Comprensión de la interpolación lineal en la animación de la interfaz de usuario

En la animación tradicional (dibujada a mano), un artista principal o principal dibuja fotogramas clave que definen el movimiento.

Un asistente, generalmente un pasante o un artista junior, luego dibuja los intermedios necesarios para la escena. El trabajo del asistente, también llamado intermediario, es hacer que las transiciones entre poses clave se vean suaves y naturales.

Los intermedios son necesarios ya que las animaciones sin ellos aparecen entrecortadas. Desafortunadamente, dibujar intermedios es un trabajo más o menos pesado. Pero estamos en el siglo XXI y tenemos computadoras que pueden manejar este tipo de tareas.

¿Recuerda lo que los maestros le dijeron en la escuela primaria acerca de que las computadoras son tontas? Es necesario indicar a las computadoras la secuencia exacta de pasos para realizar una acción. Hoy veremos una secuencia de pasos, o algoritmo, que ayuda a la computadora a dibujar los elementos intermedios necesarios para crear una animación fluida.

Usaré lienzo HTML5 y JavaScript para ilustrar el algoritmo. Sin embargo, podrá leer y comprender el artículo incluso si no está familiarizado con ellos.


Nuestro objetivo es simple, para animar una bola desde el punto A (startX, startY)a B (endX, endY).

Si esta escena se pasara a un estudio que hace animación tradicional, el artista senior dibujaría las siguientes poses clave ...

... y luego pasar la mesa de dibujo a un artista joven para que dibuje entremedios así.

Para nuestra situación, no hay estudio de animación ni tenemos artistas junior. Todo lo que tenemos es una meta, una computadora y la capacidad de escribir código.


El código HTML es simple, solo necesitamos una línea.

Esta parte del código JavaScript (que se muestra a continuación) simplemente toma s/> from the Document Object Model (DOM), gets context, and sets the width and height property of the canvas to match the viewport.

const canvas = document.getElementById(‘canvas’), context = canvas.getContext(‘2d’), width = canvas.width = window.innerWidth, height = canvas.height = window.innerHeight;

The function below draws a green solid circle of radius radius at x and y coordinates.

function drawBall(x, y, radius) { context.beginPath(); context.fillStyle = ‘#66DA79’; context.arc(x, y, radius, 0, 2 * Math.PI, false); context.fill();}

All of the above code is boilerplate to set up our animation, here’s the juicy part.

// Point Alet startX = 50, startY = 50;
// Point Blet endX = 420, endY = 380;
let x = startX, y = startY;
update();function update() { context.clearRect(0, 0, width, height); drawBall(x, y, 30); requestAnimationFrame(update);}

First of all, notice the update function being called right above its declaration. Second of all, notice requestAnimationFrame(update) which calls updaterepeatedly.

Flipbook animation is a good analogy for the kind of program we’re writing. Just like repeatedly flipping through a flipbook creates the illusion of motion, repeatedly calling the update function creates the illusion of motion for our green ball.

One thing to note about the code above is that “update” is just a name. The function could have been named anything else. Some programmers like the names nextFrame, loop, draw, or flip because the function is repeatedly called. The important part is what the function does.

On each subsequent call of update, we expect the function to draw a slightly different image on the canvas than the previous one.

Our current implementation of update draws the ball at the same exact position on each call drawBall(x, y, 30). There is no animation, but let’s change that. Below is a pen that contains the code we have written so far, you can open it and follow along.

On each iteration of update let’s go ahead and increment the value of x and y and see the kind of animation it creates.

function update() { context.clearRect(0, 0, width, height); drawBall(x, y, 30); x++; y++; requestAnimationFrame(update);}

Each iteration moves the ball forward in both the x and y directions, and repeated calls of update results in the animation as shown.

Heres’ the deal though, our intent was to move the ball from a start position to an end position. But we’re doing absolutely nothing about stopping the ball at an end position. Let’s fix that.

