How do I calculate the maximum velocity of a CSS Bezier curve?

Question

I tried to do calculate if it's possible to get the top velocity of a co-ordinate point on a CSS Bezier curve. Below is my working process.

Calculate the top velocity point in a bezier curve (4 control points):

A Bezier curve can be described using a mathematical formula.

$B(t) = (1−t)³P₀ + 3(1−t)²tP₁ + 3(1−t)t²P₂ + t³P₃$

In CSS timing function, $P₀$ is $(0, 0)$ and represents the initial time and the initial state, $P₃$ is $(1, 1)$ and represents the final time and the final state. $P$ is a vector. In other words, we can put $x$ and $y$ instead of $P$ to get corresponding coordinates.

$X = (1−t)³X₀ + 3(1−t)²tX₁ + 3(1−t)t²X₂ + t³X₃$

$Y = (1−t)³Y₀ + 3(1−t)²tY₁ + 3(1−t)t²Y₂ + t³Y₃$

Since $P₀$ is $(0, 0)$ and $P₃$ is $(1, 1)$,

$X = 3(1−t)²tX₁ + 3(1−t)t²X₂ + t³$

$Y = 3(1−t)²tY₁ + 3(1−t)t²Y₂ + t³$

If I customise my curve to use $P₁ (0.4, 0)$ and $P₃ (0.2, 1)$,

$P₁ = (0.4, 0) P₂ = (0.2, 1)$

$X = 1.6t³ - 1.8t² + 1.2t$

$Y = -2t³ + 3t²$

Calculate the rate of change of $Y$,

$dy/dt = -6t² + 6t$

$dy²/dt² = -12t + 6$

$-12t + 6 = 0$

I get $t = 0.5$ Does that make sense?

Answer 1

The velocity is decided by both $x'(t)$ and $y'(t)$ as

$V(t)=(x'(t), y'(t))$

and the velocity's magnitude is $||V(t)||=\sqrt{x'(t)^2+y'(t)^2}$.

If you want to find the $t$ value corresponding to the maximum velocity, it is the same as finding the $t$ value where $f(t)=(x'(t)^2+y'(t)^2)$ is maximum. Therefore, you shall find the root of the polynomial $f'(t)=x'(t)x''(t)+y'(t)y''(t)$. Plugging in all the $x'(t)$, $x''(t)$, $y'(t)$ and $y''(t)$, we shall find that

$f'(t)=118.08t^3-159.84t^2+60.48t-4.32$,

which has 3 roots $t_0=0.0924934$, $t_1=0.58488$ and $t_2=0.676285$.

Please note that these 3 roots only correspond to the 3 points where the velocity attains local maximum or local minimum. Plugging $t_0, t_1$ and $t_2$ to $f''(t)$, we shall find that only $t_1$ results in a negative $f''$ and therefore, $f(t)$ has a local maximum at $t=0.58488$.

To find the global maximum within t=[0,1], we still need to compare the loacl maximum against the end values at $t=0.0$ and $t=1.0$ as

$||V(t=0.0)||= 1.2$,
$||V(t=0.58488)||= 1.632338$, and
$||V(t=1.0)||= 2.4$.

Therefore, your maximum speed occurs at $t=1.0$ with a local maximum at $t=0.58488$.

Answer 2

-12t + 6 = 0 is correct, but its solution is t = 1/2 which makes sense. Note that the location of maximal slope $dy/dx$ doesn't coincide with the location of maximal y-velocity $\ dy/dt\ $ because in the example of yours the x-velocity $\ dx/dt\ $ is non-constant.

Answer 3

Assuming that a point is moving along the given Bezier segment from $P_0$ to $P_3$ and its movement is governed by some unspecified force according to expressions that define the position of the point at time $t=[0,1]$,

\begin{align} P(t)&=P_0(1-t)^3+3P_1(1-t)^2t+3P_2(1-t)t^2+P_3t^3 ,\\ P'(t)&=3(P_1-P_0)(1-t)^2+6(P_2-P_1)(1-t)t+3(P_3-P_2)t^2 ,\\ P''(t)&=6(P_0-2P_1+P_2)(1-t)+6(P_1-2P_2+P_3)t . \end{align}

Velocity of the point is a vector,

\begin{align} P'(t)&=(P'_x(t),P'_y(t)) . \end{align}

You have considered only $P'_y(t)$ part of the movement, hence the result is indeed the moment of time when the velocity in $y$ direction is maximal (btw, you should get $t=\tfrac12$, not $t=2$).

If you need to find the maximum of the absolute value of velocity vector $||P'(t)||$ for $t=[0,1]$, then you have to use

\begin{align} ||P'(t)||&=\sqrt{(P_x'(t))^2+(P_y'(t))^2} \tag{1}\label{1} . \end{align}

For example, in case when $P_0=(0,0)$, $P_1=(0.4,0)$, $P_2=(0.2,1)$, $P_3=(1,1)$, expression \eqref{1} becomes

\begin{align} ||P'(t)||=s(t) &= \sqrt{(1.2(1-t)^2-1.2t+3.6t^2)^2+(6t-6t^2)^2} \\ &= \sqrt{59.04t^4-106.56t^3+60.48t^2-8.64t+1.44} , \end{align}

It is straightforward to find that zeros of $s'(t)$ are the three roots of

\begin{align} 236.16t^3-319.68t^2+120.96t-8.64&=0 , \end{align}

approximately $t_1=0.09249340673$. $t_2=0.5848801739$. and $t_3=0.6762849560$, but the global maximum of $||P'(t)||$ on $t=[0,1]$ is reached at $t=1$.