## Bush had to face Kerry for the first time, face to face

### Bush had to face Kerry for the first time, face to face

That's interesting. After all those very distant comments of evil.
Of course Bush is still going to win, but when he was there,
seeing Kerry's eyes, real eyes, he did not seem so big after all.

He is the US's most unstable president ever, and its quite impressive
what he has accomplished, considering he was an *** ic for a
decade. He is a rolemodel of sorts, not mine of course.

I asked my students to write a note on how OpenGL determines whether
front- and back-facing. Essentially, Iwas inviting them to elaborate
on the following from my notes -- in turn copied from the Red Book:

"
Front- or back-facing depends on the sign of the polygon's area
computed in window coordinates. We can compute this area as:

a = \frac{1}{2} \sum_{i=0}^{n-1} x_i y_{i \oplus 1} - x_{i \oplus 1}
y_{i},

where $i \oplus 1 = (i+1) \mod n$ and where $x_i$ and $y_i$ are the
\emph{window} coordinates of the $i$th vertex of the $n$ vertex
polygon. $i \oplus 1 = (i+1) \mod n$ simply wraps around back to zero
at $i=n$, i.e. there is no $i=n$, so we go back to the first (zeroth
index) point.

If \verb|GL_CCW| has been specified, if $a>0$, the polygon is
considered to be front-facing; otherwise, it is back-facing.
"

Mostly, they gave answers based on the sign of the scalar product of
the viewing point and the normal to the face. Which seems reasonable;
unless OpenGL actually uses the former, and the former is easier to
compute.

Any suggestions?

TIA,

Jon C.