Predicting natural channel types in the Columbia River basin Hiroo Imaki 1, Tim Beechie 1, John Buffington 2 1. NOAA, Northwest Fisheries Science Center, 2. USFS, Rocky Mountain Research Station
Questions
Columbia River Basin Historical Ecology Project Channel Type Riparian Vegetation Salmon Distribution Salmon Distribution
Project Objectives Establish a guideline to restore stream processes for endangered salmon and biodiversity Our methods and products can be used as… – Natural channel type template for restoration – Restoration prioritization tool
Columbia River Basin Area 668,000km 2 Total stream length 444,121 km Columbia River Basin Area 668,000km 2 Total stream length 444,121 km Our common perspective
Why Channel Types?
Floodplain Channel Types StraightMeanderingBraidedIsland-braided Beechie et al. (2006)
Mountain Channel Types Montgomery and Buffington (1997) Cascade Step-pool Plane-bed Pool-riffle
Channel Type System Pool-riffle Step-pool Plane-bed Cascade Slope < Slope < 0.03 Slope < BFW< 8 m * Meandering Island-braided Braided Straight Reach Prediction Model * Mountain ChannelsFloodplain Channels Yes No Montgomery and Buffington (1997) * Hall et al. (2007) Confined Beechie and Imaki (in review) * Support Vector Machine
Biophysical Floodplain Channel Type Model
Methods Establish channel type reference reaches Calculate geomorphic characteristics GIS data preparation (DEM, stream, etc.) Build a model and predict channel types (Support Vector Machine) Field measurements 1.Slope 2.Bankfull width 3.Floodplain width 4.Relative shear stress (RSS) 5.Fine sediment supply Fine sediment Alpine sediment 6.Root strength / Wood supply ABC Flow direction Profile Reach RSS + - Relative shear stress
N 0 50 km
Prediction accuracy and error Observed channel type Predicted channel type Braided Island- braided MeanderingStraightTotalAccuracy Braided % Island-braided % Meandering % Straight % Total Accuracy 77%69%85%73% Channel type ~ Slope + acc_precip + r_stress + fine_sediment + alpine_sediment Overall accuracy = 76% Kappa = 0.69
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