In previous tutorials, we used known joint variables (i.e. servo motor angles, displacement of a linear actuator, etc.) to calculate the position and orientation of the end effector of a robotic arm (e.g. robotic gripper, robotic hand, vacuum suction cup, etc.) in 3D space. This is called forward kinematics.

Forward kinematics asks the question: Where is the end effector of a robot (e.g. gripper, hand, vacuum suction cup, etc.) located in space given that we know the angles of the servo motors?

Inverse kinematics is the forward kinematics problem in reverse. We know the position and orientation we want the end effector of a robotic arm to have, and we want to find the values of the joint variables that generate that desired position and orientation of the end effector.

• Joint Variables —–> Pose of the End effector of a Robotic Arm = Forward Kinematics
• Pose of the End effector of a Robotic Arm —–> Joint Variables = Inverse Kinematics

Inverse kinematics asks the question: What do the angles of the servo motors need to be given our desired position and orientation of the end effector of a robotic arm (e.g. gripper, hand, vacuum suction cup, etc.)?

Cover Image Source: Wikipedia.

Now that we’ve seen how to fill in Denavit-Hartenberg Tables, let’s see how we can use these tables and Python code to calculate homogeneous transformation matrices for robotic arms. Our goal is to then use these homogeneous transformation matrices to determine the position and orientation of the end effector of a robotic arm.

Example 1 – Cartesian Robot

Let’s start by calculating the homogeneous transformation matrix from frame 0 to frame 1.

Here was our derivation of the Denavit-Hartenberg table for the cartesian robot. Let’s enter that one in code. Here is the program in Python:

import numpy as np # Scientific computing library

# Project: Coding Denavit-Hartenberg Tables Using Python - Cartesian Robot
#          This only looks at frame 0 to frame 1.
# Author: Addison Sears-Collins
# Date created: August 21, 2020

# Link lengths in centimeters
a1 = 1 # Length of link 1
a2 = 1 # Length of link 2
a3 = 1 # Length of link 3

# Initialize values for the displacements
d1 = 1 # Displacement of link 1
d2 = 1 # Displacement of link 2
d3 = 1 # Displacement of link 3

# Declare the Denavit-Hartenberg table. 
# It will have four columns, to represent:
# theta, alpha, r, and d
# We have the convert angles to radians.
d_h_table = np.array([[np.deg2rad(90), np.deg2rad(90), 0, a1 + d1],
                      [np.deg2rad(90), np.deg2rad(-90), 0, a2 + d2],
                      [0, 0, 0, a3 + d3]]) 

# Create the homogeneous transformation matrix
homgen_0_1 = np.array([[np.cos(d_h_table[0,0]), -np.sin(d_h_table[0,0]) * np.cos(d_h_table[0,1]), np.sin(d_h_table[0,0]) * np.sin(d_h_table[0,1]), d_h_table[0,2] * np.cos(d_h_table[0,0])],
                      [np.sin(d_h_table[0,0]), np.cos(d_h_table[0,0]) * np.cos(d_h_table[0,1]), -np.cos(d_h_table[0,0]) * np.sin(d_h_table[0,1]), d_h_table[0,2] * np.sin(d_h_table[0,0])],
                      [0, np.sin(d_h_table[0,1]), np.cos(d_h_table[0,1]), d_h_table[0,3]],
                      [0, 0, 0, 1]])  

# Print the homogeneous transformation matrix from frame 1 to frame 0.
print(homgen_0_1)

Run the program.

Here is the output:

The output is saying that the rotation matrix from frame 0 to frame 1 is as follows:

This is exactly what I got on an earlier post where I derived the rotation matrix from frame 0 to frame 1 for a cartesian robot.

You can also see the displacement vector from frame 1 to frame 0, the 3×1 vector on the right side of the matrix. The vector makes sense because it is saying that the position of frame 1 relative to frame 0 in the z₀ direction is 2 units (because a₁ = 1 and d₁ = 1):

Now, let’s add the two other homogeneous transformation matrices: from frame 1 to 2 and from frame 2 to 3.

