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

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

$a_0, a_1. a_2, a_3, a_4, a_5$ 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.

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

Then, differentiating the equation (1) with t,

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

So, when time is 0,

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

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

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

So, when time is 0,

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

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):

$x(T)=a_0+a_1T+a_2T^2+a_3T^3+a_4T^4+a_5T^5=x_e$ -- (7)

$x'(T)=a_1+2a_2T+3a_3T^2+4a_4T^3+5a_5T^4=v_e$ -- (8)

$x''(T)=2a_2+6a_3T+12a_4T^2+20a_5T^3=a_e$ -- (9)

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

$Ax=b$

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

We can get all unknown parameters now

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

$x(t) = a_0+a_1t+a_2t^2+a_3t^3+a_4t^4+a_5t^5$ --(10)

$y(t) = b_0+b_1t+b_2t^2+b_3t^3+b_4t^4+b_5t^5$ --(11)

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.

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

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

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