Physics, asked by Dhirajsen8764, 1 year ago

Explain the importance of homogeneous coordinates in modeling of robotic manipulators.

Answers

Answered by devanshyadav0011
4

Homogeneous coordinates are ubiquitous in computer graphics because they allow common vector operations such as translation, rotation, scaling and perspective projection to be represented as a matrix by which the vector is multiplied.

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Answered by gungunraman2003
0

The homogeneous transformation matrix

The transformation $ T_i$, for each $ i$ such that $ 1 < i \leq m$, is

$\displaystyle T_i = Q_{i-1} R_i = \begin{pmatrix}\cos\theta_i & -\sin\theta_i &...

...{i-1} & \cos\alpha_{i-1} & \cos\alpha_{i-1}d_i  0 & 0 & 0 & 1 \end{pmatrix} .$ (3.56)

This can be considered as the 3D counterpart to the 2D transformation matrix, (3.52). The following four operations are performed in succession:

Translate by $ d_i$ along the $ z_i$-axis.

Rotate counterclockwise by $ \theta _i$ about the $ z_i$-axis.

Translate by $ a_{i-1}$ along the $ x_{i-1}$-axis.

Rotate counterclockwise by $ \alpha _{i-1}$ about the $ x_{i-1}$-axis.

As in the 2D case, the first matrix, $ T_1$, is special. To represent any position and orientation of $ {\cal A}_1$, it could be defined as a general rigid-body homogeneous transformation matrix, (3.50). If the first body is only capable of rotation via a revolute joint, then a simple convention is usually followed. Let the $ a_0,\alpha_0$ parameters of $ T_1$ be assigned as $ a_0 = \alpha_0 = 0$ (there is no $ z_0$-axis). This implies that $ Q_0$ from (3.55) is the identity matrix, which makes $ T_1 =

R_1$.

The transformation $ T_i$ for $ i

> 1$ gives the relationship between the body frame of $ {\cal A}_i$ and the body frame of $ {\cal A}_{i-1}$. The position of a point $ (x,y,z)$ on $ {\cal A}_m$ is given by

$\displaystyle T_1 T_2 \cdots T_m \begin{pmatrix}x  y  z  1  \end{pmatrix} .$

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