D3D sprites

Hmm... Does nobody have any idea? I still don''t know how to draw my billboards.

ArgoN

I would do it like this:

1. Get the View matrix & strip away the translation part
2. Transpose the resulting rotation matrix to get the inverse
3. Concatenate this inverted rotation matrix with a rotateZ matrix to get the z roll
4. Concatenate the resulting matrix with the translation matrix for the sprite
5. Use this matrix as the world matrix when rendering the sprite

Don''t change the view matrix, it is better to multiply the world matrix with the inverse of the view matrix instead (as I do above)

- WitchLord

My problem is that (due to my class structure) I don''t have access to the view roll angle when drawing a sprite. If there would be a way to strip the translation and the Z-rotation from the view matrix, everything would be fine...

ArgoN

I''m not sure I understand what z rolling you want to do. Is it that you want the sprite edges to always be aligned with the screen edges? If you do like I wrote without adding the rotateZ matrix you get that behaviour.

1. Get the View matrix & strip away the translation part
2. Transpose the resulting rotation matrix to get the inverse
3. Concatenate the resulting matrix with the translation matrix for the sprite
4. Use this matrix as the world matrix when rendering the sprite





- WitchLord

No, that''s what I don''t want. Like I did it above, the sprites edges are aligned to screen edges. Thats how it shouldn''t be.

Maybe my idea of what I try to do is nonsense. I''ll try to explain: I''m trying to create a little 3D space shooter to gain experiences in 3D programming. I need something that I can use for e.g. the engine''s thrust or a cloud trail. So I thought I can use sprites for it. The sprite should always face the viewer (which it does yet) so that it look like an object that is 3D but looks the same from all angles you''re looking on it. But if the viewer is rolling (rotating at his local Z axis), the sprite must also rotate the same so that it looks like it keeps it''s orientation in space. If the sprites edges are always aligned to screen edges, it looks like it would turn with the viewer when rolling.

Isn''t that the way sprites are supposed to behave?

ArgoN

The simplest way would be to store the camera parameters so that they are globally available. But if you don''t, you can extract the camera roll angle by examining the camera up vector of the view matrix. The up vector can be retrieved from the the second column. (the structure of the viewmatrix)

To get the roll vector you examine the angle that would be need to roll the vector back to its position before the camera roll, which is on the yz-plane. Depending on the pitch of the camera the vector should either point upwards or downwards.

As you can imagine, it requires a lot of computations to extract the angles from the matrix. If you still want to know how it is done I can tell you how it''s done, but you have to ask nicely

- WitchLord

Ok, lets see if I get it right


// 1) Get the current view matrix
D3DXMATRIX mat;
m_pD3DDevice->GetTransform(D3DTRANSFORMSTATE_VIEW, mat);

// 2) Strip translations
mat(3, 0) = 0;
mat(3, 1) = 0;
mat(3, 2) = 0;

// 3) Invert it to get a world matrix
D3DVALUE fDet;
D3DXMatrixInverse(&mat, &fDet, &mat);

// 4) Get the up vector from it and strip pitching (z-value)
D3DXVECTOR3 vecViewUp;
vecViewUp.x = mat(0, 1);
vecViewUp.y = mat(1, 1);
vecViewUp.z = 0;
D3DXVec3Normalize(&vecViewUp, &vecViewUp);

// 5) Calc angle to world y axis (viewers roll)
D3DXVECTOR3 vecWorldUp(0, 1, 0);
D3DVALUE fCosRollAngle = D3DXVec3Dot(&vecViewUp, &vecWorldUp);
D3DVALUE fRollAngle = acosf(fCosRollAngle);

// 6) Fix left/right rotation
if (vecViewUp.x > 0)
fRollAngle = -fRollAngle;

// 7) Apply Z rotation
D3DXMATRIX matRoll;
D3DXMatrixRotationZ(&matRoll, fRollAngle);
D3DXMatrixMultiply(&mat, &matRoll, &mat);

// 8) Apply sprite coordinates
D3DXMATRIX matPos;
D3DXMatrixIdentity(&matPos);
matPos(3, 0) = m_vecPos.x;
matPos(3, 1) = m_vecPos.y;
matPos(3, 2) = m_vecPos.z;
D3DXMatrixMultiply(&mat, &mat, &matPos);

// 9) Set world matrix
m_pD3DDevice->SetTransform(D3DTRANSFORMSTATE_WORLD, mat);


Phew... that works, but isn't that a bit oversized for a simple sprite calculation? I thought sprites are supposed to be fast .

Hmmm, the code above could be optimized (e.g. the dot product of (x,y,z)*(0,1,0) can be simplified to y . And setting up the z-rotation matrix could also be faster because I already have the cosine of the angle).

But I still think it's oversized. Are sprites really that 'difficult' to calculate? Maybe I could get rid of the matrix inversion by modifying the view matrix instead.

Everything would be only little easier if I had the view roll angle: I could get rid of steps 4 to 6, but my class structure doesn't allow to easily retrieve the roll angle.

I thought about the above solution before, but I think it's pretty much for a simple sprite. If you know a better solution, I would love to ask you nicely


Edited by - ArgoN on June 22, 2000 7:04:50 PM

You can get rid of the matrix inversion by using a transpose instead. Because the matrix only contains rotation information this will be the same as inversion.

I think you are also extracting the wrong vector from the matrix, because you have inverted the matrix the up vector will now be found on the second row instead.

You have to remember that much of this information can be shared among the sprite renderings, you''ll only have to compute their rotation matrices once for all of them.



- WitchLord

Oops, you''re right with the wrong vector.

Ok, thats working for now. I''ll try to optimize it a bit more and it should be fast enough. Thanks for your help!

ArgoN

Why not
1)Just get the 3d center point the sprite should be at.
2)Scale a transformed vertex sprite by the distance of the camera
to the point.
3)Draw the sprite.

The sprite will always be pointing up, scaled, and
by using a ''billboard'' can be testing into your z-buffer,
transparent, etc.

I''d just like to say thank you very much to all who responded in this post since I''ve been struggling with this very problem for a while now.