# appendix a solutions to selected problems - boston...

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Appendix A Solutions to Selected Problems Solutions to Chapter 3 Problems .................................................................................................1 Solutions to Chapter 6 Problems .................................................Error! Bookmark not defined. Solutions to Chapter 3 Problems Problem 1. Let )(f ! , 11 !"!# , be a function of one variable, ),( !"#$ and ),( LLL !"#$ are two unit vectors. Show that )q(sin4!d)f( L

2!

L =!!!"+

.

Here

L!! is the scalar product of two vectors, and q(x), 1x0 !! , is a function of one

variable defined as

!! "+#$+#=x

0

1

x

)]dx)f((,x)f(),([21d)(f

21q(x) ,

where 1)x,()x,( =!"+!# , and

2

2

1x

x1arccos!

1x)(,!

!= .

Solution. Let !

+

"""#"2!

LL d)f()(g . (1)

In the following derivations we assume that 0

L=! , that is, vector

L! belongs to ZX plane.

This assumption does not limit the generality of the derivations as one always can substitute with (

L!"! ) without changing the domain of the integration (i.e., the upper hemisphere). For

convenience of derivations let us transform the original system of coordinates to a new one, where

L! is aligned with Z-direction (Fig. 1). This can be accomplished with rotation by angle

L!" around the Y-axis as specified with the following linear transform,

2

,R)(AR L!"#$ where

!!!

"

#

$$$

%

&

''

'('

)'(

LL

LL

L

cos0sin

010

sin0cos

)(A

and )z,y,x(R = and )z,y,x(R !!!=! are vectors in the original and new coordinate system, respectively. Note the minus sign for angle

L! . This is due to the fact that positive angles are

counted outward from Z-direction. One example of transformation is for the Z-axis, which is transformed from )1,0,0(Z = in the original system into )cos,0,sin(Z LL !!"=# in the new system. To determine the integration domain in the new coordinate system, let

)cos,sinsin,cos(sinv !"!"!= be a unit vector in the new system. The v vector falls into the upper hemisphere of the original system if and only if .0coscossinsin0cossinsinzv

LL!"#"+$"#+$"#"%&'

Therefore,

.sinsin

coscoscos

L

L

!!

!!"# (2)

Note that the angular domain of the upper hemisphere in the original coordinate system is

],0[ !"# and ]2,0[ !"# . This domain is specified differently in the new coordinate system as shown in Fig. 2. Namely, three sectors with respect to ! can be identified: ]2/,0[ L!"#$! (shown in Red), ]2/,2/[ LL !+"!#"$! (shown in green), and ],2/[ L !"+!#" (shown in blue). The corresponding variations of ! can be derived from the algebraic analysis of Eq. (2). The final expression for the angular domain in the new coordinate system is:

Figure. 1 Coordinate system transform via rotation by angle

L!" around the Y-axis.

Z

Y

X

L!

L!"

3

!!!!!!

"

!!!!!!

#

$

%&'

()+(&)*))

))

.

],,2(,1

sinsin

coscos

];sinsin

coscosarccos2,

sinsin

coscos[arccos

],2,

2[,1

sinsin

coscos1

];2,0[

),2,0[,1

sinsin

coscos

LL

L

L

L

L

L

LLL

L

LL

L

(3)

In a view that in the new coordinate system the direction

L! coincides with Z! , the argument of

function )(f L !! in Eq. (1) simplifies, i.e., !="" cosL . Therefore, !!

""

""#$

%"=%%%&%+

sinsin

coscoscos

2!

LL

L

L

d)osc(fd)f()(g

! !!!"+#

"$#

%&

'()

*""""

$#

%&

'()

*""""

#"$#

+"$"++"$"=

L

L

L

L

L

L

L 2

2

sinsin

coscosarccos2

sinsin

coscosarccos

2

0

2

0

d)dcos()osc(fd)dcos()osc(f

)d(sin)ins(f22

L

!

"

#

##"=

!"

"#$

%&'

()*

+,-

.""""

/0"+L

0 L

L )(sindsinsin

coscosarccos22)ins(f

Figure. 2 The upper hemisphere of the original system of coordinates XYZ in a new system of coordinates ZYX !!! . Vector

L! is

aligned with axis Z! of the new system.

L2

!+"

X!

L!

!

L2

!"#

L!

L2

!"# 0

2

!

Z!

X Z

4

.)(sindsinsin

coscosarccos2)ins(-f

L

0 L

L!"

