Transcript Chapter 10

Chapter 10
Geostrophic Balance
(1) Wind and Height Gradients
On a contour analysis of a pressure surface:
Strong winds
associated with contours
close together; weak
winds with contours
farther apart.
Winds tend to blow
(nearly) parallel to height
contours.
Winds tend to be oriented such that higher heights are 90o to
the right (in the northern Hemisphere) of the direction toward
which the winds are blowing. (left in S. H.)
Winds that follow this are called Geostrophic Winds.
(2) The Geostrophic Equation
Expresses the magnitude of the wind speed
as a function of the geopotential height
gradient on a constant pressure surface.
go
vg 
h Z
f
Where, go = gravitational
acceleration,
Coriolis parameter = 2Ωsin(lat).

Ω = 7.292 x 10-5 s-1
f ranges from -1.4 x 10-4 s-1 at South Pole to
+1.4 x 10-4 s-1 at the North Pole.
f ≈ about 1 x 10-4 s-1 for mid-latitudes.
f=
Consider: If the height gradient is 3
decameters (30 meters) per degree of
latitude, how strong would the geostrophic
wind be?
go
9.8ms 2
30m
1o lat
m  1.943kt  51.29kt
vg 
h Z 



26.4
s
f
1 104 s 1 1o lat 1.11 105 m
1m s
Notice: 1o latitude = 1.11 x 105 m

This is an average value at mid-latitudes.
If the x-axis is oriented
parallel to the geostrophic
wind, the height gradient
must be oriented in the
negative y-direction with
lower heights in the
positive y-direction.
The magnitude of the
height gradient is the
magnitude of:
Z
To get the wind going iny
the positive x-direction, it
must be multiplied by -1.

ug  
go Z
f y
vg 
go Z
f x
These equations
can be used to
compute the
components of the
geostrophic wind in
all circumstances,
not just when the x
component is
parallel to the wind
direction.
Magnitude


31
u g 2  v g 2 Direction arctan  90o  180o  310.7o
36 
The u-component of the
geostrophic wind is dependent u g   go Z
f y
on the height gradient with
respect to y.
The v-component is dependent
go Z
vg 
on the height gradient with

f x
respect to x.
The geostrophic wind is parallel to the height
gradient; i.e., 90o to the right of the height
gradient (in the northernhemisphere).
Remember cross products.
a  b  a b sinangle between

If a horizontal vector is crossed with a vertical
vector, the result will be a horizontal vector at
right angles with the original horizontal vector.
If the vertical vector is a unit vector, the
magnitude of the resulting cross product
vector will equal the magnitude of the original
horizontal vector.
Thus we have:
go
vg 
k  h Z
f

The index finger is pointing
upward, parallel to the unit
vector k.
The palm and three fingers are
pointing “curled” toward
increasing height gradient (the
vector pointing toward
increasing height gradient).
The thumb is pointing out from
the page parallel to the
geostrophic wind vector.
Right hand Rule: All fingers point in direction of k. Curl fingers
in direction of increasing height gradient vector. Thumb points in
direction of resultant of cross product.
An equivalent geostrophic equation is:


f k  v g  goh Z
Meaning, crossing the vertical unit
vector with the geostrophic winds
results in a vector oriented toward lower
heights.
(3) Geostrophic Divergence
Remember, divergence is defined as:
u v
h  v 

x y
And, the two components of the Geostrophic wind
are:
g Z
g Z
ug   o
vg  o
f y
f x

Then if we take the derivative of ug with respect to x
and the derivative of vg with respect to y, we can get
thedivergence byadding them together.
u g
 go Z 
  

x
x  f y 


v g go Z 
 

y  f x 
Adding the two together gives:
u g v g
 g Z   go Z 

   o
 

x
y
x  f y  y  f x 
Since go is a constant, it can be brought out
and since f is a function only of y (latitude),

then using the chain rule gives:
u g v g
go  Z  go  Z 
Z  1 


  go
 
 
x
y
f x y  f y x 
x y  f 

Since x and y are independent variables, the
order of differentiation can be reversed and
the first two terms cancel each other.
Thus, the divergence of the geostrophic wind
equation reduces to:
u g v g
Z  1 

