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Nine Minutes: Building a Climate Model That Can See a Single Hillside

Adrian Dunkley September 15, 2026 14 min read

Hurricane Melissa came ashore in western Jamaica on 28 October 2025 with sustained winds near 185 mph. The Planning Institute of Jamaica put total damage and loss at J$1.952 trillion, about US$12.2 billion, equal to 56.7 per cent of the country's 2024 gross domestic product. The PIOJ's own projection is three to five years to get back to the output we had before it.

The Caribbean Catastrophe Risk Insurance Facility paid out US$70.8 million on Jamaica's tropical cyclone policy, its largest payment in its history, and US$21.1 million on the excess rainfall policy. US$91.9 million against US$12.2 billion of loss. Three quarters of one per cent.

I have spent the last three years at the Department of Physics at UWI Mona building climate models, and the reason sits underneath both of those numbers. It has nothing to do with whether anyone saw Melissa coming. Everyone saw Melissa coming. The gap is between knowing a hurricane is coming and knowing what it will do to a particular place.

The Model Stops at the Coastline

Take the regional climate projections that Caribbean governments actually use for planning. They run at 25 to 50 km grid spacing. At 50 km, between 12 and 20 degrees north and 64 and 56 degrees west, there is no land grid point at all. Antigua, Dominica, Saint Lucia, Barbados, Grenada: all ocean. The nearest land the model knows about is 500 km away in Puerto Rico or Venezuela. Cantet and colleagues documented this in Tellus A in 2014 and it has not changed.

Jamaica is big enough to survive. The island fills four to six cells depending on where the lattice falls, so the model knows we exist. What it does not know is our shape. Blue Mountain Peak reaches 2,256 m. A 50 km cell holds the average of everything inside it, and along the line of the trade winds that average works out to about 851 m. The mountain becomes a bump.

That is not a cosmetic loss, because in Jamaica the shape is the climate. The north-eastern slopes of the Blue Mountains take 3,000 to 5,000 mm of rain a year. The southern coastal plains of St Catherine and Clarendon take under 1,500 mm. Twenty kilometres apart, a factor of three to four in annual rainfall, and in the model both get the same number.

Grid resolution and a small islandThree panels showing Jamaica under a 50 km climate-model grid, a 6.3 km km-scale grid, and a 1 km field. At 50 km the island fills four to five cells and its rainfall gradient disappears; at 1 km the four-to-one gradient between the north-eastern mountain slopes and the southern coastal plains is resolved.What grid spacing does to an islandJamaica, 10,991 km². Same island, three grids. Coastline simplified; rainfall field is a schematic (see caption).A50 km · CORDEX class56%76%71%65%57%52%50 km6 cells are more than half land. 4 contain the island but count as ocean.B6.3 km · km-scale global50 kmAbout 280 land cells. The main ranges appear, the valleys do not.C1 km · what decisions need50 kmAbout 11,000 land cells. The 4:1 rainfall gradient is resolved.3,000–5,000 mmunder 1,500 mm2,256 m850annual rainfall (mm)5,200counted as land (≥50% land)counted as ocean
Figure 1. Jamaica under three grids. At the 50 km spacing typical of CORDEX-class regional projections, six cells are more than half land and four more contain the island but are counted as ocean. Further east the same spacing is worse: between 12–20°N and 64–56°W there is no land grid point at all, so every island in the Lesser Antilles is sea to the model, and the nearest land is roughly 500 km away in Puerto Rico and Venezuela (Cantet, Déqué, Palany and Maridet, Tellus A, 2014). Panel C paints a rainfall field built from an orographic relation and scaled to the published climatology, which puts 3,000–5,000 mm a year on the north-eastern Blue Mountain slopes and under 1,500 mm on the southern coastal plains of St Catherine and Clarendon (State of the Jamaican Climate, Climate Studies Group Mona). It is a schematic, not model output.

