Issue 062 - Wildfire response - Lift energy and fuel burn
How much energy does it take to drop 240,000 liters of water on a wildfire?
AP reported that four helicopters made about 170 water drops in one day near Villa de Leyva, Colombia, releasing roughly 240,000 liters of water as more than 400 people joined the wildfire response.
Earth, climate, and environmentAbout 1 minute
Sources checked September 23, 2026. Figures and circumstances may have changed.
The problem
If helicopters dropped 240,000 liters of water over about 170 drops, about how much minimum mechanical energy was required just to lift that water to a plausible operating height?
Use your own assumptions about water density, average water per drop, vertical lift height, helicopter efficiency, and aviation-fuel energy.
Then ask the more important follow-up: how does that minimum water-lift energy compare with the realistic fuel burned by helicopters that must hover, climb, fly, maneuver, dip, and return for repeated drops?
Because Fermi problems target an order of magnitude, I normally use no more than two significant digits and write most calculations in scientific notation; the Fermi reference explains both conventions.
Grounding facts
The Guardian's AP story reports about 170 water drops and 240,000 L of water delivered by four helicopters. That averages:
water per drop
~= 2.4 x 10^5 L / 1.7 x 10^2 drops
~= 1.4 x 10^3 L/drop
~= 1.4 tonnes/dropThat payload is plausible for medium helicopter firefighting. Infobae reported that one Colombian Air Force UH-60L Black Hawk made about 35 drops, delivering about 14,000 gallons, or roughly 53,000 L, over about five hours. That is about 1,500 L per drop.
For fuel checks, the FAA cites Jet A at about 43 MJ/kg. U.S. Army aviation references put UH-60 fuel burn around 150 to 175 gallons per hour, which is roughly 450 to 530 kg/hour using jet fuel near 0.8 kg/L.
After checking sources
Checked answer and calculation
For the narrow question, treat the water as the thing being lifted. Use water density as 1 kg/L and choose a simple 400 m lift height, close to Matt's first-pass altitude assumption:
water mass
~= 2.4 x 10^5 L x 1 kg/L
~= 2.4 x 10^5 kg
minimum lift energy
~= m g h
~= 2.4 x 10^5 kg x 10 m/s^2 x 4 x 10^2 m
~= 9.6 x 10^8 J
~= 1 billion JSo the minimum gravitational energy is on the order of 1 billion joules. Per drop, that is only:
energy per drop
~= 1 x 10^9 J / 1.7 x 10^2
~= 6 x 10^6 J/dropIf the helicopter system were 30% efficient at converting fuel energy into useful lifting of the water, the fuel-energy floor would be:
fuel energy for water lift only
~= 1 x 10^9 J / 0.3
~= 3 x 10^9 J
jet fuel mass
~= 3 x 10^9 J / 4.3 x 10^7 J/kg
~= 7 x 10^1 kgThat literal lift-only fuel number, about 70 kg of jet fuel, is not the operational answer. It is just a physics floor. Helicopters burn a lot of fuel simply staying airborne and moving back and forth.
Using the Infobae-reported Black Hawk pace as a real-world check:
fuel per helicopter-hour
~= 150 gal/hour x 3.8 L/gal x 0.8 kg/L
~= 4.6 x 10^2 kg/hour
helicopter-hours for all drops
~= 170 drops / (35 drops / 5 hours)
~= 24 helicopter-hours
operational fuel
~= 24 hours x 4.6 x 10^2 kg/hour
~= 1.1 x 10^4 kg fuel
operational fuel energy
~= 1.1 x 10^4 kg x 4.3 x 10^7 J/kg
~= 5 x 10^11 JSo the clean answer has two layers: the water's minimum lift energy is about 1 billion joules, but the realistic helicopter fuel energy for the day's water-drop operation is probably closer to hundreds of billions of joules. The setup is a nice trap: `mgh` answers the minimum water-lift question, but it does not describe what a helicopter operation actually costs energetically.
