Föhn Winds & the Rain Shadow

Air that crosses a mountain and rains on the way up arrives on the far side warmer than it started. This works out by how much, and shows you where the heat came from.

Worked example — Föhn Winds & the Rain Shadow

Lee-side temperature	6.8
Föhn warming	+9.0
Cloud base	249
Crest temperature	-12.3
Rained out	3.66
Lee-side humidity	31%
Cooling inside the cloud	6.48

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

A dry crossing is free; the warming is paid for entirely by the rain 🖖

Try the dry crossing preset: 20 °C air with a dewpoint of −5 °C, over a 3,000 m ridge. The lee station reads 20.0 °C and the föhn warming reads +0.0. Drag the crest anywhere from 300 m to 3,000 m and it stays at +0.0, because that air would not reach saturation until 3,064 m. Nothing condenses, so the descent hands back exactly what the climb took. Now the Chinook preset: 8 °C, dewpoint 6 °C, the same 3,000 m ridge. It leaves 3.66 g of water behind for every kilogram of air, and Calgary at 1,045 m reads 6.8 °C where the unrained air would have arrived at −2.2 °C. Those 9.0 °C are the rain's doing.

The same ridge warms summer air more than three times as much as winter air 🖖

Set the crest to 2,000 m, the windward station at sea level, and the dewpoint one degree under the temperature, so the cloud starts almost at the ground. At 27 °C the Cooling inside the cloud card reads 3.95 °C/km and the lee side gains 10.8 °C. At −8 °C the same card reads 8.12 °C/km and the gain is 3.1 °C. Saturated air at 27 °C carries eleven times the vapour that saturated air at −8 °C does, condenses far more of it per kilometre climbed, and the latent heat released very nearly cancels the cooling from expansion. Cold air has almost nothing to give up, so it cools at nearly the dry rate: wind the temperature down to −30 °C and the card reads 9.4 against a dry 9.75. Over the whole temperature range this tool allows, on that same ridge, the card runs from 3.2 to 9.4 °C/km. The 6 °C/km of the textbooks sits near the middle of that and is nowhere in particular.

The mountain has no say in where the cloud base sits 🖖

The Chinook air is 8 °C with a dewpoint of 6 °C, a depression of two degrees, and the cloud base reads 249 m. Drag the crest from 1,000 m to 6,000 m and that card does not move a metre: the crest never enters the calculation. What sets it is the gap between temperature and dewpoint, because the rising air cools at 9.75 °C/km while its dewpoint falls at about 1.7, so the gap between them closes at roughly 8 °C per kilometre. Over every state from −20 °C to 35 °C and depressions from 1 to 20 degrees, this tool puts the cloud base between 116 and 131 m per degree of depression. Forecasters and glider pilots use 125 and are rarely far wrong. It is also why a cloud base looks ruled flat across a landscape while the ground beneath it is anything but.

Problems solved in full

Pacific air at 8 °C crossing a 3,000 m crest to Calgary 6 steps
Pacific air reaches the coast at 8 °C with a dewpoint of 6 °C, crosses a 3,000 m crest of the Rockies, and sinks to Calgary at 1,045 m. What does the thermometer in Calgary read, and where did the extra heat come from? This is the tool's opening state.
1. The dewpoint is the measurement of how much vapour the air is carrying, and the temperature has nothing to do with it. Convert it: the saturation vapour pressure at 6 °C, then the mixing ratio at the station pressure of 1013.25 hPa. Carry that 5.79 g/kg through everything that follows, because a rising parcel keeps it until it starts raining.
  
  $e_{s} (6^{\circ} C) = 6.112 exp (\frac{17.67 \times 6}{6 + 243.5}) = 9.35 hPa r_{0} = \frac{0.622 \times 9.35}{1013.25 - 9.35} = 5.79 g k g^{- 1}$
2. Lift the air. Unsaturated, it cools at g/c_p, which is 9.75 °C per kilometre and is a property of dry air rather than of the weather. Its capacity to hold vapour falls with it and falls faster, while the vapour it is actually carrying does not change at all, so somewhere the two meet. Solve for the height where the saturation mixing ratio has dropped to the 5.79 g/kg the parcel started with. The Cloud base card agrees: 249 m.
  
