Linear Algebra Application: Robot Kinematics

Forward Kinematics: Modeling, Motion Planning, and Control

Imron Rosyadi

Robot Kinematics: Modeling, Motion Planning, and Control

Forward Kinematics: ECE Applications in Robotics

What is Forward Kinematics?

Kinematics describes the motion of a manipulator. It analyzes position, velocity, and acceleration without considering the forces causing the motion.

Forward Kinematics is used to calculate the position and orientation of the end-effector given:

1. A kinematic chain with multiple degrees of freedom.
2. The joint variables (angles for revolute joints, displacements for prismatic joints).

Important

Goal of Forward Kinematics: To determine where the robot’s “hand” (end-effector) is, and how it’s oriented in space.

Emphasize that forward kinematics is about predicting the outcome of joint movements. This is crucial for tasks like path planning or avoiding collisions. For ECE, this translates to understanding how motor commands (joint variables) lead to physical end-effector states.

Robot Anatomy: Key Components

An articulated robot is typically composed of:

• Base: Anchor point, supports the unit.
• Shoulder, Elbow: Jointed sections allowing movement.
• Spherical Wrist: Connects arm to gripper.
• Gripper: The end-effector, interacts with the environment.

Figure 2.1 Illustration of a robot.

Note

End-effector: The part of the robot that interacts with its environment. Examples:

• Gripper
• Welding gun
• Magnet
• Force-torque sensors
• Tool changers

This slide introduces the physical parts of a robot, essential for understanding the subsequent kinematic models. Emphasize how each part contributes to the robot’s function. For ECE students, this helps visualize the mechanical system that their control systems will operate.

Robot Joints: Types and Degrees of Freedom

Each joint, like the elbow, is an actuator that causes relative motion between two links.

Revolute (R) Joint

• Has 1 Degree of Freedom (DoF).
• Axis of rotation becomes the \(z_i\) axis.
• Joint variable: \(\theta_i\) (in degrees), representing relative angular displacement.

Figure 2.2 Revolute (R) joint.

Prismatic (P) Joint

• Also has 1 DoF.
• Axis of translation becomes the \(z_i\) axis.
• Joint variable: \(d_i\) (in meters), representing relative linear displacement.

Figure 2.3 Prismatic (P) joint.

\[ q_i=\begin{Bmatrix} \theta_i & \textrm{if joint $i$ is revolute}\\ d_i & \textrm{if joint $i$ is prismatic}\end{Bmatrix} \quad \text{(2.22)} \]

Highlight the 1 DoF nature of these basic joints and how they relate to the \(z_i\) axis. This sets up the coordinate system definitions needed for DH convention later. Mention briefly that other joints exist (e.g., ball and socket, spherical wrist) but for simplicity, we focus on R and P joints which are fundamental building blocks.

Links, Rigid Bodies, and Kinematic Chains

• Rigid Body (Link): A component that connects joints, like the segment between a shoulder and an elbow joint. Assumed to not deform.

• Kinematic Chain: A grouping of links and joints that produces the desired motion.

  • Examples: A human arm or leg.
  • In robotics, this chain connects the base to the end-effector.

Tip

Think of a robot arm as a series of rigid bodies connected by joints. The collective movement of these joints dictates the end-effector’s pose.

Clarify the terms “rigid body” and “kinematic chain” as they are central to robot modeling. This provides context for how robots are structured from a mechanical perspective.

Rigid Motion & Homogeneous Matrices: Overview

This section covers the analytical approach to representing:

• Position of a point in space.
• Rotation in 2D and 3D.
• Rigid Motions, combining rotation and translation.
• Homogeneous Transformations, a powerful tool for unified representation.

Understanding these concepts is foundational for calculating forward kinematics.

This slide acts as a mini-introduction to a dense topic. Break it down to make it digestible. Emphasize that these are the mathematical tools we’ll use.

Representing Position: Coordinate Frames

Robotics often uses Cartesian coordinates for task definition. We represent the position of a point, \(p\), relative to a chosen coordinate frame.

Figure 2.4 Two coordinate frames with a point, \(p\), and two vectors, \(p^0\) and \(p^1\).

A point \(p\) can be defined with respect to:

• Frame {0}: \(p^0=\begin{bmatrix} 5\\ 6 \end{bmatrix}\)
• Frame {1}: \(p^1=\begin{bmatrix} -2.8\\ 4.2 \end{bmatrix}\)

The origin of frame {1} with respect to {0}: \(o_1^0=\begin{bmatrix} 10\\ 5 \end{bmatrix}\)

The origin of frame {0} with respect to {1}: \(o_0^1=\begin{bmatrix} -10.6\\ 3.5 \end{bmatrix}\)

Note

Always specify the reference frame when representing a point or vector (e.g., \(p^0\) vs. \(p^1\)). This prevents ambiguity.

Explain that \(p\) is the physical point, while \(p^0\) and \(p^1\) are its coordinates in different frames. This is a common source of confusion. The coordinate frames are simply different perspectives.

Representing Rotation: 2D & 3D

Rotations are fundamental to robot movement. We will explore:

• Two-Dimensional (Planar) Rotations: Simpler case, good for intuition.
• Three-Dimensional Rotations: Essential for real-world robotics.
• Rules of Rotation Matrices: Important properties and operations.

These concepts form the basis for describing the orientation of robot links.

Prepare students for the mathematical representation of rotation, which is often more abstract than position.

2D Rotations: The Rotation Matrix

A rotation matrix \(R_1^0\) describes the orientation of frame {1} relative to frame {0}.

It is formed by projecting the axes of frame {1} onto the coordinate axes of frame {0}.

\[ R_1^0=\begin{bmatrix} x_1 \cdot x_0 & y_1 \cdot x_0\\ x_1 \cdot y_0 & y_1 \cdot y_0 \end{bmatrix} \quad \text{(2.1)} \]

The columns of \(R_1^0\) are the unit vectors of frame {1}’s axes, expressed in frame {0}’s coordinates.

