Shadowing Effect

電機四 許泓崴 B02901023

S

Assumptions

  • We already have a target region
     
  • Consider that the absorber can completely absord light at any angle

S

S

Target region

Revised region

v1

v2

O

S

B

A

O

S

v1

v2

O

S

B

A

S

A

B

R

P

h

H

PA :AB=1:r
PA:AB=1:rPA :AB=1:r

C

AC = \frac{rPA}{2+r}
AC=rPA2+rAC = \frac{rPA}{2+r}
h = \frac{rH}{2+r}
h=rH2+rh = \frac{rH}{2+r}

If

Then

v1

v2

O

S

B

A

R

\vec{SO} = a\vec{v1} + b\vec{v2}
SO⃗=av1⃗+bv2⃗\vec{SO} = a\vec{v1} + b\vec{v2}
\vec{OA} =\alpha \vec{v1}
OA⃗=αv1⃗\vec{OA} =\alpha \vec{v1}
\vec{OB} = \beta \vec{v2}
OB⃗=βv2⃗\vec{OB} = \beta \vec{v2}
R = O + \frac{r}{2+r}\left( -a\vec{v1} + b\vec{v2} + \vec{PS} \right)
R=O+r2+r(−av1⃗+bv2⃗+PS⃗)R = O + \frac{r}{2+r}\left( -a\vec{v1} + b\vec{v2} + \vec{PS} \right)
r = -\frac{\alpha}{a+\alpha}
r=−αa+αr = -\frac{\alpha}{a+\alpha}

Input : 

\alpha
α\alpha

Output : 

R
RR

R is in a line!

v1

v2

O

S

B

A

R

H=PS
H=PSH=PS
\frac{r}{2+r}H=h
r2+rH=h\frac{r}{2+r}H=h
R = O + \frac{h}{H}\left( -a\vec{v1} + b\vec{v2} + \vec{PS} \right)
R=O+hH(−av1⃗+bv2⃗+PS⃗)R = O + \frac{h}{H}\left( -a\vec{v1} + b\vec{v2} + \vec{PS} \right)

Input : 

h
hh

Output : 

R
RR

Type 1

v1

v2

O

S

B

A

R

H=PS
H=PSH=PS
\frac{r}{2+r}H=h
r2+rH=h\frac{r}{2+r}H=h
R = O + \frac{h}{H}\left( -a\vec{v1} + b\vec{v2} + \vec{PS} \right)
R=O+hH(−av1⃗+bv2⃗+PS⃗)R = O + \frac{h}{H}\left( -a\vec{v1} + b\vec{v2} + \vec{PS} \right)

Input : 

h
hh

Output : 

R
RR

Type 1

S

v

O

S

\vec{OA} = \alpha \vec{v}
OA⃗=αv⃗\vec{OA} = \alpha \vec{v}
R = A + \frac{h}{H}\vec{PA}
R=A+hHPA⃗R = A + \frac{h}{H}\vec{PA}

Input : 

\alpha, h
α,h\alpha, h

Output : 

R
RR

R is in a line!

A

R

=O+\alpha \vec{v}+\frac{h}{H}\left(\vec{PO}+\alpha \vec{v} \right)
=O+αv⃗+hH(PO⃗+αv⃗)=O+\alpha \vec{v}+\frac{h}{H}\left(\vec{PO}+\alpha \vec{v} \right)
=O+\frac{h}{H}\vec{PO}+\left(1+\frac{h}{H}\right)\alpha\vec{v}
=O+hHPO⃗+(1+hH)αv⃗=O+\frac{h}{H}\vec{PO}+\left(1+\frac{h}{H}\right)\alpha\vec{v}

B

=B+\left(1+\frac{h}{H}\right)\alpha\vec{v}
=B+(1+hH)αv⃗=B+\left(1+\frac{h}{H}\right)\alpha\vec{v}

Type 2

v

B

S

\vec{OA} = \alpha \vec{v}
OA⃗=αv⃗\vec{OA} = \alpha \vec{v}
R = P + \frac{H-h}{H}\vec{PA}
R=P+H−hHPA⃗R = P + \frac{H-h}{H}\vec{PA}

Input : 

\alpha, h
α,h\alpha, h

Output : 

R
RR

R is in a line!

