appendix a solutions to selected problems - boston...

Click here to load reader

Post on 07-Sep-2018

233 views

Category:

Documents

0 download

Embed Size (px)

TRANSCRIPT

  • 1

    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

    !""##!#

    !""##!#$

    !"%"#!%

    !"%"#!%=!