Why is the lake turquoise, and why does a glacier ending in deep water retreat so fast?
Each button runs ten model years headless and lists the result.
The science
Meltwater. Degree-day melt, M = DDF · max(T, 0), DDF = 7 mm w.e. K⁻¹ d⁻¹ [перевірити за джерелом], over a 14 km² glacier: Qin = M · Ag, plus a 0.4 m³ s⁻¹ base flow. Air temperature is a synthetic cycle (mean 0.5 °C, amplitude 9.5 K) plus your warming.
Calving and flotation. The thickness at which ice floats is Hf = (ρw/ρi) hw ≈ 1.09 hw. The calving speed follows a Brown-type law, uc = k hw with k = 0.7 yr⁻¹ (model value for a lake; tidewater glaciers are faster [перевірити за джерелом]), multiplied by up to 3 when the front is within 1.5 Hf. The terminus moves by dx/dt = ui − uc − (M − Mref)/α with ice speed ui = 80 m yr⁻¹ and surface slope α = 0.05. The bed is over-deepened just behind the front, so a retreat can run away (calving instability).
Lake level. AL dh/dt = Qin − Qout + Vcalved·ρi/ρw with a broad-crested weir Qout = 1.7 b h3/2, b = 45 m, AL = 3.5 km².
Sediment. Stokes settling, ws = (ρs − ρw) g d² / (18 μ), μ = 1.5 mPa s (cold water), ρs = 2650 kg m⁻³. Fine flour (4 µm) falls 1 m in about 1.2 days, coarse silt (40 µm) in about 17 minutes. Lake-wide boxes: V dC/dt = Qin Cin − Qout C − AL ws C. The picture shows a 96 × 48 upwind advection-diffusion-settling grid in a 3 m surface layer with a prescribed, divergence-free flow (not a hydrodynamic model). Secchi depth ≈ 1.7 / k with extinction k = 0.1 + 35 C (C in kg m⁻³) [model value].
Simplifications. One lake box, no thermal stratification, no subglacial hydrology, constant ice speed, prescribed plume flow, synthetic temperatures. All coefficients are illustrative model values, not a calibrated reconstruction of any real lake. Assets built with Blender; scene by Three.js.