  $Γ_{d} = \frac{g}{c _{p}} = \frac{9.80665}{1005.7} = 9.75 K k m^{- 1} r_{s} (8 - Γ_{d} z, p (z)) = r_{0} ⟹ z = 249 m, T = 5.6^{\circ} C$
3. Above the cloud base the air is still cooling, but every metre of ascent condenses more vapour and gives back some latent heat, so it cools more slowly. The rate depends on how much vapour is left, which means it changes all the way up; the tool recomputes it every metre and its average over these 2,751 m is 6.48 °C/km.
  
  $Γ_{s} = \frac{g ( 1 + \frac{L r _{s}}{R _{d} T} )}{c _{p} + \frac{L ^{2} r _{s} ε}{R _{d} T ^{2}}} \overline{Γ_{s}} = 6.48 K k m^{- 1}, T_{crest} = 5.6 - 6.48 \times 2.751 = - 12.3^{\circ} C$
4. At −12.3 °C and 701 hPa the air can hold 2.13 g/kg and no more. Everything above that has already left as rain and snow on the windward slope. This is the number the whole problem turns on.
  
  $r_{s} (- 12.3^{\circ} C, 701 hPa) = 2.13 g k g^{- 1} Δ r = 5.79 - 2.13 = 3.66 g k g^{- 1}$
5. Now let it down the far side. It is unsaturated the instant it starts warming, so the whole 1,955 m runs at the dry rate again. Compare with the temperature the same air would have had at 1,045 m if it had never saturated, which is a plain dry descent from 8 °C.
  
  $T_{lee} = - 12.3 + 9.75 \times 1.955 = 6.8^{\circ} C T_{dry} = 8 - 9.75 \times 1.045 = - 2.2^{\circ} C ⟹ Δ T = + 9.0^{\circ} C$
6. Where did 9.0 °C come from? Every gram of vapour that condensed released its latent heat into the air around it. Multiply the water lost by L and divide by the specific heat, and you get 9.21 °C: the right size, and 2.4% too big. The gap is real, not rounding. The heat is released around 700 hPa and cashed in at 900 hPa, and the quantity actually conserved along a saturated ascent is equivalent potential temperature rather than T + Lr/c_p at any one level.
  
  $\frac{L Δ r}{c _{p}} = \frac{2.530 \times 1 0 ^{6} \times 3.66 \times 1 0 ^{- 3}}{1005.7} = 9.21 K \frac{9.21}{8.99} = 1.024$
Answer

Calgary reads 6.8 °C, which is 9.0 °C warmer than the same air arriving without having rained. The exchange rate is about 2.5 °C for every gram of water per kilogram of air left behind on the windward slope, and it is L/c_p with a couple of percent shaved off for the altitude the heat was released at. And it tells you what the tool will not: the rain is the energy source, so the same wind in a dry spell, with nothing left in the air to condense, brings the wind and none of the warmth.
The ceiling on a föhn wind, and why no mountain reaches past it 6 steps
The Winter air, Scandes preset lifts −8 °C air with a dewpoint of −9 over a 1,500 m ridge, and the lee station gains 2.5 °C. Work out where that gain comes from, then say how high the ridge would have to be for this air to deliver a Chinook's nine degrees.
1. The warming is a product of two things and the height of the mountain is neither of them. Inside the cloud the air cools at 7.97 °C/km instead of the dry 9.75, and coming down the far side it warms at the dry rate the whole way, so what it keeps is the gap between the two rates multiplied by the depth of cloud it climbed through. Cloud base 121 m, crest 1,500 m: 1.379 km of cloud at 1.78 °C/km of advantage.
  
