Trajectory for Manipulation, ProbMAN 5

Precursor

What is manipulation?

• Prehensile manipulation: grasp and move, We have manipulator arm, we have screws (Mostly talked)
• Nonprehensile manipulation: move without grasp, We have assembly line, there are passive fences to guide the object (Less talked)

Geometry of Manipulation

Curves (Path and Trajectory), and parameterization

• Path is a continuous function from some variable to the space. This variable is called parameter. It can be time, arc length, etc.
• Trajectory is a path, plus a velocity for each point on the path.

Arc Length Parameterization

$$ \begin{aligned} s(t_2) - s(t_1) &= \int_{t1}^{t2}(x'(t), x'(t))^{1/2}dt \\ u(t) &\dot{=} \frac{1}{||x'(t)||}x'(t) \\ \end{aligned} $$ When $t=s$, i.e. parameter is arc length, $$ u(s) = \frac{dx}{ds} $$ Since $u(s)$ is a unit vector, we can write $$ \frac{d}{ds}(u\cdot u) = 0 \Rightarrow u \cdot \frac{du}{ds} = 0 $$

Unsigned Curvature

Kappa is the unsigned curvature. N1 is the (principal) normal vector. $$ \begin{aligned} \kappa(s) &= (\frac{du}{ds} \cdot \frac{du}{ds})^{1/2} = (\frac{d^2x}{ds^2} \cdot \frac{d^2x}{ds^2})^{1/2} = ||u'(s)|| \\ n_1(s) &\dot{=} \frac{1}{\kappa(s)}\frac{du}{ds} \\ \end{aligned} $$

Example: Circle $$ \begin{aligned} x_1(s) &= r\cos(\frac{s}{r}) \\ x_2(s) &= r\sin(\frac{s}{r}) \\ \end{aligned} $$ Actually this needs experience to know this perfect parameterization. $$ $$ The second part is the calculate the unit tangent vector. Only this derivative is naturally unit. (by good parameterization) $$ \begin{aligned} & u_1(s) = x_1'(s) = -\sin(\frac{s}{r}) \\ & u_2(s) = x_2'(s) = \cos(\frac{s}{r}) \\ & u_1^2(s) + u_2^2(s) = 1 \\ & u_1'(s) = -\frac{1}{r}\cos(\frac{s}{r}) \\ & u_2'(s) = -\frac{1}{r}\sin(\frac{s}{r}) \\ \end{aligned} $$ And we can see that $$ u'(s) = \frac{1}{r}\begin{pmatrix} -c \\ -s \end{pmatrix} \\ \therefore \kappa(s) = ||u'(s)|| = \frac{1}{r} $$

Signed Curvature

$$ \begin{aligned} k(s) = \sigma(s)\kappa(s) \end{aligned} $$

Example: curve $$ \begin{aligned} y &= x^3 \\ x(t) &= t \\ y(t) &= t^3 \\ x'(t) &= 1 \\ y'(t) &= 3t^2 \\ \kappa(t) &= \sqrt{x'^2(t) + y'^2(t)} = \sqrt{1 + 9t^4} \\ \sigma(t) &= \frac{y''(t)x'''(t) - x''(t)y'''(t)}{\kappa^3(t)} = \frac{6t^2}{\kappa^3(t)} \\ \end{aligned} $$