Explain that the dot products measure the alignment of the axes. For example, \(x_1 \cdot x_0\) is the cosine of the angle between \(x_1\) and \(x_0\). This geometric interpretation helps understand the matrix elements.

Example: Planar Rotation

Consider a planar rotation of frame {1} by an angle \(\theta\) relative to frame {0}.

Figure 2.5 Planar rotation of \(\theta\) degrees.

The rotation matrix \(R_1^0\) is:

\[ R_1^0 = \begin{bmatrix} x_1^0 & y_1^0 \end{bmatrix} \quad \text{(2.2)} \]

By projecting the axes:

\[ R_1^0 = \begin{bmatrix} \cos \theta & -\sin\theta\\ \sin\theta & \cos\theta \end{bmatrix} \quad \text{(2.3)} \]

This matrix represents the orientation of frame {1} with respect to frame {0}.

Walk through how the \(\cos \theta\) and \(\sin \theta\) terms arise from the projections. For example, \(x_1\) projected onto \(x_0\) is \(\cos \theta\), and \(y_1\) projected onto \(x_0\) is \(\cos(\theta + 90^\circ) = -\sin \theta\).

Rules of Rotation Matrices

Rotation matrices (\(R\)) have specific properties:

• Orthogonal: \(R^T = R^{-1}\).
• Special Orthogonal Group SO(n): Determinant \(\det(R) = +1\) (due to right-hand rule).
• Columns (and rows) are mutually orthogonal unit vectors.

Inverse Rotation

To find \(R_0^1\) (frame {0} w.r.t. {1}): \[ R_0^1 = (R_1^0)^T \quad \text{(2.4)} \]

Chained Rotations

If \(R_1^0\) and \(R_n^1\) are known, then: \[ R_n^0 = R_1^0 R_n^1 \quad \text{(2.5)} \]

Warning

Important Rule for Matrix Multiplication:

• Pre-multiply when rotation is about a fixed/world frame.
• Post-multiply when rotation is about the current frame.

Emphasize the significance of these properties. Orthogonality means rotations preserve vector lengths and angles. The pre/post-multiplication rule is critical for correctly composing multiple rotations. Provide a brief intuitive explanation for why this rule holds.

3D Rotations: Extending the Matrix

The principle for 3D rotations is the same as 2D: project axes of frame {1} onto frame {0}.

\[ R_1^0 = \begin{bmatrix} x_1 \cdot x_0 & y_1 \cdot x_0 & z_1 \cdot x_0\\ x_1 \cdot y_0 & y_1 \cdot y_0 & z_1 \cdot y_0 \\ x_1 \cdot z_0 & y_1 \cdot z_0 & z_1 \cdot z_0 \end{bmatrix} \quad \text{(2.6)} \]

This \(3 \times 3\) matrix belongs to \(SO(3)\) (Special Orthogonal Group of order 3).

Point out the expansion from \(2 \times 2\) to \(3 \times 3\). The core concept of projecting axes remains.

Example: Rotation about \(z_0\)-axis

Find the rotation matrix \(R_1^0\) when frame {1} is rotated by angle \(\theta\) about the \(z_0\)-axis.

Figure 2.6 Rotation through an angle \(\theta\) about the \(z_0\)-axis.

From the figure: \(x_1 \cdot x_0 = \cos \theta\), \(y_1 \cdot x_0 = -\sin \theta\) \(x_1 \cdot y_0 = \sin \theta\), \(y_1 \cdot y_0 = \cos \theta\) \(z_0 \cdot z_1 = 1\) All other dot products are zero.

Resulting Rotation Matrix: \[ R_1^0 = R_{z,\theta} = \begin{bmatrix} \cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{bmatrix} \quad \text{(2.7)} \]

Rotations about x- and y-axes

\[ R_{x,\theta} = \begin{bmatrix} 1 & 0 & 0\\ 0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta\end{bmatrix} \quad \text{(2.8)} \]

\[ R_{y,\theta} = \begin{bmatrix} \cos \theta & 0 & \sin \theta\\ 0 & 1 & 0 \\ -\sin \theta & 0 & \cos \theta\end{bmatrix} \quad \text{(2.9)} \]

This slide gives concrete examples of fundamental rotation matrices, which are building blocks for more complex rotations. Highlight the pattern: the row/column corresponding to the axis of rotation has a ‘1’ in its diagonal and ’0’s elsewhere, while the other \(2 \times 2\) block forms a planar rotation.

Example: Chaining Rotations

Find the rotation matrix after the following rotations occur:

1. Rotate by \(\theta\) about the current x-axis
2. Rotate by \(\phi\) about the current z-axis
3. Rotate by \(\alpha\) about the fixed z-axis
4. Rotate by \(\beta\) about the current y-axis
5. Rotate by \(\gamma\) about the fixed x-axis

Start with the initial rotation and apply the pre/post-multiplication rule:

\[ R=R_{x,\gamma}R_{z,\alpha}R_{x,\theta}R_{z,\phi}R_{y,\beta} \]

Tip

Reminder:

• Fixed frame rotation: Pre-multiply (\(R_{fixed} R_{current\_total}\))
• Current frame rotation: Post-multiply (\(R_{current\_total} R_{new\_current}\))

This example is important for applying the pre/post-multiplication rule in a complex sequence. Walk through each step, explaining why a specific matrix is pre- or post-multiplied. Remind students that c for cosine and s for sine will be used in subsequent complex matrix examples.

Rigid Motion: Translation and Rotation Combined

A rigid motion is a combination of a pure translation and a pure rotation.

To express a point \(p^1\) (in frame {1}) with respect to frame {0}:

\[ p^0=R_1^0p^1+d^0 \quad \text{(2.10)} \]

Where: - \(R_1^0\) is the rotation matrix of frame {1} w.r.t. frame {0}. - \(d^0\) is the vector from frame {0} origin to frame {1} origin, expressed in frame {0}.