One extra grounding comparison: if 240,000 L were spread evenly across the roughly 290 hectares reported as affected by the fire, it would be a film less than a tenth of a millimeter deep. Aerial water drops are targeted suppression, not rain over the whole burn area.
Before checking sources
Matt's original estimate
This is the unverified estimate Matt wrote before checking sources, not the checked answer.
I assumed:
- water has a density of about 10^3 kg / m^3 or about 1 kg / L
- fuel is about 4.5 x 10^7 J/kg
- the helicopter's mass is about 4 tons
- it reaches an altitude of about 400 m or 1200 ft once collecting water and while moving to the drop point
- helicopters are about 30% energy efficient
If the above assumptions are true:
240,000 L of water dropped across 170 drops means about 1.4 x 10^3 L per drop, or about the same kg of water per drop.
Drop mass + helicopter mass gives a combined lift mass of about 5.4 x 10^3 kg. By mgh, that's:
lift energy per drop
~= 5.4 x 10^3 kg x 10 m/s^2 x 400 m
~= 2.2 x 10^7 JThis is not counting energy costs to move the load to the appropriate position before dropping.
If the helicopter is 30% efficient:
fuel energy per drop
~= 2.2 x 10^7 J / 0.3
~= 7.3 x 10^7 J
fuel mass per drop
~= 7.3 x 10^7 J / 4.5 x 10^7 J/kg
~= 1.6 kg fuel/drop170 drops x 1.6 kg of fuel per drop means it takes about 2,700 kg of fuel to lift that total water mass up to travel altitude.
Reasoning score
Matt's reasoning score: 55 / 100
Higher is better: earn points for useful facts, a sound reasoning approach, correct math, and a final estimate close to the sourced answer. The owl meter shows percent full of it: 100 minus the reasoning score.
Useful facts: 20/30. Water density and jet-fuel energy were strong pegs. The 400 m height is a reasonable Fermi choice for a water-lift floor, though the real operation was at mountainous terrain around 8,000 ft elevation. The helicopter fuel-burn peg was missing from the first pass but appeared in the reflection.
Reasoning approach: 15/30. `mgh` is the right model for the minimum lift-energy floor, but adding the helicopter mass and treating the setup as mostly a lift problem missed the real helicopter-energy bottleneck: hover, flight, maneuvering, and repeated cycles.
Math: 0/10. The final multiplication had a misplaced decimal: 170 x 1.6 kg is about 270 kg, not 2,700 kg.
Final estimate: 20/30. The implied mechanical energy was within an order of magnitude of the water-lift floor, and the written fuel total was also within an order of magnitude of a rough operational fuel-burn check, though partly because of canceling errors.
Post-check reflection
Matt's reflection
First I made an error in the final calculation, misplaced decimal that put my answer an order of magnitude higher than it should have been.
My bigger mistake was not properly pushing back on the setup implied by the question. I was sort of proud of myself for trying to take into account the mass of the helicopter, which wasn't suggested to be important in the question - but the actual energy consumption of a helicopter is much higher than simple mgh would suggest, because it is consuming a lot of energy just maintaining altitude, much less when it is climbing. The actual answer would have been two orders of magnitude higher than my correct calculation, and I should have been able to anticipate that. It probably would have been better if I just estimated the hourly energy consumption just to hover, and multiply that across the length of the workday.
Like ignoring that a 5 ton helicopter is probably on the lighter side, I estimate they consume about 100 kg of fuel every hour just hovering, so an 8 hour mission would require about 800 kg of fuel. I might double that estimate to approximate fuel consumed to lift and transport water to the fire, which would put me at about 1,600 kg of fuel. That's closer to the assumed correct answer, but still on the lower end of the likely range.
Recommended memory peg
For helicopter water drops, remember: 1,000 L of water is 1 tonne, and lifting 1 tonne by 100 m costs about 1 MJ of gravitational energy. Also keep a separate operational peg: a UH-60-scale helicopter can burn roughly 500 kg of fuel per flight-hour, so flight time can dominate the energy budget.
Reader results
Bars show how submitted estimates sort into the answer choices from the gut-check prompt.