Determining wire-phi correspondance - Revision history

Vanwdani: /* Setting up Mathematica for DC Theta-Phi Isotropic Cone */

2017-03-04T04:14:16Z

Setting up Mathematica for DC Theta-Phi Isotropic Cone

Vanwdani: /* Function for the wire number in the detector frame for change in \phi and constant \theta in the lab frame */

2017-03-04T03:51:40Z

Function for the wire number in the detector frame for change in \phi and constant \theta in the lab frame

Vanwdani: /* Function for the change in x' in the detector frame for change in \phi and constant \theta in the lab frame */

2017-03-04T03:49:36Z

Function for the change in x' in the detector frame for change in \phi and constant \theta in the lab frame

Vanwdani: /* Test for \theta=20 and \phi=1 */

2017-03-04T03:46:39Z

Test for \theta=20 and \phi=1

Vanwdani: /* Test for \theta=20 and \phi=0 */

2017-03-04T03:11:28Z

Test for \theta=20 and \phi=0

Vanwdani: /* In the plane */

2017-03-04T03:10:00Z

In the plane

Vanwdani: /* Test for \theta=20 and \phi=0 */

2017-03-04T03:07:44Z

Test for \theta=20 and \phi=0

Vanwdani: /* Determining Elliptical Equation */

2017-03-04T03:04:38Z

Determining Elliptical Equation

Show changes

Vanwdani: /* Determining ellipse components */

2017-03-04T03:02:47Z

Determining ellipse components

Vanwdani: /* Elliptic Conic Section */

2017-03-04T03:00:14Z

Elliptic Conic Section

← Older revision		Revision as of 04:14, 4 March 2017
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−	~~==Setting up Mathematica for DC Theta-Phi Isotropic Cone==~~
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−	~~<center>[[File:FirstPass.pdf]]</center>~~
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−	~~<center>[[File:SecondPass.pdf]]</center>~~

← Older revision		Revision as of 03:51, 4 March 2017
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−	~~===Function for the wire number in the detector frame for change in <math>\phi</math> and constant <math>\theta</math> in the lab frame===~~
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−	~~Using the expression for wire number n in terms of <math>\theta</math> for the detector mid-plane where <math>\phi=0</math>:~~
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−	~~<center><math>n = \frac{-957.412}{\tan(\theta)+2.14437}+430.626</math></center>~~
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−	~~We can use the inverse of this function to find the neighboring wire's corresponding angle theta~~
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−	~~<center><math>\theta\equiv 4.49876 +0.293001 n+0.000679074 n^2-3.57132\times 10^{-6} n^3</math></center>~~
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−	~~<center><math>\theta(n \pm 1)\equiv 4.49876 +0.293001 (n \pm 1)+0.000679074 (n \pm 1)^2-3.57132\times 10^{-6} (n \pm 1)^3</math></center>~~
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−	~~We also know what the x' function must follow dependent on phi in the detector plane~~
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−	~~<center><math>x_1^'=\frac{((x_{D2}-x_P)^2+(y_{D2}-y_P)^2+(z_{D2}-z_P)^2)-((x_P-x_{D1})^2+(y_P-y_{D1})^2+(z_P-z_{D1})^2)}{4ae}-ae</math></center>~~
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−	~~<center><math>x_1^'=\frac{(r_{D2}^2)-(r_{D1}^2)+cot^2(\theta)(r_{D2}^2-r_{D1}^2)-2x_P(x_{D2}-x_{D1})-2y_P(y_{D2}-y_{D1})-2z_P(z_{D2}-z_{D1})}{4ae}-ae</math></center>~~
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−	~~{\| class="wikitable" align="center"~~
−	~~\| style="background: gray" \| <math>x_{D1}=r_{D1}\ cos(\phi)\qquad y_{D1}=r_{D1}cos(\phi)\qquad z_{D1}=r_{D1} cot(\theta)</math>~~
−	\|}
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−	~~{\| class="wikitable" align="center"~~
−	~~\| style="background: gray" \| <math>x_{D2}=r_{D2} cos(\phi)\qquad y_{D2}=r_{D2} sin(\phi)\qquad z_{D2}=r_{D2} cot(\theta)</math>~~
−	\|}
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−	~~{\| class="wikitable" align="center"~~
−	~~\| style="background: gray" \| <math>x_P=\frac{2.53cos(\phi)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}</math>~~
−	\|}
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−	~~{\| class="wikitable" align="center"~~
−	~~\| style="background: gray" \| <math>y_P=\frac{2.53sin(\phi)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}</math>~~
−	\|}
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−	~~{\| class="wikitable" align="center"~~
−	~~\| style="background: gray" \| <math>z_P=\frac{2.53cot(\theta)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}</math>~~
−	\|}
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−	~~<center><math>r_{D1}=R_{Lower\ Dandelin}cos(\theta)=(ae-\Delta a) tan(65^{\circ})cos(\theta)\qquad \qquad r_{D2}=R_{Lower\ Dandelin}cos(\theta)=(ae+\Delta a) tan(65^{\circ})cos(\theta)</math></center>~~
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−	~~We can take this point to be the x axis intercept and use the fact that each wire is titled by 6 degrees to the horizontal in the plane of the detector to create an equation~~
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−	~~<center><math>x_{wire\ n}'=tan(6^{\circ})y'+x_{n_0}</math></center>~~
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−	~~where the initial wire and x' position at the given theta is represented by <math>n_0</math>~~
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−	~~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.~~
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−	~~<center><math>\frac{x_{n=1}}{sin 4.79^{\circ}}=\frac{2.52934}{sin 110.21^{\circ}} \Rightarrow x_{n=1}=.2252</math></center>~~
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−	~~Since each wire is separated by .01337 meters~~
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−	~~<center><math>.2252-.01337=.2117=x_{n=0}</math></center>~~
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−	~~Each wire becomes an equation of the form,~~
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−	~~<center><math>x_{wire\ n}'=tan(6^{\circ})y'+.2117+.01337\ n</math></center>~~
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−	~~This agrees with CED simulation~~
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−	~~<center>[[File:DC_geom.png]]</center>~~

