Interpolation

Lie Group Interpolation

The typical linear interpolation scheme:

\[ x = (1 - \alpha) x_1 + \alpha x_2, \quad \alpha \in \left[0, 1 \right], \]

will not work on \(SO(3)\) and \(SE(3)\) because this interpolation scheme does not satisfy closure (i.e., the result is no longer in the group):

\[ \begin{align} (1 - \alpha) \mathbf{R}_1 + \alpha \mathbf{R}_2 \notin SO(3) \\ (1 - \alpha) \mathbf{T}_1 + \alpha \mathbf{T}_2 \notin SE(3), \\ \end{align} \]

for some values of \(\alpha \in \left[0, 1 \right]\), \(\mathbf{R}_1, \mathbf{R}_2 \in SO(3)\) and \(\mathbf{T}_1, \mathbf{T}_2 \in SE(3)\).s

Rotation Interpolation

One of the interpolations schemes is:

\[ \mathbf{R} = \left( \mathbf{R}_2 \mathbf{R}^T_1 \right)^\alpha \mathbf{R}_1, \quad \alpha \in \left[0, 1 \right], \]

where \(\mathbf{R}, \mathbf{R}_1, \mathbf{R}_2 \in SO(3)\). We see that when \(\alpha = 0\), we have \(\mathbf{R} = \mathbf{R}_1\), and \(\mathbf{R} = \mathbf{R}_2\) when \(\alpha = 1\). With this definition, we can guarantee close, i.e., \(\mathbf{R} \in SO(3)\) for all \(\alpha \in \left[0, 1 \right]\).

Proof. Let \(\mathbf{R}_{21} = \mathbf{R}_2 \mathbf{R}^T_1 = \exp (\left[ \boldsymbol{\rho} \right]_\times)\). Then, we have:

\[ \mathbf{R}^\alpha_{21} = \exp \left( \left[ \boldsymbol{\rho} \right]_\times \right)^\alpha = \exp \left( \alpha \left[ \boldsymbol{\rho} \right]_\times \right) \in SO(3). \]

One way to interepret the interpolation scheme is that it enforces a constant angular velocity, \(\boldsymbol{\omega}\). If the rotation matrix is a function of time, \(\mathbf{R}(t)\), then the scheme is:

\[ \mathbf{R}(t) = \left( \mathbf{R}(t_2) \mathbf{R}(t_1)^T \right)^\alpha \mathbf{R}(t_1), \quad \alpha = \frac{t - t_1}{t_2 - t_1}. \]

Defining the constant angular velocity as:

\[ \boldsymbol{\omega} = \frac{1}{t_2 - t_1} \boldsymbol{\rho}, \]

we get:

\[ \mathbf{R}(t) = \exp((t - t_1) \left[ \boldsymbol{\omega} \right]_\times) \mathbf{R}(t_1), \]

which is the solution to the Poisson's equation:

\[ \dot{\mathbf{R}}(t) = \left[ \boldsymbol{\omega} \right]_\times \mathbf{R}(t). \]