透视投影-OpenGL

Question

I am confused about the position of objects in opengl .The eye position is 0,0,0 , the projection plane is at z = -1 . At this point , will the objects be in between the eye position and and the plane (Z =(0 to -1)) ? or its behind the projection plane ? and also if there is any particular reason for being so?

Answer 1

First of all, there is no eye in modern OpenGL. There is also no camera. There is no projection plane. You define these concepts by yourself; the graphics library does not give them to you. It is your job to transform your object from your coordinate system into clip space in your vertex shader.

I think you are thinking about projection wrong. Projection doesn't move the objects in the same sense that a translation or rotation matrix might. If you take a look at the link above, you can see that in order to render a perspective projection, you calculate the x and y components of the projected coordinate with R = V(ez/pz), where ez is the depth of the projection plane, pz is the depth of the object, V is the coordinate vector, and R is the projection. Almost always you will use ez=1, which makes that equation into R = V/pz, allowing you to place pz in the w coordinate allowing OpenGL to do the "perspective divide" for you. Assuming you have your eye and plane in the correct places, projecting a coordinate is almost as simple as dividing by its z coordinate. Your objects can be anywhere in 3D space (even behind the eye), and you can project them onto your plane so long as you don't divide by zero or invalidate your z coordinate that you use for depth testing.

Answer 2

There is no "projection plane" at z=-1. I don't know where you got this from. The classic GL perspective matrix assumes an eye space where the camera is located at origin and looking into -z direction.

However, there is the near plane at z<0 and eveything in front of the near plane is going to be clipped. You cannot put the near plane at z=0, because then, you would end up with a division by zero when trying to project points on that plane. So there is one reasin that the viewing volume isn't a pyramid with they eye point at the top but a pyramid frustum .

This is btw. also true for real-world eyes or cameras. The projection center lies behind the lense, so no object can get infinitely close to the optical center in either case.

The other reason why you want a big near clipping distance is the precision of the depth buffer. The whole depth range between the front and the near plane has to be mapped to some depth value with a limited amount of bits, typically 24. So you want to keep the far plane as close as possible, and shift away the near plane as far as possible. The non-linear mapping of the screen-space z coordinate makes this even more important, as that the precision is non-uniformely distributed over that range.