R

A

=P+\frac{H-h}{H}\left(\vec{PO}+\alpha \vec{v} \right)
=P+H−hH(PO⃗+αv⃗)=P+\frac{H-h}{H}\left(\vec{PO}+\alpha \vec{v} \right)
=P+\frac{H-h}{H}\vec{PO}+\frac{H-h}{H}\alpha\vec{v}
=P+H−hHPO⃗+H−hHαv⃗=P+\frac{H-h}{H}\vec{PO}+\frac{H-h}{H}\alpha\vec{v}

O

=B+\frac{H-h}{H}\alpha\vec{v}
=B+H−hHαv⃗=B+\frac{H-h}{H}\alpha\vec{v}

Type 3

S

R_{1}, R_{3}
R1,R3R_{1}, R_{3}

Input : 

h
hh

Output : 

R_{1},R_{2},R_{3},R_{4}
R1,R2,R3,R4R_{1},R_{2},R_{3},R_{4}

R4

R3

R2

R1

R_{4}
R4R_{4}
R_{2}
R2R_{2}

Type 1

Type 2

Type 3

Actually, we only need R2 and R4 in rectangle

S

About Type

v3

v1

v2

(v_{1}\cdot v_{3})\times (v_{2} \cdot v_{3}) \leq 0
(v1⋅v3)×(v2⋅v3)≤0(v_{1}\cdot v_{3})\times (v_{2} \cdot v_{3}) \leq 0

Type 2

v_{1}\cdot v_{3} \geq 0 \wedge v_{2} \cdot v_{3} \geq 0
v1⋅v3≥0∧v2⋅v3≥0v_{1}\cdot v_{3} \geq 0 \wedge v_{2} \cdot v_{3} \geq 0
v_{1}\cdot v_{3} \leq 0 \wedge v_{2} \cdot v_{3} \leq 0
v1⋅v3≤0∧v2⋅v3≤0v_{1}\cdot v_{3} \leq 0 \wedge v_{2} \cdot v_{3} \leq 0

Type 3

Type 1

Result

Problem size Execution time(seconds)
10,000 0.000196
1,000,000 0.01539
100,000,000 0.992516
10,000,000,000 95.78

Weakness

  • Hard to produce
     
  • Effect around the corner

S

Weakness

S

S

A

B

P

H

A

B

L

I = I_{0}\left(\tau_{1}\tau_{2} e^{-\frac{4\pi k}{\lambda}L}\right)^{2}
I=I0(τ1τ2e−4πkλL)2I = I_{0}\left(\tau_{1}\tau_{2} e^{-\frac{4\pi k}{\lambda}L}\right)^{2}

L is too small at the corner

Weakness

S

If we use rectangle,
we might loss some area

S

\alpha
α\alpha
\beta
β\beta
\gamma
γ\gamma
\zeta
ζ\zeta

(x, y)

(c, d)

S

\alpha^{\prime}
α′\alpha^{\prime}
\beta^{\prime}
β′\beta^{\prime}
\gamma
γ\gamma
\zeta
ζ\zeta
\delta
δ\delta

(x, y)

(c, d)

No improvement in experiment

S

\delta
δ\delta

(x, y)

(c, d)

\delta^{\prime}
δ′\delta^{\prime}

(a, b)