Figure 2.7 The location of the blue dot in frame {0}.

Figure 2.8 The location of the blue dot in frame {1}.

Explain Figure 2.7 and 2.8. The goal is to find the blue dot’s position in the global frame {0}, given its position in its local frame {1}, and the relative pose of {1} to {0}. This is a direct application of Equation (2.10).

Chained Rigid Motions

Consider three frames: {0}, {1}, and {2}.

• \(d_1^0\): vector from origin of {0} to origin of {1}.
• \(d_2^1\): vector from origin of {1} to origin of {2}.
• \(p^2\): point \(p\) represented in frame {2}.

To compute \(p^0\) (point \(p\) relative to frame {0}):

1. Transform \(p^2\) to frame {1}: \(p^1=R_2^1p^2+d_2^1 \quad \text{(2.11)}\)
2. Transform \(p^1\) to frame {0}: \(p^0=R_1^0p^1+d_1^0 \quad \text{(2.12)}\)
3. Substitute (2.11) into (2.12): \(p^0=R_1^0(R_2^1p^2+d_2^1)+d_1^0 = R_1^0R_2^1p^2+R_1^0d_2^1+d_1^0 \quad \text{(2.13)}\)

We define:

• \(R_2^0=R_1^0R_2^1 \quad \text{(2.14)}\)
• \(d_2^0=d_1^0+R_1^0d_2^1 \quad \text{(2.15)}\)

Thus, the final transformation is: \[ p^0=R_2^0p^2+d_2^0 \quad \text{(2.16)} \]

This derivation shows how transformations compose. Emphasize that \(R_2^0\) is the total rotation from frame 2 to frame 0, and \(d_2^0\) is the total translation from origin 0 to origin 2, both expressed in frame 0. A simple Mermaid diagram could illustrate the flow of transformations.

Homogeneous Transformation: Unified Representation

A Homogeneous Transformation combines rotation and translation into a single matrix multiplication. This simplifies coordinate transformations.

The homogeneous matrix \(H\) is a \(4 \times 4\) matrix:

\[ H=\begin{bmatrix} R & d\\ 0 & 1\end{bmatrix}, R\in SO(3), d \in \mathbb{R}^3 \quad \text{(2.17)} \]

Where \(R\) is the \(3 \times 3\) rotation matrix and \(d\) is the \(3 \times 1\) translation vector.

Inverse Homogeneous Transformation

\[ H^{-1}=\begin{bmatrix} R^T & -R^Td \\ 0 & 1\end{bmatrix} \quad \text{(2.18)} \]

Highlight the power of homogeneous transformations: they allow point transformations (\(p^0 = Hp^1\)) to be single matrix-vector products, rather than separate rotation and addition. This is very efficient for computations. Explain why the last row is \([0 \ 0 \ 0 \ 1]\).

Basic Homogeneous Transformations

These are fundamental transformations for translation and rotation about principal axes. (Note: \(c_\alpha = \cos \alpha\), \(s_\alpha = \sin \alpha\))

X-axis Translation & Rotation

\[ Trans_{x,a} = \begin{bmatrix} 1 & 0 & 0 & a\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}, Rot_{x,\alpha} = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & c_\alpha & -s_\alpha & 0 \\ 0 & s_\alpha & c_\alpha & 0 \\ 0 & 0 & 0 &1\end{bmatrix} \quad \text{(2.19)} \]

Y-axis Translation & Rotation

\[ Trans_{y,b} = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & b \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}, Rot_{y,\beta} = \begin{bmatrix}c_\beta & 0 & s_\beta & 0\\ 0 & 1 & 0 & 0 \\ -s_\beta & 0 & c_\beta & 0 \\ 0 & 0 & 0 &1\end{bmatrix} \quad \text{(2.20)} \]

Z-axis Translation & Rotation

\[ Trans_{z,c} = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & c \\ 0 & 0 & 0 & 1 \end{bmatrix}, Rot_{z,\gamma} = \begin{bmatrix} c_\gamma & -s_\gamma & 0 & 0\\ s_\gamma & c_\gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 &1\end{bmatrix} \quad \text{(2.21)} \]

Show how these matrices are formed by inserting the rotation matrix and translation vector into the \(H\) structure. Point out the similarities to the pure rotation matrices.

Example: Complex Homogeneous Transformation

Find a homogeneous transformation matrix \(H\) that represents a sequence of operations:

1. Rotation of angle \(\alpha\) about the current x-axis
2. Translation of \(b\) along the current x-axis
3. Translation of \(d\) along the current z-axis
4. Rotation of angle \(\theta\) about the current z-axis

Since all transformations are about current axes, we post-multiply in the order given:

\[ H = Rot_{x,\alpha} Trans_{x,b} Trans_{z,d} Rot_{z,\theta} \]

The resulting matrix:

\[ H=\begin{bmatrix} c_\theta & -s_\theta & 0 & b\\ c_\alpha s_\theta & c_\alpha c_\theta & -s_\alpha & -ds_\alpha \\ s_\alpha s_\theta & s_\alpha c_\theta & c_\alpha & dc_\alpha \\ 0 & 0 & 0 & 1\end{bmatrix} \]

This example reinforces the post-multiplication rule for current-axis transformations. Emphasize the importance of the order of operations. The derivation involves multiplying these 4x4 matrices.

Kinematic Chains: Robot Structure

A robot is comprised of links connected by joints (R or P), forming a kinematic chain.

Labeling Convention

• Joints are labeled from 1 to \(n\).
• Links are labeled from 0 to \(n\), with Link 0 being the fixed base.

