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| − | ===Compute <math>q_y</math>=== | + | ====Compute <math>q_y</math>==== |
| | :<math> | | :<math> |
| | \begin{alignat}{2} | | \begin{alignat}{2} |
Revision as of 15:12, 15 April 2019
Compute Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q_y}
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} \mathbf{q} & = \begin{bmatrix} q_x \\ q_y \\ q_z \end{bmatrix} \\ & = \frac{2 \pi}{\lambda} \begin{bmatrix} \sin \theta_f \cos \alpha_f \\ \cos \theta_f \cos \alpha_f - 1 \\ \sin \alpha_f \end{bmatrix} \end{alignat} }
So:
Scratch/working (contains errors)
As a check of these results, consider:
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} q & = \sqrt{ q_x^2 + q_y^2 + q_z^2 } \\ & = \frac{2 \pi}{\lambda} \sqrt{ \sin^2 \theta_f \cos^2 \alpha_f + \left ( \cos \theta_f \cos \alpha_f - 1 \right )^2 + \sin^2 \alpha_f } \\ \left( \frac{q}{k} \right)^2 & = (\sin \theta_f)^2 (\cos \alpha_f)^2 + \left ( \cos \theta_f \cos \alpha_f - 1 \right )^2 + (\sin \alpha_f)^2 \\ & = \left(\frac{x/d}{\sqrt{1+(x/d)^2}} \right)^2 \left(\cos \alpha_f \right)^2 + \left ( \cos \theta_f \cos \alpha_f - 1 \right )^2 + \left( \frac{z \cos \theta_f /d }{\sqrt{1+(z \cos \theta_f /d)^2}} \right)^2 \\ & = \left(\frac{x}{\sqrt{d^2+x^2}} \right)^2 \left(\cos \alpha_f \right)^2 + \left ( \cos \theta_f \cos \alpha_f - 1 \right )^2 + \left( \frac{z \cos \theta_f }{\sqrt{d^2+z^2 \cos^2 \theta_f }} \right)^2 \\ & = \frac{x^2}{d^2+x^2} \left(\cos \alpha_f \right)^2 + \left ( \cos \theta_f \cos \alpha_f - 1 \right )^2 + \frac{z^2 \cos^2 \theta_f }{d^2+z^2 \cos^2 \theta_f } \\ & = \frac{x^2}{d^2+x^2} \frac{d^4}{d^2+z^2 \cos^2 \theta_f} + \left ( \cos \theta_f \frac{d^2}{\sqrt{d^2+z^2 \cos^2 \theta_f}} - 1 \right )^2 + \frac{z^2 \cos^2 \theta_f }{d^2+z^2 \cos^2 \theta_f } \end{alignat} }
Where we used:
![{\displaystyle {\begin{alignedat}{2}\sin(\arctan[u])&={\frac {u}{\sqrt {1+u^{2}}}}\\\sin \theta _{f}&=\sin(\arctan[x/d])\\&={\frac {x/d}{\sqrt {1+(x/d)^{2}}}}\\&={\frac {x}{\sqrt {d^{2}+x^{2}}}}\end{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc9628f0d308053600f4e02d8f80c69fc9f356d0)
And, we further note that:
![{\displaystyle {\begin{alignedat}{2}\cos(\arctan[u])&={\frac {1}{\sqrt {1+u^{2}}}}\\\cos \theta _{f}&={\frac {1}{\sqrt {1+(x/d)^{2}}}}\\&={\frac {d^{2}}{\sqrt {d^{2}+x^{2}}}}\end{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/584da470743b7feac2b55988526dabce5b4313c4)
Continuing:
