Difference between revisions of "DC Super Layer 1:Layer 1"

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<center><math>\underline{\textbf{Navigation}}</math>
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[[CED_Verification_of_DC_Angle_Theta_and_Wire_Correspondance|<math>\vartriangleleft </math>]]
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[[VanWasshenova_Thesis#Determining_wire-theta_correspondence|<math>\triangle </math>]]
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[[DC_Binning_Based_On_Wire_Numbers|<math>\vartriangleright </math>]]
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</center>
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 +
 +
=DC Super Layer 1:Layer 1=
 +
 +
 +
From the ced simulations
 +
 +
 
{|  border=1 align=center
 
{|  border=1 align=center
 
   |+ Table 1: Superlayer 1 Change of Coordinates From Wire 1 to Wire 2(cm)
 
   |+ Table 1: Superlayer 1 Change of Coordinates From Wire 1 to Wire 2(cm)
Line 36: Line 51:
  
  
Using the geometric construction for determining angle theta to wire 2:
+
While the wire placement is off, the difference in x and z position is matched by the calculated values.  Using the geometric construction for determining angle theta to wire 2:
  
<center><math>\tan{\theta}=\frac{x+\Delta x}{z+\Delta z}\Rightarrow \theta=\arctan{\frac{x+\Delta x}{z+\Delta z}}</math></center>
+
<center><math>\tan{\theta_2}=\frac{x_1+\Delta x}{z_1+\Delta z}\Rightarrow \theta_2=\arctan{\frac{x_1+\Delta x}{z_1+\Delta z}}</math></center>
  
  
<center><math>\arctan{\frac{20.42+1.212}{243.39-0.5652}}=\arctan{\frac{21.632}{242.8248}}=\arctan{0.0891}=5.09^{\circ}</math></center>
+
<center><math>\arctan{\frac{22.2092+1.212}{242.5779-0.5652}}=\arctan{\frac{23.4212}{242.0127}}=\arctan{0.0967}=5.53^{\circ}=\tan{\theta_2}</math></center>
  
  
Line 47: Line 62:
  
  
<center><math>\tan{\theta}=\frac{x-\Delta x+n\Delta x}{z-\Delta z+n\Delta z}\Rightarrow \theta=\arctan{\frac{x+n\Delta x}{z+n\Delta z}}</math></center>
+
<center><math>\tan{\theta_n}=\frac{x-\Delta x+n\Delta x}{z-\Delta z+n\Delta z}\Rightarrow \theta_n=\arctan{\frac{x+n\Delta x}{z+n\Delta z}}</math></center>
  
 
where
 
where
  
<center><math>x=20.42-1.212\approx 19.21cm</math></center>
+
<center><math>x_0=22.2092-1.212\approx 20.9972cm</math></center>
  
<center><math>z=243.39+0.5652\approx 243.96cm</math></center>
 
  
 +
<center><math>z_0=242.5779+0.5652\approx 243.1431cm</math></center>
  
<center><math>\theta=\arctan{\frac{19.21+n1.212}{243.39-n0.5652}}</math></center>
 
  
 +
<center><math>\theta_n=\arctan{\frac{20.9972+n1.212}{243.1431-n0.5652}}</math></center>
  
<center><math>\tan{\theta}=\frac{19.21+n1.212}{243.39-n0.5652}\Rightarrow \tan{ \theta}=-2.14437 + \frac{957.412}{430.626 - n}\Rightarrow 430.626 - n = \frac{957.412}{\tan{ \theta}+2.14437}</math></center>
 
  
 +
<center><math>\tan{\theta_n}=\frac{20.9972+n1.212}{243.1431-n0.5652}\Rightarrow \tan{ \theta_n}=-2.14437 + \frac{959.637}{430.189 - n}\Rightarrow 430.189 - n = \frac{959.637}{\tan{ \theta_n}+2.14437}</math></center>
  
<center><math>\Rightarrow n = \frac{-957.412}{\tan{ \theta}+2.14437}+430.626</math></center>
+
 
