Talk:Geometry:WAXS 3D
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Revision as of 15:25, 13 January 2016 by
KevinYager
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We define:
d
′
=
x
2
+
d
2
+
z
2
=
‖
v
1
‖
(
v
2
y
)
=
(
d
cos
θ
g
−
z
sin
θ
g
)
(
v
2
y
)
2
=
(
d
cos
θ
g
−
z
sin
θ
g
)
2
=
d
2
cos
2
θ
g
−
d
z
cos
θ
g
sin
θ
g
+
z
2
sin
2
θ
g
{\displaystyle {\begin{alignedat}{2}d^{\prime }&={\sqrt {x^{2}+d^{2}+z^{2}}}=\|\mathbf {v} _{1}\|\\(v_{2y})&=(d\cos \theta _{g}-z\sin \theta _{g})\\(v_{2y})^{2}&=(d\cos \theta _{g}-z\sin \theta _{g})^{2}\\&=d^{2}\cos ^{2}\theta _{g}-dz\cos \theta _{g}\sin \theta _{g}+z^{2}\sin ^{2}\theta _{g}\end{alignedat}}}
And calculate:
q
2
=
[
(
q
x
)
2
+
(
q
y
)
2
+
(
q
z
)
2
]
(
q
k
)
2
d
′
2
=
[
(
x
cos
ϕ
g
−
sin
ϕ
g
(
d
cos
θ
g
−
z
sin
θ
g
)
)
2
+
(
x
sin
ϕ
g
+
cos
ϕ
g
(
d
cos
θ
g
−
z
sin
θ
g
)
−
d
′
)
2
+
(
d
sin
θ
g
+
z
cos
θ
g
)
2
]
=
[
(
x
cos
ϕ
g
−
sin
ϕ
g
(
v
2
y
)
)
2
+
(
x
sin
ϕ
g
+
cos
ϕ
g
(
v
2
y
)
−
d
′
)
2
+
(
d
sin
θ
g
+
z
cos
θ
g
)
2
]
=
[
x
2
cos
2
ϕ
g
−
x
cos
ϕ
g
sin
ϕ
g
(
v
2
y
)
+
sin
2
ϕ
g
(
v
2
y
)
2
+
x
2
sin
2
ϕ
g
+
x
sin
ϕ
g
cos
ϕ
g
(
v
2
y
)
−
d
′
x
sin
ϕ
g
+
x
sin
ϕ
g
cos
ϕ
g
(
v
2
y
)
+
cos
2
ϕ
g
(
v
2
y
)
2
−
d
′
cos
ϕ
g
(
v
2
y
)
−
d
′
x
sin
ϕ
g
−
d
′
cos
ϕ
g
(
v
2
y
)
+
d
′
2
+
d
2
sin
2
θ
g
+
d
z
sin
θ
g
cos
θ
g
+
z
2
cos
2
θ
g
]
{\displaystyle {\begin{alignedat}{2}q^{2}&=[(q_{x})^{2}+(q_{y})^{2}+(q_{z})^{2}]\\\left({\frac {q}{k}}\right)^{2}d^{\prime 2}&={\begin{alignedat}{2}[&\left(x\cos \phi _{g}-\sin \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})\right)^{2}\\&+\left(x\sin \phi _{g}+\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})-d^{\prime }\right)^{2}\\&+\left(d\sin \theta _{g}+z\cos \theta _{g}\right)^{2}]\end{alignedat}}\\&={\begin{alignedat}{2}[&\left(x\cos \phi _{g}-\sin \phi _{g}(v_{2y})\right)^{2}\\&+\left(x\sin \phi _{g}+\cos \phi _{g}(v_{2y})-d^{\prime }\right)^{2}\\&+\left(d\sin \theta _{g}+z\cos \theta _{g}\right)^{2}]\end{alignedat}}\\&={\begin{alignedat}{2}[&x^{2}\cos ^{2}\phi _{g}-x\cos \phi _{g}\sin \phi _{g}(v_{2y})+\sin ^{2}\phi _{g}(v_{2y})^{2}\\&+x^{2}\sin ^{2}\phi _{g}+x\sin \phi _{g}\cos \phi _{g}(v_{2y})-d^{\prime }x\sin \phi _{g}\\&+x\sin \phi _{g}\cos \phi _{g}(v_{2y})+\cos ^{2}\phi _{g}(v_{2y})^{2}-d^{\prime }\cos \phi _{g}(v_{2y})\\&-d^{\prime }x\sin \phi _{g}-d^{\prime }\cos \phi _{g}(v_{2y})+d^{\prime 2}\\&+d^{2}\sin ^{2}\theta _{g}+dz\sin \theta _{g}\cos \theta _{g}+z^{2}\cos ^{2}\theta _{g}]\end{alignedat}}\\\end{alignedat}}}
