[osg-users] Is the Z-Buffer linear or non-linear?
Hello,
I am completely confused about the Z-Buffer. I thought it is non-linear but now
I read a lot in forums and now I am not sure about that. The reason about this
question is I want render linear depth values into a texture. I think, I can do
it with this shader:
varying float depth;
uniform float near;
uniform float far;
void main(void)
{
vec4 viewPos = gl_ModelViewMatrix * gl_Vertex;
depth = (-viewPos.z-near)/(far-near);
gl_Position = ftransform();
};
varying float depth;
void main(void)
{
gl_FragDepth = depth;
}
But the poor think is all other shader will be overwritten.
Now I use this:
_rendertarget-setInternalFormat(GL_DEPTH_COMPONENT);
_cam-setClearMask(GL_DEPTH_BUFFER_BIT);
_cam-setRenderTargetImplementation(osg::Camera::FRAME_BUFFER_OBJECT);
_cam-attach(osg::Camera::DEPTH_BUFFER, _rendertarget);
_cam-setRenderOrder(::osg::Camera::PRE_RENDER, 0);
But this is nonlinear, isn't it?
I hope someone can help me about this.
Thanke you very much.
Martin
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Re: [osg-users] Is the Z-Buffer linear or non-linear?
Hello,
I think I got it. Correct me if I am wrong, if I use an orthogonal projection I
get linear Z-Buffer values. If I use a perspective projection I get non-linear
values.
The Z-Buffer value (z') of a perspective projection is:
z' = (far + near) / (far - near) + 1/z ((-2 * far * near)/(far-near))
The Z-Buffer value (z') of a orthogonal projection is:
z' = 2 * ((z-near)/(far-near))-1
So, I use a orthogonal projection therefore I get a linear Z-Buffer.
I hope that is right.
Cheers
Martin
Original-Nachricht
Datum: Wed, 24 Aug 2011 10:01:23 +0200
Von: Martin Großer grosser.mar...@gmx.de
An: osg-users@lists.openscenegraph.org
Betreff: [osg-users] Is the Z-Buffer linear or non-linear?
Hello,
I am completely confused about the Z-Buffer. I thought it is non-linear
but now I read a lot in forums and now I am not sure about that. The reason
about this question is I want render linear depth values into a texture. I
think, I can do it with this shader:
varying float depth;
uniform float near;
uniform float far;
void main(void)
{
vec4 viewPos = gl_ModelViewMatrix * gl_Vertex;
depth = (-viewPos.z-near)/(far-near);
gl_Position = ftransform();
};
varying float depth;
void main(void)
{
gl_FragDepth = depth;
}
But the poor think is all other shader will be overwritten.
Now I use this:
_rendertarget-setInternalFormat(GL_DEPTH_COMPONENT);
_cam-setClearMask(GL_DEPTH_BUFFER_BIT);
_cam-setRenderTargetImplementation(osg::Camera::FRAME_BUFFER_OBJECT);
_cam-attach(osg::Camera::DEPTH_BUFFER, _rendertarget);
_cam-setRenderOrder(::osg::Camera::PRE_RENDER, 0);
But this is nonlinear, isn't it?
I hope someone can help me about this.
Thanke you very much.
Martin
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Re: [osg-users] Is the Z-Buffer linear or non-linear?
Hi
The z-buffer is not linear ( in my experience) its weighted to give higher
resolution closer to the near clip plane
See
http://www.sjbaker.org/steve/omniv/love_your_z_buffer.html
Gordon Tomlinson
3D Technical Product Manager
Overwatch®
An Operating Unit of Textron Systems
__
-Original Message-
From: osg-users-boun...@lists.openscenegraph.org
[mailto:osg-users-boun...@lists.openscenegraph.org] On Behalf Of Martin Großer
Sent: Wednesday, August 24, 2011 7:58 AM
To: OpenSceneGraph Users
Subject: Re: [osg-users] Is the Z-Buffer linear or non-linear?
Hello,
I think I got it. Correct me if I am wrong, if I use an orthogonal projection I
get linear Z-Buffer values. If I use a perspective projection I get non-linear
values.
The Z-Buffer value (z') of a perspective projection is:
z' = (far + near) / (far - near) + 1/z ((-2 * far * near)/(far-near))
The Z-Buffer value (z') of a orthogonal projection is:
z' = 2 * ((z-near)/(far-near))-1
So, I use a orthogonal projection therefore I get a linear Z-Buffer.
I hope that is right.