Here is the code:

import numpy as np # Scientific computing library

# Project: Coding Denavit-Hartenberg Tables Using Python - Cartesian Robot
# Author: Addison Sears-Collins
# Date created: August 21, 2020

# Link lengths in centimeters
a1 = 1 # Length of link 1
a2 = 1 # Length of link 2
a3 = 1 # Length of link 3

# Initialize values for the displacements
d1 = 1 # Displacement of link 1
d2 = 1 # Displacement of link 2
d3 = 1 # Displacement of link 3

# Declare the Denavit-Hartenberg table. 
# It will have four columns, to represent:
# theta, alpha, r, and d
# We have the convert angles to radians.
d_h_table = np.array([[np.deg2rad(90), np.deg2rad(90), 0, a1 + d1],
                      [np.deg2rad(90), np.deg2rad(-90), 0, a2 + d2],
                      [0, 0, 0, a3 + d3]]) 

# Homogeneous transformation matrix from frame 0 to frame 1
i = 0
homgen_0_1 = np.array([[np.cos(d_h_table[i,0]), -np.sin(d_h_table[i,0]) * np.cos(d_h_table[i,1]), np.sin(d_h_table[i,0]) * np.sin(d_h_table[i,1]), d_h_table[i,2] * np.cos(d_h_table[i,0])],
                      [np.sin(d_h_table[i,0]), np.cos(d_h_table[i,0]) * np.cos(d_h_table[i,1]), -np.cos(d_h_table[i,0]) * np.sin(d_h_table[i,1]), d_h_table[i,2] * np.sin(d_h_table[i,0])],
                      [0, np.sin(d_h_table[i,1]), np.cos(d_h_table[i,1]), d_h_table[i,3]],
                      [0, 0, 0, 1]])  

# Homogeneous transformation matrix from frame 1 to frame 2
i = 1
homgen_1_2 = np.array([[np.cos(d_h_table[i,0]), -np.sin(d_h_table[i,0]) * np.cos(d_h_table[i,1]), np.sin(d_h_table[i,0]) * np.sin(d_h_table[i,1]), d_h_table[i,2] * np.cos(d_h_table[i,0])],
                      [np.sin(d_h_table[i,0]), np.cos(d_h_table[i,0]) * np.cos(d_h_table[i,1]), -np.cos(d_h_table[i,0]) * np.sin(d_h_table[i,1]), d_h_table[i,2] * np.sin(d_h_table[i,0])],
                      [0, np.sin(d_h_table[i,1]), np.cos(d_h_table[i,1]), d_h_table[i,3]],
                      [0, 0, 0, 1]])  

# Homogeneous transformation matrix from frame 2 to frame 3
i = 2
homgen_2_3 = np.array([[np.cos(d_h_table[i,0]), -np.sin(d_h_table[i,0]) * np.cos(d_h_table[i,1]), np.sin(d_h_table[i,0]) * np.sin(d_h_table[i,1]), d_h_table[i,2] * np.cos(d_h_table[i,0])],
                      [np.sin(d_h_table[i,0]), np.cos(d_h_table[i,0]) * np.cos(d_h_table[i,1]), -np.cos(d_h_table[i,0]) * np.sin(d_h_table[i,1]), d_h_table[i,2] * np.sin(d_h_table[i,0])],
                      [0, np.sin(d_h_table[i,1]), np.cos(d_h_table[i,1]), d_h_table[i,3]],
                      [0, 0, 0, 1]])  

homgen_0_3 = homgen_0_1 @ homgen_1_2 @ homgen_2_3

# Print the homogeneous transformation matrices
print("Homogeneous Matrix Frame 0 to Frame 1:")
print(homgen_0_1)
print()
print("Homogeneous Matrix Frame 1 to Frame 2:")
print(homgen_1_2)
print()
print("Homogeneous Matrix Frame 2 to Frame 3:")
print(homgen_2_3)
print()
print("Homogeneous Matrix Frame 0 to Frame 3:")
print(homgen_0_3)
print()

Now, run the code.

