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OpenGL OpenGL Perspective Projection and the View Plane

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Hi guys,

I'm having a little trouble wrapping my mind completely around the perspective projection and the relationship to the view plane. A little background:[list]
[*]I'm working on a 3D Software Renderer.
[*]I'm reading Lengyel's "Mathematics for 3D Game Programming and Computer Graphics" (I've added a link at the end to a preview in Google Books, has a good chunk of what I'm reading at least, pp. 111 and on.)
[/list]
So, I understand that a basic perspective projection can be achieved with (obviously the matrix is over kill):

[ 1 0 0 0 ]
[ 0 1 0 0 ]
[ 0 0 1 0 ]
[ 0 0 1/e 1 ]

Where e is the distance to the view-plane(focal length) as a function of the horizontal FOV.

This would correctly project points/vertices, but would not perform any clipping.

I know that the OpenGL projection matrix transforms all of the points into homogeneous clip-space (transforming the view-frustum to a cuboid), and makes clipping to the frustum much more simple. (The matrix can be seen on pp. 124 linked below, labeled Mfrustum)

I also see how the focal length determines the normals from the frustum planes, as well as the aspect ratio's role in the frustum plane normals.

What I don't understand completely, is that, both e(focal length) and the aspect ratio seem to have no contribution to the perspective projection matrix. Is the actual projection a final step after transforming to clip-space (and doing the clipping)? Is OpenGL doing something extra behind the scenes? (recall, I'm working on a software renderer) More over, what is the view-plane's relation to the view-frustum?

I hope my question makes sense, if not I will be happy to rephrase or elaborate. I appreciate and knowledge you can share.


Book Preview:

[url="http://books.google.com/books?id=bfcLeqRUsm8C&pg=PA115&lpg=PA115&dq=frustum+plane+normal&source=bl&ots=FqTvf6tUfB&sig=habQb5--rTggjW1ilsbDvv-4eD8&hl=en&sa=X&ei=5nxjUO_fO-XW2AX2zIDAAg&ved=0CEYQ6AEwBDgK#v=onepage&q=frustum%20plane%20normal&f=false"]http://books.google.... normal&f=false[/url] Edited by RasterGuy

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The focal length and aspect ratio are determined from the parameters used in the equation in the book. For example, the aspect ration is [i](r-l)/(t-b)[/i] and the focal length (as you appear to be defining it) is [i]n/(t-b)[/i].

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[quote name='Brother Bob' timestamp='1348700869' post='4984154']
The focal length and aspect ratio are determined from the parameters used in the equation in the book. For example, the aspect ration is [i](r-l)/(t-b)[/i] and the focal length (as you appear to be defining it) is [i]n/(t-b)[/i].
[/quote]

Thanks Brother Bob,

I guess I need to go over it some more. It seems like, in the book, the near plane is actually replacing the view/projection plane. Is that right? I apologize if this is obvious, but I seem to stumbling on it for one reason or another.

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There really is no [i]the[/i] projection plane, but [i]a[/i] projection plane. Within the view volume, you can slide the projection plane as you like along the depth axis without changing what is projected onto the plane relative the size of the view volume at the depth where the projection plane is positioned. You can think of the near plane as the projection plane, but you can also think of the far plane as the projection plane, or almost [i]any[/i] plane along the depth axis whether it's inside or outside the actual view volume.

If you want to think of a concrete projection plane to visualize how everything works, the near plane is probably the easiest plane to think of as the projection plane.

Keep in mind here that talking about an absolute focal length is not really relevant as it says nothing about how the view volume looks. The focal length only makes sense when you also have the size of a projection plane. In that case, the relevant number is the ratio between the focal length and the size of the projection plane. For a camera analogy, a 50mm focal length lens says nothing about what the picture from it will look like (I'm talking about perspective effects only), unless you also say that you either have, say, a 24mm or 35mm film/sensor. The ratio between the two is related to the tangent of the field of view.

If you look at the equation I mentioned about the focal length, you will see that it is a normalized focal length. The value [i]n/(b-t)[/i] is the ratio of the distance to the near plane and the size of the near clip plane (in the vertical direction in this case). This is the general case where you can define the size of the near clip plane and the distance of it from the view point. In your case, you have an implicit size of the projection plane equal to 2 (from -1 to +1), and your parameter [i]e[/i] is in relation to this size.

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Thanks Bob,

I think that is what I was trying to come to. That helps a lot, starting to see the forest now, I think.

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I'm pretty sure you are missing only one thing, because it is an obscure one...
everything in the projected vector is divided by w afterward by the rasterizer.

so, the projection matrix actually doesn't do anything really, apart from copying z into w. then the real "perspective division", like we call it, is done in a hidden step afterward.
you can do it yourself if you are coding a projection by hand on CPU (or even in shader) by doing Vproj = Vworld * WorldViewProjMatrix; Vproj /= Vproj.w;

there you go.

it is really easy to understand actually : the farther the points, the closer to the center of the view. that is how one creates parallax effects. (think of scolling planes with different speeds according to distances, or also, think of convergence lines in a drawing of a cube. further points are closer together.

then the focal enters in the equation as a multiplier for the perspective effect. the larger the FOV the stronger the "gather to the center division effect" will be. that is all :)

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Thank you Lightness,

That definitely makes sense and kind of ties everything together. I really appreciate your input =)

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