Chapter 9
Winds and Pressure
What we know so far:
 Air pressure is the weight of air
above a given level
 Pressure decreases as we go aloft
Consider
1
2
4
3
Same Temp.
Same Pressure
cold warm
Same Pressure
AND Same mass
- Warmer  less dense 
grows taller
- Colder  more dense 
shrinks.
So…
 It takes a shorter column of
cold, more dense air to exert
the same surface pressure as a
taller, warmer column of less
dense air
But…
 Warm air aloft is normally
associated with high pressure;
cold air with low pressure.
“Why?!? ” you ask.
THE IDEAL GAS LAW
Recall:
PV=nRT
P = pressure of gas
V = volume of gas
n = moles of gas
R = specific gas constant
T = temperature of gas
The Meteorology Version…
P = (n / V) R T
n =
V
where,
mass = density = 
volume
So…
p =  Rd T
p = pressure of air (Pascals)
 = density of air (kg/m3)
Rd = specific gas constant of
dry air (287.06 J kg-1 K-1)
T = temperature of air (in
Kelvins)
Rd = is a proportionality constant.
1. p  T x 
 Hold T constant
 Therefore, if p increases, then 
must increase also
 At same temperature, higher
pressure is more dense
 If you’re also at same elevation,
then in this case it follows that
Air over high pressure is more
dense than that over a low
pressure area.
 Surface pressure rises when the
wind places more mass overhead
(net convergence aloft).
 Surface pressure decreases when
wind removes mass from column
over head (net divergence).
2. p  T x 
 now hold p constant.
 As temperature increases, 
decreases
 Therefore at a given pressure:
- Cold air is more dense
- Warm air is less dense
An Ideal Gas Law Example
Given:
p = 500 mb = 500 hPa =
50000 Pa
 = 0.690 kg / m3
Rd = 287.06 J kg-1 K-1
Find:
T at 500 mb
Begin with: p =  Rd T
1) Plug in:
50000 Pa = 0.690 (kg / m3) x
287.05 (J / kg K) x (T)
2) And rearrange:
50000 (kg / m s2)
T = -----------------------------198.06 (kg / m3)( kg m2 / s2 kg K)
T = 252.4 K
Instrumentation
 Mercury barometers are the most
accurate
 Aneroid barometers
- Barograph
- Altimeter
 Electronic barometers
Station vs. Sea Level Pressure
Station
Sea-level
pressure
station
pressure
ocean
1. READ station pressure
2. CORRECT for:
 Temperature
 Gravity
 Instrument error (surface
tension, friction / viscosity)
3. REDUCE to sea-level
Why???
Pressure changes s more rapidly in
vertical than in horizontal. Small
changes in elevation yield large
changes in pressure.
1/2km
--- 900 mb
1000 mb
NY
900 mb
Miami
1600 km
Consider an isolated mountain
925mb
1000mb
1000mb
L
1000mb
925mb
Without reduction, there always
appears to be a low over the mountain
High heights on a constant pressure
chart are proportional to higher than
normal pressures at a particular
altitude.
PRESSURE UNITS
 Inches of mercury (Hg)
 Millibars (mb)
 Hectopascals (hPa)
Sea level standards
29.92 in.
1013.25 mb
1013.25 hPa
Newton’s Laws of Motions
Newton’s First Law – Consider a
body on which no NET force acts. If
the body is at rest, it will remain at
rest. If the body is moving with a
constant velocity it will continue to
do so.
Newton’s Second Law -  F = m a
The vector sum of all the forces
that act on the body is equal to the
product of the body’s mass and
acceleration.
(Special Case of 1st Law)
Acceleration – change in
velocity (magnitude OR
direction [speed])
Newton’s Third Law –
MA
A
FAB
F AB = -FBA
FAB
mB
B
Basic Forces Influencing the
Wind
1. Pressure Gradient Force
2. Coriolis Force
1. Pressure Gradient Force
 Pressure Gradient - the amount
of pressure change that occurs
over a given distance.
 P. G = change in pressure
Distance
 P. G. = p /d
 Example
500 km
1004mb
1021mb
P G = 1021 mb – 1004 mb
500 km
= 17 mb
500 km
 large p over a short d implies a
steep pressure gradient.
 Small p over the same d
implies a weak pressure gradient
L
-an open valve, and higher
fluid pressure will force water
through conduit until
equilibrium is attained.