Here is the output:

The homogeneous matrix from frame 0 to frame 3 is:

Let’s pull out the rotation matrix:

Let’s go through the matrix to see if it makes sense. Here is the diagram:

In the first column of the matrix, we see that the value on the third row is 1. It is saying that the x₃ vector is in the same direction as z₀. If we look at the diagram, we can see that this is in fact the case.

In the second column of the matrix, we see that the value on the first row is -1. It is saying that the y₃ vector is in the opposite direction as x₀. If we look at the diagram, we can see that this is in fact the case.

In the third column of the matrix, we see that the value on the second row is -1. It is saying that the z₃ vector is in the opposite direction as y₀. If we look at the diagram, we can see that this is in fact the case.

Finally, in the fourth column of the matrix, we have the displacement vector. Let’s pull that out.

The displacement vector portion of the matrix says that the position of the end effector of the robot is 2 units along x₀, -2 units along y₀, and 2 units along z₀. These results make sense given the link lengths and displacements were 1.

Example 2 – Six Degree of Freedom Robotic Arm

Let’s take a look at another example, a six degree of freedom robotic arm.

Here is the code:

import numpy as np # Scientific computing library

# Project: Coding Denavit-Hartenberg Tables Using Python - 6DOF Robotic Arm
#          This code excludes the servo motor that controls the gripper.
# Author: Addison Sears-Collins
# Date created: August 22, 2020

# Link lengths in centimeters
a1 = 1 # Length of link 1
a2 = 1 # Length of link 2
a3 = 1 # Length of link 3
a4 = 1 # Length of link 4
a5 = 1 # Length of link 5
a6 = 1 # Length of link 6

# Initialize values for the joint angles (degrees)
theta_1 = 0 # Joint 1
theta_2 = 0 # Joint 2
theta_3 = 0 # Joint 3
theta_4 = 0 # Joint 4
theta_5 = 0 # Joint 5

# Declare the Denavit-Hartenberg table. 
# It will have four columns, to represent:
# theta, alpha, r, and d
# We have the convert angles to radians.
d_h_table = np.array([[np.deg2rad(theta_1), np.deg2rad(90), 0, a1],
                      [np.deg2rad(theta_2), 0, a2, 0],
                      [np.deg2rad(theta_3), 0, a3, 0],
                      [np.deg2rad(theta_4 + 90), np.deg2rad(90), a5,0],
                      [np.deg2rad(theta_5), 0, 0, a4 + a6]]) 

# Homogeneous transformation matrix from frame 0 to frame 1
i = 0
homgen_0_1 = np.array([[np.cos(d_h_table[i,0]), -np.sin(d_h_table[i,0]) * np.cos(d_h_table[i,1]), np.sin(d_h_table[i,0]) * np.sin(d_h_table[i,1]), d_h_table[i,2] * np.cos(d_h_table[i,0])],
                      [np.sin(d_h_table[i,0]), np.cos(d_h_table[i,0]) * np.cos(d_h_table[i,1]), -np.cos(d_h_table[i,0]) * np.sin(d_h_table[i,1]), d_h_table[i,2] * np.sin(d_h_table[i,0])],
                      [0, np.sin(d_h_table[i,1]), np.cos(d_h_table[i,1]), d_h_table[i,3]],
                      [0, 0, 0, 1]])  

# Homogeneous transformation matrix from frame 1 to frame 2
i = 1
homgen_1_2 = np.array([[np.cos(d_h_table[i,0]), -np.sin(d_h_table[i,0]) * np.cos(d_h_table[i,1]), np.sin(d_h_table[i,0]) * np.sin(d_h_table[i,1]), d_h_table[i,2] * np.cos(d_h_table[i,0])],
                      [np.sin(d_h_table[i,0]), np.cos(d_h_table[i,0]) * np.cos(d_h_table[i,1]), -np.cos(d_h_table[i,0]) * np.sin(d_h_table[i,1]), d_h_table[i,2] * np.sin(d_h_table[i,0])],
                      [0, np.sin(d_h_table[i,1]), np.cos(d_h_table[i,1]), d_h_table[i,3]],
                      [0, 0, 0, 1]])  

