Hail Formation and Forecasting
How large will the hail be and where will it fall? Hailstone growth depends on updraft strength, temperature profile, and liquid water content. This model derives terminal velocity equations, implements accretion models, demonstrates wet vs dry growth regimes, and forecasts maximum hail size from storm parameters.
Prerequisites: terminal velocity, accretion rate, wet growth, hail size prediction
1. The Question
Will this supercell produce baseball-sized hail?
Hail definition:
Ice particles ≥5 mm diameter (pea-sized or larger).
Damage thresholds:
Pea (6 mm): Minimal crop damage
Quarter (25 mm, 1 inch): Severe criteria, car dents
Golf ball (44 mm, 1.75 inch): Car windshields broken
Baseball (70 mm, 2.75 inch): Roof damage, vehicle totaled
Softball (114 mm, 4.5 inch): Structural damage
Record: 8 inch (203 mm), Vivian SD 2010
Annual losses (USA): ~$1-2 billion
Applications:
- Agriculture (crop insurance)
- Aviation safety
- Severe weather warnings
- Building/vehicle damage assessment
- Climate studies
2. The Conceptual Model
Terminal Velocity
Balance drag and gravity:
\[F_d = F_g\] \[\frac{1}{2} \rho_a C_d A v_t^2 = m g\]Where:
- $\rho_a$ = air density (kg/m³)
- $C_d$ = drag coefficient (~0.6 for sphere)
- $A$ = cross-sectional area (m²)
- $v_t$ = terminal velocity (m/s)
- $m$ = mass (kg)
- $g$ = 9.81 m/s²
For sphere:
\[A = \pi r^2, \quad m = \frac{4}{3}\pi r^3 \rho_h\]Solving:
\[v_t = \sqrt{\frac{8 r g \rho_h}{3 C_d \rho_a}}\]Hail (ice density $\rho_h = 900$ kg/m³):
\(v_t \approx 9 \sqrt{r}\) (m/s, $r$ in meters)
Example: 5 cm diameter (r = 0.025 m)
\[v_t = 9 \sqrt{0.025} = 9 \times 0.158 = 1.42 \times 9 \approx 14 \text{ m/s}\]Updraft must exceed 14 m/s to suspend this hailstone!
Growth Regimes
Dry growth:
Temperature < -40°C or low liquid water content.
Collected droplets freeze instantly.
Opaque ice (trapped air bubbles).
Wet growth:
Warmer temperature (-10 to 0°C) or high liquid water content.
Droplets form liquid layer before freezing.
Clear ice (no air, denser).
Transition:
Depends on accretion rate vs heat removal rate.
Spongy hail:
Alternating dry/wet layers (trajectories through different cloud zones).
Accretion Rate
Mass growth:
\[\frac{dm}{dt} = E \cdot A \cdot W \cdot v\]Where:
- $E$ = collection efficiency (~0.8)
- $A$ = cross-sectional area
- $W$ = liquid water content (g/m³)
- $v$ = relative velocity (terminal fall + updraft)
For suspended hailstone ($v \approx$ updraft speed):
\[\frac{dm}{dt} = 0.8 \cdot \pi r^2 \cdot W \cdot w\]Typical: $W = 1-5$ g/m³, $w = 20-50$ m/s
Example: $r = 0.02$ m, $W = 3$ g/m³, $w = 30$ m/s
\[\frac{dm}{dt} = 0.8 \times 3.14 \times 0.02^2 \times 3 \times 30 = 0.091 \text{ g/s}\]Residence time in updraft: 5 minutes = 300 s
\[\Delta m = 0.091 \times 300 = 27.3 \text{ g}\]Final mass: ~50-100 g (golf ball to baseball)
3. Building the Mathematical Model
Maximum Hail Size
Balance condition:
Hailstone suspended when $v_t = w$ (updraft).
From terminal velocity:
\[r_{max} = \frac{w^2}{81 g}\](Using $v_t = 9\sqrt{r}$, solved for $r$)
Simplified:
\(r_{max} = \frac{w^2}{800}\) (m, $w$ in m/s)
Example: $w = 40$ m/s
\[r_{max} = \frac{1600}{800} = 2 \text{ cm} = 20 \text{ mm diameter}\]Quarter-sized hail (severe threshold)
For softball (57 mm radius):
\[w = \sqrt{800 \times 0.057} = \sqrt{45.6} = 67.5 \text{ m/s}\]Extreme updrafts required!
MESH (Maximum Expected Size of Hail)
Radar-based estimate:
Uses vertically integrated reflectivity.
\[MESH = 2.54 \times W^{0.5}\]Where $W$ = integrated reflectivity (g/m²).
Severe Hail Index (SHI):
\[SHI = 0.1 \int_{H_0}^{H_{-20}} (Z - Z_0) \, dh\]Where integration from freezing level to -20°C level.
MESH from SHI:
\[MESH = 2.54 \times SHI^{0.5}\]MESH = 25 mm: Severe hail likely
MESH = 50 mm: Very large hail
MESH > 75 mm: Giant hail (baseball+)
Hail Swath
Falling trajectory:
Hail falls from updraft maximum.
Lateral displacement:
\[\Delta x = \frac{v_t}{w} \times H \times \frac{U}{w}\]Where:
- $H$ = fall distance
- $U$ = storm-relative wind
Typical: 5-20 km downwind of updraft maximum
Swath width: 1-10 km (updraft width + spread)
4. Worked Example by Hand
Problem: Predict maximum hail size and swath location.
Storm parameters:
- Maximum updraft: 45 m/s
- Updraft top: 12 km
- Freezing level: 4 km
- Storm motion: 240° at 15 m/s
- Environmental wind at 6 km: 270° at 20 m/s
Radar:
- Peak reflectivity: 65 dBZ
- Reflectivity top: 14 km
- SHI: 150
Calculate maximum hail size and impact location relative to updraft.
Solution
Step 1: Maximum suspended hail
\[r_{max} = \frac{w^2}{800} = \frac{45^2}{800} = \frac{2025}{800} = 2.53 \text{ cm}\]Diameter: 5.06 cm = 2 inch (golf ball)
Step 2: MESH from radar
\[MESH = 2.54 \times 150^{0.5} = 2.54 \times 12.25 = 31.1 \text{ mm}\]1.25 inch (between quarter and golf ball)
Slightly lower than maximum (radar underestimates largest stones)
Step 3: Terminal velocity for 2-inch hail
\[v_t = 9 \sqrt{0.0253} = 9 \times 0.159 = 14.3 \text{ m/s}\]Step 4: Fall time
From updraft top (12 km) to ground:
Assume average updraft weakens: 30 m/s upper, 20 m/s lower
Ascent slowed but falls eventually.
Fall distance after release: 12 - 4 = 8 km (below freezing)
Fall time:
\[t = \frac{8000}{14.3} = 560 \text{ seconds} \approx 9 \text{ minutes}\]Step 5: Storm-relative wind
At 6 km: 270° at 20 m/s
Storm: 240° at 15 m/s
Relative: ~15 m/s from west (simplified)
Step 6: Lateral displacement
\[\Delta x = 15 \times 560 = 8400 \text{ m} = 8.4 \text{ km}\]Hail falls 8-9 km east of updraft maximum
Step 7: Impact assessment
Golf ball hail expected:
- Car windshields broken
- Roof damage possible
- Crop total loss
Swath: 8-10 km east of storm core, width 2-3 km
Warning: Large hail imminent, take shelter
5. Computational Implementation
Below is an interactive hail growth simulator.
Max hail diameter: -- mm
Hail category: --
Terminal velocity: -- m/s
Damage potential: --
Observations:
- Hail growth accelerates (larger stones collect more droplets)
- Updraft strength limits maximum size
- High liquid water content accelerates growth
- Longer residence time produces larger hail
- Terminal velocity increases with square root of diameter
- Severe threshold (25mm) reached in strong updrafts
Key insights:
- Extreme updrafts (>60 m/s) required for softball hail
- Growth rate nonlinear (larger stones grow faster)
- Minutes in updraft sufficient for destructive hail
- Terminal velocity determines suspension requirement
6. Interpretation
Crop Damage Assessment
Hail impact on agriculture:
Pea-sized (6-13 mm):
- Leaf damage, some defoliation
- 10-30% yield loss
Dime-quarter (13-25 mm):
- Stem breakage, severe defoliation
- 30-60% loss
Golf ball+ (>44 mm):
- Total crop destruction
- 80-100% loss
Economic:
2017 Colorado hailstorm: $2.2 billion crop damage (corn, wheat)
Insurance:
Crop insurance covers hail explicitly (common peril).
Payouts based on damage assessment (percentage loss).
Hail Suppression
Cloud seeding:
Inject silver iodide or other nuclei.
Theory: Increase ice particles → compete for water → smaller hail
Evidence: Mixed/inconclusive
Alberta Hail Project (1956-1985): 15-20% reduction claimed
Recent studies: No significant effect detected
Operational: Some programs continue (insurance-funded)
Vehicle/Property Damage
Hail damage costs (USA):
$1-2 billion annually
Golf ball hail:
- Car windshields: $300-800 replacement
- Body damage: $1,000-5,000 repair
- Total vehicle loss: Possible with repeated impacts
Roof damage:
- Asphalt shingles: Severe damage >44 mm
- Metal roofs: Denting >25 mm
- Solar panels: Crack/break risk
Building codes:
Some regions require hail-resistant roofing (impact-rated).
7. What Could Go Wrong?
Giant Hail Not Detected
Radar MESH underestimates largest stones.
Reasons:
- Hail concentration in small area
- Beam averaging
- Attenuation
Example - 2010 Vivian SD:
8-inch hailstone (record), but radar MESH ~75 mm (3 inches)
4× underestimate!
Solution: Spotter reports, damage surveys verify
Wet Hail Misidentified
Wet growth hail (clear ice) has higher density.
Radar reflectivity similar to smaller dry hail.
Can be more damaging (heavier, denser) than radar suggests.
Hail-Rain Confusion
Large raindrops can produce moderate reflectivity.
Without dual-pol: Hard to distinguish
Dual-polarization: ZDR, ρHV differentiate
ZDR:
- Rain: +2 to +4 dB (oblate drops)
- Hail: 0 to -2 dB (tumbling, irregular)
ρHV:
- Rain: >0.98
- Hail: 0.90-0.95 (mixed shapes)
Hail Melting
Long fall distance (high cloud tops) → melting
Surface reports: Rain, but hail aloft
Melting level critical:
Low (2-3 km): Hail reaches surface
High (4+ km): Melts before impact
Wet-bulb zero height better predictor than simple 0°C level
8. Extension: Hail Climatology
Geographic distribution:
Hail Alley (USA): Wyoming, Colorado, Nebraska
Argentina Pampas: World’s highest hail frequency
Northern India: Large hail events
Factors:
- High terrain (orographic lift)
- Strong CAPE
- Dry air aloft (promotes supercells)
Seasonal:
Spring-early summer peak (transitional seasons, strong shear + instability)
Trends:
Some evidence of:
- Fewer hail days
- But more extreme events (larger stones)
- Difficult to detect trends (sparse observations)
Satellite potential:
GOES-16 Overshooting Top detection correlates with large hail
9. Math Refresher: Drag Force
Stokes vs Newton Drag
Stokes (laminar, small Re):
\[F_d = 6 \pi \mu r v\]Newton (turbulent, large Re):
\[F_d = \frac{1}{2} \rho C_d A v^2\]Reynolds number:
\[Re = \frac{\rho v D}{\mu}\]Hail: Re ~ 10⁴-10⁵ → Newton drag applies
Sphere drag coefficient:
$C_d \approx 0.4-0.6$ (depends on surface roughness)
Summary
- Hail growth requires strong updrafts exceeding terminal velocity of growing stones
- Maximum hail size scales with updraft velocity squared via r_max = w²/800 formula
- Terminal velocity increases as square root of radius for ice spheres
- Wet growth produces clear dense ice while dry growth creates opaque spongy hail
- MESH radar product estimates hail size from integrated reflectivity
- Severe hail threshold 25mm (1 inch) causes significant crop and property damage
- Golf ball hail (44mm) breaks windshields requiring 35+ m/s updrafts
- Record 8-inch hail required extreme updrafts exceeding 70 m/s
- Dual-polarization radar improves hail detection via ZDR and correlation coefficient
- Annual agricultural and property losses exceed $1-2 billion in USA alone