Moving Items Along Bezier Curves with CSS Animation (Part 2: Time Warp)

This is a follow-up of my earlier article. I realized that there is another way of achieving the same effect.

This article has lots of nice examples and explanations, the basic idea is to make very simple @keyframe rules, usually just a linear movement, then use timing function to distort the time, such that the motion path becomes the desired curve.

I'd like to call it the "time warp" hack.


How does it work?

Recall that a cubic Bezier curve is defined by this formula:

\[B(t) = (1-t)^3P_0+3(1-t)^2tP_1+3(1-t)t^2P_2+t^3P_3,\ 0 \le t \le 1.\]

In the 2D case, \(B(t)\) has two coordinates, \(x(t)\) and \(y(t)\). Define \(x_i\) to the be x coordinate of \(P_i\), then we have:

\[x(t) = (1-t)^3x_0+3(1-t)^2tx_1+3(1-t)t^2x_2+t^3x_3,\ 0 \le t \le 1.\]

So, for our animated element, we want to make sure that the x coordiante (i.e. the "left" CSS property) is \(x(t)\) at time \(t\). 

Because \(x(0)=x_0\) and \(x(1)=x_3\), we know that the @keyframes rule must be defined as

@keyframes move-x {
  from { left: x0; }
  to { left: x3; }
Now to determine the timing function, suppose the function is
cubic-bezier(u1, v1, u2, v2)
Note that this is again a 2D cubic Bezier curve, defined by four points \((0, 0), (u_1, v_1), (u_2, v_2), (1, 1)\). And the function for each coordinate would be:

\[ u(t) = 3(1-t)^2tu_1 + 3(1-t)t^2u_2 + t^3 \]
\[ v(t) = 3(1-t)^2tv_1 + 3(1-t)t^2v_2 + t^3 \]

Recall that, according to the CSS spec, at any time \(t\), the animate value \(\text{left}(t)\) is calculated as:

\[ \text{left}(t) = x_0 + v(t')(x_3 - x_0), \text{where}\ u(t')=t \]

Our first step is to set \(u_1=1/3\) and \(u_2=2/3\), such that \(u(t)=t\) for all \(t\).

Then we set:
\[ v_1 = \frac{x_1-x_0}{x_3-x_0}, v_2 = \frac{x_2-x_0}{x_3-x_0}  \]

This way we have

\[ v(t) = \frac{x(t) - x_0}{x_3-x_0} \]

Combining everything together, we know that if we set the animation-timing-function as

\[ \text{cubic-bezier}(\frac{1}{3}, \frac{x_1-x_0}{x_3-x_0}, \frac{2}{3},  \frac{x_2-x_0}{x_3-x_0}) \]

then we have \(\text{left}(t)=x(t)\) as desired.

Simliarly we can define @keyframes and animation-timing-function for \(y(t)\), then our CSS animation is completed.

Note: obviously the method does not work when \(x_0=x_3\) or \(y_0=y_3\), but in practice we can add a tiny offset in such cases.

Animation Timing

Observe that \(u(t)\) controls the mapping between the animation progress and the variable \(t\) of the curve. \(u_1=1/3\) and \(u_2=2/3\) are chosen to achieve the default linear timing. We can tweak the values of \(u_1\) and \(u_2\) to alter the timing.

Note that the methods from the previous article supports any timing functions, including "steps()" and "cubic-bezier()".

It's easy to see that a "cubic-bezier(u1, 1/3, u2, 2/3)" timing function for the previous article would be the same as setting the same values of \(u_1\) and \(u_2\) for the "time warp" version. In other words, animation timing is limited here, we have only the input progress mapping, but not the output progress mapping.

Of course the reason is we are already using the output progress mapping for the time warp effect.


plus said...
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plus said...

I don't quite understand the meaning of modifying these parameters cubic-bezier(1, v1, 0, v2), does this mean that the x0 point is t=1 and t=0 after reaching the x1 point
Suppose cubic-bezier (1, 0.33, 0, 0.66) matches the image results https://cubic-bezier.com/#1,.33,0,.66 this site?

Lu Wang said...


> Suppose cubic-bezier (1, 0.33, 0, 0.66) matches the image results https://cubic-bezier.com/#1,.33,0,.66 this site?

Yes that's correct.

I'm not sure what you mean by "cubic-bezier(1, v1, 0, v2)", and what are t, x0 and x1. Can you please elaborate?

On the other hand, as you can see in your example on cubic-bezier.com, (1, v1) and (0, v2) are the coordinates of the control points.