Vector from Angle

Of course. You don''t even actually need the angle, but simple the point (since that point essentially contains the angle within it). In fact, that point *is* the vector itself, starting from the origin. If you then want a unit vector, just normalize that.

The Artist Formerly Known as CmndrM

http://chaos.webhop.org

What if that point is at the origin?

That's it. Thank you so much. Actually I figured it out but I just wasn't sure why whatever I was doing was working, thanks for laying out the mathematical foundation.

[edited by - kelaklub on August 29, 2003 1:13:53 PM]

Think about the parametric equation of a circle...(looking DOWN the Y acis)

x = a + (r * sin(theta) )
z = b + (r * cos(theta) )

where x and z are your vector points (in 2D) and a and b are the position of the point, or 0 and 0 if at the origin and r is the radius of the circle.

you have to picture 0 degrees as "North" or UP, or maybe (0, 1) as a 2D vector in the XZ plane. Then, imagine rotating a vector by hand, or drawing it on paper, about the origin by whatever angle. The new position of the end of the vector is the x and z value computed above.

This example rotates around the Y axis....it can be used in exactly the same way with a little alteration for rotation about all 3 axis....everything is just done in 2D steps.

Hope that helps...and hope I havn''t made a stoopid mistake! ;o)

see how you get on,

GCoder

The AP is correct. I mispoke, as there really is no "starting point" per se of a vector. However, there *is* one using the geometric interpretation of a vector, and that is what I meant by that.

The Artist Formerly Known as CmndrM

http://chaos.webhop.org

isn''t it

x = cosf(a);
y = -sinf(a);

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No. It''s not. Using the standard use of the angle theta (measured counterclockwise), y = sin theta, not -sin.

The Artist Formerly Known as CmndrM

http://chaos.webhop.org

d3dx uses clockwise rotation

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