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Energy Height⚠ unverified

Aerospace / Performance · Compute the energy height (specific energy) of the aircraft

Parameters

InputSymbolUnitDefaultDescription
hhm1.0Altitude
VVm/s1.0True airspeed
ggm/s**29.81Gravitational acceleration, in metres per second squared (m/s**2). Default is 9.81
OutputSymbolUnitDescription
resulthemEnergy height, in metres (m)

The science & history

Understanding the Parameters

Derivation (Approaching a Proof)

Write the aircraft's total mechanical energy as the sum of potential and kinetic energy:

$$E = \underbrace{mgh}_{\text{potential}} + \underbrace{\tfrac12 mV^2}_{\text{kinetic}}.$$

Energy height is defined as this total energy per unit weight ($W = mg$):

$$h_e \equiv \frac{E}{W} = \frac{mgh + \tfrac12 mV^2}{mg}.$$

The mass cancels in both terms:

$$h_e = h + \frac{\tfrac12 V^2}{g} = h + \frac{V^2}{2g}. \qquad\blacksquare$$

The result is a length because energy per unit weight is $(\text{J})/(\text{N}) = (\text{N}\cdot\text{m})/\text{N} = \text{m}$.

The rate — where the physics lives. Differentiate $h_e$ with respect to time:

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

From the flight-path energy balance (Excess Power, Rate of Climb), the right-hand side equals the excess power per unit weight:

$$\frac{dh_e}{dt} = \frac{(T - D)V}{W} = P_s.$$

This is the key result: the rate of change of energy height is the specific excess power, set entirely by thrust minus drag. A maneuver (pure trade of $h$ for $V$) leaves $h_e$ unchanged; only the propulsion/drag balance moves it. An aircraft in a sustained turn with $T < D$ has $P_s < 0$ and is bleeding energy height — descending its energy state — no matter how it distributes the loss between altitude and speed.

Dimensional check. $$h + \frac{V^2}{2g} = \text{m} + \frac{(\text{m}/\text{s})^2}{\text{m}/\text{s}^2} = \text{m} + \frac{\text{m}^2/\text{s}^2}{\text{m}/\text{s}^2} = \text{m} + \text{m} = \text{m}. \checkmark$$ The kinetic term correctly reduces to a length.

History and Development

Related Concepts: Excess Power, Rate of Climb, Time To Climb, Turn Radius, Kinetic Energy, Potential Energy, Best Climb Speed

Notes: Specific energy (per unit weight) ⇒ output is a length. $V^2/2g$ is the "kinetic altitude" — the zoom height if all speed traded for altitude. A maneuver conserves $h_e$; only thrust−drag changes it. Its rate $\dot h_e = P_s = (T-D)V/W$ is the specific excess power — the decisive quantity of Boyd's energy–manoeuvrability theory. $g$ exposed as input but constant.

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