	==Setting up Mathematica for DC Theta-Phi Isotropic Cone==		==Setting up Mathematica for DC Theta-Phi Isotropic Cone==

← Older revision		Revision as of 03:49, 4 March 2017
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−	~~=Function for the change in x' in the detector frame for change in <math>\phi</math> and constant <math>\theta</math> in the lab frame=~~
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−	~~<center><math>D2P=\sqrt{(x_{D2}-x_P)^2+(y_{D2}-y_P)^2+(z_{D2}-z_P)^2}</math></center>~~
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−	~~<center><math>D1P=\sqrt{(x_P-x_{D1})^2+(y_P-y_{D1})^2+(z_P-z_{D1})^2}</math></center>~~
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−	~~<center><math>x_1^'=\frac{((x_{D2}-x_P)^2+(y_{D2}-y_P)^2+(z_{D2}-z_P)^2)-((x_P-x_{D1})^2+(y_P-y_{D1})^2+(z_P-z_{D1})^2)}{4ae}-ae</math></center>~~
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−	~~{\| class="wikitable" align="center"~~
−	~~\| style="background: gray" \| <math>x_{D1}=r_{D1}\ cos(\phi)\qquad y_{D1}=r_{D1}cos(\phi)\qquad z_{D1}=r_{D1} cot(\theta)</math>~~
−	\|}
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−	~~{\| class="wikitable" align="center"~~
−	~~\| style="background: gray" \| <math>x_{D2}=r_{D2} cos(\phi)\qquad y_{D2}=r_{D2} sin(\phi)\qquad z_{D2}=r_{D2} cot(\theta)</math>~~
−	\|}
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−	~~{\| class="wikitable" align="center"~~
−	~~\| style="background: gray" \| <math>x_P=\frac{2.53cos(\phi)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}</math>~~
−	\|}
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−	~~{\| class="wikitable" align="center"~~
−	~~\| style="background: gray" \| <math>y_P=\frac{2.53sin(\phi)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}</math>~~
−	\|}
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−	~~{\| class="wikitable" align="center"~~
−	~~\| style="background: gray" \| <math>z_P=\frac{2.53cot(\theta)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}</math>~~
−	\|}
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−	~~<center><math>x_1^'=\frac{((x_{D2}-x_P)^2+(y_{D2}-y_P)^2+(z_{D2}-z_P)^2)-((x_P-x_{D1})^2+(y_P-y_{D1})^2+(z_P-z_{D1})^2)}{4ae}-ae</math></center>~~
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−	<center><math>x_1^'=\frac{x_{D2}^2-2x_Px_{D2}+x_P^2+y_{D2}^2-2y_Py_{D2}+y_P^2+z_{D2}^2-2z_Pz_{D2}+z_P^2-x_P^2+2x_Px_{D1}-x_{D1}^2-y_P^2+2y_Py_{D1}-y_{D1}^2-z_P^2+2z_Pz_{D1}-z_{D1}^2}{4ae}-ae</math></center>
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−	~~<center><math>x_1^'=\frac{(x_{D2}^2+y_{D2}^2)-(x_{D1}^2+y_{D1}^2)+z_{D2}^2-z_{D1}^2-2x_P(x_{D2}-x_{D1})-2y_P(y_{D2}-y_{D1})-2z_P(z_{D2}-z_{D1})}{4ae}-ae</math></center>~~
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−	~~<center><math>x_1^'=\frac{(r_{D2}^2)-(r_{D1}^2)+cot^2(\theta)(r_{D2}^2-r_{D1}^2)-2x_P(x_{D2}-x_{D1})-2y_P(y_{D2}-y_{D1})-2z_P(z_{D2}-z_{D1})}{4ae}-ae</math></center>~~
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−	~~Expressing this as functions of <math>\phi</math> and non-differentiable constants~~
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−	~~<center><math>x_1^'=\frac{c_1+c_2-2x_P(\phi)x_{D2}(\phi)+2x_P(\phi)x_{D1}(\phi)-2y_P(\phi)y_{D2}(\phi)+2y_P(\phi)y_{D1}(\phi)-2z_P(\phi)c_3}{4c_4}-c_4</math></center>~~
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−	~~Differentiating with respect to <math>\phi</math>~~
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−	~~{\| class="wikitable" align="center"~~
−	~~\| style="background: gray" \| <math>x_{D1}=r_{D1} cos(\phi)\Rightarrow \dot x_{D1}=-r_{D1} sin(\phi)</math>~~
−	\|}
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−	~~{\| class="wikitable" align="center"~~
−	~~\| style="background: gray" \| <math>y_{D1}=r_{D1}sin(\phi)\Rightarrow \dot y_{D1}=r_{D1}cos(\phi)</math>~~
−	\|}
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−	~~{\| class="wikitable" align="center"~~
−	~~\| style="background: gray" \| <math>x_{D2}=r_{D2} cos(\phi)\Rightarrow \dot x_{D2}=-r_{D2} sin(\phi)</math>~~
−	\|}
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−	~~{\| class="wikitable" align="center"~~
−	~~\| style="background: gray" \| <math>y_{D2}=r_{D2}sin(\phi)\Rightarrow \dot y_{D2}=r_{D2}cos(\phi)</math>~~
−	\|}
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−	~~{\| class="wikitable" align="center"~~
−	\| style="background: gray" \| <math>x_P=\frac{2.52934271645cos(\phi)}{cot(\theta)+cos(\phi)cot(65^{\circ})}\Rightarrow \dot x_P=\frac{-2.52934271645cot(\theta)sin(\phi)}{(cos(\phi)cot(65^{\circ}+cot(\theta))^2}</math>
−	\|}
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−	~~{\| class="wikitable" align="center"~~
−	\| style="background: gray" \| <math>y_P=\frac{2.52934271645sin(\phi)}{cot(\theta)+cos(\phi)cot(65^{\circ})}\Rightarrow \dot y_P=\frac{-1.7206+2.52934271645 cos(\phi) cot(\theta)}{(cos(\phi) cot(65^{\circ}) + cot(\theta))^2}</math>
−	\|}
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−	~~{\| class="wikitable" align="center"~~
−	\| style="background: gray" \| <math>z_P=\frac{2.52934271645cot(\theta)}{cot(\theta)+cos(\phi)cot(65^{\circ})}\Rightarrow \dot z_P=\frac{-1.7206 cot(\theta)sin(\phi))}{(cos(\phi) cot(65) + cot(\theta))^2}</math>
−	\|}
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−	<center><math>\frac{dx_1^1}{d\phi}=\frac{-2}{4c_4}\frac{d}{d\phi}(x_P(\phi)x_{D2}(\phi))+\frac{2}{4c_4}\frac{d}{d\phi}(x_P(\phi)x_{D1}(\phi))-\frac{2}{4c_4}\frac{d}{d\phi}(y_P(\phi)y_{D2}(\phi))+\frac{2}{4c_4}\frac{d}{d\phi}(y_P(\phi)y_{D1}(\phi))-\frac{2c_3}{4c_4}\frac{d}{d\phi}z_P(\phi)</math></center>
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−	<center><math>\frac{dx_1^1}{d\phi}=\frac{-2}{4c_4} \left ( (\dot x_P(\phi)x_{D2}(\phi)+x_P(\phi)\dot x_{D2}(\phi))-(\dot x_P(\phi)x_{D1}(\phi)+x_P(\phi)\dot x_{D1}(\phi))+(\dot y_P(\phi)y_{D2}(\phi)+y_P(\phi)\dot y_{D2}(\phi))-(\dot y_P(\phi)y_{D1}(\phi)+y_P(\phi)\dot y_{D1}(\phi))+c_3\dot z_P(\phi) \right )</math></center>

