What a payload costs in flight time

The extra hover power and the flight time lost when a multirotor carries a payload, from its mass and any electrical draw, by momentum theory.

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20.6min

With the payload it hovers for about 20.6 min, 8.1 min less than without it.

Hover time without it
Flight time lost
Hover power with the payload
Power added
Power added (%)

Provenance

Computed by: drone.power.payload-impact 1.0.0, core 0.1.0
Model: Lift power = (m·g0)^1.5 / √(2ρA) / (FM·η), so it grows with mass to the power 1.5; hover power = lift power + avionics (+ payload draw when carried); time = usable energy / power
Accuracy: Momentum-theory estimate. It ignores the motors' thrust limit and any drag the payload adds in forward flight, so check the maximum takeoff mass in the manual.
Notes: 1 shown with the answer
Cites: Leishman, J. G., Cambridge University Press, Principles of Helicopter Aerodynamics

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Takeoff mass Everything on board, like 1.4 kg · plain numbers mean kg Rotors Like 4 Rotor diameter Like 9.4 in · plain numbers mean in Payload mass Added to the takeoff mass, like 0.3 kg · plain numbers mean kg Usable energy After reserve, like 72 Wh · plain numbers mean Wh

More options 7

Payload electrical poweroptional What the payload draws from the flight battery, like 8 W; default 0 · plain numbers mean W Figure of meritoptional Rotor efficiency vs ideal, 0.4 to 0.8, default 0.6 Motor and speed-controller efficiencyoptional 0.5 to 1, default 0.85 Avionics poweroptional Flight controller, radios, and so on, like 10 W; default 0 · plain numbers mean W Altitudeoptional Pressure altitude of the site, like 5000 ft; default sea level · plain numbers mean ft Air temperatureoptional Like 30 °C; default ISA for the altitude · plain numbers mean °C Density altitudeoptional Instead of altitude and temperature, like 8000 ft · plain numbers mean ft

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How we got thisFormula, worked example, sources, and proof

Model: Lift power = (m·g0)^1.5 / √(2ρA) / (FM·η), so it grows with mass to the power 1.5; hover power = lift power + avionics (+ payload draw when carried); time = usable energy / power

Show your work

Lift power with the payload
lift power grows with mass^1.5
150.6 W × (1.7 kg / 1.4 kg)^1.5 = 201.5 W
Hover power with the payload
lift + avionics + payload draw
201.5 W + 0 W + 8 W = 209.5 W
Hover time with the payload
time = usable energy / power
72 Wh / 209.5 W = 20.6 min

The same steps an agent gets from the MCP server with explain: true.

Accuracy: Momentum-theory estimate. It ignores the motors' thrust limit and any drag the payload adds in forward flight, so check the maximum takeoff mass in the manual.

Worked example: The 1.4 kg quad with a 0.3 kg, 8 W payload on 72 Wh. Source: Momentum theory (Leishman 2006, chapter 2) at both masses. It is golden test vector v001, and every build checks the tool still gives its answer within its tolerance.

You enter

Takeoff mass: 1.4 kg
Payload mass: 0.3 kg
Payload electrical power: 8 W
Rotor diameter: 9.4 in
Rotors: 4
Usable energy: 72 Wh

You get

Hover time with the payload: 20.6 min
Hover time without it: 28.7 min
Flight time lost: 8.1 min
Hover power with the payload: 209.5 W
Power added: 58.9 W
Power added (%): 39.1

Review: Not yet independently reviewed by a Part 107 remote pilot.

Last verified: 2026-09-18, when a maintainer last confirmed this tool's sources at the issuer. See the sources ledger.

Status: version 1.0.0, core 0.1.0. See this tool in the verification report.

Checked against: 6 golden test vectors (download the test vectors, each with its source and tolerance). See how results are checked and every source.

Sources

Principles of Helicopter Aerodynamics, Leishman, J. G., Cambridge University Press, 2nd edition. Chapter 2 (momentum theory: ideal hover power T^1.5/√(2ρA), figure of merit).

Experimental means this tool has not yet met the stable bar: at least 20 golden vectors, differential tests, and an independent worked example.

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