MT11B Weather Systems Analysis
Dr E Highwood, December 2003
EC7 Vectors and gradient winds
EC7.1 Vectors refresher course!
Quantities such as wind and temperature gradient need both a magnitude and direction to be fully
specified, i.e. they are vectors.
In the horizontal plane, i.e. across the Earth’s surface, we define a set of perpendicular axes as follows.
The x direction is defined as West to East and the y direction is defined as South to North (see
diagram below). Any vector lying on the Earth’s surface can be made up by combining a vector in the
x direction with a vector in the y direction – you just need to alter the relative sizes of the x and y
direction vectors. This is the same as saying that any vector in the x-y plane can be split into two
perpendicular components, one in the x-direction and one in the y-direction. This is very useful in
meteorology.
There are two types of problem you will meet in this course.
a) Given a wind in terms of speed and bearing, you will need to work out the components of this wind
in our standard x and y directions.
e.g. the surface geostrophic wind is 15 ms-1, bearing 240. Find the components in the x and y
directions (sometimes called the “zonal wind component” and the “meridional wind component”).
Writing the full geostrophic wind as Vg (a vector quantity), we then define the component in the Westto-East (or x) (zonal) direction as having magnitude ug and direction x . Similarly, the component in the
South-North (meridional) direction has magnitude vg and direction y. (NB You will find this
convention throughout meteorology, x and u are in the west-to-east (zonal) direction and v and y
are in the south-north (meridional) direction.)
MT11B Weather Systems Analysis
Dr E Highwood, December 2003
So, to get back to our example:
Vg=ugi+ vgj
Now in the diagram above, we must imagine we are“making up” V in terms of the two components at
right angles to each other. Therefore we end up with a right angled triangle made up of our vectors. We
can then use the simple Pythagoras theorem to finish the problem, i.e.(paraphrasing!)
“`The square of the length of the side opposite the right angle is equal to the sum of the squares of the
lengths of the other two sides'
Therefore,
Vg

 u g  vg
2
2
And
u g V cos 
vg V sin 
Now since:
Bearing = 270 - 
(this is very useful and worth remembering if you haven't already)
in our example we have =30 and this gives
ug= 13 ms-1
and
vg=7.5 ms-1.
b) The second type of problem works in reverse. For example, you might be given the low level wind
and the thermal wind shear (relates the wind at different heights to the temperature structure of the
atmosphere T5) and be required to find the upper level wind.
In this case we’ll start with the same surface level geostrophic wind as in a) but written as:
(x-component, y-component) i.e. (13,7.5).
Imagine we now have a thermal wind in the west-east direction (magnitude 26.3 ms-1), as there might
MT11B Weather Systems Analysis
Dr E Highwood, December 2003
be when the south is warm and the north is cold (we’ll see why in lecture T5). Using the thermal wind
equation, a change in temperature (and therefore a thermal wind) in the N-S direction implies a change
in wind strength in the x direction.
Thus the upper level wind will have the same meridional component but a larger zonal component than
at the surface. We add the thermal wind onto the x component so that the upper level wind is described
by (13.7+26.3=39.3,7.5). To find the complete wind description we use the same principle of vector
addition and Pythagoras theorem as before. So
Vg

 u g  vg
2
2
Giving us V2=39.32 + 7.52, i.e. V=40 ms-1. We can also find the angle  by remembering that tan  =
length of side opposite the angle/ length of side adjacent to angle, i.e.
 = tan-1 (vg/ug).
You can then use this angle to work out the bearing.
EC7.2 Gradient Wind balance examples:
Worked Example: A wind of 25 m s-1 is observed at 90 kPa and 48N, circulating around a depression
whose centre is 300 km away. Ignoring friction, calculate the geostrophic wind at the observing station
and compare it with the actual wind. If the height of the 90 kPa surface at the station is 1200 m,
estimate its height in the centre of the depression.
Step 1. Draw it!
MT11B Weather Systems Analysis
Dr E Highwood, December 2003
Step 2. Find the first equation you need to use.
Flow is circular and if we ignore friction the observed wind will be equal to the gradient wind, i.e.
U=25ms-1.
From the gradient wind relationship, we have:
U g  U  U 2 ( fr ) .
Step 3. Calculate the Coriolis parameter
f  2sin   1. 084  10-4 s-1. And then fr is 32.5 m s-1.
Step 4. Substitute into the gradient wind balance equation.
In this case, U g is 44.2 m s-1, which is 77% stronger than the gradient wind.
Step 5. To find the height at the centre of the depression we need to know the geopotential height
gradient which we can get now that we know the geostrophic wind.
 Z  r  fU g g , which works out as 4.8810-4.
Step 6. The central height is therefore 1200  3  105  Z  r , i.e., 1054 m.
Some more to try (break each into steps to make it easier):
1) An air parcel is situated at a latitude of 50N. It is circulating anti-clockwise at a speed of 20 m s-1 in
a circle of radius 100 km. Calculate the centripetal and Coriolis accelerations it experiences. By
comparing the Coriolis acceleration and the centripetal acceleration, deduce the magnitude of the
gradient of geopotential height in the neighbourhood of the parcel.
2) A forecaster has analysed a chart showing a major anticyclone centred at 45N. She has plotted
isobars every 0.4 kPa. They are separated by a distance of 100 km, and they have a radius of
curvature of 250 km. Discuss whether this analysis is physically reasonable.