $J$.
- Finally, draw the unique ellipse which is tangent to $ABCD$ at $F$, $G$, $H$, and $J$.
Given the coordinates of $A$, $B$, $C$ and $D$, what algorithm can be used to find the resulting ellipse?
I'm not 100% sure that there is always a unique ellipse tangent to $ABCD$ at $F$, $G$, $H$ and $J$, but it definitely seems like there is. There's always a unique ellipse which is tangent at $F$ and which passes through $G$, $H$, and $J$; and from experimentation, it looks like this ellipse is always tangent at $G$, $H$ and $J$ as well.
In the case where $ABCD$ is a square, everything is especially simple. The resulting ellipse is the circle inscribed in the square. I suspect that every case is simply the image of this case under some type of perspective transformation which preserves ellipses. But I don't know how to prove this, or how to make use of this fact.
geometry
asked Jan 9 at 3:40
Tanner SwettTanner Swett
4,2241639
$endgroup$
add a comment |
$begingroup$
There's a certain drawing exercise designed to improve a drawing student's understanding of perspective and ability to draw shapes freehand. (The actual exercise is described by Irshad Karim on Drawabox.com, on the pages "Ghosted Planes" and "Ellipses in Planes".)
The exercise consists of drawing a convex quadrilateral, then drawing a window-like structure inside the quadrilateral, and finally drawing the ellipse implied by this structure.
Ideally, the result will look similar to this diagram:
The exact steps of the exercise are:
- Draw any convex quadrilateral $ABCD$.
- Draw the diagonals of the quadrilateral; call their intersection point $E$.
- Draw a line segment passing through $E$, which is concurrent to the edges $AD$ and $BC$. (Line segments are concurrent if they are all parallel, or if, when they are extended to lines, the resulting lines all intersect at a single point.) One endpoint should lie on $AB$, and be labeled $F$; the other endpoint should lie on $CD$, and be labeled $G$.
- Likewise, draw a line segment passing through $E$, which is concurrent to the edges $AB$ and $CD$. One endpoint lies on $AD$ and is labeled $H$, the other lies on $BC$ and is labeled $J$.
- Finally, draw the unique ellipse which is tangent to $ABCD$ at $F$, $G$, $H$, and $J$.
Given the coordinates of $A$, $B$, $C$ and $D$, what algorithm can be used to find the resulting ellipse?
I'm not 100% sure that there is always a unique ellipse tangent to $ABCD$ at $F$, $G$, $H$ and $J$, but it definitely seems like there is. There's always a unique ellipse which is tangent at $F$ and which passes through $G$, $H$, and $J$; and from experimentation, it looks like this ellipse is always tangent at $G$, $H$ and $J$ as well.
In the case where $ABCD$ is a square, everything is especially simple. The resulting ellipse is the circle inscribed in the square. I suspect that every case is simply the image of this case under some type of perspective transformation which preserves ellipses. But I don't know how to prove this, or how to make use of this fact.
geometry
asked Jan 9 at 3:40
Tanner SwettTanner Swett
4,2241639
$endgroup$
$begingroup$
Interesting question, although I think the point of the original freehand-drawing exercise is not to use an algorithm.
$endgroup$
– David K
Jan 9 at 13:16
add a comment |
$begingroup$
There's a certain drawing exercise designed to improve a drawing student's understanding of perspective and ability to draw shapes freehand. (The actual exercise is described by Irshad Karim on Drawabox.com, on the pages "Ghosted Planes" and "Ellipses in Planes".)
The exercise consists of drawing a convex quadrilateral, then drawing a window-like structure inside the quadrilateral, and finally drawing the ellipse implied by this structure.
Ideally, the result will look similar to this diagram:
The exact steps of the exercise are:
- Draw any convex quadrilateral $ABCD$.
- Draw the diagonals of the quadrilateral; call their intersection point $E$.
- Draw a line segment passing through $E$, which is concurrent to the edges $AD$ and $BC$. (Line segments are concurrent if they are all parallel, or if, when they are extended to lines, the resulting lines all intersect at a single point.) One endpoint should lie on $AB$, and be labeled $F$; the other endpoint should lie on $CD$, and be labeled $G$.
- Likewise, draw a line segment passing through $E$, which is concurrent to the edges $AB$ and $CD$. One endpoint lies on $AD$ and is labeled $H$, the other lies on $BC$ and is labeled $J$.
- Finally, draw the unique ellipse which is tangent to $ABCD$ at $F$, $G$, $H$, and $J$.
Given the coordinates of $A$, $B$, $C$ and $D$, what algorithm can be used to find the resulting ellipse?
I'm not 100% sure that there is always a unique ellipse tangent to $ABCD$ at $F$, $G$, $H$ and $J$, but it definitely seems like there is. There's always a unique ellipse which is tangent at $F$ and which passes through $G$, $H$, and $J$; and from experimentation, it looks like this ellipse is always tangent at $G$, $H$ and $J$ as well.
In the case where $ABCD$ is a square, everything is especially simple. The resulting ellipse is the circle inscribed in the square. I suspect that every case is simply the image of this case under some type of perspective transformation which preserves ellipses. But I don't know how to prove this, or how to make use of this fact.
geometry
asked Jan 9 at 3:40
Tanner SwettTanner Swett
4,2241639
$endgroup$
There's a certain drawing exercise designed to improve a drawing student's understanding of perspective and ability to draw shapes freehand. (The actual exercise is described by Irshad Karim on Drawabox.com, on the pages "Ghosted Planes" and "Ellipses in Planes".)
The exercise consists of drawing a convex quadrilateral, then drawing a window-like structure inside the quadrilateral, and finally drawing the ellipse implied by this structure.
Ideally, the result will look similar to this diagram:
The exact steps of the exercise are:
- Draw any convex quadrilateral $ABCD$.
- Draw the diagonals of the quadrilateral; call their intersection point $E$.
- Draw a line segment passing through $E$, which is concurrent to the edges $AD$ and $BC$. (Line segments are concurrent if they are all parallel, or if, when they are extended to lines, the resulting lines all intersect at a single point.) One endpoint should lie on $AB$, and be labeled $F$; the other endpoint should lie on $CD$, and be labeled $G$.
- Likewise, draw a line segment passing through $E$, which is concurrent to the edges $AB$ and $CD$. One endpoint lies on $AD$ and is labeled $H$, the other lies on $BC$ and is labeled $J$.
- Finally, draw the unique ellipse which is tangent to $ABCD$ at $F$, $G$, $H$, and $J$.
Given the coordinates of $A$, $B$, $C$ and $D$, what algorithm can be used to find the resulting ellipse?
I'm not 100% sure that there is always a unique ellipse tangent to $ABCD$ at $F$, $G$, $H$ and $J$, but it definitely seems like there is. There's always a unique ellipse which is tangent at $F$ and which passes through $G$, $H$, and $J$; and from experimentation, it looks like this ellipse is always tangent at $G$, $H$ and $J$ as well.
In the case where $ABCD$ is a square, everything is especially simple. The resulting ellipse is the circle inscribed in the square. I suspect that every case is simply the image of this case under some type of perspective transformation which preserves ellipses. But I don't know how to prove this, or how to make use of this fact.
geometry
geometry
asked Jan 9 at 3:40
Tanner SwettTanner Swett
4,2241639
asked Jan 9 at 3:40
Tanner SwettTanner Swett
4,2241639
asked Jan 9 at 3:40
Tanner SwettTanner Swett
4,2241639
asked Jan 9 at 3:40
Tanner SwettTanner Swett
4,2241639
asked Jan 9 at 3:40
Tanner SwettTanner Swett
4,2241639
4,2241639
$begingroup$
Interesting question, although I think the point of the original freehand-drawing exercise is not to use an algorithm.
$endgroup$
– David K
Jan 9 at 13:16
add a comment |
$begingroup$
Interesting question, although I think the point of the original freehand-drawing exercise is not to use an algorithm.
$endgroup$
– David K
Jan 9 at 13:16
$begingroup$
Interesting question, although I think the point of the original freehand-drawing exercise is not to use an algorithm.
$endgroup$
– David K
Jan 9 at 13:16
$begingroup$
Interesting question, although I think the point of the original freehand-drawing exercise is not to use an algorithm.
$endgroup$
– David K
Jan 9 at 13:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is a projective transformation that takes the quadrilateral to a square. Draw a circle inscribed in the square, transform back and you get an ellipse inscribed in the quadrilateral.
The answers to this question describe how to find a projective transformation which takes the quadrilateral to a square: Mapping Irregular Quadrilateral to a Rectangle
edited Jan 9 at 4:04
Tanner Swett
4,2241639
answered Jan 9 at 3:57
Robert IsraelRobert Israel
320k23209459
$endgroup$
$begingroup$
This is the help that I needed, thank you! I went ahead and edited your answer by adding a link to another question with more information; please feel free to revert my edit.
$endgroup$
– Tanner Swett
Jan 9 at 4:06
$begingroup$
@TannerSwett I find this answer the best of the descriptions of how to construct the required homography.
$endgroup$
– amd
Jan 9 at 5:28
add a comment |
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1 Answer
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active
oldest
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1 Answer
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active
oldest
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$begingroup$
There is a projective transformation that takes the quadrilateral to a square. Draw a circle inscribed in the square, transform back and you get an ellipse inscribed in the quadrilateral.
