Library / Driveline and transmission / DRV-03
Driveline and transmission
Engine power required at a given speed
Engine power needed to hold a speed against a given total resistance, allowing for the losses in the driveline.
- Ref
- DRV-03
- Rev
- v1
- Field
- Driveline and transmission
- Result
- Pe,req [kW]
At the ground, the power needed is the force times the speed. The engine must supply more by the factor 1/η to cover driveline losses, so Pe,req = Freq v/η. With the force in kN and the speed in m/s the result is in kW. The relation applies to steady motoring at a speed above zero. A nominal top-gear ratio does not by itself set an achievable top speed: the engine must also be able to supply this much power at the speed in question.
Engine power needed to hold a speed against a given total resistance, allowing for the losses in the driveline. Power at the ground is force times speed; the engine must supply more by the factor 1/η.
Calculator
- Power delivered at the ground, F v120 kW
- Power lost in the driveline13.33 kW
- Engine power in horsepower (mechanical)178.8 hp
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- Holding 8 kN at 15 m/s takes 120 kW at the wheels and 133.3 kW from the engine, with 13.33 kW lost in the driveline.
- The numbers show why a high-speed climb is power-limited: at 8 kN the engine needs 222.2 kW at 25 m/s, but only 44.44 kW at 5 m/s.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (kW) | Local step | Change in result |
|---|---|---|---|---|---|---|
| Freq Tractive force required | 0.5 to 60 kN | 8.333 | 1,000 | 991.7 | +10% | +10 % |
| v Vehicle speed | 1 to 40 m/s | 8.889 | 355.6 | 346.7 | +10% | +10 % |
| η Driveline efficiency | 0.5 to 1 | 120 | 240 | 120 | +10% | −9.09 % |
Sweep of Vehicle speed
| Vehicle speed (m/s) | Engine power required (kW) |
|---|---|
| 1 | 8.889 |
| 5.875 | 52.22 |
| 10.75 | 95.56 |
| 15.63 | 138.9 |
| 20.5 | 182.2 |
| 25.38 | 225.6 |
| 30.25 | 268.9 |
| 35.13 | 312.2 |
| 40 | 355.6 |
Sweep one variable
Variables and how each one contributes
Engine power required
What the equation gives you. Everything else feeds it.
Tractive force required
The total tractive force required, in kN (see the total tractive force entry). Raises the power in proportion. Raising it by 10 % (to 8.8) moves the result by +10 %, from 133.3 to 146.7 kW.
Vehicle speed
The vehicle speed, in m/s. 15 m/s is 54 km/h. Raises the power in proportion, at fixed force. In practice the force also rises with speed through air drag. Raising it by 10 % (to 16.5) moves the result by +10 %, from 133.3 to 146.7 kW.
Driveline efficiency
The driveline efficiency from engine to ground, a plain number between 0 and 1: about 0.85 to 0.95 for a manual gearbox. Lowers the engine power needed: a better driveline needs less engine power for the same road load. Raising it by 0.05 (to 0.95) moves the result by −5.26 %, from 133.3 to 126.3 kW.
Worked example with the values above
- Power at the ground: F v = 8 kN × 15 m/s = 120 kW.
- Divide by the efficiency: 120 / 0.9.
- Pe,req = 133.3 kW.
Assumptions and limits
- Steady motion at a speed above zero, with the total force known.
- Constant driveline efficiency over the operating point.
Common mistakes
- Entering the force in N and the speed in m/s but reading the result as kW. Use kN to get kW.
- Forgetting the efficiency and sizing the engine for the ground power alone.
Statistics and uncertainty
Driveline efficiency varies with load, gear and temperature, so a nominal 0.9 might be 0.8 to 0.95 in practice. A 5-point drop in efficiency raises the engine power needed here by 5.9 %. Add margin for altitude and heat derating.
Where it is used
- Checking whether an engine can reach a target speed on a slope.
- Sizing the engine for a duty cycle.
Maths you need
- Power: force times speed
- Efficiency: output over input