MTR 2400 Introduction to Atmospheric Science
Lab Assignment #9
Upper Level Heights and Winds
Introduction
Upper-level characteristics such as temperature, heights, and wind are generally
shown on constant pressure surfaces. Common levels for these quasi-horizontal
surfaces are 850-mb and 700-mb in the lower troposphere, 500-mb in the midtroposphere, and 300-mb and 200-mb in the upper troposphere.
Recall that in the upper atmosphere, frictional forces are negligible, so that the
two main forces controlling air motions (horizontal winds) are the pressure
gradient force and the Coriolis force. The atmosphere is usually close to a
balance between these forces (geostrophic balance) resulting in predictable wind
patterns.
Geostrophic winds are directed parallel to height contours, with lower heights to
the left. In the mid-latitudes this usually is characterized by generally westerly
winds, since lower heights are located to the north where the air column is colder.
The strength of the geostrophic winds is determined primarily by the pressure
gradient. When height contours are closer together (i.e. a stronger pressure
gradient), the geostrophic wind is stronger.
Geostrophic Wind Calculation
Pressure Gradient Force = gZ/d
where g is acceleration due to gravity (g = 9.8ms-2),
Z is the height difference between 2 locations on a constant pressure map,
and d is the horizontal distance between the 2 locations.
Coriolis Force = (2sin)V = fV
where f = 2sin = (1.46 x 10-4 s-1) sin
and  is the latitude
Geostrophic Wind results when these 2 forces are balanced (equal and opposite).
fVg = gZ/d
Vg = gZ/fd
Example of Geostrophic Wind Calculation
The 570 and 564 height contours on a 500-mb chart are 400km apart. This
distance is measured approximately perpendicular to the contours. Determine the
speed of the geostrophic wind. Use a latitude of 40N.
Solution:
f = (1.46 x 10-4 s-1)sin = (1.46 x 10-4 s-1)sin(40)
= (1.46 x 10-4 s-1)(.643) = 9.38 x 10-5 s-1
Vg = gZ/fd = (9.8ms-2)(60m)/( 9.38 x 10-5 s-1)(400km)
Convert 400km to 4 x 105 m
Vg = (9.8ms-2)(60m)/( 9.38 x 10-5 s-1)(4 x 105 m) = 15.7 m s-1
Convert ms-1 to knots
Vg = (15.7 m s-1)/(.515ms-1/kt) = 30.4 kt
MTR 2400 Introduction to Atmospheric Science
Lab Assignment #9
Upper Level Heights and Winds
1. The three figures (below and on the next page) show 850-mb heights, 700-mb
heights and 500-mb heights at 12Z on March 23, 2006.
a) Use the charts to estimate to the nearest ten meters the heights at each
level for the three locations marked with dots.
Location
500-mb
700-mb
850-mb
A
B
C
b) Assuming that winds are geostrophic, draw arrows indicating the
direction of the wind.
c) At the 500-mb level which of these locations is experiencing the fastest
winds?
C
A
B
a) 500-mb Chart (in decameters; i.e. 552 = 5520 m)
A
C
B
b) 700-mb Heights (in meters with first digit omitted; i.e. 940 = 2940 m or
060 = 3060m)
C
A
B
c) 850-mb Heights (in meters with first digit omitted; i.e. 530 = 1530m )
2. The map below indicates the 300-mb heights for 12Z 22 March, 2006. The map
on the next page indicates the 300-mb winds at the same time.
a) Isopleth the height field by drawing in contours at 120-m intervals from
9000 m (8880 m, 9000 m, 9120m, etc.) Use interpolation to estimate the
position of lines between two points. For example, suppose you have
points indicating 9100 and 9180 m. The 9120 m height contour should
pass between the two points, but also should be closer to the 9100 m point.
The line should be ¼ of the distance from the 9100 m point towards the
9180 m point.
b) Draw 50, 75 and 100 knot isotachs for the map on the next page. Recall
that the “flags” represent 50 knots, the long lines 10 knots and the short
lines 5 knots.
c) Is the jet stream where you would expect it, based on the height contours?
3. Estimate the geostrophic wind speed (in knots) at points A and B on this map of 500mb heights. Use latitudes of 43N for point A and 33N for point B
A
B
0
500
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2000
2500 km