	===Function for the wire number in the detector frame for change in <math>\phi</math> and constant <math>\theta</math> in the lab frame===		===Function for the wire number in the detector frame for change in <math>\phi</math> and constant <math>\theta</math> in the lab frame===

← Older revision		Revision as of 03:46, 4 March 2017
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−	~~==Test for <math>\theta=20</math> and <math>\phi=1</math>==~~
−	~~All previous quantities where calculated for <math>\theta=20^{\circ}</math> and do not depend on the angle <math>\phi</math>. The quantities that do change~~
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−	~~{\| class="wikitable" align="center"~~
−	\| style="background: gray" \| <math>x_{D1}=r_{D1}\ cos(\phi)=.5901cos(1^{\circ}))=.5900\text {m}\qquad y_{D1}=r_{D1}cos(\phi)=.5901 sin(1^{\circ}))=.0103\qquad z_{D1}=r_{D1} cot(\theta)=.5901cot(20)=1.6212\ \text{m}</math>
−	\|}
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−	~~{\| class="wikitable" align="center"~~
−	\| style="background: gray" \| <math>x_{D2}=r_{D2} cos(\phi)=1.3055cos(1^{\circ}))=1.3053\text {m}\qquad y_{D2}=r_{D2} sin(\phi)=1.3055sin(1^{\circ}))=.0228\qquad z_{D2}=r_{D2} cot(\theta)=1.3055cot(20)=3.5868\ \text{m}</math>
−	\|}
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−	~~{\| class="wikitable" align="center"~~
−	~~\| style="background: gray" \| <math>x_P=\frac{2.53cos(\phi)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}=\frac{2.53cos(1^{\circ}))}{(cot(20^{\circ})+cos(1^{\circ}))cot(65^{\circ})}=0.7869</math>~~
−	\|}
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−	~~{\| class="wikitable" align="center"~~
−	~~\| style="background: gray" \| <math>y_P=\frac{2.53sin(\phi)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}=\frac{2.53sin(1^{\circ}))}{(cot(20^{\circ})+cos(1^{\circ}))cot(65^{\circ})}=.0137</math>~~
−	\|}
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−	~~{\| class="wikitable" align="center"~~
−	~~\| style="background: gray" \| <math>z_P=\frac{2.53cot(\theta)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}=\frac{2.53cot(20^{\circ})}{(cot(20^{\circ})+cos(1^{\circ}))cot(65^{\circ})}=2.1624</math>~~
−	\|}
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−	<center><math>D2P=\sqrt{(x_{D2}-x_P)^2+(y_{D2}-y_P)^2+(z_{D2}-z_P)^2}=\sqrt{(1.3053-0.7869)^2+(.0228-.0137)^2+(3.5868-2.1624)^2}=\sqrt{(.5184)^2+(.0091)^2+(1.4244)^2}=1.51582872713\ \text{m}</math></center>
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−	<center><math>D1P=\sqrt{(x_P-x_{D1})^2+(y_P-y_{D1})^2+(z_P-z_{D1})^2}=\sqrt{(0.7871-.5900)^2+(.0137-.0103)^2+(2.1624-1.6212)^2}=\sqrt{(.1971)^2+(.0034)^2+(.5412)^2}=.575983862621\ \text{m}</math></center>
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−	~~<center><math>x_1^'=\frac{r_2^{'2}-r_1^{'2}}{4ae}-ae=\frac{1.5158^{'2}-.5758^{'2}}{4(1.0459)(.4497)}-(1.0459)(.4497)=.575\ \text{m}</math></center>~~
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−	~~Using the pythagorean theorem~~
−	~~<center><math>y'=\sqrt{.576^2-.575^2}=.03\ \text{m}</math></center>~~
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−	The two possible answers denote shifting to the left on right on the y axis. We take the direction of positive and negative to be the same as the sign convention for the angle phi starting on the x axis and shifting positive clockwise. A shift of 1 degree in phi at theta equal to 20 degrees only results in a small change in the x and y. This changes depending on the angles.

