Hydrostatic pressure and thrust
p = ρ g h grows linearly with depth. On a vertical face of height h the resultant per metre of dam is P = (1/2) ρ g h², acting at h/3 above the base (centroid of the pressure triangle). 100 m of water pushes with about 49 MN on every metre of dam: the thrust grows with the square of the depth.
Gravity dam: weight versus water
A concrete gravity dam stays put because of its own weight W = γc A. The section is analysed as a rigid body per metre length: water thrust P, weight W, uplift U (water in the foundation cracks pushing up under the base), tailwater thrust and weight of water on the sloping faces. Uplift is linear from the heel pressure ρ g h to the toe pressure ρ g h_t and is reduced by a drainage curtain (efficiency E, at 6 % of B from the heel). Sliding: FS = (c B + μ ΣV) / ΣH. Overturning about the toe: FS = ΣM_restoring / ΣM_overturning with uplift counted as overturning. The resultant should stay inside the middle third of the base so no tension opens at the heel; base stresses are σ = ΣV/B (1 ± 6e/B), and a no-tension redistribution σ = 2ΣV/(3x̄) when it leaves the kern. Earthquake: pseudo-static inertia α W plus Westergaard hydrodynamic pressure 0.583 α ρ g h² at 0.4 h. Not modelled: silt, ice, temperature, internal stresses, foundation compliance, progressive cracking, three-dimensional action.
Textbook triangle and the minimum base
For a triangle with a vertical upstream face, full uplift and no tailwater, FS_slide = μ B (S−1)/H and FS_over = 2 S B² / (H² + 2 B²) with S = ρc/ρ. The middle-third condition gives B ≥ H/√(S−1). Real dams use less width than the sliding limit by drains, cohesion and a sloped base; the experiment shows how much each helps.
Arch dam
An arch carries the pressure sideways. For a thin ring of radius r the hoop force is N = p r and the stress σ = p r / t, so the thickness for a constant stress rises with depth, t = p r / σ_allow. The end thrust N goes into the rock of the valley walls. Concrete volume across a uniform gorge of width W is proportional to θ / sin²θ (r = W/(2 sin θ)); setting the derivative to zero gives tan θ = 2θ, θ = 66.78°, a central angle of 133.6°. Simplifications: the dam is treated as independent rings (membrane action only), whereas real arch dams share load between arches and vertical cantilevers (trial-load or finite-element design), and the abutment rock must also be checked.
Spillway
Free overflow over a crest of effective length L and head H: Q = C L H^1.5 with C ≈ 2.2 (m^0.5/s) for an ogee crest. Under a partly open radial gate the flow is limited by the orifice, Q = Cd b a √(2g(H − a/2)); the model takes the smaller of the two, continuous at a = H. Several gates add up.
Hydropower
P = η ρ g Q H_net, with the net head reduced by penstock friction h_f = (f L/D + ΣK) v²/(2g) (Darcy-Weisbach). The Francis efficiency follows a parabola in load fraction, peaking at 0.93 near 85 % load; the generator is 98.5 % efficient. The runner is held at synchronous speed n = 120 f / poles, shown here at a tenth of real speed. Specific speed n_s = n √P / H^1.25 (kW, m, rpm) of 50 to 400 selects a Francis runner.
Reservoir and flood routing
Water balance: A(h) dh/dt = Q_in − Q_spill(h) − Q_turbine, integrated with fourth-order Runge-Kutta. The surface area A(h) is measured from the real valley mesh by flood-filling the terrain below each level, then scaled by a factor of 40 for the part of the reservoir that continues beyond the visible frame. Tailwater follows a simple rating curve. The flood hydrograph is a single-peak Nash-type curve. Automatic gate control opens the gates to pass the surplus inflow plus a correction towards the normal pool.
Educational model, not a design tool. The valley, the fishway and the power station are schematic; a real fishway for a dam of this height would be a fish lift. No company or product is depicted. Everything is shown in normal operation.