Is it a vector problem?

the dot product is the way to go with vectors. the dot product returns the cosine of the vectors, so in your case:

u = B-A
v = C-A

normalize(u)
normalize(v)

dot = u*v
acos(dot) = angle

where * is the dot product

[edited by - sphinx23 on December 31, 2003 7:04:03 AM]

I'd say you need to research a bit more on vector and trigonometry.

in 2D, if you have two directional vectors u and v, and you want the angle(u, v)

cosine(angle(u, v)) = u . v / (|u| * |v|)
sine (angle(u, v)) = u x v / (|u| * |v|)

and

u . v = u.x*v.x + u.y*v.y
u x v = u.x*v.y - u.y*v.x

and to find angle(u, v)

angle(u, v) = arctangent(sine(angle(u, v)) / cosine(angle(u, v)))
angle(u, v) = arctangent((u x v) / (u . v))

I do specify in 2D, as in 3D, the cross product is slightly different.


with atan() (arctangent in C/C++), and acos() and asin(), you have to be careful of the signs of (u x v) and (u . v). The range of the angle returned will be between [0, PI], or [-PI/2, PI/2] (or if you prefer between [0, 180 degrees], or [-90, 90] degrees).

in C and C++, you have the function atan2(y, x), which takes care of the signs of y anx and return the angle between u and v in the range [-PI, PI] ([-180, 180 degrees]).

so, in short, in C

float Dot  (Vector2 u, Vector2 v) 
{ 
    return u.x*v.x + u.y*v.y; 
}
float Cross(Vector2 u, Vector2 v) 
{ 
    return u.x*v.y - u.y*v.x; 
}
Vector2 Sub(Vector2 u, Vector2 v) 
{ 
    return Vector(u.x-v.x, u.y-v.y); 
}

float RadiansToDegrees(float rad) 
{ 
    return rad * 57.29578f; 
}

float AngleInDegrees(Vector2 A, Vector2 B, Vector2 C)
{
    Vector2 u = Sub(B, A);
    Vector2 v = Sub(C, A);

    float dot   = Dot  (u, v);
    float cross = Cross(u, v);
    
    float radians = atan2(cross, dot);
    
    return RadiansToDegrees(radians);
}

[edited by - oliii on December 31, 2003 7:57:06 AM]

I've tried your way sphinx and it works like a charm!
Oliii, is there any advantage to your way? The way I've done it does seem a litlle messier, but will I get the same results?

[edited by - desertcube on December 31, 2003 8:00:45 AM]

His method expresses the angle in terms of an arctangent so he can break tangent down into sine and cosine, which both have special vector operations. It''s the same thing really, if you understand the theory.

the acos() only returns an angle in the range [0, Pi], which maybe not enough (depends what you want). And I don't normalise, which is faster, but that's a moot point.

and also, it will fail if u . v > 1.0f (like u . v = 1.0000000001f). But that's also debatable.

[edited by - oliii on January 1, 2004 6:12:47 PM]

quote:

---

Original post by oliii
the acos() only returns an angle in the range [0, Pi], which maybe not enough (depends what you want). And I don''t normalise, which is faster, but that''s a moot point.

and also, it will fail if u . v > 1.0f (like u . v = 1.0000000001f). But that''s also debatable.

[edited by - oliii on January 1, 2004 6:12:47 PM]

---

You''re not normalizing but you are dividing the by the product of the magnitude which is almost the same thing. Timing wise it doesn''t seem like either is faster.

Thanks for all your replies, I''ve got the problem fixed. It''s not really a time critical part of the program, so it doesn''t matter which one is faster, but I''ve used the angle(u, v) = arctangent((u x v) / (u . v)) way as it gives me angles over 90 degrees.
Thanks again.