MT11B Weather Systems Analysis
Dr E Highwood, November 2003
Lecture D3 Gradient Wind Balance
Reading: FWC section 7.11
D3.1 Gradient wind balance:
Recap: Geostrophic balance: If the motion of an air parcel is in geostrophic balance,
then the pressure gradient force acting on a fluid parcel exactly balances the Coriolis
force. This can only be true for parcels of air moving in straight lines at uniform
velocity, without friction, across the Earth's surface.
If the parcel is accelerating in any way, the pressure gradient and Coriolis forces
cannot be in balance. An imbalance is required to provide the necessary acceleration
(Newton’s 2nd law of motion).
In the atmosphere, we often see air parcels in circular motion around a centre of
low or high pressure. An air parcel is subject to a centripetal acceleration in
order to maintain a circular path.
In the case of circular motion round a low pressure centre:
Centripetal acceleration =
The balance of forces in this case must be:
Or mathematically:
U 2 1 dp
dZ

 fU  g
 fU
r
 dr
dr
depending on the vertical co-ordinate system.
,
MT11B Weather Systems Analysis
Dr E Highwood, November 2003
It is simpler to write this relationship in terms of a hypothetical “geostrophic wind
speed” U g where:
fU g 
1 dp
dZ
.
g
 dr
dr
Then, the gradient wind balance can be expressed as:
U2
 fU g  fU .
r
(Eq. 1)
And the gradient wind is sub-geostrophic (by up to 20% in vigorous extra-tropical
cyclones).
“Cyclostrophic balance” occurs for very large Ug where the centripetal acceleration
is entirely provided by the PGF (i.e. CF is small). In what situations might this be
important and why?
In the case of circular motion around a high pressure centre:
PGF away from centre, CF towards centre.
Centripetal force must still be towards the
centre so CF > PGF.
U2
Then for an anticyclone: fU  fU g 
r
Which can be rearranged to give:
U 2  ( fr)U  ( fr)U g  0
MT11B Weather Systems Analysis
Dr E Highwood, November 2003
Now, this can be made to look exactly like the equation for a cyclone (Eq 1) if we
define r as negative for an anticyclonic situation. Then we can solve the quadratic
equation in the usual way to give the roots
1
 fr fr  4U g  2
 .
U 
 1 
2
2
fr 
(Eq. 2)
in either case, so long as r is defined as negative in the case of anticyclones.
There are two things we need to know in order to use this equation practically.
1. The term under the square root must be positive or there will be no real roots, so
that in the case of the anticyclone, where r is negative,
Ug 
4U g
f r
 1,
rf
, and there is a maximum geostrophic wind speed which cannot be exceeded in
4
an anticyclone.
Substituting this criteria back into the equation (2) gives a maximum speed of U=2Ug.
Q. What use is this to us when analysing charts for anticyclones?
2. The decision whether to use the + or – root depends on the situation. Bearing in
mind that U is a speed and therefore must be both real and positive:
Cyclone: r is +ve, so to get U +ve we must take +ve root
Anticyclone: r is –ve, and
1
4U g
fr
 1 , so if we took the +ve root the system would
have negative absolute angular momentum, which is something that only happens
occasionally near the equator. But taking the –ve root allows the system to have
positive angular momentum when r is negative.
MT11B Weather Systems Analysis
Dr E Highwood, November 2003
Revision Summary
Cyclones
Anticyclones
How does U compare
to Ug ?
Is there a maximum
value to Ug or U?
What does this mean
for analyses that you
draw?
D3.2- Effects of friction:
The friction force at the ground generally acts in the opposite direction to the velocity
vector, acting to slow air parcels down. A balance of forces between CF, PGF and FD
can be achieved if the parcel moves at an angle to the isobars, drifting towards low
pressure (Remember Lecture AP2).
Simplified expression for friction
force is known as “Rayleigh
friction”:
FD  
v
D
where D is ?
We can work out a value for D and for the windspeed by resolving forces.
MT11B Weather Systems Analysis
Dr E Highwood, November 2003
If the angle between the velocity vector and the isobars is , then resolving forces
parallel to the isobars gives:
FD cos  Fc sin  or tan  
FD
1
.

Fc
f D
Resolving forces at right angles to the direction of motion determines the
magnitude of the velocity:
Fp cos   Fc or U  U g cos  .
A more accurate parametrization would have the friction force increasing as the square
of the wind speed.
NB A parametrization is just a simplified expression used to represent complex physics.