Appendix B. Solutions to Chapter 2 Problems

Transcription

1 Appendix B Solutions to Chapter Problems Problem Problem 4 Problem 3 5 Problem 4 6 Problem 5 6 Problem 6 7 Problem 7 8 Problem 8 8 Problem 9 Problem Problem 3 Problem et f ( ) be a function of one variable ( θ ϕ) and ( θ ϕ ) are two unit vectors Show that f( ) 4 q(sinθ ) + Here is the scalar product of two vectors and q(x) x is a function of one variable defined as where α ( x) + β( x) and x q(x) f ( )d + [ α( x)f() + β( x)f( )]d x β( x) x arccos x Solution et g ( ) f( ) () +

2 In the following derivations we assume that ϕ that is vector belongs to ZX plane This assumption does not limit the generality of the derivations as one always can substitute φ with ( ϕ ϕ ) without changing the domain of the integration (ie the upper hemisphere) For convenience of derivations let us transform the original system of coordinates to a new one where is aligned with Z-direction (Fig ) This can be accomplished with rotation by angle around the Y-axis as specified with the following linear transform θ where R Â( θ ) R Â ( θ ) and R (x y z) and R (x y z ) are vectors in the original and new coordinate system respectively Note the minus sign for angle θ 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 Z ( ) in the original system into Z ( sin θ cos θ ) in the new system θ Z Figure Coordinate system transform via rotation by angle θ around the Y-axis Y X To determine the integration domain in the new coordinate system let v (cosϕ sin ϕ ) be a unit vector in the new system The v vector falls into the upper hemisphere of the original system if and only if Therefore v z sin θ cos ϕ + sin ϕ + cos θ cos θ cosϕ ()

3 Note that the angular domain of the upper hemisphere in the original coordinate system is θ [ ] and ϕ [ ] This domain is specified differently in the new coordinate system as shown in Fig Namely three sectors with respect to θ can be identified: θ [ / θ ] (shown in Red) θ [ / θ / + θ] (shown in green) and θ [ / + θ ] (shown in blue) The corresponding variations of ϕ can be derived from the algebraic analysis of Eq () The final expression for the angular domain in the new coordinate system is: > < θ [ θ ϕ [ ]; θ [ θ + θ ] ϕ [arccos arccos θ ( + θ ] ϕ ) ]; (3) Z θ Z θ θ X X Figure The upper hemisphere of the original system of coordinates XYZ in a new system of coordinates X Y Z Vector is aligned with axis Z of the new system + θ In a view that in the new coordinate system the direction coincides with Z the argument of function f ( ) in Eq () simplifies ie cos θ Therefore g ( ) f( + ) cosϕ sinθ f () 3

4 θ f () ( d) θ f (sinθ) d(sinθ) + θ + +θ θ f () ( d) arccos sinθ arccos cos θ cos θ f (sinθ) arccos d() θ sinθ sinθ cos θ cos θ + f (-sinθ) arccos d() sin sin (4) θ θ Note that in the last step of derivations we substituted θ with / θ and accounted for the fact that cos θ sin( / θ) cos( / θ) and arccos( ψ) arccos( ψ) Finally let x sin( θ ) sin(θ) and β( x) arccos arccos x α ( x) β( x) x Substituting the new variables in Eq (4) and changing the limits of integration according to the variable we finally have g( ) x f ( )d + x [ α( x)f() + β( x)f( )]d 4q(x) Problem etting f () show that + Solution According to the results of Problem z f ( ) 4 d + [ α( x) + β( x)] d x + + Taking into account that α ( λ x) + β ( λ x) (cf Problem ) we have 4

5 + z d d + d d x Problem 3 Show that the leaf albedo for the bi-ambertian model is d d ( ) ρ d + τ d 4 Solution The bi-ambertian model for diffuse leaf scattering phase function is (cf Chapter Eq (9)) ρ ) τ d ( ( )( )( ) < d d ( () ) > Note that the angles and are fixed in the integral of interests and integration over whole sphere can be spitted into two parts namely 4 d ( ) d ( ) + d ( )( ) > ( )( ) < ( ) () Taking into account Eq () we have ( ) τ d d ( )( ) > ( )( ) > ( ) ρ d d ( )( ) < ( )( ) < (3a) (3b) In view that the angles and are fixed the integrals on the right hand side of Eqs (3a) and (3b) are identical and equal to the integral over hemisphere ( + or ) The last integral was evaluated in the Problem namely Combining Eqs ()-(4) we finally have + (4) 5