 go
 
x
y
x y  f 
Since f changes so slowly with latitude, the
right-hand side is typically much smaller in

magnitude
than any typical values of
observed divergence of the actual wind.
Thus, divergence of the geostrophic wind is
essentially zero. It is non-divergent.
Therefore:
(1) the primary wind patterns in the atmosphere
are not, by themselves, associated with upward or
downward motion.
(2) departures from geostrophic balance are
required for strong upward and downward motion.
(3) it is very difficult to see convergence and
divergence on constant pressure maps.
(4) Because geostrophic winds are not divergent
or convergent, constrictions on contours means
the air must flow faster since the air, in
geostrophic balance, is essentially confined
between the contours.
If the wind obeys
geostrophic
balance, air
between two
height contours at
a particular level
must stay between
those two height
contours and stay
on that level as
well.
(4) Geostrophic Advection
Remember, advection can be expressed as a
dot product of two vectors with one of the
vectors being wind (v)and the other the “del”
of a scalar quantity (A) such as temperature
(the gradient of A).
The result is a scalar quantity.
v   h A   v  h A cos

This can be written as:
v   h A  v cos h A

Meaning, advection is proportional to the
projection of the wind (⎮v⎮) onto the
gradient (⎮∇hA⎮) multiplied by the
magnitude of that gradient(⎮∇hA⎮).
For geostrophic wind, it is written:
v   h A   v g  h A cos

Meaning, the projection of the gradient
(⎮∇hA⎮)onto the direction of the
geostrophic wind (⎮vg⎮) multiplied by the
magnitude of that wind.
v   h A   v g  h A cos

So, the advection of “A” by the
geostrophic wind, vg, is directly
proportional to:
(1) the magnitude of the geostrophic wind which is inversely proportional to the
spacing between height contours.
(2) the magnitude of the gradient of “A” which is inversely proportional to the
spacing between successive contours
(isopleths) of “A.”
Then, the smaller the areas bounded by the
height contours and isopleths of “A”, the
greater the advection.
Thus, we could say that
the advection of “A” by
the geostrophic wind is
inversely proportional to
the area of the
parallelogram formed by
the height contours and
the isopleths of “A.”
We could also
say that the
advection is
greatest when
the “spatial
density of
intersections” of
contours and
isopleths of “A”
is greatest.
(5) Streamfunctions
Any vector that is non-divergent can be
expressed in terms of a scalar quantity known
as a stream function; Ψ.
v  k  
The geostrophic
wind vector is
written as:
Therefore, the
geostrophic stream
function must 
be:
go
vg 
k  h Z
f
go
g 
Z
f
If “f” is considered constant.
Then, if the wind is non-divergent as it
is for geostrophic wind:
It is not necessary to plot the wind vectors
to see the wind pattern. Contours of the
stream function will represent the wind.
Where the stream function contours are close
together, the wind speed is strong; and farther
apart where the wind is weak.
All stream function contours (streamlines)
should parallel the height contours.
Adding arrows to the height contours produces
streamlines (stream function contours).
(6) Nature of Geostrophic Balance
Consider the horizontal forces acting on
a volume of air above the near surface
region (where friction becomes
significant).
The forces are the Pressure Gradient
Force, F   m dP F   m dP
x
y
 dx
 dy
The Coriolis Force.


The Pressure Gradient Force is directed
away from high pressure and toward low
pressure.
In other words, it is opposite to the pressure
gradient (which is directed from low pressure
toward high pressure).
A gradient
vector
always
points in the
direction of
the
greatest rate of increase.
The Coriolis Force - keeps the air from
moving directly from high pressure to low
pressure.
Its magnitude is proportional to the Coriolis
parameter:
2 sin
Its magnitude is proportional to the horizontal wind
2v sin
speed.
Its direction in the northern/southern hemisphere
o to the right/left of whatever direction the
is 90
horizontal wind is blowing.
Its magnitude
is so weak it would take several

