I tried Euler angles transformation in this case, but I could not find the correct sequence of xyz to get target orientation.
P.S. I am following the convention of x->y->z for Euler angles transformation.
P.S. Euler angles transformation: each rotation takes place relative to original frame.
The rotation of z axis can only switch y and x, but it won't lead to final target.
What is the right way?
final answer:
#Rooscannon, thanks you very much!
Rotate (45, 180 , 90)
Here the maths
http://www.staff.city.ac.uk/~sbbh653/publications/euler.pdf
getting the rotation matrix is very simple in your use case. The columns of the 3X3 rotation matrix are the rotated unit vectors x,y,z in the new position. For example x was originally (1,0,0) and after rotation (0,-1,0). y = (-sqrt(2)/2, 0, -sqrt(2)/2), z = ( sqrt(2)/2, 0, -sqrt(2)/2).
So your rotation matrix will be something like
0 -0.7 0.7
R = -1 0 0
0 -0.7 -0.7
Use exact values though. If insecure the determinant should be 1 and the norm of each column and row should be 1.
Rotate 135 on y. And then rotate -90 on z to get the desired angel.
Your final rotation will be :
(0,135,-90)
Hope this helps.
Related
I've been trying to figure out the 2D rotation value as seen from orthographic "top" view for a 3D object with XYZ rotation values in Maya. Maybe another way to ask this could be: I want to figure out the 2D rotation of a 3D obj's direction.
Here is a simple image to illustrate my question:
I've tried methods like getting the twist value of an object using quaternion (script pasted below), to this post I've found: Component of a quaternion rotation around an axis.
If I set the quaternion's X and Z values to zero, this method works half way. I can get the correct 2D rotation even when obj is rotated in both X and Y axis, but when rotated in all 3 axis, the result is wrong.
I am pretty new to all the quaternion and vector calculations, so I've been having difficulty trying to wrap my head around it.
;)
def quaternionTwist(q, axisVec):
axisVec.normalize()
# Get the plane the axisVec is a normal of
orthonormal1, orthonormal2 = findOrthonormals(axisVec)
transformed = rotateByQuaternion(orthonormal1, q)
# Project transformed vector onto plane
flattened = transformed - ((transformed * axisVec) * axisVec)
flattened.normalize()
# Get angle between original vector and projected transform to get angle around normal
angle = math.acos(orthonormal1 * flattened)
return math.degrees(angle)
q = getMQuaternion(obj)
# Zero out X and Y since we are only interested in Y axis.
q.x = 0
q.z = 0
up = om2.MVector.kYaxisVector
angle = quaternionTwist(q, up)
Can you get the (x,y,z) coordinates of the rotated vector? Once you have them use the (x,y) values to find the angle with atan2(y,x).
I'm not familiar with the framework you're using, but if it does what it seems, I think you're almost there. Just don't zero out the X and Z components of the quaternion before calling quaternionTwist().
The quaternions q1 = (x,y,z,w) and q2 = (0, y, 0, w) don't represent the same rotation about the y-axis, especially since q2 written this way becomes unnormalized, so what you're really comparing is (x,y,z,w) with (0, y/|q2|, 0, w/|q2|) where |q2| = sqrt(y^2 + w^2).
Here is a working code for Maya using John Alexiou's answer:
matrix = dagPath.inclusiveMatrix() #OpenMaya dagPath for an object
axis = om2.MVector.kZaxisVector
v = (axis * matrix).normal()
angle = math.atan2(v.x, v.z) #2D angle on XZ plane
I have two 3D direction (normalized) vector A and B. I am looking for the Euler angles to rotate A into B. I know it has many solution because it is possible to rotate a normal vector anywhere with just using two axis like X and Y or roll and pitch. I have to find the solution where the Z rotation is Zero.
I would like to create a function like this:
Vector3 dir1 (0, 1, 0);
Vector3 someRotation(Pi / 4, Pi / 4, 0);
Vector3 dir2 = dir1.rotateXYZ(someRotation);
Vector3 xyRotation = dir1.eulerToDirection(dir2);
// now I expect that the eulerToDirection fv calculated the X, Y rotation from the vectors at Z = 0
// so xyRotation.x == Pi / 4 && xyRotation.y == Pi / 4 && xyRotation.z == 0 is true
// aside from the floating point error
Of corse the some rotation not always 0 at the Z. It is just for the example
First use atan2(z2-z1, y2-y1) to find the angle to rotate around the X-axis that aligns the y's and the z's. Then, use acosof the dot product between the just rotated vector and the final vector. This will be the angle needed for the rotation around Y. Depending on how you implemented the rotations, you might need to flip some signs.
I am trying to programmatically visualise a vector point but I want to clarify my output result.
If a vector p = i = [1,0,0] rotate by 90 degree about the x-axis, then quaternion q is: q = cos(45) + [1,0,0]*sin(45) = 0.707 + 0.707*i.
pn = qpq-1;
Now calculate pn: (0.707+0.707*i)(i)(0.707-0.707*i) = i.
So, the rotated vector pn = [1,0,0]. Which is p=pn.
Is p=pn correct? If it is can anyone explain it? or is this a special property of quaternions?
In the example you provided, you basically rotate a vector around itself (i.e. the axis of rotation is equal to the rotated vector, in this case [1,0,0]). As said in the comments, rotating a vector around itself, leaves it intact, regardless of the rotation angle.
Try your example where the rotated vector is along the y-axis [0,1,0], and the rotation axis is [1,0,0]. Maybe this will help you visualize some basic rotations.
Also, be mindful that a rotation of vector v using a unit quaternion q is given by:
Imaginary{q * [0, v_x, v_y, v_z] * conjugate(q)}
I Have the orthographic projection of a unit cube with one of its vertex at origin as shown above. I have the x,y (no z) co ordinates of the projections. I would like to compute the angle of rotation of the plane to get the second orthographic projection from the first one (maybe euler angles??)
Is there any other easy way to compute this?
UPDATE:
Could I use this rotation matrix to get a system of equations in cos, sin angles and the x,y and x',y' and solve them easily? Or is there any easier way to get the angles back? (Am I on the right direction to solve this? )
First method
Use this idea to generate equations:
a1, a2 and a3 are coordinates in the original system, x y are the coordinates you get from the end-result and z is a coordinate you don’t know. This generates 2 equations for every point of the cube. E.g for point 0 with coordinates (-1, -1, 1) these are:
Do this for the 4 front points of the cube and you get 8 equations. Now add the fact that this is a rotation matrix -> the determinant is 1 and you have 9 equations. Solve these with any of the usual algorithms for solving equation systems and you have the transformation matrix. Getting the axis and angle from that is easy via google: http://www.euclideanspace.com/maths/geometry/rotations/conversions/matrixToAngle/
Second method
Naming your points 0, 1, 2, 3 a, b, c, d respectively, you can get the z coordinates of the vectors between them (e.g. b-a) with this idea:
you will still have to sort out if b3-a3 is positive, though. One way to do that is to use the centermost point as b (calculate distance from the center for all points, use the one with the minimal distance). Then you know for sure that b3-a3 is positive (if z is positive towards you).
Now assume that a is (0,0,0) in your transformed space and you can calculate all the point positions by adding the appropriate vectors to that.
To get the rotation you use the fact that you know where b-a did point in your origin space (e.g. (1,0,0)). You get the rotation angle via dot product of b-a and (1,0,0) and the rotation axis via cross product between those vectors.
I have two squares, S1 = (x1,y1,x2,y2) and S2 = (a1,b1,a2,b2)
I'm looking for the A transformation matrix with which
A * S1 = S2
As far as I see, A is an affine 3x3 matrix, so I have 9 unknown values.
How can I calculate these values?
thanks and best,
Viktor
There are really only four unknown values here. A rotation angle, a scale factor and an x and y translation. Of your three by three matrix the bottom row is always 0,0,1 which reduces you to six unknowns. The right hand column will be Tx,Ty,1 which are your translations (and the 1 we already know about).
The two by two "matrix" left will be your rotation and scaling. This will (off the top of my head) be something like:
ACos(B), -Asin(B)
ASin(B), aCos(B)
So in total:
ACos(B), -Asin(B), Tx
ASin(B), ACos(B), Ty
0 , 0 , 1
You extend your co-ordinate matrices with the 1 on the end of each co-ordinate to give 2x3 matrices and they then multiply to give you the four equations you need to solve for the four variables. That is left as an exercise for the reader.
A transformation matrix is a factor of scaling matrix Ss, transition matrix St and rotation matrix Sr.
Assume the old point is Po is (Xo,Yo) and as vector will be represented as (Xo Yo 1)' same for the new point Pn
Then Pnv =SsStSrPov
Where Sx is
Sx 0 0
0 Sy 0
0 0 1
St is
1 0 Tx
0 1 Ty
0 0 1
Sr is
Cos(th) -Sin(th) 0
Sin(th) Cos(th) 0
0 0 1
Now back to your question. if two point are giving to represent a rectangle we can just find the parameter of two matrix and the third one will be an identity matrix.
Rect1 is represented as Top-Left point P11 and Bottom-Right Point P12
Rect2 is represented as Top-Left point P21 and Bottom-Right Point P22
S=Ss*St
Sx 0 Tx
0 Sy Ty
0 0 1
Now you have 4 missing parameters and 4 set of equations
P21=S*P11
P22=S*P12
X[P21] =Sx*X[P11]+Tx
Y[P21] =Sy*Y[P11]+Ty
X[P22] =Sx*X[P12]+Tx
Y[P22] =Sy*Y[P12]+Ty
Solve it and you'll get your answer.
and if you have transition and rotation then
S=Sr*St.
