104 105 106 107 108 Modified mean [-] 104 105 106 107 Standard deviation [-] γ D Kbed Kblock Tr z0 τc ∆z τcb Total volume eroded [m 3 ] 100 101 102 103 Modified mean [-] 100 101 102 103 Standard deviation [-] γ D Kbed Kblock Tr z0 τc ∆z τcb Final max. elevation [m] 100 101 102 103 104 Modified mean [-] 100 101 102 103 Standard deviation [-] γ D Kbed Kblock Tr z0 τc ∆z τcb Final mean elevation [m] 10-1 100 101 102 Modified mean [-] 100 101 102 Standard deviation [-] γ D Kbed Kblock Tr z0 τc ∆z τcb Final stdev elevation [m] 10-1 100 101 102 Modified mean [-] 10-1 100 101 102 Standard deviation [-] γ D Kbed Kblock Tr z0 τc ∆z τcb Final max. slope [-] 10-3 10-2 10-1 100 Modified mean [-] 10-3 10-2 10-1 100 Standard deviation [-] γ D Kbed Kblock Tr z0 τc ∆z τcb Final mean slope [-] 10-2 10-1 100 101 Modified mean [-] 10-2 10-1 100 101 Standard deviation [-] γ D Kbed Kblock Tr z0 τc ∆z τcb Final stdev slope [-] 103 104 105 Modified mean [-] 103 104 105 106 Standard deviation [-] γ D Kbed Kblock Tr z0 τc ∆z τcb Total blocks delivered [# ] 1) CIRES and Dept. of Geological Sciences, CU Boulder 2) EarthLab, CU Boulder 0.0 0.2 0.4 0.6 0.8 1.0 x / L 0.0 0.5 1.0 1 f c 0.0 0.5 1.0 0.0 0.2 0.4 0.6 0.8 1.0 x / L 0.0 0.5 1.0 1 f c 0.0 0.5 1.0 0.0 0.2 0.4 0.6 0.8 1.0 x / L 0.0 0.2 0.4 0.6 0.8 1.0 Elevation / total baselevel lowering 0.0 0.2 0.4 0.6 0.8 1.0 x / L 0.0 0.2 0.4 0.6 0.8 1.0 Elevation / total baselevel lowering 0.0 0.2 0.4 0.6 0.8 1.0 x / L 0.0 0.2 0.4 0.6 0.8 1.0 Elevation / total baselevel lowering A B1 C1 B2 C2 Legend Constant normalized elevations of 0.2, 0.4, and 0.6 1-fc Fraction of bed exposed at 400 ky τ* Ratio of bed shear stress to total shear stress at 400 ky mean τ* = 0.63 mean 1-fc = 0.95 mean τ* = 0.45 mean 1-fc = 0.89 τ* τ* γ = 0 m-2 γ = 0.2 m-2 γ = 1.0 m-2 γ = 0.2 m-2 γ = 1.0 m-2 1 2 3 4 5 6 10-5 10-4 10-3 10-2 Baselevel lowering rate [m/yr] 1 2 3 4 5 6 10-5 10-4 10-3 10-2 Baselevel lowering rate [m/yr] γ = 5 m −2 γ = 4 m −2 γ = 3 m −2 γ = 2 m −2 γ = 1 m −2 γ = 0.5 m −2 γ = 0.2 m −2 γ = 0.1 m −2 c= 1.5 c= 1.0 c= 0.5 c= 0.25 Slope exponent n [-] 10-4 10-3 10-2 10-1 100 Steady State Mean Slope [-] y = 1401.8x1.3613, R2 = 0.99 No blocks, constant τc = 100 Pa Blocks, γ = 5 m −2 No blocks, variable τc fit No blocks, constant τc = 0 Pa 10-5 10-4 10-3 10-2 Baselevel Lowering Rate [m/yr] 10-4 10-3 10-2 10-1 100 Steady State Mean Slope [-] y = 1401.8x1.3613, R2 = 0.99 Blocks, γ = 5 m −2 No blocks, τc = 300000E + 40 Pa Blocks, γ = 2 m −2 No blocks, τc = 120000E + 30 Pa Blocks, γ = 0.5 m −2 No blocks, τc = 70000E + 5 Pa No blocks, constant τc = 0 Pa 0.0 0.2 0.4 0.6 0.8 1.0 ∆Z = 0.00001 m/yr 0.0 0.2 0.4 0.6 0.8 1.0 x / L 0.0 0.2 0.4 0.6 0.8 1.0 ∆Z = 0.001 m/yr 0.0 0.2 0.4 0.6 0.8 1.0 x / L Elevation / total baselevel lowering Block model Threshold model Block model Threshold model 0.6 0.8 1.0 1.2 1.4 Steady state φ 10-5 10-4 10-3 10-2 Baselevel lowering rate [m/yr] 0.6 0.8 1.0 1.2 1.4 Steady state φ 10-5 10-4 10-3 10-2 Baselevel lowering rate [m/yr] τc fit to γ = 0 m −2 τc fit to γ = 0.5 m −2 τc fit to γ = 2 m −2 τc fit to γ = 5 m −2 c= 1.5 c= 1.0 c= 0.5 c= 0.25 D0 dy dx qb q high fc low fc 10-4 10-3 10-2 10-1 100 y = 998.9x1.3615, R2 = 0.99 y = 1072x1.3617, R2 = 0.99 10-5 10-4 10-3 10-2 Baselevel lowering rate [m/yr] 10-4 10-3 10-2 10-1 100 y = 1401x1.3613, R2 = 0.99 10-5 10-4 10-3 10-2 Baselevel lowering rate [m/yr] y = 2543x1.3514, R2 = 0.99 γ = 5 m −2 γ = 4 m −2 γ = 3 m −2 γ = 2 m −2 γ = 1 m −2 γ = 0.5 m −2 γ = 0.2 m −2 γ = 0.1 m −2 γ = 0 m −2 c= 1.5 c= 1.0 c= 0.5 c= 0.25 Steady state mean slope [-] 1) The interaction between blocks, discharge variability, and slope adjustment sets channel response to block delivery. 2) Blocks significantly alter topographic scaling at channel slopes below the critical slope for block transport. 3) The "n" exponent values depend on block effects and proximity to the critical block transport slope. 4) The complex influence of blocks on bed cover and form drag is well-described by an erosion th