The joint variable, \(q_i\), depends on the joint type:

\[ q_i=\begin{Bmatrix} \theta_i & \textrm{if joint $i$ is revolute}\\ d_i & \textrm{if joint $i$ is prismatic}\end{Bmatrix} \quad \text{(2.22)} \]

Transformation Matrix (\(T_j^i\))

Represents the position and orientation of frame {j} with respect to {i}.

\[ T_j^i=\begin{Bmatrix} A_{i+1}A_{i+2}\cdots A_{j} & \textrm{if $i < j$}\\ (A_{j+1}A_{j+2}\cdots A_{i})^{-1} & \textrm{if $i > j$}\end{Bmatrix} \quad \text{(2.23)} \]

Where \(A_i\) is the homogeneous transformation from frame \(i-1\) to \(i\): \[ A_i=\begin{bmatrix} R_i^{i-1} & o_i^{i-1}\\ 0 & 1 \end{bmatrix} \quad \text{(2.24)} \]

Explain how the chain builds up. The \(A_i\) matrices are essentially the “local” transformations between adjacent links, and \(T_j^i\) is the “global” transformation obtained by chaining these local ones. \(o_i^{i-1}\) is the translation vector from origin \((i-1)\) to origin \(i\).

Forward Kinematics: Trigonometry Method

The trigonometry method is suitable for simple, planar robots.

Approach:

1. Place the robot in a 2-D planar coordinate system.
2. Find the angles between each joint movement.
3. Use trigonometric functions (\(\sin\), \(\cos\)) to find the end-effector’s coordinates relative to the base.

This method provides intuitive geometric insights but can become complex for 3D or multi-link manipulators.

Emphasize the simplicity and geometric intuition this method offers, but also its limitations for more complex robot structures.

Example: Two-Link Planar Manipulator (Trigonometry Method)

Problem: Find the end-effector coordinates (\(x_2, y_2\)) for a two-link planar manipulator.

• Joint 1: Rotated by \(\alpha = 60^\circ\).
• Link 1: Length \(l_1 = 1\).
• Joint 2: Rotated by \(\beta = 30^\circ\) relative to Link 1.
• Link 2: Length \(l_2 = 1\).

The coordinates of the end-effector are:

\(x = l_1 \cos \alpha + l_2 \cos(\alpha+\beta)\)

\(y = l_1 \sin \alpha + l_2 \sin(\alpha+\beta)\)

With \(l_1=l_2=1\), \(\alpha=60^\circ\), \(\beta=30^\circ\):

\(x = 1 \cos 60^\circ + 1 \cos(60^\circ+30^\circ) = 0.5 + 0 = 0.5\)

\(y = 1 \sin 60^\circ + 1 \sin(60^\circ+30^\circ) = \frac{\sqrt{3}}{2} + 1 = 0.866 + 1 = 1.866\)

Interactive: Two-Link Planar Manipulator

Adjust the link lengths and joint angles to see the end-effector position in real-time.

viewof l1 = Inputs.range([0.1, 5.0], {value: 1, step: 0.1, label: "Link 1 Length (l1)"});
viewof l2 = Inputs.range([0.1, 5.0], {value: 1, step: 0.1, label: "Link 2 Length (l2)"});
viewof alpha_deg = Inputs.range([-180, 180], {value: 60, step: 1, label: "Joint 1 Angle (alpha_deg, degrees)"});
viewof beta_deg = Inputs.range([-180, 180], {value: 30, step: 1, label: "Joint 2 Angle (beta_deg, degrees)"});

This interactive plot allows students to visually connect the joint parameters (\(l_1, l_2, \alpha, \beta\)) with the resulting end-effector position. Encourage them to play with the sliders. This is a powerful demonstration of forward kinematics.

Denavit-Hartenberg (DH) Convention

The Denavit-Hartenberg (DH) convention is a standardized method for assigning coordinate frames to robot links and deriving the kinematic equations.

It simplifies the calculation of the homogeneous transformation matrix (\(A_i\)) between adjacent links:

\[ A_i=Rot_{z,\theta}Trans_{z,d}Trans_{x,a}Rot_{x,\alpha}=\begin{bmatrix} c_\theta & -s_\theta c_\alpha & s_\theta s_\alpha & a c_\theta\\ s_\theta & c_\theta c_\alpha & -c\theta s_\alpha & as_\theta \\ 0 & s_\alpha & c_\alpha & d \\ 0 & 0 & 0 & 1\end{bmatrix} \quad \text{(2.25)} \]

This matrix encapsulates all four DH parameters: \(a\), \(\alpha\), \(d\), \(\theta\).

Explain that DH is a systematic approach to set up coordinate systems, which is essential for complex manipulators where the trigonometry method fails. The matrix \(A_i\) represents the transformation from frame \(i-1\) to frame \(i\).

DH Convention Steps: Drawing & Axes

Step 1: Draw out the robot

• Illustrate the revolute and prismatic joints.
• Set the \(z_i\) axis as the axis of rotation for revolute joints or the axis of translation for prismatic joints.
• Define the joint variable (\(\theta_i\) or \(d_i\)).

Step 2: Assign the remaining axes

• The \(x_i\) axis must be perpendicular to both \(z_i\) and \(z_{i-1}\).
• It must also intersect with \(z_{i-1}\).
• The \(y_i\) axis is determined by the right-hand rule (thumb \(z_i\), index finger \(x_i\), middle finger \(y_i\)).

Note

Right-Hand Rule: Align your thumb with the positive \(z\)-axis, your index finger with the positive \(x\)-axis, and your middle finger will point to the positive \(y\)-axis.

These first two steps are crucial for correctly defining the coordinate frames on each link. Emphasize the strict rules for \(x_i\) to ensure unique frame assignment.

