THE EQUILIBRIUM SHAPE OF SELF-FORMED CHANNELS IN NONCOHESIVE ALLUVIUM Victor M. Ponce and Shurui Jiang San Diego State University [201107] Abstract. The Lane et al. (1959) theory for the equilibrium shape of self-formed channels in noncohesive alluvium has been revisited, with all assumptions and derivations clearly stated. The results are used to estimate self-formed top-width/maximum-depth ratios as a function of: (1) the friction angle of the noncohesive material forming the bed, and (2) the lift-to-drag force ratio acting on the particles. The findings may be used as a point-of-start in the study of unsteady alluvial channel morphology. 1. INTRODUCTION Observations of self-formed channels in noncohesive materials indicate that they tend to have a shape of their own. A mechanistic approach to determine the shape has been the subject of numerous studies, most notable, that of Lane et al. (1959). The objective of this paper is to fully document the Lane et al. findings, which although restricted to the equilibrium case, provide a point-of start in the study of unsteady alluvial channel morphology. 2. LANE ET AL.'S ASSUMPTIONS The size of the cross-section for stream channels in equilibrium, formed in noncohesive materials, is dependent on the maximum threshold discharge. Using principles of mechanics, Lane et al. (1959) showed that the shape of this equilibrium cross section is a function of the following parameters: (1) the friction angle φ of the material forming the channel bed, and (2) the lift-to-drag force ratio β acting on a particle. The main assumptions used by Lane et al. (op. cit.) in deriving their theory are given here: The side slope of the channel, at or above the water surface, is equal to the angle of repose φ of the noncohesive alluvial material (Fig. 1). The particles at the channel boundary are at the condition of incipient motion. The lift and drag forces acting on each particle, together with the downward component of the gravity force, are balanced exactly by the frictional force developed between particles. The lift and drag forces are assumed to be proportional to the tractive force. The particles are held in place by virtue of their submerged weight resolved in a direction normal to the bed. The tractive force acts in the direction of the flow and is equal to the weight of the column of water above the area on which the force acts. The channel is prismatic, with negligible secondary currents and absence of backwater effects. Fig. 1  Definition sketch and force diagram. 3. THE LANE ET AL. METHOD REVISITED Under the assumption of incipient motion on the periphery of the channel, the drag force Fd, and the downward component of the submerged weight of the particle, W sinθ, are balanced by the friction force developed between particles. The angle θ is the local side slope angle and W is the submerged weight of the particle. The friction force is the product of the normal force and the tangent of the friction angle φ. The normal force is W cosθ - Fl , where W cosθ is the normal (to the side slope) component of submerged weight and Fl is the lift force on the particle. The friction angle φ is the angle of repose of the noncohesive material. The balance of forces yields: W 2 sin2 θ + Fd 2  =  [ ( W cos θ  -  Fl ) tan φ ] 2 (1) It is assumed that Fd = δτ, in which τ is the local bed shear stress or tractive force and δ is a proportionality constant. Furthermore, it is assumed that Fl = βFd, where β is the lift-to-drag force ratio. Substituting these relationships into Eq. 1 yields: τ 2 + (W/δ)2 sin2 θ  =  [ (W/δ) cos θ  -  βτ ] 2 tan2 φ (2) At the channel centerline: θ = 0 and τ = τo (Fig. 1). Substituting this condition into Eq. 2 yields: τo2 = [ (W/δ)  -  βτo ] 2 tan2 φ (3) Solving for W/δ :      W          1 + β tan φ     ___  =  _______________ τo      δ                  tan φ (4) The quantity τo, the tractive force at the channel centerline, is equal to: τo  =  γ Yo S (5) in which γ = unit weight of water, Yo = maximum flow depth, and S = channel bed slope. At a distance ε from the channel centerline, the depth of flow is y. The weight of fluid in a column of unit area and depth Y is: γY. The normal component of this fluid weight is: γ Y cosθ. Thus, the tractive force is: τ  =  γ Y S cosθ (6) Substituting Eq. 5 into Eq. 4 yields:      W          1 + β tan φ     ___  =  _______________  γ Yo S      δ                 tan φ (7) Substituting Eq. 7 into Eq. 2 yields:               1 + β tan