aoss 401, fall 2007 lecture 3 september 10, 2007 richard b. rood (room 2525, srb) 734-647-3530 derek...
DESCRIPTION
Weather National Weather Service –http://www.nws.noaa.gov/http://www.nws.noaa.gov/ –Model forecasts: 7loop.html 7loop.html Weather Underground –http://www.wunderground.com/cgi- bin/findweather/getForecast?query=ann+arborhttp://www.wunderground.com/cgi- bin/findweather/getForecast?query=ann+arbor –Model forecasts: ?model=NAM&domain=US ?model=NAM&domain=USTRANSCRIPT
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AOSS 401, Fall 2007Lecture 3
September 10, 2007Richard B. Rood (Room 2525, SRB)
Derek Posselt (Room 2517D, SRB)[email protected]
734-936-0502
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Class News
• Ctools site (AOSS 401 001 F07)– A PDF of my Meeting Maker Calendar is Posted.
• It has when I am in and out of town.• It has my cell phone number.• When I am out of town, I plan to be available for my Tues-
Thursday office hours.• Write or call
• Homework has been posted– Under “resources” in homework folder
• Due Wednesday (September 12, 2007)
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Weather
• National Weather Service– http://www.nws.noaa.gov/– Model forecasts:
http://www.hpc.ncep.noaa.gov/basicwx/day0-7loop.html
• Weather Underground– http://www.wunderground.com/cgi-bin/findweather/getForecast?
query=ann+arbor– Model forecasts: http://
www.wunderground.com/modelmaps/maps.asp?model=NAM&domain=US
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Outline
• Review
• Coriolis Force
• Vertical structure and vertical coordinate
Should be review. So we are going fast.You have the power to slow us down.
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From last time
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Our momentum equation
rzaagp
dtd ruu
2
2
02
)()(1
+ other forces
Now using the text’s convention that the velocity is u = (u, v, w).
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Apparent forces
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Two coordinate systems
xy
z
x’
y’
z’
Can describe the velocity and forces (acceleration) in either coordinate system.
dtdor
dtd 'xx
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Two coordinate systems
y
zz’ axis is the same as z, and there is rotation of the x’ and y’ axis
z’
y’
x’
x
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Apparent forces• With one coordinate system moving relative to the other,
we have the velocity of a particle relative to the coordinate system and the velocity of one coordinate system relative to the other.
• This velocity of one coordinate system relative to the other leads to apparent forces. They are real, observable forces to the observer in the moving coordinate system.– The apparent forces that are proportional to rotation and the
velocities in the inertial system (x,y,z) are called the Coriolis forces.
– The apparent forces that are proportional to the square of the rotation and position are called centrifugal forces.
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Centrifugal force of Earth
• Vertical component incorporated into re-definition of gravity.
• Horizontal component does not need to be considered when we consider a coordinate system tangent to the Earth’s surface, because the Earth has bulged to compensate for this force.
• Hence, centrifugal force does not appear EXPLICITLY in the equations.
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Apparent forces:A physical approach
• Coriolis Force• http://climateknowledge.org/figures/AOSS
401_coriolis.mov
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Two coordinate systems
y
zz’ axis is the same as z, and there is rotation of the x’ and y’ axis
z’
y’
x’
x
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One coordinate system related to another by:
T
zztytxytytxx
2
')cos()sin(')sin()cos('
T is time needed to complete rotation.
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Circle Basics
ω
θ
s = rθ
r (radius)
Arc length ≡ s = rθ
dtdsv
dtddtdr
dtds
... Magnitude
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Angular momentum
• Like momentum, angular momentum is conserved in the absence of torques (forces) which change the angular momentum.
• This comes from considering the conservation of momentum of a body in constant body rotation in the polar coordinate system.
• If this seems obscure or is cloudy, need to review a introductory physics text.
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Angular speed
rvω
ΔθΔvr (radius)
v
v
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What direction does the Earth’s centrifugal force point?
Ω
R
Earth
R Direction away from axis of rotation
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Magnitude of R the axis of rotation
Ω
R
Earth
R=acos()
Φ = latitudea
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Tangential coordinate system
Ω
R
Earth
Place a coordinate system on the surface.
x = east – west (longitude)y = north – south (latitude)z = local vertical
Φa
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Angle between R and axes
Ω
R
Earth
Φ = latitudea
Φ
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Assume magnitude of vector in direction R
Ω
R
Earth
Φ = latitudea
Vector of magnitude B
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Vertical component
Ω
R
Earth
Φ = latitudea
z component = Bcos()
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Meridional component
Ω
R
Earth
Φ = latitudea y component = Bsin()
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Earth’s angular momentum (1)
Ω
R
Earth
Φ = latitudea
What is the speed of this point due only to the rotation of the Earth?
Rv
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Earth’s angular momentum (2)
Ω
R
Earth
Φ = latitudea
Angular momentum is
RL v
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Earth’s angular momentum (3)
Ω
R
Earth
Φ = latitudea
Angular momentum due only to rotation of Earth is
2RL
RL
v
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Earth’s angular momentum (4)
Ω
R
Earth
Φ = latitudea
Angular momentum due only to rotation of Earth is
)(cos22
2
aL
RL
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Angular momentum of parcel (1)
Ω
R
Earth
Φ = latitudea
Assume there is some x velocity, u. Angular momentum associated with this velocity is
uRLu
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Total angular momentum
Ω
R
Earth
Φ = latitudea
Angular momentum due both to rotation of Earth and relative velocity u is
)(
))cos()(cos()cos()(cos
2
22
2
RuRL
uaaLuaaL
uRRL
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Displace parcel south (1)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
Let’s imagine we move our parcel of air south (or north). What happens? Δy
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Displace parcel south (2)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
We get some change ΔR
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Displace parcel south (3)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
But if angular momentum is conserved, then u must change.