"#$

%&'

()*

+,-

.""""

"+ (4)

Note that in the last step of derivations we substituted ! with !"# 2/ , and accounted for the fact that )2/sin(cos !"#=! , )2/cos(sin !"#=! , and )arccos()arccos( !"#=!" . Finally, let

)sin(x L!= , )sin(!=" , and

,1x

x1arccos!

1cossin

sincosarccos!

1x)(,2

2

L

L

!

!="

#

$%&

'((

((=

).x,(1)x,( !"#=!$ Substituting the new variables in Eq. (4) and changing the limits of integration according to the variable , we finally have

.q(x)4)]dx)f((,x)f(),([2d)(f2)g(x

0

1

x

L !! "=#+$%"+$"=&

Problem 2. Letting !=!)(f , show that .d

2

L!="""#

+!

Solution. According to the results of Problem 1

.d)]x,()x,([d214d)(fd

1

x

z

022

L

!!"

#

$$%

&'''(+')+''*=+',+++ - ---

++ **

Taking into account that 1)x,()x,( =!"+!# (cf. Problem 1), we have

!! !! "=##"=$$%

&

''(

)##+##"=***

+"

1

0

1

x

z

02

L.d2dd2d

Problem 3. Show that the leaf albedo for the bi-Lambertian model is dL,dL,LdL,

4!

),(d +=!!"!#!$ .

5

Solution. The bi-Lambertian model for diffuse leaf scattering phase function is (cf. Chapter 3, Eq. (9))

!"

!#

$

>%%&%%%%

Problem 5. Prove Equation (14). Solution. Equation (14) in Chapter 3 states: ,

21),r(Gd

21

2

=!!" #

+"

(1)

6

where (cf. Chapter 3, Eq. (2)) ,)(gd

21),r(G L2

LLL !!!!"=! #

+"

(2a)

.1)(gd2!1

LL

2!

L =!!"+

(2b)

Combining Eq. (1)-(2) and result of Problem 2, we have !

+"

##"2

),r(Gd21

! !+ +" "

####"

#"

=

2

L

2

LLL )(gd21d

21

! !+ +" "

#####"

=

2 2

LLLL2d)(gd

41

.2

12

4

12

=!"!!

=

Problem 6. Show that the geometry factor ),r(G ! for spherically distributed leaf normals depends neither r nor ! and is equal to 0.5. Solution. Recall (cf. Chapter 3, Eq. (2)), .)(gd

21),r(G L2

LLL !!!!"=! #

+"

Recall also, that for spherically distributed leaf normals 1)(g LL =! (cf. Chapter 3, Eq. (5f)). Combining this result with the result of the Problem 2, we have .

21

21d

21),r(G L2

L =!"!=###

!=# $

+!

Problem 7. Show that =G for horizontal leaves, and !"= sin)/2(G for vertical leaves. Solution. Recall (cf. Chapter 3, Eq. (2)), .)(gd

21),r(G L2

LLL !!!!"=! #

+"

7

Further, let )cos,sinsin,cos(sin !"!"!=# and )cos,sinsin,cos(sin LLLLLL !"!"!=# . Therefore, ( ) ( ) ( ) ( ) ( ) ( )!"!+#!"#!+#!"#!=$$ coscossinsinsinsincossincossin

LLLLLL

.coscos)cos(sinsin LLL !!+"#"$!!= For horizontal leaves the derivations for ),r(G ! are as follows. The probability density of leaf normal orientation (cf. Chapter 3, Eq. (4)) is

.sin

)0()(g

L

LLL !

"!#=$

Therefore,

!!!"##$

=! % %$ $

LLL

2/

0

2

0

LLL )(gddsin21),r(G

! !" "

##+$%$##

%#'#$#

"=

2/

0

2

0

LLLL

LLLL coscos)cos(sinsinsin

)0(sindd

21

.cos !"= Similar derivations can be performed for vertical leaves. In this case the probability density of leaf normal orientation is

.sin

)2/()(g

L

LLL !

"#!$=%

Therefore,

! !" "

##+$%$##

"%#'#$#

"=(

2/

0

2

0

LLLL

LLLL coscos)cos(sinsinsin

)2/(sindd

21),r(G

.sin20)cos(sin1d212

0

LL !"=+#$#%!%#

"= &

"

8

Problem 8. Let the polar angle, L! , and azimuth,

L! , of leaf normals be independent (see Eq.

