实现缓动函数

Question

我正在尝试移植并实现我发现的缓动功能

编辑

：我粘贴了错误的缓动功能，抱歉！这是正确的：

Math.easeOutQuart = function (t, b, c, d) {
    t /= d;
    t--;
    return -c * (t*t*t*t - 1) + b;
};

---

我使用的语言不是 Flash 或 Actionscript。这是我的代码：

ease:{outquart:{function(t as float,b as float,c as float,d as float) as float
        t=t/d
        t=t-1
        return -c * (t*t*t*t - 1) + b
    end function}}

我正在循环中调用该函数：

EDIT2 - 调用函数。

m.move 设置为 1 或 -1 表示移动方向，或 -5 +5 表示移动 5 个长度。 setspritemoves 被尽可能频繁地调用，目前它与系统调用的速度一样快，但我可以在毫秒计时器上触发调用。

setspritemoves:function()
                if m.move=1 then
                m.duration=1
                    if m.ishd then
                        for i=0 to m.spriteposx.count()-1
                            m.moveto[i]=m.spriteposx[i]+m.move*324
                        next i
                    else
                        for i=0 to m.spriteposx.count()-1
                            m.moveto[i]=m.spriteposx[i]+m.move*224
                        next i
                    end if                          
                else if m.move=5 then
                    m.duration=5
                    if m.ishd then
                        for i=0 to m.spriteposx.count()-1
                            m.moveto[i]=m.spriteposx[i]+m.move*324
                        next i
                    else
                        for i=0 to m.spriteposx.count()-1
                            m.moveto[i]=m.spriteposx[i]+m.move*224
                        next i
                    end if      
                else if m.move=-1 then
                m.duration=1
                    if m.ishd then
                        for i=0 to m.spriteposx.count()-1
                            m.moveto[i]=m.spriteposx[i]-m.move*324
                        next i
                    else
                        for i=0 to m.spriteposx.count()-1
                            m.moveto[i]=m.spriteposx[i]-m.move*224
                        next i
                    end if      
                else if m.move=-5 then
                    m.duration=5
                    if m.ishd then
                        for i=0 to m.spriteposx.count()-1
                            m.moveto[i]=m.spriteposx[i]-m.move*324
                        next i
                    else
                        for i=0 to m.spriteposx.count()-1
                            m.moveto[i]=m.spriteposx[i]-m.move*224
                        next i
                    end if
                end if
                end function

m.moveto[i] 是目标 x 坐标，m.time 是整数 I 增量，m.duration 是我假设的完成更改所需的时间，m.spriteposx 是当前位置我正在移动的物体。 [i] 是当前精灵。

如果我想在 1/2 秒内移动 345 个像素，时间的增量值应该是多少，持续时间应该是多少？

在我所有的实验中，我要么超出了很大的因素，要么只移动了几个像素。

目前，m.time 每次迭代都会增加 1，m.duration 为 100。我尝试了各种值，但似乎没有一个能始终如一地工作。

Answer 1

为什么你没有复制 1-1 的逻辑？补间是一个简单的算法，它只是以四次方式将坐标从 b 映射到 b+c，即 b + c*t^4< /code> 其中 t 获取区间 [0,1] 中的值。通过替换可以看出，当 t=0 时，该值为初始值 b，当 t->1 时，位置为所需的b+c。

这就是 t \= d 行的原因，因此 d 是任意持续时间，而 t 是自动画开始以来经过的时间得到上述范围内的值。但你已经完成了 t=t-1 并拍摄了底片等。为什么？

例如，在 0.5 秒内移动 345px，假设 px 是测量单位，您将有一个初始位置 b 和 c=345。 d=0.5 并且您将动画分割为您选择的长度间隔（取决于运行动画的机器的功率。移动设备不如台式机强大，因此您选择在这种情况下合理的帧率）。假设我们选择 24 fps，因此我们将间隔分割为 0.5*24 = 12 帧，并每 1/24^th 秒调用一次该函数，每次其中 t 取值 1/24、2/24 等。如果不是以秒为单位而是以帧为单位工作更舒服，则 d=12 和t 取值 1,2,...,12。无论哪种方式，计算都是相同的。

这是一个很好的示例（单击该框来运行演示），请随意修改这些值：

http://jsfiddle。净/nKhxw/

Answer 2

贝塞尔函数

借自 http: //blog.greweb.fr/2012/02/bezier-curve-based-easing-functions-from-concept-to-implementation/

/**
* KeySpline - use bezier curve for transition easing function
* is inspired from Firefox's nsSMILKeySpline.cpp
* Usage:
* var spline = new KeySpline(0.25, 0.1, 0.25, 1.0)
* spline.get(x) => returns the easing value | x must be in [0, 1] range
*/
function KeySpline (mX1, mY1, mX2, mY2) {

  this.get = function(aX) {
    if (mX1 == mY1 && mX2 == mY2) return aX; // linear
    return CalcBezier(GetTForX(aX), mY1, mY2);
  }

  function A(aA1, aA2) { return 1.0 - 3.0 * aA2 + 3.0 * aA1; }
  function B(aA1, aA2) { return 3.0 * aA2 - 6.0 * aA1; }
  function C(aA1)      { return 3.0 * aA1; }

  // Returns x(t) given t, x1, and x2, or y(t) given t, y1, and y2.
  function CalcBezier(aT, aA1, aA2) {
    return ((A(aA1, aA2)*aT + B(aA1, aA2))*aT + C(aA1))*aT;
  }

