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HLSL - Billboarding a mesh


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#1 relaxok   Members   -  Reputation: 133

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Posted 22 January 2014 - 12:49 AM

Guys:

 

I have a vertex shader that I use for all my non-terrain models.  My per-mesh structure has a flags uint where i can set things like billboard, unlit, etc.  Some models have single meshes that need to be billboarded, like a light halo that looks the same from any angle.  However, I can't figure out how to successfully billboard the mesh.

 

Below is my vertex shader.. in the billboard section I've tried various combinations of matrices as you can see.. based on MJP's response in this topic:  

 

http://www.gamedev.net/topic/507808-hlsl-billboard/


PixelInputType VS(VertexInputType input, uint InstanceID : SV_InstanceID )
{	
	PixelInputType output;
    
	// Fix w
        input.position.w = 1.0f;

	// Calculate animation matrices
	matrix bonematrix = ( float( input.boneweight0 ) / 255.0f ) *  BoneMatrices[ input.boneindex0 ];
	bonematrix += ( float( input.boneweight1 ) / 255.0f ) *  BoneMatrices[ input.boneindex1 ];
	bonematrix += ( float( input.boneweight2 ) / 255.0f ) *  BoneMatrices[ input.boneindex2 ];
	bonematrix += ( float( input.boneweight3 ) / 255.0f ) *  BoneMatrices[ input.boneindex3 ];

	// World matrix from instance buffer
	matrix worldMatrixInstance = InstanceLocations[ InstanceID ];

	// Transform via combined bones + world matrix
	output.position = mul(input.position, bonematrix);
	output.position = mul(output.position, worldMatrixInstance );
	
	// Animate normal the same way
	output.normal = mul(input.normal, bonematrix);
	output.normal = mul(output.normal, worldMatrixInstance );

	// Camera view
        output.position = mul(output.position, viewMatrix);
	float4 cameraPosition = output.position;

	// vertex color
	output.vertex_color = input.vertex_color;

	// billboard?
	if( flags & PER_MESH_FLAG_BILLBOARD )
	{
		matrix boneWorldMat = mul( bonematrix, worldMatrixInstance );
		matrix boneWorldViewMat = mul( boneWorldMat, viewMatrix );
		//output.position = input.position + float4( boneWorldMat._41, boneWorldMat._42, boneWorldMat._43, 1.0f );
		output.position = input.position + float4( boneWorldViewMat._41, boneWorldViewMat._42, boneWorldViewMat._43, 1.0f );
		//output.position = input.position + float4( viewMatrix._41, viewMatrix._42, viewMatrix._43, 1.0f );
	}
		
	output.position = mul(output.position, projectionMatrix);
	 
	output.tex = input.tex;

	// Fog calc
	output.fogFactor = saturate((fogEnd - cameraPosition.z) / (fogEnd - fogStart));

	output.instance = InstanceID;
    
        return output;
}

I am using a y-up system, if it matters.

 

Please note there are no issues with the shader outside of the billboarding so please no recommendations on other things I can do differently.  I just want to know how to billboard within the context of what you see here.

 

Thanks..


Edited by relaxok, 22 January 2014 - 03:05 PM.


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#2 8up tuotlo   Members   -  Reputation: 120

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Posted 23 January 2014 - 07:27 PM

Out.Pos = mul(float4(J + mul(Wm,Vm)[3], 1.0f), Pm);

 

float4 Out.Pos - result

float3 J - triangle in object space what must to bee bilboarded

float4x4 Wm - object to world space matrix

float4x4 Vm - world to view space matrix

float4x4 Pm - view to screen space matrix

 

And this tringle will always be aligned to screen

 

But triangel must be aligned to screen.

Screen coordinates is:

X wide

Y heght

Z deep

 

----------------------------

| Y                             |

| ^                              |

| |                               |
| | Z                            |

| |------------> X           |

----------------------------

 

Culling may to make your triangle invizible


Edited by 8up tuotlo, 24 January 2014 - 12:18 AM.


