Difference between revisions of "The Wires"

From New IAC Wiki
Jump to navigation Jump to search
(Created page with "We can parametrize the equations for the wires and wire midpoints to express the equation in vector form. In the y'-x' plane the general equation follows the relationship: <cen…")
 
 
(11 intermediate revisions by the same user not shown)
Line 1: Line 1:
 +
<center><math>\underline{\textbf{Navigation}}</math>
 +
 +
[[Points_of_Intersection|<math>\vartriangleleft </math>]]
 +
[[VanWasshenova_Thesis#Determining_wire-theta_correspondence|<math>\triangle </math>]]
 +
[[Right_Hand_Wall|<math>\vartriangleright </math>]]
 +
 +
</center>
 +
 +
 
We can parametrize the equations for the wires and wire midpoints to express the equation in vector form.  In the y'-x' plane the general equation follows the relationship:
 
We can parametrize the equations for the wires and wire midpoints to express the equation in vector form.  In the y'-x' plane the general equation follows the relationship:
  
−
<center><math>x'=y' tan 6^{\circ}+x_0</math></center>
+
<center><math>x'=y'\ tan\ 6^{\circ}+x_0</math></center>
 
 
 
where <math>x_0</math> is the point where the line crosses the x axis.
 
where <math>x_0</math> is the point where the line crosses the x axis.
  
−
<center><math>y' \Rightarrow  {y tan 6^{\circ}+x_0, y, 0}</math></center>
+
<center><math>y' \Rightarrow  {y\ tan\ 6^{\circ}+x_0, y, 0}</math></center>
  
  
 
In this form we can easily see that the components of x and y , in the y'-x' plane are
 
In this form we can easily see that the components of x and y , in the y'-x' plane are
  
−
<center><math>x' = y sin 6^{\circ}+x_0</math></center>
+
<center><math>x' = y\ sin\ 6^{\circ}+x_0</math></center>
−
<center><math>y' = y cos 6^{\circ}</math></center>
+
 
 +
 
 +
<center><math>y' = y\ cos\ 6^{\circ}</math></center>
 
 
 
The parameterization has reduced two equations with two variables, to two equations which depend on one variable.  Working in the y-x plane, we will undergo a positive rotation,
 
The parameterization has reduced two equations with two variables, to two equations which depend on one variable.  Working in the y-x plane, we will undergo a positive rotation,
  
−
R(Subscript[\[Theta], yx])=(cos 6\[Degree] -sin 6\[Degree] 0
 
−
sin 6\[Degree] cos 6\[Degree] 0
 
−
0 0 1
 
  
−
)
+
<center><math>R(\theta_{yx})=\begin{bmatrix}
 +
cos\ 6^{\circ} &-sin\ 6^{\circ} & 0 \\
 +
sin\ 6^{\circ} & cos\ 6^{\circ} &0 \\
 +
0 &0 & 1
 +
\end{bmatrix}</math></center>
  
  
−
(Components of
+
<center><math>\begin{bmatrix}
−
same vector
+
Components\ of \\
−
in new system
+
same\ vector \\
−
 
+
in\ new\ system
−
)=(Passive
+
\end{bmatrix}
−
transformation
+
=\begin{bmatrix}
 +
Passive \\
 +
transformation \\
 
matrix
 
matrix
 +
\end{bmatrix}\cdot
 +
\begin{bmatrix}
 +
Components\ of \\
 +
vector\ in \\
 +
original\ system
 +
\end{bmatrix}</math></center>
  
−
) . (Components of
 
−
vector in
 
−
original system
 
  
−
) (New
 
−
basis
 
−
vectors
 
  
−
)=(Active
 
−
transformation
 
−
matrix
 
  
−
) . (original
+
<center><math>
−
basis
+
\begin{bmatrix}
−
vectors
+
x'' \\
−
 
+
y'' \\
−
)
 
−
 
−
 
−
(x''
 
−
y''
 
 
z''
 
z''
−
 
+
\end{bmatrix}=
−
)=(cos 6\[Degree] -sin 6\[Degree] 0
+
\begin{bmatrix}
−
sin 6\[Degree] cos 6\[Degree] 0
+
cos\ 6^{\circ} &-sin\ 6^{\circ} & 0 \\
−
0 0 1
+
sin\ 6^{\circ} & cos\ 6^{\circ} &0 \\
−
 
+
0 &0 & 1
−
) . (x'
+
\end{bmatrix}\cdot
−
y'
+
\begin{bmatrix}
 +
x' \\
 +
y' \\
 
z'
 
z'
 +
\end{bmatrix}</math></center>
 +
  
−
) (x'
+
<center><math>
−
y'
+
\begin{bmatrix}
−
z'
+
x'' \\
 +
y'' \\
 +
z''
 +
\end{bmatrix}=
 +
\begin{bmatrix}
 +
cos\ 6^{\circ} &-sin\ 6^{\circ} & 0 \\
 +
sin\ 6^{\circ} & cos\ 6^{\circ} &0 \\
 +
0 &0 & 1
 +
\end{bmatrix}\cdot
 +
\begin{bmatrix}
 +
y'\ sin\ 6^{\circ}+x_0 \\
 +
y'\ cos\ 6^{\circ} \\
 +
0
 +
\end{bmatrix}</math></center>
  
−
)=(cos 6\[Degree] sin 6\[Degree] 0
 
−
-sin 6\[Degree] cos 6 \[Degree] 0
 
−
0 0 1
 
  
−
) . (x''
 
−
y''
 
−
z''
 
