Circular Cross Sections: Difference between revisions

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<center><math>\textbf{\underline{Navigation}}</math>
<center><math>\underline{\textbf{Navigation}}</math>


[[Conic_Sections|<math>\vartriangleleft </math>]]
[[Conic_Sections|<math>\vartriangleleft </math>]]
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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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<center><math>\textbf{\underline{Navigation}}</math>
<center><math>\underline{\textbf{Navigation}}</math>


[[Conic_Sections|<math>\vartriangleleft </math>]]
[[Conic_Sections|<math>\vartriangleleft </math>]]

Latest revision as of 20:22, 15 May 2018

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

[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

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


Using the relation

[math]\displaystyle{ sin(90^{\circ}-\theta)=cos(\theta) }[/math]


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

[math]\displaystyle{ \vartriangleleft }[/math] [math]\displaystyle{ \triangle }[/math] [math]\displaystyle{ \vartriangleright }[/math]