# Homogeneous transformation matrix from frame 2 to frame 3
i = 2
homgen_2_3 = np.array([[np.cos(d_h_table[i,0]), -np.sin(d_h_table[i,0]) * np.cos(d_h_table[i,1]), np.sin(d_h_table[i,0]) * np.sin(d_h_table[i,1]), d_h_table[i,2] * np.cos(d_h_table[i,0])],
                      [np.sin(d_h_table[i,0]), np.cos(d_h_table[i,0]) * np.cos(d_h_table[i,1]), -np.cos(d_h_table[i,0]) * np.sin(d_h_table[i,1]), d_h_table[i,2] * np.sin(d_h_table[i,0])],
                      [0, np.sin(d_h_table[i,1]), np.cos(d_h_table[i,1]), d_h_table[i,3]],
                      [0, 0, 0, 1]])  

# Homogeneous transformation matrix from frame 3 to frame 4
i = 3
homgen_3_4 = np.array([[np.cos(d_h_table[i,0]), -np.sin(d_h_table[i,0]) * np.cos(d_h_table[i,1]), np.sin(d_h_table[i,0]) * np.sin(d_h_table[i,1]), d_h_table[i,2] * np.cos(d_h_table[i,0])],
                      [np.sin(d_h_table[i,0]), np.cos(d_h_table[i,0]) * np.cos(d_h_table[i,1]), -np.cos(d_h_table[i,0]) * np.sin(d_h_table[i,1]), d_h_table[i,2] * np.sin(d_h_table[i,0])],
                      [0, np.sin(d_h_table[i,1]), np.cos(d_h_table[i,1]), d_h_table[i,3]],
                      [0, 0, 0, 1]])  

# Homogeneous transformation matrix from frame 4 to frame 5
i = 4
homgen_4_5 = np.array([[np.cos(d_h_table[i,0]), -np.sin(d_h_table[i,0]) * np.cos(d_h_table[i,1]), np.sin(d_h_table[i,0]) * np.sin(d_h_table[i,1]), d_h_table[i,2] * np.cos(d_h_table[i,0])],
                      [np.sin(d_h_table[i,0]), np.cos(d_h_table[i,0]) * np.cos(d_h_table[i,1]), -np.cos(d_h_table[i,0]) * np.sin(d_h_table[i,1]), d_h_table[i,2] * np.sin(d_h_table[i,0])],
                      [0, np.sin(d_h_table[i,1]), np.cos(d_h_table[i,1]), d_h_table[i,3]],
                      [0, 0, 0, 1]])  

homgen_0_5 = homgen_0_1 @ homgen_1_2 @ homgen_2_3 @ homgen_3_4 @ homgen_4_5 

# Print the homogeneous transformation matrices
print("Homogeneous Matrix Frame 0 to Frame 5:")
print(homgen_0_5)

Now, run the code:

Here is the output:

In matrix form, this is:

Does this make sense?  Here is the diagram:

In the first column of the matrix, we see that the value on the third row is 1. It is saying that the x₅ vector is in the same direction as z₀. If we look at the diagram, we can see that this is in fact the case.

In the second column of the matrix, we see that the value on the second row is -1. It is saying that the y₅ vector is in the opposite direction as y₀. If we look at the diagram, we can see that this is in fact the case.

In the third column of the matrix, we see that the value on the first row is 1. It is saying that the z₅ vector is in the same direction as x₀. If we look at the diagram, we can see that this is in fact the case.

