我想找到一个点+方向和一条直线的交点。

• 点位于多边形的中心，线是多边形的边。
• 方向是多边形最长线段的弧度旋转。

为了说明我的问题，我做了一个截图：
image of the current state
你可以看到中心点，右边多边形的最长线段+最长线段的起点，都是黑色的。

• 紫色和红色的点是我的算法找到的交点。
• 底部和右侧的交叉点是正确的，但在左侧它找到了2个交叉点，其中一个是正确的。
• 顶部交点未到达多边形的边。
• 紫色的点应该是顶部和底部的交点
• 较小的红点应位于多边形的左侧和右侧，因为您可以看到它们部分混合在一起。

我的代码：
第一个循环计算最长线段和该线段的旋转。第二个循环根据最长线段的旋转检查垂直和水平方向上的相交

let sqLongestSide = 0;
let longestSideXDiff = 0;
let longestSideYDiff = 0;

for (let i = 1; i < coordinates.length; i++) {
    let pointFrom = coordinates[i - 1];
    let pointTo = coordinates[i];

    const xDiff = pointFrom[0] - pointTo[0];
    const yDiff = pointFrom[1] - pointTo[1];

    const sqDistance = Math.sqrt(xDiff * xDiff + yDiff * yDiff);

    const roundedDistance = Math.round(sqDistance * 100) / 100;
    const roundedLongestSide = Math.round(sqLongestSide * 100) / 100;

    if (roundedDistance > roundedLongestSide) {
        sqLongestSide = sqDistance;
        longestSideXDiff = xDiff;
        longestSideYDiff = yDiff;
    }
}

const rotation = Math.atan(longestSideYDiff / longestSideXDiff);

for (let i = 1; i < coordinates.length; i++) {

    let pointFrom = coordinates[i - 1];
    let pointTo = coordinates[i];

    const intersectionTopAndBottom = intersectionPoint(anchor, rotation, [pointFrom, pointTo]);

    if (intersectionTopAndBottom) {
        setPointStyle(drawContext, "#FF00FF", 20);
        drawContext.drawPoint(new Point(intersectionTopAndBottom));
        drawContext.drawLineString(new LineString([anchor, intersectionTopAndBottom]));
    }

    const intersectionLeftAndRight = intersectionPoint(anchor, (rotation + Math.PI / 2), [pointFrom, pointTo]);

    if (intersectionLeftAndRight) {
        setPointStyle(drawContext, "#FF0000", 10);
        drawContext.drawPoint(new Point(intersectionLeftAndRight));
        setLineStyle(drawContext, "#FF0000", 5);
        drawContext.drawLineString(new LineString([anchor, intersectionLeftAndRight]));
    }

求交集的函数如下所示：

function intersectionPoint(
  point: Coordinate,
  theta: number,
  line: Coordinate[]
): Coordinate {
  const x0 = Math.round(point[0] * 100) / 100;
  const y0 = Math.round(point[1] * 100) / 100;

  const x1 = Math.round(line[0][0] * 100) / 100;
  const y1 = Math.round(line[0][1] * 100) / 100;
  const x2 = Math.round(line[1][0] * 100) / 100;
  const y2 = Math.round(line[1][1] * 100) / 100;

  // Check if the line is vertical or horizontal
  if (x1 === x2) {
    // The line is vertical, so the intersection point is simply the point with the same x-coordinate as the line
    return [x1, y0];
  } else if (y1 === y2) {
    // The line is horizontal, so the intersection point is simply the point with the same y-coordinate as the line
    return [x0, y1];
  }

  // Convert the line to slope-intercept form
  const slope = (y2 - y1) / (x2 - x1);
  const intercept = y1 - slope * x1;

  // Convert the point and direction to slope-intercept form
  const slope2 = Math.tan(theta);
  const intercept2 = y0 - slope2 * x0;

  // Find the intersection point of the two lines
  const x = (intercept2 - intercept) / (slope - slope2);
  const y = slope * x + intercept;

  return [x, y];
}

我只需要4个交叉点，但得到更多的这个特定的多边形，但简单的矩形，我得到正确的结果。
编辑：
只是为了弄清楚。我需要对射线应用一个旋转。这是在第一个循环中计算的最长线段的旋转。
@Blindman67你的解决方案给了我正确的结果为第一个多边形，我只需要交叉"十字"旋转。如下图：
correct
该算法给出以下结果：
not correct