Why Shower Curtains Billow Inward

Equivalent forms

\Delta p = -\tfrac{1}{2}\rho v_{in}^2\quad(v_{out}=0)

p_{in} + \tfrac12\rho v_{in}^2 = p_{out} + \tfrac12\rho v_{out}^2

F = \Delta p \cdot A

A daily annoyance explained by the same equation that lifts airplanes.

Symbol	Name	SI unit	Dimension	Range
$\rho$	Air Density Density of the bathroom air	kg/m³	M·L⁻³	0.8 – 1.4
$v_{in}$	Inside Air Speed Speed of air dragged down by the spray	m/s	L·T⁻¹	0 – 6
$v_{out}$	Outside Air Speed Speed of the still bathroom air	m/s	L·T⁻¹	0 – 6
$\Delta p$	Pressure Differenceoutput Net pressure pushing the curtain inward	Pa	M·L⁻¹·T⁻²	-50 – 50

Unit systems

Symbol	SI	CGS	Imperial
$\rho$	kg/m³	g/cm³	slug/ft³
$v_{in}$	m/s	cm/s	ft/s
$v_{out}$	m/s	cm/s	ft/s
$\Delta p$	Pa	ba	psi

Where it holds

Qualitative Bernoulli estimate; the true effect is a combination of this pressure drop and a buoyant horizontal vortex set up by warm, humid shower air. Good for order-of-magnitude intuition, not a precise force.

Dimensional analysis

[\tfrac12 \ $\rho v^2] = (M\cdot L^{-3})(L\cdot T^{-1})^{2} = M\cdot L^{-1}\cdot T^{-2} = Pa \checkmark$ (pressure)

Discovery

David Schmidt · 2001

Engineer David Schmidt ran a full CFD simulation of a shower and won a tongue-in-cheek Ig Nobel Prize in 2001 for showing a buoyancy-driven vortex (plus Bernoulli pressure drop) sucks the curtain inward — settling decades of bathroom debate.

Try this

Turn on the shower and the curtain lunges at your legs — what force is pulling it in?

Inside-the-shower air is set moving at about 2 m/s by the spray while the bathroom air outside is still (ρ = 1.2 kg/m³). Estimate the pressure difference Bernoulli predicts across the curtain.

Research status: well studied

Real-world applications

Curtain and liner design (weighted/magnetic hems)
Ventilation and entrainment intuition
Teaching Bernoulli with a household example
HVAC draft and door-billow analysis

Common misconceptions

The spray physically pushes the curtain — droplets mostly miss it; the air pressure does the work
It is purely Bernoulli — a buoyancy-driven vortex contributes strongly in a hot shower
Cold showers billow as much — they billow less because the buoyant-vortex term is weaker

Experimental verification

Schmidt's CFD reproduced the inward billow and identified the driving vortex; simple home tests (cold vs hot shower, weighted vs free hem) confirm the predicted dependence on air speed and buoyancy.

Derivation

Treat the curtain as a streamline boundary.

Bernoulli on each side at the same height gives p_in +

\tfrac{1}{2}\rho v

_

in^{2} = p

_out +

\tfrac{1}{2}\rho v

_

out^{2}

.

With

v_out \approx 0

the inside pressure is lower

by \Delta p = \tfrac{1}{2}\rho v

_

in^{2}

, and the net inward force

is \Delta p

times the curtain area.

Limiting cases

v_{in} \to 0⟶ \Delta p \to 0No air motion, no pressure difference — the curtain hangs straight.

Heavy/weighted curtain⟶ No billowA magnetized or weighted hem resists the small Bernoulli force, which is why hotels use them.

Hot shower⟶ Stronger effectBuoyant warm air sets up a vortex that adds to the Bernoulli suction, intensifying the lunge.

What if…

What if you weight the curtain hem?

The few-newton Bernoulli force can't overcome the added weight, so the curtain stays put — the standard fix.

What if the shower runs hot?

Buoyant warm air spins up a vortex that adds to the suction, so hot showers billow noticeably harder than cold ones.

1

Inward pressure of a running shower

Given ·

\rho:: 1.2
v {in}:: 2
v {out}:: 0

Find · \Delta p

Steps

Bernoulli: $\Delta p = \tfrac{1}{2}\rho (v$ _ $out^{2} - v$ _ $in^{2})$
$\Delta p = \tfrac{1}{2}(1.2)(0^{2} - 2^{2}) = -2.4\,\mathrm{Pa}$
Over a $1.5\,\mathrm{m}^{2}$ curtain that $is \sim 3.6\,\mathrm{N}$ — small, but the curtain is light

Result ·

\Delta p = \tfrac{1}{2}\rho (v

_

out^{2} - v

_

in^{2}) = \tfrac{1}{2}(1.2)(0 - 4) = -2.4\,\mathrm{Pa}

;

the \sim 2.4\,\mathrm{Pa}

lower inside pressure pulls the curtain in.

Bernoulli's Equation→Venturi Effect→Continuity Equation→