#3 relaxok   Members   -  Reputation: 133

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Posted 11 February 2014 - 03:46 PM

Out.Pos = mul(float4(J + mul(Wm,Vm)[3], 1.0f), Pm);

 

float4 Out.Pos - result

float3 J - triangle in object space what must to bee bilboarded

float4x4 Wm - object to world space matrix

float4x4 Vm - world to view space matrix

float4x4 Pm - view to screen space matrix

 

And this tringle will always be aligned to screen

 

But triangel must be aligned to screen.

Screen coordinates is:

X wide

Y heght

Z deep

 

----------------------------

| Y                             |

| ^                              |

| |                               |
| | Z                            |

| |------------> X           |

----------------------------

 

Culling may to make your triangle invizible

 

It seems like that's what I'm doing?

 

Latest attempt:

// billboard?
	if( flags & PER_MESH_FLAG_BILLBOARD )
	{
		matrix boneWorldMatrix = mul( bonematrix, worldMatrixInstance );
		matrix boneWorldViewMatrix = mul( boneWorldMatrix, viewMatrix );
		
		float3 position = input.position + float3( boneWorldViewMatrix._41, boneWorldViewMatrix._42, boneWorldViewMatrix._43 );
		output.position = mul( float4( position, 1.0f ), projectionMatrix );
	}
	else
	{	
		output.position = mul(output.position, projectionMatrix);
	}


#4 8up tuotlo   Members   -  Reputation: 120

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Posted 11 February 2014 - 06:04 PM

Yes, it is.

BUT your triangle must already be placed in CMERA coordinates in vertex buffer(your mesh).

it means thad what up side of your triangle is y+;

front side z+( or z-) i forget ( try two variants; 

and right side is x+;

 

THAN THIS works so

 

if( flags & PER_MESH_FLAG_BILLBOARD )
    {
        matrix boneWorldMatrix = mul( bonematrix, worldMatrixInstance );    
        matrix boneWorldViewMatrix = mul( boneWorldMatrix, viewMatrix );  
        
        float3 position
= input.position + float3( boneWorldViewMatrix._41, boneWorldViewMatrix._42, boneWorldViewMatrix._43 );

        now we TRANSLATE our triangle from Object space to World space ONLY translate no rotations (and this is the bilboarding)
        output.position = mul( float4( position, 1.0f ), projectionMatrix ); standart projection

    }
    else
    {    
        output
.position = mul(output.position, projectionMatrix);
    }    

 

we do NOT make rotations

thet means thad what if our triangle were ben rotatet from the CMERA we do not see it

screen tays on z+ (or z-) and takes look an another side (this is CMERA SPACE)

 

Wm takes our verties from ObjectSpce to WorldSpace

Vm takes our vertices from WorldSpace to ViewSpace - transfer vertices without rotation its (BILLBOARDING) but triangle must be already placed correctly

Pm takes our vertices from ViewSpace to ScreenSpace



#5 relaxok   Members   -  Reputation: 133

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Posted 12 February 2014 - 02:57 PM

        now we TRANSLATE our triangle from Object space to World space ONLY translate no rotations (and this is the bilboarding)

 

 

Isn't the line:

 

float3 position = input.position + float3( boneWorldViewMatrix._41, boneWorldViewMatrix._42, boneWorldViewMatrix._43 );

 

supposed to do exactly that?  adjust by the translation only..



#6 8up tuotlo   Members   -  Reputation: 120

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Posted 16 February 2014 - 07:19 PM

Yes it is...

 

you have billboarding working correctly or its any problems?



#7 L. Spiro   Crossbones+   -  Reputation: 13576

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Posted 16 February 2014 - 08:40 PM

To billboard an object you are supposed to multiply it by the camera matrix, not the view matrix.  The view matrix is the inverse of the camera matrix and it is what you send in an MVP (model-view-projection) matrix.

You should not be manually adjusting positions etc.  It’s nothing more than a single matrix multiply.

 

 

L. Spiro


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#8 relaxok   Members   -  Reputation: 133

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Posted 17 February 2014 - 01:06 AM

To billboard an object you are supposed to multiply it by the camera matrix, not the view matrix.  The view matrix is the inverse of the camera matrix and it is what you send in an MVP (model-view-projection) matrix.