What the Cell Throws Away

The physics here is old and simple, which is what makes the loss so annoying.

A parcel of air leaves the sea on the north-east coast at 28 °C with a dewpoint of 24 °C. The trade wind pushes it up the slope. It cools at 9.8 K per kilometre until temperature and dewpoint meet, which given a 4 K dewpoint depression happens at about 500 m. That is cloud base, and you can see it on any clear morning in Portland. Above it the air is saturated and cools more slowly, around 5 K per kilometre in air this warm, arriving at the summit near 14 °C.

On the way up, the mixing ratio falls from 19.0 to 13.3 grams of water per kilogram of air. The 5.6 grams that went missing fell as rain on the windward slope. The parcel that comes over the top has been wrung dry, so on the way down it warms at the full dry rate and reaches the leeward plain near 36 °C.

That is why Portland is the wettest parish and St Elizabeth is where we get droughts. It is one mechanism, it happens over about twenty kilometres, and a 50 km cell has exactly one elevation, so it has exactly one lapse-rate path. The ascent, the rain and the drying all collapse into a single number.

Orographic rainfall across the Blue MountainsA 64 km cross-section along the trade-wind flow through Blue Mountain Peak. Air rises over the John Crow and Blue Mountain ranges, saturates at 500 m, cools at the saturated lapse rate to 14 degrees at the 2,256 m summit, sheds 5.6 grams of water per kilogram of air, then descends and warms at the dry rate. One 50 km cell replaces the whole profile with a single mean elevation of 851 m and a single mean rainfall.The mechanism a 50 km cell cannot holdCross-section along the trade-wind flow through Blue Mountain Peak, Jamaica. Idealised single-parcel trajectory.05001,0001,5002,0002,500metres0102030405060distance along flow (km)lifting condensation level, 500 mone 50 km cell sees mean elevation 851 mtrade wind from ENEsea level: T 28 °C, dewpoint 24 °Cdry ascentΓd 9.8 K/kmsaturated ascent, Γs ≈ 5 K/kmmixing ratio 19.0 → 13.3 g/kg, so 5.6 g/kg rains outdry descent, Γd 9.8 K/kmnothing left to condense, so the lee runs dryJohn Crow MtnsBlue Mountainssummit 2,256 m · 14.3 °Cleeward plain · 35.9 °CSchematic rainfall along the same transect, scaled to the published climatology2,5005,0000mm/yrcell mean 3,096 mmlee 850 mm, 3.6 times below the cell mean
Figure 2. A section along the trade-wind flow through Blue Mountain Peak. A parcel leaving the sea at 28 °C with a dewpoint of 24 °C saturates at 500 m, cools at about 5 K per kilometre to 14.3 °C at the 2,256 m summit, and sheds 5.6 grams of water per kilogram of air on the way up. Descending the far side it warms at the dry rate and reaches the leeward plain near 36 °C. One 50 km cell replaces that whole profile with a mean elevation of 851 m and a single rainfall number 3.6 times what the leeward plain actually gets. Saturation vapour pressure from Bolton (1980); the rainfall strip uses the same schematic field. A real parcel is a blend of air that crossed the summit and air that went around it, so the problem needs three dimensions.

The Load Is Not the Wind

Now move from rain to damage, because this is where the resolution problem turns into a body count.

Wind does not push on a building in proportion to its speed. It pushes in proportion to dynamic pressure:

q = ½ ρ V²

With warm, moist air at about 1.15 kg m−3, Melissa's 185 mph landfall wind gives q = 3.93 kPa. A Category 2 wind of 110 mph gives 1.39 kPa. The stronger storm is 68 per cent faster and applies 2.83 times the load.

Take that to a roof. Building codes give a net uplift coefficient of about 1.08 over the field of a low-slope roof and up to 2.18 at the corners. At Melissa's landfall wind that is 4.25 kPa over the field, which on a 100 m² roof is 425 kN, about 43 tonnes-force trying to lift the roof off the walls. At the corners it is roughly double that per square metre, which is why corners go first and why hurricane straps are specified where they are.

