Circular Cross Sections: Difference between revisions
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Created page with "==Circular Conic Section== If the conic is an circle, e=0. This implies <center><math>e=\frac{\sin (\beta)}{\sin (\alpha)}=\frac{\sin (25^{\circ})}{\sin (90-\theta)}=0</math><…" |
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<center><math>\underline{\textbf{Navigation}}</math> | |||
[[Conic_Sections|<math>\vartriangleleft </math>]] | |||
[[VanWasshenova_Thesis#Determining_wire-theta_correspondence|<math>\triangle </math>]] | |||
[[Elliptical_Cross_Sections|<math>\vartriangleright </math>]] | |||
</center> | |||
==Circular Conic Section== | ==Circular Conic Section== | ||
If the conic is an circle, e=0. This implies | If the conic is an circle, e=0. This implies | ||
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<center><math>sin(90^{\circ}-\theta)=cos(\theta)</math></center> | <center><math>sin(90^{\circ}-\theta)=cos(\theta)</math></center> | ||
<center><math>\frac{sin (25^{\circ})}{0}=cos( \theta) =\infty</math></center> | <center><math>\frac{sin (25^{\circ})}{0}=cos( \theta) =\infty</math></center> | ||
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The sector angle will never be perpendicular to the plane of the light cone, so this is not a physical possibility. | The sector angle will never be perpendicular to the plane of the light cone, so this is not a physical possibility. | ||
---- | |||
<center><math>\underline{\textbf{Navigation}}</math> | |||
[[Conic_Sections|<math>\vartriangleleft </math>]] | |||
[[VanWasshenova_Thesis#Determining_wire-theta_correspondence|<math>\triangle </math>]] | |||
[[Elliptical_Cross_Sections|<math>\vartriangleright </math>]] | |||
</center> | |||
Latest revision as of 20:22, 15 May 2018
[math]\displaystyle{ \vartriangleleft }[/math] [math]\displaystyle{ \triangle }[/math] [math]\displaystyle{ \vartriangleright }[/math]
Circular Conic Section
If the conic is an circle, e=0. This implies
Using the relation
The sector angle will never be perpendicular to the plane of the light cone, so this is not a physical possibility.
[math]\displaystyle{ \vartriangleleft }[/math] [math]\displaystyle{ \triangle }[/math] [math]\displaystyle{ \vartriangleright }[/math]