Timeline for How to derive a perspective projection matrix from its components?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Aug 2, 2019 at 8:17 | comment | added | Tare | Well spotted. I fixed it | |
| Aug 2, 2019 at 8:17 | history | edited | Tare | CC BY-SA 4.0 |
fixed a typo in a formula
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| Aug 2, 2019 at 3:20 | comment | added | Meowmere |
\frac{p_x}{d} = \frac{v_x}{v_x} \implies p_x = \frac{v_x\cdot d}{v_z} = \frac{v_x}{v_x \cdot \tan(\frac{\alpha}{2})}. Probably you mean \frac{v_x}{v_z}?
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| Sep 1, 2018 at 14:56 | comment | added | Doug Voss | I did a more in depth derivation of the perspective projection matrix here: dovo329.github.io/DeriveOpenGLPerspectiveProjectionMatrix | |
| Feb 15, 2018 at 17:22 | vote | accept | Manh Nguyen Huu | ||
| Feb 15, 2018 at 8:04 | comment | added | Tare | Although I guess it's possible, there is no matrix multiplication that I know of that does this. I also don't see the reason why you would want to add another multiplication, that achieves nothing else than the single matrix but adds an overhead of calculations? | |
| Feb 14, 2018 at 15:32 | comment | added | Manh Nguyen Huu | Thank you very much, I understand more thoroughly now. 1 question though, so the third row, mapping [-1, 1], aren't there any matrices multiplication that can get to that row? What I mean is a combination of scaling, translating, shearing, etc | |
| Feb 14, 2018 at 10:24 | comment | added | Tare | I have added the complete derivation of a simpler version, plus the explanation on the differences to the matrix you wanted to achieve. hope that helps. | |
| Feb 14, 2018 at 10:23 | history | edited | Tare | CC BY-SA 3.0 |
Added the complete derivation of the projection matrix
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| Feb 13, 2018 at 22:17 | comment | added | Manh Nguyen Huu | I'm not sure how can I derive this projection matrix. Can you explain it more clearly? | |
| Feb 13, 2018 at 7:57 | history | answered | Tare | CC BY-SA 3.0 |