Cheers
Martin
Original-Nachricht
Datum: Wed, 24 Aug 2011 10:01:23 +0200
Von: Martin Großer grosser.mar...@gmx.de
An: osg-users@lists.openscenegraph.org
Betreff: [osg-users] Is the Z-Buffer linear or non-linear?
Hello,
I am completely confused about the Z-Buffer. I thought it is non-linear
but now I read a lot in forums and now I am not sure about that. The reason
about this question is I want render linear depth values into a texture. I
think, I can do it with this shader:
varying float depth;
uniform float near;
uniform float far;
void main(void)
{
vec4 viewPos = gl_ModelViewMatrix * gl_Vertex;
depth = (-viewPos.z-near)/(far-near);
gl_Position = ftransform();
};
varying float depth;
void main(void)
{
gl_FragDepth = depth;
}
But the poor think is all other shader will be overwritten.
Now I use this:
_rendertarget-setInternalFormat(GL_DEPTH_COMPONENT);
_cam-setClearMask(GL_DEPTH_BUFFER_BIT);
_cam-setRenderTargetImplementation(osg::Camera::FRAME_BUFFER_OBJECT);
_cam-attach(osg::Camera::DEPTH_BUFFER, _rendertarget);
_cam-setRenderOrder(::osg::Camera::PRE_RENDER, 0);
But this is nonlinear, isn't it?
I hope someone can help me about this.
Thanke you very much.
Martin
--
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Re: [osg-users] Is the Z-Buffer linear or non-linear?
The transform below for the orthogonal projection is indeed linear, mapping
[n,f] to [-1,1] in linear fashion.
If you want to linearize depth with a perspective projection (i.e. survive the
perspective divide between the vertex and fragment shader), here's a way to do
that. The example uses HLSL but it gets the point across...
http://www.mvps.org/directx/articles/linear_z/linearz.htm
Another interesting discussion related to linear depth using W-buffers can be
found at
http://www.mvps.org/directx/articles/using_w-buffers.htm
Just make sure you have support for W buffers...
-Shayne
-Original Message-
From: osg-users-boun...@lists.openscenegraph.org
[mailto:osg-users-boun...@lists.openscenegraph.org] On Behalf Of Martin Großer
Sent: Wednesday, August 24, 2011 5:58 AM
To: OpenSceneGraph Users
Subject: Re: [osg-users] Is the Z-Buffer linear or non-linear?
Hello,
I think I got it. Correct me if I am wrong, if I use an orthogonal projection I
get linear Z-Buffer values. If I use a perspective projection I get non-linear
values.
The Z-Buffer value (z') of a perspective projection is:
z' = (far + near) / (far - near) + 1/z ((-2 * far * near)/(far-near))
The Z-Buffer value (z') of a orthogonal projection is:
z' = 2 * ((z-near)/(far-near))-1
So, I use a orthogonal projection therefore I get a linear Z-Buffer.
I hope that is right.
Cheers
Martin
Original-Nachricht
Datum: Wed, 24 Aug 2011 10:01:23 +0200
Von: Martin Großer grosser.mar...@gmx.de
An: osg-users@lists.openscenegraph.org
Betreff: [osg-users] Is the Z-Buffer linear or non-linear?
Hello,
I am completely confused about the Z-Buffer. I thought it is non-linear
but now I read a lot in forums and now I am not sure about that. The reason
about this question is I want render linear depth values into a texture. I
think, I can do it with this shader:
varying float depth;
uniform float near;
uniform float far;
void main(void)
{
vec4 viewPos = gl_ModelViewMatrix * gl_Vertex;
depth = (-viewPos.z-near)/(far-near);
gl_Position = ftransform();
};
varying float depth;
void main(void)
{
gl_FragDepth = depth;
}
But the poor think is all other shader will be overwritten.
Now I use this:
_rendertarget-setInternalFormat(GL_DEPTH_COMPONENT);
_cam-setClearMask(GL_DEPTH_BUFFER_BIT);
_cam-setRenderTargetImplementation(osg::Camera::FRAME_BUFFER_OBJECT);
_cam-attach(osg::Camera::DEPTH_BUFFER, _rendertarget);
_cam-setRenderOrder(::osg::Camera::PRE_RENDER, 0);
But this is nonlinear, isn't it?
I hope someone can help me about this.
Thanke you very much.
Martin
--
NEU: FreePhone - 0ct/min Handyspartarif mit Geld-zurück-Garantie!
Jetzt informieren: http://www.gmx.net/de/go/freephone
___
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