想实现缓动函数中的动画效果,发现很多都是贝塞尔方程实现的
所以现在需要实现它
贝塞尔曲线三阶方程
首先我们要重写Interpolator 中的getInterpolation()
方法
@Override
public float getInterpolation(float input) {
}
input 时间因子是介于0、1之间的,返回的值也是介于0,1之间的
也就是贝塞尔方程的x,y也是介于0,1之间
所以起点(0,0)终点(1,1)
简化后的三阶方程
思路是这样的已知x可求 t,根据t得出y
public static double cubicEquation(double t, double p1, double p2) {
double u = 1 - t;
double tt = t * t;
double uu = u * u;
double ttt = tt * t;
return 3 * uu * t * p1 + 3 * u * tt * p2 + ttt;
}
// 近似求解t
double tempX;
for (int i = mLastI; i < 4096; i++) {
t =i * STEP_SIZE;
tempX = cubicEquation(t, point1.x, point2.x);
if (tempX >= input) {
mLastI = i;
break;
}
}
value = cubicEquation(t, point1.y, point2.y);
这样很多缓动动画效果都可以实现了
通过这个网址可以编辑和查看动画效果
public class CubicBezierInterpolator implements Interpolator {
private int mLastI = 0;
private static final float STEP_SIZE = 1.0f / 4096;
private final PointF point1 = new PointF();
private final PointF point2 = new PointF();
public CubicBezierInterpolator(float x1, float y1, float x2, float y2) {
point1.x = x1;
point1.y = y1;
point2.x = x2;
point2.y = y2;
}
@Override
public float getInterpolation(float input) {
float t = input;
//如果重新开始要重置缓存的i。
if (input == 0) {
mLastI = 0;
}
// 近似求解t
double tempX;
for (int i = mLastI; i < 4096; i++) {
t = i * STEP_SIZE;
tempX = cubicEquation(t, point1.x, point2.x);
if (tempX >= input) {
mLastI = i;
break;
}
}
double value = cubicEquation(t, point1.y, point2.y);
//如果结束要重置缓存的i。
if (input == 1) {
mLastI = 0;
}
return (float) value;
}
public static double cubicEquation(double t, double p1, double p2) {
double u = 1 - t;
double tt = t * t;
double uu = u * u;
double ttt = tt * t;
return 3 * uu * t * p1 + 3 * u * tt * p2 + ttt;
}
}
补充:一般方程式
n阶方程
最后的最后 Android里源码里也有一个实现贝塞尔插值器的利用的是对曲线上点的枚举,不过控制点是固定的
然后精确度就是枚举数组的大小
/** * A pre-baked bezier-curved interpolator for indeterminate progress animations. */
final class BakedBezierInterpolator implements Interpolator {
private static final BakedBezierInterpolator INSTANCE = new BakedBezierInterpolator();
public final static BakedBezierInterpolator getInstance() {
return INSTANCE;
}
/** * Use getInstance instead of instantiating. */
public BakedBezierInterpolator() {
super();
}
/** * Lookup table values. * Generated using a Bezier curve from (0,0) to (1,1) with control points: * P0 (0,0) * P1 (0.4, 0) * P2 (0.2, 1.0) * P3 (1.0, 1.0) * <p/> * Values sampled with x at regular intervals between 0 and 1. */
private static final float[] VALUES = new float[]{
0.0f, 0.0002f, 0.0009f, 0.0019f, 0.0036f, 0.0059f, 0.0086f, 0.0119f, 0.0157f, 0.0209f,
0.0257f, 0.0321f, 0.0392f, 0.0469f, 0.0566f, 0.0656f, 0.0768f, 0.0887f, 0.1033f, 0.1186f,
0.1349f, 0.1519f, 0.1696f, 0.1928f, 0.2121f, 0.237f, 0.2627f, 0.2892f, 0.3109f, 0.3386f,
0.3667f, 0.3952f, 0.4241f, 0.4474f, 0.4766f, 0.5f, 0.5234f, 0.5468f, 0.5701f, 0.5933f,
0.6134f, 0.6333f, 0.6531f, 0.6698f, 0.6891f, 0.7054f, 0.7214f, 0.7346f, 0.7502f, 0.763f,
0.7756f, 0.7879f, 0.8f, 0.8107f, 0.8212f, 0.8326f, 0.8415f, 0.8503f, 0.8588f, 0.8672f,
0.8754f, 0.8833f, 0.8911f, 0.8977f, 0.9041f, 0.9113f, 0.9165f, 0.9232f, 0.9281f, 0.9328f,
0.9382f, 0.9434f, 0.9476f, 0.9518f, 0.9557f, 0.9596f, 0.9632f, 0.9662f, 0.9695f, 0.9722f,
0.9753f, 0.9777f, 0.9805f, 0.9826f, 0.9847f, 0.9866f, 0.9884f, 0.9901f, 0.9917f, 0.9931f,
0.9944f, 0.9955f, 0.9964f, 0.9973f, 0.9981f, 0.9986f, 0.9992f, 0.9995f, 0.9998f, 1.0f, 1.0f
};
private static final float STEP_SIZE = 1.0f / (VALUES.length - 1);
@Override
public float getInterpolation(float input) {
long a=System.nanoTime();
if (input >= 1.0f) {
return 1.0f;
}
if (input <= 0f) {
return 0f;
}
int position = Math.min(
(int) (input * (VALUES.length - 1)),
VALUES.length - 2);
float quantized = position * STEP_SIZE;
float difference = input - quantized;
float weight = difference / STEP_SIZE;
float result=VALUES[position] + weight * (VALUES[position + 1] - VALUES[position]);
Log.e("time1=",System.currentTimeMillis()-a+"");
return result;
}
}
杏树林研发 倪圣文