Difference between revisions of "Circular Cross Sections"

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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.
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----
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<center><math>\textbf{\underline{Navigation}}</math>
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[[Conic_Sections|<math>\vartriangleleft </math>]]
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[[VanWasshenova_Thesis#Determining_wire-theta_correspondence|<math>\triangle </math>]]
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[[Elliptical_Cross_Sections|<math>\vartriangleright </math>]]
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</center>

Revision as of 18:42, 15 May 2017

[math]\textbf{\underline{Navigation}}[/math]

[math]\vartriangleleft [/math] [math]\triangle [/math] [math]\vartriangleright [/math]

Circular Conic Section

If the conic is an circle, e=0. This implies

[math]e=\frac{\sin (\beta)}{\sin (\alpha)}=\frac{\sin (25^{\circ})}{\sin (90-\theta)}=0[/math]


Using the relation

[math]sin(90^{\circ}-\theta)=cos(\theta)[/math]
[math]\frac{sin (25^{\circ})}{0}=cos( \theta) =\infty[/math]


The sector angle will never be perpendicular to the plane of the light cone, so this is not a physical possibility.




[math]\textbf{\underline{Navigation}}[/math]

[math]\vartriangleleft [/math] [math]\triangle [/math] [math]\vartriangleright [/math]