我能够从 Helmut Sennewald 的 LTSpic 电位器代码中翻译出来。
但数学不是我的事,所以我不确定是否有可能扭转这一点,如果给定点
ln 段算法将在原始直线段上返回相应的值。现在我 super 天真的解决方案是扫描东西以获得答案。这是原始的工作前向代码。
//power function slider
public float projectOntoSegment(float val, float min, float max) {
//val is location as percentage on segment
float range = max - min;
float tap_point = 0.2f;//knee point as percentage
float tap_val = range * 0.05f; //how much at knee:lets say 5% of scale
double exp = Math.log(tap_val / range) / Math.log(tap_point);
double ratio = Math.pow(val, exp);
float eff_val = min + (float) (range * ratio);
//eff_val = min + (range*val);//linear
return eff_val;
}
最佳答案
我在一些stackoverflow和运气的帮助下找到了解决方案。为什么甚至需要这个?因为在对数 slider 的初始化阶段需要它。以下是方法和示例用法。用户给出对数刻度的初始值。它仍然只是想要的值(value)而不是真实的。要获得有效值,需要的值必须首先在 slider 上反向映射为等效线性值,然后馈入正向对数映射。这会产生良好而流畅的全方位 slider 体验。
//ln power log
public float projectOntoLogCurve(float input, float min, float max, float kneePoint, float kneeMount) {
//input is location as percentage on segment
float range = max - min;
double exp = Math.log(kneeMount) / Math.log(kneePoint);
double ratio = Math.pow(input, exp);
float output = min + (float) (range * ratio);
return output;
}
//**********************************************
//reverse ln power log
public float projectBackFromLogCurve(float ln_input, float min, float max, float kneePoint, float kneeMount) {
//ln_input is logaritmic location on log curve
float range = max - min;
float z = (ln_input - min) / range;
double exp = Math.log(kneeMount) / Math.log(kneePoint);
float output = (float) Math.pow(z, 1.0 / exp);
return output;// projection from log curve to linear curve
}
像这样使用:
float min = -20.0f;
float max = 100.0f;
float def_val = 42.0f;
float knee_point = 0.2f; //choose point as percentage on slider
float knee_mount = 0.05f; //choose level value at that point
float initVal = projectBackFromLogCurve(def_val, min, max, knee_point, knee_mount);
this.value = initVal;
float start = projectOntoLogCurve(this.value, min, max, knee_point, knee_mount);
//now you have valid init value for slider
setEffectiveValue(start);
关于audio - 如何反转 slider 的这个 ln 幂函数?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/59690660/