# Quintic polynomials planning

Motion planning with quintic polynomials.

It can calculate 2D path, velocity, and acceleration profile based on quintic polynomials.

## Quintic polynomials for one dimensional robot motion

We assume a one-dimensional robot motion $$x(t)$$ at time $$t$$ is formulated as a quintic polynomials based on time as follows:

(1)$x(t) = a_0+a_1t+a_2t^2+a_3t^3+a_4t^4+a_5t^5$

$$a_0, a_1. a_2, a_3, a_4, a_5$$ are parameters of the quintic polynomial.

It is assumed that terminal states (start and end) are known as boundary conditions.

Start position, velocity, and acceleration are $$x_s, v_s, a_s$$ respectively.

End position, velocity, and acceleration are $$x_e, v_e, a_e$$ respectively.

So, when time is 0.

(2)$x(0) = a_0 = x_s$

Then, differentiating the equation (1) with t,

(3)$x'(t) = a_1+2a_2t+3a_3t^2+4a_4t^3+5a_5t^4$

So, when time is 0,

(4)$x'(0) = a_1 = v_s$

Then, differentiating the equation (3) with t again,

(5)$x''(t) = 2a_2+6a_3t+12a_4t^2$

So, when time is 0,

(6)$x''(0) = 2a_2 = a_s$

so, we can calculate $$a_0, a_1, a_2$$ with eq. (2), (4), (6) and boundary conditions.

$$a_3, a_4, a_5$$ are still unknown in eq (1).

We assume that the end time for a maneuver is $$T$$, we can get these equations from eq (1), (3), (5):

(7)$x(T)=a_0+a_1T+a_2T^2+a_3T^3+a_4T^4+a_5T^5=x_e$
(8)$x'(T)=a_1+2a_2T+3a_3T^2+4a_4T^3+5a_5T^4=v_e$
(9)$x''(T)=2a_2+6a_3T+12a_4T^2+20a_5T^3=a_e$

From eq (7), (8), (9), we can calculate $$a_3, a_4, a_5$$ to solve the linear equations: $$Ax=b$$

$\begin{split}\begin{bmatrix} T^3 & T^4 & T^5 \\ 3T^2 & 4T^3 & 5T^4 \\ 6T & 12T^2 & 20T^3 \end{bmatrix}\begin{bmatrix} a_3\\ a_4\\ a_5\end{bmatrix}=\begin{bmatrix} x_e-x_s-v_sT-0.5a_sT^2\\ v_e-v_s-a_sT\\ a_e-a_s\end{bmatrix}\end{split}$

We can get all unknown parameters now.

## Quintic polynomials for two dimensional robot motion (x-y)

If you use two quintic polynomials along x axis and y axis, you can plan for two dimensional robot motion in x-y plane.

(10)$x(t) = a_0+a_1t+a_2t^2+a_3t^3+a_4t^4+a_5t^5$
(11)$y(t) = b_0+b_1t+b_2t^2+b_3t^3+b_4t^4+b_5t^5$

It is assumed that terminal states (start and end) are known as boundary conditions.

Start position, orientation, velocity, and acceleration are $$x_s, y_s, \theta_s, v_s, a_s$$ respectively.

End position, orientation, velocity, and acceleration are $$x_e, y_e. \theta_e, v_e, a_e$$ respectively.

Each velocity and acceleration boundary condition can be calculated with each orientation.

$$v_{xs}=v_scos(\theta_s), v_{ys}=v_ssin(\theta_s)$$

$$v_{xe}=v_ecos(\theta_e), v_{ye}=v_esin(\theta_e)$$