Brake Mean Effective Pressure   Volumetric Flow Rate  
  Indicated Mean Effective Pressure   Friction Mean Effective Pressure  
Compression/Expansion Ratio   Combustion   Pumping Losses   Mean Piston Speed   Total Bore Area   Volumetric Efficiency
  Gas Dynamics  

Since all engines are producing work by manipulating a fluid (air in our case), we can relate the power (P) to the pressure (p) and the volumetric flow rate (Q):

P = pQ (1)

Because in an engine p and Q are varying constantly throughout a cycle, we must used average values.

For any engine, you can find the power at the crankshaft output with the brake mean effective differential pressure (BMEPd) and the measured volumetric flow rate (Qm):

P = BMEPd Qm (2)

In addition, for a reciprocating engine or a rotary combustion engine you can find the power at the crankshaft output with the brake mean effective pressure (BMEP) and the theoritical volumetric flow rate (Qth):

P = BMEP Qth (3)


If you want your engine to produce 100 hp (74 600 W) and you know that similar engines produce 10 bar (1 000 000 Pa) of BMEPd, then you will need to displace 158 cfm (0.075 m³/s) of air.

This is true no matter what type of engine it is. It could be a turboshaft engine, a Wankel engine, a piston engine (4-stroke, 2-stroke; SI or CI) with or without forced induction or any other type of engine you can think of.


For any reciprocating engine or rotary combustion engine, you can define its torque ( T ) with the following: (more)


Where Vθ is the amount of air displaced per angular displacement of the output shaft (m³/rad).

The reason torque is introduced in this page about power, is to show that:

Engine torque does not represent the engine's potential

In fact, if you want to know the potential of your vehicle, all you need to know is the power output of the engine, not the torque (see acceleration simulator).

For example, let's say we have an engine that displaces 3 L (0.003 m³) of air for each cycle, with a BMEP of 10 bar (1 000 000 Pa). If the engine is a 4-stroke, it completes a cycle in 2 revolutions (4π rad) of the crankshaft.

Let's also imagine that this engine is connected to a wheel via a gearbox with a gear ratio of 3:1. So every time the engine makes 3 revolutions, the wheel makes only 1 revolution. That means that the wheel makes 2/3 revolution (4π/3 rad) for every engine cycle.

So we can find the engine crankshaft torque (Te) and the wheel torque (Tw) with equation (4):

Te  = 
(1 000 000)(0.003)
Tw  = 
(1 000 000)(0.003)
Te  =  239 N.m (= 176 lb.ft)
Tw  =  716 N.m (= 528 lb.ft)

Even if the engine was a 2-stroke − and completed one cycle in 1 revolution (2π rad) − we could change the gear ratio to 1.5:1 and the wheel would still see one cycle completed in 2/3 revolution, hence:

Te  = 
(1 000 000)(0.003)
Tw  = 
(1 000 000)(0.003)
Te  =  477 N.m (= 352 lb.ft)
Tw  =  716 N.m (= 528 lb.ft)

So even if the engine torque is twice as big,

the wheel displacement and the wheel torque are the same for one engine cycle

Hence, the vehicle wouldn't see a difference.