问题
I have the following piece of code
int steps = 10;
for (int i = 0; i <= steps; i++) {
float t = i / float(steps);
console.log( "t " + t );
}
That out puts numbers in a linear fashion like this { 0, 0.1, 0.2, ..., 0.9, 1.0 } I would like apply the cubic (in or out) easing equation so the output numbers increase or decrease gradually
UPDATE
Not sure if my implementation if correct but I am getting curve as expected
float b = 0;
float c = 1;
float d = 1;
for (int i = 0; i <= steps; i++) {
float t = i / float(steps);
t /= d;
float e = c * t * t * t + b;
console.log( "e " + e );
//console.log( "t " + t );
}
回答1:
EasIn Cubic Function
/**
* @param {Number} t The current time
* @param {Number} b The start value
* @param {Number} c The change in value
* @param {Number} d The duration time
*/
function easeInCubic(t, b, c, d) {
t /= d;
return c*t*t*t + b;
}
EaseOut Cubic Function
/**
* @see {easeInCubic}
*/
function easeOutCubic(t, b, c, d) {
t /= d;
t--;
return c*(t*t*t + 1) + b;
}
Here you can find othere useful equations: http://www.gizma.com/easing/#cub1
Put this code in a while, as you've don before and you will have your output cubic decreasing numbers.
回答2:
You can use code from jQuery Easing plugin: http://gsgd.co.uk/sandbox/jquery/easing/
/*
* t: current time
* b: begInnIng value
* c: change In value
* d: duration
*/
easeInCubic: function (x, t, b, c, d) {
return c*(t/=d)*t*t + b;
},
easeOutCubic: function (x, t, b, c, d) {
return c*((t=t/d-1)*t*t + 1) + b;
},
easeInOutCubic: function (x, t, b, c, d) {
if ((t/=d/2) < 1) return c/2*t*t*t + b;
return c/2*((t-=2)*t*t + 2) + b;
}
来源:https://stackoverflow.com/questions/7346805/help-formulating-a-cubic-easing-equation