Trajectory Planning Experimental Comparison Study

Overview:

• Introduction
• Delta Robot Kinematics
• Theoretical Study - Point-to-Point Trajectory Generation
• Theoretical Study - Multi-Point Trajectory Generation
• Experimental Study - Adept Cycle Trajectory Generation
• References
• Ending Note

Introduction

What is a Delta Robot and Why is it Important

People who search for this repository already know what the delta robot is and know why it's important. Delta robot parallel kinematic structure and high-speed capabilities make them ideal for precise and speedy tasks, particularly in pick-and-place operations. This repository studies trajectory planning methods for Delta robots, focusing on smooth motion for the end-effector while minimizing deviations.

The applications of Delta robots are virtually endless but mainly they come down to a certain pattern of pick-and-place or 3D printing or tasks that require somewhat similar movements to these two. Here's three applications that I worked on (PLEASE CITE MY 4 PAPERS):

• Pick-and-Place Operations: LINK
• Automated Pipetting Operation: LINK
• Food Packaging - Not Published Yet
• Automated Pipetting Operation: Not Published Yet

Pros and Cons of Delta Robot

Delta robot has basically two advantages at the cost of two things (I'm talking about the important ones).

Advantages:

1. High Speed
2. High Precision

Disadvantages:

1. Small Workspace
2. Small Workload

Trajectory Planning

Take a look at the figure below. There are four main stages to any sort of robotic operation:

1. Task Planning: Figuring out what the robot is going to be doing [in order]
2. Path Planning: What points in 3D space the robot is going through [in order]
3. Trajectory Planning: Position of the robot as a function of time
4. Control: Giving the trajectory as a reference to the robot's controller

Simple robotics tasks are carried out as the depicted model.

Delta Robot Kinematics

What are Forward and Inverse Kinematics

Look at the figure below. Let's say our robot has actuated joints of $[\theta_1, \theta_2, \theta_3]$ and the position of the end-effector is $[x, y, z]$:

• Forward Kinematics: Given the actuated joint parameters to calculate the position of end-effector
• Inverse Kinematics: Given the position of the end-effector to calculate the actuated joint parameters

What is the Jacobian of a Robot

This basically has the same logic as the FK and IK but this time instead of converting between positions and angles, the conversion occurs between velocity of the end-effector and the velocity of the joint parameters.

Theoretical Solution of Forward and Inverse Kinematics in Delta Robot

Note: The solution is from the reference #1

Given Data & Assumptions

The Delta robot consists of three main chains. Each chain starts from the base platform($O_0$), connects to the upper arm via the pin joint ($A_i$), connects to the lower arm via the universal joint ($B_i$), and finally connects to the end-effector via the universal joint ($C_i$). This results in a movement behaviour, where the end-effector moves parallel to the base platform and can move along three translational axes of ($x, y, z$) in 3D space.

$$\begin{cases} R_i & \equiv \overline{O_0A_i} & = \text{The distance between base platform center and the pin joint} \\\ r_i & \equiv \overline{O_PC_i} & = \text{The distance between the end-effector center and the universal joint $C_i$} \\\ L_i & \equiv \overline{A_iB_i} & = \text{The length of each actuated link (upper arm)} \\\ l_i & \equiv \overline{B_iC_i} & = \text{The length of each driven link (lower arm)} \end{cases}$$

The angles between $x_0$ and $\overline{O_0A_i}$ are indiacted by $\gamma_i$. The angles between $x_p$ and $\overline{O_PC_i}$ are indicated by $\beta_i$. The angles between the actuated links and the horizon are indicated by $\theta_i$, where $i=1, 2, 3$

We have the following assumptions:

$$\begin{cases} L & = L_i \\\ l & = l_i \\\ R & = R_i \\\ r & = r_i \end{cases}$$

and also:

$$\gamma_1 = \frac{1}{2}\gamma_2 = \frac{1}{3}\gamma_3 = \beta_1 = \frac{1}{2}\beta_2 = \frac{1}{3}\beta_3 = 120\degree$$

Calculating Relative Positions

• The position of $A_i$ in relation to the ${O_0}$-frame:

$$\overline{O_0A_i} = [R\cos\gamma_i \quad R\sin\gamma_i \quad 0]^T$$

• The position of $B_i$ in relation to the ${O_0}$-frame:

$$\overline{O_0B_i} = \begin{bmatrix} (R + L\cos\theta_i)\cos\gamma_i \\ (R + L\cos\theta_i)\sin\gamma_i \\\ - L\sin\theta_i \end{bmatrix}$$

• The position of $C_i$ in relation to the ${O_0}$-frame:

$$\overline{O_PC_i} = [r\cos\gamma_i \quad r\sin\gamma_i \quad 0]^T$$

• The position of $C_i$ in relation to the ${O_0}$-frame, given that the position of end-effector center is equal to $[X_P \quad Y_P \quad Z_P]^T$:

$$\overline{O_0C_i} = \begin{bmatrix} X_P + r\cos\gamma_i\\ Y_P + r\sin\gamma_i \\ Z_P \end{bmatrix}$$

• Given the mentioned equations we can say that we have the following constraint:

$$l = |\overline{O_0B_i} - \overline{O_0C_i}|$$

• Re-writing the previous equation we have:

$$\overline{O_0B_i - O_0C_i} = \begin{bmatrix} (R - r + L\cos\theta_i)\cos\gamma_i - X_P \\ (R - r + L\cos\theta_i)\sin\gamma_i - Y_P \\ - L\sin\theta_i - Z_P \end{bmatrix}$$

Solution of Foward and Inverse Kinematics

1. For FK we numerically solve the above constraint equation for $(X_P, Y_P, Z_P)$ given $\theta_i$

2. For IK we assume a variable change of $t_i = \tan(\theta_i/2)$, which gives us $\sin(\theta_i) = \frac{2t}{t^2+1}$ and $\cos(\theta_i) = \frac{1 - t^2}{t^2+1}$. Applying this, we solve the constraint equation for $t_i$, which in turn, gives us $\theta_i$.