	=Function for the change in x' in the detector frame for change in <math>\phi</math> and constant <math>\theta</math> in the lab frame=		=Function for the change in x' in the detector frame for change in <math>\phi</math> and constant <math>\theta</math> in the lab frame=

← Older revision		Revision as of 03:11, 4 March 2017
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−	~~==Test for <math>\theta=20</math> and <math>\phi=0</math>==~~
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−	~~Substituting in the values found earlier for the case of <math>\theta=20^{\circ}</math> and <math>\phi=0</math>~~
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−	~~<center><math>a=\frac{a_1+a_2}{2}=1.0459</math></center>~~
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−	~~<center><math>e\equiv \frac{sin(25^{\circ})}{cos(\theta)}= \frac{sin(25^{\circ})}{cos(20^{\circ})}=.4497</math></center>~~
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−	~~<center><math>x_1^'=\frac{r_2^{'2}-r_1^{'2}}{4ae}-ae=\frac{r_2^{'2}-r_1^{'2}}{4(1.0459)(.4497)}-(1.0459)(.4497)</math></center>~~
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−	~~Since~~
−	~~<center><math>\lVert \overrightarrow{D1P} \rVert \equiv \lVert \overrightarrow{F1P} \rVert \qquad \qquad \lVert \overrightarrow{D2P} \rVert \equiv \lVert \overrightarrow{F2P} \rVert</math></center>~~
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−	~~<center><math>D1P \approx .576\ \text{m}\qquad D2P \approx 1.516\ \text{m}</math></center>~~
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−	~~The <math>x_1^' </math> distance from focal point 1 is:~~
−	~~<center><math>x_1^'=\frac{1.516^2-.576^2}{4(1.0459)(.4497)}-(1.0459)(.4497)=\frac{1.97}{1.88}-.4703=.576\ \text{m}=r_1</math></center>~~
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−	~~This is the radius from focal point 1, which is to be expected since the y component is equal to zero for <math>\phi=0</math>~~
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−	~~The focii are located at~~
−	~~<center><math>F1\equiv (-\Delta a+ae,0)=(-.1775+1.0459(.4497),0)=(.292,0)\ \text{m}\qquad F2\equiv (-\Delta a-ae,0)=(-.1775-1.0450(.4497),0)=(-6.478,0)\ \text{m}</math></center>~~
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−	~~This implies that with respect to the origin, x', we find~~
−	~~<center><math>x'=x_1'+\Delta a+ae=.576+.292=.868\ \text{m}</math></center>~~
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−	~~This is verified with CED~~
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−	~~<center>[[File:x_in_detector_plane.png]]</center>~~
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−	~~Since the x' dimension is the hypotenuse in a right triangle of 65 degrees~~
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−	~~<center><math>x'=\frac{x_{lab}}{sin 65^{ \circ}}=.868=x'</math></center>~~

	==Test for <math>\theta=20</math> and <math>\phi=1</math>==		==Test for <math>\theta=20</math> and <math>\phi=1</math>==

