javascript - 与正方形碰撞后如何保持圆周速度？

问题描述

我正在开发一个游戏，其中玩家是一个圆圈，而瓷砖是正方形。用户使用键盘移动头像（圆圈），并且不应与瓷砖（正方形）发生碰撞。

另外，如果他们碰到角落，我希望圆圈沿着正方形滑动，这样如果玩家一直按下键向同一方向移动，他们将沿着正方形滑动而不是卡在上面。

我已经开发了我在这里面临的问题的完整再现：

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

class Vec2 {
  constructor(x, y) {
    this.x = x || 0;
    this.y = y || 0;
  }

  distance(v) {
    let x = v.x - this.x;
    let y = v.y - this.y;

    return Math.sqrt(x * x + y * y);
  }

  magnitude() { 
    return Math.sqrt(this.x * this.x + this.y * this.y);
  }

  dot(v) { 
    return this.x * v.x + this.y * v.y;
  }

  normalize() {
    let magnitude = this.magnitude();
    
    return new Vec2(this.x / magnitude, this.y / magnitude);
  }
  
  multiply(val) {
    return typeof val === "number" ? new Vec2(this.x * val, this.y * val) : new Vec2(this.x * val.x, this.y * val.y);
  }

  subtract(val) {
    return typeof val === "number" ? new Vec2(this.x - val, this.y - val) : new Vec2(this.x - val.x, this.y - val.y);
  }

  add(val) {
    return typeof val === "number" ? new Vec2(this.x + val, this.y + val) : new Vec2(this.x + val.x, this.y + val.y);
  }
}

function clamp(value, min, max) {
  return Math.min(Math.max(value, min), max);
}

function drawCircle(xCenter, yCenter, radius) {
  ctx.beginPath();
  ctx.arc(xCenter, yCenter, radius, 0, 2 * Math.PI);
  ctx.fill();
}

function drawSquare(x, y, w, h) {
  ctx.beginPath();
  ctx.rect(x, y, w, h);
  ctx.stroke();
}

function circleRectangleCollision(cX, cY, cR, rX, rY, rW, rH) {
  let x = clamp(cX, rX, rX + rW);
  let y = clamp(cY, rY, rY + rH);

  let cPos = new Vec2(cX, cY);

  return cPos.distance(new Vec2(x, y)) < cR;
}

function getCircleRectangleDisplacement(rX, rY, rW, rH, cX, cY, cR, cVel) {
  let circle = new Vec2(cX, cY);

  let nearestX = Math.max(rX, Math.min(cX, rX + rW));
  let nearestY = Math.max(rY, Math.min(cY, rY + rH));    
  let dist = new Vec2(cX - nearestX, cY - nearestY);

  let tangentVel = dist.normalize().dot(cVel);

  // The original answer had `cVel.subtract(tangentVel * 2);` here
  // but that was giving me issues as well
  return cVel.add(tangentVel);
}

let circlePos = new Vec2(150, 80);
let squarePos = new Vec2(240, 110);

let circleR = 50;

let squareW = 100;
let squareH = 100;

let circleVel = new Vec2(5, 0);

draw = () => {
  ctx.fillStyle = "#b2c7ef";
  ctx.fillRect(0, 0, 800, 800); 

  ctx.fillStyle = "#ffffff";

  drawCircle(circlePos.x, circlePos.y, circleR);
  drawSquare(squarePos.x, squarePos.y, squareW, squareH);
}

update = () => {
  draw();

  if (circleRectangleCollision(circlePos.x, circlePos.y, circleR, squarePos.x, squarePos.y, squareW, squareH)) {
    circleVel = getCircleRectangleDisplacement(squarePos.x, squarePos.y, squareW, squareH, circlePos.x, circlePos.y, circleR, circleVel);
  }

  circlePos = circlePos.add(circleVel);
}

setInterval(update, 30);

canvas { display: flex; margin: 0 auto; }

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

如果您运行该代码段，您会看到圆圈正确地围绕正方形移动，但之后它会向下和向右移动。我不确定为什么会这样。之后它应该保持完全笔直并向右移动。

不幸的是，我数学不是很好，所以我很难弄清楚为什么会这样。我通过这个答案了解了主要算法，但也参考了以下答案：一二三

我注意到的另一个问题是，如果将circlePosfrom的 y 位置更改80为240，那么它仍然会沿着正方形的顶部滑动，而不是沿着正方形的底部走更自然的滑动路径。如果可能的话，我也想解决这个问题。

此外，理想情况下，如果圆圈直接撞到瓷砖上，则根本不应该有任何滑动，如果这是有道理的话。在这种情况下，它应该卡在正方形上。

标签： javascriptalgorithmmathcollision-detectiongame-physics

解决方案

我建议进行以下更改：

在你的类中定义另外两个方法：

  crossProductZ(v) {
    return this.x * v.y - v.x * this.y;
  }
  
  perpendicular() {
    return new Vec2(this.y, -this.x);
  }

getCircleRectangleDisplacement将return语句替换为：

return dist.perpendicular().normalize()
           .multiply(cVel.magnitude() * Math.sign(cVel.crossProductZ(dist)));

