Symmetric bezier curve with unsymmetric control points
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I'm solving this problem about Cubic Bezier Curve which have $4$ control points $C_0,C_1,C_2,C_3$.
$$
C_0 = (0,0),: C_1 = (x_1,y_1),: C_2=(x_2,y_2),: C_3=(1,0)
$$
where $0 < x_1 < x_2 < 1$ and $y_1, y_2>0$.
If $C_1$ and $C_2$ are symmetric with respect to axis $x = 0.5$ (such as $C_1 = (0.25,1)$, $C_2 = (0.75,1)$), obviously whole bezier curve is symmetric (w.r.t. $x=0.5$). But I'm still confusing if converse of this statement is still correct. Is any symmetric (cubic) bezier curve can be existed with two unsymmetrical control points $C_1,C_2$?
I tried with reduction, if $C_1, C_2$ is not symmetric and the image $B(C_0,C_1,C_2,C_3)$ is symmetric, there should be $C_1^prime$ that symmetric with respect to $C_1$ but $C_1^prime := C2 cdot B(C_0,C_1,C_1^prime,C_3)$ is symmetric and when I move just one control point $C_1^prime$ to $C2$, I'm not sure that symmetric property will broke or not.
bezier-curve
edited yesterday
Coolwater
723720
asked 2 days ago
Donguk Jung
61
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Donguk Jung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
up vote
1
down vote
favorite
I'm solving this problem about Cubic Bezier Curve which have $4$ control points $C_0,C_1,C_2,C_3$.
$$
C_0 = (0,0),: C_1 = (x_1,y_1),: C_2=(x_2,y_2),: C_3=(1,0)
$$
where $0 < x_1 < x_2 < 1$ and $y_1, y_2>0$.
If $C_1$ and $C_2$ are symmetric with respect to axis $x = 0.5$ (such as $C_1 = (0.25,1)$, $C_2 = (0.75,1)$), obviously whole bezier curve is symmetric (w.r.t. $x=0.5$). But I'm still confusing if converse of this statement is still correct. Is any symmetric (cubic) bezier curve can be existed with two unsymmetrical control points $C_1,C_2$?
I tried with reduction, if $C_1, C_2$ is not symmetric and the image $B(C_0,C_1,C_2,C_3)$ is symmetric, there should be $C_1^prime$ that symmetric with respect to $C_1$ but $C_1^prime := C2 cdot B(C_0,C_1,C_1^prime,C_3)$ is symmetric and when I move just one control point $C_1^prime$ to $C2$, I'm not sure that symmetric property will broke or not.
bezier-curve
edited yesterday
Coolwater
723720
asked 2 days ago
Donguk Jung
61
New contributor
Donguk Jung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm solving this problem about Cubic Bezier Curve which have $4$ control points $C_0,C_1,C_2,C_3$.
$$
C_0 = (0,0),: C_1 = (x_1,y_1),: C_2=(x_2,y_2),: C_3=(1,0)
$$
where $0 < x_1 < x_2 < 1$ and $y_1, y_2>0$.
If $C_1$ and $C_2$ are symmetric with respect to axis $x = 0.5$ (such as $C_1 = (0.25,1)$, $C_2 = (0.75,1)$), obviously whole bezier curve is symmetric (w.r.t. $x=0.5$). But I'm still confusing if converse of this statement is still correct. Is any symmetric (cubic) bezier curve can be existed with two unsymmetrical control points $C_1,C_2$?
I tried with reduction, if $C_1, C_2$ is not symmetric and the image $B(C_0,C_1,C_2,C_3)$ is symmetric, there should be $C_1^prime$ that symmetric with respect to $C_1$ but $C_1^prime := C2 cdot B(C_0,C_1,C_1^prime,C_3)$ is symmetric and when I move just one control point $C_1^prime$ to $C2$, I'm not sure that symmetric property will broke or not.
bezier-curve
edited yesterday
Coolwater
723720
asked 2 days ago
Donguk Jung
61
New contributor
Donguk Jung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm solving this problem about Cubic Bezier Curve which have $4$ control points $C_0,C_1,C_2,C_3$.
$$
C_0 = (0,0),: C_1 = (x_1,y_1),: C_2=(x_2,y_2),: C_3=(1,0)
$$
where $0 < x_1 < x_2 < 1$ and $y_1, y_2>0$.
If $C_1$ and $C_2$ are symmetric with respect to axis $x = 0.5$ (such as $C_1 = (0.25,1)$, $C_2 = (0.75,1)$), obviously whole bezier curve is symmetric (w.r.t. $x=0.5$). But I'm still confusing if converse of this statement is still correct. Is any symmetric (cubic) bezier curve can be existed with two unsymmetrical control points $C_1,C_2$?
I tried with reduction, if $C_1, C_2$ is not symmetric and the image $B(C_0,C_1,C_2,C_3)$ is symmetric, there should be $C_1^prime$ that symmetric with respect to $C_1$ but $C_1^prime := C2 cdot B(C_0,C_1,C_1^prime,C_3)$ is symmetric and when I move just one control point $C_1^prime$ to $C2$, I'm not sure that symmetric property will broke or not.
bezier-curve
bezier-curve
edited yesterday
Coolwater
723720
asked 2 days ago
Donguk Jung
61
New contributor
Donguk Jung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Coolwater
723720
asked 2 days ago
Donguk Jung
61
New contributor
Donguk Jung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Coolwater
723720
edited yesterday
Coolwater
723720
edited yesterday
Coolwater
723720
723720
asked 2 days ago
Donguk Jung
61
New contributor
Donguk Jung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Donguk Jung
61
asked 2 days ago
Donguk Jung
61
61
New contributor
Donguk Jung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Donguk Jung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Donguk Jung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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Donguk Jung is a new contributor. Be nice, and check out our Code of Conduct.
Donguk Jung is a new contributor. Be nice, and check out our Code of Conduct.
Donguk Jung is a new contributor. Be nice, and check out our Code of Conduct.
Donguk Jung is a new contributor. Be nice, and check out our Code of Conduct.
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