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zuraneur
12-11-2014, 12:03 PM
Good Evening everybody.

I am currently re encoding opengl 1.5 and a problem appears with the light process.

On the specification there is a blurry part :

if P 1 and P 2 are (homogeneous, with four coordinates) points then let P1 P 2 be the unit vector that points from P1 to P2 .

Note that if P2 has a zero w coordinate and P1 has non-zero w coordinate, then P1 P2 is the unit vector corresponding to the direction specified by the x, y, and z coordinates of P2 ; if P1 has a zero w coordinate and P2 has a non-zero w coordinate then P1 P2 is the unit vector that is the negative of that corresponding to the direction specified by P1 . If both P1 and P2 have zero w coordinates, then P1 P2 is the unit vector obtained by normalizing the direction corresponding to P2 − P1 .

I guess that if a something has 0 w coordinate, it is a direction, otherwise it is a position (Isn't it ? there is very important)
Does it mean that if i have two vertex , one for my triangle position P1(2, 4, 6, 1) and another for the light position P2(1, 5, 9, 0),
if i have to compute the vertex P1P2, then i will only considere that P1P2 is P1P2(1, 5, 9, 0) and i will have to normalize it ?

Is the mindset to consider that something that is not a direction is a position , and above all that :
2 positions stay a position, 2 direction stay a direction , a direction and a position is a direction ?


I expect that my questions were clear, but it is really important to me to understand it.

Thank for your time passed to my post.

zuraneur.

Aleksandar
12-11-2014, 02:57 PM
if P 1 and P 2 are (homogeneous, with four coordinates) points then let P1 P 2 be the unit vector that points from P1 to P2 .
[\QUOTE]
P2-P1 is a vector from point P1 to point P2.

[QUOTE=zuraneur;1262944]Note that if P2 has a zero w coordinate...[\QUOTE]
Then P2 is not a point, but a vector (a direction)!

[QUOTE=zuraneur;1262944]I guess that if a something has 0 w coordinate, it is a direction, otherwise it is a position (Isn't it ? there is very important)
Yes it is.


Does it mean that if i have two vertex , one for my triangle position P1(2, 4, 6, 1) and another for the light position P2(1, 5, 9, 0),
if i have to compute the vertex P1P2, then i will only considere that P1P2 is P1P2(1, 5, 9, 0) and i will have to normalize it ?
I don't understand what you want to achieve? P2 is not a vertex. Formally, P1P2=P2-P2=(1, 5, 9, 0)-(2, 4, 6, 1)=(-1, 1, 3, -1). In Cartesian coordinate system, that point has coordinates (1,-1,-3).


Is the mindset to consider that something that is not a direction is a position , and above all that :
2 positions stay a position, 2 direction stay a direction , a direction and a position is a direction ?
As you can see from the previous calculation, a position and direction give a position.

zuraneur
12-11-2014, 03:29 PM
Hello Aleksandar and thank you for your answer.


[QUOTE=zuraneur;1262944]
if P 1 and P 2 are (homogeneous, with four coordinates) points then let P1 P 2 be the unit vector that points from P1 to P2 .
[\QUOTE]
P2-P1 is a vector from point P1 to point P2.

I don't understand what you want to achieve? P2 is not a vertex. Formally, P1P2=P2-P2=(1, 5, 9, 0)-(2, 4, 6, 1)=(-1, 1, 3, -1). In Cartesian coordinate system, that point has coordinates (1,-1,-3).



I'm trying to translate what the specification tell me to do when I have to compute a vector such as P1P2 on lighting formula.
For example, my highlight point position is P1 (1, 2, 3, 1) and my light has a parameter "position" with four coordinates, so it involve that his w coordinate will always be 1 because that's a position parameters.

But users can change fours of them, so my light "position" could become a direction.

That's why opengl specification told me how to react if i have to compute a position and a direction.
That's what I don't understand.
According to specification :

Note that if P2 has a zero w coordinate and P1 has non-zero w coordinate, then P1 P2 is the unit vector corresponding to the direction specified by the x, y, and z coordinates of P2

Does it means that if i have to build a vector with a position and a direction, then i will only build my result with direction coordinates ?
It is what the previous sentence means to me, but i must be wrong.

That's why I told you with position P1(1, 2, 3, 1), and direction P2(4, 5, 6, 0), If i build P1P2, i will just have P1P2(4, 5, 6, 0).

I expect you could enlighten me.

GClements
12-11-2014, 04:51 PM
Does it means that if i have to build a vector with a position and a direction, then i will only build my result with direction coordinates ?


For homogeneous vectors P1 and P2, the direction from P1 to P2 is defined as


D = P2/P2.w - P1/P1.w


In the case where a vector has a zero w component, the division cannot be performed. However, we can take account of the fact that scaling all four components of a homogeneous vector by the same factor doesn't change the Euclidean vector which it represents, and convert the above to:


D' = P1.w*P2.w*D = P1.w*P2 - P2.w*P1

Now, we no longer need to perform division, only multiplication. If either vector's w component is zero, that vector disappears from the equation, and we're left with either P1 (if P2.w is zero) or -P2 (if P1.w is zero).

So, what the specification is saying is that the direction is defined as:


// calculate direction from P1 to P2, i.e. P2-P1
vec4 direction(vec4 P1, vec4 P2)
{
if (P1.w == 0 && P2.w == 0)
return P2 - P1;
else if (P2.w == 0)
return P2;
else if (P1.w == 0)
return -P1;
else
return P2/P2.w - P1/P1.w;
}

Note that in all four cases, the result will have w equal to zero (in the last case, both operands to the subtraction will have w equal to 1).

zuraneur
12-13-2014, 01:01 PM
Thanks a lot for these answers, they really make me aware that I have to study what projection geometry really are, then I can continue to code to implement light.

Gentlemen, have a nice day.