Skip to content

Instantly share code, notes, and snippets.

@gre
Last active March 19, 2024 01:19
Show Gist options
  • Save gre/1650294 to your computer and use it in GitHub Desktop.
Save gre/1650294 to your computer and use it in GitHub Desktop.
Simple Easing Functions in Javascript - see https://github.com/gre/bezier-easing
/*
* This work is free. You can redistribute it and/or modify it under the
* terms of the Do What The Fuck You Want To Public License, Version 2,
* as published by Sam Hocevar. See the COPYING file for more details.
*/
/*
* Easing Functions - inspired from http://gizma.com/easing/
* only considering the t value for the range [0, 1] => [0, 1]
*/
EasingFunctions = {
// no easing, no acceleration
linear: t => t,
// accelerating from zero velocity
easeInQuad: t => t*t,
// decelerating to zero velocity
easeOutQuad: t => t*(2-t),
// acceleration until halfway, then deceleration
easeInOutQuad: t => t<.5 ? 2*t*t : -1+(4-2*t)*t,
// accelerating from zero velocity
easeInCubic: t => t*t*t,
// decelerating to zero velocity
easeOutCubic: t => (--t)*t*t+1,
// acceleration until halfway, then deceleration
easeInOutCubic: t => t<.5 ? 4*t*t*t : (t-1)*(2*t-2)*(2*t-2)+1,
// accelerating from zero velocity
easeInQuart: t => t*t*t*t,
// decelerating to zero velocity
easeOutQuart: t => 1-(--t)*t*t*t,
// acceleration until halfway, then deceleration
easeInOutQuart: t => t<.5 ? 8*t*t*t*t : 1-8*(--t)*t*t*t,
// accelerating from zero velocity
easeInQuint: t => t*t*t*t*t,
// decelerating to zero velocity
easeOutQuint: t => 1+(--t)*t*t*t*t,
// acceleration until halfway, then deceleration
easeInOutQuint: t => t<.5 ? 16*t*t*t*t*t : 1+16*(--t)*t*t*t*t
}
@Rkokie
Copy link

Rkokie commented Jan 15, 2020

I see you're using the "getBezierAngle" function, which doesn't calculate the position in the bezier curve but the angle the curve is at, at a certain point in the curve.

@thednp
Copy link

thednp commented Jan 15, 2020

I used the other function as well, I still don't get the animation to complete, it's like half the distance and wrong coordinates.

@Rkokie
Copy link

Rkokie commented Jan 15, 2020

I'm curious what you're trying to do that isn't fulfilled with the function I posted. Could you whip up an example in codepen or jsfiddle?

@thednp
Copy link

thednp commented Jan 15, 2020

As you can see, I'm looking for a simple and working easing function for cubic bezier. Something like tween.js could use such.

@thednp
Copy link

thednp commented Jan 15, 2020

@Rkokie
To tell you the truth I'm trying to fix some problem with @gre 's script, join us here.

@jwdunn1
Copy link

jwdunn1 commented Jan 15, 2020

Not sure how well it works, but try this (found in Ocanvas):

cubicBezier: function (x1, y1, x2, y2, time) {

  // Inspired by Don Lancaster's two articles
  // http://www.tinaja.com/glib/cubemath.pdf
  // http://www.tinaja.com/text/bezmath.html

  // Set start and end point
  var x0 = 0,
    y0 = 0,
    x3 = 1,
    y3 = 1,

    // Convert the coordinates to equation space
    A = x3 - 3*x2 + 3*x1 - x0,
    B = 3*x2 - 6*x1 + 3*x0,
    C = 3*x1 - 3*x0,
    D = x0,
    E = y3 - 3*y2 + 3*y1 - y0,
    F = 3*y2 - 6*y1 + 3*y0,
    G = 3*y1 - 3*y0,
    H = y0,

    // Variables for the loop below
    t = time,
    iterations = 5,
    i, slope, x, y;

  // Loop through a few times to get a more accurate time value, according to the Newton-Raphson method
  // http://en.wikipedia.org/wiki/Newton's_method
  for (i = 0; i < iterations; i++) {

    // The curve's x equation for the current time value
    x = A* t*t*t + B*t*t + C*t + D;

    // The slope we want is the inverse of the derivate of x
    slope = 1 / (3*A*t*t + 2*B*t + C);

    // Get the next estimated time value, which will be more accurate than the one before
    t -= (x - time) * slope;
    t = t > 1 ? 1 : (t < 0 ? 0 : t);
  }

  // Find the y value through the curve's y equation, with the now more accurate time value
  y = Math.abs(E*t*t*t + F*t*t + G*t * H);

  return y;
}

The time parameter varies from 0 through 1, returning the corresponding y value.

