SME
Climatology
General circulation
Use this for “Air leaves the equator at rest with the ground and drifts poleward keeping its angular momentum. How fast a westerly wind does it make at 30° N?”
Diagram
Air leaves the equator turning with the ground at ΩR = 465 m/s and keeps its angular momentum.
At
φ
=
30°
the
ground
moves
ΩR
cos
φ
=
402
m/s
(grey)
and
the
air
gains
u
=
ΩR
sin²
φ
cos φ
=
134
m/s
eastward.
This is an example. Type your own values to start a new problem.
u (m/s)
=
ΩR (m/s)
×
sin²(
°)
cos(
°)
Formulas
u
=
ΩR
×
sin
(
φ
°)
2
÷
cos(
φ
°)
134.1185
=
464.6
×
sin
(30°)
2
÷
cos(30°)
Assumptions
•
Air leaves the equator at rest with the ground and keeps its angular momentum, so it turns east as its distance from the axis shrinks.
•
Earth’s rim speed at the equator is ΩR = 7.292 × 10⁻⁵ rad/s × 6,371 km = 464.6 m/s.
•
Real Hadley flow loses momentum to friction and eddies, so real winds are far slower than u.
Step-by-step
WE KNOW
φ = 30°, ΩR = 464.6 m/s
FIND
eastward wind (u)
Step 1 · Find eastward wind (u)
Use u = ΩR × sin(φ°)² ÷ cos(φ°)
The
ring
of
air
shrinks
to
cos
φ
of
its
size
at
the
equator,
so
it
spins
faster
than
the
ground
beneath
it.
u
=
464.6
×
sin
(30°)
2
÷
cos(30°)
sin
(30°)
2
=
0.25
cos(30°) = 0.866
u
=
134.1185
m/s
Check
Put the numbers back in. Both sides should match.
134.1185
=
464.6
×
sin
(30°)
2
÷
cos(30°)
✓
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Coriolis parameter and inertial circles
Topic check
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