I'm working on a 3d sketch where I draw an L-system tree and am trying to save the locations of the points at the termination of each branch.  I am trying to get these points by using modelX,Y,Z  but the behavior seems very wrong.  It appears that there is a bug, as described in issue 148:

http://code.google.com/p/processing/issues/detail?id=148

anyway, is anyone here aware of a workaround for this?  the drawing I am doing relies heavily on matrix transforms.  I think I could try and find some way to keep track of the position as I go along, but that seems somewhat difficult.  In the picture below, the red cubes are drawn where the modelX function *reports* the terminal points on the branches should be.  

 treeForFlakes2.zip

the above is a link to the processing file.  my general code for drawing the tree is based on the L-system tree code from the examples:

http://processing.org/learning/topics/tree.html

2. void branch(float h, float s) {
3.   // Each branch will be H* the size of the previous one
4.   h *= 0.66;
5.   s *= 0.75;
6.   strokeWeight(s);
7. 
   
8.   if (h > 30) {
9.     //branch 1
10.     pushMatrix();    // Save the current state of transformation (i.e. where are we now)
11.     rotateX(thetaX);   // Rotate by theta
12.     rotateY(thetaY);
13.     rotateZ(thetaZ);
14. 
    
15.     line(0, 0, 0, -h);  // Draw the branch
16.     translate(0, -h); // Move to the end of the branch
17.     branch(h, s);       // Ok, now call myself to draw two new branches!!
18.     popMatrix();     // Whenever we get back here, we "pop" in order to restore the previous matrix state
19. 
    
20.     //branch 2
21.     strokeWeight(s);
22.     pushMatrix();
23.     rotateX(-thetaX);
24.     rotateY(-thetaY);
25.     rotateZ(thetaZ);
26. 
    
27.     line(0, 0, 0, -h);
28.     translate(0, -h);
29.     branch(h, s);
30.     popMatrix();
31. 
    
32.     //branch 3
33.     strokeWeight(s);
34.     pushMatrix();
35.     rotateX(thetaX);
36.     rotateY(-thetaY);
37.     rotateZ(-thetaZ);
38. 
    
39.     line(0, 0, 0, -h);
40.     translate(0, -h);
41.     branch(h, s);
42.     popMatrix();
43.   }
44.   else {
45.     fCount ++;
46.     
47.     flowersPos[fCount][0] = modelX(0,0,0);
48.     flowersPos[fCount][1] = modelY(0,0,0);
49.     flowersPos[fCount][2] = modelZ(0,0,0);
50. 
    
51.     rotateX(flowerRots[fCount]);
52.     strokeWeight(1);
53.     drawPolyFlower(flowerNums[fCount],h/2);
54.     if (fCount > hiFCount) {
55.       hiFCount = fCount;
56.     }
57.   }
58. }

that is the tree drawing function, and in the else{} code, it saves the position of each terminal branch to an array.

Thanks much for any insight!