幫助制定一個三次緩和方程

Question

我有以下代碼

int steps = 10;
for (int i = 0; i <= steps; i++) {
    float t = i / float(steps);
    console.log( "t  " + t );
}

那個數字以線性方式排列，如{0,0.1,0.2，...，0.9,1.0}我想應用立方（in或out）緩動方程，使輸出數逐漸增加或減少

---

UPDATE

不確定我的實現是否正確，但我得到了預期的曲線

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 );
}

Answer 1

EasIn立體功能

/**
 * @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立方功能

/**
 * @see {easeInCubic}
 */
function easeOutCubic(t, b, c, d) {
   t /= d;
   t--;
   return c*(t*t*t + 1) + b;
}

在這里您可以找到其他有用的公式： http ： //www.gizma.com/easing/#cub1

把這段代碼放一段時間，就像你以前一樣，你的輸出立方數會減少。

Answer 2

你可以使用jQuery Easing插件中的代碼： 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;
}