vector math question. Help pls!!

Newbie & Debugging Questions
#1

My question is I have two objects. ObjA at posA and ObjB at posB. ObjA has a normal(facing) and want to know how much deg to turn in order to face ObjB.

vectorBtoA = posA - posB
arccos(dotproduct(normalize(ObjA’s facing)), normalize(vectorBtoA)))

what I get is a degree but don’t have a sign to indicate which way to turn. Any suggestion or better approach?
Thank you very much.

#2 
Vector ab = objectB.getPosition().minus(objectA.getPosition());

ab.normalize();

Vector normalA = objectA.getDirection();

normalA.normalize();

double theta = Math.arccos(normalA.dot(ab));

// I've got no idea what this operation is called;
// it's kinda like a cross product... but gives a scalar, not a vector

double cross = ab.x * normalA.y - normalA.x * ab.y;

if(cross<0) theta = -theta;
#3

[quote]// I’ve got no idea what this operation is called;
// it’s kinda like a cross product… but gives a scalar, not a vector

double cross = ab.x * normalA.y - normalA.x * ab.y;

if(cross<0) theta = -theta;
[/quote]
Abuse: It’s still the cross product :wink:
You are generating just the z component of the cross product from the x and y. You can tell whether the y is to the right/left of the x by the sign.

Hata: Do you really need the angle & the sign?
If you just wish to orient the object A to face object B, then the best way is to build the orientation matrix directly, otherwise you will be doing a lot of normalising & trig to get the same result.

For this code, I will assume X is the axis for the ‘facing’ vector, and Y is up:

(Pseudocode)



// Build the true x axis:
vectorX = posA-posB;
vectorX.normalize();

// Now build a true z axis:
vectorY.set(0.0f, 1.0f, 0.0f);
vectorZ.cross( vectorX, vectorY );
vectorZ.normalise();

// Now build the true Y axis:
vectorY.cross( vectorZ, vectorX);

The orientation matrix is actually these three vectors packed into the matrix, either as rows (vectorX, VectorY, VectorZ) or as columns depending on your coordinate scheme.

  • Dom
#4

i think Abuse’s answer is quite smart and it certainly solves this 2d problem :smiley:

thank you Crystalsquid also, i think what u r talking about is orthogonal martix, rite? the reason i asked this question is becos i do want to find out the degree and the sign.

Thank you all.

#5

Ah, ok.

You can also do this in one step, as the size of the cross product result is the sin of the angle:


Vector ab = objectB.getPosition().minus(objectA.getPosition()); 
 
ab.normalize(); 
 
Vector normalA = objectA.getDirection(); 
 
normalA.normalize();  
 
double cross = ab.x * normalA.y - normalA.x * ab.y; 

double theta = Math.arcsin(cross);
  • Dom
#6

that is even better. thanks :slight_smile: