Two links

If there are two links, $ {\cal A}_1$ and $ {\cal A}_2$, then the C-space can be nicely visualized as a square with opposite faces identified. Each coordinate, $ \theta_1$ and $ \theta_2$, ranges from 0 to $ 2 \pi$, for which $ 0 \sim 2 \pi$. Suppose that each link has length $ 1$. This yields $ a_1 = 1$. A point $ (x,y) \in {\cal A}_2$ is transformed as

$\displaystyle \begin{pmatrix}\cos\theta_1 & -\sin\theta_1 & 0  \sin\theta_1 &...
...& 0  0 & 0 & 1  \end{pmatrix} \begin{pmatrix}x  y  1  \end{pmatrix} .$ (4.58)

To obtain polynomials, the technique from Section 4.2.2 is applied to replace the trigonometric functions using $ a_i = \cos\theta_i$ and $ b_i=\sin\theta_i$, subject to the constraint $ a_i^2 + b_i^2 = 1$. This results in

$\displaystyle \begin{pmatrix}a_1 & -b_1 & 0  b_1 & a_1 & 0  0 & 0 & 1  \e...
...& 0  0 & 0 & 1  \end{pmatrix} \begin{pmatrix}x  y  1  \end{pmatrix} ,$ (4.59)

for which the constraints $ a_i^2 + b_i^2 = 1$ for $ i = 1, 2$ must be satisfied. This preserves the torus topology of $ {\cal C}$, but now the C-space is embedded in $ {\mathbb{R}}^4$. The coordinates of each point are $ (a_1,b_1,a_2,b_2) \in {\mathbb{R}}^4$; however, there are only two degrees of freedom because each $ a_i,b_i$ pair must lie on a unit circle.

Multiplying the matrices in (4.59) yields the polynomials, $ f_1, f_2 \in {\mathbb{R}}[a_1,b_1,a_2,b_2]$,

$\displaystyle f_1 = x a_1 a_2 - y a_1 b_2 - x b_1 b_2 + y a_2 b_1 + a_1$ (4.60)

and

$\displaystyle f_2 = -y a_1 a_2 + x a_1 b_2 + x a_2 b_1 - y b_1 b_2 + b_1 ,$ (4.61)

for the $ x$ and $ y$ coordinates, respectively. Note that the polynomial variables are configuration parameters; $ x$ and $ y$ are not polynomial variables. For a given point $ (x,y) \in {\cal A}_2$, all coefficients are determined.

Steven M LaValle 2020-08-14