The answers to this question describe how to find a projective transformation which takes the quadrilateral to a square: Mapping Irregular Quadrilateral to a Rectangle
edited Jan 9 at 4:04
Tanner Swett
4,2241639
answered Jan 9 at 3:57
Robert IsraelRobert Israel
320k23209459
$endgroup$
$begingroup$
This is the help that I needed, thank you! I went ahead and edited your answer by adding a link to another question with more information; please feel free to revert my edit.
$endgroup$
– Tanner Swett
Jan 9 at 4:06
$begingroup$
@TannerSwett I find this answer the best of the descriptions of how to construct the required homography.
$endgroup$
– amd
Jan 9 at 5:28
add a comment |
$begingroup$
There is a projective transformation that takes the quadrilateral to a square. Draw a circle inscribed in the square, transform back and you get an ellipse inscribed in the quadrilateral.
The answers to this question describe how to find a projective transformation which takes the quadrilateral to a square: Mapping Irregular Quadrilateral to a Rectangle
edited Jan 9 at 4:04
Tanner Swett
4,2241639
answered Jan 9 at 3:57
Robert IsraelRobert Israel
320k23209459
$endgroup$
$begingroup$
This is the help that I needed, thank you! I went ahead and edited your answer by adding a link to another question with more information; please feel free to revert my edit.
$endgroup$
– Tanner Swett
Jan 9 at 4:06
$begingroup$
@TannerSwett I find this answer the best of the descriptions of how to construct the required homography.
$endgroup$
– amd
Jan 9 at 5:28
add a comment |
$begingroup$
There is a projective transformation that takes the quadrilateral to a square. Draw a circle inscribed in the square, transform back and you get an ellipse inscribed in the quadrilateral.
The answers to this question describe how to find a projective transformation which takes the quadrilateral to a square: Mapping Irregular Quadrilateral to a Rectangle
edited Jan 9 at 4:04
Tanner Swett
4,2241639
answered Jan 9 at 3:57
Robert IsraelRobert Israel
320k23209459
$endgroup$
There is a projective transformation that takes the quadrilateral to a square. Draw a circle inscribed in the square, transform back and you get an ellipse inscribed in the quadrilateral.
The answers to this question describe how to find a projective transformation which takes the quadrilateral to a square: Mapping Irregular Quadrilateral to a Rectangle
edited Jan 9 at 4:04
Tanner Swett
4,2241639
answered Jan 9 at 3:57
Robert IsraelRobert Israel
320k23209459
edited Jan 9 at 4:04
Tanner Swett
4,2241639
edited Jan 9 at 4:04
Tanner Swett
4,2241639
edited Jan 9 at 4:04
Tanner Swett
4,2241639
4,2241639
answered Jan 9 at 3:57
Robert IsraelRobert Israel
320k23209459
answered Jan 9 at 3:57
Robert IsraelRobert Israel
320k23209459
answered Jan 9 at 3:57
Robert IsraelRobert Israel
320k23209459
320k23209459
$begingroup$
This is the help that I needed, thank you! I went ahead and edited your answer by adding a link to another question with more information; please feel free to revert my edit.
$endgroup$
– Tanner Swett
Jan 9 at 4:06
$begingroup$
@TannerSwett I find this answer the best of the descriptions of how to construct the required homography.
$endgroup$
– amd
Jan 9 at 5:28
add a comment |
$begingroup$
This is the help that I needed, thank you! I went ahead and edited your answer by adding a link to another question with more information; please feel free to revert my edit.
$endgroup$
– Tanner Swett
Jan 9 at 4:06
$begingroup$
@TannerSwett I find this answer the best of the descriptions of how to construct the required homography.
$endgroup$
– amd
Jan 9 at 5:28
$begingroup$
This is the help that I needed, thank you! I went ahead and edited your answer by adding a link to another question with more information; please feel free to revert my edit.
$endgroup$
– Tanner Swett
Jan 9 at 4:06
$begingroup$
This is the help that I needed, thank you! I went ahead and edited your answer by adding a link to another question with more information; please feel free to revert my edit.
$endgroup$
– Tanner Swett
Jan 9 at 4:06
$begingroup$
@TannerSwett I find this answer the best of the descriptions of how to construct the required homography.
$endgroup$
– amd
Jan 9 at 5:28
$begingroup$
@TannerSwett I find this answer the best of the descriptions of how to construct the required homography.
$endgroup$
– amd
Jan 9 at 5:28
add a comment |
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$begingroup$
Interesting question, although I think the point of the original freehand-drawing exercise is not to use an algorithm.
$endgroup$
– David K
Jan 9 at 13:16