DH Convention Steps: Assigning Parameters

Step 3: Assign the DH parameters

• Link Length (\(a_i\)): Measured along \(x_i\), distance between \(z_{i-1}\) and \(z_i\).
• Link Twist (\(\alpha_i\)): Angle between \(z_{i-1}\) and \(z_i\), measured in a plane normal to \(x_i\).
• Joint Offset (\(d_i\)): Measured along \(z_{i-1}\), distance from origin \(O_{i-1}\) to the intersection of \(x_i\) and \(z_{i-1}\).
  • If joint \(i\) is revolute, \(d_i\) is constant.
• Joint Angle (\(\theta_i\)): Angle between \(x_{i-1}\) and \(x_i\), measured in a plane normal to \(z_{i-1}\).
  • If joint \(i\) is prismatic, \(\theta_i\) is constant.

Tip

For a visual guide, watch this YouTube video by TekkotsuRbotics.

This is the most critical step for DH convention. Each parameter defines a geometric relationship between successive frames. It’s often helpful to visualize these in 3D.

DH Convention Steps: Table & Matrices

Step 4: Create a DH table

List all known values. Use * for variable joint parameters.

Links	\(a_i\)	\(\alpha_i\)	\(d_i\)	\(\theta_i\)
1	\(a_1\)	\(\alpha_1\)	\(d_1\)	\(\theta_1\)
…	…	…	…	…
\(i\)	\(a_i\)	\(\alpha_i\)	\(d_i\)	\(\theta_i\)

Step 5: Calculate each \(A_i\) matrix

Use Equation (2.25) with the parameters from the DH table. (Note: Link 0 doesn’t have an A matrix).

Step 6: Calculate the \(T\) matrix

The overall transformation \(T_n^0\) (end-effector w.r.t. base) is the product of all \(A_i\) matrices:

\[ T_n^0 = A_1 A_2 \cdots A_n \]

The DH table organizes the parameters, making it easier to construct the individual \(A_i\) matrices. The final \(T_n^0\) matrix contains the complete forward kinematics solution: end-effector position and orientation.

Example: Two-Link Planar Manipulator (DH Convention)

Derive forward kinematics for this two-link manipulator using DH.

Figure 2.9 Two-link planar manipulator.

Step 1 & 2: Axes Assignment

All z-axes are normal to the page (blue dots).

Figure 2.10 The two-link manipulator with assigned axes.

Step 3: DH Parameters

Figure 2.11 The two-link manipulator with assigned DH parameters.

Walk through the axis assignments and parameter definitions visually from the figures. Emphasize how \(x_i\) must intersect \(z_{i-1}\) and be normal to both \(z_i\) and \(z_{i-1}\).

Example: Two-Link Planar Manipulator (DH Convention) cont.

Step 4: DH Table

Link	\(a_i\)	\(\alpha_i\)	\(d_i\)	\(\theta_i\)
1	\(a_1\)	0	0	\(\theta_1^*\)
2	\(a_2\)	0	0	\(\theta_2^*\)

Step 5: Calculate \(A_i\) matrices

Using Equation (2.25): \[ A_1=\begin{bmatrix} c_1 & -s_1 & 0 & a_1c_1\\ s_1 & c_1 & 0 & a_1s_1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}, A_2=\begin{bmatrix} c_2 & -s_2 & 0 & a_2c_2\\ s_2 & c_2 & 0 & a_2s_2 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]

Step 6: Calculate \(T_2^0\) matrix

\[ T_2^0=A_1A_2=\begin{bmatrix} c_{12} & -s_{12} & 0 & a_1c_1+a_2c_{12}\\ s_{12} & c_{12} & 0 & a_1s_1+a_2s_{12} \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \] Here, \(c_{12} = \cos(\theta_1+\theta_2)\) and \(s_{12} = \sin(\theta_1+\theta_2)\).

The end-effector coordinates (\(x,y\)) are the last column’s first two entries: \(x=a_1c_1+a_2c_{12}\), \(y=a_1s_1+a_2s_{12}\).

Compare these results with the trigonometry method for the same manipulator. They should be identical, illustrating that DH is a formalized way to achieve the same kinematic solution.

Forward Kinematics: Screw Theory

Screw Theory is an alternative to the Denavit-Hartenberg convention. It describes rigid motion as a twist about a screw axis.

Key Ideas:

• Only two coordinate systems are needed:
  • Space {s} frame: Fixed in space.
  • Body {b} frame: End-effector frame, moves with the robot.
• Uses the Product of Exponentials (PoE) formula for transformation matrix \(T(\theta)\).

Multiplication Rules

• If using {s} frame: Pre-multiplication (\(e^{\hat{S}_1\theta_1} \dots M\))
• If using {b} frame: Post-multiplication (\(M \dots e^{\hat{B}_n\theta_n}\))

Introduce screw theory as a more modern and often more elegant approach for certain types of analysis, particularly in advanced robotics and motion planning. Emphasize the concepts of space and body frames.

Screw Theory: Twists & Screw Axis

A twist defines a body’s instantaneous linear and angular velocity. It is described by a screw axis, which has:

• A point \(q\) on the axis.
• A unit vector \(s\) in the direction of the axis.
• A pitch \(h\).

Figure 2.12 Screw axis representation.

The pitch (\(h\)) is the ratio of linear to angular speed: \[ h=pitch=\frac{\textrm{linear speed}}{\textrm{angular speed}} \quad \text{(2.26)} \]

The screw axis \(S\) represents angular velocity (\(\omega\)) and linear velocity (\(v\)) components: \[ S=\begin{bmatrix} S_\omega\\ S_v \end{bmatrix}=\begin{bmatrix} \textrm{angular velocity when } \dot{\theta}=1\\ \textrm{linear velocity of the origin when } \dot{\theta}=1 \end{bmatrix} \quad \text{(2.27)} \]

Figure 2.13 Screw axis representation with the linear velocity of the origin defined.

The full twist \(V\) is: \[ V=\begin{bmatrix} \omega\\ v \end{bmatrix}=S\dot{\theta} \quad \text{(2.28)} \]

This slide defines the core concepts of screw theory: the screw axis and the twist. Explain how the pitch relates to rotation and translation along the axis. \(S_\omega\) is the unit direction vector for angular velocity, and \(S_v\) is the linear velocity of the origin when \(\dot{\theta}=1\).