hours for the Coriolis force to cause a substantial
change in the wind direction if it were the only
force present.
Consider a parcel of air moving parallel to the
contours on an upper-air chart.
The pressure gradient force is oriented roughly
90o to the direction the parcel is moving and
toward lower heights.
The Coriolis force is oriented 90o to the direction
the parcel is moving and toward higher heights.
If the speed of the wind is just right, the pressure
gradient force and the Coriolis force are of the
same magnitude and pointed opposite to each
other.
(7) Attaining
Coriolis Nirvana
If the forces are not in balance, if something
changes; e.g., the pressure gradient force,
the parcel will accelerate in the direction of
the stronger force.
How does the parcel stay in geostrophic
balance?
Consider the ways the wind might be
out of geostrophic balance.
(1) Component of wind across contours is
zero but the component along the contours
is too weak.
(2) Component of wind across contours is
zero but the component along the contours
is too strong.
(3) Component of wind along the contours is just
right but there is a component of wind across
contours from high to low pressure (heights).
Component of wind along contours is just right, but
(4) there is a component of wind across contours
from low to high pressure (heights).
Consider case (1),
the wind speed is
too weak.
Departures from geostrophic balance
cause a reorientation of the forces
which causes the air motion to speed up
or slow down and move back toward
geostrophic balance.
The time spent going to fast is the same
as going to slow, so in the average, the
air is in geostrophic balance.
(8) Nature of the Coriolis Force
Consider a ball of mass, m, attached to a string and
whirled through a circle of radius r at a constant
angular velocity w.
v is the tangential velocity
of the ball.
ΔΘ is the angle through
which the ball moves (the
change in the angle
measured from the zero
angle position).
r is the radius from the
axis of rotation to the ball.
Δv is the change in velocity (acceleration - change in
direction) of the ball as it moves in a circular path.
v + Δv is the new velocity of the ball (I.e., because it
changed direction).
From the point of view of an observer in fixed space,
the speed of the ball is constant, but its direction of
travel is continuously changing, so its velocity is not
constant; i.e., it has an acceleration.
To compute the
acceleration, we consider
the change in velocity,
v, which occurs for a
time increment, t, during
which the ball rotates
through an angle .
 is also the angle between the vectors v and v +
v.
v  v 
Remember: S = rΘ
If we divide by t and note that in the limit as
, v is directed
t  0toward the axis of rotation, (it
is a -v), then:
dv
d
 v
dt
dt
Since, tangential and angular velocity are:
d
v  wr and
w
dt
Then,
dv
2
 w r
dt
dv
2
 w r
dt
Or, since
And,
v
v  wr then, w 
r
v 2 
dv
v2
a
  2 r  
dt
r
r 

The acceleration of the ball (directed toward
the axis of rotation) is equal to the square its

linear
velocity divided by the distance from
the axis of rotation.
Therefore, when viewed from fixed
coordinates, the motion is one of
uniform acceleration directed toward
the axis of rotation and equal to the
square of the angular velocity times the
distance from the axis of rotation. This
acceleration is the Centripetal
Acceleration.
The resulting force on the ball to
produce this acceleration toward the
axis of rotation is the Centripetal
Force.
Now, suppose the motion is observed in a
coordinate system rotating with the ball; e.g.,
the earth.
Now, the ball appears stationary (if it is
rotating at the same rate as the earth, but the
centripetal force is still acting on the ball,
namely the pull of the string.
In order to apply Newton’s second law to
describe the motion relative to this rotating
coordinate system, we must include an
additional (apparent) force, the Centrifugal
Force, which just balances the force of the
string on the ball. It is acting outward, away
from the axis of rotation.
Thus, the Centrifugal Force is equivalent to
the inertial reaction of the ball on the string and
is equal but opposite to the centripetal
acceleration.
In a fixed system, the rotating ball undergoes
constant Centripetal acceleration in response
to the force exerted by the string.
In a moving system, the ball is stationary and
the force exerted by the string is balanced by a
Centrifugal force.
The Centrifugal force can be considered the
force necessary to balance all the other forces
acting on the ball.
Since, for air at rest on Earth’s surface, M =
wr,
Where M represents the linear wind velocity
and direction vector, (Note: here we are letting
M represent the wind whereas previously, we
were letting v represent the velocity of the ball.
In the following, we are letting v represent a
component of the wind.)
Then
and
2
M

r
Fcentrifugal
dM M


  2r
dt
r
m
Remember, for centripetal acceleration:

v ball
w  
r
2
dv
v
 w 2r   ball
dt
r
To get the components in the x- and y-direction,
remember that
M 2  u2  v 2
and to get the sign right, (Centrifugal force
going in the proper direction), consider the
following low pressure center.