 Fluid pressure equal in all
directions
 Fluid at tube in A 1 has a higher
pressure than at B 2
 Fluid flows from A to B
1. The greater the pressure
difference
2. The greater the force
3. The faster the fluid moves
Pressure Gradient Force – a net force
acting on a parcel due to a
horizontal change in pressure
 it is directed perpendicular to
the isobars, an always FROM
HIGH TO LOW
2. Coriolis Force
 Formulated by Gaspard Coriolis,
1800’s
 An apparent force which acts to
turn moving objects to the
right
 Consider 2 rotating platforms
A
B
 A has no rotation and no
deflection
 B has rotation and deflection
 Draw a line from A to B, it is a
straight line
 Draw a line toward B’ from A’.
Your hand moves straight but
ends up at B”
 Think of A’ as North Pole
 Think of edge as equator
 Parcel actually moves in a
straight path, but EARTH
TURNS UNDER it.
 Since Earth is our frame of
reference, the parcel flow
appears to turn right.
 This happens for ALL motion
North, South, East, and West.
Controls on the Coriolis
Force
1. The rotation of the Earth ()
 Angular velocity
 Constant
 7.292 * 10-5 radians / second
2. An object’s speed
 The magnitude of V  C.F.
 Stronger V, more deflection
 Weaker V, less deflection
3. Latitude
 Zero at equator (small
deviations)
 Maximum at poles (large
deviations)
 Does not vary along a latitude
circle.
Coriolis Force = 2 m  V sin 
m = mass of object (in kg)
 = angular velocity of earth
(7.292 * 10-5 radians / second)
V = velocity of object
(m/s), e.g (10m/s)
sin  = sine of latitude, e.g.
(40)
C.F. = 2 ( 1 kg)(7.292 * 10 –5/sec)(10m/sec)(sin 40)
= kg * m/s2 = Newton = Unit of
force
If we assume it is the C. F. for a
standard 1 kg mass (m), then we
have
C. F. = 2 m  V sin 
m
m
C. F. = 2  V sin 
=2 (1/sec)(m/sec)( )
=m/s2
Therefore, Coriolis acceleration!
Geostrophic Wind
Balance of C.F. and P.G.F.
PGF
C.F.
Wind blows parallel to isobars
(evenly-spaced) at constant speed.
How do we achieve this balance?
1. Place a parcel in a pressure
gradient; it will begin to move
due to the PGF
2. The parcel accelerates, and the
C.F. does in proportion.
3. C. F. just balances PGF
Acceleration ceases, since NET force
is now zero (C.F. and PGF balance.)
F = m a = 0
 no net force
 Newton’s first law holds here,
in the absence of friction
 The body is now in motion
It is not acted on by any net force.
So…
It continues parallel to isobars in
balance between C.F. and PGF
Expression for Vg
Vg =
1
p
 (2  sin ) d
or…
= 1 p
f d
Gradient Winds
 Wind that blows at constant speed
parallel to curved isobars / heights
 Recall, now, that acceleration
implies a change in speed or
direction or both
 So we have a geostrophic wind in
curved flow
BUT…
Curved flows – involve extra force;
because change in direction is an
acceleration
F=ma
Therefore, there must be another
force.
Due to this acceleration (centripetal
accerleration) the extra force is
knows as the centripetal force.
L
P. G. F. > C. F.
So, Net force is the inward
directed centripetal force.
H
C. F. > P. G. F.
So, net force is the inward
directed centripetal force.
Cyclostrophic – balance in between
PGF and centripetal (tropics where
C.F. is WEAK)
Friction
 Boundary layer; 0 to ~1000m
 Drag on the wind by the
underlying surface
 Slows wind
 Reduced Coriolis
Process:
Friction added  V decreases 
C.F. decreases  PGF no change
 Turn to the left
Geostrophic, NO friction
Total wind, WITH friction
 larger over rough terrain,
35 to 40
 smaller over smooth
terrain, 10 to 15
H
L
As we go aloft, friction
becomes less important.
(Greater distance from land surface.)
Average  = 30
Buys Ballot’s Law – If we
stand with our backs to
the wind, then turn
clockwise about 30,
lower pressure will be on
our left.
The HYDROSTATIC
Assumption:
 In the vertical, there is usually a
balance between gravity and the
vertical PGF
PGFvert + g = 0
PGFvert = -g
(1/)(p/z) = -g
p/z = -g
z = 1000m
 = 1.18 kg /m3
g = 9.81 m / s2
p = -(1000m)( 1.18 kg /m3)( 9.81
m / s2)
= -11575.8 (kg / m s2)
= -11575.8 N/m2 = Pa
=-115.758 mb = -116 mb