Finally, in the fourth column of the matrix, we have the displacement vector. This portion of the matrix says that the position of the end effector of the robot is 4 units along x₀, 0 units along y₀, and 2 units along z₀. These results make sense given that we set all the link lengths to 1.

References

Credit to Professor Angela Sodemann from whom I learned these important robotics fundamentals. Dr. Sodemann teaches robotics over at her website, RoboGrok.com. While she uses the PSoC in her work, I use Arduino and Raspberry Pi since I’m more comfortable with these computing platforms. Angela is an excellent teacher and does a fantastic job explaining various robotics topics over on her YouTube channel.

Once we have filled in the Denavit-Hartenberg (D-H) parameter table for a robotic arm, we find the homogeneous transformation matrices (also known as the Denavit-Hartenberg matrix) by plugging the values into the matrix of the following form, which is the homogeneous transformation matrix for joint n (i.e. the transformation from frame n-1 to frame n).

where R is the 3×3 submatrix in the upper left that represents the rotation from frame n-1 (e.g. frame 0) to frame n (e.g. frame 1), and T is the 3×1 submatrix in the upper right that represents the translation (or displacement) from frame n-1 to frame n.

For example, suppose we have a SCARA robotic arm.

We draw the kinematic diagram for this arm.

We then use that diagram to find the Denavit-Hartenberg parameters:

We now want to find the homogeneous transformation matrix from frame 0 to frame 1, so we look at the first row of the Denavit-Hartenberg parameter table labeled ‘Joint 1’ (remember that a joint in our case is either a servo motor that produces rotational motion or a linear actuator that produces linear motion). 

In this case, n=1, and we would look at the first row of the D-H table (Joint 1) and plug in the appropriate values into the following matrix:

For the homogeneous transformation matrix from frame 1 to 2, we would look at the second row of the D-H table (Joint 2) and plug in the corresponding values into the following matrix.

For the homogeneous transformation matrix from frame 2 to 3, we would look at the third row of the table (Joint 3) and plug in the corresponding values into the following matrix.

Then, to find the homogeneous transformation matrix from the base frame (frame 0) to the end-effector frame (frame 3), we would multiply all the transformation matrices together.

homgen_0_3 = (homgen_0_1)(homgen_1_2)(homgen_2_3)

Let’s take a look at some examples.

Example 1 – Cartesian Robot

We have four reference frames, so we need to have three rows in our Denavit-Hartenberg parameter table.

Let’s begin by looking at the first column of the table. 

Find θ

For the Joint 1 row (Servo 0), we are going to focus on the relationship between frame 0 and frame 1. θ is the angle from x₀ to x₁ around z₀.

If you look at the diagram, take the thumb of your right hand and stick it in z₀ direction. Your fingers curl in the direction of positive rotation. The angle from x₀ to x₁ around z₀ is 90 degrees in the counterclockwise (i.e. positive) direction. This angle won’t change regardless of how the linear actuator at frame 0 moves in the d₁ direction.

Now, let’s look at the Joint 2 row. We take a look at the rotation between frame 1 and frame 2. θ is the angle from x₁ to x₂ around z₁.

If you look at the diagram, take the thumb of your right hand and stick it in z₁ direction. Your fingers curl in the direction of positive rotation. The angle from x₁ to x₂ around z₁ is 90 degrees in the counterclockwise (i.e. positive) direction. This angle won’t change regardless of how the linear actuator at frame 1 moves in the d₂ direction.

Now, let’s look at the Joint 3 row. We take a look at the rotation between frame 2 and frame 3. θ is the angle from x₂ to x₃ around z₂.

If you look at the diagram, x₂ and x₃ both point in the same direction. The axes are therefore aligned. When the robot is in motion, there is only linear motion along z₂. The angle from x₂ to x₃ around z₂ will remain 0, so let’s put that in the third row of our table.

Find α

Let’s start with the Joint 1 row of the table.

α is the angle from z₀ to z₁ around x₁.

