Optimization help

[ThePuska]

I want to compare a direction given as (yaw, pitch), from the point of view of object A (vector From) to the line that goes from object A to object B. I came up with a few ways to do this and below is one of them. I only use the return value for one comparison, to determine whether object A is looking at object B.

But as you can see, it's really costly. Any ideas how to optimize it, or any other approaches to the problem?

function DirectionDifference(From, To: Vector; Pitch, Yaw: Single): Single;
Local
	X:    Vector;
	XDir: Vector;
	Dir:  Vector;
	D:    Single;
{

	X = To - From;
	D = sqrt( X.x*X.x + X.y*X.y + X.z*X.z );
	XDir.x = X.x/D;
	XDir.y = X.y/D;
	XDir.z = X.z/D;

	Dir.x = sin(Yaw);
	Dir.y = cos(Yaw);
	Dir.z = sin(Pitch)

	X = XDir - Dir;

	D = X.x*X.x + X.y*X.y + X.z*X.z;

	return D;

}

Functionality of the algorithm isn't confirmed but you should get the general idea. Using unit spheres, I'm making the aiming vectors (XDir being the line from A-B and Dir being the looking direction of A) equally long and then calculate the distance between the points on the sphere. The return value is 2 if there's a 90 degree difference and 4 if there's a 180 degree difference between the angles.

[Cathbadh]

Yes. It's called the dot-product.

The dot product is equal to the sum of the products of respective components of 2 vectors:

dot(A, B) = A.x*B.x + A.y*B.y + A.z*B.z

But geometrically, it is also equal to the product of the magnitudes and the cosine of the angle between them.

dot(A, B) = |A| * |B| * cos(angle)

You can find the dot product easy. You are trying to get the angle, so solve for that.

angle = acos(dot(A, B) / (|A| * |B|))

You can optimize this further if you can assume that either or both of the vectors are of unit length.

[editline]03:01AM[/editline]

You need to fix this bit of code:

Dir.x = sin(Yaw);
Dir.y = cos(Yaw);
Dir.z = sin(Pitch)

In order for Dir to be a representative unit vector, it needs to be:

Dir.x = cos(Pitch)cos(Yaw);
Dir.y = cos(Pitch)sin(Yaw);
Dir.z = sin(Pitch);

[ThePuska]

[QUOTE=Cathbadh;17174144]Yes. It's called the dot-product.

The dot product is equal to the sum of the products of respective components of 2 vectors:

dot(A, B) = A.x*B.x + A.y*B.y + A.z*B.z

But geometrically, it is also equal to the product of the magnitudes and the cosine of the angle between them.

dot(A, B) = |A| * |B| * cos(angle)

You can find the dot product easy. You are trying to get the angle, so solve for that.

angle = acos(dot(A, B) / (|A| * |B|))

You can optimize this further if you can assume that either or both of the vectors are of unit length.

[editline]03:01AM[/editline]

You need to fix this bit of code:

Dir.x = sin(Yaw);
Dir.y = cos(Yaw);
Dir.z = sin(Pitch)

In order for Dir to be a representative unit vector, it needs to be:

Dir.x = cos(Pitch)cos(Yaw);
Dir.y = cos(Pitch)sin(Yaw);
Dir.z = sin(Pitch);

[/QUOTE]

Thanks. I'm supposed to know this stuff already, but I don't know why I didn't think of the scalar product.