Composite Bézier curve

Given two cubic Bézier curves with control points ${\displaystyle [\mathbf {P} _{0},\mathbf {P} _{1},\mathbf {P} _{2},\mathbf {P} _{3}]}$ and ${\displaystyle [\mathbf {P} _{3},\mathbf {P} _{4},\mathbf {P} _{5},\mathbf {P} _{6}]}$ respectively, the constraints for ensuring continuity at ${\displaystyle \mathbf {P} _{3}}$ can be defined as follows:

• ${\displaystyle C^{0}/G^{0}}$ (positional continuity) requires that they meet at the same point, which all Bézier splines do by definition. In this example, the shared point is ${\displaystyle \mathbf {P} _{3}}$

• ${\displaystyle C^{1}}$ (velocity continuity) requires the neighboring control points around the join to be mirrors of each other. In other words, they must follow the constraint of ${\displaystyle \mathbf {P} _{4}=2\mathbf {P} _{3}-\mathbf {P} _{2}}$

• ${\displaystyle G^{1}}$ (tangent continuity) requires the neighboring control points to be collinear with the join. This is less strict than ${\displaystyle C^{1}}$ continuity, leaving an extra degree of freedom which can be parameterized using a scalar ${\displaystyle \beta _{1}}$. The constraint can then be expressed by ${\displaystyle \mathbf {P} _{4}=\mathbf {P} _{3}+(\mathbf {P} _{3}-\mathbf {P} _{2})\beta _{1}}$

While the following continuity constraints are possible, they are rarely used with cubic Bézier splines, as other splines like the B-spline or the β-spline[7] will naturally handle higher constraints without loss of local control.

• ${\displaystyle C^{2}}$ (acceleration continuity) is constrained by ${\displaystyle \mathbf {P} _{5}=\mathbf {P} _{1}+4(\mathbf {P} _{3}-\mathbf {P} _{2})}$. However, applying this constraint across an entire cubic Bézier spline will cause a cascading loss of local control over the tangent points. The curve will still pass through every third point in the spline, but control over its shape will be lost. In order to achieve ${\displaystyle C^{2}}$ continuity using cubic curves, it's recommended to use a cubic uniform B-spline instead, as it ensures ${\displaystyle C^{2}}$ continuity without loss of local control, at the expense of no longer being guaranteed to pass through specific points

• ${\displaystyle G^{2}}$ (curvature continuity) is constrained by ${\displaystyle \mathbf {P} _{5}=\mathbf {P} _{3}+(\mathbf {P} _{3}-\mathbf {P} _{2})(2\beta _{1}+\beta _{1}^{2}+\beta _{2}/2)+(\mathbf {P} _{1}-\mathbf {P} _{2})\beta _{1}^{2}}$, leaving two degrees of freedom compared to ${\displaystyle C^{2}}$, in the form of two scalars ${\displaystyle \beta _{1}}$ and ${\displaystyle \beta _{2}}$. Higher degrees of geometric continuity is possible, though they get increasingly complex[8]

• ${\displaystyle C^{3}}$ (jolt continuity) is constrained by ${\displaystyle \mathbf {P} _{6}=\mathbf {P} _{3}+(\mathbf {P} _{3}-\mathbf {P} _{0})+6(\mathbf {P} _{1}-\mathbf {P} _{2}+\mathbf {P} _{3}-\mathbf {P} _{2})}$. Applying this constraint to the cubic Bézier spline will cause a complete loss of local control, as the entire spline is now fully constrained and defined by the first curve's control points. In fact, it is arguably no longer a spline, as its shape is now equivalent to extrapolating the first curve indefinitely, making it not only ${\displaystyle C^{3}}$ continuous, but ${\displaystyle C^{\infty }}$, as joins between separate curves no longer exist

Considering only the 90-degree unit-circular arc in the first quadrant, we define the endpoints ${\displaystyle \mathbf {A} }$ and ${\displaystyle \mathbf {B} }$ with control points ${\displaystyle \mathbf {A'} }$ and ${\displaystyle \mathbf {B'} }$, respectively, as:

${\displaystyle {\begin{aligned}\mathbf {A} &=[0,1]\\\mathbf {A'} &=[\mathbf {k} ,1]\\\mathbf {B'} &=[1,\mathbf {k} ]\\\mathbf {B} &=[1,0]\\\end{aligned}}}$

From the definition of the cubic Bézier curve, we have:

${\displaystyle \mathbf {C} (t)=(1-t)^{3}\mathbf {A} +3(1-t)^{2}t\mathbf {A'} +3(1-t)t^{2}\mathbf {B'} +t^{3}\mathbf {B} }$

With the point ${\displaystyle \mathbf {C} (t=0.5)}$ as the midpoint of the arc, we may write the following two equations:

${\displaystyle {\begin{aligned}\mathbf {C} &={\frac {1}{8}}\mathbf {A} +{\frac {3}{8}}\mathbf {A'} +{\frac {3}{8}}\mathbf {B'} +{\frac {1}{8}}\mathbf {B} \\\mathbf {C} &={\sqrt {1/2}}={\sqrt {2}}/2\end{aligned}}}$

Solving these equations for the x-coordinate (and identically for the y-coordinate) yields:

${\displaystyle {\frac {0}{8}}\mathbf {+} {\frac {3}{8}}\mathbf {k} +{\frac {3}{8}}+{\frac {1}{8}}={\sqrt {2}}/2}$

${\displaystyle \mathbf {k} ={\frac {4}{3}}({\sqrt {2}}-1)\approx 0.5522847498}$

Note however that the resulting Bézier curve is entirely outside the circle, with a maximum deviation of the radius of about 0.00027. By adding a small correction to intermediate points such as

${\displaystyle {\begin{aligned}\mathbf {A'} &=[\mathbf {k} +0.0009,1-0.00103]\\\mathbf {B'} &=[1-0.00103,\mathbf {k} +0.0009],\end{aligned}}}$

the magnitude of the radius deviation to 1 is reduced by a factor of about 3, to 0.000068 (at the expense of the derivability of the approximated circle curve at endpoints).

We may compose a circle of radius ${\displaystyle R}$ from an arbitrary number of cubic Bézier curves.[11] Let the arc start at point ${\displaystyle \mathbf {A} }$ and end at point ${\displaystyle \mathbf {B} }$, placed at equal distances above and below the x-axis, spanning an arc of angle ${\displaystyle \theta =2\phi }$:

${\displaystyle {\begin{aligned}\mathbf {A} _{x}&=R\cos(\phi )\\\mathbf {A} _{y}&=R\sin(\phi )\\\mathbf {B} _{x}&=\mathbf {A} _{x}\\\mathbf {B} _{y}&=-\mathbf {A} _{y}\end{aligned}}}$

The control points may be written as:[12]

${\displaystyle {\begin{aligned}\mathbf {A'} _{x}&={\frac {4R-\mathbf {A} _{x}}{3}}\\\mathbf {A'} _{y}&={\frac {(R-\mathbf {A} _{x})(3R-\mathbf {A} _{x})}{3\mathbf {A} _{y}}}\\\mathbf {B'} _{x}&=\mathbf {A'} _{x}\\\mathbf {B'} _{y}&=-\mathbf {A'} _{y}\end{aligned}}}$

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  Eight-segment quadratic polybezier (red) approximating a circle (black) with control points
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  Four-segment cubic polybezier (red) approximating a circle (black) with control points