4. Homogeneous Transformations
In order to position the end-effector of a robot at a specific point in the workspace, the robot must go through a series of rotations and translations of its links about the joints. This requires creating a composite homogeneous transformation matrix, which is the result of matrix multiplications. It is important to note that matrix multiplication is non-commutative, meaning the order of the multiplications matters. To create the composite matrix, the following rules should be followed:
-
Initially, the fixed and mobile coordinate systems are coincident, so the homogeneous transformation matrix is an identity matrix.
-
If the mobile frame (OABC) is rotating or translating about one of the axes (OXYZ) of the fixed frame, the previous result matrix should be "pre-multiplied" by the current homogeneous rotation or translation matrix.
-
If the mobile coordinate system (OABC) is rotating or translating about one of its own axes, the previous result matrix should be "post-multiplied" by the present homogeneous transformation matrix.
|
Example 1 |
Determine the homogeneous transformation matrix to represent a rotation of \(40^\circ\) about OX-axis and a translation of \(7\) units along the OB-axis of the mobile frame. |
|
Solution:
\[\begin{align*}
H(x,\alpha) &= H(x, 40^{\circ}) \\
H(x,\beta) &= H(B, 7) \\
H &= H(x, 40^{\circ}) \cdot I \cdot H(B, 7) \\
&=\left[\begin{array}{cccc}1 & 0 & 0 & 0 \\0 & \cos 40^{\circ} & -\sin 40^{\circ} & 0 \\0 & \sin 40^{\circ} & \cos 40^{\circ} & 0 \\0 & 0 & 0 & 1\end{array}\right] \cdot \left[\begin{array}{cccc}1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1\end{array}\right] \cdot \left[\begin{array}{cccc}1 & 0 & 0 & 0 \\0 & 1 & 0 & 7 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1\end{array}\right] \\
&=\left[\begin{array}{cccc}1 & 0 & 0 & 0 \\0 & \cos 40^{\circ} & -\sin 40^{\circ} & 0 \\0 & \sin 40^{\circ} & \cos 40^{\circ} & 0 \\0 & 0 & 0 & 1\end{array}\right] \cdot \left[\begin{array}{cccc}1 & 0 & 0 & 0 \\0 & 1 & 0 & 7 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1\end{array}\right] \\
&=\left[\begin{array}{cccc}1 & 0 & 0 & 0 \\0 & 0.7660 & -0.6428 & 0 \\0 & 0.6428 & 0.7660 & 0 \\0 & 0 & 0 & 1\end{array}\right] \cdot \left[\begin{array}{cccc}1 & 0 & 0 & 0 \\0 & 1 & 0 & 7 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1\end{array}\right] \\
&=\left[\begin{array}{cccc}1 & 0 & 0 & 0 \\0 & 0.7660 & -0.6428 & 5.3620 \\0 & 0.6428 & 0.7660 & 4.4996 \\0 & 0 & 0 & 1\end{array}\right]\\
\end{align*}\]
|
The following is a Python code implementation of the above calculation (Click to reveal). This code implementation provides 3 outputs: a matrix in surd form, 4 decimal numerical values, and a latex format.
|
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Example 2 |
Determine the overall homogeneous transformation matrix for the following sequence of operations. Rotation of \(50^\circ\) OX-axis, translation of \(4\) units along OX-axis, translation of \(-6\) units along OC-axis, and rotation of \(25^\circ\) about OB-axis. |
|
Solution:
\[\begin{align*}
H &= H(x, 4) \cdot H(x, 50^{\circ}) \cdot I \cdot H(c, -6) \cdot H(b, 25^{\circ}) \\
&=\left[\begin{array}{cccc}1 & 0 & 0 & 4 \\0 & 1 & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1\end{array}\right]
\left[\begin{array}{cccc}1 & 0 & 0 & 0 \\0 & 0.6428 & -0.7660 & 0 \\0 & 0.7660 & 0.6428 & 0 \\0 & 0 & 0 & 1\end{array}\right] \\
& \times \left[\begin{array}{cccc}1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & 1 & -6 \\0 & 0 & 0 & 1\end{array}\right]
\left[\begin{array}{cccc}0.9063 & 0 & 0.4226 & 0 \\0 & 1 & 0 & 0 \\-0.4226 & 0 & 0.9063 & 0 \\0 & 0 & 0 & 1\end{array}\right] \\
&= \left[\begin{matrix}0.9063 & 0 & 0.4226 & 4.0\\0.3237 & 0.6428 & -0.6943 & 4.596\\-0.2717 & 0.766 & 0.5826 & -3.857\\0 & 0 & 0 & 1.0\end{matrix}\right]
\end{align*}\]
|
The following is a Python code implementation of the above calculation (Click to reveal). This code implementation provides 3 outputs: a matrix in surd form, 4 decimal numerical values, and a latex format.
