在肩膀和手之间创建肘关节

Question

我一直在尝试制作一个函数，returns 肘关节的位置基于臂段长度以及肩膀和手的位置。但是，尽管我尽了最大努力，我的方法还是行不通！

这是我到目前为止尝试过的方法：

var x = 0;
var y = 0;

var r = 0;
while(dist(x, y, p1x, p1y) > segLength-1 && dist(x, y, p1x, p1y) < segLength+1 && dist(x, y, p2x, p2y) > segLength-1 && dist(x, y, p2x, p2y) < segLength + 1){
    r ++;
    
    x = 200 + (cos(r) * segLength);
    y = 200 + (sin(r) * segLength);
}

我知道在这个距离上有两个不同的点，我想选择最有意义的一个作为肘关节（通常是下一个）

我正在使用 Khan Academy 的 Processing 环境，所以我不能使用 ES6 语法（遗憾）。

Answer 1

Circle-circle intersection points 帮助了我。

下面是工作代码，做我最初想要它做的事情（画一只手臂）

// from stack overflow
//{
// Let EPS (epsilon) be a small value
var EPS = 0.0000001;

// Let a point be a pair: (x, y)
function Point(x, y) {
  this.x = x;
  this.y = y;
}

// Define a circle centered at (x,y) with radius r
function Circle(x,y,r) {
  this.x = x;
  this.y = y;
  this.r = r;
}

// Due to double rounding precision the value passed into the Math.acos
// function may be outside its domain of [-1, +1] which would return
// the value NaN which we do not want.
function acossafe(x) {
  if (x >= +1.0){return 0;}
  if (x <= -1.0){return 360;}
  return Math.acos(x);
}

// Rotates a point about a fixed point at some angle 'a'
function rotatePoint(fp, pt, a) {
  var x = pt.x - fp.x;
  var y = pt.y - fp.y;
  var xRot = x * Math.cos(a) + y * Math.sin(a);
  var yRot = y * Math.cos(a) - x * Math.sin(a);
  return new Point(fp.x+xRot,fp.y+yRot);
}

// Given two circles this method finds the intersection
// point(s) of the two circles (if any exists)
function circleCircleIntersectionPoints(c1, c2) {

    var r, R, d, dx, dy, cx, cy, Cx, Cy;
    
    if (c1.r < c2.r) {
        r  = c1.r;  R = c2.r;
        cx = c1.x; cy = c1.y;
        Cx = c2.x; Cy = c2.y;
    } else {
        r  = c2.r; R  = c1.r;
        Cx = c1.x; Cy = c1.y;
        cx = c2.x; cy = c2.y;
    }
    
    // Compute the vector <dx, dy>
    dx = cx - Cx;
    dy = cy - Cy;
    
    // Find the distance between two points.
    d = Math.sqrt( dx*dx + dy*dy );
    
    // There are an infinite number of solutions
    // Seems appropriate to also return null
    if (d < EPS && abs(R-r) < EPS){return [];}
    
    // No intersection (circles centered at the 
    // same place with different size)
    else if (d < EPS){
        return [];
    }
    
    var x = (dx / d) * R + Cx;
    var y = (dy / d) * R + Cy;
    var P = new Point(x, y);
    
    // Single intersection (kissing circles)
    if (abs((R+r)-d) < EPS || abs(R-(r+d)) < EPS){return [P];}
    
    // No intersection. Either the small circle contained within 
    // big circle or circles are simply disjoint.
    if ( (d+r) < R || (R+r < d) ){return [];}
    
    var C = new Point(Cx, Cy);
    var angle = acossafe((r*r-d*d-R*R)/(-2.0*d*R));
    var pt1 = rotatePoint(C, P, +angle);
    var pt2 = rotatePoint(C, P, -angle);
    return [pt1, pt2];

}
//}\

function drawArm(startX, startY, endX, endY, segLength){
    var Circle_1 = new Circle(startX, startY, segLength);
    var Circle_2 = new Circle(endX, endY, segLength);
    
    noFill();
    
    /*
    strokeWeight(2);
    ellipse(Circle_1.x, Circle_1.y, Circle_1.r*2, Circle_1.r*2);
    ellipse(Circle_2.x, Circle_2.y, Circle_2.r*2, Circle_2.r*2);
    */
    
    var points = circleCircleIntersectionPoints(Circle_1, Circle_2);
    
    /*
    for(var i = 0; i<points.length; i++){
        strokeWeight(8);
        point(points[i].x, points[i].y);
    }
    */
    
    points.sort(function(a, b){
        return b.y - a.y;
    });
    
    line(startX, startY, points[0].x, points[0].y);
    line(endX, endY, points[0].x, points[0].y);
}

draw = function() {
    background(255, 255, 255);
    
    drawArm(50, 200, 200, 200, 100);
};

可以进行一些小的修改以防止错误

在肩膀和手之间创建肘关节

Creating an elbow joint between shoulder and hand

javascript

ecmascript-5

khan-academy