3). Show that

),,(),r(gd),r(G LL1

0

LL != "

where

LLcos != , != cos , and

! ""#=2!

0

LLLLL d||)(h21)(, .

Solution. Since the polar angle,

L! , and azimuth angle,

L , of leaf normals are independant, the

following representation of the probability density of leaf normal orientation is valid (cf. Chapter 3, Eq. (3)): ),(h)(g)(g LLLLLL =! where L Lg ( ) and L Lh ( )/2! are the probability density functions of leaf normal inclination and azimuth, respectively. Therefore, !!!!

"= #

+"

L

2

LLLL ),r(gd21),r(G

!!""#

= $ $#

LLLL

1

0

L

2

0

LL )(h),r(gdd21

!!""#

= $ $#

LLL

1

0

2

0

LLLL )(hd),r(gd21

.),(),r(gd211

0

LLLL! "#=

Problem 9. Show that in canopies where leaf normals are distributed uniformly along the azimuthal coordinate [i.e., 1)(h LL =! ], ),( L! can be reduced to

!"

!#$

%&&'+&%

(=

otherwise,sin11)(2/1)/!(2

,sinsin if,)(,

,t2L

2tL

LLL

L

where the branch angle

t! is )cotcotarccos( L!"!# .

9

Solution. Recall (cf. Problem 6), .coscos)cos(sinsin LLLL !!+"#"$!!=%% Therefore,

!!""#

$%

L

2

0

LLLL )(hd21),(

!"

##+$%$##$"

=

2

0

LLLL coscos)cos(sinsind21

!"

##+$%$##$"

=

0

LLLL coscos)cos(sinsind1

( )!"

##+$%##$"

=

0

LLLL ctgctgcossinsind1 (1)

Note that in the above derivations we took into account that 1)(h LL =! and also symmetry of function )b)cos(a( +!" in the interval ]2,0[ !"# . In the derivations to follow we need to consider two cases. First, consider the case, when 1ctgctg L !"" . In this case the expression

)ctgctg(cos LL !!+" in the Eq. (1) does not change the sign over the whole interval of integration, ],0[ ! . Therefore,

( )!"

##+$%##$"

&0

LLLLL ctgctgcossinsind1)(,

( )!"

##+$%##$"

=

0

LLLL ctgctgcossinsind1

.coscosL

!!= (2a) Now consider the second case, when 1ctgctg L

)ctgctg(cos LL !!+" in the Eq. (1) is monotonically decreasing in the interval ],0[ !"# . The value of *!=! , where the integrand changes the sign is given by 0ctgctgcos L* =!!+" [ ].ctgctgarccos L* !!"=#$ Now we can evaluate Eq. (1) by splitting the interval of integration into two parts, where the integrand has the opposite signs,

10

( )!"

##+$%##$"

&0

LLLLL ctgctgcossinsind1)(,

( ) ( )!!"

#

#

$$+#$$#%$$+#$$#"

=*

*

ctgcoscossinsindcoscoscossinsind1 LLLL0

LLLL

( ) .2coscossinsinsin21 *L

*L

!"#$$+#$$!

= (2b)

Combined, the Eq. (2a)-(2b) present the solution to the problem. Problem 10. Show that in canopies with constant leaf normal inclination but uniform orientation along the azimuth [cf. Eq. (4)], ),()(G L!= Solution. For canopies with independent polar angle,

L! , and azimuth,

L! , of leaf normals (cf.

Problem 7), we have

),,(),r(gd),r(G LL1

0

LL != " (1)

where

LLcos != , != cos , and

.d||)(h21)(,

2!

0

LLLLL ! ""#= (2)

The condition of constant leaf normal inclination and uniform orientation along azimuth implies

),(sin

)(),r(g *L

L

*L

LL !"=#

#!#"=

.1)(h LL =! (3) Combining Eqs. (1)-(3), we have

),(),r(gd),r(G LL1

0

LL != "

),,()(d L*L1

0

L !"#= $

.),( *!=

11

Problem 11. Using Eq. (6), derive the )(, L in the case of heliotropic orientations. Solution. Recall (cf. Problem 7),

.d||)(h21)(,

2!

0

LLLLL ! ""#$ (1)

In the case when the leaf azimuths have a preferred orientation with respect to the solar azimuth orientation (heliotropism) the

Lh function may be modeled as follows (Chapter 3, Eq. (6)),

.)(cos

!