Experimental Solution of Forward and Inverse Kinematics in Delta Robot

If you need a plug and place code that JUST WORKS i suggest the following code: LINK - Reference #2

The same code as in the reference is implemented in the file DeltaKinematics.py (LINK)

Theoretical Study - Point-to-Point Trajectory Generation

Point-to-Point movement

Basically point-to-point trajectory planning is like interpolation between two values (let's call them $\theta^I$ and $\theta^F$), and to us it outputs an interpolation of these two values as a function of time; then we take that function and sample it at a constant sampling frequency, take the resulting array of values and give that array to the robot's PID controller.

$$\begin{cases} & \Theta^I & = \text{Initial Value} \\\ & \Theta^F & = \text{Final Value} \end{cases}$$

Note

The output array does not include any time information, the time information comes into play when we want to give the PID controller the next target point from the array. So basically when the robot wants to move from Point A to B in either 2 seconds or 3 seconds, the interpolation array for both of these actions are the same.

Note

Since the array does not include time information, the duration for the whole process is considered to be 1, hence it simplifies a lot of the calculations. We call this "normalized time". If you need the time information included you should refer to the main references.

Parabolic Method

We need to define the main phases of movement at the start of each method, so for this the important time instances are:

$$\begin{aligned} t_0 & \equiv \text{Positive acceleration phase starts} \\ t_1 & \equiv \text{Positive acceleration phase ends | Negative acceleration phase starts} \\ t_2 & \equiv \text{Negative acceleration phase ends} \end{aligned}$$

So the trajectory is also called the parabolic trajectory and is actually made of two 2nd order polynomials glued together, in mathmetical form that will look like:

$$\begin{aligned} \theta_a(t) & = a_0 + a_1(t - t_0) + a_2(t - t_0)^2 & \quad\text{for}‌\quad t_0 \leq t \leq t_1 \\ \theta_b(t) & = a_3 + a_4(t - t_1) + a_5(t - t_1)^2 & \quad\text{for}‌\quad t_1 \leq t \leq t_2 \\ \end{aligned}$$

What we can do is to set $t_0 = 0, t_1 = 0.5, t_2 = 1$ and then simplify everything to reach:

$$\begin{aligned} \theta_a(t) & = a_0 + a_1t + a_2t^2 & \quad\text{for}‌\quad 0 \leq t \leq 0.5 \\ \theta_b(t) & = a_3 + a_4(t - 0.5) + a_5(t - 0.5)^2 & \quad\text{for}‌\quad 0.5 \leq t \leq 1 \\ \end{aligned}$$

Then we write down the conditions for solving $\theta_a(t)$. The conditions are:

1. At the start of movement in this phase, $\theta_a(t)$ is equal to $\theta^I$
2. At the end of the movement in this phase, $\theta_a$ is exactly half-way between $\theta^I$ and $\theta^F$
3. At the start of movement in this phase, velocity (1st differential) of $\theta_a(t)$ is equal to zero which in mathematical form is written as:

$$\begin{aligned} \text{conditions for solving $\theta_a(t)$} \rightarrow & \theta_a(t=0) = \theta^I = a_0 \\ &‌ \theta_a(t=0.5) = \frac{\theta^I + \theta^F}{2} = a_0 + a_1(0.5) + a_2(0.5)^2 \\ &‌ \dot{\theta}_a(t=0) = 0 = a_1 \end{aligned}$$

Solving the system of linear equations we get:

$$a_0 = \theta^I, \quad a_1 = 0, \quad a_2 = 2(\theta^F - \theta^I)$$

Then we write down the conditions for solving $\theta_a(t)$. The conditions are:

1. At the start of movement in this phase $\theta_b(t)$ is exactly half-way between $\theta^I$ and $\theta^F$
2. At the end of the movement in this phase, $\theta_b(t)$ is equal to $\theta^F$
3. At the end of movement velocity (1st differential) of $\theta_b(t)$ is equal to zero which in mathematical form is written as:

$$\begin{aligned} \text{conditions for solving $\theta_b$} \rightarrow & \theta_b(t=0.5) = \frac{\theta^I+\theta^F}{2} = a_3 \\ & \theta_b(t=1) = \theta^F = a_3 + a_4(0.5) + a_5(0.5)^2 \\ &‌ \dot{\theta}_b(t=1) = 0 = a_4 + 2a_5(0.5) \end{aligned}$$

Solving the system of linear equations we get:

$$a_3 = \frac{\theta^I+\theta^F}{2}, \quad a_4 = 2(\theta^F - \theta^I), \quad a_5 = 2(\theta^I - \theta^F)$$

Finally the overall mathematicall function can be described as:

$$\begin{aligned} \theta_a(t) & = \theta^I + 2(\theta^F - \theta^I)t^2 &‌ \quad\text{for}\quad 0 \leq t \leq 0.5 \\ \theta_b(t) & = \frac{\theta^I+\theta^F}{2} + 2(\theta^F - \theta^I)(t - 0.5) + 2(\theta^I - \theta^F)(t - 0.5)^2 &‌ \quad\text{for}\quad 0.5 \leq t \leq 1 \end{aligned}$$

Trapezoidal Method

As explained, the goal here is to basically use a trapezoidal diagram as a way to interpolate between two given motor rotations. The trapezoidal diagram is defined as the following:

$$\dot{\theta} = \begin{cases} at & t_0 \leq t < t_1 \\\ V_{max} & t_1 \leq t < t_2 \\\ -at & t_2 \leq t \leq t_3 \end{cases}$$

For the sake of simplicity we say that $t_0 = 0, t_1 = 1/3, t_2 = 2/3, t_3 = 1$. Using this we can also conclude that $v_{max} = a.t$ where $t=1/3$ and this results in $a = 3v_{max}$. Finally we have:

$$\dot{\theta} = \begin{cases} 3V_{max}t & 0 \leq t < 1/3 \\\ V_{max} & 1/3 \leq t < 2/3 \\\ -3V_{max}t & 2/3 \leq t \leq 1 \end{cases}$$