Future work

  • Consider rectangle only
  • Consider path length in object
  • Consider threshold
V = 0, I\leq I_{l}
V=0,I≤IlV = 0, I\leq I_{l}
=1, I\geq I_{h}
=1,I≥Ih=1, I\geq I_{h}
=\frac{I-I_{l}}{I_{h}-I_{l}}, I_{l} < I< I_{h}
=I−IlIh−Il,Il<I<Ih=\frac{I-I_{l}}{I_{h}-I_{l}}, I_{l} < I< I_{h}
I = I_{0}\left(\tau_{1}\tau_{2} e^{-\frac{4\pi k}{\lambda}L}\right)^{2}
I=I0(τ1τ2e−4πkλL)2I = I_{0}\left(\tau_{1}\tau_{2} e^{-\frac{4\pi k}{\lambda}L}\right)^{2}
I_{l}
IlI_{l}
I_{l}
IlI_{l}
I_{0}
I0I_{0}
1
11
0
00
I
II
V
VV

New model

  • Consider rectangle object only
  • Introduce decay and refraction
  • Put a constraint in h
I_{r} = f(h)\leq I_{l}
Ir=f(h)≤IlI_{r} = f(h)\leq I_{l}
h\geq f^{-1}(I_{l})
h≥f−1(Il)h\geq f^{-1}(I_{l})
h
hh

Uniformly distribute samples to evaluate the difference between refernce

Cost Evaluation

cost=\sum\limits_{\forall s \in S}\lvert V(s) - T(s)\rvert
cost=∑∀s∈S∣V(s)−T(s)∣cost=\sum\limits_{\forall s \in S}\lvert V(s) - T(s)\rvert

(a,b)

(c,d)

Optimal solution

Black:rect

Red:f^{-1}(rect)

Green:f(rect)

Optimal solution must exist between green and red rectangles

z

h

\phi
ϕ\phi

L

L = \frac{n_{r}(h-z)}{cos\phi}
L=nr(h−z)cosϕL = \frac{n_{r}(h-z)}{cos\phi}
n_{r} = \frac{n_{abs}}{n_{air}}
nr=nabsnairn_{r} = \frac{n_{abs}}{n_{air}}
\sqrt{\frac{I}{I_{0}}} = \frac{2n_{abs}}{n_{abs}+n_{air}}\frac{2n_{air}}{n_{abs}+n_{air}}e^{-\frac{4\pi k}{\lambda}L}=\frac{4e^{-\frac{4\pi k}{\lambda}L}}{(1+\frac{1}{n_{r}})(1+n_{r})}
II0=2nabsnabs+nair2nairnabs+naire−4πkλL=4e−4πkλL(1+1nr)(1+nr)\sqrt{\frac{I}{I_{0}}} = \frac{2n_{abs}}{n_{abs}+n_{air}}\frac{2n_{air}}{n_{abs}+n_{air}}e^{-\frac{4\pi k}{\lambda}L}=\frac{4e^{-\frac{4\pi k}{\lambda}L}}{(1+\frac{1}{n_{r}})(1+n_{r})}

z

h

\phi
ϕ\phi

L

L = \frac{d}{\sqrt{1-\left( \frac{\cos{\phi}}{n_r}\right)^{2}}}
L=d1−(cosϕnr)2L = \frac{d}{\sqrt{1-\left( \frac{\cos{\phi}}{n_r}\right)^{2}}}
\sqrt{\frac{I}{I_{0}}} =\frac{4e^{-\frac{4\pi k}{\lambda}L}}{(1+\frac{1}{n_{r}})(1+n_{r})}
II0=4e−4πkλL(1+1nr)(1+nr)\sqrt{\frac{I}{I_{0}}} =\frac{4e^{-\frac{4\pi k}{\lambda}L}}{(1+\frac{1}{n_{r}})(1+n_{r})}

d

d < \frac{\sqrt{n_{r}^{2}-\cos^{2}{\phi}}}{\cos{\phi}}(h-z)
d<nr2−cos2ϕcosϕ(h−z)d < \frac{\sqrt{n_{r}^{2}-\cos^{2}{\phi}}}{\cos{\phi}}(h-z)

z

h

\phi
ϕ\phi

L

L = \frac{n_{r}(h+z)}{cos\phi}
L=nr(h+z)cosϕL = \frac{n_{r}(h+z)}{cos\phi}
\sqrt{\frac{I}{I_{0}}} =\frac{4e^{-\frac{4\pi k}{\lambda}L}}{(1+\frac{1}{n_{r}})(1+n_{r})}
II0=4e−4πkλL(1+1nr)(1+nr)\sqrt{\frac{I}{I_{0}}} =\frac{4e^{-\frac{4\pi k}{\lambda}L}}{(1+\frac{1}{n_{r}})(1+n_{r})}