  $Δ T = (Γ_{d} - \overline{Γ_{s}}) (z_{crest} - z_{cloud}) = (9.751 - 7.969) \times 1.379 = 2.46^{\circ} C$
2. So raise the crest. The cloud gets deeper, which helps, and the cooling inside it climbs at the same time: 7.97 °C/km at a 1,500 m crest, 8.39 at 3,000, 8.95 at 6,000, 9.22 at 9,000. It is heading for the dry rate, because air this cold has less and less vapour left to condense the higher it goes.
  
  $\overline{Γ_{s}} : 1500 m : 7.97 3000 : 8.39 6000 : 8.95 9000 : 9.22 K k m^{- 1} ⟶ Γ_{d}$
3. The two factors pull against each other and the second one wins. Drag the crest and the warming runs 2.46, 3.91, 4.69 and 4.74 °C at those same four heights. Six times the mountain does not buy twice the warming.
  
  $Δ T : 1500 m : 2.46 3000 : 3.91 6000 : 4.69 9000 : 4.74^{\circ} C$
4. Behind that sits a hard ceiling, and it belongs to the air rather than to the terrain. Every gram of vapour that condenses releases its latent heat into the parcel, nothing takes it back on the descent, so the most the air can gain is all of its water converted to heat: L·r₀ / c_p. This parcel carries 1.91 g of vapour per kilogram, worth 4.75 °C.
  
  $Δ T_{m a x} = \frac{L r _{0}}{c _{p}} = \frac{2.5 \times 1 0 ^{6} \times 1.910 \times 1 0 ^{- 3}}{1005.7} = 4.75^{\circ} C$
5. Set the crest to 9,000 m, higher than any mountain on Earth, and the föhn warming reads +4.7. The Rained out card reads 1.91 g/kg, which is everything the air arrived with, and what remains at the crest is 0.0003 g/kg. There is no more fuel.
  
  $z_{crest} = 9000 m : Δ r = 1.910 = r_{0}, r_{crest} = 0.0003 g k g^{- 1}$
6. The Chinook air is sixteen degrees warmer and carries three times the water: 5.79 g/kg, a ceiling of 14.4 °C, and 9.0 °C actually delivered over a 3,000 m crest. Saturation vapour pressure roughly doubles for every ten degrees, so warm air turns up at the mountain with a far larger tank.
  
  $Chinook air : r_{0} = 5.792 ⟹ Δ T_{m a x} = 14.40^{\circ} C, Δ T (3000 m) = 8.99^{\circ} C$
Answer

No ridge is high enough. This air tops out near 4.7 °C, and the tool's tallest crest of 9,000 m collects 4.74 of it. The first 1,379 m of cloud is worth 2.46 degrees, more than half the total; the remaining 7,500 m is worth the other 2.28.

A föhn is a water engine and the mountain only decides how much of the tank gets emptied. That is the practical reading of the Rained out card: the rain is the fuel rather than a by-product. It also says where to look for the large ones. A modest ridge standing in warm humid air beats a much higher one in cold dry air, and the tool has both cases on its own preset row. South föhn, Innsbruck gains 9.4 °C over 2,800 m, while this Scandinavian air over the same 2,800 m crest would gain 3.8.

$Δ T = (Γ_{d} - \overline{Γ_{s}}) Δ z_{cloud} \leq \frac{L r _{0}}{c _{p}}$

By the 42Math team · Updated 2026-08-26

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Chinook, Calgary - Pacific air at 8 °C crosses a 3,000 m crest and reaches Calgary, 1,045 m, at 6.8 °C: 9.0 °C above the same air with no rain.
South föhn, Innsbruck - Po valley air at 12 °C arrives in Innsbruck at 17.7 °C with the humidity down to 30%, which is the Alpine south föhn the word comes from.
Monsoon, Western Ghats - Warm monsoon air leaves 2.36 g/kg on the Western Ghats and cools inside the cloud at only 4.05 °C/km.
Winter air, Scandes - Winter air at −8 °C has little vapour to give: the cloud cools at 7.97 °C/km and the lee side gains just 2.5 °C.
Dry crossing - Air at 20 °C with a dewpoint of −5 °C never reaches its cloud base under a 3,000 m crest, so the lee station reads 20.0 °C and the warming is exactly zero.