The square law has a second consequence that matters more for my work than the first. If the model's predicted wind is 20 per cent wrong, the predicted load is 44 per cent wrong. Forecast error does not pass through to structural consequence unchanged; it gets amplified on the way.

Wind load scales with the square of wind speedA plot of dynamic pressure against wind speed, showing that the load at 185 miles per hour is 2.8 times the load at 110 miles per hour, and a diagram of the pressure coefficients acting on a building, with roof corners carrying the largest suction. Because load goes as the square of speed, a 20 per cent error in the predicted wind becomes a 44 per cent error in the predicted load.Why a 20% wind error is a 44% load errorDynamic pressure q = ½ρV², with ρ = 1.15 kg m⁻³ for warm, moist tropical air. Coefficients follow the usual building-code zones.ALoad against wind speed01234kPa04080120160200sustained wind (mph)Cat 1Cat 2Cat 3Cat 4Cat 5110 mph 1.39 kPa185 mph 3.93 kPa2.83× the load of Cat 2Melissa at Jamaican landfall, 28 Oct 2025BWhere the load actswindward GCₚ +0.83.1 kPaleeward −0.52.0 kParoof field −1.084.2 kPacorner −2.188.6 kPa425 kN of uplift on a 100 m² roof at 185 mphabout 43 tonnes trying to lift the roof off the walls. Corners let go first.Because q ∝ V², forecast error is amplified before it reaches the structure:a 10% error in wind becomes 21% in load, a 20% error becomes 44%, a 30% error becomes 69%. This is why resolving the wind field over terrain matters more than it looks.
Figure 3. Dynamic pressure and the square law. q = ½ρV² with ρ = 1.15 kg m−3 for warm, moist tropical air. At the 185 mph sustained wind recorded at Hurricane Melissa's Jamaican landfall on 28 October 2025, q is 3.93 kPa, which is 2.83 times the load at 110 mph. The pressure coefficients are the standard building-code zones: the roof-field value works out to 425 kN, about 43 tonnes-force, on a 100 m² roof, and corners carry roughly twice that per unit area. Because load goes as the square of speed, a 20 per cent error in the predicted wind is a 44 per cent error in the predicted load.

From a Field to a House

A kilometre-scale wind and rainfall field is an input, not an answer. What I want out of it is a three-dimensional scene: terrain, buildings, vegetation, drainage, and the storm running over all of it.

Two things happen in that scene that a coarse field cannot produce. The first is topographic speed-up. Air accelerating over a ridge of moderate slope gains around 40 per cent in speed at the crest, which by the square law is roughly double the load. A house on a ridge and an identical house on the plain a kilometre away are in two different storms, and only one of them is in the storm the forecast described.

The second is water routing, which is where the nine minutes comes from. Peak discharge from a small catchment follows the rational method, Q = CiA. Channel velocity follows Manning. For a 4 km² catchment under 80 mm of rain an hour, that is 62 m³ s−1 moving at 5.3 m s−1. A settlement 2.8 km down the gully has about nine minutes from the moment that water starts moving.