z

h

\phi
ϕ\phi

L

d

d < \frac{\sqrt{n_{r}^{2}-\cos^{2}{\phi}}}{\cos{\phi}}(z+h)
d<nr2−cos2ϕcosϕ(z+h)d < \frac{\sqrt{n_{r}^{2}-\cos^{2}{\phi}}}{\cos{\phi}}(z+h)
L = \frac{d}{\sqrt{1-\left( \frac{\cos{\phi}}{n_r}\right)^{2}}}
L=d1−(cosϕnr)2L = \frac{d}{\sqrt{1-\left( \frac{\cos{\phi}}{n_r}\right)^{2}}}

z

h

\phi
ϕ\phi

L

d

\sqrt{\frac{I}{I_{0}}} =\frac{4e^{-\frac{4\pi k}{\lambda}L}}{(1+\frac{1}{n_{r}})(1+n_{r})}
II0=4e−4πkλL(1+1nr)(1+nr)\sqrt{\frac{I}{I_{0}}} =\frac{4e^{-\frac{4\pi k}{\lambda}L}}{(1+\frac{1}{n_{r}})(1+n_{r})}

h

\phi
ϕ\phi

L

L = \frac{2h}{\sqrt{1-\left(\frac{sin\phi}{n_{r}} \right)^{2}}}
L=2h1−(sinϕnr)2L = \frac{2h}{\sqrt{1-\left(\frac{sin\phi}{n_{r}} \right)^{2}}}
\sqrt{\frac{I}{I_{0}}} =\frac{4e^{-\frac{4\pi k}{\lambda}L}}{(1+\frac{1}{n_{r}})(1+n_{r})}
II0=4e−4πkλL(1+1nr)(1+nr)\sqrt{\frac{I}{I_{0}}} =\frac{4e^{-\frac{4\pi k}{\lambda}L}}{(1+\frac{1}{n_{r}})(1+n_{r})}

z

h

\phi
ϕ\phi

L

L = \frac{d\frac{\sin{\phi}}{n_{r}}}{\sqrt{1-\left(\frac{sin\phi}{n_{r}} \right)^{2}}}
L=dsinϕnr1−(sinϕnr)2L = \frac{d\frac{\sin{\phi}}{n_{r}}}{\sqrt{1-\left(\frac{sin\phi}{n_{r}} \right)^{2}}}

d

z

h

\phi
ϕ\phi

L

d

\sqrt{\frac{I}{I_{0}}} =\frac{4e^{-\frac{4\pi k}{\lambda}L}}{(1+\frac{1}{n_{r}})(1+n_{r})}
II0=4e−4πkλL(1+1nr)(1+nr)\sqrt{\frac{I}{I_{0}}} =\frac{4e^{-\frac{4\pi k}{\lambda}L}}{(1+\frac{1}{n_{r}})(1+n_{r})}

Problems

Evaluation point

Source

R = \frac{n_{abs}-n_{air}}{n_{abs}+n_{air}}
R=nabs−nairnabs+nairR = \frac{n_{abs}-n_{air}}{n_{abs}+n_{air}}
R = \frac{4n_{abs}n_{air}}{\left(n_{abs}+n_{air}\right)^{2}}\frac{2n_{abs}}{n_{abs}+n_{base}}
R=4nabsnair(nabs+nair)22nabsnabs+nbaseR = \frac{4n_{abs}n_{air}}{\left(n_{abs}+n_{air}\right)^{2}}\frac{2n_{abs}}{n_{abs}+n_{base}}
n_{abs}
nabsn_{abs}
n_{air}
nairn_{air}
n_{base}
nbasen_{base}

It seems that reflected light is the majoring effect

Shifting is a problem

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