  // Returns dx/dt given t, x1, and x2, or dy/dt given t, y1, and y2.
  function GetSlope(aT, aA1, aA2) {
    return 3.0 * A(aA1, aA2)*aT*aT + 2.0 * B(aA1, aA2) * aT + C(aA1);
  }

  function GetTForX(aX) {
    // Newton raphson iteration
    var aGuessT = aX;
    for (var i = 0; i < 4; ++i) {
      var currentSlope = GetSlope(aGuessT, mX1, mX2);
      if (currentSlope == 0.0) return aGuessT;
      var currentX = CalcBezier(aGuessT, mX1, mX2) - aX;
      aGuessT -= currentX / currentSlope;
    }
    return aGuessT;
  }
}

常见曲线的别名：

{
    "ease":        [0.25, 0.1, 0.25, 1.0], 
    "linear":      [0.00, 0.0, 1.00, 1.0],
    "ease-in":     [0.42, 0.0, 1.00, 1.0],
    "ease-out":    [0.00, 0.0, 0.58, 1.0],
    "ease-in-out": [0.42, 0.0, 0.58, 1.0]
}

应该是轻松制作自己的曲线...

Answer 3

谢谢你，乔尼！

以下是如何实现 Bezier 缓动函数：C 或 Objective-C for iOS

// APPLE ORIGINAL TIMINGS:
//    linear        (0.00, 0.00), (0.00, 0.00), (1.00, 1.00), (1.00, 1.00)
//    easeIn        (0.00, 0.00), (0.42, 0.00), (1.00, 1.00), (1.00, 1.00)
//    easeOut       (0.00, 0.00), (0.00, 0.00), (0.58, 1.00), (1.00, 1.00)
//    easeInEaseOut (0.00, 0.00), (0.42, 0.00), (0.58, 1.00), (1.00, 1.00)
//    default       (0.00, 0.00), (0.25, 0.10), (0.25, 1.00), (1.00, 1.00)

+(double)defaultEase_Linear:(double)t
{
    return t;
}

// Замедление в начале
+(double)defaultEase_In:(double)t
{
    return [AnimationMath easeBezier_t:t

                              point0_x:0
                              point0_y:0

                              point1_x:0.42
                              point1_y:0

                              point2_x:1
                              point2_y:1

                              point3_x:1
                              point3_y:1];
}

// Замедление в конце
+(double)defaultEase_Out:(double)t
{
    return [AnimationMath easeBezier_t:t

                              point0_x:0
                              point0_y:0

                              point1_x:0
                              point1_y:0

                              point2_x:0.58
                              point2_y:1

                              point3_x:1
                              point3_y:1];
}

+(double)defaultEase_InOut:(double)t
{
    return [AnimationMath easeBezier_t:t

                              point0_x:0
                              point0_y:0

                              point1_x:0.42
                              point1_y:0

                              point2_x:0.58
                              point2_y:1

                              point3_x:1
                              point3_y:1];
}

+(double)defaultEase_default:(double)t
{
    return [AnimationMath easeBezier_t:t

                              point0_x:0
                              point0_y:0

                              point1_x:0.25
                              point1_y:0.1

                              point2_x:0.25
                              point2_y:1.0

                              point3_x:1
                              point3_y:1];
}


// For *better understanding* there is p1 and p2, because it is a Bezier curve from 0,0 to 1,0. So, you can remove p1 and p2 from this method, it is just for better understanding what's going on here

double ease_bezier_A(double aA1, double aA2) { return 1.0 - 3.0 * aA2 + 3.0 * aA1; }
double ease_bezier_B(double aA1, double aA2) { return 3.0 * aA2 - 6.0 * aA1; }
double ease_bezier_C(double aA1)      { return 3.0 * aA1; }

// Returns x(t) given t, x1, and x2, or y(t) given t, y1, and y2.
double ease_bezier_calc(double aT, double aA1, double aA2) {
    return ((ease_bezier_A(aA1, aA2)*aT + ease_bezier_B(aA1, aA2))*aT + ease_bezier_C(aA1))*aT;
}

// Returns dx/dt given t, x1, and x2, or dy/dt given t, y1, and y2.
double ease_bezier_get_slope(double aT, double aA1, double aA2) {
    return 3.0 * ease_bezier_A(aA1, aA2)*aT*aT + 2.0 * ease_bezier_B(aA1, aA2) * aT + ease_bezier_C(aA1);
}

double ease_bezier_get_t_for_x(double aX, double mX1, double mX2) {
    // Newton raphson iteration
    double aGuessT = aX;
    for (int i = 0; i < 4; ++i) {
        double currentSlope = ease_bezier_get_slope(aGuessT, mX1, mX2);
        if (currentSlope == 0.0) return aGuessT;
        double currentX = ease_bezier_calc(aGuessT, mX1, mX2) - aX;
        aGuessT -= currentX / currentSlope;
    }
    return aGuessT;
}



// Objective-C
// For ***better understanding*** there is p1 and p2, because it is a Bezier curve from 0,0 to 1,0. So, you can remove p1 and p2 from this method, it is just for better understanding what's going on here
// p1_x always = 0
// p1_y always = 0
// p2_x always = 1.0
// p2_y always = 1.0
+(double)easeBezier_t:(double)t
             point0_x:(double)point0_x point0_y:(double)point0_y
             point1_x:(double)point1_x point1_y:(double)point1_y
             point2_x:(double)point2_x point2_y:(double)point2_y
             point3_x:(double)point3_x point3_y:(double)point3_y
{
    if (point0_x != 0 || point0_y != 0 || point3_x != 1 || point3_y != 1) {
        [NSException raise:@"Error! Your bezier is wrong!!!" format:@""];
    }

    double v = ease_bezier_calc(ease_bezier_get_t_for_x(t, point1_x, point2_x), point1_y, point2_y);

    return v;
}