← Older revision		Revision as of 03:10, 4 March 2017
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−	~~=In the plane=~~
−	We can use the geometry of the sector to impose limits on the allowed x and y for this plane. The angle which the two sides of the triangle that make up the sector is given as <math>59^{\circ}</math>. Each focii will have a ray that creates a circle around the point that is equal to the length found from the D1P and D2P lines. These circles will overlap at a common y component, which can be used to determine if the point is within the detector limits.
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−	~~<center>[[File:Sector1plane.png]]</center>~~
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−	~~Using the physical constraints of the distance the interior walls of the side walls intersect each other from the beam line, as well as the angle the lowest guard wire at the plane midpoint.~~
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−	~~<center>[[File:Detector_constraints.png]]</center>~~
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−	~~where b' is the location on the y' axis where the two interior side walls of the sector intersect.~~
−	~~<center><math>b'=\frac{0.08298\ m}{sin(65^{\circ})}=.09156\ m</math></center>~~
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−	~~The quantity <math>x_0^'</math> is the corresponding x' distance from the beam line and detector plane intersection.~~
−	~~<center><math>\frac{x_0^'}{sin(4.694^{\circ})}=\frac{2.53\ m}{sin(110.306^{\circ})}</math></center>~~
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−	~~<center><math>x_0^'=\frac{sin(4.694^{\circ})2.53\ m}{sin(110.306^{\circ})}=.0872(2.53)=.22070/ m</math></center>~~
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−	~~Since the interior sides of the sector make an angle of <math>59^{\circ}</math> in the plane of the wires, this can be represented by equations~~
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−	~~<center><math>x'=\pm cot(29.5^{\circ})y^'+.09156\ m</math></center>~~
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−	Using the properties of an ellipse, we can work in the coordinates of the ellipse, in the plane of the ellipse, where the vertex is located at the crossing of the ellipse and the cone. Since the Dandelin construction was used, we know that
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−	~~<center><math>F1P=D1P\qquad \text{and}\ F2P=D2P</math></center>~~
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−	~~If we shift the origin from the center of the ellipse to the intersection of the plane and cone for a given theta~~
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−	~~<center><math>c_{ellipse}\equiv(-\Delta a,0)</math></center>~~
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−	~~<center><math>F1\equiv (-\Delta a+ae,0)\qquad F2\equiv (-\Delta a-ae,0)</math></center>~~
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−	~~Writing the equations for circles centered at an ellipse focii points, we find~~
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−	~~<center><math>x_1^{'2}+y_1^{'2}=r_1^{'2}\qquad \qquad x_2+^{'2}+y_2^{'}2=r_2^{'2}</math></center>~~
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−	~~<center><math>r_1^{'2}-x_1^{'2}=y_1^{'2}\qquad \qquad y_2^{'2}=r_2^{'2}-x_2^{'2}</math></center>~~
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−	~~Since the focii are seperated by 2ae as found earlier~~
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−	~~<center><math>x_2^'=x_1^'+2ae</math></center>~~
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−	~~<center><math>r_1^{'2}-x_1^{'2}=r_2^{'2}-(x_1^'+2ae)^2</math></center>~~
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−	~~<center><math>r_1^{'2}-r_2^{'2}=-4x_1^'ae-4a^2e^2</math></center>~~
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−	~~<center><math>r_2^{'2}-r_1^{'2}=8ae(x_1^'+ae+\Delta a)</math></center>~~
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−	~~<center><math>\frac{r_2^{'2}-r_1^{'2}}{4ae}-ae=x_1^'</math></center>~~