@thednp
Copy link

thednp commented Jan 15, 2020

@jwdunn1 nop, but I thank you for the code.

@jwdunn1
Copy link

jwdunn1 commented Jan 15, 2020

The UnitBezier function in this CodePen tracks somewhat closely to the CSS cubic-bezier transition timing function: https://codepen.io/jwdunn/pen/VJGzNm

@thednp
Copy link

thednp commented Jan 16, 2020

@Jdunn1 confirmed, the UnitBezier kicks some punch and it moves pretty fast as well, thank you.

@thednp
Copy link

thednp commented Jan 16, 2020

@jwdunn1 if you're looking for an ES6 version, here's one.

@MisterSirCode
Copy link

Tip, you should convert them to arrow expressions, and dont minify, this is 2020, we dont need minification :D

ALSO you seemed to forget the "var" part of the EasingFunctions variable.

var EasingFunctions = {
 // no easing, no acceleration
 linear: t => {
  return t;
 },
 // accelerating from zero velocity
 easeInQuad: t => {
  return t * t;
 },
 // decelerating to zero velocity
 easeOutQuad: t => {
  return t * (2 - t);
 },
 // acceleration until halfway, then deceleration
 easeInOutQuad: t => {
  return t < 0.5 ? 2 * t * t : -1 + (4 - 2 * t) * t;
 },
 // accelerating from zero velocity
 easeInCubic: t => {
  return t * t * t;
 },
 // decelerating to zero velocity
 easeOutCubic: t => {
  return --t * t * t + 1;
 },
 // acceleration until halfway, then deceleration
 easeInOutCubic: t => {
  return t < 0.5 ? 4 * t * t * t : (t - 1) * (2 * t - 2) * (2 * t - 2) + 1;
 },
 // accelerating from zero velocity
 easeInQuart: t => {
  return t * t * t * t;
 },
 // decelerating to zero velocity
 easeOutQuart: t => {
  return 1 - --t * t * t * t;
 },
 // acceleration until halfway, then deceleration
 easeInOutQuart: t => {
  return t < 0.5 ? 8 * t * t * t * t : 1 - 8 * --t * t * t * t;
 },
 // accelerating from zero velocity
 easeInQuint: t => {
  return t * t * t * t * t;
 },
 // decelerating to zero velocity
 easeOutQuint: t => {
  return 1 + --t * t * t * t * t;
 },
 // acceleration until halfway, then deceleration
 easeInOutQuint: t => {
  return t < 0.5 ? 16 * t * t * t * t * t : 1 + 16 * --t * t * t * t * t;
 }
};

@gre
Copy link
Author

gre commented Jan 17, 2020

@skylerspark done! it's a snippet, up to you to var it or const it or module.exports= it or export default it ;)

@mhaidarhanif
Copy link

@gre Very useful! Thank you very much, Gaëtan!

@littlefinger1
Copy link

Very helpful!

@temm1210
Copy link

really awesome and useful!! Thank you!!

@aarongeorge
Copy link

aarongeorge commented Oct 5, 2020

Great gist @gre!

I thought it might be helpful to those who wanted to extend the curves to see how that could be done using higher order functions and increasing the power (Thanks to @lindell for the original snippet! This version however removes the unnecessary Math.abs, reuses easeIn for the easeOut function and changes linear to reference easeIn rather than easeInOut to reduce the complexity for each call (although this definitely isn't an exercise in performance))

const easeIn = p => t => Math.pow(t, p)
const easeOut = p => t => 1 - easeIn(p)(1 - t)
const easeInOut = p => t => t < .5 ? easeIn(p)(t * 2) / 2 : easeOut(p)(t * 2 - 1) / 2 + .5
const Easings = {
	linear: easeIn(1),
	easeInQuad: easeIn(2),
	easeOutQuad: easeOut(2),
	easeInOutQuad: easeInOut(2),
	easeInCubic: easeIn(3),
	easeOutCubic: easeOut(3),
	easeInOutCubic: easeInOut(3),
	easeInQuart: easeIn(4),
	easeOutQuart: easeOut(4),
	easeInOutQuart: easeInOut(4),
	easeInQuint: easeIn(5),
	easeOutQuint: easeOut(5),
	easeInOutQuint: easeInOut(5)
}

