SPEC.md

Double Joint Specification

Specification to implement Double Joint algorithm.

A Note About Voxel Size

Normally, voxel size is $1\text{x}1\text{x}1$ cube. But the size of voxels at joint is $1\text{x}\sqrt{2}\text{x}1$. When you rotate the arm 90 degrees at 45 degrees offset, it will rotate the diagonal at 45 degrees slope. Because the length of diagonal is $\sqrt{2}$, then the height must be $\sqrt{2}$.

Other voxels will not be affected, non-joint or single joint (think elbow).

The Original Method

This section covers the original method, as per reference.

1. Find the rotation matrix

First of all, we need to find the rotation matrix, defined by $r_x$ and $r_y$. The axis of rotation $\overrightarrow{X}$ is rotated to$\overrightarrow{X'}$ by $r_x$. Then the plane is rotated by $r_y$. Here is the entire rotation matrix:

$$\mathbf{R} = \begin{bmatrix} \cos(r_y) + \sin(r_x)^2*(1 - \cos(r_y)) & -\cos(r_x)*\sin(r_y) & \sin(r_x)*\cos(r_x)*(1 - \cos(r_y)) \\\ \cos(r_x)*\sin(r_y) & \cos(r_y) & -\sin(r_x)*\sin(r_y) \\\ \sin(r_x)*\cos(r_x)*(1 - \cos(r_y)) & \sin(rx)*\sin(r_y) & \cos(r_y) + \cos(r_x)^2*(1 - \cos(r_y)) \\\ \end{bmatrix}$$

2. Find the rotated cut plane

The cut plane formed by $A'B'C'D'$ is rotated to $ABCD$. We only need the Y coordinate, since we define X and Z as the same (vector $\overrightarrow{AA'}$ is vertical).

$$\begin{array}{rcl} \begin{bmatrix} A''_x \\ A''_z \end{bmatrix} & = & \begin{bmatrix} \mathbf{R}_{11} & \mathbf{R}_{13} \\ \mathbf{R}_{31} & \mathbf{R}_{33} \end{bmatrix}^{-1} \begin{bmatrix} A'_x \\ A'_z \end{bmatrix} \\\ A_y & = & \begin{bmatrix} \mathbf{R}_{21} & \mathbf{R}_{23} \end{bmatrix} \begin{bmatrix} A''_x \\A''_z \end{bmatrix} \end{array}$$

3. Interpolate and scale points

Since we have the rotated cut plane, we can interpolate the Y scaling factor for any XZ pair. Use it to scale the original arm (red dashed line) to transformed one (black line).

4. Generate the opposite end

Now we have half of the joint. To generate the other half, duplicate, flip, and rotate.

Alternative Method

There is another way of finding the rotated plane, the core of Double Joint. As seen from the blue triangle, $\overline{A'A}$ is sine of $r_y$. Equation 1 shows to find it, start from $\overline{OA'}$ ( $O$ is not the origin). Statement 2 shows relation to axis of rotation $\overrightarrow{X}$. To find $\overrightarrow{X}$, we use equation 3. Equation 4 is used to find signed distance $\left| OA' \right|$. And finally we combine all equations to gain equation 5.

$$\begin{array}{rcll} \dfrac{ \left\| A'A \right\| }{ \left\| OA' \right\| } & = & \text{tan} \left( r_y \right) & \text{(1)} \\\ \overrightarrow{OA'} & \bot & \overrightarrow{X} & \text{(2)} \\\ \overrightarrow{X} & = & \begin{bmatrix} \text{cos} \left( r_x \right) \\ \text{sin} \left( r_x \right) \end{bmatrix} & \text{(3)} \\\ \left\| OA' \right\| & = & A'_y X_x - A'_x X_y X & \text{(4)} \\\ A'A & = & \left( A'_y X_x - A'_x X_y \right) \text{tan} \left( r_y \right) & \text{(5)} \end{array}$$