Fx CN
VV
By writing the equations as:
 s 
m
r
and using the sign below, the
UU
direction of the force is proper. Fy CN
 s 
m
r
Direction and sign
the Centrigugal Forces
should have.
2
F y centrifugal = + U
r
M = +V
M = -U
Fx centrifugal= -
V2
r
L
M = -V
Fx centrifugal= +
M = +U
2
F y centrifugal = - U
r
V2
r
Effective Gravity
A particle of unit mass at rest on the
surface of the earth observed in a
reference frame rotating with the earth, is
subject to a centrifugal force, 2R,
where  is the angular speed of rotation
of the earth and R is the distance of the
particle from the axis of rotation.
Thus, the weight of the
particle of mass, m, at
rest on the Earth’s
surface will generally be
less than the
gravitational force, mg*,
because the centrifugal
force partly balances the
gravitational force.
g  g *  R
2
The spheroid shape of the Earth makes g
(the Effective gravity) directed normal to the
level surface.
Coriolis Force: east-west motion
Consider an object moving eastward in the
northern hemisphere with velocity U with
respect to the Earth’s surface.
For an observer NOT on the Earth, its
tangential (linear) velocity is:
v  R(object s angular velocit)y R  U
Since it is moving in a circular path, its
Centripetal acceleration is toward the axis of
rotation of magnitude:

R  U
R
2
or
R U 
R
2
Expanding gives:
R  U 
R
2
2
 2 R2  2RU  U 2
U

 2 R  2U 
R
R
For an observer on the Earth, the Earth’s
surface seems stationary, so there is no
tangential velocity due to the Earth’s
motion, only due to the object’s motion.
The centripetal acceleration on the
object would be just:
U2
U2

or magnitude
R
R
The discrepancy between the magnitude of
2

R

U

U2
2
the two is:
  R  2U 
R
 R  2U
2
R
U2

R
The Centrifugal Force (acceleration) we
know is:
2
 R
The Coriolis Force (acceleration) is:
2U
also directed outward, like the Centrifugal Force.
This outward directed
Coriolis Force
(acceleration) can be
divided into components
in the vertical and
parallel to the Earth’s
surface along the
meridional directions
(North - South).
The horizontal force causes a change in the direction
of movement of the object (as perceived by someone
on a moving coordinate system).
That change in direction for our object moving
eastward with velocity U, is toward the south,
i.e., to the right of the direction of motion in the
northern hemisphere. (In other words, it is
causing a change in the north-south velocity of
the parcel).
The vertical component is much smaller than
the gravitational acceleration (force) and its
only effect is to cause a very minor change in
the apparent weight of an object.
The term
is called the
f c  2sin
Coriolis parameter.
It is 
positive in NH and
negative in SH.
Objects moving north or south.
Consider an object, of
unit mass, initially at rest
on the surface of the
Earth at latitude . It
starts moving southward
and moves .
As the object moves equatorward, it will
conserve its angular momentum in the
absence of torques in the east-west direction.
Its initial angular
momentum,
L, is:
2
  w R, where w is
initially equal to zero.
For other than unit mass
it is:
L  I   w   m  w R
2
Since the distance to the axis of rotation, R, increases
to R + R for the object moving equatorward, a
relative westward velocity, must develop, (w becomes
smaller - or negative), if the object is to conserve its
angular momentum.
We can write the following showing conservation of
angular momentum.