6 4 d ( ) τ d + ρ d τ d + ρ d Problem 4 Prove Equation (4) Solution Equation (4) in Chapter states: + G(r ) () where (cf Chapter Eq ()) G(r ) + + g ( ) g ( ) Combining Eq ()-() and result of Problem we have (a) (b) G(r ) g g ( ) ( + + ) Problem 5 Show that the geometry factor G(r ) for spherically distributed leaf normals depends neither r nor and is equal to 5 Solution Recall (cf Chapter Eq ()) G(r ) + g ( ) Recall also that for spherically distributed leaf normals g ( ) (cf Chapter Eq (5f)) Combining this result with the result of the Problem we have 6

7 G(r ) + Problem 6 Show that G µ for horizontal leaves and G ( / ) for vertical leaves Solution Recall (cf Chapter Eq ()) G(r ) + g ( ) Further let (cosϕ sin ϕ ) and ( cosϕ sin ϕ ) Therefore ( cosϕ ) ( cosϕ) + ( sin ϕ ) ( sin ϕ) + ( )( ) cos( ϕ ϕ) + For horizontal leaves the derivations for G(r ) are as follows The probability density of leaf normal orientation (cf Chapter Eq (4)) is Therefore g ( δ( θ ) ) G (r ) / / dθ g ( ) dθ cos θ µ δ( θ ) cos( ϕ ϕ) + Similar derivations can be performed for vertical leaves In this case the probability density of leaf normal orientation is Therefore g ( δ( θ / ) ) 7

8 G (r ) / dθ cos( ϕ δ( θ / ) ϕ) + cos( ϕ ϕ) + Problem 7 et the polar angle θ and azimuth 3) Show that ϕ of leaf normals be independent (see Eq G(r µ ) dµ g (r µ ) ψ( µ µ ) where µ cos θ µ cos θ and ψ(µ µ ) h (φ ) dφ Solution Since the polar angle θ and azimuth angle φ of leaf normals are independant the following representation of the probability density of leaf normal orientation is valid (cf Chapter Eq (3)): g ( ) g where g (µ ) and h (φ )/ are the probability density functions of leaf normal inclination and azimuth respectively Therefore + (µ G (r µ ) d g (r ) )h (φ dµ g (r µ )h ( ϕ ) dµ g (r µ ) h ( ϕ ) d g (r ) ( µ µ ψ µ µ ) ) Problem 8 Show that in canopies where leaf normals are distributed uniformly along the azimuthal coordinate [ie h ( ϕ ) ] ψ µ µ ) can be reduced to ( 8

9 ψ(µµ µ µ if µ µ sinθ sinθ ) µµ (ϕ / ) + (/ ) µ t µ sin ϕ t otherwise where the branch angle ϕ is arccos( cot θ cot θ ) t Solution Recall (cf Problem 6) Therefore ψ( µ µ ) ϕ ϕ d h ( ) cos( ϕ cos( ϕ cos( ϕ ϕ) + ϕ) + ϕ) + ( cosϕ + ctgθ ctgθ) () Note that in the above derivations we took into account that h ( ϕ ) and also symmetry of function ( a cos( ξ) + b) in the interval ξ [ ] In the derivations to follow we need to consider two cases First consider the case when ctgθ ctgθ In this case the expression (cosϕ + ctg θctg θ) in the Eq () does not change the sign over the whole interval of integration [ ] Therefore ψ(µ µ ) ( cosϕ + ctgθ ctgθ) ( cosϕ + ctgθ ctgθ) cos θ cos θ (a) Now consider the second case when ctgθ ctgθ < Here the value of the expression (cosϕ + ctg θctg θ) in the Eq () is monotonically decreasing in the interval ϕ [ ] The value of ϕ ϕ where the integrand changes the sign is given by 9

10 cosϕ + ctgθ ctgθ ϕ arccos[ ctgθ ctg θ ] Now we can evaluate Eq () by splitting the interval of integration into two parts where the integrand has the opposite signs ψ(µ µ ) ϕ ( cosϕ + ctgθ ctgθ) ( cosϕ + ) ( cosϕ + ctgθ) ϕ ( ϕ ) sin sin θ ϕ + cos θ cos θ (b) Combined the Eq (a)-(b) present the solution to the problem Problem 9 Show that in canopies with constant leaf normal inclination but uniform orientation along the azimuth [cf Eq (4)] G( µ ) ψ( µ µ ) Solution For canopies with independent polar angle θ and azimuth Problem 7) we have ϕ of leaf normals (cf G(r µ ) dµ g (r µ ) ψ( µ ) () µ where µ cos θ µ cos θ and ψ(µµ ) h (φ ) dφ () The condition of constant leaf normal inclination and uniform orientation along azimuth implies Combining Eqs ()-(3) we have δ( θ θ ) g (r ) ( µ δ µ µ ) h ( ϕ ) (3) G(r µ ) dµ g (r µ ) ψ( µ µ )

