How to calculate a Bézier curve with only start and end points?
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This animation from Wikipedia shows basically what I want to accomplish, however - I'm hoping to have it flipped around, where it starts progressing more towards the destination and "up" (in this image), and then arcs more directly to the end point. However, I only have access to a starting point and ending point, what I am hoping to do is be able to determine the other points by specifying a "height" (or width, whatever you want to call it), to determine how high the arc actually goes.
Help or direction would be appreciated.
geometry algorithms bezier-curve
edited Dec 25 '18 at 21:30
Glorfindel
3,42981830
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add a comment |
$begingroup$
This animation from Wikipedia shows basically what I want to accomplish, however - I'm hoping to have it flipped around, where it starts progressing more towards the destination and "up" (in this image), and then arcs more directly to the end point. However, I only have access to a starting point and ending point, what I am hoping to do is be able to determine the other points by specifying a "height" (or width, whatever you want to call it), to determine how high the arc actually goes.
Help or direction would be appreciated.
geometry algorithms bezier-curve
edited Dec 25 '18 at 21:30
Glorfindel
3,42981830
$endgroup$
1
$begingroup$
Can you post a sketch of what you are trying to accomplish? Also, you could look into quadratic Bézier curves.
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– Rahul
Oct 28 '10 at 4:32
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Well, a (cubic) Bézier requires four points, so as it stands, you still have two degrees of freedom for your problem. You might have to think about how to position those other two points to get what you want.
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– J. M. is not a mathematician
Oct 28 '10 at 4:37
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en.wikipedia.org/wiki/B%C3%A9zier_curve#Generalization In a parametric equation, $x=B(t)$, and $y=B(t)$
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– Mateen Ulhaq
Oct 28 '10 at 5:16
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If I'm reading this correctly, he wants a cubic bezier identical to the one in the picture but reflected across the $y$ axis, and scaled vertically (fixed at the start and end points). He wants to be able to have the two other points in the bezier a function of the height (vertical scale) of the curve.
$endgroup$
– Justin L.
Oct 28 '10 at 5:48
add a comment |
$begingroup$
This animation from Wikipedia shows basically what I want to accomplish, however - I'm hoping to have it flipped around, where it starts progressing more towards the destination and "up" (in this image), and then arcs more directly to the end point. However, I only have access to a starting point and ending point, what I am hoping to do is be able to determine the other points by specifying a "height" (or width, whatever you want to call it), to determine how high the arc actually goes.
Help or direction would be appreciated.
geometry algorithms bezier-curve
edited Dec 25 '18 at 21:30
Glorfindel
3,42981830
$endgroup$
This animation from Wikipedia shows basically what I want to accomplish, however - I'm hoping to have it flipped around, where it starts progressing more towards the destination and "up" (in this image), and then arcs more directly to the end point. However, I only have access to a starting point and ending point, what I am hoping to do is be able to determine the other points by specifying a "height" (or width, whatever you want to call it), to determine how high the arc actually goes.
Help or direction would be appreciated.
geometry algorithms bezier-curve
geometry algorithms bezier-curve
edited Dec 25 '18 at 21:30
Glorfindel
3,42981830
edited Dec 25 '18 at 21:30
Glorfindel
3,42981830
edited Dec 25 '18 at 21:30
Glorfindel
3,42981830
edited Dec 25 '18 at 21:30
Glorfindel
3,42981830
edited Dec 25 '18 at 21:30
Glorfindel
3,42981830
3,42981830
asked Oct 28 '10 at 4:18
Frustrated Guy
1
$begingroup$
Can you post a sketch of what you are trying to accomplish? Also, you could look into quadratic Bézier curves.
$endgroup$
– Rahul
Oct 28 '10 at 4:32
$begingroup$
Well, a (cubic) Bézier requires four points, so as it stands, you still have two degrees of freedom for your problem. You might have to think about how to position those other two points to get what you want.