One obvious solution is to only increment the coordinates when they are smaller than endX and endY values. This way, once the ball crosses endX, endY its coordinates will stop updating and the ball will stop.

function update() { context.clearRect(0, 0, width, height); drawBall(x, y, 30); if(x <= endX && y <= endY) { x++; y++; } requestAnimationFrame(update);}

There’s an error in this approach though. Do you see it?

The problem here is that you cannot make the ball reach any end coordinate you want just by incrementing x and y values by 1. For example, consider end coordinates (500, 500), of course if you start at (0, 0) and add 1 to x and y, they will eventually get to(500, 500). But what if I chose (432, 373) as end coordinates?

Using the above approach, you can only reach points lying in a straight line 45 degrees from the horizontal axis.

Now you can use trigonometry and fancy math to calculate precise amounts that x and y should be incremented by to reach any coordinate you want. But you don’t need to do that when you have linear interpolation.

Approach with linear interpolation

Here’s what linear interpolation function a.k.a lerp looks like.

function lerp(min, max, fraction) { return (max — min) * fraction + min;}

To understand what linear interpolation does, consider a slider with a min value on the left end and a max value on the right end.

The next thing we need to choose is fraction. lerp takes fraction and converts that to a value between min and max.

When I put 0.5 in the lerp formula — no surprises — it translates to 50. This is exactly the halfway point between 0 (min) and 100 (max).

Similarly, if we choose another value for fraction say 0.85

And if we let fraction = 0, lerp will output 0 (min) and on fraction = 1, lerp will produce 100 (max).

I chose 0 and 100 as min and max to keep this example simple, but lerp will work for any arbitrary choice of min and max.

For values of fraction between 0 and 1, lerp allows you to interpolate between min and max. Or in other words, traverse between min and max values, where choosing 0 for fraction puts you at min, choosing 1 puts you at max and for any other value between 0 and 1, puts you anywhere between min and max. You can also see min and max as key poses, like in traditional animation, and lerp outputs as inbetweens ;-).

Alright, but what if someone gives a value outside the bounds of 0 and 1 as fraction to lerp? You see the formula for lerp is extremely straightforward with most basic of mathematical operations. There’s no trick or bad values here, just imagine extending the slider in both directions. Whatever value for fraction is supplied, lerpwill produce a logical result. We shouldn’t pay much thought to bad values here though, what we should think about is how all of this maps to animating the ball.

If you’re following along go ahead and change the update function to match the following code. Also don’t forget to add in the lerp function we defined at the beginning of this section.

function update() { context.clearRect(0, 0, width, height); drawBall(x, y, 30); x = lerp(x, endX, 0.1); y = lerp(y, endY, 0.1); requestAnimationFrame(update);}

Here’s a pen of what our program looks like now. Try clicking around :)

Smooth right? Here’s how lerp helps to improve the animation.

In the code, notice the variables x and y — which are initially set to startX and startY— mark the current position of the ball in any frame. Also my choice of 0.1 as fraction is arbitrary, you can choose any fractional value you want. Keep in mind that your choice of fraction affects the speed of the animation.

In every frame x and endX are taken as min and max and interpolated with 0.1 as fraction to obtain a new value forx. Similarly y and endY are used as min and max to obtain a new value for y using 0.1 as fraction.

The ball is then drawn at the newly calculated(x, y) coordinate.

These steps are repeated until x becomes endX and y becomes endY in which case min = max. When min and max become equal lerp throws the exact same value(min/max) for any further frames thus stopping the animation.

And that is how you use linear interpolation to smoothly animate a ball.

This short article covers a lot. We started by defining terms like key poses and inbetweens. Then we tried a trivial approach for drawing inbetweens and noticed its limitations. Finally, with linear interpolation, we were able to achieve our intent.

I hope all the math made sense to you. Feel free to play with linear interpolation even more. This article was inspired by Rachel Smith’s post on CodePen. Rachel’s post has many more examples, be sure to check it out.

Looking for more? I publish regularly on my blog at See you there, have a good one!