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Volgogradetzzz

Deriving projection matrix

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Hello. I want to understand how projection works, so I want to derive projection matrix. Also I want it to be in normalized form. Finding [b]x[/b] and [b]y[/b] is not hard, but there're problems with finding [b]z[/b] coordinate. I have l, r, t, b (left, right, top, bottom planes - they are not necessary for [b]z[/b]) and n, f (near, far planes). How can I bound new projected [b]z[/b] with real [b]z[/b] in space?
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Hello again and sorry for my persistence. But I'm really confused. Below is a picture with all parameters for my projection matrix. After multiplying matrix by vector and dividing by vector [b]w[/b] I receive [0, 0.5, 0.75. 1] normalized vectors coordinates. x = 0 and y = 0.5 - it's ok - and it can be seen from picture. But z = 0.75 looks strange for me - from picture it should be 0.5! Is it correct that z = 0.75? And if it is - why?
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Your Z-range is not linear, so a vertex halfway between the near and far plane will not have a normalized Z=0.5. The greater the ratio between the near and the far plane distances, the more prominent is the non-linearity.

As a rule of thumb, your projected Z is 0.5 if the eye Z is two times the distance of the near clip plane. That is, if near=1, and far=100, then eye Z=2 yields a projected Z=0.5. So half your Z-range is cramped into the first 1 unit, while the other half is spread out over the last 98 units. This is why depth buffer precision deteriorates rapidly over distance if your near plane is too close. In my example, half the depth buffer resolution is allocated for eye Z=1 to 2, and the other half of the resolution from eye Z=2 to 100.

Note that this rule of thumb is fairly accurate as the near to far plane ratio above at least 10, so you cannot really apply it in your case since they are too close. But Z=0.75 is correct in your case at least.
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