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.   
  
(x''
+
\begin{bmatrix}
y''
+
x'' \\
 +
y'' \\
 
z''
 
z''
 
+
\end{bmatrix}=
)=(cos 6\[Degree] -sin 6\[Degree] 0
+
\begin{bmatrix}
sin 6\[Degree] cos 6\[Degree] 0
+
cos\ 6^{\circ} & -sin\ 6^{\circ} & 0 \\
0 0 1
+
sin\ 6^{\circ} & cos\ 6^{\circ} & 0 \\
 
+
0 & 0 & 1
) . (x'
+
\end{bmatrix} \cdot
y'
+
\begin{bmatrix}
 +
x' \\
 +
y' \\
 
z'
 
z'
 +
\end{bmatrix}
 +
  
 
)
 
)

Revision as of 14:19, 28 April 2017

[math]x=-y\ cot\ 29.5^{\circ}+0.09156[/math]

Parameterizing this

[math]r\mapsto {-y\ cot\ 29.5^{\circ}+0.09156,y,0}[/math]


[math]t\mapsto {t\ cos\ 29.5^{\circ}+0.09156,-t\ sin\ 29.5^{\circ},0}[/math]


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

[math]y''=0.09156\ sin\ 6^{\circ}-t\ sin\ 23.5^{\circ} \Rightarrow t=\frac{-(y''-0.09156 sin 6^{\circ})}{sin 23.5^{\circ}}[/math]


Substituting this into the expression for x

[math]x''=0.09156\ cos\ 6^{\circ}+t\ cos\ 23.5^{\circ}[/math]


[math]x''=0.09156\ cos\ 6 ^{\circ}+\frac{-(y''-0.09156 sin 6 ^{\circ})}{sin 23.5^{\circ}}(cos 23.5^{\circ})[/math]


[math]x''=0.091058+\frac{y''-.0095706 }{-0.398749} (.917060)[/math]


[math]x''=0.091058+(y''-.0095706 ) (-2.299843)[/math]


[math]x''=-2.299843\ y''+.022011+.091058[/math]


[math]x''=-2.299843\ y''+.113069[/math]