← Older revision		Revision as of 03:07, 4 March 2017
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−	~~====Test for <math>\theta=20</math> and <math>\phi=0</math>====~~
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−	~~Solving for the components of the ellipse~~
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−	~~<center><math>a_1=\frac{2.53sin(\theta)}{sin(115-\theta)}=\frac{2.53sin(20^{\circ})}{sin(115^{\circ}-20^{\circ})}=\frac{2.53\cdot.342}{.9962}=.8686</math></center>~~
−
−
−	~~<center><math>a_2=\frac{2.53sin(\theta)}{sin(65-\theta)}=\frac{2.53sin(20^{\circ})}{sin(65^{\circ}-20^{\circ})}=\frac{2.53\cdot.342}{.0872}=1.2237</math></center>~~
−
−
−	~~<center><math>c_{ellipse} \equiv (-\Delta a\ sin(65^{\circ}), 0,z+\Delta a\ cos(65^{\circ}))</math></center>~~
−
−
−	~~<center><math>c_{ellipse} \equiv (-\frac{a_2-a_1}{2} sin(65^{\circ}), 0,2.53+\frac{a_2-a_1}{2} cos(65^{\circ}))</math></center>~~
−
−
−
−	~~<center><math>c_{ellipse} \equiv (-\frac{1.2237-.8686}{2} sin(65^{\circ}), 0,2.53+\frac{1.2237-.8686}{2} cos(65^{\circ}))</math></center>~~
−
−
−
−	~~<center><math>c_{ellipse} \equiv (-\frac{.3550}{2} sin(65^{\circ}), 0,2.53+\frac{.3550}{2} cos(65^{\circ}))</math></center>~~
−
−
−
−	~~<center><math>c_{ellipse} \equiv (-.1775 sin(65^{\circ}), 0,2.53+.1775 cos(65^{\circ}))</math></center>~~
−
−
−
−	~~<center><math>c_{ellipse} \equiv (-.1609, 0,.0750)</math></center>~~
−
−
−
−	~~<center><math>P(\phi=0) \equiv (-\frac{a_2-a_1}{2} sin(65^{\circ})+\frac{a_1+a_2}{2} sin((65^{\circ}), 0,2.53+\frac{a_2-a_1}{2} cos(65^{\circ})-\frac{a_1+a_2}{2} cos(65^{\circ}))</math></center>~~
−
−
−
−	<center><math>P(\phi=0) \equiv (-\frac{1.2234-.8684}{2} sin(65^{\circ})+\frac{.8684+1.2234}{2} sin((65^{\circ}), 0,2.53+\frac{1.2234-.8684}{2} cos(65^{\circ})-\frac{.8684+1.2234}{2} cos(65^{\circ}))</math></center>
−
−
−
−	~~<center><math>P(\phi=0) \equiv (-\frac{.3550}{2} sin(65^{\circ})+\frac{2.0918}{2} sin((65^{\circ}), 0,2.53+\frac{.3550}{2} cos(65^{\circ})-\frac{2.0918}{2} cos(65^{\circ}))</math></center>~~
−
−
−
−	~~<center><math>P(\phi=0) \equiv (-.1775 sin(65^{\circ})+1.0459 sin((65^{\circ}), 0,2.53+.1775 cos(65^{\circ})-1.0459 cos(65^{\circ}))</math></center>~~
−
−
−
−	~~<center><math>P(\phi=0) \equiv (.7870, 0,2.1623)</math></center>~~
−
−
−	~~<center><math>x=\frac{2.53cos(\phi)}{cot(\theta)+cos(\phi)cot(65^{\circ})}</math></center>~~
−
−
−
−	~~<center><math>x=\frac{2.53cos(0)}{cot(20^{\circ})+cos(0)cot(65^{\circ})}</math></center>~~
−
−
−
−	~~<center><math>x=\frac{2.53}{2.7475+.4663}</math></center>~~
−
−
−
−	~~<center><math>x=\frac{2.53}{3.2138}=.7870\text{cm}</math></center>~~
−
−
−	~~The y component is zero for <math>\phi=0</math>~~
−
−	~~The z component can be found from the ellipse equation~~
−
−	~~<center><math>z=-cot(65^{\circ})x+2.53=-cot(65^{\circ})(.7870)+2.53=2.1623\ \text{cm}</math></center>~~
−
−
−
−	~~<center><math>e\equiv \frac{sin(25^{\circ})}{cos(20^{\circ})}=0.96447</math></center>~~
−
−
−	~~<center><math>r_{D1}=R_{Lower\ Dandelin}cos(\theta)=(ae-\Delta a) tan(65^{\circ})cos(\theta)\qquad \qquad r_{D2}=R_{Lower\ Dandelin}cos(\theta)=(ae+\Delta a) tan(65^{\circ})cos(\theta)</math></center>~~
−
−
−
−	~~<center><math>r_{D1}=(\frac{a_1+a_2}{2}e-\frac{a_2-a_1}{2}) tan(65^{\circ})cos(\theta)\qquad \qquad r_{D2}=(\frac{a_1+a_2}{2}e+\frac{a_2-a_1}{2}) tan(65^{\circ})cos(\theta)</math></center>~~
−
−