这个想法是圆应该垂直于通过圆心和命中点（即dist）的线移动。垂线上当然有两个方向：应该dist是与当前速度矢量在同一侧的方向。这样，圆圈将选择正方形的右侧。

该移动的幅度应该等于当前速度的幅度（因此速度没有变化，只是方向上的变化）。

最后，还要对update函数进行以下更改：

  let nextCirclePos = circlePos.add(circleVel);
  if (circleRectangleCollision(nextCirclePos.x, nextCirclePos.y, circleR, squarePos.x, squarePos.y, squareW, squareH)) {
    let currentVel = getCircleRectangleDisplacement(squarePos.x, squarePos.y, squareW, squareH, circlePos.x, circlePos.y, circleR, circleVel);
    nextCirclePos = circlePos.add(currentVel);
  }
  circlePos = nextCirclePos;

这里的想法是，我们首先像往常一样进行移动（circleVel），看看这是否意味着碰撞。在那种情况下，我们不会采取行动。相反，我们从当前位置获得位移。

而且，我们从不更新currentVel. 这将保证一旦障碍物消失，运动将像以前一样继续。

在下面的代码段中进行了这些更改。此外，我在圆圈的路径中添加了第二个正方形，一旦圆圈不在视线范围内，我添加了第二次运行，圆圈采用不同的路径：

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

class Vec2 {
  constructor(x, y) {
    this.x = x || 0;
    this.y = y || 0;
  }

  distance(v) {
    let x = v.x - this.x;
    let y = v.y - this.y;

    return Math.sqrt(x * x + y * y);
  }

  magnitude() { 
    return Math.sqrt(this.x * this.x + this.y * this.y);
  }

  dot(v) { 
    return this.x * v.x + this.y * v.y;
  }

  normalize() {
    let magnitude = this.magnitude();
    
    return new Vec2(this.x / magnitude, this.y / magnitude);
  }
  
  multiply(val) {
    return typeof val === "number" ? new Vec2(this.x * val, this.y * val) : new Vec2(this.x * val.x, this.y * val.y);
  }

  subtract(val) {
    return typeof val === "number" ? new Vec2(this.x - val, this.y - val) : new Vec2(this.x - val.x, this.y - val.y);
  }

  add(val) {
    return typeof val === "number" ? new Vec2(this.x + val, this.y + val) : new Vec2(this.x + val.x, this.y + val.y);
  }
  
  crossProductZ(v) {
    return this.x * v.y - v.x * this.y;
  }
  
  perpendicular() {
    return new Vec2(this.y, -this.x);
  }
}

function clamp(value, min, max) {
  return Math.min(Math.max(value, min), max);
}

function drawCircle(xCenter, yCenter, radius) {
  ctx.beginPath();
  ctx.arc(xCenter, yCenter, radius, 0, 2 * Math.PI);
  ctx.fill();
}

function drawSquare(x, y, w, h) {
  ctx.beginPath();
  ctx.rect(x, y, w, h);
  ctx.stroke();
}

function circleRectangleCollision(cX, cY, cR, rX, rY, rW, rH) {
  let x = clamp(cX, rX, rX + rW);
  let y = clamp(cY, rY, rY + rH);

  let cPos = new Vec2(cX, cY);

  return cPos.distance(new Vec2(x, y)) < cR;
}

function getCircleRectangleDisplacement(rX, rY, rW, rH, cX, cY, cR, cVel) {
  let circle = new Vec2(cX, cY);

  let nearestX = clamp(cX, rX, rX + rW);
  let nearestY = clamp(cY, rY, rY + rH);
  let dist = new Vec2(cX - nearestX, cY - nearestY);

  return dist.perpendicular().normalize().multiply(cVel.magnitude() * Math.sign(cVel.crossProductZ(dist)));
}

let circlePos = new Vec2(100, 80);
let squarePosList = [new Vec2(240, 110), new Vec2(480, -50)];

let circleR = 50;

let squareW = 100;
let squareH = 100;

let circleVel = new Vec2(5, 0);

draw = () => {
  ctx.fillStyle = "#b2c7ef";
  ctx.fillRect(0, 0, 800, 800); 

  ctx.fillStyle = "#ffffff";

  drawCircle(circlePos.x, circlePos.y, circleR);
  for (let squarePos of squarePosList) {
    drawSquare(squarePos.x, squarePos.y, squareW, squareH);
  }
}

update = () => {
  draw();

  let nextCirclePos = circlePos.add(circleVel);
  for (let squarePos of squarePosList) {
    if (circleRectangleCollision(nextCirclePos.x, nextCirclePos.y, circleR, squarePos.x, squarePos.y, squareW, squareH)) {
      let currentVel = getCircleRectangleDisplacement(squarePos.x, squarePos.y, squareW, squareH, circlePos.x, circlePos.y, circleR, circleVel);
      nextCirclePos = circlePos.add(currentVel);
      break; // we only deal with one collision (otherwise it becomes more complex)
    }
  }
  circlePos = nextCirclePos;
  if (circlePos.x > 800 + circleR) { // Out of view: Repeat the animation but with a diagonal direction
       circlePos = new Vec2(100, 400);
       circleVel = new Vec2(3.6, -3.6);
  }
}

let interval = setInterval(update, 30);

canvas { display: flex; margin: 0 auto; }

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

注意：您在碰撞和位移函数中有一些代码重复。他们都计算几乎相同的东西。这可以优化。

javascript - 与正方形碰撞后如何保持圆周速度？

问题描述

解决方案

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