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Wielder

OpenGL 2 vectors, axis of rotation, dot product

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I have 2 vectors in my OpenGL. I need to rotate the first across an (existing, arbitrary) axis so that it is on top of the second. I can find out the angle between them (dot product) and I already have the axis I need to rotate on. However, I don't know in which direction to rotate. Dependant on where the first vector is (if its to the right of the second vector, or to the left, looking down the axis of rotation from 0,0,0), it needs to be rotated by the angle or by -angle. Is there a simple way through mathematics to find out which?

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How do you know what the axis is?

To find the axis I would use the cross product, n = a x b. Depending on which "side" of a b is, the sign of n will change. So your issue of "angle or -angle" is resolved.

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Quote:
Original post by Wielder
Okay, I need more help :)

When you say n = a x b, how do you calculate n as a scalar (to know the sign)? When I do cross products I end up with a vector.
If you calculate the axis as cross(a,b), the 'correct' direction of rotation will take care of itself. If the axis was pre-calculated in some other way (perhaps the normal to a plane in which the vectors are known to lie), you can compute the sign of the angle as sign(dot(cross(a,b),axis)).

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I'm still having troubles :/ My goal is to implement 3d rotation but I can't seem to work this out!

My first post for help was in this thread earlier
http://www.gamedev.net/community/forums/topic.asp?topic_id=393767

Quote:
I have a point (x,y,z) in space. The point can be anywhere in space. I need to draw an object (basic stretched cube), which starts out facing (1,0,0), to face that the x,y,z point in space using a rotation matrix. The object's roll should remain in a consistent position (ie. I can't have the cube flip around so that its top faces down after the rotation).

What is the best way to do this? I tried using cross and dot products to find an axis rotation between the objects, and the angle, then rotate using that. But that brings up the problem that the cube spins in strange directions even though it faces the right way.

I also tried finding out the angle on the XY plane and the XZ plane and then rotating across known axes once by each, but I think I have problems with my algorithm because it doesn't end up facing the point it needs to some of the time.

I'm using OpenGL implemented in C++. Any ideas would be apreciated."


I haven't made any real progress since then, I think my approach is just flawed.

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Okay gonna give you my code:

bool srf::MakeSrf(point orgpt, vector vect1, vector vect2, int order)
{
// The parameters are as follows:
//
// orgpt: the origin of the srf
// vect1: the defining vector for the first axis
// vect2: the second defining line which defines the 2nd and 3rd axes
// order: the axis ordering
// 0 = XYZ
// 1 = XZY
// 2 = YXZ
// 3 = YZX
// 4 = ZXY
// 5 = ZYX
//

point v1, v2, v3, v4; // The four vectors that are required for srf
// generation. Note that although these are
// vectors, we actually use points for brevity
float v1len, v3len, v4len; // Lengths of the initial x, y and z axes; used
// for normalisation

// Generate v1. This is done by subtracting vect1.start from vect1.end
// (normalising to the origin)
v1.x = vect1.end.x - vect1.start.x;
v1.y = vect1.end.y - vect1.start.y;
v1.z = vect1.end.z - vect1.start.z;
// if v1 has 0 length then normalise will produce div by 0 error
// Normalise v1
v1.normalize();

// Same with v2 and vect2
v2.x = vect2.end.x - vect2.start.x;
v2.y = vect2.end.y - vect2.start.y;
v2.z = vect2.end.z - vect2.start.z;
// if v2 has 0 length then normalise will produce div by 0 error
// Normalise v2
v2.normalize();

// v3 is now generated by obtaining the cross product v1^v2. This is our
// non-normalised 2nd axis (iniintial y axis, prior to reordering)
v3 = v1 ^ v2;
// Quick check here that we have actually generated a normal, and that the
// initial vectors provided were not parallel, coincident or reflected -
// this we can tell from all three components of v3 being equal to zero

if (v3.x == 0 && v3.y == 0 && v3.z == 0)
// Non-normal vector. Big problem
return false;

// Continue. v4 is normal to v1 and v3, and is the non-normalised, initial z
// axis.
v4 = v1 ^ v3;
// We now have all three axes but they are non-normalised. So we will now
// perform the normalisation operation.
// Normalise v3
v3.normalize();
// Normalise v4
v4.normalize();
// Now insert the three axes v1, v3 and v4 into our srf result object
// according to the desired ordering
//
// 0 = XYZ
// 1 = XZY
// 2 = YXZ
// 3 = YZX
// 4 = ZXY
// 5 = ZYX
//

switch (order)
{
case 0:
xaxis.SetPoint(&v1); // X
yaxis.SetPoint(&v3); // Y
zaxis.SetPoint(&v4); // Z
break;
case 1:
xaxis.SetPoint(&v1); // X
yaxis.SetPoint(&v4); // Z
zaxis.SetPoint(&v3); // Y
break;
case 2:
xaxis.SetPoint(&v3); // Y
yaxis.SetPoint(&v1); // X
zaxis.SetPoint(&v4); // Z
break;
case 3:
xaxis.SetPoint(&v3); // Y
yaxis.SetPoint(&v4); // Z
zaxis.SetPoint(&v1); // X
break;
case 4:
xaxis.SetPoint(&v4); // Z
yaxis.SetPoint(&v1); // X
zaxis.SetPoint(&v3); // Y
break;
case 5:
xaxis.SetPoint(&v4); // Z
yaxis.SetPoint(&v3); // Y
zaxis.SetPoint(&v1); // X
break;
default:
return false; // Out of bounds
}

origin.SetPoint(&orgpt);

return true;
}



This function takes 2 vectors as input, the first one defines one of the axis (which one depends on the order specified), the second one gives the orientation (just pass an up vector or something, you can work it out).
Its part of a class called srf (sub-coordinate reference frame). An srf is basically a matrix, and the conversion to a matrix is easy:


WWMatrix srf::ToMatrix()
{
WWMatrix mat;

mat.SetSize(4);
mat.MakeIdentity();

mat.Set(0, 0, xaxis.x);
mat.Set(1, 0, xaxis.y);
mat.Set(2, 0, xaxis.z);
mat.Set(0, 1, yaxis.x);
mat.Set(1, 1, yaxis.y);
mat.Set(2, 1, yaxis.z);
mat.Set(0, 2, zaxis.x);
mat.Set(1, 2, zaxis.y);
mat.Set(2, 2, zaxis.z);

mat.Set(0, 3, origin.x);
mat.Set(1, 3, origin.y);
mat.Set(2, 3, origin.z);

return(mat);
}



So you just take that matrix and apply it to the object you want to face towards the point.
Hope this helps!

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