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.. include:: ../../abbreviation.txt
.. _bezier-geometry-ressources-page:
===============
Bézier Curves
===============
Definitions
-----------
A Bézier curve is defined by a set of control points :math:`\mathbf{P}_0` through
:math:`\mathbf{P}_n`, where :math:`n` is called its order (:math:`n = 1` for linear, 2 for
quadratic, 3 for cubic etc.). The first and last control points are always the end points of the
curve;
In the following :math:`0 \le t \le 1`.
Linear Bézier Curves
---------------------
Given distinct points :math:`\mathbf{P}_0` and :math:`\mathbf{P}_1`, a linear Bézier curve is simply
a straight line between those two points. The curve is given by
.. math::
\begin{align}
\mathbf{B}(t) &= \mathbf{P}_0 + t (\mathbf{P}_1 - \mathbf{P}_0) \\
&= (1-t) \mathbf{P}_0 + t \mathbf{P}_1
\end{align}
and is equivalent to linear interpolation.
Quadratic Bézier Curves
-----------------------
A quadratic Bézier curve is the path traced by the function :math:`\mathbf{B}(t)`, given points
:math:`\mathbf{P}_0`, :math:`\mathbf{P}_1`, and :math:`\mathbf{P}_2`,
.. math::
\mathbf{B}(t) = (1 - t)[(1 - t) \mathbf{P}_0 + t \mathbf{P}_1] + t [(1 - t) \mathbf{P}_1 + t \mathbf{P}_2]
which can be interpreted as the linear interpolant of corresponding points on the linear Bézier
curves from :math:`\mathbf{P}_0` to :math:`\mathbf{P}_1` and from :math:`\mathbf{P}_1` to
:math:`\mathbf{P}_2` respectively.
Rearranging the preceding equation yields:
.. math::
\begin{align}
\mathbf{B}(t) &= (1 - t)^{2} \mathbf{P}_0 + 2(1 - t)t \mathbf{P}_1 + t^{2} \mathbf{P}_2 \\
&= (\mathbf{P}_0 - 2\mathbf{P}_1 + \mathbf{P}_2) t^2 +
(-2\mathbf{P}_0 + 2\mathbf{P}_1) t +
\mathbf{P}_0
\end{align}
This can be written in a way that highlights the symmetry with respect to :math:`\mathbf{P}_1`:
.. math::
\mathbf{B}(t) = \mathbf{P}_1 + (1 - t)^{2} ( \mathbf{P}_0 - \mathbf{P}_1) + t^{2} (\mathbf{P}_2 - \mathbf{P}_1)
Which immediately gives the derivative of the Bézier curve with respect to `t`:
.. math::
\mathbf{B}'(t) = 2(1 - t) (\mathbf{P}_1 - \mathbf{P}_0) + 2t (\mathbf{P}_2 - \mathbf{P}_1)
from which it can be concluded that the tangents to the curve at :math:`\mathbf{P}_0` and
:math:`\mathbf{P}_2` intersect at :math:`\mathbf{P}_1`. As :math:`t` increases from 0 to 1, the
curve departs from :math:`\mathbf{P}_0` in the direction of :math:`\mathbf{P}_1`, then bends to
arrive at :math:`\mathbf{P}_2` from the direction of :math:`\mathbf{P}_1`.
The second derivative of the Bézier curve with respect to :math:`t` is
.. math::
\mathbf{B}''(t) = 2 (\mathbf{P}_2 - 2 \mathbf{P}_1 + \mathbf{P}_0)
Cubic Bézier Curves
-------------------
Four points :math:`\mathbf{P}_0`, :math:`\mathbf{P}_1`, :math:`\mathbf{P}_2` and
:math:`\mathbf{P}_3` in the plane or in higher-dimensional space define a cubic Bézier curve. The
curve starts at :math:`\mathbf{P}_0` going toward :math:`\mathbf{P}_1` and arrives at
:math:`\mathbf{P}_3` coming from the direction of :math:`\mathbf{P}_2`. Usually, it will not pass
through :math:`\mathbf{P}_1` or :math:`\mathbf{P}_2`; these points are only there to provide
directional information. The distance between :math:`\mathbf{P}_1` and :math:`\mathbf{P}_2`
determines "how far" and "how fast" the curve moves towards :math:`\mathbf{P}_1` before turning
towards :math:`\mathbf{P}_2`.
Writing :math:`\mathbf{B}_{\mathbf P_i,\mathbf P_j,\mathbf P_k}(t)` for the quadratic Bézier curve
defined by points :math:`\mathbf{P}_i`, :math:`\mathbf{P}_j`, and :math:`\mathbf{P}_k`, the cubic
Bézier curve can be defined as an affine combination of two quadratic Bézier curves:
.. math::
\mathbf{B}(t) = (1-t) \mathbf{B}_{\mathbf P_0,\mathbf P_1,\mathbf P_2}(t) +
t \mathbf{B}_{\mathbf P_1,\mathbf P_2,\mathbf P_3}(t)
The explicit form of the curve is:
.. math::
\begin{align}
\mathbf{B}(t) &= (1-t)^3 \mathbf{P}_0 + 3(1-t)^2t \mathbf{P}_1 + 3(1-t)t^2 \mathbf{P}_2 + t^3\mathbf{P}_3 \\
&= (\mathbf{P}_3 - 3\mathbf{P}_2 + 3\mathbf{P}_1 - \mathbf{P}_0) t^3 +
3(\mathbf{P}_2 - 2\mathbf{P}_1 + \mathbf{P}_0) t^2 +
3(\mathbf{P}_1 - \mathbf{P}_0) t +
\mathbf{P}_0
\end{align}
For some choices of :math:`\mathbf{P}_1` and :math:`\mathbf{P}_2` the curve may intersect itself, or
contain a cusp.
The derivative of the cubic Bézier curve with respect to :math:`t` is
.. math::
\mathbf{B}'(t) = 3(1-t)^2 (\mathbf{P}_1 - \mathbf{P}_0) + 6(1-t)t (\mathbf{P}_2 - \mathbf{P}_1) + 3t^2 (\mathbf{P}_3 - \mathbf{P}_2)
The second derivative of the Bézier curve with respect to :math:`t` is
.. math::
\mathbf{B}''(t) = 6(1-t) (\mathbf{P}_2 - 2 \mathbf{P}_1 + \mathbf{P}_0) + 6t (\mathbf{P}_3 - 2 \mathbf{P}_2 + \mathbf{P}_1)
Recursive definition
--------------------
A recursive definition for the Bézier curve of degree :math:`n` expresses it as a point-to-point
linear combination of a pair of corresponding points in two Bézier curves of degree :math:`n-1`.
Let :math:`\mathbf{B}_{\mathbf{P}_0\mathbf{P}_1\ldots\mathbf{P}_n}` denote the Bézier curve
determined by any selection of points :math:`\mathbf{P}_0`, :math:`\mathbf{P}_1`, :math:`\ldots`,
:math:`\mathbf{P}_{n-1}`.
The recursive definition is
.. math::
\begin{align}
\mathbf{B}_{\mathbf{P}_0}(t) &= \mathbf{P}_0 \\[1em]
\mathbf{B}(t) &= \mathbf{B}_{\mathbf{P}_0\mathbf{P}_1\ldots\mathbf{P}_n}(t) \\
&= (1-t) \mathbf{B}_{\mathbf{P}_0\mathbf{P}_1\ldots\mathbf{P}_{n-1}}(t) +
t \mathbf{B}_{\mathbf{P}_1\mathbf{P}_2\ldots\mathbf{P}_n}(t)
\end{align}
The formula can be expressed explicitly as follows:
.. math::
\begin{align}
\mathbf{B}(t) &= \sum_{i=0}^n b_{i,n}(t) \mathbf{P}_i \\
&= \sum_{i=0}^n {n\choose i}(1 - t)^{n - i}t^i \mathbf{P}_i \\
&= (1 - t)^n \mathbf{P}_0 +
{n\choose 1}(1 - t)^{n - 1}t \mathbf{P}_1 +
\cdots +
{n\choose n - 1}(1 - t)t^{n - 1} \mathbf{P}_{n - 1} +
t^n \mathbf{P}_n
\end{align}
where :math:`b_{i,n}(t)` are the Bernstein basis polynomials of degree :math:`n` and :math:`n
\choose i` are the binomial coefficients.