$endgroup$
– J. M. is not a mathematician
Oct 28 '10 at 4:37
$begingroup$
en.wikipedia.org/wiki/B%C3%A9zier_curve#Generalization In a parametric equation, $x=B(t)$, and $y=B(t)$
$endgroup$
– Mateen Ulhaq
Oct 28 '10 at 5:16
$begingroup$
If I'm reading this correctly, he wants a cubic bezier identical to the one in the picture but reflected across the $y$ axis, and scaled vertically (fixed at the start and end points). He wants to be able to have the two other points in the bezier a function of the height (vertical scale) of the curve.
$endgroup$
– Justin L.
Oct 28 '10 at 5:48
add a comment |
1
$begingroup$
Can you post a sketch of what you are trying to accomplish? Also, you could look into quadratic Bézier curves.
$endgroup$
– Rahul
Oct 28 '10 at 4:32
$begingroup$
Well, a (cubic) Bézier requires four points, so as it stands, you still have two degrees of freedom for your problem. You might have to think about how to position those other two points to get what you want.
$endgroup$
– J. M. is not a mathematician
Oct 28 '10 at 4:37
$begingroup$
en.wikipedia.org/wiki/B%C3%A9zier_curve#Generalization In a parametric equation, $x=B(t)$, and $y=B(t)$
$endgroup$
– Mateen Ulhaq
Oct 28 '10 at 5:16
$begingroup$
If I'm reading this correctly, he wants a cubic bezier identical to the one in the picture but reflected across the $y$ axis, and scaled vertically (fixed at the start and end points). He wants to be able to have the two other points in the bezier a function of the height (vertical scale) of the curve.
$endgroup$
– Justin L.
Oct 28 '10 at 5:48
1
1
$begingroup$
Can you post a sketch of what you are trying to accomplish? Also, you could look into quadratic Bézier curves.
$endgroup$
– Rahul
Oct 28 '10 at 4:32
$begingroup$
Can you post a sketch of what you are trying to accomplish? Also, you could look into quadratic Bézier curves.
$endgroup$
– Rahul
Oct 28 '10 at 4:32
$begingroup$
Well, a (cubic) Bézier requires four points, so as it stands, you still have two degrees of freedom for your problem. You might have to think about how to position those other two points to get what you want.
$endgroup$
– J. M. is not a mathematician
Oct 28 '10 at 4:37
$begingroup$
Well, a (cubic) Bézier requires four points, so as it stands, you still have two degrees of freedom for your problem. You might have to think about how to position those other two points to get what you want.
$endgroup$
– J. M. is not a mathematician
Oct 28 '10 at 4:37
$begingroup$
en.wikipedia.org/wiki/B%C3%A9zier_curve#Generalization In a parametric equation, $x=B(t)$, and $y=B(t)$
$endgroup$
– Mateen Ulhaq
Oct 28 '10 at 5:16
$begingroup$
en.wikipedia.org/wiki/B%C3%A9zier_curve#Generalization In a parametric equation, $x=B(t)$, and $y=B(t)$
$endgroup$
– Mateen Ulhaq
Oct 28 '10 at 5:16
$begingroup$
If I'm reading this correctly, he wants a cubic bezier identical to the one in the picture but reflected across the $y$ axis, and scaled vertically (fixed at the start and end points). He wants to be able to have the two other points in the bezier a function of the height (vertical scale) of the curve.
$endgroup$
– Justin L.
Oct 28 '10 at 5:48
$begingroup$
If I'm reading this correctly, he wants a cubic bezier identical to the one in the picture but reflected across the $y$ axis, and scaled vertically (fixed at the start and end points). He wants to be able to have the two other points in the bezier a function of the height (vertical scale) of the curve.
$endgroup$
– Justin L.
Oct 28 '10 at 5:48
add a comment |
1 Answer
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active
oldest
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I had a related problem, where I knew the four points (start, end, two control points) and needed to generate the height (which it turns out is called the Sagitta). Here's my question:
Find sagitta of a cubic Bézier-described arc
My maths isn't strong enough to work it backwards, but you may be able to decode it from one of the very helpful answers there.
edited Apr 13 '17 at 12:19
Community♦
1
answered May 23 '11 at 21:29
GeoffGeoff
1435
$endgroup$
$begingroup$
I suspect that in the past 7 months the OP has either solved his problem or moved on.
