我有很多安排在三维空间的精灵，他们的父容器已旋转应用。 我要如何扭转精灵3D旋转，它们总是面对摄像机（ActionScript 3的）？

i have lots of sprites arranged in 3D space, and their parent container has rotations applied. How do i reverse the sprites 3D rotation, that they always face the camera (Actionscript 3)?

继承人code来测试它：

heres a code to test it:

package{
import flash.display.Sprite;
import flash.events.Event;
public class test extends Sprite{

var canvas:Sprite = new Sprite();
var sprites:Array = []

public function test(){
    addChild(canvas)
    for (var i:int=0;i<20;i++){
        var sp:Sprite = new Sprite();
        canvas.addChild(sp);
        sp.graphics.beginFill(0xFF0000);
        sp.graphics.drawCircle(0,0,4);
        sp.x = Math.random()*400-200;
        sp.y = Math.random()*400-200;
        sp.z = Math.random()*400-200;
        sprites.push(sp);
    }
    addEventListener(Event.ENTER_FRAME,function():void{
        canvas.rotationX++;
        canvas.rotationY = canvas.rotationY+Math.random()*2;
        canvas.rotationZ++;
        for (var i:int=0;i<20;i++){
            //this is not working...
            sprites[i].rotationX = -canvas.rotationX
            sprites[i].rotationY = -canvas.rotationY
            sprites[i].rotationZ = -canvas.rotationZ
        }
    })
}
}
}

我猜我已经做了一些魔法与精灵的rotation3D矩阵... 我试图执行这个脚本：的http：// ughzoid .word press.com / 2011/02/03 / Papervision3D的-sprite3d / ，但有这么成功 感谢您的帮助。

I am guessing i have to do some magic with the rotation3D matrices of the sprites... I've tried to implement this script: http://ughzoid.wordpress.com/2011/02/03/papervision3d-sprite3d/ , but had so success Thanks for help.

推荐答案

要做到这一点最简单的方法是清算的变换矩阵的旋转部分。典型的同质转型看起来像这样

The easiest way to do this is "clearing" the rotational part of the transform matrix. Your typical homogenous transformation looks like this

| xx xy xz xw |
| yx yy yz yw |
| zx zy zz zw |
| wx wy wz ww | 

与WX = WY = WZ = 0，WW = 1。如果你把你会看到，其实这个矩阵是由一个3×3子矩阵的定义旋转仔细一看，有3子向量的翻译和均匀行0 0 0 1

with wx = wy = wz = 0, ww = 1. If you take a closer look you'll see that in fact this matrix is composed of a 3x3 submatrix defining the rotation, a 3 subvector for the translation and a homogenous row 0 0 0 1

| R       T |
| (0,0,0) 1 |

对于要保留翻译，但摆脱了旋转的广告牌/雪碧，即R =一，在某些情况下是scaleing应用的身份需要进行调整为好。

For a billboard/sprite you want to keep the translation, but get rid of the rotation, i.e. R = I. In case some scaleing was applied the identity needs to be scaled as well.

这给出了以下recipie：

This gives the following recipie:

D =开方（xx²+yx²+zx²）

d = sqrt( xx² + yx² + zx² )

| d 0 0 T.x |
| 0 d 0 T.y |
| 0 0 d T.z |
| 0 0 0   1 |

加载这个矩阵可以绘制相机对准精灵。

Loading this matrix allows you to draw camera aligned sprites.