1),(h2!1

L2

LL !!= (2)

Recall also (cf. Problem 6), .coscos)cos(sinsin LLLL !!+"#"$!!=%% (3) Combining Eq. (1)-(3), we have

! ""#$2!

0

LLLLL d||)(h21)(,

! ""+##""$###%=2!

0

LLLL2

L |coscos)-cos(sinins|)--(cosd1

! ""+#""$##%=2!

0

LL2 |coscoscossinins|)-(cosd1

,|)tgctgccos(sinins|)-(cosd12!

0

LL

2

! ""+#$""%##&= (4)

where

L!"!=# . To evalute the above integral we need to consider two cases (cf. Problem 8).

First, consider the case, when 1ctgctg L !"" . In this case the integrand does not change the sign over whole interval of integration. Therefore,

! ""+#""$##%&2!

0

LL2

L |)soccoscossinins|)-(cosd1)(,

[ ]! ""+#""$##%=2!

0

LL2 coscoscossinins)-(cosd1

.)-(cosdosccoscos)-(cosdsinins12!

0

2L

2!

0

2L !! "##$$+#"##$$%= (5)

12

The evaluation of two definite integrals in Eq. (5) requires derivations of corresponding indefinite integrals,

1I and

2I :

!"!+!##=!"!!##$ %% cos2)2-cos(21

dsininscos)-(cosdsininsI L2

L1

.6

)2-sin(3

2

)2-sin(ins

2

sinins L!"

#$%

& '(+

'(+(

))= (6a)

!!"#+

#$$="##$$%2

)2-cos(21dosccos)-(cosdosccosI L

2L2

.2

)2-sin(2

2

osccos L!"

#$%

& '(+(

))= (6b)

Therefore, the corresponding definite integrals over interval ]2,0[ !"# are 0I 2

01=

!="

=" and .osccosI

L

2

02!!"=

"=#

=# (7)

Combining Eqs. (5)-(7), we have .coscos)(, LLL !""= (8) Now consider the second case, when 1ctgctg L

"#!

"

"+++#+

!=

2

221

2

21021 *

*

*

*

IIIIII1

[ ] [ ]*

*

2

21

2

021II2II

1!"#

!

#+$"+

#= (10)

,I2II2I1*

*

*

*

2

2

2

02

2

1

2

01

!"#

!

#!"#

!

#$"+$"

#= (10)

13

where

,)2cos()3sin(31)2cos(sinsin2sinsinI2I ***L

2

1

2

01

*

*

!"#$

%& '(+'(+()**=)+

(+,

(

, (11a)

( ).)2cos()2sin(2coscosI2I **L2

2

2

02

*

*

!"+#$"%&&=%$"$#

"

# (11b)

Combining Eqs. (10)-(11), we have

!"#$

%& '(+'(+()

*

++= )2cos()3sin(

31)2cos(sinsin2

sinsin)(, ***LL

( ) ,)2cos()2sin(2coscos **L !"+#$"%#

&&+ (12)

where *! is given by Eq. (9). Overall, Eqs. (8) and (12) present the complete solution to the problem. Finally, note that in the special case of dia-heliotropic distribution, ( 0=! ), the Eq. (12) reduces to

( ) ,)2sin(2coscos)3sin(31sin3

sinsin)(, **L**LL !+"#!$"

%%+&'()

*+ !+!$

"

%%=

and in the case of para-heliotropic distribution ( 2/!=" ), to

( ) .)2sin(2coscos)3sin(31sin2

sinsin)(, **L**LL !"#"!$#

%%+&'()

*+ !"!$

#

%%=

Problem 12. Calculate area scattering phase function for diffuse radiation, )(d !"!#$ , in the case of uniform leaf normal distribution, 1gL = . Solution. Recall (Chapter 3, Eq. (16)-(17)), ,)()()( ddL,ddL,d !"!#+!"!#=!"!# +$ (1a) where,

.)()()(gdd2!1)( LLL,LL

2!

0

L

1

0

Ld !!"!!##=!$!"% && (1b)

14

The (+) in the above definition indicates that the L

! integration is over that portion of the interval ]2,0[ ! for which the integrand is either positive (+) or negative (). We need to calculate phase function for the case of uniform leaf normal orientation, 1gL = . In view of the symmetry of the integrand with respect to

L! , the integral over the upper hemisphere is equal to that over

lower hemisphere and,

.)()(d2!1

21)()(d

2!1)( LL

4

LLL

2

Ld!"