Take your right thumb and point it in the x₁ direction. Curl your fingers counterclockwise from z₀ to z₁. The angle is 90 degrees.

Let’s go to the Joint 2 row of the table.

α is the angle from z₁ to z₂ around x₂. Take your right thumb, and stick it in the x₂ direction. We would have to rotate 90 degrees in the clockwise (i.e. negative) direction to go from z₁ to z₂.

Let’s go to the Joint 3 row of the table.

α is the angle from z₂ to z₃ around x₃. In the kinematic diagram, you can see that this angle is 0 degrees, so we put 0 in the table.

Find r

Let’s start with the Joint 1 row of the table.

r is the distance between the origin of frame 0 and the origin of frame 1 along the x₁ direction. You can see in the diagram that this distance is 0.

Now let’s look at the Joint 2 row. r is the distance between the origin of frame 1 and the origin of frame 2 along the x₂ direction. You can see in the diagram that this distance is 0.

Now let’s look at the Joint 3 row. r is the distance between the origin of frame 2 and the origin of frame 3 along the x₃ direction. You can see in the diagram that this distance is 0.

Find d

Now let’s find d. We’ll start on the first row of the table as usual.

d is the distance from x₀ to x₁ along the z₀ direction. You can see that this distance is a₁ + d₁. a₁ is the distance between the origin of frame 0 and the origin of frame 1 when the linear actuator (i.e. motor that generates linear motion) is in its home position, unextended. d₁ is the distance the actuator extends beyond the home position.

Now let’s take a look at frame 1 to frame 2. d is the distance from x₁ to x₂ along the z₁ direction. You can see that this distance is a₂ + d₂.

Now let’s take a look at frame 2 to frame 3. d is the distance from x₂ to x₃ along the z₂ direction. You can see that this distance is a₃ + d₃.

To check your work, all of the link lengths and all of the joint variables in the kinematic diagram should appear in the Denavit-Hartenberg table.

Example 2 – Articulated Robot

We have four reference frames, so we need to have three rows in our Denavit-Hartenberg parameter table.

Let’s begin by looking at the first column of the table. 

Find θ

For the Joint 1 row (Servo 0), we are going to focus on the relationship between frame 0 and frame 1. θ is the angle from x₀ to x₁ around z₀.

If you look at the diagram, take the thumb of your right hand and stick it in z₀ direction. Your fingers curl in the direction of positive rotation. The angle from x₀ to x₁ around z₀ is θ₁ degrees in the counterclockwise (i.e. positive) direction.

Now, let’s look at the Joint 2 row. We take a look at the rotation between frame 1 and frame 2. θ is the angle from x₁ to x₂ around z₁.

If you look at the diagram, take the thumb of your right hand and stick it in z₁ direction. Your fingers curl in the direction of positive rotation. The angle from x₁ to x₂ around z₁ is θ₂ degrees in the counterclockwise (i.e. positive) direction.

Now, let’s look at the Joint 3 row. We take a look at the rotation between frame 2 and frame 3. θ is the angle from x₂ to x₃ around z₂.

If you look at the diagram, x₂ and x₃ both point in the same direction. The axes are therefore aligned. When the robot is in motion, the angle from x₂ to x₃ around z₂ will be θ₃ degrees, so let’s put that in the third row of our table.

Find α

Let’s start with the Joint 1 row of the table.

α is the angle from z₀ to z₁ around x₁.

Take your right thumb and point it in the x₁ direction. Curl your fingers counterclockwise from z₀ to z₁. The angle is 90 degrees.

Let’s go to the Joint 2 row of the table.

α is the angle from z₁ to z₂ around x₂. Take your right thumb, and stick it in the x₂ direction. The angle between z₁ and z₂ is 0 degrees since both point in the same direction.

Let’s go to the Joint 3 row of the table.

α is the angle from z₂ to z₃ around x₃. In the kinematic diagram, you can see that this angle is 0 degrees, so we put 0 in the table.