$endgroup$
– Peter Taylor
May 23 '11 at 21:43
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I had a related problem, where I knew the four points (start, end, two control points) and needed to generate the height (which it turns out is called the Sagitta). Here's my question:
Find sagitta of a cubic Bézier-described arc
My maths isn't strong enough to work it backwards, but you may be able to decode it from one of the very helpful answers there.
edited Apr 13 '17 at 12:19
Community♦
1
answered May 23 '11 at 21:29
GeoffGeoff
1435
$endgroup$
$begingroup$
I suspect that in the past 7 months the OP has either solved his problem or moved on.
$endgroup$
– Peter Taylor
May 23 '11 at 21:43
add a comment |
$begingroup$
I had a related problem, where I knew the four points (start, end, two control points) and needed to generate the height (which it turns out is called the Sagitta). Here's my question:
Find sagitta of a cubic Bézier-described arc
My maths isn't strong enough to work it backwards, but you may be able to decode it from one of the very helpful answers there.
edited Apr 13 '17 at 12:19
Community♦
1
answered May 23 '11 at 21:29
GeoffGeoff
1435
$endgroup$
$begingroup$
I suspect that in the past 7 months the OP has either solved his problem or moved on.
$endgroup$
– Peter Taylor
May 23 '11 at 21:43
add a comment |
$begingroup$
I had a related problem, where I knew the four points (start, end, two control points) and needed to generate the height (which it turns out is called the Sagitta). Here's my question:
Find sagitta of a cubic Bézier-described arc
My maths isn't strong enough to work it backwards, but you may be able to decode it from one of the very helpful answers there.
edited Apr 13 '17 at 12:19
Community♦
1
answered May 23 '11 at 21:29
GeoffGeoff
1435
$endgroup$
I had a related problem, where I knew the four points (start, end, two control points) and needed to generate the height (which it turns out is called the Sagitta). Here's my question:
Find sagitta of a cubic Bézier-described arc
My maths isn't strong enough to work it backwards, but you may be able to decode it from one of the very helpful answers there.
edited Apr 13 '17 at 12:19
Community♦
1
answered May 23 '11 at 21:29
GeoffGeoff
1435
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
1
answered May 23 '11 at 21:29
GeoffGeoff
1435
answered May 23 '11 at 21:29
GeoffGeoff
1435
answered May 23 '11 at 21:29
GeoffGeoff
1435
1435
$begingroup$
I suspect that in the past 7 months the OP has either solved his problem or moved on.
$endgroup$
– Peter Taylor
May 23 '11 at 21:43
add a comment |
$begingroup$
I suspect that in the past 7 months the OP has either solved his problem or moved on.
$endgroup$
– Peter Taylor
May 23 '11 at 21:43
$begingroup$
I suspect that in the past 7 months the OP has either solved his problem or moved on.
$endgroup$
– Peter Taylor
May 23 '11 at 21:43
$begingroup$
I suspect that in the past 7 months the OP has either solved his problem or moved on.
$endgroup$
– Peter Taylor
May 23 '11 at 21:43
add a comment |
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1
$begingroup$
Can you post a sketch of what you are trying to accomplish? Also, you could look into quadratic Bézier curves.
$endgroup$
– Rahul
Oct 28 '10 at 4:32
$begingroup$
Well, a (cubic) Bézier requires four points, so as it stands, you still have two degrees of freedom for your problem. You might have to think about how to position those other two points to get what you want.
$endgroup$
– J. M. is not a mathematician
Oct 28 '10 at 4:37
$begingroup$
en.wikipedia.org/wiki/B%C3%A9zier_curve#Generalization In a parametric equation, $x=B(t)$, and $y=B(t)$
$endgroup$
– Mateen Ulhaq
Oct 28 '10 at 5:16
$begingroup$
If I'm reading this correctly, he wants a cubic bezier identical to the one in the picture but reflected across the $y$ axis, and scaled vertically (fixed at the start and end points). He wants to be able to have the two other points in the bezier a function of the height (vertical scale) of the curve.
$endgroup$
– Justin L.
Oct 28 '10 at 5:48