|
4.1. Composite Homogeneous Transformations
|
Example 3 |
A robotic work cell has a camera within the setup. The camera can see the origin of the six-joint robot fixed to a base. A cube placed on the work cell table is also seen by the camera. The homogeneous transformation matrix \(H_1\) maps the camera to the centre of the cube. The origin of the base coordinate system as seen from the camera is represented by the homogeneous transformation matrix \(H_2\). |
|
\[\begin{aligned}
H_1 =
\begin{bmatrix}
0 & 1 & 0 & 2\\
1 & 0 & 0 & 8\\
0 & 0 & -1 & 7\\
0 & 0 & 0 & 1
\end{bmatrix},
H_2 =
\begin{bmatrix}
1 & 0 & 0 & -8\\
0 & -1 & 0 & 15\\
0 & 0 & -1 & 6\\
0 & 0 & 0 & 1
\end{bmatrix}
\end{aligned}\]
Solution:
The position and orientation of the cube with respect to the base coordinate system.
\[\begin{aligned}
& { }^\text{camera}H_{\text {cube }}=H_{1}=\left[\begin{array}{rrrr}0 & 1 & 0 & 2 \\1 & 0 & 0 & 8 \\0 & 0 & -1 & 7 \\0 & 0 & 0 & 1\end{array}\right] \\
& { }^\text{camera}H_{\text {base }}=H_{2}=\left[\begin{array}{rrrr}1 & 0 & 0 & -8 \\0 & -1 & 0 & 15 \\0 & 0 & -1 & 6 \\0 & 0 & 0 & 1\end{array}\right] \\
& \text {To find }{ }^{\text{base}} H_{\text {cube }} \text { by chain product rule, } \\
{ }^{\text{base}} H_{\text {cube }} &= { }^\text{base}H_{\text {camera}} \text{ . } { }^\text{camera}H_{\text {cube}} = (H_{2})^{-1} \text{ . } H_{1} \\
& =\left[\begin{matrix}1 & 0 & 0 & 8\\0 & -1 & 0 & 15\\0 & 0 & -1 & 6\\0 & 0 & 0
& 1\end{matrix}\right]
\left[\begin{matrix}0 & 1 & 0 & 2\\1 & 0 & 0 & 8\\0 & 0 & -1 & 7\\0 & 0 & 0 &
1\end{matrix}\right]\\
&=\left[\begin{matrix}0 & 1 & 0 & 10\\-1 & 0 & 0 & 7\\0 & 0 & 1 & -1\\0 & 0 & 0
& 1\end{matrix}\right] \\
&\text{The position of the cube is given by } \left[\begin{matrix}10\\7\\-1\end{matrix}\right] \\
&\text{The orientation } [n, s, a]=\left[\begin{matrix}0 & 1 & 0\\-1 & 0 & 0\\0 & 0 & 1\end{matrix}\right] \\
\end{aligned}\]
The position and orientation of the camera with respect to the robot’s base coordinate system.