Twists: Finite and Infinite Pitch

Finite Pitch:

• Both rotation and translation occur.
• \(\|S_\omega\|=1\)
• \(\dot{\theta}\) represents rotation speed.

Infinite Pitch (Prismatic Joint):

• Purely linear motion, no rotation.
• \(S_\omega=0\)
• \(\|S_v\|=1\)
• \(\dot{\theta}\) represents linear speed.

Body Twist (\(V_b\))

Defined in the {b} frame (end-effector): \[ V_b=(\omega_b,v_b)=S\dot{\theta} \quad \text{(2.29)} \]

Spatial Twist (\(V_s\))

Defined in the {s} frame (space/fixed): \[ V_s=(\omega_s,v_s)=S\dot{\theta} \quad \text{(2.30)} \]

Distinguish between finite and infinite pitch, connecting the latter to prismatic joints. Emphasize the concepts of body twist (local to end-effector) and spatial twist (global).

Example: Simple Twist

Consider a zero-pitch screw, with a pure rotation. The screw is pointing out of the screen (black dot). Imagine the dotted line rotates counterclockwise. Find the angular and linear velocity.

This corresponds to: \[ S_\omega=\begin{bmatrix} 0\\ 0 \\ 1 \end{bmatrix}, S_v=\begin{bmatrix} 0\\ -2 \\ 0 \end{bmatrix} \]

Explain how \(S_\omega\) points out of the page (\(z\)-direction) and \(S_v\) is the velocity of the origin if it were rotating around this axis. The “-2” in \(S_v\) suggests the origin is at \((0,2)\) and rotating around \(z\)-axis.

Screw Theory: Practical Procedure

To find forward kinematics using screw theory:

Step 1: Establish {s} and {b} frames

• {s} frame: Basic coordinate system fixed in space.
• {b} frame: End-effector coordinate system, moves with the robot.

Step 2: Determine \(M\)

The transformation matrix \(M\) of the end-effector frame when all joint variables \(\theta = 0^\circ\).

Step 3: Find Screw Axes

Determine the {s} frame screw axes \(S_1, \dots, S_n\) or {b} frame screw axes \(B_1, \dots, B_n\) for each of the \(n\) joint axes when \(\theta = 0^\circ\).

Step 4: Evaluate the Product of Exponentials (PoE)

For {s} frame (space frame): \[ T(\theta)=e^{\hat{S}_1\theta_1}e^{\hat{S}_2\theta_2}\cdots e^{\hat{S}_n\theta_n}M \quad \text{(2.31)} \]

For {b} frame (body frame): \[ T(\theta)=Me^{\hat{B}_1\theta_1}e^{\hat{B}_2\theta_2}\cdots e^{\hat{B}_n\theta_n} \quad \text{(2.32)} \]

This procedure systematizes the application of screw theory. \(\hat{S}_i\) and \(\hat{B}_i\) are the \(4 \times 4\) matrix representations of the screw axes. The exponential map \(e^{\hat{S}\theta}\) transforms a twist into a homogeneous transformation.

Example: Screw Theory in {s} frame

Consider an RPR (Revolute-Prismatic-Revolute) robot in its’ zero-frame orientation and then in a rotated orientation. Derive the forward kinematics using Screw Theory.

Zero-frame orientation: \(\theta=\begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}\)

Rotated orientation: \(\theta=\begin{bmatrix} 0 \\ 0.5 \\ 45^\circ\end{bmatrix}\)

Step 1 & 2: {s} and {b} frames, and \(M\)

Analyzing the zero-frame, the matrix \(M\) (initial end-effector pose in space frame) is: \[ M=\begin{bmatrix} 1 & 0 & 0 & 3\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]

Step 3: {s} frame Screw Axes (\(S_i\))

\[ S_1=\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0\end{bmatrix}, S_2=\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0\end{bmatrix}, S_3=\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \\ -2 \\ 0\end{bmatrix} \]

Explain how \(M\) is derived: The rotation part is identity as {b} is aligned with {s} at zero configuration. The translation part (\(d\)) comes from the end-effector’s position relative to the space frame at \(\theta=0\).

For the screw axes:

\(S_1\): Revolute about Z-axis at origin (no linear velocity).

\(S_2\): Prismatic along X-axis (no angular velocity, unit linear velocity in X).

\(S_3\): Revolute about Z-axis, but not at the origin (so has an \(v = -\omega \times q\) linear component).

Chapter Summary

We’ve explored three methods for solving Forward Kinematics:

1. Trigonometry Method:
   • Best for simple, 2D planar robots.
   • Intuitive geometric understanding.
2. Denavit-Hartenberg (DH) Convention:
   • Most commonly used, systematic approach.
   • Requires defining four parameters (\(a, \alpha, d, \theta\)) for each link-joint pair.
   • Results in a chain of homogeneous transformations.
3. Screw Theory:
   • Alternative, more modern approach.
   • Focuses on twists about screw axes.
   • Uses Product of Exponentials (PoE) formula.
   • Requires defining fixed space frame {s} and end-effector frame {b}.

Each method has its strengths and is chosen based on robot complexity and analysis requirements.

Recap the key takeaways for each method, highlighting their appropriate use cases. Emphasize that all methods aim to achieve the same goal: determining the end-effector’s pose.

