Got minimum angle between 2 vectors, but need direction as well...

[Alluvian]

I know I should be able to find this in google, but my google fu is failing, HARD.

So, the magic of dotproduct can get an angle between two vectors really easily (using points here just for simplicity), I managed to find that much on google. I don't think I need to bother with a full vector class, I am only using this stuff sparingly right now.

p2pang = function (p1:Point, p2:Point, inradians) {
    if (inradians==undefined){inradians = false;}
    var cosA:Number = p2pdot(p1, p2) / (p1.length * p2.length);
    var A : Number = Math.acos( cosA );
    if ( inradians == false ) {A = A / Math.PI * 180;}
    return A;
}
p2pdot = function(p1:Point, p2:Point){
    return ( p1.x * p2.x + p1.y * p2.y );
}

So great, works nice, no awkward problems with -180/180 issues, but I want to be able to know which vector is to which side of which. Basically, what rotation do I have to apply to p1 to get to p2. I need the +/- for what I am doing. The above just returns an always positive value.

Is there an easy way to add this to the function above?

[dandylion13]

I'm not sure I really understand your question but...

... what's the value that it's returning?
If it's returning from 0 to 360, couldn't you just subtract 180 to give you to the -180 to 180 range?

Sorry if I didn't understand your question

[Alluvian]

It is returning 0 to 180.

If p1 is 1,0 and p2 is 0,-1 it returns 90. That is the angle between the vectors.
If p1 is 0,-1 and p2 is 1,0 it also returns 90. That is still the angle between the vectors.

But I want the first case to return -90 and the second to return 90. Basically, what is the shortest rotation around the origin needed to make to p1 so it has the same angle as p2.

[dandylion13]

Could you use this formula to help you find the angle of each vector?

vectorAngle = Math.atan2(vy, vx) * 180/Math.PI

(you will need to know the vx and vy values).

[tonypa]

To know which side of p1 is p2 you use dot product of p1 normal (normal is perpendicular or 90 degrees from original vector) to p2. Right hand normal of p1:

p1r.x = -p1.y;
p1r.y = p1.x;

If p1r dot product with p2 is positive (> 0), it means p2 is on the right hand side of p1 and you need to turn p1 clockwise to reach p2. If the dot product is negative, turn anti-clockwise.

[Alluvian]

I thought the normal vector of a vector was the unit length 1 vector with the same angle?

Anyway, the dot product of the perpendicular worked awesome, thank you very much!

For anyone's information, the final function with sign is:

function p2pang (p1:Point, p2:Point, inradians) : Number {
    if (inradians==undefined){inradians = false;}
    var cosA:Number = p2pdot(p1, p2) / (p1.length * p2.length);
    var A : Number = Math.acos( cosA );
    p1r = new Point(-p1.y, p1.x);
    if (p2pdot(p1r, p2) > 0) {sign=1;} else {sign=-1;}
    if ( inradians == false ) {A = A / Math.PI * 180;}
    return A * sign;
}

[dandylion13]

I thought the normal vector of a vector was the unit length 1 vector with the same angle?

No, that's a "normalized" vector. A vector's "normal" is a vector that lies perpendicular to it (at a 90 degree angle).

... don't worry, it's confusing!

[Alluvian]

I get it, just forgot the term for the perp vector. Thanks.