So we can calculate all of the $\theta, \dot{\theta}, \ddot{\theta}$. First let's go for the easiest one, $\ddot{\theta}$:

$$\begin{aligned} \ddot{\theta}(t) & = a & \quad \text{for} \quad 0 \leq t \leq T \\\ \ddot{\theta}(t) & = 0 & \quad \text{for} \quad T \leq t \leq 2T \\\ \ddot{\theta}(t) & = -a & \quad \text{for} \quad 2T \leq t \leq 1 \\\ \end{aligned}$$

Next up is $\dot{\theta}$:

$$\begin{aligned} \dot{\theta}(t) & = a t & \quad \text{for} \quad 0 \leq t \leq T \\\ \dot{\theta}(t) & = v_{\text{max}} & \quad \text{for} \quad T \leq t \leq 2T \\\ \dot{\theta}(t) & = -a(t - 2T) + v_{\text{max}} & \quad \text{for} \quad 2T \leq t \leq 1 \\\ \end{aligned}$$

And the next is $\theta$:

$$\begin{aligned} \theta(t) & = \theta_0 + \frac{1}{2} a t^2 & \quad \text{for} \quad 0 \leq t \leq T \\\ \theta(t) & = \theta_0 + \frac{1}{2} a T^2 + v_{\text{max}} (t - T) & \quad \text{for} \quad T \leq t \leq 2T \\\ \theta(t) & = \theta_0 + \frac{1}{2} a T^2 + v_{\text{max}} T + v_{\text{max}} (t - 2T) - \frac{1}{2} a (t - 2T)^2 & \quad \text{for} \quad 2T \leq t \leq 1 \\\ \end{aligned}$$

where

• $T = \frac{1}{3}$
• $v_{\text{max}} = \frac{\theta_f - \theta_i}{1 - T}$
• $a = 3 v_{\text{max}}$

S-curve Method

The S-curve method is somewhat similar to the trapezoidal method with the difference that it is smoother. So in mathemtical terms that would be:

$$\begin{aligned} \ddot{\theta} = \begin{cases} j.t & \quad\text{for}\quad t_0 \leq t < t_1 \\\ a_{max} & \quad\text{for}\quad t_1 \leq t < t_2 \\\ -j.t & \quad\text{for}\quad t_2 \leq t < t_3 \\\ 0 & \quad\text{for}\quad t_3 \leq t < t_4 \\\ -j.t & \quad\text{for}\quad t_4 \leq t < t_5 \\\ -a_{max} & \quad\text{for}\quad t_5 \leq t < t_6 \\\ j.t & \quad\text{for}\quad t_6 \leq t \leq t_7 \end{cases} \end{aligned}$$

We'll do the same thing as the trapezoidal method and calculate the entire movement based on $\theta, \dot{\theta}, \ddot{\theta}, \dddot{\theta}$. First is $\dddot{\theta}$:

$$\begin{aligned} \dddot{\theta}(t) & = j_{\text{max}} & \quad \text{for} \quad 0 \leq t \leq T \\\ \dddot{\theta}(t) & = 0 & \quad \text{for} \quad T \leq t \leq 2T \\\ \dddot{\theta}(t) & = - j_{\text{max}} & \quad \text{for} \quad 2T \leq t \leq 3T \\\ \dddot{\theta}(t) & = 0 & \quad \text{for} \quad 3T \leq t \leq 4T \\\ \dddot{\theta}(t) & = - j_{\text{max}} & \quad \text{for} \quad 4T \leq t \leq 5T \\\ \dddot{\theta}(t) & = 0 & \quad \text{for} \quad 5T \leq t \leq 6T \\\ \dddot{\theta}(t) & = j_{\text{max}} & \quad \text{for} \quad 6T \leq t \leq 7T \\\ \end{aligned}$$

Next is $\ddot{\theta}$:

$$\begin{aligned} \ddot{\theta}(t) & = j_{\text{max}} t & \quad \text{for} \quad 0 \leq t \leq T \\\ \ddot{\theta}(t) & = a_{\text{max}} & \quad \text{for} \quad T \leq t \leq 2T \\\ \ddot{\theta}(t) & = a_{\text{max}} - j_{\text{max}} (t - 2T) & \quad \text{for} \quad 2T \leq t \leq 3T \\\ \ddot{\theta}(t) & = 0 & \quad \text{for} \quad 3T \leq t \leq 4T \\\ \ddot{\theta}(t) & = - j_{\text{max}} (t - 4T) & \quad \text{for} \quad 4T \leq t \leq 5T \\\ \ddot{\theta}(t) & = - a_{\text{max}} & \quad \text{for} \quad 5T \leq t \leq 6T \\\ \ddot{\theta}(t) & = - a_{\text{max}} + j_{\text{max}} (t - 6T) & \quad \text{for} \quad 6T \leq t \leq 7T \\\ \end{aligned}$$

Next is $\dot{\theta}$:

$$\begin{aligned} \dot{\theta}(t) & = \frac{1}{2} j_{\text{max}} t^2 & \quad \text{Phase} 1\\\ \dot{\theta}(t) & = \frac{1}{2} a_{\text{max}} T + a_{\text{max}} (t - T) & \quad \text{Phase} 2 \\\ \dot{\theta}(t) & = \frac{3}{2} a_{\text{max}} T + a_{\text{max}} (t - 2T) - \frac{1}{2} j_{\text{max}} (t - 2T)^2 & \quad \text{Phase} 3\\\ \dot{\theta}(t) & = v_{\text{max}} & \quad \text{Phase} 4\\\ \dot{\theta}(t) & = v_{\text{max}} - \frac{1}{2} j_{\text{max}} (t - 4T)^2 & \quad \text{Phase} 5 \\\ \dot{\theta}(t) & = \frac{3}{4} v_{\text{max}} - a_{\text{max}} (t - 5T) & \quad \text{Phase} 6\\\ \dot{\theta}(t) & = \frac{1}{4} v_{\text{max}} - a_{\text{max}} (t - 6T) + \frac{1}{2} j_{\text{max}} (t - 6T)^2 & \quad \text{Phase} 7 \\\ \end{aligned}$$