Knowing this, you can re-write @gre's gist like so:

const Easings = {
	linear = t => t,
	easeInQuad = t => Math.pow(t, 2),
	easeOutQuad = t => 1 - Math.pow(1 - t, 2),
	easeInOutQuad = t => t < .5 ? Math.pow(t * 2, 2) / 2 : (1 - Math.pow(1 - (t * 2 - 1), 2)) / 2 + .5,
	easeInCubic = t => Math.pow(t, 3),
	easeOutCubic = t => 1 - Math.pow(1 - t, 3),
	easeInOutCubic = t => t < .5 ? Math.pow(t * 2, 3) / 2 : (1 - Math.pow(1 - (t * 2 - 1), 3)) / 2 + .5,
	easeInQuart = t => Math.pow(t, 4),
	easeOutQuart = t => 1 - Math.pow(1 - t, 4),
	easeInOutQuart = t => t < .5 ? Math.pow(t * 2, 4) / 2 : (1 - Math.pow(1 - (t * 2 - 1), 4)) / 2 + .5,
	easeInQuint = t => Math.pow(t, 5),
	easeOutQuint = t => 1 - Math.pow(1 - t, 5),
	easeInOutQuint = t => t < .5 ? Math.pow(t * 2, 5) / 2 : (1 - Math.pow(1 - (t * 2 - 1), 5)) / 2 + .5
}

Although it is much more verbose, hopefully it's easier to understand for those who don't understand the pre-decrement tricks

Finally you can take the previous example and substitute parts of the equations to reference each other (cleaning it up a little bit)

const linear = t => t
const easeInQuad = t => Math.pow(t, 2)
const easeOutQuad = t => 1 - easeInQuad(1 - t)
const easeInOutQuad = t => t < .5 ? easeInQuad(t * 2) / 2 : easeOutQuad(t * 2 - 1) / 2 + .5
const easeInCubic = t => Math.pow(t, 3)
const easeOutCubic = t => 1 - easeInCubic(1 - t)
const easeInOutCubic = t => t < .5 ? easeInCubic(t * 2) / 2 : easeOutCubic(t * 2 - 1) / 2 + .5
const easeInQuart = t => Math.pow(t, 4)
const easeOutQuart = t => 1 - easeInQuart(1 - t)
const easeInOutQuart = t => t < .5 ? easeInQuart(t * 2) / 2 : easeOutQuart(t * 2 - 1) / 2 + .5
const easeInQuint = t => Math.pow(t, 5)
const easeOutQuint = t => 1 - easeInQuint(1 - t)
const easeInOutQuint = t => t < .5 ? easeInQuint(t * 2) / 2 : easeOutQuint(t * 2 - 1) / 2 + .5

@therealparmesh
Copy link

Great thread!

@gre
Copy link
Author

gre commented Apr 20, 2021

@aarongeorge thanks a lot! this is super useful! specifically, recently I was wondering how to generate it to any pow, I used this to something unrelated to easing but to do "distribution" => https://greweb.me/plots/109 (making my lines reaching more the edges)

@aarongeorge
Copy link

@gre glad I could return the favour! Full credit to @lindell for the original Higher Order implementation. I just wanted to break it down and tweak a few things so it benefits more people.

Copy link

ghost commented Nov 24, 2021

An alternative (works somehow)

eval(function(p,a,c,k,e,d){e=function(c){return c.toString(36)};if(!''.replace(/^/,String)){while(c--){d[c.toString(a)]=k[c]||c.toString(a)}k=[function(e){return d[e]}];e=function(){return'\\w+'};c=1};while(c--){if(k[c]){p=p.replace(new RegExp('\\b'+e(c)+'\\b','g'),k[c])}}return p}('f d={c:3(n){0 n},g:3(n){0 n*n},a:3(n){0 n*(2-n)},7:3(n){0 n<.5?2*n*n:(4-2*n)*n-1},9:3(n){0 n*n*n},b:3(n){0--n*n*n+1},e:3(n){0 n<.5?4*n*n*n:(n-1)*(2*n-2)*(2*n-2)+1},k:3(n){0 n*n*n*n},h:3(n){0 1- --n*n*n*n},i:3(n){0 n<.5?8*n*n*n*n:1-8*--n*n*n*n},l:3(n){0 n*n*n*n*n},j:3(n){0 1+--n*n*n*n*n},m:3(n){0 n<.5?6*n*n*n*n*n:1+6*--n*n*n*n*n}};',24,24,'return|||function|||16|easeInOutQuad||easeInCubic|easeOutQuad|easeOutCubic|linear|EasingFunctions|easeInOutCubic|var|easeInQuad|easeOutQuart|easeInOutQuart|easeOutQuint|easeInQuart|easeInQuint|easeInOutQuint|'.split('|'),0,{}))

@nukadelic
Copy link

visual cheat sheet https://easings.net/

@phil-green-CTI
Copy link

This is really good but the easing functions are causing me trouble. I would like the first 75% of the count to go very quickly, and then decelerate to a crawl. I what would the proper function be for this? I imagine something like:

const easeOut = t<.75 ? t => t : [SOMETHING] ;

but I don't know what to put there.

Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment