Drawing Euler Angles rotational model on a 2d image

Question

I'm currently attempting to draw a 3d representation of euler angles within a 2d image (no opengl or 3d graphic windows). The image output can be similar to as below.

Essentially I am looking for research or an algorithm which can take a Rotation Matrix or a set of Euler angles and then output them onto a 2d image, like above. This will be implemented in a C++ application that uses OpenCV. It will be used to output annotation information on a OpenCV window based on the state of the object.

I think I'm over thinking this because I should be able to decompose the unit vectors from a rotation matrix and then extract their x,y components and draw a line in cartesian space from (0,0). Am i correct in this thinking?

EDIT: I'm looking for an Orthographic Projection. You can assume the image above has the correct camera/viewing angle.

Any help would be appreciated.

Thanks,

EDIT: The example source code can now be found in my repo. Header: https://bitbucket.org/jluzwick/tennisspindetector/src/6261524425e8d80772a58fdda76921edb53b4d18/include/projection_matrix.h?at=master

Class Definitions: https://bitbucket.org/jluzwick/tennisspindetector/src/6261524425e8d80772a58fdda76921edb53b4d18/src/projection_matrix.cpp?at=master

It's not the best code but it works and shows the steps necessary to get the projection matrix described in the accepted answer.

Also here is a youtube vid of the projection matrix in action (along with scale and translation added): http://www.youtube.com/watch?v=mSgTFBFb_68

Answer 1

Here are my two cents. Hope it helps.

If I understand correctly, you want to rotate 3D system of coordinates and then project it orthogonally onto a given 2D plane (2D plane is defined with respect to the original, unrotated 3D system of coordinates).

"Rotating and projecting 3D system of coordinates" is "rotating three 3D basis vectors and projecting them orthogonally onto a 2D plane so they become 2D vectors with respect to 2D basis of the plane". Let the original 3D vectors be unprimed and the resulting 2D vectors be primed. Let {e1, e2, e3} = {e1..3} be 3D orthonormal basis (which is given), and {e1', e2'} = {e1..2'} be 2D orthonormal basis (which we have to define). Essentially, we need to find such operator PR that PR * v = v'.

While we can talk a lot about linear algebra, operators and matrix representation, it'd be too long of a post. It'll suffice to say that :

1. For both 3D rotation and 3D->2D projection operators there are real matrix representations (linear transformations; 2D is subspace of 3D).
2. These are two transformations applied consequently, i.e. PR * v = P * R * v = v', so we need to find rotation matrix R and projection matrix P. Clearly, after we rotated v using R, we can project the result vector vR using P.
3. You have the rotation matrix R already, so we consider it is a given 3x3 matrix. So for simplicity we will talk about projecting vector vR = R * v.
4. Projection matrix P is a 2x3 matrix with i-th column being a projection of i-th 3D basis vector ei onto {e1..2'} basis.

Let's find P projection matrix such as a 3D vector vR is linearly transformed into 2D vector v' on a 2D plane with an orthonormal basis {e1..2'}.

A 2D plane can be easily defined by a vector normal to it. For example, from the figures in the OP, it seems that our 2D plane (the plane of the paper) has normal unit vector n = 1/sqrt(3) * ( 1, 1, 1 ). We need to find a 2D basis in the 2D plane defined by this n. Since any two linearly independent vectors lying in our 2D plane would form such basis, here are infinite number of such basis. From the problem's geometry and for the sake of simplicity, let's impose two additional conditions: first, the basis should be orthonormal; second, should be visually appealing (although, this is somewhat a subjective condition). As it can be easily seen, such basis is formed trivially in the primed system by setting e1' = ( 1, 0 )' = x'-axis (horizontal, positive direction from left to right) and e2' = ( 0, 1 )' = y'-axis (vertical, positive direction from bottom to top).

Let's now find this {e1', e2'} 2D basis in {e1..3} 3D basis.

1. Let's denote e1' and e2' as e1" and e2" in the original basis. Noting that in our case e1" has no e3-component (z-component), and using the fact that n dot e1" = 0, we get that e1' = ( 1, 0 )' -> e1" = ( -1/sqrt(2), 1/sqrt(2), 0 ) in the {e1..3} basis. Here, dot denotes dot-product.
2. Then e2" = n cross e1" = ( -1/sqrt(6), -1/sqrt(6), 2/sqrt(6) ). Here, cross denotes cross-product.

The 2x3 projection matrix P for the 2D plane defined by n = 1/sqrt(3) * ( 1, 1, 1 ) is then given by:

( -1/sqrt(2)    1/sqrt(2)        0     )
( -1/sqrt(6)   -1/sqrt(6)    2/sqrt(6) )

where first, second and third columns are transformed {e1..3} 3D basis onto our 2D basis {e1..2'}, i.e. e1 = ( 1, 0, 0 ) from 3D basis has coordinates ( -1/sqrt(2), -1/sqrt(6) ) in our 2D basis, and so on.

To verify the result we can check few obvious cases:

1. n is orthogonal to our 2D plane, so there should be no projection. Indeed, P * n = P * ( 1, 1, 1 ) = 0.
2. e1, e2 and e3 should be transformed into their representation in {e1..2'}, namely corresponding column in P matrix. Indeed, P * e1 = P * ( 1, 0 ,0 ) = ( -1/sqrt(2), -1/sqrt(6) ) and so on.

To finalize the problem. We now constructed a projection matrix P from 3D into 2D for an arbitrarily chosen 2D plane. We now can project any vector, previously rotated by rotation matrix R, onto this plane. For example, rotated original basis {R * e1, R * e2, R * e3}. Moreover, we can multiply given P and R to get a rotation-projection transformation matrix PR = P * R.

P.S. C++ implementation is left as a homework exercise ;).

Answer 2

The rotation matrix will be easy to display,

A Rotation matrix can be constructed by using a normal, binormal and tangent.

You should be able to get them back out as follows:-

Bi-Normal (y') : matrix[0][0], matrix[0][1], matrix[0][2]
Normal    (z') : matrix[1][0], matrix[1][1], matrix[1][2]
Tangent   (x') : matrix[2][0], matrix[2][1], matrix[2][2]

Using a perspective transform you can the add perspective (x,y) = (x/z, y/z)

To acheive an orthographic project similar to that shown you will need to multiply by another fixed rotation matrix to move to the "camera" view (45° right and then up)

You can then multiply your end points x(1,0,0),y(0,1,0),z(0,0,1) and center(0,0,0) by the final matrix, use only the x,y coordinates.

center should always transform to 0,0,0

You can then scale these values to draw to you 2D canvas.