 +
<center><math>\Rightarrow n = \frac{-959.637}{\tan{ \theta_n}+2.14437}+430.189</math></center>
  
 
Using Mathematica, a series expansion about n=0 can be found:
 
Using Mathematica, a series expansion about n=0 can be found:
  
 
<pre>
 
<pre>
In[1]:= Series[ArcTan[(19.21+1.212 x)/(243.39-0.5652 n)],{n,0,4}]
+
In[2]:= Series[ArcTan[(20.9972+1.212 n)/(243.1431-0.5652 n)],{n,0,4}]
  
Out[1]= 0.0787635+0.00513098 n+9.83726*10^-6 n^2-2.61675*10^-8 n^3-2.22826*10^-10 n^4+O[n]^5
+
Out[2]= 0.0861437+0.00514708 n+9.67686*10^-6 n^2-2.72596*10^-8 n^3-2.22159*10^-10 n^4+O[n]^5
 
</pre>
 
</pre>
  
<center><math>\theta\approx 0.0787635+0.00513098 n+9.83726\times 10^{-6} n^2-2.61675\times 10^{-8} n^3</math></center>
+
<center><math>\theta\approx 0.0861437+0.00514708 n+9.67686 \times10^{-6} n^2-2.72596 \times 10^{-8} n^3</math></center>
  
  
Line 78: Line 94:
  
 
<pre>
 
<pre>
In[2]:= 180*(0.07876354504106763`+0.005130982891104289` n+9.837257652182922`*^-6 n^2-2.616751708660921`*^-8 n^3-2.2282595631625944`*^-10 n^4+O[n]^5)/3.14159625359
+
In[3]:= 180(0.08614365821719583`+0.005147076698801422` n+9.676859262994711`*^-6 n^2-2.7259639443492433`*^-8 n^3-2.2215881081233252`*^-10 n^4+O[n]^5)/3.14159265359
  
Out[2]= 4.51281+0.293983 n+0.000563633 n^2-1.49929*10^-6 n^3-1.2767*10^-8 n^4+O[n]^5
+
Out[3]= 4.93567+0.294906 n+0.000554443 n^2-1.56186*10^-6 n^3-1.27288*10^-8 n^4+O[n]^5
 
</pre>
 
</pre>
  
  
<center><math>\theta\approx 4.51281+0.293983 n+0.000563633 n^2-1.49929\times10^{-6} n^3</math></center>
+
<center><math>\theta\approx 4.93567+0.294906 n+0.000554443 n^2-1.56186*10^{-6} n^3</math></center>
 +
 
 +
 
  
 +
This tells us that the expression for theta will follow a function that comes from a series expansion.  Using Mathematica, a line can be fitted to the data collected on the wire number to angle theta correspondence.
  
  
This tells us that the expression for theta will follow a function that comes from a series expansion.  To account for instrument and measurement errors, angle measurements from ced can be used to find a better fit.
 
  
 +
{|  border=1 align=center
 +
  |+ Table 4: Calculated Superlayer 1 Wire-Angle Theta Correspondence in Degrees
 +
|- style="font-weight:bold; text-align:center;"
 +
  !  style=" border:1px solid gray;"|Wire Number
 +
  ! style=" border:1px solid gray;"|Layer 1
  
Using Mathematica, a line can be fitted to the data collected on the wire number to angle theta correspondence.
+
|-
 +
  | style="border:1px solid gray;"|'''1'''
 +
  | style="border:1px solid gray;"|5.23
 +
|-
 +
  | style="border:1px solid gray;"|''' 2'''
 +
  | style="border:1px solid gray;"|5.53
 +
|-
 +
  | style="border:1px solid gray;"|'''77''' 
 +
  | style="border:1px solid gray;"|29.80
 +
|-
 +
  | style="border:1px solid gray;"|'''78''' 
 +
  | style="border:1px solid gray;"|30.13
 +
|-
 +
  | style="border:1px solid gray;"|'''79''' 
 +
  | style="border:1px solid gray;"|30.46
 +
|-
 +
  | style="border:1px solid gray;"|'''110'''
 +
  | style="border:1px solid gray;"|40.45
 +
|-
 +
  | style="border:1px solid gray;"|'''111'''
 +
  | style="border:1px solid gray;"|40.76
 +
|-
 +
  | style="border:1px solid gray;"|''' 112'''
 +
  | style="border:1px solid gray;"|41.07
 +
|}
  
