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)?
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
}
})
}
}
}
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.
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 |
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 |
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.
This gives the following recipie:
d = sqrt( xx² + yx² + zx² )
| d 0 0 T.x |
| 0 d 0 T.y |
| 0 0 d T.z |
| 0 0 0 1 |
Loading this matrix allows you to draw camera aligned sprites.
Apologies in advance for this being a suggestion, not a solution:
If your intent is simulate 3D spherical rotations, your easiest route would be to forego the 2.5D API and just use simple scaling and sin/cos calculations for the positioning.
A fun place to start: http://www.reflektions.com/miniml/template_permalink.asp?id=329
I've solved it using Wikipedia, matrices and black magic. I choose to implement custom rotation instead of inverting the rotation of all objects.
Heres the code if anyones interested:
package{
import flash.display.Sprite;
import flash.events.Event;
public class test extends Sprite{
private var canvas:Sprite = new Sprite();
private var sprites:Array = []
private var rotx:Number=0,roty:Number=0,rotz:Number=0;
private var mm:Matrix3 = new Matrix3();
public function test(){
addChild(canvas);
canvas.x = canvas.y = 230
for (var i:int=0;i<30;i++){
var sp:Sprite = new Sprite();
canvas.addChild(sp);
sp.graphics.beginFill(0xFF0000);
sp.graphics.drawCircle(0,0,2);
sp.x = Math.random()*200-100;
sp.y = Math.random()*200-100;
sp.z = Math.random()*200-100;
sprites.push(sp);
rotx=0.06; //from top to bottom
//roty=0.1; //from right to left
rotz=0.1; //clockwise
mm.make3DTransformMatrix(rotx,roty,rotz);
}
addEventListener(Event.ENTER_FRAME,function():void{
for (var i:int=0;i<sprites.length;i++){
var s:Sprite = sprites[i];
mm.rotateByAngles(s);
}
})
}
}
}
and the matrix3 class:
public class Matrix3{
private var da:Vector.<Number>; // rows
public function make3DTransformMatrix(rotx:Number,roty:Number,rotz:Number):void{
var cosx:Number = Math.cos(rotx);
var cosy:Number = Math.cos(roty);
var cosz:Number = Math.cos(rotz);
var sinx:Number = Math.sin(rotx);
var siny:Number = Math.sin(roty);
var sinz:Number = Math.sin(rotz);
da = new <Number>[
cosy*cosz, -cosx*sinz+sinx*siny*cosz, sinx*sinz+cosx*siny*cosz,
cosy*sinz, cosx*cosz+sinx*siny*sinz , -sinx*cosz+cosx*siny*sinz,
-siny , sinx*cosy , cosx*cosy ];
}
public function rotateByAngles(d:DisplayObject):void{
var dx:Number,dy:Number,dz:Number;
dx = da[0]*d.x+da[1]*d.y+da[2]*d.z;
dy = da[3]*d.x+da[4]*d.y+da[5]*d.z;
dz = da[6]*d.x+da[7]*d.y+da[8]*d.z;
d.x = dx;
d.y = dy;
d.z = dz;
}
}
}