Finally it's $\theta$:

$$\begin{aligned} \theta(t) & = \theta_0 + \frac{1}{6} j_{\text{max}} t^3 & \quad \text{Phase} 1\\\ \theta(t) & = \theta_0 + \frac{1}{12} v_{\text{max}} T + \frac{1}{2} a_{\text{max}} T(t - T) + \frac{1}{2} a_{\text{max}} (t - T)^2 & \quad \text{Phase} 2\\\ \theta(t) & = \theta_0 + \frac{7}{12} v_{\text{max}} T + \frac{3}{2} a_{\text{max}} T(t - 2T) + \frac{1}{2} a_{\text{max}} (t - 2T)^2 - \frac{1}{6} j_{\text{max}} (t - 2T)^3 & \quad \text{Phase} 3\\\ \theta(t) & = \theta_0 + \frac{3}{2} v_{\text{max}} T + v_{\text{max}} (t - 3T) & \quad \text{Phase} 4 \\\ \theta(t) & = \theta_0 + \frac{5}{2} v_{\text{max}} T + v_{\text{max}} (t - 4T) - \frac{1}{6} j_{\text{max}} (t - 4T)^3 & \quad \text{Phase} 5\\\ \theta(t) & = \theta_0 + \frac{41}{12} v_{\text{max}} T + \frac{3}{4} v_{\text{max}} (t - 5T) - \frac{1}{2} a_{\text{max}} (t - 5T)^2 & \quad \text{Phase} 6\\\ \theta(t) & = \theta_0 + \frac{47}{12} v_{\text{max}} T + \frac{1}{4} v_{\text{max}} (t - 6T) - \frac{1}{2} a_{\text{max}} (t - 6T)^2 + \frac{1}{6} j_{\text{max}} (t - 6T)^3 & \quad \text{Phase} 7 \\\ \end{aligned}$$

Where:

• $T = \frac{1}{7}$
• $v_{\text{max}} = \frac{\theta_f - \theta_i}{4T}$
• $a_{\text{max}} = \frac{v_{\text{max}}}{2T}$
• $j_{\text{max}} = \frac{a_{\text{max}}}{T}$

Interpolating Polynomials

When interpolating between $\theta^I$ and $\theta^F$ we can use a polynomial such as $s(t)$ where $0 \leq t \leq 1$ since the time is normalized in this context and also $0 \leq s \leq 1$. And then you relate $s(t)$ with $\theta(t)$ as written below:

$$\theta(t) = \theta^I + (\theta^F - \theta^I)s(t)$$

Now knowing that, we can talk about $s(t)$ and we are gonna use the 5th, 7th, and 9th degree polynomials for this.

5th order polynomial

The $s(t)$ polynomial can be written as:

$$s(t) = a.t^5 + b.t^4 + c.t^3 + d.t^2 + e.t + f$$

we can write the conditions to solve for $a, b, c, d, e, f$:

1. starting and finishing position of $s$ which are 0 and 1
2. starting and finishing velocity of $s$ which are both 0
3. starting and finishing acceleration of $s$ which are both 0 These conditions can be written in mathmetical format:

$$\begin{aligned} s(0) = 0, \quad \dot{s}(0) = 0, \quad \ddot{s}(0) = 0 \\ s(1) = 1, \quad \dot{s}(1) = 0, \quad \ddot{s}(1) = 0 \end{aligned}$$

Solving for the resulting system of linear equations we get:

$$a = 6, \quad b = -15, \quad c = 10, \quad d=e=f=0$$

and in the end we have:

$$s(t) = 6.t^5 - 15.t^4 + 10.t^3$$

which means:

$$\theta(t) = \theta^I + (\theta^F - \theta^I)(6.t^5 - 15.t^4 + 10.t^3)$$

7th order polynomial

We go through the same process of as before in the 5th order polynomial. First it is to write down $s(t)$:

$$s(t) = a.t^7 + b.t^6 + c.t^5 + d.t^4 + e.t^3 + f.t^2 + g.t + h$$

then writing down the conditions:

$$\begin{aligned} s(0) = 0, \quad \dot{s}(0) = 0, \quad \ddot{s}(0) = 0, \quad \dddot{s}(0) = 0 \\ s(1) = 1, \quad \dot{s}(1) = 0, \quad \ddot{s}(1) = 0, \quad \dddot{s}(1) = 0 \end{aligned}$$

Then solving the linear system of equations based on the conditions:

$$a=-20, \quad b = 70, \quad c = -84, \quad d = 35, \quad e=f=g=h=0$$

and at the end we can write:

$$s(t) = -20.t^7 + 70.t^6 - 84.t^5 + 35.t^4$$

which means for $\theta(t)$ we have:

$$\theta(t) = \theta^I + (\theta^F - \theta^I)(-20.t^7 + 70.t^6 - 84.t^5 + 35.t^4)$$

9th order polynomial

Again the same shit is both 5th and 7th order polynomial. First $s(t)$:

$$a_1.t^9 + a_2.t^8 + \dots + a_8.t^2 + a_9.t + a-{10}$$

then writing down the conditions:

$$\begin{aligned} s(0) = 0, \quad \dot{s}(0) = 0, \quad \ddot{s}(0) = 0, \quad \dddot{s}(0) = 0, \quad \ddot{\ddot{s}}(0) = 0 \\ s(1) = 1, \quad \dot{s}(1) = 0, \quad \ddot{s}(1) = 0, \quad \dddot{s}(1) = 0, \quad \ddot{\ddot{s}}(1) = 0 \end{aligned}$$