!#$

!%

!&'

(()(((=(()(((=(*()+ ,,-+-

(1c)

The integrand depends on three vectors: ! , !" ,

L! . In order to simplify integration, let us

choose the coordinate system where X-Y plane coincides with the plane of vectors !" and ! ( 0==! ). Therefore, ,)0,sin,(cos !!=" ,)0,sin,(cos !"!"=#" (2) .)cos,sinsin,cos(sin LLLLLL !"!"!=# Taking into account Eq. (2) and performing trigonometric transformations, we have .)cos()cos(sin))(( LLL2LL !"#!!#!$=%%%"% (3) Let !"!#$% , an angle between ! and !" , )arccos( !"!=# . Let !"#!$%

L, and

Ldd !=" . Therefore,

)cos()cos(d)1(d41

2

0

1

1

L2

L !"+!!#$=% &&

$

#

[ ])2cos()cos(21d)1(d

41

2

0

L2

1

1

L !+"+!"#$= %%

$

#

[ ] .)ycos()cos(21dy)1(d

41

2

0

L2

1

1

L +!"#= $$

#

"

(4)

Note, the integrand in Eq. (4) is changing sign two times over the interval ]2;0[ ! : !"#= )cos()ycos( .y !"= Namely, the integrand is positive over ];0[ !"# , negative over ];[ !+"!#" and again positive over ]2;[ !"+! . Therefore,

15

!"

!#$

!%

!&'

+(++()( **+

(++

(,+

+

2

0

dy)]ycos()[cos(dy)]ycos()[cos(21)(I

!"#$

+"=0

dy)]ycos()[cos(

,)sin()cos()( !+!!"#= (5a) and

!"+#

"$#

$ +"%" dy)]ycos()[cos(21)(I

!"

#$"

+#= dy)]ycos()[cos(

).sin()cos( !"!!= (5b) Combining Eq. (4) and (5), we have,

.3

)(I)1(d

4

)(I1

1

L2

L !

"=#

!

"=$

#

% (6)

Finally, substituting Eq. (6) into Eq. (1), we have )()()( ddL,ddL,d !"!#+!"!#=!"!#

+$

{ })]sin()cos()[()]sin()cos([31

dL,dL, !+!!"#+!+!!"#=

.)cos(3

)]cos()[sin(

3

dL,d,L!+!!"!

#

$=

Problem 13. Given the direction of incoming, ! , and reflected, !" , fluxes at the leaf surface ( 1=!"=! ) show that the direction of leaf normals ),(n LLL ! in the case of specular reflection is given by

,)1(2

L!!"#

#"=

.cos1cos1

sin1sin1tan

22

22

L

!""!##"

!""!##"=!

Solution. Consider the geometry of interaction of solar beams with a leaf surface in the case of specular reflection as shown in Fig. (1). Given the incoming, , and reflected, ! , directions,

16

the angle between them is given by cos !=" . The leaf normal in the case of specular reflection is

nL !"

!"= . (1)

The norm !" can be calculated as follows (cf. Fig. 1(b)):

.)1(22cos12

2sin2 !"!#=$#=$="# (2)

Let ,)cos,cossin,sin(sin !"!"!=# (3a) ,)cos,cossin,sin(sin !"#"!"#"!"=$" (3b) .)cos,cossin,sin(sinn LLLLLL !"!"!= (3c) Combining Eqs. (1)-(3) we will get formulas for the thee components of vector

Ln , namely,

,)1(2

sinsinsinsinsinsin LL

!"!#

$"%"#$%=$% (4a)

,)1(2

cossincossincossin LL

!"!#

$"%"#$%=$% (4b)

.)1(2

coscoscos L!"!#

$"#$=$ (4c)

The Eq. (4c) directly evaluates

LLcos!= . The expression for

L! can be derived dividing Eq.

(4a) by (4b), namely,

Figure. 1 The geometry of interaction of radiation with leaf. Panel (a) shows direc-tion of leaf surface, leaf nor-mal (

Ln ), incoming (! ) and

reflected (!" ) beams. Panel (b) is a schematic plot to eva-luate the norm of !" .

!"

!

Ln

!

1=!

1=!"

(a) (b)

!"#!

17

.cos1cos1

sin1sin1

cossincossin

sinsinsinsintan

22

22

L

!""##!#

!""##!#$

!"%"#!%

!"%"#!%=!