Name _________________________
Date __________________________
Meteorology 430
Fall 2010
Lab 2
Review of Basic Techniques
(Due Beginning of Class, Friday 10 September)
1. All labs are to be kept in a three hole binder. Turn in the binder when
you have finished the Lab.
2. Show all PROCEDURE. No credit given if only
answer is provided.
3. Unless otherwise noted, you may work together. But remember, YOU
are the one who will be responsible for understanding the material for
exams and when you are out in the profession. So STRIVE to
understand what you are doing. Don't let someone else do your
thinking for you.
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1. Determine the value of a degree in km at 40oN latitude (trigonometry).
a = radius of earth = 6370 km
r = radial distance to the axis of rotation which,
by simple geometry, = a cos phi whereis the latitude
Circumference = 2π R
Distance (km) of 1o = (Circumference)/360 o
For latitude R = a and a degree is (2π 6370 km )/360 o = 111.3
km
For longitude R = r and a degree is (2π 6370 km cos 40) =
85.3 km
2
2. (a)
Give the PHYSICAL INTERPRETATION of each term of
equation (cartesian coordinate) below, where V is the three
dimensional wind vector:
A
(b)
(c)
(d)
(e)
B
C
Name each term in equation.
Solve the equation for the local tendency.
Expand term "C".
Assume that dT/dt = 0 for the chart given and that there is no
vertical motion. Determine the local temperature tendency at the
station reporting the wind of 225o, 25 knots, shown in the diagram
below. External grid points indicated by crosses are 100 km distant
from station, nominally placed at center of implied x-y coordinate
cross. Isotherms are north-south.
3
(a) A is the temperature change experienced by the moving parcel, as if a
thermometer moved with the air parcel.. B is the temperature change measured at
a location fixed with respect to the surface of the earth. C is the contribution to
the local temperature change made by cold or warm advection.
(b)
Term A
dT
dt is the rate of the change of temperature
following the air parcel or Lagrangian derivative.
In other words, this term measures the temperature change in the air parcel
itself caused by a myriad of sources/sinks (radiative,
sensible htg/cooling etc)
Term B
¶T
¶t is the rate of change of temperature
observed locally (e.g., by a thermometer
fixed with respect to the surface of the
earth)
Term C
(c)
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(d)
(e)
¶T
¶t
= dT
dt
- u ¶T
¶x
- v ¶T
¶y
- w ¶T
¶z (3)
but dT/dt and - w ¶T ¶z = 0 so
¶T
¶t
= -u ¶T
¶x
- v ¶T
(4)
¶y
u and v are the horizontal wind components, but there is no north/south
temperature gradient in this case. Hence,
¶T
¶t
= -u¶T
¶x
5
where u = 25 knots (sin 45) = 17.68 knots = 32.5 km h -1
The temperature change will be entirely due to advection,
-1.6o C h-1
6
3. (a) Physically interpret the terms to the right of the equals sign in
equation (2) (basic calculus)
(b)
Equations (3a,b) are the geostrophic wind components. Put (3a,b) into (2)
to determine the divegence of the geostrophic wind. Assume
g and
f are constants. I realize we have not discussed divergence, or horizontal
divergence. But this question involves you working with the mathematical
operations, not the meteorological interpretation.
4. (a) Physically interpret the terms to the right of the equals sign in
equation (2) (basic calculus)
¶u ¶v
DIVH = +
¶x ¶y
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(2)
g ¶z
ug = f ¶y
g ¶z
vg =
f ¶x
(b)
(a)
(3a, b)
Equations (3a,b) are the geostrophic wind components. Put (3a,b)
into (2) to determine the divegence of the geostrophic wind. Assume
g and f are constants.
∂u/∂x is the horizontal shear of the west wind along the x axis or
the variation of the west wind along the x axis or the variation of the west
wind speed along the x axis.
∂v/∂y is the horizontal shear of the south wind along the y axis or
the variation of the south wind along the y axis or the variation of the
south wind speed along the y axis.
(b)
DIVH =
ug = -
¶u ¶v
+
¶x ¶y
g ¶z
f ¶y
(1)
(2a, b)
g ¶z
vg =
f ¶x
Put Eq 2(a,b) into (1)
DIVH = -
æ g ¶z ö
¶ç
è f ¶y ø
¶x
æ g ¶z ö
è f ¶x ø
+
¶y
¶
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(3a)
DIVH = -
æ ¶z
¶ç ö
g è ¶y ø
f
¶x
æ ¶z ö
g è ¶x ø
+
f ¶y
¶
(3b)
Equation (3b) states that the horizontal divergence of the geostrophic wind is related to
the variation of north-south slope of an isobaric surface along the x-axis and to the variation of
the west-east slope of an isobaric surface along the y-axis. However, the second derivatives are
identical but opposite sign, so the horizontal divergence of the geostrophic wind
DIVh = 0
(4)
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4.
Obtain a plot of the NAM initialization of 500 mb heights and
absolute vorticity AND surface pressure with 1000-500 mb
thickness at initialization time for 12 UTC 31 August 2010.
(a) Write out the script you executed on the command line to
obtain this.
(b) Append -p on the command line after the command to print
a black and white copy to turn in with this lab.
5.
(a)
(b)
6.
Use the script mod_grib -p to print out a copy of the 500 mb
heights for 12 UTC 31 August 2010.
Annotate troughs and ridges as we did in Metr 201 (and Metr
400).
For the sounding distributed (KOAX 12 UTC 31 August 2010),
determine the
(a) LCL, LFC and CAPE and CIN areas for 12 UTC (first copy);
(b) CT, and CCL (second copy)
Make sure you draw everything neatly and use the correct color
conventions.
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