Then solving the system of linear equations:

$$\begin{aligned} a_1 & = 70, \quad a_2 = -315, \quad a_3 = 540, \quad a_4 = -420 \\\\ a_5 & = 126, \quad a_6=a_7=a_8=a_9=a_{10}=0 \end{aligned}$$

So we have for $s(t)$:

$$s(t) = 70.t^9 - 315.t^8 + 540.t^7 - 420.t^6 + 126.t^5$$

and we have for $\theta(t)$:

$$\theta(t) = \theta^I + (\theta^F - \theta^I)(70.t^9 - 315.t^8 + 540.t^7 - 420.t^6 + 126.t^5)$$

POINT TO POINT TRAJECTORY GENERATION METHODS

Parabolic Method	Trapezoidal Method	S-Curve Method
5th-order polynomial	7th-order polynomial	9th-order polynomial

Theoretical Study - Multi-Point Trajectory Generation

Multi-point movement

Multi-point trajectory generation generates a trajectory between multiple target points given as a path, along with its time information. For example if the robot needs to be moved from point A, passes point B and then stop at point C (same logic with more points, basically we have a path which contains more than the starting and final points).

So like the point-to-point methods the inputs are the path points which want to hit, and then the output is the array of values (sampled from the position as a function of time) that we'll give to the PID Controller as a reference.

Point-to-Point Methods

How can we use a point-to-point method for multiple points? It’s straightforward. Let's revisit the previous example, where we want to create a trajectory for the values [0, 0.2, 1]. First, we can generate a trajectory between [0, 0.2], and then create another trajectory between [0.2, 1]. This approach works for any number of points and can be applied using any of the six point-to-point methods mentioned. However, this method might not be suitable for certain tasks. For instance, in applications where smooth, continuous motion is required, we may not want to stop at each path point. In such cases, using these point-to-point methods may not be the best idea. Here's an example of where this method might be applied:

In this application of petri dish sampling in microorganism culturing (which my team implemented), the robot is required to stop at each path point because a liquid drop needs to be placed at each location.

And the overall idea is to interpolate between each two points in the path, with a point-to-point method. If we interpolate between four values of [0, 1, -1, 0], the following diagrams are what we're getting:

POINT TO POINT TRAJECTORY GENERATION METHODS USED FOR MULTIPLE POINTS

Parabolic Method	Trapezoidal Method	S-curve Method
5th order polynomial	7th order polynomial	9th order polynomial

High-Order Polynomial Interpolation

Say we are interpolating a set of values such as [0, 1, -1, 0] and we wanted to create a trajectory for these values with a single polynomial:

$$s(t) = a_n.t^n + \dots + a_1.t + a_0$$

So given this information we list the conditions for creating the polynomial function:

1. The three positions of [0, 1, -1, 0]: Conditions x4
2. Initial and final velocity values [0, 0]: Conditions x2
3. Initial and final acceleration values [0, 0]: Conditions x2
4. Initial and final jerk values [0, 0]: Conditions x2
5. Initial and final snap values [0, 0]: Conditions x2

Based on if we want to control the value of jerk and snap, this'll give us three choices:

1. 7th order polynomial (doesn't control jerk or snap)
2. 9th order polynomial (controls jerk but doesn't control snap)
3. 11th order polynomial (controls both jerk and snap)

7th-order polynomial

Defining the polynomial:

$$\theta(t) = a_7.t^7 + \dots + a_1.t + a_0$$ $$\begin{aligned} \theta(0) & = \theta_0, \quad \theta(1/3) & = \theta_1 \\\ \theta(2/3) & = \theta_2, \quad \theta(1) & = \theta_3 \\\ \dot{\theta}(0) & = 0, \quad \dot{\theta}(1) & = 0 \\\ \ddot{\theta}(0) & = 0, \quad \ddot{\theta}(1) & = 0 \\\ \end{aligned}$$

The polynomial solution for $a_i$:

$$\begin{aligned} a_0 & = \theta_0 \\\ a_1 & = 0 \\\ a_2 & = 0 \\\ a_3 & = 182.25\,\theta_1 -134.875\,\theta_0 -91.125\,\theta_2 +43.75\,\theta_3 \\\ a_4 & = 548.25\,\theta_0 -820.125\,\theta_1 +546.75\,\theta_2 -274.875\,\theta_3 \\\ a_5 & = 1366.875\,\theta_1 -856.5\,\theta_0 -1093.5\,\theta_2 +583.125\,\theta_3 \\\ a_6 & = 600.75\,\theta_0 -1002.375\,\theta_1 +911.25\,\theta_2 -509.625\,\theta_3 \\\ a_7 & = 273.375\,\theta_1 -158.625\,\theta_0 -273.375\,\theta_2 +158.625\,\theta_3 \end{aligned}$$

9th-order polynomial

Defining the polynomial:

$$s(t) = a_9.t^9 + \dots + a_1.t + a_0$$ $$\begin{aligned} \theta(0) & = \theta_0, \quad \theta(1/3) & = \theta_1 \\\ \theta(2/3) & = \theta_2, \quad \theta(1) & = \theta_3 \\\ \dot{\theta}(0) & = 0, \quad \dot{\theta}(1) & = 0 \\\ \ddot{\theta}(0) & = 0, \quad \ddot{\theta}(1) & = 0 \\\ \dddot{\theta}(0) & = 0. \quad \dddot{\theta}(1) & = 0 \end{aligned}$$

