Difference between revisions of "Test for Theta at 20 degrees and Phi at 0"
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(Created page with "====Test for <math>\theta=20</math> and <math>\phi=0</math>==== Solving for the components of the ellipse <center><math>a_1=\frac{2.53sin(\theta)}{sin(115-\theta)}=\frac{2.53s…") |
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− | + | <center><math>\underline{\textbf{Navigation}}</math> | |
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+ | [[Determing_Elliptical_Equations|<math>\vartriangleleft </math>]] | ||
+ | [[VanWasshenova_Thesis#Determining_wire-theta_correspondence|<math>\triangle </math>]] | ||
+ | [[In_the_Detector_Plane|<math>\vartriangleright </math>]] | ||
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+ | </center> | ||
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+ | =Test for <math>\theta=20</math> and <math>\phi=0</math>= | ||
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<center><math>D1P=\sqrt{(x_P-x_{D1})^2+(y_P-y_{D1})^2+(z_P-z_{D1})^2}=\sqrt{(0.7872-.5901)^2+(0-0)^2+(2.1623-1.6212)^2}=\sqrt{(.1971)^2+(.5411)^2}=\sqrt{.0388+.2927}=.575879865944\ \text{m}</math></center> | <center><math>D1P=\sqrt{(x_P-x_{D1})^2+(y_P-y_{D1})^2+(z_P-z_{D1})^2}=\sqrt{(0.7872-.5901)^2+(0-0)^2+(2.1623-1.6212)^2}=\sqrt{(.1971)^2+(.5411)^2}=\sqrt{.0388+.2927}=.575879865944\ \text{m}</math></center> | ||
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+ | ---- | ||
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+ | <center><math>\underline{\textbf{Navigation}}</math> | ||
+ | |||
+ | [[Determing_Elliptical_Equations|<math>\vartriangleleft </math>]] | ||
+ | [[VanWasshenova_Thesis#Determining_wire-theta_correspondence|<math>\triangle </math>]] | ||
+ | [[In_the_Detector_Plane|<math>\vartriangleright </math>]] | ||
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+ | </center> |
Latest revision as of 20:24, 15 May 2018
Test for and
Solving for the components of the ellipse
The y component is zero for
The z component can be found from the ellipse equation
The height to the first directrix circle is
and the x and y components are
The height to the second directrix circle is
and the x and y components are
The distance between the two point
This should be equal to 2a
This gives the components for a point on the directrix circles as