A zero-dimensional variety

Now a kinematic closure constraint will be imposed. Fix the point $(1,0)$ in the body frame of ${\cal A}_2$ at $(1,1)$ in ${\cal W}$. This yields the constraints

$$f_1 = a_1 a_2 - b_1 b_2 + a_1 = 1$$ (4.62)

and

$$f_2 = a_1 b_2 + a_2 b_1 + b_1 = 1 ,$$ (4.63)

by substituting $x=1$ and $y=0$ into (4.60) and (4.61). This yields the variety

$$V(a_1 a_2 - b_1 b_2 + a_1 - 1,a_1 b_2 + a_2 b_1 + b_1 - 1, a_1^2 + b_1^2 - 1,a_2^2 + b_2^2 - 1) ,$$ (4.64)

which is a subset of ${\mathbb{R}}^4$. The polynomials are slightly modified because each constraint must be written in the form $f = 0$.

Figure 4.22: Two configurations hold the point $p$ at $(1,1)$.

Although (4.64) represents the constrained configuration space for the chain of two links, it is not very explicit. Without an explicit characterization (i.e., a parameterization), it complicates motion planning. From Figure 4.22 it can be seen that there are only two solutions. These occur for $\theta_1 = 0$, $\theta_2 = \pi/2$ and $\theta_1 = \pi/2$, $\theta_2 = -\pi/2$. In terms of the polynomial variables, $(a_1,b_1,a_2,b_2)$, the two solutions are $(1,0,0,1)$ and $(0,1,0,-1)$. These may be substituted into each polynomial in (4.64) to verify that 0 is obtained. Thus, the variety represents two points in ${\mathbb{R}}^4$. This can also be interpreted as two points on the torus, ${\mathbb{S}}^1 \times {\mathbb{S}}^1$.

It might not be surprising that the set of solutions has dimension zero because there are four independent constraints, shown in (4.64), and four variables. Depending on the choices, the variety may be empty. For example, it is physically impossible to bring the point $(1,0) \in {\cal A}_2$ to $(1000,0) \in {\cal W}$.