\[\begin{aligned}
{ }^\text{base}H_{\text {camera}} &=\left(H_{2}\right)^{-1} \cdot H\left(z, 60^{\circ}\right)_{\text {camera}} \\
H\left(z, 60^{\circ}\right)_{\text{camera}} &=\left[\begin{array}{cccc}
\cos 60^{\circ} & -\sin 60^{\circ} & 0 & 0 \\
\sin 60^{\circ} & \cos 60^{\circ} & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{array}\right] \\
{ }^{\text {base }} \mathrm{H}_{\text {camera }} &=\left[\begin{matrix}1 & 0 & 0 & 8\\0 & -1 & 0 & 15\\0 & 0 & -1 & 6\\0 & 0 & 0
& 1\end{matrix}\right]
\left[\begin{matrix}\frac{1}{2} & - \frac{\sqrt{3}}{2} & 0 & 0\\\frac{\sqrt{3}}{2} & \frac{1}{2} & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{matrix}\right]\\
&=\left[\begin{matrix}\frac{1}{2} & - \frac{\sqrt{3}}{2} & 0 & 8\\- \frac{\sqrt{3}}{2} & - \frac{1}{2} & 0 & 15\\0 & 0 & -1 & 6\\0 & 0 & 0 & 1\end{matrix}\right] \\
&\text{The position of the camera after the change is given by } \\
&\left[\begin{matrix}8\\15\\6\end{matrix}\right] \\
&\text{The orientation of the camera with respect to the base } \\
&[n, s, a]=\left[\begin{matrix}\frac{1}{2} & - \frac{\sqrt{3}}{2} & 0\\- \frac{\sqrt{3}}{2} & - \frac{1}{2} & 0\\0 & 0 & -1\end{matrix}\right] \\
\end{aligned}\]
The position and orientation of the cube with respect to the robot’s base coordinate system?
\[\begin{aligned}
{ }^{\text{base}} H_{\text{cube-2}} &={ }^{\text {base }} H_{\text {cube }} \text{ . } H\left(x, 60^{\circ}\right) \text{ . } H (y, 5) \\
{ }^{\text{base}} H_{\text{cube-2}} &=\left[\begin{matrix}0 & 1 & 0 & 10\\-1 & 0 & 0 & 7\\0 & 0 & 1 & -1\\0 & 0 & 0
& 1\end{matrix}\right]
\left[\begin{matrix}1 & 0 & 0 & 0\\0 & \frac{1}{2} & - \frac{\sqrt{3}}{2} & 0\\0 & \frac{\sqrt{3}}{2} & \frac{1}{2} & 0\\0 & 0 & 0 & 1\end{matrix}\right]
\left[\begin{matrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 5\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1
\end{matrix}\right]\\
&=\left[\begin{matrix}0 & \frac{1}{2} & - \frac{\sqrt{3}}{2} & \frac{25}{2}\\-1
& 0 & 0 & 7\\0 & \frac{\sqrt{3}}{2} & \frac{1}{2} & -1 + \frac{5 \sqrt{3}}{2}\\0 & 0 & 0 & 1\end{matrix}\right]
\\
&\text{The position of the object with respect to the base } \\
&= \left[\begin{matrix}\frac{25}{2}\\7\\-1 + \frac{5 \sqrt{3}}{2}\end{matrix}\right] \\
&\text {The orientation, }[n, s, a]=\left[\begin{matrix}0 & \frac{1}{2} & - \frac{\sqrt{3}}{2}\\-1 & 0 & 0\\0 & \frac{\sqrt{3}}{2} & \frac{1}{2}\end{matrix}\right] \\
\end{aligned}\]
|
The following is a Python code implementation of the above calculation (Click to reveal).
|
|
Example 4 |
A six-joint robotic manipulator equipped with a digital television camera is capable of continuously monitoring the position and orientation of an object. The position and orientation of the object with respect to the camera are expressed by a matrix \((T_1)\), the origin of the robot’s base coordinate with respect to the camera is given by \((T_2)\), and the position and orientation of the gripper with respect to the base coordinate frame are given by \((T_3)\). |
|
\[\begin{aligned}
T_1 =
\left[\begin{matrix}0 & 1 & 0 & 4\\1 & 0 & 0 & 3\\0 & 0 & -1 & 7\\0 & 0 & 0 &
1\end{matrix}\right],
T_2 =
\left[\begin{matrix}1 & 0 & 0 & -15\\0 & -1 & 0 & 5\\0 & 0 & -1 & 6\\0 & 0 & 0
& 1\end{matrix}\right],
T_3 =
\left[\begin{matrix}1 & 0 & 0 & 7\\0 & 1 & 0 & 2\\0 & 0 & 1 & 3\\0 & 0 & 0 & 1
\end{matrix}\right]
\end{aligned}\]
Solution:
The position and orientation of the object with respect to the base coordinate system.