  
 
Declaring the data set:
 
Declaring the data set:
 
<pre>
 
<pre>
In[1]:= data1={{1,4.79},{2,5.09},{62,24.42},{63,24.76},{64,25.10},{78,29.79},{111,40.50},{112,40.82}}
+
In[4]:= data1={{1,5.23},{2,5.53},{77,29.80},{78,30.13},{79,30.46},{110,40.45},{111,40.76},{112,41.07}}
  
Out[1]= {{1,4.79},{2,5.09},{62,24.42},{63,24.76},{64,25.1},{78,29.79},{111,40.5},{112,40.82}}
+
Out[4]= {{1,5.23},{2,5.53},{77,29.8},{78,30.13},{79,30.46},{110,40.45},{111,40.76},{112,41.07}}
 
</pre>
 
</pre>
  
Line 104: Line 151:
 
Testing for a linear fit:
 
Testing for a linear fit:
 
<pre>
 
<pre>
In[2]:= line1=Fit[data1,{1,n},n]
+
In[5]:= line1=Fit[data1,{1,n},n]
  
Out[2]= 4.39975 + 0.32469 n
+
Out[5]= 4.90443 +0.323148 n
 
</pre>
 
</pre>
  
Examing the range limits for the angle theta for layer 1:
+
Examining the range limits for the angle theta for layer 1:
  
  
Line 115: Line 162:
 
Taking the difference of the upper and lower limits in theta,  
 
Taking the difference of the upper and lower limits in theta,  
  
<center><math>\Delta degrees=40.82^{\circ}-4.79^{\circ}=36.03^{\circ}</math></center>
+
 
 +
<center><math>\theta=\arctan{\frac{20.9972+n1.212}{243.1431-n0.5652}}</math></center>
 +
 
 +
 
 +
<center><math>\theta_{n=1}=\arctan{\frac{20.9972+(1)1.212}{243.1431-(1)0.5652}}=5.23\ \ \ \ \ \ \theta_{n=112}=\arctan{\frac{20.9972+(112)1.212}{243.1431-(112)0.5652}}=41.07</math></center>
 +
 
 +
 
 +
 
 +
 
 +
<center><math>\Delta degrees=41.07^{\circ}-5.23^{\circ}=35.84^{\circ}</math></center>
  
  
 
Dividing by the change in wire numbers (112-1=111), we find
 
Dividing by the change in wire numbers (112-1=111), we find
  
<center><math>\frac{\Delta degrees}{\Delta wire\ number}=\frac{36.03^{\circ}}{111\ wires}\approx\  \frac{0.325^{\circ}}{wire\ number}</math></center>
+
<center><math>\frac{\Delta degrees}{\Delta wire\ number}=\frac{35.84^{\circ}}{111\ wires}\approx\  \frac{0.323^{\circ}}{wire\ number}</math></center>
  
This would imply that if the wires were evenly placed, their change in angle theta would increase by the factor of .325 degrees for each increase in wire number, starting obviously with wire 1 at 4.79 degrees.  In addition, this implies that the bin spacing for each wire would be around .325 degrees in width.
+
This would imply that if the wires were evenly placed, their change in angle theta would increase by the factor of .323 degrees for each increase in wire number, starting obviously with wire 1 at 5.23 degrees.  In addition, this implies that the bin spacing for each wire would be around .323 degrees in width.
  