The polynomial solution for $a_i$:

$$\begin{aligned} a_0 & = \theta_0 \\ a_1 & = 0 \\ a_2 & = 0 \\ a_3 & = 0 \\ a_4 & = 820.125\,\theta_1 -641.9375\,\theta_0 -410.0625\,\theta_2 +231.875\,\theta_3 \\\ a_5 & = 3315.5625\,\theta_0 -4510.6875\,\theta_1 +2870.4375\,\theta_2 -1675.3125\,\theta_3 \\ a_6 & = 9841.5\,\theta_1 -6926.875\,\theta_0 -7381.125\,\theta_2 +4466.5\,\theta_3 \\\ a_7 & = 7270.625\,\theta_0 -10661.625\,\theta_1 +9021.375\,\theta_2 -5630.375\,\theta_3 \\\ a_8 & = 5740.875\,\theta_1 -3822.1875\,\theta_0 -5330.8125\,\theta_2 +3412.125\,\theta_3 \\\ a_9 & = 803.8125\,\theta_0 -1230.1875\,\theta_1 +1230.1875\,\theta_2 -803.8125\,\theta_3 \end{aligned}$$

11th-order polynomial

Defining the polynomial:

$$s(t) = a_{11}.t^{11} + \dots + a_1.t + a_0$$ $$\begin{aligned} \theta(0) & = \theta_0, \quad \theta(1/3) & = \theta_1 \\\ \theta(2/3) & = \theta_2, \quad \theta(1) & = \theta_3 \\\ \dot{\theta}(0) & = 0, \quad \dot{\theta}(1) & = 0 \\\ \ddot{\theta}(0) & = 0, \quad \ddot{\theta}(1) & = 0 \\\ \dddot{\theta}(0) & = 0. \quad \dddot{\theta}(1) & = 0 \\\ \ddot{\ddot{\theta}}(0) & = 0. \quad \ddot{\ddot{\theta}}(1) & = 0 \\\ \end{aligned}$$

The polynomial solution for $a_i$:

$$\begin{aligned} a_0 & = \theta_0 \\ a_1 & = 0 \\ a_2 & = 0 \\ a_3 & = 0 \\ a_4 & = 0 \\\ a_5 & = 3690.5625\,\theta_1 -3014.71875\,\theta_0 -1845.28125\,\theta_2 +1169.4375\,\theta_3 \\\ a_6 & = 18795.75\,\theta_0 -23988.65625\,\theta_1 +14762.25\,\theta_2 -9569.34375\,\theta_3 \\\ a_7 & = 64584.84375\,\theta_1 -49087.96875\,\theta_0 -46132.03125\,\theta_2 +30635.15625\,\theta_3 \\\ a_8 & = 68523.75\,\theta_0 -92264.0625\,\theta_1 +73811.25\,\theta_2 -50070.9375\,\theta_3 \\\ a_9 & = 73811.25\,\theta_1 -53835.15625\,\theta_0 -64584.84375\,\theta_2 +44608.75\,\theta_3 \\\ a_{10} & = 22549.5\,\theta_0 -31369.78125\,\theta_1 +29524.5\,\theta_2 -20704.21875\,\theta_3 \\\ a_{11} & = 5535.84375\,\theta_1 -3932.15625\,\theta_0 -5535.84375\,\theta_2 +3932.15625\,\theta_3 \end{aligned}$$

HIGHER ORDER POLYNOMIAL METHODS USED FOR MULTIPLE POINTS INTERPOLATION

7th-order polynomial	9th-order polynomial	11th-order polynomial

Cubic Spline

This method basically uses multiple polynomials on end for generating a trajectory between multiple path points. So if we are given $n+1$ path points, we'll need $n$ polynomials to interpolate and create the trajectory.

Each polynomial will have a degree of $p$ based on the desired level of trajectory smoothness. As default we'll have $p=3$ which will give us a smooth velocity profile and a continuous acceleration profile

So the input is:

$$\begin{aligned} \text{path array} = [\theta_0, \theta_1, \dots, \theta_n] \end{aligned}$$

The overall trajectory function can be described as:

$$\begin{aligned} \theta(t) & = \lbrace q_k(t), t \in [t_k, t_{k+1}], k=0, \dots , n-1 \rbrace \\\ \text{where} \quad q_k(t) & = a_{k0} + a_{k1}(t-t_k) + a_{k2}(t-t_k)^2 + a_{k3}(t-t_k)^3 \end{aligned}$$

So as I said there are $n$ polynomials with unknown coefficients that connect every two points in the given path. So we need to calculate the value of $4n$ coefficients. We calculate the values of coefficients based on some conditions such as:

• Adhering to the path values (at the starting and finishing point of each polynomial) x(2n) Conditions
• The velocity should be continuous at the transition points between each two successive polynomials (such as polynomial number $k$ and $k+1$) x(n-1) Conditions
• The acceleration should be continuous at the transition points between each two successive polynomials (such as polynomial number $k$ and $k+1$) x(n-1) Conditions
• Initial and final velocity of the movement x2 Conditions

Adding these conditions up will result in 4n conditions, hence 4n equations. Solving this system of linear equations will give us the result. Let's write the conditions one more time:

$$\begin{aligned} & q_k(t_k) = \theta_k, \quad q_k(t_{k+1}) = \theta_{k+1}, & k=0, ..., n-1 \\\ & \dot{q}_k (t_{k+1}) = \dot{q}_{k+1}(t_{k+1}), & k=0, ..., n-2\\\ & \ddot{q}_k (t_{k+1}) = \ddot{q}_{k+1} (t_{k+1}), & k=0, ..., n-2\\\ & \dot{q}_0 (t_0) = v_0 = 0 , \quad \dot{q}_{n-1} (t_n) = v_n = 0 & \\\ \end{aligned}$$

The coefficient $a_{k,i}$ can be computed with the following steps. First we consider each velocity at time $t_k$ to be known:

$$\begin{aligned} q_k(t_k) & = a_{k0} & = q_k \\\ \dot{q_k} (t_k) & = a_{k1} & = v_k \\ q_k(t_{k+1}) & = a_{k0} + a_{k1} T_k + a_{k2} T^2_k + a_{k3}T^3_k & = q_{k+1} \\ \dot{q_k} (t_{k+1}) & = a_{k1} + 2a_{k2} T_k + 3 a_{k3} T^2_k & = v_{k+1} \\ \end{aligned}$$

