如何计算围绕其中心旋转的矩形的边界框？

Question

我想要计算围绕其中心旋转的矩形的边界框。我已阅读 this question，虽然 MarkusQ 的回答一般有效，但效率不足以满足我的需求。我正在尝试让 Troubadour 的答案起作用，但它似乎只在旋转原点位于拐角而不是中心时才起作用。

是否可以调整他的解决方案来处理以中心为中心旋转原点的矩形？

我已经完全重现了以下问题：

let canvas = document.querySelector("canvas");
let ctx = canvas.getContext("2d");  

function drawRectangle(rX, rY, rW, rH) {
  ctx.beginPath();
  ctx.rect(rX, rY, rW, rH);
  ctx.stroke();
}

function degreesToRadians(degrees) { return degrees * (Math.PI / 180); }

function rotateCanvas(radians, centerX, centerY) {
  ctx.translate(centerX, centerY);
  ctx.rotate(radians);
  ctx.translate(-centerX, -centerY);
}

function drawRotatedRectangle(rX, rY, rW, rH, rAngle) {
  let rXCenter = rX + rW / 2;
  let rYCenter = rY + rH / 2;

  rotateCanvas(rAngle, rXCenter, rYCenter);
  drawRectangle(rX, rY, rW, rH);
  rotateCanvas(-rAngle, rXCenter, rYCenter);
}

function computeAABBCenter(x, y, w, h, theta) {
  const ux = Math.cos(theta) * 0.5; // half unit vector along w
  const uy = Math.sin(theta) * 0.5;
  const wx = w * ux;
  const wy = w * uy; // vector along w
  const hx = h * -uy;
  const hy = h * ux; // vector along h
  
  // all point from top left CW
  const x1 = x - wx - hx;
  const y1 = y - wy - hy;
  const x2 = x + wx - hx;
  const y2 = y + wy - hy;
  const x3 = x + wx + hx;
  const y3 = y + wy + hy;
  const x4 = x - wx + hx;
  const y4 = y - wy + hy;
  
  return {
    x1: Math.min(x1, x2, x3, x4),
    y1: Math.min(y1, y2, y3, y4),
    x2: Math.max(x1, x2, x3, x4),
    y2: Math.max(y1, y2, y3, y4),
  };
}

let rX = 100;
let rY = 100;
let rW = 100;
let rH = 50;
let rA = 0.707;

drawRotatedRectangle(rX, rY, rW, rH, rA);

let bb = computeAABBCenter(rX, rY, rW, rH, rA);

drawRectangle(bb.x1, bb.y1, bb.x2 - bb.x1, bb.y2 - bb.y1);

body { margin: 0; overflow: hidden; }

<canvas width="800" height="800"></canvas>

如您所见，边界框矩形不正确。这是它目前的样子，它应该是这样的：

Answer 1

边界框的大小（）

H = w * Abs(Sin(Fi)) + h * Abs(Cos(Fi))
W = w * Abs(Cos(Fi)) + h * Abs(Sin(Fi))

有了rXCenter, rYCenter，你可以找到获取边界坐标

x0  = rXCenter - W/2
y0  = rYCenter - H/2

Answer 2

感觉你想多了。在这里，我展示了如何绘制三个围绕我制作的蓝色中心点旋转的矩形。

const wrapper = document.getElementById('wrapper');
const canvas = document.getElementById('canvas');
const wCS = window.getComputedStyle(wrapper, null)
const pl = parseFloat(wCS.getPropertyValue('padding-left'));
const pt = parseFloat(wCS.getPropertyValue('padding-top'));
const pr = parseFloat(wCS.getPropertyValue('padding-right'));
const pb = parseFloat(wCS.getPropertyValue('padding-bottom'));
canvas.width = wrapper.clientWidth - pl - pr;
canvas.height = wrapper.clientHeight - pt - pb;

const ctx = canvas.getContext('2d');
// setup a rectangle
let cornerLeft = 70;
let cornerTop = 70;
let width = 180;
let height = 60;
let horizontalCenter = cornerTop + (width / 2);
let verticalCenter = cornerTop + (height / 2);
let centerPoint = {
  horz: horizontalCenter,
  vert: verticalCenter
};
let colorBlue = '#0000FF';
let colorLime = '#40FF40';
let colorPurple = '#4040FF';
let colorOrange = '#FFA500';

ctx.fillStyle = colorBlue;
ctx.fillRect(centerPoint.horz, centerPoint.vert, 8, 8); // fill in the pixels at center point

ctx.strokeStyle = colorPurple;
ctx.strokeRect(cornerTop, cornerLeft, width, height); // horizontal purple box

ctx.strokeStyle = colorOrange;
ctx.translate(centerPoint.horz, centerPoint.vert); // center point note positive
ctx.rotate(Math.PI / 2); // vertical
ctx.translate(-centerPoint.horz, -centerPoint.vert); // center point - note negative
ctx.strokeRect(cornerTop, cornerLeft, width, height); // vertical orange box

ctx.strokeStyle = colorLime;
ctx.translate(-centerPoint.horz * (1.375), -centerPoint.vert * (0.5)); // center point - note negative
ctx.rotate((Math.PI / 2) / -2); //back left by 45 deg left
ctx.translate(centerPoint.horz * (.025), centerPoint.vert * (2.75)); // center point note positive
ctx.strokeRect(cornerTop, cornerLeft, width, height);

#wrapper {
  width: 400px;
  height: 400px;
  padding: 2rem;
  border: 1px solid blue;
}

#canvas {
  border: 1px solid orange;
}

<div id="wrapper"><canvas id="canvas"></canvas></div>

如何计算围绕其中心旋转的矩形的边界框？

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