HTML5 Canvas - 與球物理故障的碰撞

Question

我正在使用牛頓方程來制作我目前正在研究的這個程序中的球，當它們相互碰撞時會“分裂”，但有時它們會相互卡住，這會造成很多麻煩。

.

這是我的代碼：

<center>
<canvas id="canvas" style="border: 2px solid black; cursor: crosshair;" width="1000"                 height="500"></canvas>
</center>

<script>
var canvas = document.getElementById("canvas")
var ctx = canvas.getContext("2d")

var w = canvas.width
var h = canvas.height

var ball = []

var gravity = 0.3
var force = 0.2

var mouse = {
d: false,
x1: 0,
y1: 0,
x2: 0,
y2: 0,
}




window.onmousedown = function(e) {
mouse.d = true
mouse.x1 = mouse.x2 = e.pageX - canvas.getBoundingClientRect().left
mouse.y1 = mouse.y2 = e.pageY - canvas.getBoundingClientRect().top
}
window.onmousemove = function(e) {
if (mouse.d) {
    mouse.x2 = e.pageX - canvas.getBoundingClientRect().left
    mouse.y2 = e.pageY - canvas.getBoundingClientRect().top
} else {
    mouse.x1 = mouse.x2 = e.pageX - canvas.getBoundingClientRect().left
    mouse.y1 = mouse.y2 = e.pageY - canvas.getBoundingClientRect().top
}
}
window.onmouseup = function() {
if (mouse.d) {
    mouse.d = false

    var dx = (mouse.x1 - mouse.x2);
    var dy = (mouse.y1 - mouse.y2);
    var mag = Math.sqrt(dx * dx + dy * dy);

    ball.push({
        x: mouse.x1,
        y: mouse.y1,
        r: Math.floor(Math.random() * 20) + 10,
        vx: dx / mag * -(mag * force),
        vy: dy / mag * -(mag * force),
        b: 0.7,
    })
}
}
document.onselectstart = function() {return false}
document.oncontextmenu = function() {return false}


setInterval(update, 1000 / 60)
function update() {
ctx.clearRect(0, 0, w, h)

ctx.beginPath()
ctx.moveTo(mouse.x1, mouse.y1)
ctx.lineTo(mouse.x2, mouse.y2)
ctx.stroke()
ctx.closePath()

for (i = 0; i < ball.length; i++) {
    ball[i].vy += gravity
    ball[i].x += ball[i].vx
    ball[i].y += ball[i].vy

    if (ball[i].x > w - ball[i].r) {
        ball[i].x = w - ball[i].r
        ball[i].vx *= -ball[i].b
    }
    if (ball[i].x < ball[i].r) {
        ball[i].x = ball[i].r
        ball[i].vx *= -ball[i].b
    }
    if (ball[i].y > h - ball[i].r) {
        ball[i].y = h - ball[i].r
        ball[i].vy *= -ball[i].b
    }
    if (ball[i].y < ball[i].r) {
        ball[i].y = ball[i].r
        ball[i].vy *= -ball[i].b
    }

    for (j = i + 1; j < ball.length; j++) {
        var dx = ball[i].x - ball[j].x
        var dy = ball[i].y - ball[j].y
        var dist = Math.sqrt(dx * dx + dy * dy)
        if (Math.abs(dx) + Math.abs(dy) != 0 && dist <= ball[i].r + ball[j].r) {
            var angle = Math.atan2(dy, dx)

            var sp1 = Math.sqrt(ball[i].vx*ball[i].vx + ball[i].vy*ball[i].vy);
            var sp2 = Math.sqrt(ball[j].vx*ball[j].vx + ball[j].vy*ball[j].vy);

            var dir1 = Math.atan2(ball[i].vy, ball[i].vx);
            var dir2 = Math.atan2(ball[j].vy, ball[j].vx);

            var vx1 = sp1 * Math.cos(dir1 - angle);
            var vy1 = sp1 * Math.sin(dir1 - angle);
            var vx2 = sp2 * Math.cos(dir2 - angle);
            var vy2 = sp2 * Math.sin(dir2 - angle);

            var fvx1 = ((ball[i].r - ball[j].r) * vx1 + (2 * ball[j].r) * vx2) / (ball[i].r + ball[j].r);
            var fvx2 = ((2 * ball[i].r) * vx1 + (ball[j].r - ball[i].r) * vx2) / (ball[i].r + ball[j].r);
            var fvy1 = vy1;
            var fvy2 = vy2;