\[\begin{aligned}
% \left[{ }^{\text{base}}T_\text{object} \right] \text{ by chain product rule, } \\
{ }^\text{base}T_{\text {object}} &={ }^{\text {base}} T_{\text {camera}} . { }^{\text {camera}} T_{\text {object}} \\
&=\left[T_{2}\right]^{-1} \cdot\left[T_{1}\right] \\
&=\left[\begin{matrix}1 & 0 & 0 & 15\\0 & -1 & 0 & 5\\0 & 0 & -1 & 6\\0 & 0 & 0
& 1\end{matrix}\right]
\left[\begin{matrix}0 & 1 & 0 & 4\\1 & 0 & 0 & 3\\0 & 0 & -1 & 7\\0 & 0 & 0 &
1\end{matrix}\right] \\
&=\left[\begin{matrix}0 & 1 & 0 & 19\\-1 & 0 & 0 & 2\\0 & 0 & 1 & -1\\0 & 0 & 0
& 1\end{matrix}\right] \\
&\text{The position vector} =\left[\begin{matrix}19\\2\\-1\end{matrix}\right] \\
&\text{The orientation matrix, } \\
&[n, s, a]=\left[\begin{matrix}0 & 1 & 0\\-1 & 0 & 0\\0 & 0 & 1\end{matrix}\right] \\
\end{aligned}\]
The position and orientation of the object with respect to the gripper.
\[\begin{aligned}
{ }^\text {gripper} T_{\text {object}} &= { }^\text{gripper}T_{\text {base }} \cdot{ }^{\text {base }} T_{\text {object }}=\left[T_{3}\right]^{-1}\left(\left[T_{2}\right]^{-1}\left[T_{1}\right]\right) \\
& =\left[\begin{matrix}1 & 0 & 0 & -7\\0 & 1 & 0 & -2\\0 & 0 & 1 & -3\\0 & 0 & 0
& 1\end{matrix}\right]
\left[\begin{matrix}0 & 1 & 0 & 19\\-1 & 0 & 0 & 2\\0 & 0 & 1 & -1\\0 & 0 & 0
& 1\end{matrix}\right] \\
&= \left[\begin{matrix}0 & 1 & 0 & 12\\-1 & 0 & 0 & 0\\0 & 0 & 1 & -4\\0 & 0 & 0
& 1\end{matrix}\right] \\
&\text{The position vector } =\left[\begin{matrix}12\\0\\-4\end{matrix}\right] \\
&\text{The orientation vectors, }\\
&[n, s, a]=\left[\begin{matrix}0 & 1 & 0\\-1 & 0 & 0\\0 & 0 & 1\end{matrix}\right] \\
\end{aligned}\]
|
The following is a Python code implementation of the above calculation (Click to reveal).
|
|
Example 5 |
A triangular prism with the coordinates of its vertices \(A(1, 3, 0)\), \(B(-1, 3, 0)\), \(C(-1, 3, 2)\), \(D(1, 3, 2)\), \(E(1, 5, 2)\) and \(F(-1, 5, 2)\) relative to the fixed reference frame \(OXYZ\) is shown in Figure 7. The prism is moved to the new position with a rotation of \(+60^\circ\) about the \(X\) axis, a rotation of \(-60^\circ\) about the \(Z\) axis and a translation of 5 units in the \(Y\) direction. |
|
Figure 6. Prism bot.
|
Solution:
The homogeneous transformation that describes the change in position of the prism.
