这个问题非常类似于:Quadratic bezier curve: Y coordinate for a given X?。但这是立方的...
我正在使用getBezier函数来计算贝塞尔曲线的Y坐标。贝塞尔曲线始终在(0,0)处开始,并始终在(1,1)处结束。
我知道X值,所以我尝试将其作为百分比插入(我是白痴)。但这显然没有用。您能提供解决方案吗?这是一个白痴证明功能是必要的。喜欢:
function yFromX (c2x,c2y,c3x,c3y) { //c1 = (0,0) and c4 = (1,1), domainc2 and domainc3 = [0,1]
//your magic
return y;
}
最佳答案
由于问题是如此有限(函数x(t)是单调的),我们可能可以避免使用一种非常便宜的解决方案-二进制搜索。
var bezier = function(x0, y0, x1, y1, x2, y2, x3, y3, t) {
/* whatever you're using to calculate points on the curve */
return undefined; //I'll assume this returns array [x, y].
};
//we actually need a target x value to go with the middle control
//points, don't we? ;)
var yFromX = function(xTarget, x1, y1, x2, y2) {
var xTolerance = 0.0001; //adjust as you please
var myBezier = function(t) {
return bezier(0, 0, x1, y1, x2, y2, 1, 1, t);
};
//we could do something less stupid, but since the x is monotonic
//increasing given the problem constraints, we'll do a binary search.
//establish bounds
var lower = 0;
var upper = 1;
var percent = (upper + lower) / 2;
//get initial x
var x = myBezier(percent)[0];
//loop until completion
while(Math.abs(xTarget - x) > xTolerance) {
if(xTarget > x)
lower = percent;
else
upper = percent;
percent = (upper + lower) / 2;
x = myBezier(percent)[0];
}
//we're within tolerance of the desired x value.
//return the y value.
return myBezier(percent)[1];
};
当然,这超出了您的约束范围,可能会破坏某些输入。
关于javascript - 给定x三次方贝塞尔曲线的y坐标,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/7348009/