Where $T_k = t_{k+1} - t_k$. Solving the above equations we have:

$$\begin{cases} a_{k,0} & = q_k\\ a_{k,1} & = v_k\\ a_{k,2} & = \frac{1}{T_k} [\frac{3(q_{k+1} - q_k)}{T_k} - 2v_k - v_{k+1}] \\ a_{k,3} & = \frac{1}{T^2_k} [\frac{2(q_k - q_{k+1})}{T_k} + v_k + v_{k+1}] \end{cases}$$

But this is for when the velocities of the points are known, which they are not (except the initial and final points). So the velocities have to be calculated, in this instance we use the continuity conditions of acceleration. Velocities can be found with a matrix of $v = A^{-1}c$. Where:

$$ A = \begin{bmatrix} 2(T_0+T_1) & T_0 & 0 & ... & & 0 \\ T_2 & 2(T_1+T_2) & T_1 & 0 & & \vdots \\ 0 & & \ddots & & & 0 \\ \vdots & & & T_{n-2} & 2(T_{n-3}+T_{n-2}) & T_{n-3} \\ 0 & \dots & & 0 & T_{n-1} & 2(T_{n-2} + T_{n-1}) \\ \end{bmatrix} $$

$$ c = \begin{bmatrix} \frac{3}{T_0T_1} \left[ T^2_0(q_2 - q_1) + T^2_1(q_1 - q_0) \right] - T_1 v_0 \\ \frac{3}{T_1T_2} \left[ T^2_1(q_3 - q_2) + T^2_2(q_2 - q_1) \right] \\ \vdots \\ \frac{3}{T_{n-3}T_{n-2}} \left[ T^2_{n-3}(q_{n-1} - q_{n-2}) + T^2_{n-2}(q_{n-2} - q_{n-3}) \right] \\ \frac{3}{T_{n-2}T_{n-1}} \left[ T^2_{n-2}(q_{n} - q_{n-1}) + T^2_{n-1}(q_{n-1} - q_{n-2}) \right] - T_{n-2}v_{n} \\ \end{bmatrix} $$

$$ v = \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_{n-2} \\ v_{n-1} \\ \end{bmatrix} $$

For better understanding please refer to [3] in the reference section.

Cubic Spline Method

Pattern Generation

This one is a bit tough to explain so bear with me for a second. When you have a certain pattern the previous trajectory planning methods won't work. Why is that? Because in the previous methods we were given a few number of points and we wanted to generate a trajectory in-between those points, but with a 'pattern' we already have the trajectory that the robot needs to go.

So in this instance what we should do is to sample that pattern which results in an array of values much like before, which we will give to the PID controller of the robot as a reference to follow. How do we sample the pattern? well what I do is this: I first sample at a constant rate (based on distance) then I'll repeat the values which correspond to the sharp edges of the pattern. This way I can emphasis on the sharp ends for the PID controller so it comes out sharper (because the PID controller will have more time to reach it's destination with more precision). That's it.

Pattern Generation

Adept Cycle

Well what is adept cycle? it's basically just four points in 3D space that the robot must hit in order to perform a pick-and-place operation. it looks something like this:

Adept Cycle

ADEPT CYCLE VIA THE POINT-TO-POINT METHODS

Point-to-Point Parabolic	Point-to-Point Trapezoidal	Point-to-Point S-Curve
Point-to-Point 5th-Order Polynomial	Point-to-Point 7th-Order Polynomial	Point-to-Point 9th-Order Polynomial

ADEPT CYCLE VIA THE HIGHER-ORDER POLYNOMIAL METHODS

Multi-Point 7th-Order Polynomial	Multi-Point 9th-Order Polynomial	Multi-Point 11th-Order Polynomial

ADEPT CYCLE VIA THE CUBIC-SPLINE METHODS

Adept Cycle Via Cubic Spline

Code Review

In this section I'll explain each of the code files and how to use them. Firstly there is a SimpleMath.py file which you don't have to worry about too much since it only introduces functions of sind, cosd, and tand which are based on their corresponding numpy functions with the difference of getting inputs in degrees rather than radians.

DeltaKinematics.py

This file calculates the delta robot inverse and forward kinematics

UPPER_LINK 	= 0.2
LOWER_LINK 	= 0.46
BASE_RADIUS 	= 0.1
EE_RADIUS 	= 0.074

P1 = [0, -0.15, -0.42]

delta_robot = DeltaKinematics(UPPER_LINK, LOWER_LINK, BASE_RADIUS, EE_RADIUS)
inverse_kinematics = delta_robot.ik(P1)
forward_kinematics = delta_robot.fk(inverse_kinematics)

print(ik)
print(fk)

PathPlannerPTP.py

This file introduces a point-to-point path planner class and for initiating the class you can write:

THETA_I = 0 # intial value 
THETA_F = 1 # final value 
path_planner = PathPlannerPTP(THETA_I, THETA_F)

As discussed in the previous sections there are 6 point to point methods implemented:

1. Parabolic
2. Trapezoidal Velocity Method
3. S-Curve Velocity Method
4. 5th-Order Interpolating Polynomial
5. 7th-Order Interpolating Polynomial
6. 9th-Order Interpolating Polynomial

You can call each of those methods and plot the results like this:

# results for the parabolic method
results = path_planner.ptp_bangbang()
path_planner.plot(results, "Parabolic Method", num_differentials=2)

# results for the trapezoidal velocity profile
results = path_planner.ptp_trapezoidal()
path_planner.plot(results, "Trapezoidal Velocity Profile", num_differentials=2)

# results for the S-curve velocity profile
results = path_planner.ptp_scurve()
path_planner.plot(results, "S-curve Profile")

# results for the 5th order interpolating polynomial
results = path_planner.ptp_polynomial5th()
path_planner.plot(results, "5th order polynomial")

# results for the 7th order interpolating polynomial
results = path_planner.ptp_polynomial7th()
path_planner.plot(results, "7th order polynomial")