Find r

Let’s start with the Joint 1 row of the table.

r is the distance between the origin of frame 0 and the origin of frame 1 along the x₁ direction. You can see in the diagram that this distance is 0.

Now let’s look at the Joint 2 row. r is the distance between the origin of frame 1 and the origin of frame 2 along the x₂ direction. You can see in the diagram that this distance is a₂.

Now let’s look at the Joint 3 row. r is the distance between the origin of frame 2 and the origin of frame 3 along the x₃ direction. You can see in the diagram that this distance is a₃.

Find d

Now let’s find d. We’ll start on the first row of the table as usual.

d is the distance from x₀ to x₁ along the z₀ direction. You can see that this distance is a₁. a₁ is the distance between the origin of frame 0 and the origin of frame 1. No matter how the servos move, this distance will always remain a₁.

Now let’s take a look at frame 1 to frame 2. d is the distance from x₁ to x₂ along the z₁ direction. You can see that this distance is 0.

Now let’s take a look at frame 2 to frame 3. d is the distance from x₂ to x₃ along the z₂ direction. You can see that this distance is 0.

To check your work, all of the link lengths and all of the joint variables in the kinematic diagram should appear in the Denavit-Hartenberg table.

Example 3 – Six Degree of Freedom Robotic Arm

We have six reference frames, so we need to have five rows in our Denavit-Hartenberg parameter table. 

See if you can do this one all by yourself, and then use the table below to check your work. Remember the rules:

• θ is the angle from x_n-1 to x_n around z_n-1.
• α is the angle from z_n-1 to z_n around x_n.
• r (sometimes you’ll see the letter ‘a’ instead of ‘r’) is the distance between the origin of the n-1 frame and the origin of the n frame along the x_n direction.
• d is the distance from x_n-1 to x_n along the z_n-1 direction.

Example 4 – Six Degree of Freedom Collaborative Robot

We have seven reference frames, so we need to have six rows in our Denavit-Hartenberg parameter table. 

See if you can do this one all by yourself, and then use the table below to check your work. Remember the rules:

• θ is the angle from x_n-1 to x_n around z_n-1.
• α is the angle from z_n-1 to z_n around x_n.
• r (sometimes you’ll see the letter ‘a’ instead of ‘r’) is the distance between the origin of the n-1 frame and the origin of the n frame along the x_n direction.
• d is the distance from x_n-1 to x_n along the z_n-1 direction.

A few things to note:

• We have negative values for θ above due to the clockwise rotation about the z_n-1 axis.
• I also didn’t use all of the a variables in the diagram because some of these measurements are irrelevant for calculating the Denavit-Hartenberg parameter table.
• Displacement is a vector quantity, which means it has both a magnitude and a direction. You’ll therefore see negative values for a in the table above.

Example 5 – Six Degree of Freedom Industrial Robot

We have seven reference frames, so we need to have six rows in our Denavit-Hartenberg parameter table. 

See if you can do this one all by yourself, and then use the table below to check your work. Remember the rules:

• θ is the angle from x_n-1 to x_n around z_n-1.
• α is the angle from z_n-1 to z_n around x_n.
• r (sometimes you’ll see the letter ‘a’ instead of ‘r’) is the distance between the origin of the n-1 frame and the origin of the n frame along the x_n direction.
• d is the distance from x_n-1 to x_n along the z_n-1 direction.

Remember when you’re trying to find theta that rotation in the counterclockwise direction (as viewed from looking down on the axis of rotation) is positive. Otherwise, it is negative.

You point your thumb in the direction of the axis of rotation and curl your fingers. Your finger-curl direction is positive rotation.

References

Credit to Professor Angela Sodemann from whom I learned these important robotics fundamentals. Dr. Sodemann teaches robotics over at her website, RoboGrok.com. While she uses the PSoC in her work, I use Arduino and Raspberry Pi since I’m more comfortable with these computing platforms. Angela is an excellent teacher and does a fantastic job explaining various robotics topics over on her YouTube channel.