  
  
Chercking this, we can find the difference between wires 1 and 2,
+
Checking this, we can find the difference between wires 1 and 2,
  
<center><math>\frac{\Delta degree}{\Delta wire\ number}=\frac{5.09^{\circ}-4.79^{\circ} }{wire\ number}=\frac{.3^{\circ}}{wire\ number} </math></center>
+
<center><math>\frac{\Delta degree}{\Delta wire\ number}=\frac{5.5277^{\circ}-5.2311^{\circ} }{wire\ number}=\frac{.2965^{\circ}}{wire\ number} </math></center>
  
 
Similarly, finding the difference between wires 111 and 112,
 
Similarly, finding the difference between wires 111 and 112,
  
<center><math>\frac{\Delta degree}{\Delta wire\ number}=\frac{40.82^{\circ}-40.51^{\circ}}{wire\ number}=\frac{.31^{\circ}}{wire\ number}</math></center>
+
<center><math>\frac{\Delta degree}{\Delta wire\ number}=\frac{41.0740^{\circ}-40.7649^{\circ}}{wire\ number}=\frac{.0.3091^{\circ}}{wire\ number}</math></center>
 +
 
 +
 
 +
These differing values show that the bin width is not uniform in length, therefore a first order, linear fit, will not suffice.  The change in bin size, with respect to the angle theta covered, is due to the tilt of the wire plane.
 +
 
 +
 
 +
 
 +
 
 +
[[File:Bin_angle_size.png|thumb|center|500px|alt=Geometry of Wire Bins Angular Size in DC|'''Figure 5.1.1:''' For a tilted plane, that is offset from the origin, the change in angle between uniformly spaced points depends on the tilt with respect to the beam line.  The uniform distance between points, A, can be used to show the angle, <math>\alpha</math> each wire bin covers can be found using the law of sines.  The interior angle <math>\beta</math>]]
 +
 
 +
Using the law of sines,
  
 +
<center><math>\frac{\sin{\alpha}}{A}=\frac{\sin{\beta}}{B}</math></center>
  
These differing values show that the bin width is not uniform in length, therefore a first order, linear fit, will not suffice.
 
  
 +
<center><math>\Rightarrow \alpha=\arcsin{\frac{A \sin{\beta}}{B}}</math></center>
  
  
 +
The range of <math>\beta</math> runs from near to 65 to almost 180 degrees.  At <math>\theta=25^{\circ}</math> with respect to the positive z direction, a right triangle is created.  Since the sine function is periodic and reaches its maximum at 90 degrees, this implies that the opposite side to this angle reaches its minimum.  On either side of the right angle, as the angle <math>\beta</math> either increases or decreases, the function <math>\sin{\beta} will be smaller.  This implies for movement in angle <math>\alpha</math> in a counter-clockwise direction, the angle <math>\beta</math> will continue to increase, while the length of B will decrease, then increase, reaching it's minimum at <math>\theta=25^{\circ}</math>.  This also implies that <math>\alpha</math> will increase in size, then decrease, hitting its maximum at the same spot where <math>\beta</math> is at a minimum.
  
  
 
Testing for a quadratic fit:
 
Testing for a quadratic fit:
 
<pre>
 
<pre>
In[3]:= quad1=Fit[data1,{1,n,n^2},n]
+
In[6]:= quad1=Fit[data1,{1,n,n^2},n]
  
Out[3]= 4.45564 +0.32015 n+0.0000417787 n^2
+
Out[6]= 4.89326 +0.324556 n-0.0000129154 n^2
 
</pre>
 
</pre>
  
 
The quadratic fit does not work since it's first derivative
 
The quadratic fit does not work since it's first derivative
  
<center><math>\theta'\approx \Delta \theta=.325072-6.83308\ n</math></center>
+
<center><math>\theta'\approx \Delta \theta=0.324556 -0.0000258\ n</math></center>
  
does not give the same spacing between low and high values of n as seen in the ced data.  The coefficient near the first order will have to smaller than .31 to find a correlation that would agree.
+
does not give the same spacing between low and high values of n as seen in the calculated data.  The coefficient near the zeroth order for the derivative will have to be smaller than .31 to find a correlation that would agree with calculated values.
  
