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|
Line 10: |
Line 10: |
| </math> | | </math> |
| | | |
− | And solve: | + | And calculate: |
| :<math> | | :<math> |
| \begin{alignat}{2} | | \begin{alignat}{2} |
Line 22: |
Line 22: |
| & + \left( d \sin \theta_g + z \cos \theta_g \right)^2 ] \end{alignat} \\ | | & + \left( d \sin \theta_g + z \cos \theta_g \right)^2 ] \end{alignat} \\ |
| | | |
| + | & = \begin{alignat}{2} [ |
| + | & x^2 \cos^2 \phi_g - x \cos \phi_g \sin \phi_g ( v_{2y} ) + \sin^2 \phi_g ( v_{2y} )^2 \\ |
| + | & + x^2 \sin^2 \phi_g + x \sin \phi_g \cos \phi_g ( v_{2y} ) - d^{\prime} x \sin \phi_g \\ |
| + | & + x \sin \phi_g \cos \phi_g ( v_{2y} ) + \cos^2 \phi_g ( v_{2y} )^2 - d^{\prime} \cos \phi_g ( v_{2y} ) \\ |
| + | & - d^{\prime} x \sin \phi_g - d^{\prime} \cos \phi_g ( v_{2y} ) + d^{\prime 2} \\ |
| + | & + d^2 \sin^2 \theta_g + d z \sin \theta_g \cos \theta_g + z^2 \cos^2 \theta_g ] \end{alignat} \\ |
| + | \end{alignat} |
| + | </math> |
| + | Grouping and rearranging: |
| + | :<math> |
| + | \begin{alignat}{2} |
| + | \left ( \frac{q}{k} \right )^2 d^{\prime 2} |
| & = \begin{alignat}{2} [ | | & = \begin{alignat}{2} [ |
| & x^2 \cos^2 \phi_g - x \cos \phi_g \sin \phi_g ( v_{2y} ) + \sin^2 \phi_g ( v_{2y} )^2 \\ | | & x^2 \cos^2 \phi_g - x \cos \phi_g \sin \phi_g ( v_{2y} ) + \sin^2 \phi_g ( v_{2y} )^2 \\ |
Revision as of 15:14, 13 January 2016
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We define:
And calculate:
Grouping and rearranging: