Radar Meteorology
Theoretical work (Mie scattering theory) in the late 1940s
showed that “weather clutter” arose from the scattering
of electromagnetic radiation by precipitation particles
(resonant interaction between propagating EM wave and a
dielectric such as water and ice).
Today modern radars can not only detect hydrometeors
(both precipitation and cloud particles), but “clear air”
targets such as insects and large aerosol particles, as well
as changes in the index of refraction, the latter caused by
turbulent motions in the atmosphere.
RADAR-Radio Detection and Ranging
Radar is the “art of detecting by means of radio echoes the presence of
objects, determining their direction and range, recognizing their characteristics
and employing the data thus obtained”.
“Object” refers to meteorological targets such as raindrops, hailstones, cloud
ice and liquid particles and snowflakes. For the purpose of clear air detection,
insects are considered the “objects”. Birds also are readily detected and hence
are of interest.
Radar is based on the propagation of electromagnetic waves through the
atmosphere, a non-vacuum. EM waves propagate at the speed of light in a
vacuum, c = 2.998 x 108 m s-1.
Propagation speed in a non-vacuum determines the index of refraction, n = c/ν
where ν is the wave speed (Note : water and ice have different refractive
index)
Electromagnetic Waves and Their Propagation Through
the Atmosphere
Electromagnetic Waves are characterized by:
Wavelength,  [m, cm, mm, mm etc]
Frequency, ν [s-1, hertz (hz), megahertz (Mhz), gigahertz (Ghz)
where: c = λν
Polarization of electromagnetic waves
The polarization is specified by the orientation of the electromagnetic field.
The plane containing the electric field is called the plane of polarization.
Electric field will oscillate in the x,y plane with z as the propagation direction
For a monochromatic wave:
E x  E xm cos 2 ft 
E y  E ym cos  2 ft  

where f is the frequency and  is the phase difference between Exm and Eym and the coordinate x
is parallel to the horizon, y normal to x, and z in the direction of propagation.
If Eym = 0, Electric field oscillates in the x direction
and wave is said to be “horizontally polarized”
If Exm = 0, Electric field oscillates in the y direction
and wave is said to be “vertically polarized”
If Exm = Eym, and  =  or -  electric field vector
rotates in a circle and wave is circularly polarized
All other situations: E field rotates as an ellipse
How does radar scan ?
Ground/ship radar
Scanning strategies for scanning radars must take into account the
propagation path of the beam if certain operational or scientific
objectives are to be addressed. Here, 3 common NWS NEXRAD Volume
Coverage Patters (VCPs) are illustrated. NEXRADs have a 5-6 minute
scan update requirement for severe weather detection, so they vary
their VCPs and scan rates depending on the weather situation.
6 min update, slow scan rate
VCP 31
“clear air
mode”
5 min update, fast scan rate
VCP 11
“severe
weather
mode”
6 min update, slow scan rate
VCP 21
Widespread
precip
Airborne
Commercial
airplanes
Airborne
Research airplanes
Space borne
What kind of electromagnetic
pulse do we send?
Sidelobes
E
PULS
ic
Electr
Field
ANTENNA
Half-power beamwidth
TRANSMITTER
Duplexer
switch
Klystron
Amplifier
Pulse
modulator
Frequency
Mixer
Frequency
Mixer
STALO
Microwave
Oscillator
Amplifier
DISPLAY
COHO
Microwave
Oscillator
RECEIVER
Phase
Detector
Block Diagram of a Radar System
Antenna
Transmitter
106 W
Display
T/R switch
Receiver
10-14 W
Why is wavelength important?
• Longer wave length -> sensitive to larger
objects -> larger penetration ability (long
range), but require a larger antenna to obtain
enough return signal
• Shorter wave length -> sensitive to smaller
objects -> more scattering and more
attenuation of signal, require a smaller
antenna to obtain enough return signal
W and K band radars are “cloud radars”
X, C, S and L band radars are “precipitation radars”
Also - Wind Profilers (UHF & VHF; ~50 to 900 MHz; ~6 to 0.3 m)
How does electromagnetic wave
travel in the atmosphere?
Electromagnetic waves:
Interact with matter in four ways:
Reflection:
Refraction:
Scattering:
Diffraction:
Snell’s law:
n - Dn
Vr
r
n
i
Vi
n  Dn
n

sin i

sin r
Vi
Vr
Where: i is the angle of incidence
r is the angle of refraction
Vi is the velocity of light in medium n
Vr is the velocity of light in medium n - Dn
In the atmosphere, n normally decreases continuously with height…
Therefore: due to refraction, electromagnetic rays propagating
upward away from a radar will bend toward the
earth’s surface
Propagation of electromagnetic waves in the atmosphere
Speed of light in a vacuum: C
Speed of light in air: V
Refractive index: n=C/V
At sea level: n = 1.0003
In space:
n = 1.0000
c = 2.998 x 108 m s-1
The Refractive Index is related to:
n  1  7 . 76  10
1.
5
K mb
1
 Pd 
6
1  e 
2
1  e 

  5 . 6  10 K mb    0 . 375 K mb  2 
T 
T 
 T 
Density of air (a function of
dry air pressure (Pd), temperature (T), vapor pressure (e)
2. The polarization of molecules in the air
(molecules that produce their own electric field
in the absence of external forces)
The water molecule consists of three atoms,
one O and two H. Each H donates an electron
to the O so that each H carries one positive
charge and the O carries two negative
charges, creating a polar molecule – one side
of the molecule is negative and the other
positive.
Earth curvature
Electromagnetic ray propagating away from the radar will rise above the
earth’s surface due to the earth’s curvature.
Ray Path Geometry
Consider the geometry for a ray path in
the Earth’s atmosphere. Here R is the
radius of the Earth, h0 is the height of
the transmitter above the surface,  0 is
the initial launch angle of the beam,  h
is the angle relative to the local tangent
at some point along the beam (at
height h above the surface at great
circle distance s from the transmitter).
Equation governing the path of a ray in the earth’s atmosphere:
2
d h
ds
2
2
2
1 dn   dh 
1 dn 
 2
Rh  1




 
 
0
 R  h n dh   ds 
 R   R  h n dh 
where R is the radius of the earth, h is the height of the beam above the
earth’s surface, and s is distance along the earth’s surface.
To simplify this equation we will make three approximations
1. Large earth approximation
Rh  R
2. Small angle approximation
dh
ds
3. Refractive index ~ 1 in term:
1
n
 tan     1
(1)
1
2
d h
ds
2
2
1/R
1
2
1 dn   dh 
1 dn 
 2
Rh  1




 
 
0
 R  h n dh   ds 
 R   R  h n dh 
X X X
Approximate equation for the path of a ray at small angles relative to the earth’s
surface:
2
d h
ds
2

1

R
dn
dh
Or, in terms of the elevation angle of the beam
d
ds

1
R

dn
dh
(2)
Curvature of Ray Paths Relative to the Earth
An additional equation of interest is the equation that provides the great
circle distance s, from the radar, for the r, h pair (slant range, beam height),
which is
s = keR sin-1[rcos /(keR + h)]
Here ke=4/3
We can get even simpler and consider a the height of the beam at slant
range R and elevation angle  ,
h (km) = R2/17000 + R sin 
R