φ                           τ 2 + [ ______________ γ Yo S ] 2 sin2 θ  =                    tan φ        1 + β tan φ [ (_______________ γ Yo S ) cos θ  -  βτ ] 2 tan2 φ             tan φ (8) Substituting Eq. 6 into Eq. 8 yields:                                1 + β tan φ                           (γ Y S cos θ)2 + [______________ γ Yo S ] 2 sin2 θ  =                                      tan φ        1 + β tan φ [ (______________  γ Yo S ) cos θ - β (γ Y S cos θ) ] 2 tan2 φ             tan φ (9) Taking the left-hand side (LHS) of this equation and carrying out the squared terms yields:                                     1 + β tan φ                           γ 2 Y 2 S 2 cos2 θ + ( _____________ ) 2 γ 2 Yo 2 S 2 sin2 θ                                          tan φ (10) Factoring out γ 2 S 2 gives:                                       1 + β tan φ                           γ 2 S 2 [ Y 2 cos2 θ + ( _____________ ) 2 Yo 2 sin2 θ ]                                             tan φ (11) Working with the right-hand side (RHS) of Eq. 9:       1 + β tan φ [ (______________) 2 γ 2 Yo 2 S 2 cos2 θ            tan φ         1 + β tan φ - 2 (______________) β γ 2 Y Yo S 2 cos2 θ               tan φ + β 2 γ 2 Y 2 S 2 cos2 θ ] tan2 φ (12) Factoring out γ 2 S 2 cos2 θ within the brackets gives:                              1 + β tan φ γ 2 S 2 cos2 θ [ (______________) 2 Yo 2                                   tan φ          1 + β tan φ -  2 (______________) β Y Yo + β 2 Y 2 ] tan2 φ                tan φ (13) Dividing Eq. 11 (LHS) and Eq. 13 (RHS) by γ 2 S 2 yields:                        1 + β tan φ Y 2 cos2 θ + (______________) 2 Yo 2 sin2 θ  =                               tan φ                  1 + β tan φ                        1 + β tan φ cos2 θ [ (______________) 2 Yo 2 - 2 (_____________) β Y Yo                       tan φ                                 tan φ + β 2 Y 2 ] tan2 φ (14) Dividing Eq. 14 by Yo2 cos2 θ yields:     Y             1 + β tan φ (____) 2 + (______________) 2 tan2 θ  =     Yo                  tan φ       1 + β tan φ                   1 + β tan φ       β Y [ (______________) 2   -  2 (______________) _____             tan φ                           tan φ            Yo       β Y + (_____) 2 ] tan2 φ        Yo (15) from which the following equation is obtained:     Y            1 + β tan φ                       (____) 2 + (____________) 2 tan2 θ  =     Yo                tan φ     1 + β tan φ        β Y (____________  -  _____) 2 tan2 φ        tan φ             Yo (16) By definition, tanθ = - dY / dε, and calling r = β tanφ, Eq. 16 can be expressed as follows:     Y             1 + r              dY                1 + r        β Y (____) 2 + (_______) 2  ( - ____ ) 2  =  (_______  -  _____) 2 tan2 φ    Yo            tan φ             dε               tan φ         Yo (17)                                                 dY   Solving for the expression ( - ____ ) 2 yields:                                                 dε                                1 + r         β Y                         Y                           (_______  -  _____) 2 tan2 φ  -  (____) 2       dY                   tan φ        Yo                         Yo ( - ____ ) 2  =  _________________________________________       dε                                      1 + r                                             ( _______ ) 2                                                 tan φ (18) Working with the right-hand side (RHS) only, carrying out the squared term within the brackets yields:      1 + r                  1 + r        β Y                  Y                              Y [ (_______) 2  -  2 (_______) (_____)  +  β 2 (_____) 2 ] tan2 φ  -  (_____) 2      tan φ                tan φ        Yo                  Yo                            Yo ____________________________________________________________________                                                 1 + r                                             ( _______ ) 2                                                 tan φ (19) Simplifying:                       2 β tan φ    Y        β 2 tan2 φ     Y               Y        tan2 φ tan2 φ [ 1  -  __________  ___  +  ___________ (___) 2 ]  -  (___) 2 _________                        ( 1 + r )     Yo          ( 1 + r ) 2   Yo              Yo      ( 1 + r ) 2 (20) Since r = β tan φ, Eq. 20 results in:                          2 r        Y                r 2        Y               Y          tan2 φ tan2 φ [ 1  -  ________ ____  +  __________ (___) 2 ]  -  (___) 2 __________                       ( 1 + r )   Yo         ( 1 + r ) 2   Yo             Yo       ( 1 + r ) 2 (21) Factoring out tan2 φ gives:                          2 r        Y                r 2        Y             Y            1 tan2 φ [ 1  -  ________ ____  +  __________ (___) 2  -  (___) 2 _________ ]                       ( 1 + r )   Yo         ( 1 + r ) 2   Yo           Yo      ( 1 + r ) 2 (22) Simplifying further yields:                           2 r        Y              r 2 - 1        Y tan2 φ [ 1  -  _________ ____  +  ___________ (____) 2 ]                       ( 1 + r )    Yo          ( 1 + r ) 2     Yo (23) Replacing (r 2 - 1) = ( r + 1 )( r - 1 ) yields:                           2 r         Y          ( r +1 )( r - 1 )     Y tan2 φ [ 1  -  _________ ____  +  _______________ (___) 2 ]                       ( 1 + r )    Yo             ( 1 + r ) 2        Yo (24) Simplifying Eq. 24 and substituting into Eq. 18:       dY                                     2 r        Y           ( r - 1 )     Y ( - ____ ) 2  =  tan2 φ [ 1  -  ________ ____  +  _________ (____) 2 ]       dε                                   1 + r      Yo          ( 1 + r )    Yo (25) Changing (r - 1) to (1 - r ) gives:       dY                                     2 r       Y          ( 1 - r )      Y ( - ____ ) 2  =  tan2 φ [ 1  -  ________ ____  -  _________ (____) 2 ]       dε                                   1 + r      Yo         ( 1 + r )    Yo (26) Taking the square root of each side yields:       dY                                 2 r       Y          ( 1 - r )      Y ( - ____ )  =  tan φ [ 1  -  ________ ____  -  _________ (____) 2 ] 1/2       dε                               1 + r      Yo        ( 1 + r )     Yo (27)                                                                    dε   Multiplying each side of the equation by ______ yields:                                                                   dYo        dY            dε                          2 r      Y          ( 1 - r )     Y - ______  =  _____ tan φ [ 1  -  _______ ____  -  _________ (____) 2 ] 1/2     dYo          dYo                        1 + r   Yo         ( 1 + r )    Yo (28) Dividing each side by the term within brackets yields:                            dY / dYo                                           dε ___________________________________________  =  - _____ tan φ                2 r        Y           1 - r       Y                          dYo [ 1  -  (_______) ____  -  (_______) (___) 2 ] 1/2              1 + r      Yo          1 + r      Yo (29) The left-hand side of this equation follows the indefinite integral form:       dx                    1                      2ax + b ∫_______  =  - ___________ sin -1 ___________     X 1/2            ( - a ) 1/2               ( - Δ ) 1/2 (30) in which: Δ = 4ac - b 2 (30 a) X = ax 2 + bx + c (30 b) ( a < 0; Δ < 0 ) (30 c) From Eq. 29, defining:             1 - r a = - (_______)             1 + r (31)               2 r b = - (_______)             1 + r (32) c = 1 (33) Substituting the terms a, b, and c (Eqs. 31 to 33) in Eq. 30 a yields:                                     1 - r                      2 r ( - Δ ) 1/2  =  [ - [ 4 ( - ______ ) (1)  -  ( - ______ ) 2 ] ] 1/2                                     1 + r                    1 + r (34) Carrying out the squared term:                                     1 - r                 4 r 2 ( - Δ ) 1/2  =  [ - [ 4 ( - ______ )  -  ( _________ ) ] ] 1/2                                     1 + r            ( 1 + r ) 2 (35) Simplifying:                              1 - r                 4 r 2 ( - Δ ) 1/2  =  [ 4 ( ______ )  +  ( _________ ) ] 1/2                              1 + r             ( 1 + r ) 2 (36) Factoring out a 2 from under the radicand gives:                            1 - r               r 2 ( - Δ ) 1/2  =  2 [ ______  +  _________ ] 1/2                            1 + r        ( 1 + r ) 2 (37)                           1 + r Multiplying by (_______) inside the radicand yields:                           1 + r                            ( 1 - r 2 )               r 2 ( - Δ ) 1/2  =  2 [ __________  +  _________ ] 1/2                            ( 1 + r ) 2        ( 1 + r ) 2 (38) Collecting common terms yields:                          2 ( - Δ ) 1/2  =  ______                       1 + r (39) Therefore, in Eq. 30:                                                                  1 - r           Y              -2 r                                                     [ 2 ( - ( ______ ) ) (____)  +  ( ______ ) ]       dx                    1                                 1 + r          Yo            1 + r ∫_______  =  - ___________ sin -1 ______________________________________                          1 - r                                               2     X 1/2           (______) 1/2                                  ______                          1 + r                                           1 + r (40) Inverting the square root term in the denominator and simplifying further:       dx              1 + r                                     Y ∫_______  =  - (______) 1/2 sin -1 [ - ( 1 - r ) ____ - r ]     X 1/2             1 - r                                     Yo (41) Integrating the right hand side (RHS) of Eq. 29 gives:      ε - ____ tan φ + C     Yo (42) Now, equating Eqs. 41 and 42 gives:      ε                             1 + r                                     Y - ____ tan φ + C  =  - (______) 1/2 sin -1 [ - ( 1 - r ) ____ - r ]     Yo                            1 - r                                     Yo (43) The constant of integration is evaluated by recognizing that Y/Yo = 1 for ε/Yo = 0. Therefore:              1 + r C  =  - (______) 1/2 sin -1 ( - 1 )              1 - r                    (44) which reduces to:            π      1 + r C  =  (___) (______) 1/2            2      1 - r (45) The substitution of C (Eq. 45) into Eq. 43 yields:      ε                 π     1 + r                  1 + r                                     Y - ____ tan φ + ___ (______) 1/2  =  - (______) 1/2 sin -1 [ - ( 1 - r ) ____ - r ]     Yo                2     1 - r                   1 - r                                     Yo (46)                         1 + r Dividing by - (_______) 1/2 yields:                         1 - r        ε                 π     1 + r   - ____ tan φ + ___ (______) 1/2       Yo                2     1 - r                                              Y _______________________________  =  sin -1 [ - ( 1 - r ) ____ - r ]                       1 + r                                                       Yo                   - (_____) 1/2                       1 - r (47) Taking the sine of each side yields:                  ε                 π     1 + r             - ____ tan φ + ___ (______) 1/2                 Yo                2     1 - r                                  Y - sin [ _______________________________ ]  =  ( 1 - r ) ____  +  r                               1 + r                                              Yo                           - (_____) 1/2                               1 - r (48) Simplifying the left-hand side:              ε     1 - r                          π                        Y - sin [ ____ (______) 1/2  tan φ  -  ___ ]  =  ( 1 - r ) ____  +  r             Yo    1 + r                         2                       Yo (49)                     Y Solving for ____ yields:                    Yo                            ε      1 - r                          π                - sin ( ____ (______) 1/2  tan φ  -  ___ )  -  r   Y                      Yo    1 + r                          2 ____  =  __________________________________________  Yo                                      1 - r (50) From the trigonometric identity:           π sin ( ____ - θ )  =  cos θ           2 (51) where θ in this case is:            ε     1 - r θ  =  ____ (______) 1/2  tan φ           Yo    1 + r (52) it follows that:                            ε      1 - r                 cos ( ____ (______) 1/2  tan φ )  -  r   Y                      Yo    1 + r ____  =  ____________________________________  Yo                                      1 - r (53) For Y = 0, ε = To /2, in which To = channel top width. Substituting these values into Eq. 53:             To       1 - r cos [ ______ (______) 1/2  tan φ ]  -  r  = 0            2Yo     1 + r (54) Solving for r  in Eq. 54:                    To       1 - r r  =  cos [ ______ (______) 1/2  tan φ ]                   2Yo     1 + r (55) from which:                    To       1 - r cos-1 r  =  ______ (______) 1/2  tan φ                   2Yo     1 + r (56) from which:    To                2          1 + r ______  =  _________ (______) 1/2  cos-1 r    Yo             tan φ      1 - r (57) Equation 57 is the relationship derived by Lane et al. (1959) linking the top-width-to-maximum-depth ratio (a parameter describing the shape of the cross section) to the values of φ and r. Since r = β tan φ, Eq. 57 expresses