Useful Properties for Practice

These trigonometric identities are often useful in kinematic derivations:

• Pythagorean identity: \(\cos^2\theta+\sin^2\theta=1 \quad \text{(2.33)}\)

• Angle addition/subtraction (example given is incorrect, correct version is below): \(\cos(\theta_1+\theta_2)\cos\theta_1+\sin(\theta_1+\theta_2)\sin\theta_1 = \cos(\theta_2) \quad \text{(2.34)}\) (Original was \(\cos(\theta_1 + \theta_2 - \theta_1) = \cos\theta_1\), which simplifies to \(\cos(\theta_2)\))

• Sine addition/subtraction: \(\sin(\alpha\pm\beta)=\sin\alpha \cos\beta \pm \cos\alpha \sin\beta \quad \text{(2.35)}\)

• Cosine addition/subtraction: \(\cos(\alpha\pm\beta)=\cos\alpha \cos\beta \mp \sin\alpha \sin\beta \quad \text{(2.36)}\)

Briefly mention the common identities. Correct the potentially confusing equation (2.34) in the original text for clarity.

Practice Question 1: Rotation Matrix

Frame {0} and frame {1} are shown below. Find the resulting rotation matrix \(R_1^0\).

Answer:

First, identify the unit vectors of frame {1} in terms of frame {0}: \(x_1 = \begin{bmatrix} \frac{1}{\sqrt{2}}\\ 0 \\ \frac{1}{\sqrt{2}} \end{bmatrix}, y_1 = \begin{bmatrix} \frac{1}{\sqrt{2}}\\ 0 \\ -\frac{1}{\sqrt{2}} \end{bmatrix}, z_1 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}\)

Then, form \(R_1^0\) by these column vectors: \[ R_1^0 = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0\\ 0 & 0 & 1\\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0\end{bmatrix} \]

Walk through how to extract the coordinates of \(x_1, y_1, z_1\) from the image relative to frame {0}. For example, \(x_1\) has equal components in \(x_0\) and \(z_0\), and no component in \(y_0\).

Practice Question 2: Chained Rotations (Current Axes)

An x-y-z axis coordinate system experiences:

1. A rotation of angle \(\phi\) about the current y-axis.
2. A rotation of angle \(\theta\) about the current z-axis.

Find the resulting rotation matrix.

Answer:

Since both rotations are about current axes, we post-multiply: \(R = R_{y,\phi} R_{z,\theta}\)

\[ R=\begin{bmatrix} \cos\phi & 0 & \sin\phi \\ 0 & 1 & 0 \\ -\sin\phi & 0 & \cos\phi\end{bmatrix}\begin{bmatrix} \cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1\end{bmatrix} \]

\[ R=\begin{bmatrix} c_\phi c_\theta & -c_\phi s_\theta & s_\phi \\ s_\theta & c_\theta & 0 \\ -s_\phi c_\theta & s_\phi s_\theta & c_\phi \end{bmatrix} \]

This question tests the understanding of post-multiplication. Briefly explain the matrix multiplication steps if time permits.

Practice Question 3: Chained Rotations (Fixed & Current)

Find the rotation matrix for the following sequence:

1. Rotate by \(\alpha\) about the current x-axis.
2. Rotate by \(\theta\) about the fixed z-axis.
3. Rotate by \(\phi\) about the fixed y-axis.

Answer: We build the total rotation matrix \(R\) by applying the fixed/current rule: \(R = Rot_{y,\phi} Rot_{z,\theta} Rot_{x,\alpha}\)

\[ R = \begin{bmatrix} c_\phi & 0 & s_\phi \\ 0 & 1 & 0 \\ -s_\phi & 0 & c_\phi\end{bmatrix}\begin{bmatrix}c_\theta & -s_\theta & 0 \\ s_\theta & c_\theta & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & c_\alpha & -s_\alpha \\ 0 & s_\alpha & c_\alpha \end{bmatrix} \]

\[ R = \begin{bmatrix} c_\phi c_\theta & s_\phi s_\alpha - c_\phi s_\theta c_\alpha & c_\phi s_\theta s_\alpha + s_\phi c_\alpha \\ s_\theta & c_\theta c_\alpha & -c_\theta s_\alpha \\ -s_\phi c_\theta & s_\phi s_\theta c_\alpha + c_\phi s_\alpha & c_\phi c_\alpha - s_\phi s_\theta s_\alpha \end{bmatrix} \]

This is a more complex example combining both rules. It’s a good test of understanding the order of matrix multiplication based on the frame of rotation.

Practice Question 4: Homogeneous Transformation

Frame {0} and frame {1} are shown below with their respective vectors. All axes are parallel (\(z_0 || y_1, x_0 || x_1, y_0 || z_1\)). Find the position vector \(p^0\) and the homogeneous transformation matrix \(H_1^0\).

Answer: From the image, \(x_0 || x_1\), \(y_0 || z_1\), \(z_0 || y_1\). This means frame {1} is rotated 90 degrees about \(x_0\) relative to frame {0}.

So, \(R_1^0 = R_{x,90^\circ} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{bmatrix}\).

The translation vector \(d_1^0 = \begin{bmatrix} 0 \\ 3 \\ 1 \end{bmatrix}\).

The point \(p^1 = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}\).

Position vector \(p^0 = R_1^0 p^1 + d_1^0\):

\(p^0 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 3 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix} + \begin{bmatrix} 0 \\ 3 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \\ 2 \end{bmatrix}\)

Homogeneous transformation matrix \(H_1^0\): \[ H_1^0=\begin{bmatrix} R_1^0 & d_1^0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & -1 & 3 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]

This combines rotation matrices with translation vectors using the rigid motion equation and homogeneous transformation. Emphasize visual interpretation of axis alignment to find \(R_1^0\) and \(d_1^0\). (Note: \(p^0\) calculation in provided text p0 = [0,2,2] seems incorrect based on the image and \(p^1\) given.) My calculation uses the \(R_{x,90}\) and \(d_1^0\) from the image which yields [2,3,2].

Practice Question 5: DH Convention - Cylindrical Robot

Derive the forward kinematics equations using the DH Convention for the cylindrical robot shown.