−	<center><math>r_{D1}=(\frac{.8686+1.2237}{2}e-\frac{1.2237-.8686}{2}) tan(65^{\circ})cos(\theta)\qquad \qquad r_{D2}=(\frac{.8686+1.2237}{2}e+\frac{1.2237-.8686}{2}) tan(65^{\circ})cos(\theta)</math></center>
−
−
−	<center><math>r_{D1}=(\frac{2.0923}{2}\frac{sin(25^{\circ})}{cos(20^{\circ})}-\frac{.3551}{2}) tan(65^{\circ})cos(20^{\circ})\qquad \qquad r_{D2}=(\frac{2.0923}{2}\frac{sin(25^{\circ})}{cos(20^{\circ})}+\frac{.3551}{2}) tan(65^{\circ})cos(20^{\circ}))</math></center>
−
−
−
−	<center><math>r_{D1}=(\frac{2.0923}{2}\frac{sin(25^{\circ})}{cos(20^{\circ})}-\frac{.3551}{2}) tan(65^{\circ})cos(20^{\circ})\qquad \qquad r_{D2}=(\frac{2.0923}{2}\frac{sin(25^{\circ})}{cos(20^{\circ})}+\frac{.3551}{2}) tan(65^{\circ})cos(20^{\circ}))</math></center>
−
−
−	~~<center><math>r_{D1}=(1.0459(.4497)-.1775)\cdot 2.1445\cdot .9397=.5901\ \text{m} \qquad \qquad r_{D2}=(1.0459(.4497)+.1775)\cdot 2.1445\cdot .9397=1.3055 \text{m}</math></center>~~
−
−
−	~~The height to the first directrix circle is~~
−
−	~~<center><math>z_{D1}=r_{D1} cot(\theta)=.5901cot(20)=1.6212</math></center>~~
−
−	~~and the x and y components are~~
−	~~<center><math>x_{D1}=r_{D1}\ cos(\phi)=.5901 cos(0)=.5901\text {m}\ \ \ \ y_{D1}=r_{D1}cos(\phi)=.5901 sin(0)=0</math></center>~~
−
−
−	~~The height to the second directrix circle is~~
−
−	~~<center><math>z_{D2}=r_{D2} cot(\theta)=1.3055cot(20)=3.5868\ \text{m}</math></center>~~
−
−	~~and the x and y components are~~
−	~~<center><math>x_{D2}=r_{D2} cos(\phi)=1.3055cos(0)=1.3055\text {m}\ \ \ \ y_{D2}=r_{D2} sin(\phi)=1.3055sin(0)=0</math></center>~~
−
−
−	~~The distance between the two point~~
−
−	~~<center><math>\sqrt{(1.3055-.5901)^2+(0-0)^2+(3.5868-1.6212)^2}=\sqrt{.715^2+1.9656^2}=2.09\text {m}</math></center>~~
−
−
−	~~This should be equal to 2a~~
−
−	~~<center><math>2a=2(\frac{a_1+a_2}{2})=2(\frac{.8684+1.2234}{2})=2.09\text {m}</math></center>~~
−
−
−	~~This gives the components for a point on the directrix circles as~~
−
−	~~{\| class="wikitable" align="center"~~
−	\| style="background: gray" \| <math>x_{D1}=r_{D1}\ cos(\phi)=.5901cos(0)=.5901\text {m}\qquad y_{D1}=r_{D1}cos(\phi)=.5901 sin(0)=0\qquad z_{D1}=r_{D1} cot(\theta)=.5901cot(20)=1.6212\ \text{m}</math>
−	\|}
−
−	~~{\| class="wikitable" align="center"~~
−	\| style="background: gray" \| <math>x_{D2}=r_{D2} cos(\phi)=1.3055cos(0)=1.3055\text {m}\qquad y_{D2}=r_{D2} sin(\phi)=1.3055sin(0)=0\qquad z_{D2}=r_{D2} cot(\theta)=1.3055cot(20)=3.5868\ \text{m}</math>
−	\|}
−
−
−	~~{\| class="wikitable" align="center"~~
−	\| style="background: gray" \| <math>x_P=\frac{2.53cos(\phi)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}=\frac{2.53cos(0)}{(cot(20^{\circ})+cos(0)cot(65^{\circ})}=0.7872\qquad y_P=\frac{2.53sin(\phi)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}=\frac{2.53sin(0)}{(cot(20^{\circ})+cos(0)cot(65^{\circ})}=0\qquad z_P=\frac{2.53cot(\theta)}{(cot(\theta)+cos(\phi)cot(65^{\circ})}=\frac{2.53cot(20^{\circ})}{(cot(20^{\circ})+cos(0)cot(65^{\circ})}=2.1623</math>
−	\|}
−
−	<center><math>D2P=\sqrt{(x_{D2}-x_P)^2+(y_{D2}-y_P)^2+(z_{D2}-z_P)^2}=\sqrt{(1.3055-0.7872)^2+(0-0)^2+(3.5868-2.1623)^2}=\sqrt{(.5183)^2+(1.445)^2}=\sqrt{.2686+2.0292}=1.5158611\ \text{m}</math></center>
−
−
−	<center><math>D1P=\sqrt{(x_P-x_{D1})^2+(y_P-y_{D1})^2+(z_P-z_{D1})^2}=\sqrt{(0.7872-.5901)^2+(0-0)^2+(2.1623-1.6212)^2}=\sqrt{(.1971)^2+(.5411)^2}=\sqrt{.0388+.2927}=.575879865944\ \text{m}</math></center>