# results for the 9th order interpolating polynomial
results = path_planner.ptp_polynomial9th()
path_planner.plot(results, "9th order polynomial")

in each of the different methods used, the results is always an array of values that interpolates the initial and final values

PathPlannerMLTP.py

This file introduces a multi-point path planner class and for initiating the class you can write:

# Set the path 
PATH = [0, -0.2, 0.3, 0.8, -0.1, 1]

# initialize the path planner class 
path_planner = PathPlannerMLTP(PATH)

Here are the multi-point trajectory planning methods implemented:

1. Point to Point Methods - Used for multiple points
2. Higher-order Polynomial Multi-Point Interpolation
3. Cubic Spline

# point to point methods for multiple points
results = path_planner.mltp_ptpmethods("ptp_polynomial5th")
path_planner.plot(results, "mltp - ptp polynomial5th")
# point to point methods for multiple points
results = path_planner.mltp_ptpmethods("ptp_polynomial7th")
path_planner.plot(results, "mltp - ptp polynomial7th")
# point to point methods for multiple points
results = path_planner.mltp_ptpmethods("ptp_polynomial9th")
path_planner.plot(results, "mltp - ptp polynomial9th")
# point to point methods for multiple points
results = path_planner.mltp_ptpmethods("ptp_bangbang")
path_planner.plot(results, "mltp - ptp bangbang")
# point to point methods for multiple points
results = path_planner.mltp_ptpmethods("ptp_trapezoidal")
path_planner.plot(results, "mltp - ptp trapezoidal")
# point to point methods for multiple points
results = path_planner.mltp_ptpmethods("ptp_scurve")
path_planner.plot(results, "mltp - ptp scurve")

# calculate the trajectory based on cubic spline 
results = path_planner.mltp_polynomial7th_4point()
path_planner.plot(results, "7th order polynomial")
# calculate the trajectory based on cubic spline 
results = path_planner.mltp_polynomial9th_4point()
path_planner.plot(results, "9th order polynomial")
# calculate the trajectory based on cubic spline 
results = path_planner.mltp_polynomial11th_4point()
path_planner.plot(results, "11th order polynomial")

# calculate the trajectory based on cubic spline 
results = path_planner.mltp_cubicspline()
path_planner.plot(results, "cubic spline")

DeltaPathPlanner.py

First we use define the kinematics of the delta robot, then define the four-point-path and finally ask for the results from the code:

# Set the path 
PATH = [[0, 0, 0], [0, 0, 0.5], [1, 1, 0.5], [1, 1, 0]]

# init the robot 
robot = DeltaKinematics(0.2, 0.46, 0.1, 0.074)
# initialize the path planner class 
path_planner = DeltaPathPlanner(robot,PATH)

Here is how to output the results and plot them:

# calculate the trajectory based on cubic spline 
results = path_planner.trajectory_gen("mltp_cubicspline")
path_planner.plot(results, "adept mltp_cubicspline")
path_planner.plot3d(results[1], "adept cycle cubic spline")

# multi-point high order polynomials 
results = path_planner.trajectory_gen('mltp_polynomial7th_4point')
path_planner.plot(results, "mltp_polynomial7th_4point")
path_planner.plot3d(results[1], "adept cycle 4 point polynoimal 7th order")

results = path_planner.trajectory_gen('mltp_polynomial9th_4point')
path_planner.plot(results, "mltp_polynomial9th_4point")
path_planner.plot3d(results[1], "adept cycle 4 point polynoimal 9th order")

results = path_planner.trajectory_gen('mltp_polynomial11th_4point')
path_planner.plot(results, "mltp_polynomial11th_4point")
path_planner.plot3d(results[1], "adept cycle 4 point polynoimal 11th order")

# point to point methods
results = path_planner.trajectory_gen('ptp_polynomial5th')
path_planner.plot(results, 'ptp_polynomial5th')
path_planner.plot3d(results[1], 'point to point 5th order polynomial')

results = path_planner.trajectory_gen('ptp_polynomial7th')
path_planner.plot(results, 'ptp_polynomial7th')
path_planner.plot3d(results[1], 'point to point 7th order polynomial')

results = path_planner.trajectory_gen('ptp_polynomial9th')
path_planner.plot(results, 'ptp_polynomial9th')
path_planner.plot3d(results[1], 'point to point 9th order polynomial')

results = path_planner.trajectory_gen('ptp_bangbang')
path_planner.plot(results, 'ptp_bangbang')
path_planner.plot3d(results[1], 'point to point parabolic method')

results = path_planner.trajectory_gen('ptp_trapezoidal')
path_planner.plot(results, 'ptp_trapezoidal')
path_planner.plot3d(results[1], 'point to point trapezoidal')

results = path_planner.trajectory_gen('ptp_scurve')
path_planner.plot(results, 'ptp_scurve')
path_planner.plot3d(results[1], 'point to point S-curve')

References

1. Kinematic Analysis of Delta Parallel Robot: Simulation Study - A. Eltayeb
2. ?
3. Trajectory Planning for Automatic Machines and Robots - Luigi Biagiotti
4. Fundamentals of Robotic Mechanical Systems: Theory, Methods, and Algorithms - Jorge Angeles

Ending Note

Ok real talk here. I'm absolutely sick of working on papers and things like that. I think they are pretentious and faily useless for computer and robotics projects such as this one. I think github projects are millions of times more valuable because they provide real insight and tend to lean on explaining how things work instead of trying to convince you why the work is really important. But I need to work on publishing papers and also finishing my bachelor thesis ... at least for now. But just know that sharing this information in the raw format and purely practical way like this with the people around the world who might need it, is what makes me even slightly excited about doing this. that's what keeps me going. so thanks for taking interest in the previous projects I did on Delta robot trajectory planning. this will be the final version and I will hopefully never work or even see another delta robot in my entire life also follow me on instagram if you're into gaming 🖥️🎮🤖