Difference between revisions of "Elliptical Cross Sections"

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<center><math> 0<\theta<65^{\circ}</math></center>
 
<center><math> 0<\theta<65^{\circ}</math></center>
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<center><math>\textbf{\underline{Navigation}}</math>
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[[Circular_Cross_Sections|<math>\vartriangleleft </math>]]
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[[VanWasshenova_Thesis#Determining_wire-theta_correspondence|<math>\triangle </math>]]
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[[Determing_Elliptical_Components|<math>\vartriangleright </math>]]
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</center>

Revision as of 18:41, 15 May 2017

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

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

Elliptic Conic Section

If the conic is an ellipse, 0<e<1. This implies

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



[math]\frac{sin (25^{\circ})}{cos (\theta)}=e[/math]


since e must be less than 1, this sets the limit of theta at less than 65 degrees. Since the limit of [math]\theta=0[/math], this implies the minimum eccentricity will be [math]e\approx .4291[/math]


This implies that the shape made on the the plane of the sector is an ellipse for angles

[math] 0\lt \theta\lt 65^{\circ}[/math]
[math]\textbf{\underline{Navigation}}[/math]

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