A three-kilometre scene rendered from the governing relationsAn isometric view of a ridge and a coastal settlement. Wind vectors accelerate over the crest according to the standard topographic speed-up, so buildings on the ridge carry about twice the load of buildings on the plain in the same storm. The gully is routed with Manning's equation and the rational method, giving a nine-minute travel time from the crest to the settlement, and the floodplain is inundated to the computed water surface.Where a downscaled field becomes a damage estimateA 3 km × 3 km scene. Every number below is computed from the relation printed beside it, at V₀ = 45 m s⁻¹ upstream and 80 mm h⁻¹ of rain on a 4 km² catchment. Vertical scale exaggerated about 8×.ridge crestV = 64 m/s, q = 2.38 kPacoastal plainV = 46 m/s, q = 1.22 kPasame storm, 2.0× the load on the crestgullyQ = C i A = 62 m³ s⁻¹v = (1/n) R²ᐟ³ S¹ᐟ² = 5.3 m s⁻¹2.8 km of channel, so about9 minutes of warning belowLoad relative to a building on the plainunder 1.4× · light exposure1.4 to 2.2× · moderate exposureover 2.2× · severe exposureExposure ratios, not fragility curves. A fragility model needs the local building stock.What the scene is forA 50 km cell gives one wind speed and one rainfall for this whole view.At 1 km the crest and the plain separate, and so do their loads.The gully then tells you how long the settlement below has, in minutes.That is the difference between a forecast and an instruction.
Figure 4. A three-kilometre scene, rendered from the relations printed on it. Topographic speed-up puts 64 m s−1 on the ridge crest where the plain sees 46 m s−1, so a house on the crest carries twice the load of an identical house below it in the same storm. The gully carries 62 m³ s−1 under 80 mm of rain an hour on a 4 km² catchment and moves it at 5.3 m s−1, which leaves the settlement below about nine minutes. The colours are exposure ratios. Turning them into expected damage needs a fragility model fitted to the local building stock, and that is separate work that has not been done.

Nine minutes is something a parish disaster coordinator can use. A regional rainfall anomaly is not. That difference is the entire argument for doing this at sub-kilometre scale, and it is why I care more about the last step of the chain than the first.

Why This Can Be Cheap

The obvious objection is that everybody already knows finer models are better, and the reason we do not have them is that they cost too much to run. That is true of dynamical models and it is true for a specific reason.

Halving the grid spacing quarters the cell area, so the cell count goes as the inverse square. A dynamical model also has to shorten its timestep to keep the Courant condition, Δt ≤ Δx/c, or the numerics fall apart. So the step count goes as the inverse first power on top of the cell count, and refining the vertical as well takes the exponent to four. Going from 50 km to 1 km is a factor of 125,000 in two dimensions and 6.25 million in three. Then multiply by the ensemble size you need to say anything about uncertainty.

An emulator has no timestep to shorten. Its total work still grows with the cell count, so it still grows as the inverse square, but the constant in front is a forward pass rather than a full integration, and a decomposed emulator evaluates each subdomain independently. The subdomains do not wait for one another. Wall-clock time becomes a function of how many you can run at once, which is a procurement decision instead of a physical limit.

Cost against grid spacingA log-log plot of relative cost against grid spacing. A dynamical model must refine the timestep as it refines the grid to keep the Courant condition, so its cost grows as the inverse cube of grid spacing in two dimensions and the inverse fourth power in three. Going from 50 km to 1 km is a factor of 125,000 in two dimensions. A decomposed emulator pays a fixed forward pass per subdomain, so its cost grows with the number of subdomains, which is the inverse square of grid spacing, and the subdomains run in parallel.Why the dynamical route stops before it reaches the kilometreRelative cost, normalised to 1 at 50 km. Log scales on both axes.110102103104105106107relative cost502512.56.25421grid spacing (km)6,250,000×125,000×2,500×1× wall-clock50× cheaper than the 2D dynamical route at 1 kmdynamical model, 3D + CFL ∝ Δx⁻⁴dynamical model, 2D + CFL ∝ Δx⁻³emulator, total work ∝ Δx⁻²emulator, wall-clock with bricks in parallelWhy the exponent is not 2Halving the grid spacing quarters the cellarea, so cell count goes as Δx⁻².A dynamical model must also shorten thetimestep to keep the Courant condition Δt ≤ Δx / cso the step count goes as Δx⁻¹ as well.With vertical refinement it is Δx⁻⁴.An emulator has no timestep to shorten.Its total work still grows as Δx⁻², but thebricks do not wait for each other, so thewall-clock is set by how many you runat once.Compute stops being a ceiling andbecomes a line item.Scaling exponents only. Absolute wall-clock and cost depend on the backbone, the hardware and the data pipeline, and are reported per configuration.
Figure 5. Why the dynamical route stops short of the kilometre. A dynamical model has to shorten its timestep as it refines its grid, to keep Δt ≤ Δx/c, so its cost grows faster than the cell count. Going from 50 km to 1 km is a factor of 125,000 in two dimensions and 6.25 million once the vertical is refined too. An emulator has no timestep to shorten: its total work grows with the cell count alone, and the subdomains do not wait for each other, so wall-clock is set by how many you can run at once. Scaling exponents only. Absolute cost depends on the backbone, the hardware and the data pipeline, and belongs in a results table.