You should not be manually adjusting positions etc.  It’s nothing more than a single matrix multiply.

 

Why would that accomplish not rotating it?



#9 relaxok   Members   -  Reputation: 133

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Posted 18 February 2014 - 02:22 PM

My new idea was to try making a billboarding view matrix which does the translation based on the real camera rotations, but for the rotation part of the view matrix, uses the inverse matrices.

 

This results in ALMOST the correct result.. the mesh now stays 'camera facing' no matter what the camera is doing, however the translation seems to put it all over the place.  I'm wondering how this can be when the translation is the same as when not billboarding:

 

http://bh.polpo.org/quad_no_rot1.avi

 

What adjustment can I make to keep the translation correct?

void CCamera::transform( float x, float y, float z, float fOrientation )
{
	XMMATRIX matRotationX = XMMatrixRotationX( m_Pitch );
	XMMATRIX matRotationY = XMMatrixRotationY( XMConvertToRadians( fOrientation  * 360.0f ) + m_Yaw );
	XMVECTOR unitVec = XMVectorSet( 0.0f, m_Height, -m_Distance, 0.0f );
	XMVECTOR transVec = XMVector3TransformCoord( unitVec, matRotationX * matRotationY );

	m_posX = x + XMVectorGetX( transVec );
	m_posY = y + XMVectorGetY( transVec );
	m_posZ = z + XMVectorGetZ( transVec );

	XMMATRIX matTranslation = XMMatrixTranslation( m_posX, m_posY, m_posZ );
	XMVECTOR matInvDeter;
	XMMATRIX matInverse = XMMatrixInverse( &matInvDeter, matRotationX * matRotationY );
	XMMATRIX matTransInverse = XMMatrixInverse( &matInvDeter, matTranslation );

	// Billboard version
	XMVECTOR matRotBillboardDeter;
	XMMATRIX matRotationXBillboard = XMMatrixInverse( &matRotBillboardDeter, matRotationX );
	XMMATRIX matRotationYBillboard = XMMatrixInverse( &matRotBillboardDeter, matRotationY );
	XMMATRIX matInverseBillboard = XMMatrixInverse( &matInvDeterBillboard, matRotationXBillboard * matRotationYBillboard );
	
	XMStoreFloat4x4( &m_matView, matTransInverse * matInverse );
	XMStoreFloat4x4( &m_matBillboard, matTransInverse * matInverseBillboard );
	XMStoreFloat4x4( &m_matRotationX, matRotationX );
	XMStoreFloat4x4( &m_matRotationY, matRotationY );
	XMStoreFloat4x4( &m_matTranslation, matTranslation );

	calculateViewFrustum();
}

Edited by relaxok, 18 February 2014 - 02:32 PM.


#10 L. Spiro   Crossbones+   -  Reputation: 13576

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Posted 18 February 2014 - 03:13 PM

Why would that accomplish not rotating it?

Because it undoes the rotation applied by the camera…

 

My new idea was to try making a billboarding view matrix which does the translation based on the real camera rotations, but for the rotation part of the view matrix, uses the inverse matrices.

Why don’t you just quickly make the WBP (world-billboard-projection) matrix and be done with all that ridiculous math?
 
XMMATRIX mBillboard;
XMMatrixInverse( &mBillboard, m_matView ); // m_matView is the inverse of another matrix, so instead of inverting it again you could just copy that matrix.
// Only keep the upper-left 3×3.
mBillboard._41 = mBillboard._42 = mBillboard._43 = 0.0f;  // Billboard matrix complete.


XMMATRIX mBillboardView = mBillboard * m_matView; // Can be used instead of the view matrix.
XMMATRIX mBillboardViewProjection = mBillboardView * m_matProj; // Can be used instead of the view-projection matrix.
Why make it so complicated?


L. Spiro

Edited by L. Spiro, 18 February 2014 - 03:19 PM.