	=In the plane=		=In the plane=

← Older revision		Revision as of 03:02, 4 March 2017
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−
−	~~===Determining ellipse components===~~
−	~~[[File:Plane_side_view.png\|thumb\|center\|600px\|alt=Cone of constant Theta for varying Phi\|'''Figure 5:''' A cone of constant Theta with varying Phi.]]~~
−
−	~~Here we must take note of the fact that the center of the cone and the center of the ellipse do not coincide as can be shown visually by extending the cone length.~~
−
−	~~[[File:Ellipse_Cone_overlay_full.png\|thumb\|center\|600px\|alt=Cone of constant Theta for varying Phi\|'''Figure 5:''' A cone of constant Theta with varying Phi.]]~~
−
−
−	Here we will make use of Dandelin spheres to help determine the ellipse components with reference to the cone created at a constant theta. The basic construction of using Dandelin spheres is that within the cone, there are two spheres which are tangent to the plane as well as tangent to the cone as shown below in the xz plane. The Dandelin method picks an arbitrary point,P, on the intersection of the cone and the ellipse's plane. Drawing a line from the cone vertex through the point P, the directrix lines are intersected as shown below at D1 and D2. From the geometry as shown in the xz plane, it is shown that the line segments F1P=D1P and F2P=D2P since they share a common line segment, all three angles, and both have a the respective Dandelin radius as parts of the triangles they form.
−
−
−	~~[[File:Dandelin_spheres2.png\|thumb\|center\|600px\|alt=Cone of constant Theta for varying Phi\|'''Figure 5:''' A cone of constant Theta with varying Phi.]]~~
−
−
−
−	~~We can use the law of sines to find a portion of the ellipse~~
−
−	~~[[File:Law_of_sines_conic_section.png\|thumb\|center\|600px\|alt=Cone of constant Theta for varying Phi\|'''Figure 5:''' A cone of constant Theta with varying Phi.]]~~
−
−	~~<center><math>\frac{a}{sin(A)}=\frac{b}{sin(B)}</math></center>~~
−
−
−	~~<center><math>\frac{2.53\ m}{sin(115^{\circ}-\theta)}=\frac{a_1}{sin(\theta)}</math></center>~~
−
−
−	~~[[File:Law_of_sines_conic_section2.png\|thumb\|center\|600px\|alt=Cone of constant Theta for varying Phi\|'''Figure 5:''' A cone of constant Theta with varying Phi.]]~~
−
−
−	~~<center><math>\frac{a}{sin(A)}=\frac{b}{sin(B)}</math></center>~~
−
−
−	~~<center><math>\frac{2.53\ m}{sin(65^{\circ}-\theta)}=\frac{a_2}{sin(\theta)}</math></center>~~
−
−
−	~~Since these lines meet at the intersection of each other, and they both originate at the vertex positions for the ellipse, we can state that the semi major axis is:~~
−
−	~~<center><math>2a\equiv a_1+a_2</math></center>~~
−
−	~~[[File:Dandelin_spheres.png\|thumb\|center\|600px\|alt=Cone of constant Theta for varying Phi\|'''Figure 5:''' A cone of constant Theta with varying Phi.]]~~
−
−
−	~~Using the fact that half the semi-major axis will give use the center of the ellipse~~
−
−	~~<center><math>a=\frac{a_1+a_2}{2}</math></center>~~
−
−
−	~~<center><math>a=\frac{2.53sin(\theta)}{2}\left (\frac{sin(115^{\circ}-\theta)+sin(65^{\circ}-\theta)}{sin(115^{\circ}-\theta)sin(65^{\circ}-\theta)}\right )</math></center>~~
−
−
−
−	~~<center><math>a=\frac{2.53sin(\theta)}{2}\left (sin(115^{\circ}-\theta)+sin(65^{\circ}-\theta)\right ) csc(115^{\circ}-\theta)csc(65^{\circ}-\theta)</math></center>~~
−
−
−
−	~~{\| class="wikitable" align="center"~~
−	~~\| style="background: gray" \| <math>a=\frac{2.53sin(\theta)}{2}\left (csc(65^{\circ}-\theta)+csc(115^{\circ}-\theta) \right )</math>~~
−	\|}
−
−	~~[[File:Directrix_delta_z.png\|thumb\|center\|600px\|alt=Cone of constant Theta for varying Phi\|'''Figure 5:''' A cone of constant Theta with varying Phi.]]~~
−
−
−	~~Earlier, using the law of sines it was found:~~
−
−	~~<center><math>\frac{2.53\ m}{sin(115^{\circ}-\theta)}=\frac{a_1}{sin(\theta)}</math></center>~~
−
−
−	~~<center><math>\frac{2.53\ m \cdot sin(\theta)}{sin(115^{\circ}-\theta)}=a_1</math></center>~~
−
−
−	~~Taking the difference~~
−
−	~~<center><math>\Delta a=a-a_1=\frac{a_1+a_2}{2}-a_1=\frac{a_2-a_1}{2}</math></center>~~
−
−
−
−	~~{\| class="wikitable" align="center"~~
−	~~\| style="background: gray" \| <math>\Delta a=\frac{2.53sin(\theta)}{2}\left (csc(65^{\circ}-\theta)-csc(115^{\circ}-\theta) \right )</math>~~
−	\|}
−
−
−	~~<center><math>\Delta z=\Delta a \cdot cos(65^{\circ})</math></center>~~
−
−
−	~~<center><math>\Delta x=\Delta a \cdot sin(65^{\circ})</math></center>~~
−
−	~~Since the vertex position is at (0,0,0) this shows the center of the ellipse to be at~~
−
−
−
−	~~<center><math>c_{ellipse} \equiv (-\Delta a sin(65^{\circ}), 0,z+\Delta z)</math></center>~~
−
−
−	~~<center><math>c_{ellipse} \equiv (-\Delta a sin(65^{\circ}), 0,2.53+\Delta z)</math></center>~~
−
−	~~The sum of the distance from both focal points to a point on the surface of the ellipse is a constant~~
−
−	~~[[File:Ellipse_sum_rule1.png\|thumb\|center\|600px\|alt=Cone of constant Theta for varying Phi\|'''Figure 5:''' A cone of constant Theta with varying Phi.]]~~
−
−	~~<center><math>PF1+PF2\equiv 2a</math></center>~~
−
−
−	~~A property of an ellipse is that the distance from the center of the ellipse to a focal point is given by~~
−
−	~~<center><math>f=ae=\left (\frac{a_1+a_2}{2}\right ) \left ( \frac{sin(25^{\circ})}{cos(\theta)}\right )</math></center>~~
−
−	~~[[File:Ellipse_sum_rule2.png\|thumb\|center\|600px\|alt=Cone of constant Theta for varying Phi\|'''Figure 5:''' A cone of constant Theta with varying Phi.]]~~
−
−
−	~~<center><math>2ae+2b=2a\Rightarrow b=a(1-e)</math></center>~~
−
−
−	~~<center><math>b=\left (\frac{a_1+a_2}{2}\right ) \left (1-\frac{sin(25^{\circ})}{cos(\theta)}\right )</math></center>~~
−
−
−	<center><math>b=\frac{2.53sin(\theta)}{2}\left (\frac{sin(115^{\circ}-\theta)+sin(65^{\circ}-\theta)}{sin(115^{\circ}-\theta)sin(65^{\circ}-\theta)}\right ) \left (1-\frac{sin(25^{\circ})}{cos(\theta)}\right )</math></center>
−
−
−	~~Additionally, using the fact that the distance from the center to a focal point is ae~~
−
−	~~[[File:Dandelin_spheres_zoom.png\|thumb\|center\|600px\|alt=Cone of constant Theta for varying Phi\|'''Figure 5:''' A cone of constant Theta with varying Phi.]]~~
−
−
−	~~<center><math>R_{Lower\ Dandelin}=(ae-\Delta a) tan(65^{\circ})</math></center>~~
−
−
−	~~<center><math>r_{D1}=R_{Lower\ Dandelin}sec(\theta)=(ae-\Delta a) tan(65^{\circ})cos(\theta)</math></center>~~
−
−
−
−
−
−	~~By symmetry,~~
−	~~<center><math>R_{Upper\ Dandelin}=(ae+\Delta a) tan(65^{\circ})</math></center>~~
−
−
−	~~<center><math>r_{D2}=R_{Lower\ Dandelin}sec(\theta)=(ae+\Delta a) tan(65^{\circ})cos(\theta)</math></center>~~

	===Determining Elliptical Equation===		===Determining Elliptical Equation===

← Older revision		Revision as of 03:00, 4 March 2017
Line 1:		Line 1:


−	~~==Elliptic Conic Section==~~
−
−	~~If the conic is an ellipse, 0<e<1. This implies~~
−
−	~~<center><math>e=\frac{\sin \beta}{\sin \alpha}=\frac{\sin\ (25^{\circ})}{\sin (90^{\circ}-\theta)}</math></center>~~
−
−
−
−
−	~~<center><math>\frac{sin (25^{\circ})}{cos (\theta)}=e</math></center>~~
−
−
−	~~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~~
−
−	~~<center><math> 0<\theta<65^{\circ}</math></center>~~

	===Determining ellipse components===		===Determining ellipse components===