Why a Raindrop Is Not Tear-Shaped — The Weber Number and TAB Breakup

Umbrella icons and emoji all draw a raindrop as a teardrop — pointed on top, round at the bottom. No raindrop actually looks like that. Any drop wider than 2 mm gets squashed into a flat-bottomed hamburger-bun shape. Grow it more and its center inflates like a bag, then bursts.

One Weber number calls the burst. The TAB model O'Rourke and Amsden built in 1987 turns that call into a spring. Rupture is the instant the amplitude passes the radius.

Drop the Weber slider below 12 and the deformation $y$ never reaches 1. Push it past 12 and the drop splits on the first peak.

Raise Ohnesorge and the same Weber dies out early. Viscosity is what shaves that first peak.

A raindrop is not a teardrop#

Two forces compete inside a falling droplet. One is the aerodynamic force of the surrounding air hammering its surface. It presses the drop sideways and flattens it. The other is surface tension, which pulls the drop back into a sphere.

For a small drop, surface tension wins outright. That's why tiny drops are always spherical. As the drop grows, or as the relative velocity rises, the aerodynamic force climbs. At some point surface tension can no longer hold. The drop flattens, inflates, and finally shatters. We call this secondary breakup — as opposed to the primary breakup of a jet into drops.

Diesel engines, rocket injectors, agricultural sprays. Every device that atomizes a liquid runs on this phenomenon. Smaller drops mean more surface area, and faster evaporation and combustion.

The Weber number — the scale where inertia beats surface tension#

Collapse the two forces into one number and you get the Weber number (inertia / surface-tension ratio).

Here $\rho_g$ is gas density, $u_{rel}$ the relative velocity between drop and gas, $d$ the drop diameter, and $\sigma$ the surface tension. The numerator is the dynamic pressure the gas exerts; the denominator is the grip of surface tension.

When $We$ is small, the drop just oscillates politely. Once $We$ crosses a threshold, the drop breaks up. Experimentally this critical Weber number sits near 12. Water drop or fuel drop alike — above this value, none survive.

A second dimensionless group weighs in for viscosity: the Ohnesorge number (viscous / √(inertia · surface-tension)).

$\mu_l$ is the liquid viscosity and $\rho_l$ the liquid density. A viscous drop (picture a drop of honey) resists deformation strongly. Higher $Oh$ pushes the critical Weber number upward — meaning it's harder to break.

Taylor's analogy: hang the droplet on a spring#

The Weber number only tells you whether a drop breaks. Knowing when and how requires solving the drop's deformation over time. This is where G.I. Taylor's insight enters. He mapped an oscillating, distorting drop onto a spring–mass–damper oscillator.

Match the three forces one by one. The spring's restoring force is surface tension. The external force on the mass is the aerodynamic drag of the gas. The damping is the liquid's viscosity. Translate the analogy into an equation and you get a second-order ODE for the deformation $y$.

$y$ is how far the drop's equator bulges (dimensionless deformation) and $r$ is the radius. The first term on the right is the aerodynamic forcing, the second is the surface-tension restoring, and the third is the viscous damping. The constants $C_F, C_k, C_d, C_b$ are fixed by experiment; O'Rourke–Amsden used $1/3,\ 8,\ 5,\ 1/2$ respectively.

When forcing and restoring balance, the drop settles at a steady deformation $y_{eq}$.

$We_r$ is the radius-based Weber number. The key point: deformation is directly proportional to the Weber number. Double the Weber, double the squash.

The deformation parameter y and the trigger for breakup#

The TAB rule is simple. Breakup occurs the moment $y>1$. $y=1$ is when the equatorial bulge equals the radius. Once the amplitude grows that large, the drop can no longer hold its shape.

Here the nature of an undamped oscillation matters. Kick a resting oscillator with a sudden forcing and it doesn't stop at $y_{eq}$ — it overshoots, up to a maximum of $2y_{eq}$. So even a $y_{eq}$ of just 0.5 — half the deformation at steady state — can spike to $y=1$ on its first swing and burst. This overshoot lowers the critical Weber number in theory.

On the curve above, the split is whether that first peak clears 1. Ohnesorge is what shaves it.