Cos(th) -Sin(th) Tx
Sin(th) Cos(th) Ty
0 0 1
Now you have 3 missing parameters and 4 set of equations
P21=S*P11
P22=S*P12
X[P21] =Cos(th)*X[P11]-Sin(th)*Y[P11]+Tx
Y[P21] =Sin(th)*X[P11]+Cos(th)*Y[P11]+Ty
X[P22] =Cos(th)*X[P11]-Sin(th)*Y[P12]+Tx
Y[P22] =Sin(th)*X[P11]+Cos(th)*Y[P12]+Ty
Replace Cos(th) with A and Sin(th) With B and solve the equations.
X[P21] =A*X[P11]-B*Y[P11]+Tx
Y[P21] =B*X[P11]+A*Y[P11]+Ty
X[P22] =A*X[P11]-B*Y[P12]+Tx
Y[P22] =B*X[P11]+A*Y[P12]+Ty
Check if its correct A^2+B^2 =? 1 if is true then th = aCos(A)
The last part of the solution, if you'll have all three matrixes, then S=SrStSs is
Sx*sin(th) -Sx*cos(th) Tx
Sy*cos(th) Sy*sin(th) Ty
0 0 1
Now we have 5 missing variables and we need 6 different set of equations to solve it. which is mean 3 points from each rectangle.
You shouldn't have a 3x3 matrix if you're just looking to transform a 2D object. What you're looking for is a 2x2 matrix that solves A*S1=S2. This can be done in many different ways; in MATLAB, you'd do a S2/S1 (right matrix division), and generally this performs some kind of Gaussian elimination.
How can I calculate these values?
When applied to 2d/3d transformations, matrix can be represented a coordinate system, unless we are talking about projections.
Matrix rows (or columns, depending on notation) form axes of a new coordinate system, in which object will be placed placed if every object vertex is multiplied by the matrix. Last row (or columne, depending on notation) points to the center of the new coordinate system.
Standard OpenGL/DirectX transformation matrix (NOT a projection matrix):
class Matrix{//C++ code
public:
union{
float f[16];
float m[4][4];
};
};
Can be represented as combination of 4 vectors vx (x axis of the new coordinate system), vy(y axis of a new coordinate system), vz(z axis of a new coordinate system), and vp (center of the new system). Like this:
vx.x vx.y vx.z 0
vy.x vy.y vy.z 0
vz.x vz.y vz.z 0
vp.x vp.y vp.z 1
All "calculate rotation matrix", "calculate scale matrix", etc go down to this idea.
Thus, for 2d matrix, you'll have 3x3 matrix that consists of 3 vectors - vx, vy, vp, because there is no z vector in 2d. I.e.:
vx.x vx.y 0
vy.x vy.y 0
vp.x vp.y 1
To find a transform that would transform quad A into quad B, you need to find two transforms:
Transform that will move quad A into origin (i.e. at point zero), and convert it into quad of fixed size. Say, quad (rectangle) whose one vertex x = 0, y = 0, and whose vertices are located at (0, 1), (1, 0), (1, 1).
Transform that turns quad of fixed size into quad B.
You CANNOT do that it this way if opposite edges of quad are not parallel. I.e. parallelograms are fine, but random 4-sided polygons are not.
A quad can be represented by base point (vp) which can be any vertex of the quad and two vectors that define quad sizes (direction of the edge multiplied by edge's length). I.e. "up" vector and "side" vector. Which makes it a matrix:
side.x side.y 0
up.x up.y 0
vp.x vp.y 1
So, multiplying a quad (vp.x = 0, vp.y = 0, side.x = 1, side.y = 0, up.x = 0, up.y = 1) by this matrix will turn original quad into your quad. Which means, that in order to transform
quad A into quad B, you need to do this:
1) make a matrix that would transform "base 1unit quad" into quad A. Let's call it matA.
2) make a matrix that would transform "base 1 unit quad" into quad B. let's call it matB.
3) invert matA and store result into invMatA.
4) the result matrix is invMatA * matB.
Done. If you multiply quad A by result matrix, you'll get quad B. This won't work if quads have zero widths or heights, and it won't work if quads are not parallelograms.
This is hard to understand, but I cannot to make it simpler.
What do you mean by S1 = (x1,y1,x2,y2)?
Do they represent the top-left and bottom-right corners of the square?
Also, can you guarantee there's only rotation between the squares or do you need a full affine transformation which allows for scaling, skewing, and translation?
Or do you also need a perspective transformation?
Only if it's a perspective transformation, will you need 3x3 matrix with 8 dof as you've mentioned in your post.