Jump to content

  • Log In with Google      Sign In   
  • Create Account

Interested in a FREE copy of HTML5 game maker Construct 2?

We'll be giving away three Personal Edition licences in next Tuesday's GDNet Direct email newsletter!

Sign up from the right-hand sidebar on our homepage and read Tuesday's newsletter for details!


We're also offering banner ads on our site from just $5! 1. Details HERE. 2. GDNet+ Subscriptions HERE. 3. Ad upload HERE.


Deriving projection matrix


Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.

  • You cannot reply to this topic
3 replies to this topic

#1 nikitablack   Members   -  Reputation: 581

Like
0Likes
Like

Posted 05 May 2011 - 12:50 AM

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 x and y is not hard, but there're problems with finding z coordinate. I have l, r, t, b (left, right, top, bottom planes - they are not necessary for z) and n, f (near, far planes). How can I bound new projected z with real z in space?

Sponsor:

#2 nikitablack   Members   -  Reputation: 581

Like
0Likes
Like

Posted 06 May 2011 - 01:36 AM

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 w 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?

Attached Thumbnails

  • ndc.jpg


#3 Brother Bob   Moderators   -  Reputation: 8429

Like
2Likes
Like

Posted 06 May 2011 - 02:22 AM

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.

#4 nikitablack   Members   -  Reputation: 581

Like
0Likes
Like

Posted 06 May 2011 - 03:06 AM

Oh, I see now. Thanks for good explanation Brother Bob.




Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.



PARTNERS