問題描述

我正在嘗試編寫一個 Java 移動應用程序 (J2ME)，但遇到了一個問題:在我的項目中，有稱為鏡頭的移動圓圈和稱為球體的非移動圓圈.當一個球擊中球體時，它應該按照經典物理定律反彈.但是我找不到這種算法.

I'm trying to write a java mobile application (J2ME) and I got stuck with a problem: in my project there are moving circles called shots, and non moving circles called orbs. When a shot hits an orb, it should bounce off by classical physical laws. However I couldn't find any algorithm of this sort.

鏡頭的運動由 x 軸和 y 軸上的速度(像素/更新)來描述.所有關于圓的信息都是已知的:它們的位置、半徑和鏡頭的速度(在 x 軸和 y 軸上).

The movement of a shot is described by velocity on axis x and y (pixels/update). all the information about the circles is known: their location, radius and the speed (on axis x and y) of the shot.

注意:碰撞后球體不會開始移動，它會停留在原位.碰撞是兩者之間的彈性碰撞，而球體保持靜止

Note: the orb does not start moving after the collision, it stays at its place. The collision is an elastic collision between the two while the orb remains static

這里是Shot類中的碰撞解決方法:

here is the collision solution method in class Shot:

public void collision(Orb o)
{
    //the orb's center point
    Point oc=new Point(o.getTopLeft().x+o.getWidth()/2,o.getTopLeft().y+o.getWidth()/2);
    //the shot's center point
    Point sc=new Point(topLeft.x+width/2,topLeft.y+width/2);

    //variables vx and vy are the shot's velocity on axis x and y
    if(oc.x==sc.x)
    {
        vy=-vy;
        return ;
    }

    if(oc.y==sc.y)
    {
        vx=-vx;
        return ;
    }

    // o.getWidth() returns the orb's width, width is the shot's width

    double angle=0;  //here should be some sort of calculation of the shot's angle
    setAngle(angle);
}

public void setAngle(double angle)
{
    double v=Math.sqrt(vx*vx+vy*vy);
    vx=Math.cos(Math.toRadians(angle))*v;
    vy=-Math.sin(Math.toRadians(angle))*v;
}

提前感謝所有幫助者

推薦答案

在碰撞點，動量、角動量和能量被保存下來.設m1，m2為圓盤的質量，p1=(p1x,p1y)，p2=(p2x,p2y)為碰撞時圓盤中心的位置，u1,u2為碰撞前的速度，v1,v2為碰撞后的速度碰撞.然后守恒定律要求

At the point of collision, momentum, angular momentum and energy are preserved. Set m1, m2 the masses of the disks, p1=(p1x,p1y), p2=(p2x,p2y) the positions of the centers of the disks at collition time, u1, u2 the velocities before and v1,v2 the velocities after collision. Then the conservation laws demand that

0 = m1*(u1-v1)+m2*(u2-v2)
0 = m1*cross(p1,u1-v1)+m2*cross(p2,u2-v2)
0 = m1*dot(u1-v1,u1+v1)+m2*dot(u2-v2,u2+v2)

使用第一個方程消除 u2-v2

Eliminate u2-v2 using the first equation

0 = m1*cross(p1-p2,u1-v1)
0 = m1*dot(u1-v1,u1+v1-u2-v2)

第一個告訴我們 (u1-v1) 和因此 (u2-v2) 是 (p1-p2) 的倍數，脈沖交換是在法線或徑向方向，沒有切向相互作用.沖動和能量守恒現在導致相互作用常數a，因此

The first tells us that (u1-v1) and thus (u2-v2) is a multiple of (p1-p2), the impulse exchange is in the normal or radial direction, no tangential interaction. Conservation of impulse and energy now leads to a interaction constant a so that

u1-v1 = m2*a*(p1-p2)
u2-v2 = m1*a*(p2-p1)
0 = dot(m2*a*(p1-p2), 2*u1-m2*a*(p1-p2)-2*u2+m1*a*(p2-p1))

導致非零交互項 a

2 * dot(p1-p2, u1-u2) = (m1+m2) * dot(p1-p2,p1-p2) * a

現在可以使用分數來解決

which can now be solved using the fraction

b = dot(p1-p2, u1-u2) / dot(p1-p2, p1-p2)

作為

a = 2/(m1+m2) * b

v1 = u1 - 2 * m2/(m1+m2) * b * (p1-p2)
v2 = u2 - 2 * m1/(m1+m2) * b * (p2-p1)

要讓第二個圓盤靜止，設置 u2=0 并且它的質量 m2 非常大或無限大，然后第二個公式說 v2=u2=0 和第一個

To get the second disk stationary, set u2=0 and its mass m2 to be very large or infinite, then the second formula says v2=u2=0 and the first

v1 = u1 - 2 * dot(p1-p2, u1)/dot(p1-p2, p1-p2) * (p1-p2)

<小時>

也就是說，v1 是 u1 在以 (p1-p2) 為法線的平面上的反射.請注意，碰撞點的特征是 norm(p1-p2)=r1+r2 或

dot(p1-p2, p1-p2) = (r1+r2)^2

這樣分母就已經從碰撞檢測中知道了.

so that the denominator is already known from collision detection.

根據您的代碼，oc{x,y} 包含固定磁盤或球體的中心，sc{x,y} 包含中心和 {vx,vy} 移動磁盤的速度.

Per your code, oc{x,y} contains the center of the fixed disk or orb, sc{x,y} the center and {vx,vy} the velocity of the moving disk.

1. 計算 dc={sc.x-oc.x, sc.y-oc.y} 和 dist2=dc.x*dc.x+dc.y*dc.y

1.a 檢查 sqrt(dist2) 是否足夠接近 sc.radius+oc.radius.普遍的傳說說比較正方形更有效.如果 dist2 太小，微調交點的位置.

1.a Check that sqrt(dist2) is sufficiently close to sc.radius+oc.radius. Common lore says that comparing the squares is more efficient. Fine-tune the location of the intersection point if dist2 is too small.

計算 dot = dc.x*vx+dcy*vy 和 dot = dot/dist2

更新vx = vx - 2*dot*dc.x, vy = vy - 2*dot*dc.y

特殊情況包含在這些公式中，例如，對于 dc.y==0，即 oc.y==sc.y 得到 dot=vx/dc.x，所以 vx=-vx, vy=vy 結果.

The special cases are contained inside these formulas, e.g., for dc.y==0, that is, oc.y==sc.y one gets dot=vx/dc.x, so that vx=-vx, vy=vy results.

這篇關于動圈與非動圈的JAVA彈性碰撞的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！