Answer:

DH Table:

Link	\(a_i\)	\(\alpha_i\)	\(d_i\)	\(\theta_i\)
1	0	0°	\(d_1\)	\(\theta_1^*\)
2	0	-90°	\(d_2\)	0
3	0	0°	\(d_3\)	0

\(A_i\) Matrices:

\[ A_1=\begin{bmatrix} c_1 & -s_1 & 0 & 0 \\ s_1 & c_1 & 0 & 0 \\ 0 & 0 & 1 & d_1 \\ 0 & 0 & 0 & 1 \end{bmatrix} \] \[ A_2=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & d_2 \\ 0 & 0 & 0 & 1 \end{bmatrix} \] \[ A_3=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & d_3 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]

\(T_3^0\) Matrix:

\(T_3^0=A_1A_2A_3=\begin{bmatrix} c_1 & 0 & -s_1 & -d_3 s_1 \\ s_1 & 0 & c_1 & d_3 c_1 \\ 0 & -1 & 0 & d_1 + d_2 \\ 0 & 0 & 0 & 1 \end{bmatrix}\)

This is a standard application of DH for a cylindrical robot (R-P-P). Go through the DH parameters for each link and how they contribute to the \(A_i\) matrices.

Practice Question 6: DH Convention - Robot Arm

Derive the forward kinematics equation using the DH Convention for the following robot.

Answer:

DH Table:

Link	\(a_i\)	\(\alpha_i\)	\(d_i\)	\(\theta_i\)
1	0	90°	\(a_1\)	\(\theta_1^*+90^\circ\)
2	0	0°	\(a_2+a_3+d_2\)	0

\(A_i\) Matrices:

\[ A_1=\begin{bmatrix} -s_1 & 0 & c_1 & 0 \\ c_1 & 0 & s_1 & 0 \\ 0 & 1 & 0 & a_1 \\ 0 & 0 & 0 & 1 \end{bmatrix} \] \[ A_2=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & a_2 + a_3 + d_2 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]

\(T_2^0\) Matrix:

\(T_2^0 = A_1 A_2 = \begin{bmatrix} -s_1 & 0 & c_1 & (a_2 + a_3 + d_2) c_1 \\ c_1 & 0 & s_1 & (a_2 + a_3 + d_2)s_1 \\ 0 & 1 & 0 & a_1 \\ 0 & 0 & 0 & 1 \end{bmatrix}\)

This example includes a constant offset in the joint angle (e.g., \(\theta_1^*+90^\circ\)), which is a common occurrence in DH setup. Explain how this is incorporated into the DH table and \(A_i\) matrix.

Practice Question 7: Twist - Angular and Linear Velocity

Find the angular and linear velocity for the following illustration.

Answer: \[ S_\omega=\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, S_v=\begin{bmatrix} \frac{-2}{\sqrt{2}} \\ \frac{-2}{\sqrt{2}} \\ 0 \end{bmatrix} \]

Interpret the image: \(S_\omega\) is along the z-axis (out of page). \(S_v\) is the linear velocity of the origin if it were rotating around the screw axis. The screw axis is at \((2,2,0)\). The velocity of the origin \((0,0,0)\) due to rotation \(\omega=[0,0,1]^T\) around point \(q=[2,2,0]^T\) is \(v = -\omega \times q = -[0,0,1]^T \times [2,2,0]^T = -[-2,2,0]^T = [2,-2,0]^T\). The provided answer \(S_v\) is \([-2/\sqrt{2}, -2/\sqrt{2}, 0]^T\). This suggests a rotation axis passing through origin, and some linear velocity. The question is ambiguous about the origin’s velocity relative to the screw axis. Given the diagram, it seems the point \((0,0)\) is rotating around the screw axis located at some point not through origin. The provided answer for \(S_v\) implies a rotation where the origin moves along the direction \((-1,-1,0)\) for unit speed. Let’s assume the provided answer is correct, and interpret what it means. \(S_\omega = [0,0,1]^T\) means rotation about z-axis. \(S_v = [-2/\sqrt{2}, -2/\sqrt{2}, 0]^T\) means linear velocity is in (-1,-1) direction. This could be a screw axis parallel to Z, passing through a point. For the purpose of this presentation, we’ll present the answer as given in the source.

Practice Question 8: Twist - Angular and Linear Velocity

Find the angular and linear velocity for the following illustration.

Answer: \[ S_\omega=\begin{bmatrix} 0 \\ -1 \\ 0 \end{bmatrix}, S_v = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

In this case, the rotation is about the y-axis, and the axis passes through the origin, hence no linear velocity of the origin (\(S_v = 0\)).

Practice Question 9: Screw Theory - RPR Arm

Consider the following robot in its’ zero-frame orientation. Derive the forward kinematics using Screw Theory using either the space or end-effector frame.

Answer:

Initial Configuration Matrix \(M\):

\[ M=\begin{bmatrix}0 & -1 & 0 & 19 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & -1 & -3 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]

Space Frame Screw Axes (\(S_i\)):

\[ S_1=\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, S_2=\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \\ -10 \\ 0 \end{bmatrix}, S_3=\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \\ -19 \\ 0 \end{bmatrix}, S_4=\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \]

End Effector Frame Screw Axes (\(B_i\)):

\[ B_1=\begin{bmatrix} 0 \\ 0 \\ -1 \\ -19 \\ 0 \\ 0 \end{bmatrix}, B_2=\begin{bmatrix} 0 \\ 0 \\ -1 \\ -9 \\ 0 \\ 0 \end{bmatrix}, B_3=\begin{bmatrix} 0 \\ 0 \\ -1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, B_4=\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ -1 \end{bmatrix} \]

This is a comprehensive screw theory problem. \(M\) defines the initial pose of the end-effector. \(S_i\) are calculated with respect to the fixed space frame. \(B_i\) are calculated with respect to the body frame, transformed by \(M^{-1}\) from \(S_i\). Briefly discuss the difference between \(S_i\) and \(B_i\).