            ball[i].vx = Math.cos(angle) * fvx1 + Math.cos(angle + Math.PI/2) * fvy1;
            ball[i].vy = Math.sin(angle) * fvx1 + Math.sin(angle + Math.PI/2) * fvy1;
            ball[j].vx = Math.cos(angle) * fvx2 + Math.cos(angle + Math.PI/2) * fvy2;
            ball[j].vy = Math.sin(angle) * fvx2 + Math.sin(angle + Math.PI/2) * fvy2;
        }
    }


    ctx.beginPath()
    ctx.arc(ball[i].x, ball[i].y, ball[i].r, 0, Math.PI * 2, false)
    ctx.fillStyle = "black"
    ctx.fill()
    ctx.closePath()
}
}
</script>

當你產生了很多球並且它們的速度很快時，就會發生這種情況：

為什么？ 任何人都知道我該如何解決這個問題？

Answer 1

你的解決方案似乎只有基於速度的懲罰力。 這允許球體穿透，當任何球體沒有速度時，它們都不會試圖固定穿透。 要解決此問題，您需要添加基於位置的懲罰。 一個非常簡單的解決方案是使用彈簧。 計算穿透長度並使用胡克定律將相交的球體彼此推開。

最好的解決方案是使用隱式聯系解算算法。 這允許更加僵硬的接觸，但是算法更加復雜。 我建議你使用2D物理引擎來獲得快速和良好的結果： Box2D的JavaScript端口似乎最常用。

Answer 2

我非常關注@ schteppe的答案。 我想在這里告訴你這個鏈接： http ： //www.paulirish.com/2011/requestanimationframe-for-smart-animating/

看看“我為什么要用它？” 部分。 它可能會對你的問題的答案有一些暗示，即你的代碼究竟是什么問題。

順便說一句，我也一直試圖使你的代碼工作，但沒有實質性的結果。 你可以在這里查看我的小提琴（我從你的復制品）： http ： //jsfiddle.net/sukhmeetsd/joqpqp49/

var canvas = document.getElementById("canvas")
var ctx = canvas.getContext("2d")

var w = canvas.width
var h = canvas.height

var d = 5; //distance to move on collision

var ball = []

var gravity = 0.3
var force = 0.2

var mouse = {
    d: false,
    x1: 0,
    y1: 0,
    x2: 0,
    y2: 0,
}




window.onmousedown = function (e) {
    mouse.d = true
    mouse.x1 = mouse.x2 = e.pageX - canvas.getBoundingClientRect().left
    mouse.y1 = mouse.y2 = e.pageY - canvas.getBoundingClientRect().top
}
window.onmousemove = function (e) {
    if (mouse.d) {
        mouse.x2 = e.pageX - canvas.getBoundingClientRect().left
        mouse.y2 = e.pageY - canvas.getBoundingClientRect().top
    } else {
        mouse.x1 = mouse.x2 = e.pageX - canvas.getBoundingClientRect().left
        mouse.y1 = mouse.y2 = e.pageY - canvas.getBoundingClientRect().top
    }
}
window.onmouseup = function () {
    if (mouse.d) {
        mouse.d = false

        var dx = (mouse.x1 - mouse.x2);
        var dy = (mouse.y1 - mouse.y2);
        var mag = Math.sqrt(dx * dx + dy * dy);

        ball.push({
            x: mouse.x1,
            y: mouse.y1,
            r: Math.floor(Math.random() * 20) + 10,
            vx: dx / mag * -(mag * force),
            vy: dy / mag * -(mag * force),
            b: 0.7,
        })
    }
}

function getRandomColor() {
    var letters = '0123456789ABCDEF'.split('');
    var color = '#';
    for (var i = 0; i < 6; i++ ) {
        color += letters[Math.floor(Math.random() * 16)];
    }
    return color;
}

document.onselectstart = function () {
    return false
}
document.oncontextmenu = function () {
    return false
}


setInterval(update, 1000/60)

function update() {
    ctx.clearRect(0, 0, w, h)

    ctx.beginPath()
    ctx.moveTo(mouse.x1, mouse.y1)
    ctx.lineTo(mouse.x2, mouse.y2)
    ctx.stroke()
    ctx.closePath()

    for (i = 0; i < ball.length; i++) {
        ball[i].vy += gravity
        ball[i].x += ball[i].vx
        ball[i].y += ball[i].vy

