Trajectory generation in robotics using cubic polynomial method
The problem’s definition is that given the initial and final time, and also initial and final positions in an one DOF(degree of freedom) robot, one is asked to move it in a smooth way from initial to final position.
Using a cubic polynomial will make it for describing the position of the only joint of the system given by the following formula:
Position :
p(t) = a0 + a1t + a2*t² + a3*t³
Velocity :
p(t)’ = a1 + 2a2*t + 3a3*t²
Acceleration:
p(t)’’ = 2a2 + 6a3*t
Since then we need to move the system in a smooth way initial and final velocities will be zero.
Given the initial position from p(0) we conclude that a0 = (initial position value).
By considering the same equation for position for the final time given the final position we get and equation.(let’s call equation [1] )
By considering the velocity’s equation for initial time, we conclude that a1=0.
By taking in account the same equation for the final time, we will obtain another equation [2]. And the only not-known variables at this point are a2 and a3. Having 2 equations with two unknown’s we can solve and determine exactly the solutions for them.
In the end having all of the coefficients of the kinematic equations (a0,a1,a2,a3) we can graphical draw the position, velocity and acceleration of the system.
