0
$\begingroup$

I'm reading the chapter 8 of the Cg tutorial and I could not understand 8.4.1 :

Because all these coordinates lie in the plane of the same triangle, it is possible to derive plane equations for x, y, and z in terms of s and t:

A_0 x + B_0 s + C_0 t + D_0 = 0

A_1 y + B_1 s + C_1 t + D_1 = 0

A_2 z + B_2 s + C_2 t + D_2 = 0

How are these equations derived?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

In this context, deriving the plane equations does not refer to the equations themselves, but to the coefficients $A$, $B$, $C$ and $D$.

The equations are simply plane equations for a plane defined by three points, where the first plane is defined by the three points $(x_i,s_i,t_i)_{i=1..3}$, the second plane is defined by $(y_i,s_i,t_i)_{i=1..3}$ and the third plane by $(z_i,s_i,t_i)_{i=1..3}$. They can be seen as auxiliary objects in deriving equations for $x$, $y$ and $z$ in terms of $s$ and $t$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I still could not figure out why choosing plane equations, not other equations? $\endgroup$
    – Cu2S
    Commented Oct 31, 2018 at 5:26
  • $\begingroup$ Because you want to express $x$ (or $y$ or $z$) in terms of $s$ and $t$ in the simplest possible way. Intuitively you would expect this to be a linear combination of $s$ and $t$, which is precisely what you get when you rearrange the plane equation. This works if the triangle is non-degenerate in $(s,t)$-space, because then the points $(x_i,s_i,t_i)_{i=1..3}$ that determine the plane are non-collinear. $\endgroup$
    – user9485
    Commented Oct 31, 2018 at 11:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.