        if (ball[i].x > w - ball[i].r) {
            ball[i].x = w - ball[i].r
            ball[i].vx *= -ball[i].b
        }
        if (ball[i].x < ball[i].r) {
            ball[i].x = ball[i].r
            ball[i].vx *= -ball[i].b
        }
        if (ball[i].y > h - ball[i].r) {
            ball[i].y = h - ball[i].r
            ball[i].vy *= -ball[i].b
        }
        if (ball[i].y < ball[i].r) {
            ball[i].y = ball[i].r
            ball[i].vy *= -ball[i].b
        }

        for (j = i + 1; j < ball.length; j++) {
            var dx = ball[i].x - ball[j].x
            var dy = ball[i].y - ball[j].y
            var dist = Math.sqrt(dx * dx + dy * dy)
            if (Math.abs(dx) + Math.abs(dy) != 0 && dist <= ball[i].r + ball[j].r) {

                var angle = Math.atan2(dy, dx)

                var sp1 = Math.sqrt(ball[i].vx * ball[i].vx + ball[i].vy * ball[i].vy);
                var sp2 = Math.sqrt(ball[j].vx * ball[j].vx + ball[j].vy * ball[j].vy);

                var dir1 = Math.atan2(ball[i].vy, ball[i].vx);
                var dir2 = Math.atan2(ball[j].vy, ball[j].vx);

                d = Math.ceil(ball[i].r+ball[j].r-dist)/2;

                //moving them back
                ball[i].x = ball[i].x - Math.cos(dir1)*d-1;
                ball[i].y = ball[i].y - Math.sin(dir1)*d-1;
                ball[j].x = ball[j].x + Math.cos(dir2)*d+1;
                ball[j].y = ball[j].y + Math.sin(dir2)*d+1;

                //Checking for distance again
                /*dx = ball[i].x - ball[j].x;
                dy = ball[i].y - ball[j].y;
                dist = Math.sqrt(dx * dx + dy * dy);
                if (Math.abs(dx) + Math.abs(dy) != 0 && dist <= ball[i].r + ball[j].r){
                      ball[i].x = ball[i].x + Math.cos(dir1)*2*d;
                        ball[i].y = ball[i].y + Math.sin(dir1)*d*2;
                        ball[j].x = ball[j].x - Math.cos(dir2)*d*2;
                      ball[j].y = ball[j].y - Math.sin(dir2)*d*2;
                }*/

                var vx1 = sp1 * Math.cos(dir1 - angle);
                var vy1 = sp1 * Math.sin(dir1 - angle);
                var vx2 = sp2 * Math.cos(dir2 - angle);
                var vy2 = sp2 * Math.sin(dir2 - angle);

                var fvx1 = ((ball[i].r - ball[j].r) * vx1 + (2 * ball[j].r) * vx2) / (ball[i].r + ball[j].r);
                var fvx2 = ((2 * ball[i].r) * vx1 + (ball[j].r - ball[i].r) * vx2) / (ball[i].r + ball[j].r);
                var fvy1 = vy1;
                var fvy2 = vy2;

                ball[i].vx = Math.cos(angle) * fvx1 + Math.cos(angle + Math.PI / 2) * fvy1;
                ball[i].vy = Math.sin(angle) * fvx1 + Math.sin(angle + Math.PI / 2) * fvy1;
                ball[j].vx = Math.cos(angle) * fvx2 + Math.cos(angle + Math.PI / 2) * fvy2;
                ball[j].vy = Math.sin(angle) * fvx2 + Math.sin(angle + Math.PI / 2) * fvy2;
            }
        }


        ctx.beginPath()
        ctx.arc(ball[i].x, ball[i].y, ball[i].r, 0, Math.PI * 2, false)
        ctx.fillStyle = getRandomColor();
        ctx.fill();
        ctx.closePath();
    }
}

我的代碼不會讓球粘住但是它們處於不斷的攪拌狀態。 我正准備按照@schteppe的建議實施胡克定律，但后來我聽說過Box2d及其魔法。

Answer 3

只需使用重疊。 可以計算為兩個粒子的半徑之和與它們之間的距離之差。 下面的代碼是用 javascript 編寫的。 它在我的模擬中效果很好：

let overlap = particle1.r + particle2.r - distance(particle1, particle2);
particle1.x += 0.5*overlap;
particle2.x -= 0.5*overlap;