\[\begin{aligned}
%\iota
H &= H_{t}(0,5,0) H\left(z,-60^{\circ}\right) H\left(x,+60^{\circ}\right) . I_{4} \\
&=\left[\begin{array}{llll}1 & 0 & 0 & 0 \\0 & 1 & 0 & 5 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1\end{array}\right]\left[\begin{array}{cccc}\cos \left(-60^{\circ}\right) & -\sin \left(-60^{\circ}\right) & 0 & 0 \\\sin \left(-60^{\circ}\right) & \cos \left(-60^{\circ}\right) & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1\end{array}\right] \\ &\times\left[\begin{array}{cccc}1 & 0 & 0 & 0 \\0 & \cos 60^{\circ} & -\sin 60^{\circ} & 0 \\0 & \sin 60^{\circ} & \cos 60^{\circ} & 0 \\0 & 0 & 0 & 1\end{array}\right] \\
& =\left[\begin{matrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 5\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1
\end{matrix}\right]
\left[\begin{matrix}\frac{1}{2} & \frac{\sqrt{3}}{2} & 0 & 0\\- \frac{\sqrt{3}
}{2} & \frac{1}{2} & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{matrix}\right]
\left[\begin{matrix}1 & 0 & 0 & 0\\0 & \frac{1}{2} & - \frac{\sqrt{3}}{2} & 0\\0 & \frac{\sqrt{3}}{2} & \frac{1}{2} & 0\\0 & 0 & 0 & 1\end{matrix}\right] \\
& =\left[\begin{matrix}\frac{1}{2} & \frac{\sqrt{3}}{4} & - \frac{3}{4} & 0\\- \frac{\sqrt{3}}{2} & \frac{1}{4} & - \frac{\sqrt{3}}{4} & 5\\0 & \frac{\sqrt{3}}
{2} & \frac{1}{2} & 0\\0 & 0 & 0 & 1\end{matrix}\right] \\
\end{aligned}\]
The new coordinates of the vertices \(A,B,C,D,E\) of the prism.
\[\begin{aligned}
\left[\begin{array}{c}
\mathrm{A}_{x} \\
\mathrm{~A}_{y} \\
\mathrm{~A}_{z} \\
1
\end{array}\right] &=\left[\begin{matrix}\frac{1}{2} & \frac{\sqrt{3}}{4} & - \frac{3}{4} & 0\\- \frac{\sqrt{3}}{2} & \frac{1}{4} & - \frac{\sqrt{3}}{4} & 5\\0 & \frac{\sqrt{3}}
{2} & \frac{1}{2} & 0\\0 & 0 & 0 & 1\end{matrix}\right]\left[\begin{array}{l}
q_{x} \\
q_{y} \\
q_{z} \\
1
\end{array}\right] \\
\mathrm{A} &= [\mathrm{H}][q] \\
\left[\begin{array}{c}
\mathrm{A}_{x} \\
\mathrm{~A}_{y} \\
\mathrm{~A}_{z} \\
1
\end{array}\right] &= \left[\begin{matrix}\frac{1}{2} & \frac{\sqrt{3}}{4} & - \frac{3}{4} & 0\\- \frac{\sqrt{3}}{2} & \frac{1}{4} & - \frac{\sqrt{3}}{4} & 5\\0 & \frac{\sqrt{3}}
{2} & \frac{1}{2} & 0\\0 & 0 & 0 & 1\end{matrix}\right]\left[\begin{array}{l}
1 \\
3 \\
0 \\
1
\end{array}\right] = \left[\begin{matrix}\frac{1}{2} + \frac{3 \sqrt{3}}{4}\\\frac{23}{4} - \frac{\sqrt{3}}{2}\\\frac{3 \sqrt{3}}{2}\\1\end{matrix}\right] \\
[\mathrm{B}] &= [\mathrm{H}][b] \\
\left[\begin{array}{c}
\mathrm{B}_{x} \\
\mathrm{~B}_{y} \\
\mathrm{~B}_{z} \\
1
\end{array}\right] &= \left[\begin{matrix}\frac{1}{2} & \frac{\sqrt{3}}{4} & - \frac{3}{4} & 0\\- \frac{\sqrt{3}}{2} & \frac{1}{4} & - \frac{\sqrt{3}}{4} & 5\\0 & \frac{\sqrt{3}}
{2} & \frac{1}{2} & 0\\0 & 0 & 0 & 1\end{matrix}\right]\left[\begin{array}{r}
-1 \\
3 \\
0 \\
1
\end{array}\right] = \left[\begin{matrix}- \frac{1}{2} + \frac{3 \sqrt{3}}{4}\\\frac{\sqrt{3}}{2} + \frac{23}{4}\\\frac{3 \sqrt{3}}{2}\\1\end{matrix}\right] \\
{[C] } &=[H][c] \\
&=\left[\begin{matrix}\frac{1}{2} & \frac{\sqrt{3}}{4} & - \frac{3}{4} & 0\\- \frac{\sqrt{3}}{2} & \frac{1}{4} & - \frac{\sqrt{3}}{4} & 5\\0 & \frac{\sqrt{3}}