11 d ( µ δ µ µ ) ψ( µ µ ψ( µ µ ) ) Problem Using Eq (6) derive the ψ(µ µ ) in the case of heliotropic orientations Solution Recall (cf Problem 7) ψ(µµ ) h (φ ) dφ () In the case when the leaf azimuths have a preferred orientation with respect to the solar azimuth orientation (heliotropism) the h function may be modeled as follows (Chapter Eq (6)) Recall also (cf Problem 6) Combining Eq ()-(3) we have ψ(µ µ ) h (φ φ) cos (φ φ η) () cos( ϕ ϕ) + (3) h (φ ) dφ cos ( ϕ - ϕ - η) sinθsinθcos( ϕ - ϕ ) + dαcos ( α - η) sinθ sinθcosα + d cos α ( α - η) sinθsinθ (cosα + ctgθctgθ) (4) where α ϕ ϕ To evalute the above integral we need to consider two cases (cf Problem 8) First consider the case when ctgθ ctgθ In this case the integrand does not change the sign over whole interval of integration Therefore ψ(µ µ α ) d cos ( α - η) sinθsinθcosα + )

12 [ sinθ sinθcosα + θ] dαcos ( α - η) cos sinθ sin d cos ( - )cos cos cos d cos θ α α η α + θ θ α ( α - η) (5) The evaluation of two definite integrals in Eq (5) requires derivations of corresponding indefinite integrals I and I : I + cos(α - η) sinθ sinθdαcos ( α - η)cosα sinθsinθdα cosα sinθsinθ sin( α - η) sin(3α - η) sin α (6a) I + cos(α - η) dαcos ( α - η) dα sin(α - η) α + (6b) Therefore the corresponding definite integrals over interval α [ ] are Combining Eqs (5)-(7) we have α α I α θ α and I cos (7) ψ(µ µ ) cos θ µ (8) µ Now consider the second case when ctgθ ctgθ < In this case the expression under sign of integral in Eq (4) ( cos α + ctg θctg θ) will change sign two times (first at α and second at α ) over the interval α [ ] The value of α is given by ( cos + ctgθ ctgθ) α α arccos[ ctg θ ctg θ ] (9) In order to evaluate integral in Eq (4) we need to consider three intervals where integrand has constant sign: α [ α ] α [ α α ] and α [ α ] Therefore ψ(µ µ α θ θ α η α + θ θ ) d sin sin cos ( - )cos cos cos )cos ( α - η) α α [ I + I ] [ I + I ] + [ I + I ] α α

13 [ I + I ] [ I + ] α I () α α α I I + I I () α α where α I I sin sin sin sin cos( ) sin(3 θ θ α + α η + α )cos(η) 3 α α I I ( α + sin α cos(η) ) α (a) (b) Combining Eqs ()-() we have sin θ ψ(µ µ ) sin α + sin α cos(η) + sin(3α )cos(η) 3 cos θ + ( α + sin α cos(η) ) () where α is given by Eq (9) Overall Eqs (8) and () present the complete solution to the problem Finally note that in the special case of dia-heliotropic distribution ( η ) the Eq () reduces to sin θ ψ(µ µ ) 3sin sin(3 ) α + α + 3 and in the case of para-heliotropic distribution ( η / ) to ( α + sin α ) sin θ ψ(µ µ ) sin sin(3 ) α α + 3 ( α sin α ) Problem Given the direction of incoming and reflected fluxes at the leaf surface ( ) show that the direction of leaf normals n ( µ ϕ ) in the case of specular reflection is given by tan ϕ µ µ µ µ µ ( ) sin ϕ cosϕ µ µ sin ϕ cosϕ 3

14 Solution Consider the geometry of interaction of solar beams with a leaf surface in the case of specular reflection as shown in Fig () Given the incoming and reflected directions the angle between them is given by cos α The leaf normal in the case of specular reflection is n () The norm can be calculated as follows (cf Fig (b)): sin α cosα ( ) () n (a) (b) α Figure The geometry of interaction of radiation with leaf Panel (a) shows direction of leaf surface leaf normal ( n ) incoming ( ) and reflected ( ) beams Panel (b) is a schematic plot to evaluate the norm of et (sin ϕ cos ϕ cos θ) (3a) ( sin ϕ cosϕ ) (3b) n ( sin ϕ cosϕ ) (3c) Combining Eqs ()-(3) we will get formulas for the thee components of vector n namely sin ϕ sin ϕ sin ϕ ( ) (4a) cos ϕ cos ϕ cos ϕ ( ) (4b) cos cos θ θ ( ) (4c) 4

15 The Eq (4c) directly evaluates (4a) by (4b) namely cos θ µ The expression for ϕ can be derived dividing Eq tan ϕ sin ϕ sin ϕ cosϕ cosϕ µ µ sin ϕ cosϕ µ µ sin ϕ cosϕ 5