the relationship between To /Yo and the parameters φ and β. For β = 0, i.e., in the absence of lift forces, it follows that r = 0, and Eq. 57 reduces to:    To                π ______  =  _________    Yo             tan φ (58) For φ = 45°, this equation (Eq. 58) reduces to:    To ______  =  π  =  3.14    Yo (59) In the limit, as β ⇒ 1 and φ ⇒ 45°, it can be shown that (See Appendix):    To ______  ⇒  4    Yo (60) 4. SENSITIVITY ANALYSIS The sensitivity of Lane et al.'s formula for the shape of self-formed channels in noncohesive alluvium (Eq. 57) has been studied by varying the friction angle θ and the ratio β of lift-to-drag forces between realistic ranges. The friction angle φ of noncohesive material varies normally in the range 25°-45°. For illustrative purposes, values of φ were taken from 5° to 45°, at intervals of 5°. Values of β are somewhat more difficult to determine with sufficient accuracy. Apperley's (1968) detailed laboratory study concluded that the mean value of β is a function of elevation above the bed. He plotted y/ds vs. β in which y = elevation, and ds = grain diameter, and showed that β reached a maximum of 0.75 for y/ds = 0.25 (Fig. 2), decreasing to 0 for y/ds = 0.90. Above y/ds = 0.90, the mean lift becomes negative, at least up to y/ds = 1.50. Fig. 2  Mean lift-to-drag force ratio β as a function of normalized elevation along the bed y/ds (Apperley, 1968). Given the experimental evidence of Apperley and others, as discussed thoroughly by the former, it was decided to vary the lift-to-drag ratio in the range -1.0 ≤ β ≤ 1.0. Large negative values of β are not considered likely; they are shown here only for illustrative purposes. Table 1 shows calculated values of top-width/maximum-depth (To /Yo) as a function of φ and β, within the indicated parameter ranges. It is clearly seen that for small values of φ, say φ ≤ 10°, the presence of lift does not substantially affect the shape of the cross section. The converse is true for large values of φ, say in the range 25° ≤ φ ≤ 45°. For φ = 25°, there is a 15% increase in To /Yo as β varies from 0 to 1; for φ = 35°, there is a 20% increase; for φ = 45°, there is a 27% increase. Table 1.  Value of To /Yo as a function of friction angle φ and lift-to-drag force ratio β. β Friction angle φ (degrees) 5 10 15 20 25 30 35 40 45 -1.0 34.728 16.591 10.447 7.292 5.320 3.920 2.814 1.809 0.001 -0.9 34.849 16.722 10.589 7.448 5.495 4.124 3.064 2.160 1.235 -0.8 34.970 16.851 10.728 7.598 5.660 4.310 3.284 2.438 1.665 -0.7 35.090 16.978 10.862 7.743 5.817 4.484 3.480 2.672 1.971 -0.6 35.210 17.103 10.994 7.882 5.966 4.645 3.658 2.874 2.214 -0.5 35.328 17.226 11.123 8.017 6.109 4.796 3.822 3.054 2.418 -0.4 35.446 17.348 11.248 8.148 6.245 4.939 3.973 3.216 2.595 -0.3 35.563 17.467 11.371 8.274 6.375 5.074 4.113 3.364 2.752 -0.2 35.679 17.585 11.492 8.397 6.500 5.202 4.245 3.500 2.894 -0.1 35.794 17.702 11.609 8.516 6.621 5.325 4.369 3.626 3.023 0.0 35.909 17.817 11.725 8.631 6.737 5.441 4.487 3.744 3.142 0.1 36.022 17.930 11.838 8.744 6,849 5.553 4.598 3.855 3.252 0.2 36.135 18.042 11.948 8.854 6.958 5.660 4.704 3.959 3.354 0.3 36.248 18.153 12.057 8.961 7.063 5.763 4.805 4.058 3.451 0.4 36.359 18.262 12.164 9.065 7.165 5.863 4.902 4.152 3.542 0.5 36.470 18.370 12.269 9.167 7.264 5.959 4.994 4.241 3.628 0.6 36.580 18.476 12.371 9.266 7.360 6.051 5.083 4.327 3.709 0.7 36.690 18.581 12.473 9.363 7.453 6.141 5.169 4.408 3.787 0.8 36.799 18.685 12.572 9.458 7.544 6.227 5.252 4.487 3.861 0.9 36.907 18.788 12.670 9.551 7.632 6.311 5.331 4.562 3.932 1.0 37.014 18.889 12.766 9.642 7.718 6.393 5.408 4.635 4.000 Based on Apperley's study, the average value of β integrated along a depth equal to the particle size could be taken as approximately β = 0.4. Values of To /Yo rounded to two decimal places, for β = 0.4 and 25° ≤ φ ≤ 45°, are given in Table 2. These values may be taken as typical equilibrium values for the indicated friction angles. Table 2.  Value of To /Yo as a function of friction angle φ for lift-to-drag ratio β = 0.4. φ (degrees) Top-width to maximum-depth ratio To /Yo 25 7.16 30 5.86 35 4.90 40 4.15 45 3.54 5. ONLINE CALCULATION Equation 57 may be solved with the aid of the online calculator ONLINE LANE SHAPE. For instance, for Yo = 1 m, φ = 30°, and β = 0.4, the answer is: To = 5.8627 m, which is in agreement with the results shown in Tables 1 and 2. [Click on top of image to expand]. 