Grouping and rearranging:
(
q
k
)
2
d
′
2
=
[
x
2
+
(
v
2
y
)
2
−
2
d
′
x
sin
ϕ
g
+
x
sin
ϕ
g
cos
ϕ
g
(
v
2
y
)
−
d
′
cos
ϕ
g
(
v
2
y
)
−
d
′
cos
ϕ
g
(
v
2
y
)
+
d
′
2
+
d
2
sin
2
θ
g
+
d
z
sin
θ
g
cos
θ
g
+
z
2
cos
2
θ
g
]
=
[
x
2
+
(
d
2
cos
2
θ
g
−
d
z
cos
θ
g
sin
θ
g
+
z
2
sin
2
θ
g
)
−
2
d
′
x
sin
ϕ
g
+
(
x
sin
ϕ
g
−
d
′
−
d
′
)
cos
ϕ
g
(
v
2
y
)
+
d
′
2
+
d
2
sin
2
θ
g
+
d
z
sin
θ
g
cos
θ
g
+
z
2
cos
2
θ
g
]
=
[
d
′
2
+
x
2
+
d
2
+
z
2
−
2
d
′
x
sin
ϕ
g
+
(
x
sin
ϕ
g
−
2
d
′
)
cos
ϕ
g
(
v
2
y
)
]
=
2
d
′
2
−
2
d
′
x
sin
ϕ
g
+
(
x
sin
ϕ
g
−
2
d
′
)
cos
ϕ
g
(
v
2
y
)
=
?
=
?
=
?
=
2
d
′
2
−
2
d
′
x
sin
ϕ
g
+
2
d
′
cos
ϕ
g
(
d
cos
θ
g
−
z
sin
θ
g
)
=
2
d
′
(
d
′
−
x
sin
ϕ
g
+
cos
ϕ
g
(
d
cos
θ
g
−
z
sin
θ
g
)
)
(
q
k
)
2
=
2
(
1
−
x
sin
ϕ
g
+
cos
ϕ
g
(
d
cos
θ
g
−
z
sin
θ
g
)
d
′
)
{\displaystyle {\begin{alignedat}{2}\left({\frac {q}{k}}\right)^{2}d^{\prime 2}&={\begin{alignedat}{2}[&x^{2}+(v_{2y})^{2}\\&-2d^{\prime }x\sin \phi _{g}\\&+x\sin \phi _{g}\cos \phi _{g}(v_{2y})-d^{\prime }\cos \phi _{g}(v_{2y})\\&-d^{\prime }\cos \phi _{g}(v_{2y})+d^{\prime 2}\\&+d^{2}\sin ^{2}\theta _{g}+dz\sin \theta _{g}\cos \theta _{g}+z^{2}\cos ^{2}\theta _{g}]\end{alignedat}}\\&={\begin{alignedat}{2}[&x^{2}+(d^{2}\cos ^{2}\theta _{g}-dz\cos \theta _{g}\sin \theta _{g}+z^{2}\sin ^{2}\theta _{g})\\&-2d^{\prime }x\sin \phi _{g}\\&+(x\sin \phi _{g}-d^{\prime }-d^{\prime })\cos \phi _{g}(v_{2y})\\&+d^{\prime 2}+d^{2}\sin ^{2}\theta _{g}+dz\sin \theta _{g}\cos \theta _{g}+z^{2}\cos ^{2}\theta _{g}]\end{alignedat}}\\&={\begin{alignedat}{2}[&d^{\prime 2}+x^{2}+d^{2}+z^{2}\\&-2d^{\prime }x\sin \phi _{g}\\&+(x\sin \phi _{g}-2d^{\prime })\cos \phi _{g}(v_{2y})]\end{alignedat}}\\&=2d^{\prime 2}-2d^{\prime }x\sin \phi _{g}+(x\sin \phi _{g}-2d^{\prime })\cos \phi _{g}(v_{2y})\\&=?\\&=?\\&=?\\&=2d^{\prime 2}-2d^{\prime }x\sin \phi _{g}+2d^{\prime }\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})\\&=2d^{\prime }\left(d^{\prime }-x\sin \phi _{g}+\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})\right)\\\left({\frac {q}{k}}\right)^{2}&=2\left(1-{\frac {x\sin \phi _{g}+\cos \phi _{g}(d\cos \theta _{g}-z\sin \theta _{g})}{d^{\prime }}}\right)\end{alignedat}}}
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