Normalizing vector direction.

[Shockbolt]

Currently i'm working on a stabilizer for every direction, similar to the hoverballs Air Resistance.

So, right now i'm trying to let the computer calculate wich thruster gets wich power input.

I got to normalizing direction vectors, and found out:
Forward: X = 1, Y and Z are 0.
90 degrees left: X and Z are 0, Y = 1.
so logically, X and Y should be 0.5 at 45 degrees left or right.
But when i normalize the direction vector it gives 0.707 at that angle.

How exactly does this work?

[Syranide]

length is sqrt(X^2 + Y^2 + Z^2), and normalization is dividing the vector by its length.

[Shockbolt]

length is sqrt(X^2 + Y^2 + Z^2), and normalization is dividing the vector by its length.[/b]

No that wasn't the question, i understand how normalizing itself works, but i want to know why the normalized X vector is 0,707 at a 45 degrees angle, and not 0.5

0.5 would be logical if its 0 at 90 degrees, and 1 at 0 degrees.

[Black Phoenix]

[attachment=1299mgnoes.PNG]

[Syranide]

or simply sqrt(1^2 + 1^2 + 0^2) = sqrt(2) = 1.414 ... 1.414 / 2 = 0.707

[cadde]

Awesome replies from the devs as usual
Most people dont understand trig.

[Black Phoenix]

or simply sqrt(1^2 + 1^2 + 0^2) = sqrt(2) = 1.414 ... 1.414 / 2 = 0.707[/b]

Some people will not understand things that that complex