 
Testing for a polynomial of degree 3 fit:
 
Testing for a polynomial of degree 3 fit:
 
<pre>
 
<pre>
In[4]:= polynomial1=Fit[data1,{1,n,n^2,n^3},n]
+
In[7]:= polynomial1=Fit[data1,{1,n,n^2,n^3},n]
  
Out[4]= 4.49876 +0.293001 n+0.000679074 n^2-3.57132*10^-6 n^3
+
Out[7]= 4.93253 +0.297371 n+0.000566298 n^2-3.04016*10^-6 n^3
 
</pre>
 
</pre>
  
  
This fit best matches the data found in ced
+
This fit best matches the data found in the calculated data
  
<center><math>\theta\equiv 4.49876 +0.293001 n+0.000679074 n^2-3.57132\times 10^{-6} n^3</math></center>
+
<center><math>\theta\equiv 4.93253 +0.297371 n+0.000566298 n^2-3.04016 \times 10^{-6} n^3</math></center>
  
 
As discussed earlier, taking the 1st derivative of this function will give us the spacing of the bins as a function of wire number.
 
As discussed earlier, taking the 1st derivative of this function will give us the spacing of the bins as a function of wire number.
  
<center><math>\theta '\equiv 0.293001 + 0.00135815 n - 0.000010714 n^2</math></center>
+
<center><math>\theta '\equiv 0.297371 + 0.0011326 n - 9.12048 \times 10^{-6} n^2</math></center>
  
  
 
The derivative of this function then will tell us where the bin spacing is at a minimum and a maximum.
 
The derivative of this function then will tell us where the bin spacing is at a minimum and a maximum.
  
<center><math>\theta ''\equiv .001021976-1.61514\times 10^{-5} n</math></center>
+
<center><math>\theta ''\equiv 0.0011326 - 0.000018241 n</math></center>
 +
 
 +
 
 +
<center><math>0.0011326 - 0.000018241 n=0\Rightarrow Bin\ Spacing\ Maximum\ at\ n=62.0909</math></center>
 +
 
 +
 
 +
This is near the point of the plane where <math>\theta=25^{\circ}</math> as seen in the figure above.
 +
 
 +
 
 +
 
 +
----
  
  
<center><math>0.00135815 - 0.0000214279 n=0\Rightarrow Bin\ Spacing\ Maximum\ at\ n=63.3822</math></center>
 
  
Comparing this maximum, we can see from Table 3 that near the maximum of 63 are seperated by larger distances than at n=1 or n=112.  This is the midpoint of the plane as seen in the geometry file.
+
<center><math>\underline{\textbf{Navigation}}</math>
  
=[[CED_Verification_of_DC_Angle_Theta_and_Wire_Correspondance|<-Back]]=
+
[[CED_Verification_of_DC_Angle_Theta_and_Wire_Correspondance|<math>\vartriangleleft </math>]]
 +
[[CED_Verification_of_DC_Angle_Theta_and_Wire_Correspondance|<math>\triangle </math>]]
 +
[[DC_Binning_Based_On_Wire_Numbers|<math>\vartriangleright </math>]]
  
=[[DC_Binning_Based_On_Wire_Numbers|Forward->]]=
+
</center>

Latest revision as of 20:14, 15 May 2018

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

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


DC Super Layer 1:Layer 1

From the ced simulations


Table 1: Superlayer 1 Change of Coordinates From Wire 1 to Wire 2(cm)
Coordinates(cm) Layer 1 Layer 2 Layer 3 Layer 4 Layer 5 Layer 6
[math]\Delta x[/math] 1.22 1.22 1.22 1.22 1.21 1.22
[math]\Delta y[/math] 0.00 0.00 0.00 0.00 0.00 0.00
[math]\Delta z[/math] .55 .56 .57 .57 .57 .57


While the wire placement is off, the difference in x and z position is matched by the calculated values. Using the geometric construction for determining angle theta to wire 2:

[math]\tan{\theta_2}=\frac{x_1+\Delta x}{z_1+\Delta z}\Rightarrow \theta_2=\arctan{\frac{x_1+\Delta x}{z_1+\Delta z}}[/math]


[math]\arctan{\frac{22.2092+1.212}{242.5779-0.5652}}=\arctan{\frac{23.4212}{242.0127}}=\arctan{0.0967}=5.53^{\circ}=\tan{\theta_2}[/math]


This equation can be solved for a hypothetical wire 0, which will allow the wire number to be the multiplicative factor for the change from the starting position.


