Pitot Tube Airspeed

Equivalent forms

p_0 = p + \tfrac{1}{2}\rho v^2

q = \tfrac{1}{2}\rho v^2 = p_0 - p

v = \sqrt{2q/\rho}

A speedometer with no moving parts — just a hole facing the wind and Bernoulli.

Symbol	Name	SI unit	Dimension	Range
$p_0$	Stagnation Pressure Total pressure at the stagnation point in the tube nose	Pa	M·L⁻¹·T⁻²	0 – 200000
$p$	Static Pressure Ambient (free-stream) static pressure	Pa	M·L⁻¹·T⁻²	0 – 200000
$\rho$	Air Density Density of the flowing fluid	kg/m³	M·L⁻³	0.1 – 1300
$v$	Airspeedoutput Free-stream flow speed	m/s	L·T⁻¹	0 – 340

Unit systems

Symbol	SI	CGS	Imperial
$p_0$	Pa	ba	psi
$p$	Pa	ba	psi
$\rho$	kg/m³	g/cm³	slug/ft³
$v$	m/s	cm/s	ft/s

Where it holds

Incompressible Bernoulli form valid for steady, low-Mach (≲ 0.3) flow with negligible viscous loss between the static port and the stagnation point. Compressible corrections apply at higher Mach numbers.

Dimensional analysis

[\sqrt{(p_0-p)/\ $\rho }] =$ \ $\sqrt{(M\cdot L^{-1}\cdot T^{-2})/(M\cdot L^{-3})}$ = \ $\sqrt{L^{2}\cdot T^{-2}} = L\cdot T^{-1} \checkmark$ (velocity)

Discovery

Henri Pitot · 1732

Pitot built an L-shaped open tube to measure the speed of the river Seine, finding flow slower at depth than engineers assumed. Henry Darcy later refined the design; the pitot-static tube is now the primary airspeed sensor on virtually every aircraft.

Try this

Every airliner reads its speed from a bent tube and Bernoulli's equation — how?

An aircraft pitot tube measures a pressure difference (stagnation minus static) of 2000 Pa in air of density 1.2 kg/m³. What airspeed does that imply?

Research status: stable

Real-world applications

Primary aircraft airspeed indication
Wind-tunnel and HVAC duct flow measurement
Racing-car and marine speed sensing
Industrial gas-flow metering

Common misconceptions

The tube measures velocity directly — it measures a pressure difference that depends on density too
Indicated airspeed equals true airspeed — they diverge as air thins with altitude
Blockage just zeroes the reading — a frozen or blocked pitot can give dangerously wrong, sometimes increasing, indications (a known cause of accidents)

Experimental verification

Pitot-static readings are cross-checked against wind-tunnel reference speeds and GPS ground speed corrected for wind; aircraft certification requires calibrated airspeed agreement within a few knots across the envelope.

Derivation

Apply Bernoulli along a streamline from the free stream (pressure p, speed v) to the stagnation point inside the tube (pressure p_0, speed 0):

p + \tfrac{1}{2}\rho v^{2} = p_0 + 0

.

Solving for v gives

v = \surd (2(p_0 - p)/\rho )

.

Limiting cases

p_0 \to p⟶ v \to 0No dynamic pressure means still air — the tube reads zero airspeed.

High altitude (small

\rho )

⟶ True airspeed > indicatedThin air gives less dynamic pressure for the same speed, so uncorrected indicated airspeed underreads.

v \gtrsim 0.3\,c_{sound}⟶ Compressible correction neededAbove Mach

\sim 0.3

the incompressible form must be replaced by the compressible Rayleigh/St-Venant relation.

What if…

What if the static port ices over but the pitot stays clear?

The reference pressure freezes at the old value, so the indicated speed reads wrong as altitude changes — a classic instrument failure mode.

What if you fly the same speed at high altitude?

Lower density yields a smaller pressure difference, so indicated airspeed drops even though true airspeed is unchanged.

1

Airspeed from a pitot reading

Given ·

p 0:: 103325
p:: 101325
\rho:: 1.2

Find · v

Steps

Dynamic pressure $q = p_0 - p = 2000\,\mathrm{Pa}$
$v = \sqrt{2q/\rho } = \sqrt{2\times 2000/1.2}$
$v \approx 57.7\,\mathrm{m/s}$ — a typical small-aircraft cruise speed

Result ·

v = \surd (2(103325-101325)/1.2) = \sqrt{2\times 2000/1.2} = \sqrt{3333} \approx 57.7\,\mathrm{m/s} (\approx 112

knots).

Bernoulli's Equation→Venturi Effect→Mach Number→