How this instrument works
Stagnation temperature T₀ is the temperature a moving gas would reach if it were slowed to zero velocity along a path with no heat added and no friction losses — the condition that makes the flow isentropic, meaning entropy stays constant throughout. Energy bookkeeping gives it directly: the specific enthalpy of the still gas equals the enthalpy of the moving gas plus its kinetic energy per unit mass, cₚT₀ = cₚT + V²⁄2. Substitute the definition of Mach number, M = V ⁄ a, and the local speed of sound a² = γRT, and the velocity term collapses into a clean multiple of M²: T₀ ⁄ T = 1 + (γ−1)⁄2 × M².
The (γ−1)⁄2 factor is not decorative — it falls straight out of cₚ = γR ⁄ (γ−1), so a gas with a larger heat capacity per mole absorbs the same kinetic energy with a smaller temperature rise. Because the term scales with M², not M, the heating accelerates: air at Mach 1 stagnates only about 20% hotter, at Mach 2 it is 80% hotter, and at Mach 3 nearly triple its static temperature. A common misreading is to expect stagnation temperature to fall across a shock wave, since stagnation pressure clearly does; it does not. For adiabatic flow with no external work, energy is conserved through the shock, so T₀ is the same on both sides — only the irreversible jump in entropy shows up as a stagnation-pressure loss, not a temperature one.
This is why supersonic aircraft skins run hot in air that is bitterly cold. At 12 km, standard-atmosphere air sits near 216.65 K, roughly −56.5 °C, yet a surface that brings the local flow to rest — a Pitot probe tip, a wing leading edge, an engine inlet lip — approaches the stagnation temperature, not the ambient one. Concorde's nose reached around 127 °C at cruise for exactly this reason, and it is why the SR-71's titanium skin needed to shed heat measured in hundreds of degrees rather than tens. A real probe never fully arrests the flow isentropically, so instrument makers apply a recovery factor to predict what a thermometer will actually read; the isentropic ratio here is the ceiling that factor works down from.
- Enter the flow speed as a Mach number in the Mach number field — true airspeed divided by the local speed of sound.
- Set Specific heat ratio (γ, 1.4 for air) to 1.4 for air, or a different value for another working gas.
- Read Stagnation-to-static temperature ratio (T₀ ⁄ T), the factor by which stagnation temperature exceeds static temperature.
- Multiply that ratio by the actual static air temperature in kelvin to get the stagnation temperature itself.
- Lower γ toward 1.3 before trusting the result for hot combustion gas, where the default air value understates the rise.
Worked example — Mach 2 air at a Pitot probe
Set Mach number to 2 and Specific heat ratio (γ, 1.4 for air) to 1.4, its default for air. The instrument computes T₀ ⁄ T = 1 + (1.4 − 1) ⁄ 2 × 2² = 1 + 0.2 × 4 = 1.8. A gas moving at twice the local speed of sound is, at the point where it is finally brought to rest — the tip of a Pitot probe, say — 1.8 times as hot, in absolute terms, as the still air around the aircraft.
Put a number on it: at 12 km the standard atmosphere sits at 216.65 K. Multiply by the ratio this instrument returns, 1.8, and the stagnation temperature comes out at 389.97 K, about 116.8 °C — well above the boiling point of water, generated entirely by decelerating air that started off 56 degrees below freezing. That gap between ambient cold and stagnation heat is why supersonic leading edges need real thermal protection even far above the weather.
Questions
Why does stagnation temperature rise if no heat is actually added?
Because the gas already carries kinetic energy, and stopping it converts that energy into heat rather than removing it. The energy bookkeeping is cₚT₀ = cₚT + V²⁄2 — enthalpy plus kinetic energy stays constant along the deceleration, so if V drops to zero, T must rise to absorb it. No external heat source is needed; the gas is simply heating itself with its own lost motion.
Does stagnation temperature drop across a shock wave?
No — for adiabatic flow with no external work, total energy and therefore stagnation temperature are conserved straight across a shock. What actually falls is stagnation pressure, because a shock is irreversible and entropy rises. Anyone expecting T₀ to drop is usually thinking of that pressure loss, which is real, just not a temperature effect.
What specific heat ratio should I use for a gas other than air?
1.4 covers diatomic gases — air, nitrogen, oxygen — near room temperature. Monatomic gases like helium and argon run close to 1.67. Combustion gas in a jet engine's hot section is often modeled closer to 1.3, since γ falls as temperature and vibrational modes rise. Using air's 1.4 for hot exhaust will understate the true temperature ratio.
How is the Mach number in this formula actually defined?
Mach number is flow speed divided by the local speed of sound, M = V ⁄ a, where a = √(γRT) depends only on the static temperature of the gas an object moves through, not on pressure or altitude directly. That is why identical true airspeed buys a higher Mach number at cold, thin cruise altitude than near a warm runway.
Is this the actual temperature a real thermometer would read?
Not quite — it is the idealized ceiling. A real probe or skin panel rarely brings flow to a perfect, frictionless stop, so engineers apply a recovery factor r, typically 0.85 to 0.90 for a laminar boundary layer, giving a measured recovery temperature Tr = T + r(T₀ − T), a little below the full isentropic value this instrument returns.