[math]\tan{\theta_n}=\frac{x-\Delta x+n\Delta x}{z-\Delta z+n\Delta z}\Rightarrow \theta_n=\arctan{\frac{x+n\Delta x}{z+n\Delta z}}[/math]

where

[math]x_0=22.2092-1.212\approx 20.9972cm[/math]


[math]z_0=242.5779+0.5652\approx 243.1431cm[/math]


[math]\theta_n=\arctan{\frac{20.9972+n1.212}{243.1431-n0.5652}}[/math]


[math]\tan{\theta_n}=\frac{20.9972+n1.212}{243.1431-n0.5652}\Rightarrow \tan{ \theta_n}=-2.14437 + \frac{959.637}{430.189 - n}\Rightarrow 430.189 - n = \frac{959.637}{\tan{ \theta_n}+2.14437}[/math]


[math]\Rightarrow n = \frac{-959.637}{\tan{ \theta_n}+2.14437}+430.189[/math]

Using Mathematica, a series expansion about n=0 can be found:

In[2]:= Series[ArcTan[(20.9972+1.212 n)/(243.1431-0.5652 n)],{n,0,4}]

Out[2]= 0.0861437+0.00514708 n+9.67686*10^-6 n^2-2.72596*10^-8 n^3-2.22159*10^-10 n^4+O[n]^5
[math]\theta\approx 0.0861437+0.00514708 n+9.67686 \times10^{-6} n^2-2.72596 \times 10^{-8} n^3[/math]


This expression will find the angle theta in radians given the wire number. To convert from radians to degrees, we can multiply by 180 and divide by Pi.

In[3]:= 180(0.08614365821719583`+0.005147076698801422` n+9.676859262994711`*^-6 n^2-2.7259639443492433`*^-8 n^3-2.2215881081233252`*^-10 n^4+O[n]^5)/3.14159265359

Out[3]= 4.93567+0.294906 n+0.000554443 n^2-1.56186*10^-6 n^3-1.27288*10^-8 n^4+O[n]^5


[math]\theta\approx 4.93567+0.294906 n+0.000554443 n^2-1.56186*10^{-6} n^3[/math]


This tells us that the expression for theta will follow a function that comes from a series expansion. Using Mathematica, a line can be fitted to the data collected on the wire number to angle theta correspondence.


Table 4: Calculated Superlayer 1 Wire-Angle Theta Correspondence in Degrees
Wire Number Layer 1
1 5.23
2 5.53
77 29.80
78 30.13
79 30.46
110 40.45
111 40.76
112 41.07


Declaring the data set:

In[4]:= data1={{1,5.23},{2,5.53},{77,29.80},{78,30.13},{79,30.46},{110,40.45},{111,40.76},{112,41.07}}

Out[4]= {{1,5.23},{2,5.53},{77,29.8},{78,30.13},{79,30.46},{110,40.45},{111,40.76},{112,41.07}}


Testing for a linear fit:

In[5]:= line1=Fit[data1,{1,n},n]

Out[5]= 4.90443 +0.323148 n

Examining the range limits for the angle theta for layer 1:


Taking the difference of the upper and lower limits in theta,


[math]\theta=\arctan{\frac{20.9972+n1.212}{243.1431-n0.5652}}[/math]


[math]\theta_{n=1}=\arctan{\frac{20.9972+(1)1.212}{243.1431-(1)0.5652}}=5.23\ \ \ \ \ \ \theta_{n=112}=\arctan{\frac{20.9972+(112)1.212}{243.1431-(112)0.5652}}=41.07[/math]



[math]\Delta degrees=41.07^{\circ}-5.23^{\circ}=35.84^{\circ}[/math]


Dividing by the change in wire numbers (112-1=111), we find

[math]\frac{\Delta degrees}{\Delta wire\ number}=\frac{35.84^{\circ}}{111\ wires}\approx\ \frac{0.323^{\circ}}{wire\ number}[/math]

This would imply that if the wires were evenly placed, their change in angle theta would increase by the factor of .323 degrees for each increase in wire number, starting obviously with wire 1 at 5.23 degrees. In addition, this implies that the bin spacing for each wire would be around .323 degrees in width.


