Rate of Climb⚠ unverified

Understanding the Parameters

• Why divide by weight — excess power is an energy rate (joules per second); to convert an energy rate into a height rate you divide by weight (newtons), since raising a weight $W$ by height $dh$ costs energy $W\,dh$. So $P_{\text{excess}}/W$ has units of $\text{W}/\text{N} = (\text{J/s})/\text{N} = \text{m/s}$ — a vertical speed. Weight is the exchange rate between power and altitude.

• Best rate vs best angle — two different "best climbs". $V_y$ (best rate) is the speed of maximum excess power → maximum $ROC$ → fastest gain of altitude per unit time. $V_x$ (best angle) is the speed of maximum excess thrust → steepest gain of altitude per unit ground distance → used to clear an obstacle. $V_x < V_y$ for a jet; for a propeller aircraft they are closer. See Best Climb Speed.

• Climb angle vs climb rate — $ROC$ is the vertical component of velocity, $ROC = V\sin\gamma$ where $\gamma$ is the flight-path angle. A steep angle at low speed can give the same $ROC$ as a shallow angle at high speed.

• Altitude fade — as the aircraft climbs, $P_a$ drops with air density while $P_r$ changes more slowly, so the surplus shrinks and $ROC$ falls roughly linearly with altitude. The service ceiling is conventionally where $ROC$ has fallen to $100\ \text{ft/min}$ ($0.508\ \text{m/s}$); the absolute ceiling is where it reaches zero.

Derivation (Approaching a Proof)

Begin with the aircraft's energy rate, established in Excess Power:

$$P_a - P_r = (T - D)V = \frac{dE}{dt}, \qquad E = mgh + \tfrac12 mV^2.$$

Expand the energy derivative:

$$P_a - P_r = mg\frac{dh}{dt} + mV\frac{dV}{dt} = W\frac{dh}{dt} + \frac{W}{g}V\frac{dV}{dt}.$$

The climb rate is $ROC = dh/dt$. Solve for it:

$$ROC = \frac{P_a - P_r}{W} - \frac{V}{g}\frac{dV}{dt} = \frac{P_a - P_r}{W}\underbrace{\left(1 + \frac{V}{g}\frac{dV}{dh}\right)^{-1}}_{\text{acceleration factor}}.$$

For a steady climb the aircraft holds constant true airspeed, so $dV/dt = 0$, the acceleration factor is 1, and the second term vanishes:

$$ROC = \frac{P_a - P_r}{W}. \qquad\blacksquare$$

This is the registry formula, exact in the steady case. When climbing at constant indicated airspeed, true airspeed actually increases with altitude, so $dV/dh > 0$ and the acceleration factor is less than 1 — the real climb rate is a few percent below the simple formula. The correction matters most for fast, high climbing jets.

Dimensional check. $$\frac{P_a - P_r}{W} = \frac{\text{W}}{\text{N}} = \frac{\text{J}/\text{s}}{\text{N}} = \frac{\text{N}\cdot\text{m}/\text{s}}{\text{N}} = \frac{\text{m}}{\text{s}}. \checkmark$$ Power per unit weight is a velocity — specifically, a vertical one.

History and Development

• The power-available/power-required method. Predicting climb from the gap between the two power curves is one of the oldest quantitative tools in aeronautics, dating to the analyses of Frederick Lanchester and the aerodynamicists of the 1910s. It remains the textbook derivation.

• Ceilings and records. Rate of climb defined an early competitive frontier — interwar altitude records and the frantic climb performance of WWII interceptors (racing to reach bomber altitude) were exercises in maximising excess power at height, driving supercharging and, later, the turbojet.

• Into energy methods. For high-performance aircraft the simple $ROC$ was superseded by the energy-height formulation (see Energy Height), which treats altitude and speed as interchangeable forms of energy and optimises the minimum-time climb path — sometimes even diving to trade height for speed before zooming. Rate of climb is the special case of that theory with kinetic energy held fixed.

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

Notes: Exact for a steady (constant-TAS) climb — the full relation multiplies by an acceleration factor $(1 + \tfrac{V}{g}\tfrac{dV}{dh})^{-1} < 1$ when accelerating. $= V\sin\gamma$ (vertical velocity component). $V_y$ (best rate) is the max-excess-power speed; $V_x$ (best angle) the max-excess-thrust speed. Service ceiling at $ROC = 100\ \text{ft/min} = 0.508\ \text{m/s}$; absolute ceiling at $ROC = 0$. Defaults ⇒ $ROC = 10\ \text{m/s}$.