This is the part of the argument I am most confident about, because it is arithmetic rather than a claim about model skill. It is also the part that makes the goal reasonable. If the memory a device needs is set by the size of one subdomain instead of the size of a country, then in principle a phone can run the subdomain that covers its own neighbourhood. I have not demonstrated that. The scaling argument is established; the deployment is not.

What I Am Actually Building

The technical contribution is narrower than the vision, which is how it should be.

Physics-constrained downscaling already exists and works. The constraint people enforce is that the coarsened prediction should equal the coarse input, which is a statement about how the fine field relates to the coarse field across scales. That constraint has a dependency its own authors flagged: it assumes the coarse field is an unbiased average of the fine field. In a controlled experiment that is true by construction, because the coarse field was made by averaging the fine one. With a real global model driving it, it is false, and CMIP-class models have documented precipitation biases over the Caribbean.

There is a second conservation law that nobody in this literature has enforced. What leaves one subdomain through a face should equal what enters its neighbour through the same face. That constraint binds the fine field to itself rather than to the driver, so it does not inherit the driver's bias. The two laws are independent: you can satisfy either one while badly violating the other.

Then there is where the subdomain boundaries go. Every decomposed method in the field partitions on a regular lattice, because tensors are rectangular. Since the constraint acts on faces, face placement decides where the strongest physics in the model is doing its work. A face along a ridge crest separates two genuinely different air masses. A face drawn across a windward slope cuts one process in half and makes the model rebuild it through a penalty term.

Where I could be wrong. The lateral constraint is necessary and not sufficient: a field can be perfectly flux-consistent across every face and still be wrong about the weather. Terrain-aligned faces may turn out to make no measurable difference against regular tiling, and that is the piece I would bet against myself on. Jamaica's rain-gauge network is sparse and clustered, so point validation here will always be weaker than it would be in Europe. The experiments are designed so that a null result is publishable, and when the runs are done I will report the numbers whichever way they come out.

What a Government Should Ask For

If you are on the buying side of this, four questions separate a usable system from a demonstration.

  1. What grid spacing does the output actually have, on land? A product advertised at 10 km over the Caribbean may be interpolated from something much coarser. Ask what the native resolution of the underlying simulation is.
  2. What does one run cost, in currency, per simulated year? Almost nobody in this field publishes that number, and it decides whether you can run the ensembles you need or only the single run the vendor demonstrated.
  3. What happens when the driving model is biased? Ask for the degradation curve, not the headline skill score. Every operational deployment runs on a biased driver.
  4. What would make you say this model failed? If nobody can answer that, you are buying a demonstration.

The work I have described is doctoral research at the Climate Studies Group Mona, in the Department of Physics at UWI Mona, where Caribbean climate science has been done since 1994. The theory is written. The pipeline is being built. There are no results yet, and the figures in this article are either computed directly from the equations printed on them or labelled as schematics.

Melissa was a Category 5 at landfall and it cost more than half of a year's national output. The next one is not a question of whether. What we get to decide is how much of the country knows, in advance and at the level of a specific hillside, what is about to happen to it.

My UWI profile is here. The technical version of this argument, with the conservation-law derivation and the pre-registered experiment, is on climatephysicsai.com.