2个移动物体的碰撞算法

发布于 2024-09-29 19:30:30 字数 551 浏览 6 评论 0原文

这是一个 2d 游戏的数学问题。

Given 2 object ( 2 car, 2 tank, 2...), for each object i know: 
1. X, Y 
2. Actual Speed
3. Degree of movement (or radiant)

如何计算 2 个物体碰撞的“效果”

Ps：

我用这个简单的公式“移动”物体，每次“游戏循环”的时间点：（

ychange = Math.Sin(radiant) * ((carspeed / xVar));
xchange = Math.Cos(radiant) * ((carspeed / xVar));

其中 xVar 是一个“乘法器”，可以使我的车速度更慢对于碰撞的“第一个”对象

，如果我在该公式前面添加“减号”（-），我可以模拟第一个对象的良好“弹跳”..但不能模拟第二个对象。

所以我的问题是关于计算第二个对象的良好（不完美！）且更现实的“影响效果”的方法。

感谢您的帮助

原文

this is a Math problem for a 2d game.

Given 2 object ( 2 car, 2 tank, 2...), for each object i know: 
1. X, Y 
2. Actual Speed
3. Degree of movement (or radiant)

How to calculate the "effect" of a collision for the 2 object

Ps:

I "move" the object with this simple formula each tick of my "game loop":

ychange = Math.Sin(radiant) * ((carspeed / xVar));
xchange = Math.Cos(radiant) * ((carspeed / xVar));

(where xVar is a "multiplicator" to make more slow my car on the screen)

And for the "first" object which collide, if i add a "minus" (-) in front of that formula, i can simulate a good "bounce" of the first object.. but not for the second.

So my question regard a way to calculate a good (not perfect ! ) and more realistic "impact effect" for the second object.

Thanks for your help

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森林迷了鹿 2024-10-06 19:30:30

几个月前，我实现了一款斯诺克游戏。我使用以下代码使两个球发生碰撞。请注意，它们不是边界框。

该方法根据每个物体的质量、速度和角度计算结果速度和角度。

它对我的游戏来说就像一种魅力。

 float   dx = b1->x - b2->x, dy = b1->y - b2->y;
    float   d = sqrt(sqr(dx) + sqr(dy));
    float   vp1, vp2, vx1, vx2, vy1, vy2;
    vx1 = ballSpeedX(b1);
    vx2 = ballSpeedX(b2);
    vy1 = ballSpeedY(b1);
    vy2 = ballSpeedY(b2);
    vp1 = vx1 * dx / d + vy1 * dy / d;
    vp2 = vx2 * dx / d + vy2 * dy / d;

    float   distance = sqrt(sqr(dx) + sqr(dy));

//  Unit vector in the direction of the collision.
    float   ax = dx / distance, ay = dy / distance;

//  Projection of the velocities in these axes.
    float   va1 = vx1 * ax + vy1 * ay, vb1 = -vx1 * ay + vy1 * ax;
    float   va2 = vx2 * ax + vy2 * ay, vb2 = -vx2 * ay + vy2 * ax;

//  New velocities in these axes (after collision): edmass / b2->mass);
    float   vaP2 = va2 + (1.0 + D_BALL_ELASTIC_COEFFICIENT) * (va1 - va2) / (1.0 + b2->mass / b1->mass);

//  Undo the projections.
    vx1 = vaP1 * ax - vb1 * ay; vy1 = vaP1 * ay + vb1 * ax; // new vx,vy for ball 1 after collision.
    vx2 = vaP2 * ax - vb2 * ay; vy2 = vaP2 * ay + vb2 * ax; // new vx,vy for ball 2 after collision.

    b1->speed = sqrt(sqr(vx1) + sqr(vy1));
    b1->angle = acos(fabs(vx1 / b1->speed));

    b2->speed = sqrt(sqr(vx2) + sqr(vy2));
    b2->angle = acos(fabs(vx2 / b2->speed));

这段代码是用纯 C 语言编写的。
我存储运动数据的方式和你的一样。变量b1和b2是球。

我相信代码本身并不难理解，但其物理原理却并非微不足道。

希望有帮助！

Few months ago I implemented a Snooker game. I used the following code to make two balls collide. Note that they are not bounding boxes.

This method computes the resulting speed and angles based on the mass, speed and angle of each objects.

It worked like a charm at my game.

 float   dx = b1->x - b2->x, dy = b1->y - b2->y;
    float   d = sqrt(sqr(dx) + sqr(dy));
    float   vp1, vp2, vx1, vx2, vy1, vy2;
    vx1 = ballSpeedX(b1);
    vx2 = ballSpeedX(b2);
    vy1 = ballSpeedY(b1);
    vy2 = ballSpeedY(b2);
    vp1 = vx1 * dx / d + vy1 * dy / d;
    vp2 = vx2 * dx / d + vy2 * dy / d;

    float   distance = sqrt(sqr(dx) + sqr(dy));

//  Unit vector in the direction of the collision.
    float   ax = dx / distance, ay = dy / distance;

//  Projection of the velocities in these axes.
    float   va1 = vx1 * ax + vy1 * ay, vb1 = -vx1 * ay + vy1 * ax;
    float   va2 = vx2 * ax + vy2 * ay, vb2 = -vx2 * ay + vy2 * ax;

//  New velocities in these axes (after collision): edmass / b2->mass);
    float   vaP2 = va2 + (1.0 + D_BALL_ELASTIC_COEFFICIENT) * (va1 - va2) / (1.0 + b2->mass / b1->mass);

//  Undo the projections.
    vx1 = vaP1 * ax - vb1 * ay; vy1 = vaP1 * ay + vb1 * ax; // new vx,vy for ball 1 after collision.
    vx2 = vaP2 * ax - vb2 * ay; vy2 = vaP2 * ay + vb2 * ax; // new vx,vy for ball 2 after collision.

    b1->speed = sqrt(sqr(vx1) + sqr(vy1));
    b1->angle = acos(fabs(vx1 / b1->speed));

    b2->speed = sqrt(sqr(vx2) + sqr(vy2));
    b2->angle = acos(fabs(vx2 / b2->speed));

This code was written in pure C.
The way I store the movement data is just the same as yours. The variables b1 and b2 are balls.

I believe the code itself isn't hard to understand, but the physic isn't trivial.

Hope it helps!

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故人爱我别走 2024-10-06 19:30:30

最简单的方法是让它们交换速度。如果它们相同，这是有道理的。

如果您想做更多，您可以为参数选择一个随机值，该值将确定它们是否有掠过碰撞（几乎没有接触）或完全碰撞，或介于两者之间。您还可以模拟两个物体的不同质量（当摩托车撞上货运卡车时，卡车不会发生太大变化）。最后，您可以进行仅部分弹性的碰撞；也就是说，它们不会像橡皮球那样弹跳，而是会嘎吱作响并损失一些能量。这些需要一些数学知识和对物理学的理解。

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