User:Eml4500.f08.jamama.adam/Homework4

Meeting 19
The objective is to transform the set of local degrees of freedom $$\tilde{d}^{(e)}$$ to a different set degrees of freedom $$\tilde{d}^{(e)}_{4X1}$$ using an invertible matrix $$\tilde{T}^{(e)}{_{4X4}}$$.



$$\tilde{d}^{(e)}_{4X1} = \tilde{T}^{(e)}_{4X4}d^{(e)}_{4X1}$$

Lets define equation (1) bellow:

$$\tilde{d}^{(e)}_{1} = [l^{(e)}m^{(e)}]\begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ \end{Bmatrix}$$

$$\tilde{d}_{2}^{(e)} = \overrightarrow{d}_{1}^{(e)}\cdot \overrightarrow{j}$$ (comp. of $$ \overrightarrow{d}_{1}^{(e)}$$ along $$\vec\tilde{j}$$ i.e. y-axis)
 * $$= sin(\theta^{(e)}) + cos(\theta^{(e)}) d_{2}^{(e)}$$

And lets define equation (2) bellow:

$$\tilde{d}_{2}^{(e)} = [-m^{(e)}l^{(e)}]\begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ \end{Bmatrix}$$

By putting (1) and (2) in matrix form:

$$\begin{Bmatrix} \tilde{d}_{1}^{(e)}\\ \tilde{d}_{2}^{(e)}\\ \end{Bmatrix} = \underbrace{\begin{bmatrix} l^{(e)}&m^{(e)}\\ -m^{(e)}&l^{(e)}\\ \end{bmatrix}}_{R^{(e)}}\begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ \end{Bmatrix}$$

$$\underbrace{\begin{Bmatrix} \tilde{d}_{1}^{(e)}\\ \tilde{d}_{2}^{(e)}\\ \tilde{d}_{3}^{(e)}\\ \tilde{d}_{4}^{(e)}\\ \end{Bmatrix}}_{\tilde{d}_{4X1}^{(e)}} = \underbrace{\begin{bmatrix} R_{2X2}^{(e)}&0_{2X2}\\ 0_{2X2}&R_{2X2}^{(e)}\\ \end{bmatrix}}_{\tilde{T}_{4X4}^{(e)}}\underbrace{\begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ d_{3}^{(e)}\\ d_{4}^{(e)}\\ \end{Bmatrix}}_{d_{4X1}^{(e)}}$$



$$\tilde{f}^{(e)} = k^{(e)}\begin{bmatrix} 1&0&-1&0\\ 0&0&0&0\\ -1&0&1&0\\ 0&0&0&0\\ \end{bmatrix}\tilde{d}^{(e)}$$

$$\tilde{f}^{(e)}_{4X1} = \tilde{k}^{(e)}_{4X4}\tilde{d}^{(e)}_{4X1}$$