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Excess Power⚠ unverified

Aerospace / Performance · Compute the excess power

Parameters

InputSymbolUnitDefaultDescription
PaPaW1.0Power available
PrPrW1.0Power required
OutputSymbolUnitDescription
resultPexcessWExcess power, in watts (W)

The science & history

Understanding the Parameters

Derivation (Approaching a Proof)

Excess power is a definition, but its meaning follows from the energy balance of an aircraft. Consider the total mechanical energy (kinetic plus potential):

$$E = \tfrac12 m V^2 + m g h.$$

Differentiate with respect to time — the rate at which the aircraft gains energy:

$$\frac{dE}{dt} = m V\frac{dV}{dt} + m g\frac{dh}{dt}.$$

By the work–energy theorem, the net rate of work done on the aircraft along its path equals thrust minus drag times speed (lift does no work, being perpendicular to the flight path):

$$\frac{dE}{dt} = (T - D)\,V = T V - D V = P_a - P_r = P_{\text{excess}}.$$

So excess power is the rate of total energy addition to the aircraft. Dividing by weight gives the rate of gain of energy per unit weight — the specific excess power $P_s = P_{\text{excess}}/W$, the central quantity of energy-manoeuvrability theory (see Energy Height):

$$P_s = \frac{P_{\text{excess}}}{W} = \frac{dh}{dt} + \frac{V}{g}\frac{dV}{dt}.$$

If the surplus is spent entirely on altitude ($dV/dt = 0$), $P_s$ is the Rate of Climb; if spent entirely on speed (level flight), it is the acceleration capability. The pilot chooses how to divide it.

Dimensional check. $P_a - P_r = \text{W} - \text{W} = \text{W}$ ✓ — subtraction of like quantities. And $(T-D)V = \text{N}\cdot\text{m}/\text{s} = \text{W}$, confirming the energy-rate interpretation.

History and Development

Related Concepts: Rate of Climb, Energy Height, Time To Climb, Best Climb Speed, Rate Of Climb Ft Min, Drag Force, Thrust, Power From Force Velocity

Notes: Exact definition (not an approximation). $P_a = TV$ (jet) or $\eta_p P_{\text{shaft}}$ (prop), falls with altitude; $P_r = DV$ is U-shaped in speed. Equals the aircraft's total energy-addition rate $(T-D)V = dE/dt$; divided by weight it is the specific excess power $P_s$ of Boyd's E–M theory. Zero surplus = ceiling. Defaults $1.0\ \text{W}$ placeholder.

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