Month -Rain here 0 mm/moET here 0 mm/moRecycled share 0 %Inland rain vs intact 0 %FPS 0

Wind blows from the ocean (left) over 2,400 km of lowland forest (vertically exaggerated, trees not to scale). Blue dots: ocean-sourced vapour. Green dots: vapour that the forest itself evaporated and transpired. Dark clouds rain where the column loses water; the river grows with runoff. The tower marks the cell you inspect.

Rainforest Moisture-Recycling Lab

Trees do not just receive rain: they send much of it back up. Follow a parcel of ocean air across 2,400 km of forest, clear part of the forest and see where the rain, the river and the recycled share go.

Forest clearing

Atmosphere

Time and inspection

Live readouts

Annual water along the fetch (mm per year)

Recycled share of rain (%)

Seasonal cycle at the tower (mm per month)

Experiments

Each runs the whole model (3 spin-up years) for several values and prints the last year.

Independent numerical checks

Controls and missions

Drag / one finger: orbit; wheel / pinch: zoom. The tower slider chooses which 50 km cell the monthly chart and readouts describe.
Mission 1: clear 40 % starting at the coast and compare it with 40 % cleared at the far inland end. Which hurts the inland rain more, and why?
Mission 2: find the wind speed at which the recycled share at the end of the fetch is smallest.
Mission 3: raise the dry-season severity and watch the pasture soil run dry while the forest keeps transpiring.
Mission 4: how much more river runoff do you get from clearing 60 %, and what did it cost in rain?

The science

Atmospheric column. Precipitable water w (mm) is carried inland at wind speed u. Along the path, u dw/dx = E − P with evapotranspiration E adding water and rain P = w/τ removing it, where τ is a rain-out time scale (about 8 days, longer in the dry season). Inside one 50 km cell this has the exact solution w(x) = Eτ + (w_in − Eτ) e^−x/(uτ), and the rain of the cell follows from mass balance, P = E + u (w_in − w_out)/Δx.

Moisture recycling. A second, tagged water amount w_f carries only water that came from the land surface: it has the same source E but no inflow, and rain removes it in proportion to its share. The share of rain that is recycled is ρ = w_f/w (the Brubaker-type recycling ratio).

Evaporation demand. Priestley-Taylor equilibrium evaporation E_eq = α Δ/(Δ + γ) R_n/λ with Δ from the Magnus saturation curve, γ = 0.0665 kPa/K, λ = 2.45 MJ/kg and a seasonal R_n around 130 W/m². Forest transpiration is 92 % of the demand while its deep root zone (350 mm of plant-available water) holds water; pasture transpires 62 % of the demand from a shallow 100 mm bucket that runs dry in the dry season.

Interception. Wet canopy storage is 0.3 mm per unit leaf area (LAI 5 for forest, 1.2 for pasture); each rainy day (about 8 mm per event) fills it once and that water evaporates straight back: I = min(½ P, n_{rain days} · 0.3 LAI). This is a Gash-type simplification, typically 10-15 % of rain in tall forest.

Soil and rivers. S_t+1 = S_t + (P − I) − E_t; whatever exceeds capacity leaves as runoff Q. The river on the map thickens with the cumulative runoff.

Simplifications and honesty. This is a one-dimensional, steady-flow box model: no vertical structure, no convective feedbacks, no changes in wind or in rain-out efficiency with clearing, no topography, no groundwater and a fixed monthly cycle. Observed forest-loss effects on rainfall depend strongly on these missing processes and on the scale of clearing, so numbers here are illustrative, not predictions for any real basin. All constants (τ, LAI, bucket depths, 8 mm per rainy day, α) are typical or tuned values, not measurements. Trees and the 2,400 km strip are drawn exaggerated. Assets modelled in Blender; the particles, clouds, rain and river are driven by the numbers on this page.