Python — oscillate one droplet until it bursts#

Port the analogy straight into code. Integrate in dimensionless time and find the instant $y$ crosses 1.

import numpy as np
 
def weber(rho_g, u, d, sigma):
    """Gas inertia to surface-tension ratio (diameter-based)."""
    return rho_g * u**2 * d / sigma
 
def tab_breakup(We, Oh, dt=0.01, t_max=40.0):
    """Integrate the dimensionless TAB oscillator, return breakup time (dimensionless tau).
    y>1 means breakup; y stays below 1 to the end means None (survives)."""
    C_F, C_k, C_d, C_b = 1/3, 8.0, 5.0, 1/2
    omega2 = C_k                       # dimensionless natural frequency^2
    forcing = (C_F / C_b) * (We / 2)   # steady deformation y_eq = forcing/omega2 = We/24
    damp = C_d * Oh * np.sqrt(C_k)     # Ohnesorge raises the damping
    y, ydot, t = 0.0, 0.0, 0.0
    while t < t_max:
        acc = forcing - omega2 * y - damp * ydot
        ydot += acc * dt
        y += ydot * dt
        t += dt
        if y >= 1.0:
            return t                   # breakup!
    return None
 
for We in [6, 10, 12, 20, 40]:
    tau = tab_breakup(We, Oh=0.02)
    tag = f"breakup tau={tau:.2f}" if tau else "survives"
    print(f"We={We:5.1f}  ->  {tag}")

The output reads the critical Weber number straight off.

We=  6.0  ->  survives
We= 10.0  ->  survives
We= 12.0  ->  breakup tau=1.14
We= 20.0  ->  breakup tau=0.74
We= 40.0  ->  breakup tau=0.50

Slide the Weber marks up the ladder and the shape runs from vibration to bag, then sheet-thinning, then catastrophic.

Up to We=10 the overshoot falls short of 1 and the drop survives. At We=12 the first peak barely clears the line. As the Weber grows the breakup time $\tau$ shrinks — a harder hit bursts sooner. Exactly as the physics demands.

The constants in the code comment lock that threshold. With $C_F=1/3$, $C_b=1/2$ and $C_k=8$, $y_{eq}=We/24$. The undamped peak is $2y_{eq}=We/12$. It hits 1 at $We=12$. At $We=6$ the peak is $0.5$; at $We=10$ it is $10/12\approx 0.83$. Both sit under 1, matching the printed survivors. At $We=20$, $y_{eq}=20/24\approx 0.83$, and the overshoot pulls breakup forward to $\tau=0.74$. At $We=40$, $y_{eq}\approx 1.67$, so equilibrium itself is already past 1 and $\tau$ falls to $0.50$. With $dt=0.01$, $\tau=1.14$ is 114 steps and $0.50$ is 50. The damping term is $5\times 0.02\times\sqrt{8}$. Since $\sqrt{8}\approx 2.83$, that coefficient is about $0.283$. The printed $Oh=0.02$ is a thin value sitting right on top of 12.

The ladder of breakup modes#

Crossing the threshold doesn't mean every drop bursts the same way. As the Weber number climbs, the shape of the breakup changes: vibrational → bag → multimode → sheet-thinning → catastrophic. Each stage carves out a rough Weber band that experiments distinguish clearly.

The low rung on the ladder above is the bag. Bag breakup near $We \approx 12$ is the lowest-energy break — the drop's center inflates into a thin membrane and pops. By the hundreds in Weber, the surface itself peels off in waves: catastrophic breakup. The TAB model was originally aimed at the bag-breakup regime. That's why it matches well at low Weber but underpredicts penetration at extreme Weber — like high-pressure injection above 40 MPa. There you layer on KHRT or wave breakup models.

Freezing Weber 12 as an inviscid wall misses a sticky drop#

Undamped, the peak is $2y_{eq}=We/12$. At $We=12$ that peak is exactly 1. The printed run at $Oh=0.02$ still crosses, at $\tau=1.14$, even with the peak shaved a little. Push $Oh$ higher and the same 12 can fall short of 1. Nail the threshold at 12 and drop Ohnesorge, and a honey-like liquid survives while the model has already shattered it.

Deformation changes drag at the same time. If TAB cuts the breakup time from $y$ alone, the path change from drag on a flattened drop never enters. The $\tau=0.50$ at $We=40$ is already outside the bag regime, and above that this spring underpredicts penetration.