插值两个值之间的值

时间:2014-05-12 19:31:08

标签: c++ c math interpolation easing

我正在寻找一个在柔和曲线之后插值两个值之间的值的函数。

以下是一个例子:

enter image description here

float xLerp(float mMin, float mMax, float mFactor) { ... }

mFactor应介于1和0之间。

如何创建类似于我绘制的功能?

3 个答案:

答案 0 :(得分:2)

正如我在评论中所说,指数很合适:

double mBase = 5; // higher = more "curvy"
double minY = pow(mBase, mMin - mBase);
double maxY = pow(mBase, mMax - mBase);
double scale = (mMax - mMin) / (maxY - minY);
double shift = mMin - minY;
return pow(mBase, mFactor - mBase) * scale + shift;

呃,斜坡都错了,丑陋的黑客......

答案 1 :(得分:2)

抛物线会起作用。

#define ASSERT (cond) // Some assertion macro
/**
 * f(x) = a(x)^2 + 0x + c // b is zero, because no x shift.
 * f(0) = mMin == c = mMin
 * f(1) = mMax == a + mMin = mMax == a = mMax - mMin
 */
float xLerp (float mMin, float mMax, float mFactor) {
    ASSERT(0 <= mFactor && mFactor <= 1);
    float a = mMax - mMin;
    return a * mFactor * mFactor + mMin;
}

答案 2 :(得分:2)

正弦波会起作用。

#include <math.h>
#define ASSERT (cond) // some assertion macro
/**
 * f(x) = a * sin(x / t * PI) + b
 * f(0) = mMin == b = mMin
 * f(1) = mMax == a * sin(1/t * pi) + mMin == a * sin(pi/t) = mMax - mMin
 *        a = (mMax - mMin) / sin(pi/t)
 * (Let t == 1 for "normal" periodicity. 0 < t <= 1)
 * a = (mMax - mMin) / sin(pi/t) == a = mMax - mMin
 */
float xLerp (float mMin, float mMax, float mFactor){
    ASSERT(0 <= mFactor && mFactor <= 1);
    float a = mMax - mMin;
    return a * sin(mFactor * PI) + mMin;
}    
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