It is amazing how often people try to be unique, and yet they are always trying to make others be like them. - L. Spiro 2011
I spent most of my life learning the courage it takes to go out and get what I want. Now that I have it, I am not sure exactly what it is that I want. - L. Spiro 2013
I went to my local Subway once to find some guy yelling at the staff. When someone finally came to take my order and asked, “May I help you?”, I replied, “Yeah, I’ll have one asshole to go.”
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#11 8up tuotlo   Members   -  Reputation: 120

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Posted 21 February 2014 - 12:10 AM

on my mind billboarding its placing triangle fefore screen without rotations(always to us)

 

just make WorldViewMatrix

Add to triangle WorldViewMatrix[3] - translation

and multiple it on projection

 

all

 

that what you doing is too much


Edited by 8up tuotlo, 21 February 2014 - 12:22 AM.


#12 relaxok   Members   -  Reputation: 133

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Posted 25 February 2014 - 01:40 PM

 

Why would that accomplish not rotating it?

Because it undoes the rotation applied by the camera…

 

My new idea was to try making a billboarding view matrix which does the translation based on the real camera rotations, but for the rotation part of the view matrix, uses the inverse matrices.

Why don’t you just quickly make the WBP (world-billboard-projection) matrix and be done with all that ridiculous math?
 
XMMATRIX mBillboard;
XMMatrixInverse( &mBillboard, m_matView ); // m_matView is the inverse of another matrix, so instead of inverting it again you could just copy that matrix.
// Only keep the upper-left 3×3.
mBillboard._41 = mBillboard._42 = mBillboard._43 = 0.0f;  // Billboard matrix complete.


XMMATRIX mBillboardView = mBillboard * m_matView; // Can be used instead of the view matrix.
XMMATRIX mBillboardViewProjection = mBillboardView * m_matProj; // Can be used instead of the view-projection matrix.
Why make it so complicated?


L. Spiro

 

 

The reason it's all spread out is that it's not working yet, and this way I can experiment with changing individual elements.

 

I have tried everything suggested in this thread and it's still not working.  Using your suggested method in the above quote, the mesh simply doesn't appear at all, the same as most other attempts..

 

Still the closest I've gotten is using the inverses of the rotation matrices, but the vertices translate all over the place in this case.. as shown in the video i linked.



#13 8up tuotlo   Members   -  Reputation: 120

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Posted 27 February 2014 - 08:02 AM

Try to do so:

 

Get the transpose(float3x3(your WorldViewMatrix));

Get your mesh in local space, multiple him on this ^

And than multiple him as simple, no billboarded mesh;

 

and try to set culling mode off



#14 relaxok   Members   -  Reputation: 133

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Posted 04 March 2014 - 05:45 PM

Try to do so:

 

Get the transpose(float3x3(your WorldViewMatrix));

Get your mesh in local space, multiple him on this ^

And than multiple him as simple, no billboarded mesh;

 

and try to set culling mode off

 

I'm not quite following your description, sorry... but maybe this is going back to what you're referring to.. this and showing both sides of the mesh, I do get fairly close..

	// billboard?
	if( flags & PER_MESH_FLAG_BILLBOARD )
	{
		matrix worldViewMat = mul( worldMatrixInstance, viewMatrix );
		output.position = skinnedPos + float4( worldViewMat._41, worldViewMat._42, worldViewMat._43, 1.0f );
	}
	
	output.position = mul( output.position, projectionMatrix ); 

But as you can see, while the distance of the camera is accounted for fine.. the proper translation doesn't happen when the camera rotates on axes..

 

http://bh.polpo.org/billboard_rotate_issue.wmv

 

What's the reason for this?


Edited by relaxok, 04 March 2014 - 05:50 PM.


#15 8up tuotlo   Members   -  Reputation: 120

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Posted 08 March 2014 - 08:49 PM

Try to do so:

 

matrix worldViewMat = mul( worldMatrixInstance, viewMatrix );
output.position = mul(skinnedPos, worldViewMat );

output.position = mul(output.position, transpose(float3x3(worldViewMat)) );

 

or this:

 

matrix worldViewMat = mul( worldMatrixInstance, viewMatrix );       

output.position = skinnedPos + mul(float4( worldViewMat._41, worldViewMat._42, worldViewMat._43, 1.0f ), worldViewMat );






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