javascript - 在Javascript中计算二次曲线上任意点的切线斜率

Question

我从来都不是三 Angular 冠军，经过大约 4 个小时的搜索，我决定在这里提问:

我使用 Javascript 在 HTML5 Canvas 上绘制二次曲线(不是三次贝塞尔曲线)，如下所示:

this.shape.moveTo(50,80).curveTo(100,120,40,190);

其中moveTo指定第一个点的x和y，curveTo的前两个参数指定控制点的x和y，curveTo的第三个和第四个参数指定终点的x和y。

我需要创建一个函数，使我能够获得该曲线上任意点 t 的斜率，例如:

function getTangentSlope(P0,P1,P2,t) {
   blah blah blah
   return slope;
}

到目前为止，我找到的唯一解决方案是具有两个控制点的三次曲线( Find the tangent of a point on a cubic bezier curve (on an iPhone) )，或者我不理解符号( https://www.math.usm.edu/lee/mathphysarchive/?p=542 )，或者损坏的链接意味着我无法查看整个解决方案( Quadratic Bezier Curve: Calculate Tangent )。

最好还是以度为单位给出斜率。

哪位兄弟能帮帮我吗？

Answer 1

这将返回二阶和三阶曲线单位位置处的归一化切线。

参见this answer有关以下对象的更详细用法。

const geom = (()=>{

    function Vec(x,y){ 
        this.x = x;
        this.y = y;
    };
    function Bezier(p1,p2,cp1,cp2){  
        this.p1 =  p1;
        this.p2 =  p2;
        this.cp1 = cp1;
        this.cp2 = cp2;
    }    
    Bezier.prototype = {
        //======================================================================================
        // single dimension polynomials for 2nd (a,b,c) and 3rd (a,b,c,d) order bezier 
        //======================================================================================
        // for quadratic   f(t) = a(1-t)^2+2b(1-t)t+ct^2 
        //                      = a+2(-a+b)t+(a-2b+c)t^2
        // The derivative f'(t) =  2(1-t)(b-a)+2(c-b)t
        //======================================================================================
        // for cubic           f(t) = a(1-t)^3 + 3bt(1-t)^2 + 3c(1-t)t^2 + dt^3 
        //                          = a+(-2a+3b)t+(2a-6b+3c)t^2+(-a+3b-3c+d)t^3
        // The derivative     f'(t) = -3a(1-t)^2+b(3(1-t)^2-6(1-t)t)+c(6(1-t)t-3t^2) +3dt^2
        // The 2nd derivative f"(t) = 6(1-t)(c-2b+a)+6t(d-2c+b)
        //======================================================================================        
        p1 : undefined,
        p2 : undefined,
        cp1 : undefined,
        cp2 : undefined,
        tangentAsVec (position, vec ) { 
            var a, b, c, u;
            if (vec === undefined) { vec = new Vec(); }

            if (this.cp2 === undefined) {
                a = (1-position) * 2;
                b = position * 2;
                vec.x = a * (this.cp1.x - this.p1.x) + b * (this.p2.x - this.cp1.x);
                vec.y = a * (this.cp1.y - this.p1.y) + b * (this.p2.y - this.cp1.y);
            }else{
                a  = (1-position)
                b  = 6 * a * position;        // (6*(1-t)*t)
                a *= 3 * a;                   // 3 * ( 1 - t) ^ 2
                c  = 3 * position * position; // 3 * t ^ 2
                vec.x  = -this.p1.x * a + this.cp1.x * (a - b) + this.cp2.x * (b - c) + this.p2.x * c;
                vec.y  = -this.p1.y * a + this.cp1.y * (a - b) + this.cp2.y * (b - c) + this.p2.y * c;
            }   
            u = Math.sqrt(vec.x * vec.x + vec.y * vec.y);
            vec.x /= u;
            vec.y /= u;
            return vec;                 
        },      
    }
    return { Vec, Bezier,}
})()

二阶贝塞尔曲线的用法

   // ? represents any value 
   var p1 = new geom.Vec(?,?);  // start point
   var p2 = new geom.Vec(?,?);  // end point
   var cp1 = new geom.Vec(?,?); //control point

   var bez2 = new geom.Bezier(p1,p2,cp1); // create 2nd order bezier
   var t = ?;
   var tangent = bez2.tangentAsVec(t);

三阶贝塞尔曲线的用法

   // ? represents any value 
   var p1 = new geom.Vec(?,?);  // start point
   var p2 = new geom.Vec(?,?);  // end point
   var cp1 = new geom.Vec(?,?); // 1st control point
   var cp2 = new geom.Vec(?,?); // 2nd control point

   var bez3 = new geom.Bezier(p1,p2,cp1,cp2); // create 3rd order bezier
   var t = ?;
   var tangent = bez2.tangentAsVec(t);