h
STANDARD REFRACTION: What we expect the beam to do over the curved surface of
r
the earth
h
Φ0
s
Use standard atmosphere, solve Diff. Eq. describing ray path for height of beam above surface of earth
(assumes dn/dh is small):
d2h/ds2 – (2/R + 1/n * dn/dh)(dh/ds)2 – (R/a)2(1/R + 1/n * dn/dh) = 0
Where:
a= earth radius; s= arc distance; h= height above earth surface
n= refractive index; R= h + a; r= slant range along beam
Physically: Via equation for refractivity, we expect the beam to bend toward the surface since dP,e/dz < 0
and < dT/dz. However, h increases with s due to 1/R (curvature of earth’s surface, which diverges from
beam position). DEQ above expresses this relationship as it relates earth’s geometry and the assumed
refraction of the standard atmosphere to beam height and arc distance.
Doviak and Zrnic (1993) Sec. 2.2 show how this can be reduced to two equations for h and s using the 4/3
Earth radius model (4/3 Earth radius - dn/dh assumed to be constant - of order 0.25/a)
So, let ae= 4/3 a; then for convenience of computation:
h=[r2 + (ae)2 + 2raesinΦ0]1/2 – ae
s=aesin-1(rcosΦ0/[ae+h])
4/3 Earth Radius Model for Beam Propagation
(Standard Refraction/Reference Atmosphere Assumed)
Θe = elevation angle
Doviak and Zrnic (1993)
h=[r2 + (ae)2 + 2raesinθe]1/2 – ae
S=aesin-1(rcosθe/[ae+h])
To get h as a f(slant range:R), which is measured by the radar, use
this simple formula:
h (km)= R2/17000 + R sinθe (with R in km)
•
Non-Standard Refraction
•
Non-standard refraction typically occurs with the temperature distribution does not
follow the standard lapse rate (dn/dh  -1/4 (R)). As a result, radar waves may deviate
from their standard ray paths predicted by the previous model. This situation is known
as abnormal or anomalous propagation (AP).
Abnormal downward bending ------- super-refraction
(most common type of AP)
Abnormal upward bending ----------- sub-refraction
•
Super-refraction is associated most often with cold air at the surface, giving rise to a
near surface elevated temperature inversion in which the T increases with height.
Most commonly caused by radiational cooling at night, or a cold thunderstorm outflow.
•
Since T increases with height, n decreases (rapidly) with height (dn/dh is strongly
negative). Since n = c/v, v must increase with height, causing downward bending of the
ray path.
Recall Snell’s Law:
v2/v1 = sinθ2/sinθ1
n1sinθ1 = n2sin θ2
θ2
v2 > v1
n2
n1 > n2
n1
θ1
Wave (beam) is bent downward (refracted) in the atmosphere
So relative to the refractivity, what’s important here?
dN/dZ – change in
refraction with height- this causes velocity differences across the beam.
4 cases of refraction (dN/dZ):
Non-Standard
Standard:
dN/dZ ~ 0 and -40 km-1
Super:
dN/dZ < -79 km-1 and > -158 km-1
Sub:
dN/dZ > 0
Ducting:
dN/dZ < -158 km-1 (dn/dh = -1/R)
Non-Standard Refraction
Super-Refraction (most common)
dN/dZ < -79 km-1 and > -158 km-1
h
h’
Φ0
Beam is bent downward more than standard
Situations:
1. Temperature inversions (warm over cold air; stable layers)
2. Sharp decrease in moisture with height
(1) And (2) can occur in nocturnal and trade inversions, warm air advection (dry), thunderstorm
outflows, fronts etc.
Result:
1. Some increased clutter ranges (side lobes)
2. Overestimate of echo top heights (antenna has to be tilted higher to achieve same height as
standard refracted beam)- see figure above
Most susceptible at low elevation angles (e.g., typically less than 1o)
Sub-Refraction (not as common)
dN/dZ > 0 km-1
Inverted-V sounding
DP
h
T
h’
Φ0
Beam is bent upward more than standard
Situations:
1. Inverted-V sounding (typical of desert/intermountain west and lee-side of mountain ranges;
microburst sounding; late afternoon and early evening; see figure)
Result:
1. Underestimate of echo top heights (beam intersects top at elevation angles lower than in
standard refraction case)- see figure above
Most susceptible at low elevation angles (e.g., typically less than 1o)
Ducting or Trapping (common)
dN/dZ < -158 km-1
Beam is severely bent downward and may intersect the surface (especially at elevation angles less than
0.5o) or propagate long distances at relatively fixed heights in an elevated “duct”.
Situations:
1. Strong temperature inversions (surface or aloft)
2. Strong decreases in moisture with height
Result:
1. Markedly Increased clutter ranges at low elevation angles
2. Range increases to as much as 500% in rare instances (useful for tracking surface targets)
Most susceptible at low elevation angles (e.g., typically less than 1o)
Elevated ducts can be used as a strategic asset for military airborne surveillance and weapons control
radars. E.g., if a hostile aircraft is flying in a ducting layer … it could be detected a long way away, while its
radar cannot detect above or below the ducting layer. Conversely, friendly aircraft may not want to be
located in the duct.
Example of ray paths in surface ducting
Doviak and Zrnic (1993)
Modeled with 100 m deep surface inversion with dN/dz=300 km-1 and standard
thereafter.
One moral of the whole refraction story……..knowing the exact location of the beam can
be problematic. Remember this when you have the opportunity to compare the
measurements of two radars supposedly looking at the same storm volume!
Big implication of radar beam height increasing with range (under
normal propagation conditions) combined with broadening of the
radar beam: The radar cannot “see” the low level structures of
storms, nor resolve their spatial structure as well as at close ranges.
Thus, for purposes of radar applications such as rainfall estimation,
the uncertainty of the measurements increases markedly with range.
Storm 1
Storm 2
Beam Blockage in Complex Terrain
• Beam propagation is a function of the vertical refractivity gradient
(dN/dz)
– N = 77.6(p/T) - 5.6(e/T) + 3.75x105(e/T2)
• dN/dz is sensitive to p, T, e
• Thus, changes in the vertical profiles of these quantities can change
the height of the ray path as it propagates away from the radar
• This is especially important in complex terrain, because the amount
of beam blockage will change depending on the vertical refractivity
gradient
dN/dZ = -40/km
dN/dZ = -80/km
)
43
False Data
• Ground Clutter
– Portion of radar beam
hits buildings, trees,
hills
• Also can be due to dust,
aerosols in the air near
the radar
– Gives false indication
that precip is present
– Radar location is in the
black area surrounded
by blue/green
reflectivities
False Data
• Anomalous propagation (AP)
– Occurs when temperature
inversions are present in lowlevels
• Radar beam bent into
ground, returning strong
signal
– Common during early
morning hours after a clear
night
– Again, no precip really present
False Data
• Virga
– Radar detects
precip occurring at
upper levels, but
not making it to
the ground
• Precip quickly
evaporates in dry
air below cloud
– Precipitation is
thus overestimated
False Data
• Overshooting Beam
– Some precip can form
from clouds with
minimal height
– Beam may overshoot a
large portion of the
cloud, underestimating
the intensity of the
precipitation
False Data
• Storm Interference
– Storms closest to
radar may absorb or
reflect much of the
radar energy
• Leaves reduced
amount of energy
available to detect
distant storms
• Underestimates
precipitation
False Data
• Wind Shear
– Falling precip may
be displaced by the
wind as it falls
– Some regions may
be experiencing
precip where the
radar indicates
nothing, and vice
versa
Mid-Atlantic River Forecast Center (MARFC)
Height of Lowest Unobstructed Sampling Volume
Radar Coverage Map
West Gulf River Forecast Center (WGRFC)
Height of Lowest Unobstructed Sampling Volume
Radar Coverage Map
Southeast
67
Northeast
PRECIPITATION MOSAIC
RADAR COVERAGE MAP
Say you’d like to site a radar for a research experiment.
In a perfect world…you’d like to be able to take a swim after work, but AP and beam blockage
may be a problem. Sidelobes may intersect the highly reflective ocean – creating “sea clutter”
Mountains can be a problem…
1.5°
0.5°
Local effects can be a problem too – topographic maps and DEMs can help, but
still need to conduct a site survey to see trees, antennas, buildings, and overpasses.
Often times you end up in places like this…
Height of a ray due to earth’s curvature and standard atmospheric refraction
Assignment #1
Here we have reviewed the calculation of slant path of the radar beam. But I did
not describe the equation for the beam width change with the distance. Here is
the question: assuming you have a radar with beam width 1 degree. How big is
the beam width at 100 km?
There is an airplane flying at 10 km altitude with a radar sending a 1 degree
beam tangentially. At 200 km distance, a) what is the beam width? b) what are
the heights of the top and bottom of the beam respect to the Earth surface? If
there is a storm reaching 10 km at 200 km distance, can pilot see the storm on
his screen? (assuming the standard atmospheric refraction)
ELECTRIC FIELD
An Electric field exists in the presence of a charged body
ELECTRIC FIELD INTENSITY (E)
A vector quantity: magnitude and direction (Volts/meter)
MAGNITUDE OF E: Proportional to the force acting on a unit positive charge at
a point in the field
DIRECTION OF E: The direction that the force acts
The Electric Field (E) is represented by drawing the Electric Displacement Vector (D),
which takes into account the characteristics of the medium within which the Electric
Field exists.
D coul m
2
  E
, the Electric Conductive Capacity or Permittivity, is related to the ability of a
medium, such as air to store electrical potential energy.
Vacuum:
Air:
Ratio:
 0  8 . 850  10
 12
 1  8 . 876  10
 12
1
0
 1 . 003
coul
2
coul
2
joule
joule
1
1
m
m
1
1
The Electric Displacement Vector, D, is used to draw lines of force.
Units of D:
coul m
2
MAGNETIC FIELD
A Magnetic field exists in the presence of a current
MAGNETIC FIELD INTENSITY (H)
A vector quantity: magnitude and direction (amps/meter)
MAGNITUDE OF H: Proportional to the current
DIRECTION OF H: The direction that a compass needle points in a magnetic
field
The Magnetic Field (H) is represented by drawing the Magnetic Induction Vector (B),
which takes into account the characteristics of the medium within which the current
flows.
B  mH
m, the Magnetic Inductive Capacity, or Permeability, is related to the ability of a
medium, such as air, to store magnetic potential energy.
Vacuum:
Air:
Ratio:
6
2
1
m 0  1 . 260  10 joule amp m
6
2
1
m 1  1 . 260  10 joule amp m
m1
m0
 1 . 000
Magnetic Fields:
Magnetic fields associated with moving charges (electric currents)
Force  I  B
I: Current
coul
s
B: Magnetic Induction
Magnetic Field Lines are closed
loops surrounding the currents
that produce them
1
or amps