{2} & \frac{1}{2} & 0\\0 & 0 & 0 & 1\end{matrix}\right]\left[\begin{array}{r}
-1 \\
3 \\
2 \\
1
\end{array}\right] = \left[\begin{matrix}-2 + \frac{3 \sqrt{3}}{4}\\ \frac{23}{4}\\1 + \frac{3 \sqrt{3}}{2}\\1\end{matrix}\right] \\
{[D] } &=[\mathrm{H}][d] \\
&=\left[\begin{matrix}\frac{1}{2} & \frac{\sqrt{3}}{4} & - \frac{3}{4} & 0\\- \frac{\sqrt{3}}{2} & \frac{1}{4} & - \frac{\sqrt{3}}{4} & 5\\0 & \frac{\sqrt{3}}
{2} & \frac{1}{2} & 0\\0 & 0 & 0 & 1\end{matrix}\right]\left[\begin{array}{l}
1 \\
3 \\
2 \\
1
\end{array}\right] = \left[\begin{matrix}-1 + \frac{3 \sqrt{3}}{4}\\ \frac{23}{4} - \sqrt{3}\\1 + \frac{3 \sqrt{3}}{2}\\1\end{matrix}\right] \\
{[\mathrm{E}] } &=[\mathrm{H}][e] \\
&=\left[\begin{matrix}\frac{1}{2} & \frac{\sqrt{3}}{4} & - \frac{3}{4} & 0\\- \frac{\sqrt{3}}{2} & \frac{1}{4} & - \frac{\sqrt{3}}{4} & 5\\0 & \frac{\sqrt{3}}
{2} & \frac{1}{2} & 0\\0 & 0 & 0 & 1\end{matrix}\right]\left[\begin{array}{l}
1 \\
5 \\
2 \\
1
\end{array}\right] = \left[\begin{matrix}-1 + \frac{5 \sqrt{3}}{4}\\ \frac{25}{4} - \sqrt{3}\\1 + \frac{5 \sqrt{3}}{2}\\1\end{matrix}\right] \\
\end{aligned}\]
|
The following is a Python code implementation of the above calculation (Click to reveal).
|
|
Assignment 1 |
Compute the homogeneous transformation matrices for the coordinate frames situated at the points \(A(10, 10, 0)\), \(B(5, -20, 5)\) and \(C(0, 6, 6)\) with respect to the \(OX_0Y_0Z_0\) frame as shown in Figure 8. |
|
Figure 7. Coordinate frames.
|
Solution:
Please attempt this assignment. |
The following is a Python code implementation of the above calculation (Click to reveal).
|
|
Assignment 2 |
Compute the homogeneous transformation matrices for the coordinate frames attached to the corners \(A\), \(B\), \(C\) and \(D\) with respect to the base coordinate frame \(O\). |
|
Figure 8. Coordinate frames.
|
Solution:
Please attempt this assignment. |
The following is a Python code implementation of the above calculation (Click to reveal).
|
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Comments (4)
Using the knowledge of spatial mechanics and computer programming introduced in this course, you can go on to perform advanced kinematic analysis of 3D mechanisms.
These skills apply to robotics systems design, research, and development.
The notes are on point. I've gained alot of skill and visualization from the unit.
Does it mean the inverse of a 3x3 matrix same as its transpose
In spatial mechanics, the inverse and transpose of a 3x3 matrix are not the same, but they can be related in specific cases. The inverse (A⁻¹) is a matrix that, when multiplied by A, results in the identity matrix (A * A⁻¹ = I). The transpose (Aᵀ) is obtained by interchanging the rows and columns of A.
For 3x3 orthogonal matrices, often used to represent rotations, the transpose is equal to the inverse (A⁻¹ = Aᵀ) because they satisfy Aᵀ * A = I. This property is unique to orthogonal matrices, and for general 3x3 matrices, the inverse and transpose are not the same.