6. CONCLUDING REMARKS The Lane et al. (1959) theory for the equilibrium shape of self-formed channels in noncohesive alluvium has been revisited, with all assumptions and derivations clearly stated. The results are used to estimate top-width/maximum-depth cross-sectional ratios as a function of: (1) the friction angle of the noncohesive material forming the channel bed, and (2) the lift-to-drag force ratio acting on the particles. These findings may be used as a point-of-start in the study of unsteady alluvial channel morphology. APPENDIX By Jhonath Wensenber Mejia Gonzales L'Hospital's rule analysis to show that the top-width/maximum-depth ratio To /Yo (Eq. 57) is equal to 4 (Eq. 60 and Table 1) for the values of φ = 45° and β = 1. [Click on top of image to expand]. Equation 57 of the main body of this article is:    To                2          1 + r ______  =  _________ (______) 1/2  cos-1 r    Yo             tan φ      1 - r (57) The limit of the value of To /Yo  for β = 1 and φ = π /4 is:                  To                                   2          1 + β tan φ lim [  lim ( ___ ) ]  =  lim [ lim  [ (______) [ (____________) ] 1/2  cos-1(β tan φ) ] ] β↠1 φ↠π /4  Yo           β↠1 φ↠π /4    tan φ        1 - β tan φ (A.1) Applying the limit for β ↠ 1 :                  To                           2          1 + (1) tan φ lim [  lim ( ___ ) ]  =  lim  [ (______) [ (_____________) ] 1/2  cos-1((1) tan φ) ] β↠1 φ↠π /4  Yo         φ↠π /4    tan φ        1 - (1) tan φ (A.2)                  To                           2           1 + tan φ lim [  lim ( ___ ) ]  =  lim  [ (______) [ (___________) ] 1/2  cos-1( tan φ ) ] β↠1 φ↠π /4  Yo         φ↠π /4    tan φ        1 - tan φ (A.3)                  To                          (1 + tan φ) 1/2  cos-1( tan φ ) lim [  lim ( ___ ) ]  = 2 lim  [ (_____________________________) ] β↠1 φ↠π /4  Yo         φ↠π /4              tan φ  (1 - tan φ) 1/2 (A.4) The expression (1 + tan φ)1/2 = (2)1/2 is taken out of the limit because the indeterminate term "0/0" does not depend on (1 + tan φ)1/2. Therefore:                  To                                             cos-1 ( tan φ ) lim [  lim ( ___ ) ]  = 2 (2)1/2   lim  [ (_______________________) ] β↠1 φ↠π /4  Yo                      φ↠π /4       tan φ  (1 - tan φ) 1/2 (A.5) Applying L'Hospital's rule:                                                                   d                                                                 ____ [ cos-1 ( tan φ ) ]                  To                                            dφ lim [  lim ( ___ ) ]  = 2 (2)1/2   lim  _______________________________ β↠1 φ↠π /4  Yo                      φ↠π /4       d                                                           ____ [ tan φ  (1 - tan φ) 1/2 ]                                                            dφ (A.6)                                                                                   sec2 φ                                                                           - _______________                  To                                                        (1 - tan2 φ) 1/2 lim [  lim ( ___ ) ]  = 2 (2)1/2  lim  [ _________________________________________ ] β↠1 φ↠π /4  Yo                     φ↠π /4                                              tan φ sec2 φ                                                        sec2 φ (1 - tan φ) 1/2  -  _________________                                                                                                2 (1 - tan φ) 1/2 (A.7)                                                                                   1                                                                       - _______________                  To                                                    (1 - tan2 φ) 1/2 lim [  lim ( ___ ) ]  = 2 (2)1/2  lim  [ _________________________________ ] β↠1 φ↠π /4  Yo                     φ↠π /4                                       tan φ                                                        (1 - tan φ) 1/2  -  _________________                                                                                    2 (1 - tan φ) 1/2 (A.8)                                                                           1                                                              - _______________                  To                                            (1 - tan2 φ) 1/2 lim [  lim ( ___ ) ]  = 2 (2)1/2  lim  [ ________________________ ] β↠1 φ↠π /4  Yo                     φ↠π /4       2 (1 - tan φ) - tan φ                                                           ____________________                                                               2 (1 - tan φ) 1/2 (A.9)                                                                           1                                                              - _______________                  To                                            (1 - tan2 φ) 1/2 lim [  lim ( ___ ) ]  = 2 (2)1/2  lim  [ ________________________ ] β↠1 φ↠π /4  Yo                     φ↠π /4               2 - 3 tan φ                                                             __________________                                                                2 (1 - tan φ) 1/2 (A.10)                                                                                  1                                                            - ____________________________                  To                                         (1 + tan φ) 1/2  (1 - tan φ) 1/2 lim [  lim ( ___ ) ]  = 2 (2)1/2  lim  [ ___________________________________ ] β↠1 φ↠π /4  Yo                     φ↠π /4                        2 - 3 tan φ                                                                       __________________                                                                         2 (1 - tan φ) 1/2 (A.11)                                                                        1                                                           - ______________                  To                                        (1 + tan φ) 1/2 lim [  lim ( ___ ) ]  = 2 (2)1/2  lim  [ ____________________ ] β↠1 φ↠π /4  Yo                     φ↠π /4            2 - 3 tan φ                                                             _______________                                                                        2 (A.12) Substituting φ = π / 4:                                                                      1                                                    - ___________________                  To                                 [ 1 + tan(π / 4) ] 1/2 lim [  lim ( ___ ) ]  = 2 (2)1/2  [ ________________________ ] β↠1 φ↠π /4  Yo                                   2 - 3 tan(π / 4)                                                        ________________                                                                   2 (A.13)                                                            1                                                   - ___________                  To                                 (1 + 1) 1/2 lim [  lim ( ___ ) ]  = 2 (2)1/2  [ _______________ ] β↠1 φ↠π /4  Yo                                   2 - 3                                                        ______                                                            2 (A.14)                  To                                2 lim [  lim ( ___ ) ]  = 2 (2)1/2  [ _______ ] β↠1 φ↠π /4  Yo                             2 1/2 (A.15)                  To lim [  lim ( ___ ) ]  = 4 β↠1 φ↠π /4  Yo (A.16) Therefore, it is proven that To / Yo = 4 when β = 1 and φ = π / 4. PDF file REFERENCES Apperley, L. W. 1968. The effect of turbulence on sediment entrainment. Thesis submitted to the School of Engineering, University of Auckland, Auckland, New Zealand, in partial fulfillment for the requirements for the degree of Doctor of Philosophy, January. Lane, E. W., P. N. Lin, and H. K. Liu. 1959. The most efficient stable channel for comparatively clean water in noncohesive material. Report CER-59HKL, Colorado State University, Fort Collins, Colorado, April. NOTATION a = variable, Eqs. 30 and 31; b = variable, Eqs. 30 and 32; C = constant of integration; c = variable, Eqs. 30 and 33; ds = grain diameter; Fd = drag force; Fl = lift force; r = β tan φ; S = slope of channel bed; To = channel top width; W = particle weight; X = generic variable, Eq. 30; Y = flow depth at abscissa ε (Fig. 1); Yo = flow depth at channel centerline (maximum flow depth); β = ratio of lift-to-drag forces acting on a particle; γ = unit weight of water; Δ = variable, in Eq. 30; δ = proportionality constant, Fd = δτ; ε = abscissa of flow depth Y (Fig. 1); θ = local side slope angle (Fig. 1); π = 3.1416; τ = shear stress; τo = shear stress at channel centerline (maximum shear stress); and φ = friction angle, repose angle. ABOUT THE AUTHORS Victor M. Ponce and Shurui Jiang are emeritus professor and M.S. graduate, respectively, at the department of civil engineering at San Diego State University, San Diego, California, USA. Jhonath W. Mejia Gonzales is a hydraulic engineer residing in Cajamarca, Peru.