U 
2
R   
 R  R

R  R 
2
Where,
2is the angular momentum of the unit
Rat rest on the surface of the earth at
mass object
latitude .

is the change in west-east velocity which must
U because R is increasing to
occur
.
R + R
Remember that tangential velocity, U, is related
to
angular velocity by:
, so angular velocity for
the object is:
U  rw

U
w
R  R

If we expand the right side we get:
2
2
R

U
2R

R

U

R
U
R 2  R 2  2RR  R 2 


R  R
R  R
R  R

If we neglect second-order, and higher
differentials, we get:
0  2R R R U
2
and,

U  2R
2
We can see that
a
Arc length =
and R  a sin
So, U  2a sin
Dividing by a unit time
increment
gives the

horizontal acceleration in
the west-east direction.


U

 2a
sin
t
t
Note: “a” is the radius of the Earth.

But, a  is just the northward velocity
t
component,
V, so:

dU
 2V sin
dt
The magnitude of the Coriolis
acceleration,
or Coriolis Force per unit

mass, is given by
2Vwhich
sin  acts
to the right of the direction of motion of
the object. (Thus, it is causing a change
in the west-east motion of the parcel.)
(9) Acceleration
The Forces acting on an air parcel are, then:
Pressure Gradient Force
Coriolis Force
Friction (turbulent mixing)
These can be expressed as accelerations
(Force/unit mass).
The net effect (net force/accelerations) of
these expressed as components in the x- and
y-direction are:
Du Dv
,
Dt
Dt
Since these are the result of the various
forces/accelerations on the parcel, (Newton’s
2nd Law) they can be written as:
Du
Z
 g o
 2v sin  Fx
Dt
x
Du
Z
 fv  g o
 Fx
Dt
x
Dv
Z
 g o
 2u sin  Fy
Dt
y
Dv
Z
 fu  g o
 Fy
Dt
y
Acceleration = Pressure Gradient acceleration + Coriolis acceleration + Friction

If acceleration and friction are zero, these
 balance - Balance
then show geostrophic
between the Pressure Gradient acceleration
and the Coriolis acceleration.
Since wind is a vector, the equations
(Newton’s 2nd law) can be written in vector
notation as one.
Dv h
 f k  v h   go  h Z  Fh
Dt
Where, “h” means the horizontal components
ofthe vectors only.
Consider the change in the wind in time.
The acceleration of the wind (change in speed or
direction, or both) can be represented as the
difference between two wind vectors, at different
times.
The change in
the wind vector is
equal to a vector
drawn from the
end of the first
vector to the end
of the second
vector (second
vector minus the The acceleration is obtained by
first vector).
dividing the difference vector (blue)
by the time interval.
If only the direction is changing, not the
speed, the direction of the acceleration is
essentially at right angles to the original
direction.
If only the speed
is changing, the
direction of the
acceleration is
parallel to the
original direction.
(10) Gradient and Cyclostrophic Winds
If the acceleration (net force/unit mass) is
always at right angles to the wind direction,
(e.g., to the left), the air will constantly curve
to the left.
Then the Pressure Gradient force is greater
than the Coriolis force (since the Coriolis force
- in the northern hemisphere- tries to make
the air move to the right.
This is how air flows about a low pressure
(low height) region in the absence of friction.
This is Gradient Wind - the air flowing
parallel to curved isobars (contours).
It is called
“balanced” because
it flows parallel to the
contours, even
though the forces
are not in balance.
The winds about the low are weaker than in
straight isobar / contour flow.
In straight flow, the wind speed must be strong
enough for the Coriolis force to balance the
Pressure Gradient force.
In curved flow about a low, the Coriolis Force
is weaker than the Pressure Gradient force,
so the winds are weaker.
The winds are Subgeostrophic.
The contours / isobars about a low will be
close together, signifying a strong Pressure
Gradient force.
About a high pressure / height region, the
Coriolis force must be greater than the
Pressure Gradient force.
The wind speed is
greater than it would
be in a straight
isobar / contour flow.
The winds are
Supergeostrophic.
The center of a high
will be broad and
flat, signifying a
weak PGF.
Homework:
Do: 1, 2, 3, 6 (at latitude 37o), 7 (at
latitude 37oN)
Remember: Direction is from.
Be careful on angles.
Be careful on units.