  
−
)
+
<center><math>
−
+
\begin{bmatrix}
−
(x''
+
x'' \\
−
y''
+
y'' \\
 
z''
 
z''
 +
\end{bmatrix}=
 +
\begin{bmatrix}
 +
-y'\ cos\ 6^{\circ}sin\ 6^{\circ}+x_0\ cos\ 6^{\circ} +y'\ cos\ 6^{\circ}sin\ 6^{\circ}\\
 +
y'\ cos^2 6^{\circ}+x_0\sin\ 6^{\circ}+y sin^2 6^{\circ} \\
 +
0
 +
\end{bmatrix}</math></center>
 +
  
−
)=(cos 6\[Degree] -sin 6\[Degree] 0
 
−
sin 6\[Degree] cos 6\[Degree] 0
 
−
0 0 1
 
  
−
) . ( y sin 6\[Degree]+Subscript[x, 0]
+
<center><math>
−
y cos 6\[Degree]
+
\begin{bmatrix}
 +
x'' \\
 +
y'' \\
 +
z''
 +
\end{bmatrix}=
 +
\begin{bmatrix}
 +
x_0\ cos\ 6^{\circ}\\
 +
y'\ +x_0\sin\ 6^{\circ} \\
 
0
 
0
 +
\end{bmatrix}</math></center>
  
−
)
 
−
 
−
(x''
 
−
y''
 
−
z''
 
  
−
)= (-y cos 6 \[Degree] sin 6 \[Degree]+Subscript[x, 0]cos 6 \[Degree] +y cos 6 \[Degree]sin 6 \[Degree]
+
This relationship shows us that x'' is a constant in this frame while y'' can have any value, which is the horizontal line with respect to the y axis as expected.
−
y cos^2 6 \[Degree]+Subscript[x, 0]sin 6 \[Degree]+y sin^2 6 \[Degree]
+
 
−
0
 
  
−
) (x'
+
----
−
y'
 
−
z'
 
  
−
)= (x'' cos 6\[Degree]+y " sin 6\[Degree]
 
−
-x'' sin 6 \[Degree]+y " cos 6\[Degree]
 
−
0
 
  
−
)
+
<center><math>\underline{\textbf{Navigation}}</math>
−
 
−
  (x''
 
−
y''
 
−
z''
 
  
−
)= (Subscript[x, 0]cos 6 \[Degree]  
+
[[Points_of_Intersection|<math>\vartriangleleft </math>]]
−
y +Subscript[x, 0]sin 6 \[Degree]
+
[[VanWasshenova_Thesis#Determining_wire-theta_correspondence|<math>\triangle </math>]]
−
0
+
[[Right_Hand_Wall|<math>\vartriangleright </math>]]
  
−
)
+
</center>

Latest revision as of 20:32, 15 May 2018

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

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


We can parametrize the equations for the wires and wire midpoints to express the equation in vector form. In the y'-x' plane the general equation follows the relationship:

[math]x'=y'\ tan\ 6^{\circ}+x_0[/math]

where [math]x_0[/math] is the point where the line crosses the x axis.

[math]y' \Rightarrow {y\ tan\ 6^{\circ}+x_0, y, 0}[/math]


In this form we can easily see that the components of x and y , in the y'-x' plane are

[math]x' = y\ sin\ 6^{\circ}+x_0[/math]


[math]y' = y\ cos\ 6^{\circ}[/math]

The parameterization has reduced two equations with two variables, to two equations which depend on one variable. Working in the y-x plane, we will undergo a positive rotation,


[math]R(\theta_{yx})=\begin{bmatrix} cos\ 6^{\circ} &-sin\ 6^{\circ} & 0 \\ sin\ 6^{\circ} & cos\ 6^{\circ} &0 \\ 0 &0 & 1 \end{bmatrix}[/math]


[math]\begin{bmatrix} Components\ of \\ same\ vector \\ in\ new\ system \end{bmatrix} =\begin{bmatrix} Passive \\ transformation \\ matrix \end{bmatrix}\cdot \begin{bmatrix} Components\ of \\ vector\ in \\ original\ system \end{bmatrix}[/math]



[math] \begin{bmatrix} x'' \\ y'' \\ z'' \end{bmatrix}= \begin{bmatrix} cos\ 6^{\circ} &-sin\ 6^{\circ} & 0 \\ sin\ 6^{\circ} & cos\ 6^{\circ} &0 \\ 0 &0 & 1 \end{bmatrix}\cdot \begin{bmatrix} x' \\ y' \\ z' \end{bmatrix}[/math]


[math] \begin{bmatrix} x'' \\ y'' \\ z'' \end{bmatrix}= \begin{bmatrix} cos\ 6^{\circ} &-sin\ 6^{\circ} & 0 \\ sin\ 6^{\circ} & cos\ 6^{\circ} &0 \\ 0 &0 & 1 \end{bmatrix}\cdot \begin{bmatrix} y'\ sin\ 6^{\circ}+x_0 \\ y'\ cos\ 6^{\circ} \\ 0 \end{bmatrix}[/math]


[math] \begin{bmatrix} x'' \\ y'' \\ z'' \end{bmatrix}= \begin{bmatrix} -y'\ cos\ 6^{\circ}sin\ 6^{\circ}+x_0\ cos\ 6^{\circ} +y'\ cos\ 6^{\circ}sin\ 6^{\circ}\\ y'\ cos^2 6^{\circ}+x_0\sin\ 6^{\circ}+y sin^2 6^{\circ} \\ 0 \end{bmatrix}[/math]


[math] \begin{bmatrix} x'' \\ y'' \\ z'' \end{bmatrix}= \begin{bmatrix} x_0\ cos\ 6^{\circ}\\ y'\ +x_0\sin\ 6^{\circ} \\ 0 \end{bmatrix}[/math]


This relationship shows us that x is a constant in this frame while y can have any value, which is the horizontal line with respect to the y axis as expected.



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

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

Retrieved from "https://wiki.iac.isu.edu/index.php?title=The_Wires&oldid=122815"