Calculate direction angle from two vectors?

Question

Say I have two 2D vectors, one for an objects current position and one for that objects previous position. How can I work out the angular direction of travel?

This image might help understand what I'm after:

(image) http://files.me.com/james.ingham/crcvmy

Answer 1

The direction vector of travel will be the difference of the two position vectors,

d = (x1, y1) - (x, y) = (x1 - x, y1 - y)

Now when you ask for the direction angle, that depends what direction you want to measure the angle against. Is it against the x axis? Go with Radu's answer. Against an arbitrary vector? See justjeff's answer.

Edit: To get the angle against the y-axis:

tan (theta) = (x1 -x)/(y1 - y)          

the tangent of the angle is the ratio of the x-coordinate of the difference vector to the y-coordinate of the difference vector.

So

theta = arctan[(x1 - x)/(y1 - y)]

Where arctan means inverse tangent. Not to be confused with the reciprocal of the tangent, which many people do, since they're both frequently denoted tan^-1. And make sure you know whether you're working in degrees or radians.

Answer 2

If you're in C (or other language that uses the same function set) then you're probably looking for the atan2() function. From your diagram:

double theta = atan2(x1-x, y1-y);

That angle will be from the vertical axis, as you marked, and will be measured in radians (God's own angle unit).

Answer 3

Be careful to use atan2 to avoid quadrant issues and division by zero. That's what it's there for.

float getAngle(CGPoint ptA, CGPoint ptOrigin, CGPoint ptB)
{
    CGPoint A = makeVec(ptOrigin, ptA);
    CGPoint B = makeVec(ptOrigin, ptB);

    // angle with +ve x-axis, in the range (−π, π]
    float thetaA = atan2(A.x, A.y);  
    float thetaB = atan2(B.x, B.y);

    float thetaAB = thetaB - thetaA;

    // get in range (−π, π]
    while (thetaAB <= - M_PI)
        thetaAB += 2 * M_PI;

    while (thetaAB > M_PI)
        thetaAB -= 2 * M_PI;

    return thetaAB;
}

However, if you don't care about whether it's a +ve or -ve angle, just use the dot product rule (less CPU load):

float dotProduct(CGPoint p1, CGPoint p2) { return p1.x * p2.x + p1.y * p2.y; }

float getAngle(CGPoint A, CGPoint O, CGPoint B)
{
    CGPoint U = makeVec(O, A);
    CGPoint V = makeVec(O, B);

    float magU = vecGetMag(U);
    float magV = vecGetMag(V);
    float magUmagV = magU * magV;   assert (ABS(magUmagV) > 0.00001);

    // U.V = |U| |V| cos t
    float cosT = dotProduct(U, V) / magUmagV;
    float theta = acos(cosT);
    return theta;
}

Note that in either code section above, if one ( or both ) vectors are close to 0 length this is going to fail. So you might want to trap that somehow.

Answer 4

Still not sure what you mean by rotation matrices, but this is a simple case of getting an azimuth from a direction vector.

The complicated answer:

Normally you should pack a few conversion/utility functions with your 2D vectors: one to convert from X,Y (carthesian) to Theta,R (polar coordinates). You should also support basic vector operations like addition, substraction and dot product. Your answer in this case would be:

 double azimuth  =  (P2 - P1).ToPolarCoordinate().Azimuth;

Where ToPolarCoordinate() and ToCarhtesianCoordinate() are two reciprocal functions switching from one type of vector to another.

The simple one:

 double azimuth = acos ((x2-x1)/sqrt((x2-x1) * (x2-x1) + (y2-y1) * (y2-y1));
 //then do a quadrant resolution based on the +/- sign of (y2-y1) and (x2-x1)
 if (x2-x1)>0 {
   if (y2-y1)<0 {  azimuth = Pi-azimuth; } //quadrant 2
 } else 
 { if (y2-y1)> 0 {  azimuth = 2*Pi-azimuth;} //quadrant 4
    else  { azimuth = Pi + azimuth;} //quadrant 3
 }