我的加工草图中有一个线段和一个圆。我希望圆的中心点q在线段上找到最接近的点p,圆将朝着它移动。
我不太确定该如何编码(在处理中),因此任何建议都很棒!谢谢!
到目前为止,这是我的代码:

int xPos1 = 200;
int yPos1 = 200;
int xp1 = 50;
int yp1 = 50;
int xp2 = 350;
int yp2 = 50;

void setup() {
    size(400, 400);
    strokeWeight(2);
    line(xp1, yp1, xp2, yp2);
    strokeWeight(1);
}

void draw() {
    drawCircle();
}

void drawCircle() {
    fill(50, 120, 120);
    //circle
    ellipse(xPos1, yPos1, 75, 75);
    //circle center
    ellipse(xPos1, yPos1, 7, 7);
    fill(255);
    text("Q", xPos1 + 15, yPos1 + 5);
    fill(50, 120, 120);
}

最佳答案

点在直线上的投影如下:

从x = a + t * n形式的直线和点p开始。



表示距点p的直线上最近的点的向量分量为:

(a-p)-((a-p)点n)n

所以我们有:p +(a-p)-((a-p)点n)n

经过一些简化,我们得到:
a-(((a-p)点n)n



请注意,((a-p)点n)n是向量分量,代表沿线从最近点到起点(即从最近点到p到a)的位置

让我们使用PVector使生活更轻松一些。

PVector p = new PVector(200, 200);
PVector a = new PVector(50, 50);
PVector b = new PVector(350, 50);
PVector n = new PVector(350, 50); // |p2 - p1|

void setup() {
    size(400, 400);
    strokeWeight(2);
    strokeWeight(1);

    // initialize our normalized (unit length) line direction
    n.sub(a);
    n.normalize();
}

void draw() {
    drawCircle();
}

PVector getNearestPointOnLine(PVector p, PVector a, PVector n){
    // the notation turns the computation inside out,
    // but this is equivalent to the above equation
    PVector q = PVector.mult(n, -PVector.sub(a, p).dot(n));
    q.add(a);
    return q;
}

void drawCircle() {
    // lets draw everything here where we can see it
    background(255, 255, 255);
    line(a.x, a.y, b.x, b.y);

    fill(50, 120, 120);
    //circle

    // NOTE: this may require hooking up a mouse move event handler
    p.x = mouseX;
    p.y = mouseY;
    PVector q = getNearestPointOnLine(p, a, n);

    ellipse(q.x, q.y, 75, 75);
    //circle center
    ellipse(q.x, q.y, 7, 7);
    fill(0); // make text visible on white background
    text("Q", q.x + 15, q.y + 5);
    //fill(50, 120, 120);
}


参考:https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line#Vector_formulation

08-04 16:28