Checking this, we can find the difference between wires 1 and 2,

[math]\frac{\Delta degree}{\Delta wire\ number}=\frac{5.5277^{\circ}-5.2311^{\circ} }{wire\ number}=\frac{.2965^{\circ}}{wire\ number} [/math]

Similarly, finding the difference between wires 111 and 112,

[math]\frac{\Delta degree}{\Delta wire\ number}=\frac{41.0740^{\circ}-40.7649^{\circ}}{wire\ number}=\frac{.0.3091^{\circ}}{wire\ number}[/math]


These differing values show that the bin width is not uniform in length, therefore a first order, linear fit, will not suffice. The change in bin size, with respect to the angle theta covered, is due to the tilt of the wire plane.



Geometry of Wire Bins Angular Size in DC
Figure 5.1.1: For a tilted plane, that is offset from the origin, the change in angle between uniformly spaced points depends on the tilt with respect to the beam line. The uniform distance between points, A, can be used to show the angle, [math]\alpha[/math] each wire bin covers can be found using the law of sines. The interior angle [math]\beta[/math]

Using the law of sines,

[math]\frac{\sin{\alpha}}{A}=\frac{\sin{\beta}}{B}[/math]


[math]\Rightarrow \alpha=\arcsin{\frac{A \sin{\beta}}{B}}[/math]


The range of [math]\beta[/math] runs from near to 65 to almost 180 degrees. At [math]\theta=25^{\circ}[/math] with respect to the positive z direction, a right triangle is created. Since the sine function is periodic and reaches its maximum at 90 degrees, this implies that the opposite side to this angle reaches its minimum. On either side of the right angle, as the angle [math]\beta[/math] either increases or decreases, the function [math]\sin{\beta} will be smaller. This implies for movement in angle \lt math\gt \alpha[/math] in a counter-clockwise direction, the angle [math]\beta[/math] will continue to increase, while the length of B will decrease, then increase, reaching it's minimum at [math]\theta=25^{\circ}[/math]. This also implies that [math]\alpha[/math] will increase in size, then decrease, hitting its maximum at the same spot where [math]\beta[/math] is at a minimum.


Testing for a quadratic fit:

In[6]:= quad1=Fit[data1,{1,n,n^2},n]

Out[6]= 4.89326 +0.324556 n-0.0000129154 n^2

The quadratic fit does not work since it's first derivative

[math]\theta'\approx \Delta \theta=0.324556 -0.0000258\ n[/math]

does not give the same spacing between low and high values of n as seen in the calculated data. The coefficient near the zeroth order for the derivative will have to be smaller than .31 to find a correlation that would agree with calculated values.

Testing for a polynomial of degree 3 fit:

In[7]:= polynomial1=Fit[data1,{1,n,n^2,n^3},n]

Out[7]= 4.93253 +0.297371 n+0.000566298 n^2-3.04016*10^-6 n^3


This fit best matches the data found in the calculated data

[math]\theta\equiv 4.93253 +0.297371 n+0.000566298 n^2-3.04016 \times 10^{-6} n^3[/math]

As discussed earlier, taking the 1st derivative of this function will give us the spacing of the bins as a function of wire number.

[math]\theta '\equiv 0.297371 + 0.0011326 n - 9.12048 \times 10^{-6} n^2[/math]


The derivative of this function then will tell us where the bin spacing is at a minimum and a maximum.

[math]\theta ''\equiv 0.0011326 - 0.000018241 n[/math]


[math]0.0011326 - 0.000018241 n=0\Rightarrow Bin\ Spacing\ Maximum\ at\ n=62.0909[/math]


This is near the point of the plane where [math]\theta=25^{\circ}[/math] as seen in the figure above.




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