Difference between revisions of "Left Hand Wall"
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where the negative sign is applied to the sine function by the even odd relationships of cosine and sine, i.e. ( sin(-t)=-sin(t), cos(-t)=cos(t)) and the fact that the y'' component is in the 4th quadrant. | where the negative sign is applied to the sine function by the even odd relationships of cosine and sine, i.e. ( sin(-t)=-sin(t), cos(-t)=cos(t)) and the fact that the y'' component is in the 4th quadrant. | ||
| − | + | \begin{bmatrix} | |
| − | y'' | + | x'' \\ |
| + | y'' \\ | ||
z'' | z'' | ||
| − | + | \end{bmatrix}= | |
| − | + | \begin{bmatrix} | |
| − | sin 6\ | + | cos\ 6^{\circ} & -sin\ 6^{\circ} & 0 \\ |
| − | 0 0 1 | + | sin\ 6^{\circ} & cos\ 6^{\circ} & 0 \\ |
| − | + | 0 & 0 & 1 | |
| − | + | \end{bmatrix} \cdot | |
| − | y' | + | \begin{bmatrix} |
| + | x' \\ | ||
| + | y' \\ | ||
z' | z' | ||
| + | \end{bmatrix} | ||
| + | |||
) | ) | ||
Revision as of 14:19, 28 April 2017
Parameterizing this
where the negative sign is applied to the sine function by the even odd relationships of cosine and sine, i.e. ( sin(-t)=-sin(t), cos(-t)=cos(t)) and the fact that the y component is in the 4th quadrant.
\begin{bmatrix} x \\ y \\ z \end{bmatrix}= \begin{bmatrix} cos\ 6^{\circ} & -sin\ 6^{\circ} & 0 \\ sin\ 6^{\circ} & cos\ 6^{\circ} & 0 \\ 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} x' \\ y' \\ z' \end{bmatrix}
)
(x y z
)=(cos 6\[Degree] -sin 6\[Degree] 0 sin 6\[Degree] cos 6\[Degree] 0 0 0 1
) . (t cos (29.5\[Degree])+0.09156 -t sin (29.5\[Degree]) 0
)
(x y z
)= (0.09156cos 6 \[Degree]+t cos 6 \[Degree]cos (29.5\[Degree])+t sin 6 \[Degree]sin (29.5\[Degree]) -t cos 6 \[Degree]sin (29.5\[Degree])+0.09156 sin 6 \[Degree]+t cos (29.5\[Degree])sin 6 \[Degree] 0
)
(x y z
)= (0.09156cos 6 \[Degree]+t (cos 6 \[Degree]cos(29.5\[Degree])+ sin 6 \[Degree]sin (29.5\[Degree])) 0.09156 sin 6 \[Degree]-t (cos 6 \[Degree]sin (29.5\[Degree])-sin 6 \[Degree] cos (29.5\[Degree])) 0
)
(x y z
)= (0.09156cos 6 \[Degree]+t cos (6\[Degree] -29.5\[Degree]) 0.09156 sin 6 \[Degree]+t sin (6 \[Degree]-29.5\[Degree]) 0
)
(x y z
)= (0.09156cos 6 \[Degree]+t cos (-23.5\[Degree]) 0.09156 sin 6 \[Degree]+t sin (-23.5\[Degree]) 0
)
(x y z
)= (0.09156cos 6 \[Degree]+t cos (23.5\[Degree]) 0.09156 sin 6 \[Degree]-t sin (-23.5\[Degree]) 0
)
Using the equation for y we can solve for t
Substituting this into the expression for x