Difference between revisions of "Circular Cross Sections"

From New IAC Wiki
Jump to navigation Jump to search
 
(One intermediate revision by the same user not shown)
Line 1: Line 1:
−
<center><math>\textbf{\underline{Navigation}}</math>
+
<center><math>\underline{\textbf{Navigation}}</math>
  
 
[[Conic_Sections|<math>\vartriangleleft </math>]]
 
[[Conic_Sections|<math>\vartriangleleft </math>]]
Line 16: Line 16:
  
 
<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>
Line 26: Line 27:
  
  
−
<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]\underline{\textbf{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]\underline{\textbf{Navigation}}[/math]

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

Retrieved from "https://wiki.iac.isu.edu/index.php?title=Circular_Cross_Sections&oldid=122803"