Question

我需要找到从圆和矩形的交点创建的最大弧。我有圆的中心，半径和矩形的坐标，我需要找到交点与圆心的角度。

Problem Diagram

我有一个可行的代码，但它计算迭代圆周点的解决方案，我想知道是否有更优雅的方法来使用三角法计算解决方案而不是＆＃34;蛮力＆＃ 34。

那是我的代码：

import 'dart:math';

class CircleTheSquare {
  final Point     _circleCenter;
  final int       _circleRadius;
  final Rectangle _box;


  CircleTheSquare(this._circleCenter, this._circleRadius, this._box);


  Map<String, double> get arc {
    Map res = new Map();

    double angle = .0;
    double angleIn;
    double angleOut;
    double increment = 1.0;

    while (true) {
      if (angle > 360.0 && angleIn == null) {
        break;
      }

      // Finds a point of intersection (next points will be inside 
      // of the rectangle).
      if (!_isOutside(angle) && _isOutside(angle - increment)) {
        angleIn = angle;
      }

      // Finds next intersection (next points will be outside 
      // of the rectangle).
      if (angleIn != null &&
          _isOutside(angle + increment) && !_isOutside(angle)) {

        angleOut = angle;

        // Adds the arc to result only there's not a previous largest arc.
        if (res["in"] == null ||
            angleOut - angleIn > res["arc"]) {

          res["in"]  = angleIn;
          res["arc"] = angleOut - angleIn;
        }
        angleIn = null;
        angleOut = null;
      }
      angle += increment;
    }

    // If there's no intersections. 
    // -- For simplicity, we will assume that the
    //    rectangle and the circle intersect or that the circle is 
    //    inside of the rectangle).
    if (res["in"] == null) {
      res = {"in" : 0.0, "arc" : 360.0};
    }

    return res;
  }


  bool _isOutside(double a) {
    var res;

    double cx = _circleCenter.x + (_circleRadius * cos(a * (PI / 180)));
    double cy = _circleCenter.y + (_circleRadius * sin(a * (PI / 180)));

    bool hOut = cx < _box.left || cx > _box.left + _box.width;
    bool vOut = cy < _box.top || cy > _box.top + _box.height;

    if (hOut || vOut) {
      res = true;
    } else {
      res = false;
    }

    return res;
  }
}


main() {
  CircleTheSquare a = new CircleTheSquare(new Point(250, 250), 100,
                                          new Rectangle(0,0,500,500));

  print(a.arc); // {in: 0.0, arc: 360.0}

  CircleTheSquare b = new CircleTheSquare(new Point(450, 250), 100,
                                          new Rectangle(0,0,500,500));

  print(b.arc); // {in: 60.0, arc: 240.0}

  CircleTheSquare c = new CircleTheSquare(new Point(420, 420), 100,
                                          new Rectangle(0,0,500,500));

  print(c.arc); // 4 intersections, returns the largest arc:
                          // {in: 127.0, arc: 196.0}
}

Answer 1

为简单起见，移动所有坐标以使圆为零（box.left = box.left - circlecenter.x等）
找到带有矩形边的圆的交点。例如，对于左侧求解（box.left）^ 2 + y ^ 2 =半径^ 2，检查该点位于侧面，将交叉点添加到列表
按角度对交叉点进行排序（可能是由边检查顺序自动提供），找到矩形内弧线的最大角度间隔

Answer 2

如您所料，有更好，更直接的方法来解决这个问题。你可能想看看例如http://mathworld.wolfram.com/Circle-LineIntersection.html。将矩形拆分为线，获得每条线的交点，确定它们是否在实际线段内，然后找到给出最大弧的点。

找到圆和矩形之间的交点的圆弧

2 个答案: