SME
Meteorology · Problem type
Tornado winds and the Rossby number
Use this for “A tornado’s centre is 50 hPa below the air around it. How fast is its wind, and does the Coriolis force matter at a 200 m radius, 35° N?”
Chart
V(50) ≈ 64.55
Radius r = 200 m
Latitude φ = 35°
Coriolis parameter f = 8.365 × 10⁻⁵ s⁻¹
Rossby number Ro = 3,858.304
This is an example. Type your own values to start a new problem.
V (m/s)
=
Δp (hPa)
÷
ρ (kg/m³)
Ro
=
V (m/s)
÷
(
f (s⁻¹)
×
r (m)
)
φ
Latitude
example · °
Formulas
V
=
√(100
×
Δp
ρ
)
64.5497
=
√(100
×
50
1.2
)
f
=
2
×
7.292 ×
10
-5
×
sin(
φ
°)
8.365 ×
10
-5
=
2
×
7.292 ×
10
-5
×
sin(35°)
Ro
=
V
f
×
r
3,858.3039
=
64.5497
8.365 ×
10
-5
× 200
Assumptions
•
Cyclostrophic balance: the spin alone balances the pull toward the low centre, so V ≈ √(Δp ÷ ρ) with Δp in Pa.
•
Ω = 7.292 × 10⁻⁵ rad/s, and f = 2Ω sin φ.
•
Ro = V ÷ (fr) compares the spin with the Coriolis force: far above 1, Coriolis is negligible and the vortex can spin either way.
Step-by-step
WE KNOW
Δp = 50 hPa, ρ = 1.2 kg/m³, r = 200 m, φ = 35°
FIND
wind speed (V), Coriolis parameter (f), Rossby number (Ro)
Step 1 · Find wind speed (V)
Use V = √(100 × Δp ÷ ρ)
The
spin’s
pull,
ρV
²
r
,
balances
the
pressure
push,
Δp
r
,
so
ρV²
is
Δp
in
Pa;
take
the
square
root.
V
=
√(100
×
50
1.2
)
V
=
√(
5,000
1.2
)
V
=
√
4
,167
V
=
64.549722
m/s
Step 2 · Find Coriolis parameter (f)
Use f = 2 × 7.292 × 10⁻⁵ × sin(φ°)
The Coriolis parameter is twice Earth’s spin rate times the sine of the latitude.
f
=
2
×
7.292 ×
10
-5
×
sin(35°)
f
=
2
×
(7.292
×
10
-5
)
×
0.57358
f
=
0.00014584
×
0.57358
f
=
8.365039 ×
10
-5
s
-1
Step 3 · Find Rossby number (Ro)
Use Ro = V ÷ (f × r)
The
spin’s
pull,
V
2
r
,
over
the
Coriolis
force,
fV,
is
V
fr
.
Ro
=
64.549722
8.365039 ×
10
-5
× 200
Ro
=
64.5497
0.01673
Ro = 3,858.304
Check
Put the numbers back in. Both sides should match.
64.549722
=
√(100
×
50
1.2
)
✓
8.365039 ×
10
-5
=
2
×
7.292 ×
10
-5
×
sin(35°)
✓
3,858.304
=
64.549722
8.365039 ×
10
-5
× 200
✓
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