Degree elevation
----------------
A Bézier curve of degree :math:`n` can be converted into a Bézier curve of degree :math:`n + 1` with
the same shape.
To do degree elevation, we use the equality
.. math::
\mathbf{B}(t) = (1-t) \mathbf{B}(t) + t \mathbf{B}(t)
Each component :math:`\mathbf{b}_{i,n}(t) \mathbf{P}_i` is multiplied by :math:`(1-t)` and
:math:`t`, thus increasing a degree by one, without changing the value.
For arbitrary :math:`n`, we have
.. math::
\begin{align}
\mathbf{B}(t) &= (1 - t) \sum_{i=0}^n \mathbf{b}_{i,n}(t) \mathbf{P}_i +
t \sum_{i=0}^n \mathbf{b}_{i,n}(t) \mathbf{P}_i \\
&= \sum_{i=0}^n \frac{n + 1 - i}{n + 1} \mathbf{b}_{i, n + 1}(t) \mathbf{P}_i +
\sum_{i=0}^n \frac{i + 1}{n + 1} \mathbf{b}_{i + 1, n + 1}(t) \mathbf{P}_i \\
&= \sum_{i=0}^{n + 1} \mathbf{b}_{i, n + 1}(t)
\left(\frac{i}{n + 1} \mathbf{P}_{i - 1} +
\frac{n + 1 - i}{n + 1} \mathbf{P}_i\right) \\
&= \sum_{i=0}^{n + 1} \mathbf{b}_{i, n + 1}(t) \mathbf{P'}_i
\end{align}
Therefore the new control points are
.. math::
\mathbf{P'}_i = \frac{i}{n + 1} \mathbf{P}_{i - 1} + \frac{n + 1 - i}{n + 1} \mathbf{P}_i
It introduces two arbitrary points :math:`\mathbf{P}_{-1}` and :math:`\mathbf{P}_{n+1}` which are
cancelled in :math:`\mathbf{P'}_i`.
Matrix Forms
------------
.. math::
\mathbf{B}(t) = \mathbf{Transformation} \; \mathbf{Control} \; \mathbf{Basis} \; \mathbf{T}(t)
.. math::
\begin{align}
\mathbf{B^2}(t) &= \mathbf{Tr}
\begin{pmatrix}
P_{1x} & P_{2x} & P_{3x} \\
P_{1y} & P_{2x} & P_{3x} \\
1 & 1 & 1
\end{pmatrix}
\begin{pmatrix}
1 & -2 & 1 \\
0 & 2 & -2 \\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
1 \\
t \\
t^2
\end{pmatrix} \\[1em]
\mathbf{B^3}(t) &= \mathbf{Tr}
\begin{pmatrix}
P_{1x} & P_{2x} & P_{3x} & P_{4x} \\
P_{1y} & P_{2x} & P_{3x} & P_{4x} \\
0 & 0 & 0 & 0 \\
1 & 1 & 1 & 1
\end{pmatrix}
\begin{pmatrix}
1 & -3 & 3 & -1 \\
0 & 3 & -6 & 3 \\
0 & 0 & 3 & -3 \\
0 & 0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
1 \\
t \\
t^2 \\
t^3
\end{pmatrix}
\end{align}
.. B(t) = P0 (1 - 2t + t^2) +
P1 ( 2t - t^2) +
P2 t^2
Symbolic Calculation
--------------------
.. code-block:: py3
>>> from sympy import *
>>> P0, P1, P2, P3, P, t = symbols('P0 P1 P2 P3 P t')
>>> B2 = (1-t)*((1-t)*P0 + t*P1) + t*((1-t)*P1 + t*P2)
>>> collect(expand(B2), t)
P0 + t**2*(P0 - 2*P1 + P2) + t*(-2*P0 + 2*P1)
>>> B2_012 = (1-t)*((1-t)*P0 + t*P1) + t*((1-t)*P1 + t*P2)
>>> B2_123 = (1-t)*((1-t)*P1 + t*P2) + t*((1-t)*P2 + t*P3)
>>> B3 = (1-t)*B2_012 + t*B2_123
>>> collect(expand(B2), t)
P0 + t**3*(-P0 + 3*P1 - 3*P2 + P3) + t**2*(3*P0 - 6*P1 + 3*P2) + t*(-3*P0 + 3*P1)