 joule
amp
1
m
2

Maxwell’s Equations for time varying
electric and magnetic fields in free space
Simple interpretation
E 
E  

Divergence of electric field is a function
of charge density
0
B
A closed loop of E field lines will exist when
the magnetic field varies with time
t
Divergence of magnetic field =0
(closed loops)
B 0
  B  m0I   0m0
E
t
(where  is the charge density)
A closed loop of B field lines will exist in
The presence of a current and/or
time varying electric field
Electromagnetic Waves: A solution to Maxwell’s Equations
Electric and Magnetic Force Fields
Propagate through a vacuum at the speed of light:
c  3  10 m s
8
1
Electric and Magnetic Fields propagate as waves:
E ( r,  ,  , t ) 
or:
E ( r,  ,  , t ) 
A ( ,  )
r
A ( ,  )
r
r



exp  i2 t 
 i 




c
r



cos  2  t 
 




c

where:
exp( ix )  cos( x )  i sin( x )
ρ,    are coordinates, A is an amplitude factor, ν is the frequency and  is

an arbitrary phase
Time variations in charge, voltage and current in a simple Dipole Antenna
Pt. A
Pt. B
wavelength
All energy
stored in
electric field
All energy
stored in
magnetic
field
Energy is 1) stored in E, B fields, 2) radiated as EM waves, 3) Dissipated as heat in antenna
Near antenna:
More than a few
 from antenna:
Energy stored in induction fields (E, B fields) >> energy radiated
(near field)
Energy radiated >> energy stored in induction fields (far field)
Spherically Stratified Atmosphere; Ray Path Equation
Integrating (2) yields,
(dh/ds)2 = 2 (1/R + dn/dh) dh + constant (3)
Since dh/ds  for small  , (3) can be written as,
1/
2
2( h
-  02) = (h - h0)/R + n - n0
= (h/R + n) - (h0/R + n0)
Letting M = [h/R + (n-1)] x 106, we have
= (M - M0)10-6
M is the so-called modified index of refraction. M has a value of
approximately 300 at sea level.
Curvature of Ray Paths Relative to the Earth
•
If the vertical profile of M is known (say through a sounding yielding p, T and q),  h can
be calculated at any altitude h, that is, the angle relative to the local tangent.
•
Lets now consider the ray paths relative to the Earth. For the case of no atmosphere,
or if N is constant with height (dN/dh = 0), the ray paths would be straight lines relative
to the curved Earth.
d /ds = 1/R + dn/dh
•
For n varying with height,
d /ds = 1/R + dn/dh
•
1/R for n constant with height
(No atmosphere case?)
(“Flat earth” case?)
< 1/R since dn/dh < 0
For the special case where dn/dh = -1/R, d /ds = 0. Hence the ray travels around the
Earth concentric with it, at fixed radius, R + h. This is the case of a trapped wave.
“DUCTING”
Curvature of Ray Paths Relative to the Earth
For convenience, it is is easier to introduce a fictitious Earth radius,
1/R’ = 1/R + dn/dh
For typical conditions, dn/dh = -1/4 R m-1
Hence R’ = R/(1 - 1/4) = 4/3 R
This is the effective Earth radius model, to allow paths to be treated as
straight lines.
Doviak and Zrnic (1993) provide a complete expression for h vs. r, where r is
the slant range (distance along the ray).
h = {r2 + (keR)2 + 2rkeRsin }1/2 - keR
where h is beam height as slant range r,  is the elevation angle of the
antenna, and ke is 4/3 (R is the actual Earth radius).