)()(
)(
2
2
RRuuRR
RuRL
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Displace parcel south (4)(Conservation of angular momentum)
)()()( 22
RRuuRR
RuR
Expand right hand side, ignore squares and higher of difference terms.
RRuRu 2
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Displace parcel south (5)(Conservation of angular momentum)
yR )sin(
yRuyu )sin()sin(2
For our southward displacement
yauyu )sin()cos(
)sin(2
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Displace parcel south (6)(Conservation of angular momentum)
dtdy
au
dtdu ))sin(
)cos()sin(2(
Divide by Δt and take the limit
)tan(v)sin(v2 au
dtdu
Coriolis term (check with previous mathematical derivation … what is the same? What is different?
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Displace parcel south (7)(Conservation of angular momentum)
)tan(v)sin(v2 au
dtdu
What’s this? “Curvature or metric term.” It takes into account that y curves, it is defined on the surface of the Earth. More later.
Remember this is ONLY FOR a NORTH-SOUTH displacement.
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Coriolis Force in Three Dimensions(link to explicit derivation)
• Do a similar analysis displacing a parcel upwards and displacing a parcel east and west.
• This approach of making a small displacement of a parcel, using conversation, and exploring the behavior of the parcel is a common method of analysis.
• This usually relies on some sort of series approximation; hence, is implicitly linear. Works when we are looking at continuous limits.
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Coriolis Force in 3-D
So let’s collect together today’s apparent forces.
auu
dtdw
auu
dtd
auww
au
dtdu
2
2
)cos(2
)tan()sin(2v
)cos(2)tan(v)sin(v2
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Definition of Coriolis parameter (f)
)tan()sin(2
)tan(v)sin(v2
2
auu
dtdv
au
dtdu
Consider only the horizontal equations (assume w small)
For synoptic-scale systems in middle latitudes (weather) first terms are much larger than the second terms and we have
)sin(2
)sin(2v
v)sin(v2
f
fuudtd
fdtdu
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Our momentum equation
jikuu fufvgpdtd
)(1 2
+ other forces that are, more often than not, ignored
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Highs and Lows
Motion initiated by pressure gradient
Opposed by viscosity
In Northern Hemisphere velocity is deflected to the right by the Coriolis force
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The importance of rotation
• Non-rotating fluid– http://climateknowledge.org/figures/AOSS401
_nonrot_MIT.mpg• Rotating fluid
– http://climateknowledge.org/figures/AOSS401_rotating_MIT.mpg
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Vertical StructurePressure as a vertical coordinate
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Some basics of the atmosphere
Troposphere: depth ~ 1.0 x 104 m
Troposphere------------------ ~ 2Mountain
Troposphere------------------ ~ 1.6 x 10-3
Earth radius
This scale analysis tells us that the troposphere is thin relative to the size of the Earth and that mountains extend half way through the troposphere.
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Pressure altitude
Under virtually all conditions pressure (and density) decreases with height. ∂p/∂z < 0. That’s why it is a good vertical coordinate. If ∂p/∂z = 0, then utility as a vertical coordinate falls apart.
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Use pressure as a vertical coordinate?
• What do we need.– Pressure gradient force in pressure
coordinates.– Way to express derivatives in pressure
coordinates.– Way to express vertical velocity in pressure
coordinates.
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Expressing pressure gradient force
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Integrate in altitude
gzp
z
z
gdzzp
gdzzpp
)(
)()(
Pressure at height z is force (weight) of air above height z.
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Concept of geopotential
kF g
gdzd
Define a variable
such that the gradient of is equal to g. This is called a potential function.
We have assumed here that is a function of only z.
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Integrating with height
gdzd
z
z
gdzz
gdzz
gdzd
0
0
)(
0)0(
)0()(
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What is geopotential?
• Potential energy that a parcel would have if it was lifted from surface to the height z.
• It is analogous to the height of a pressure surface.– We seek to have an analogue for pressure on
a height surface, which will be height on a pressure surface.
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Linking geopotential to pressure
pRTdpdgdz
pRT
dpdpgdz
dgdz
/1Definition of specific volume
Ideal gas law
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Remembering some calculus
pdTRpTdRzz
pRTdd
dpp
pd
p
p
p
p
lnln)()(
ln
1ln
2
1
1
2
12
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Define geopotential height(assumption of constant g = g0)
1
2
ln
)(
012
0
p
ppTd
gRZZ
gzZ
Z2-Z1 = ZT ≡ Thickness - is proportional to temperature is often used in weather forecasting to determine, for instance, the rain-snow transition. (We will return to this.)
Note link of thermodynamic variables, and similarity to scale heights calculated in idealized atmospheres above.
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Getting pressure gradient in pressure coordinates
x
z
Constant pressure p0
Constant pressure p0+Δp
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Getting pressure gradient in pressure coordinates
x
z
Constant pressure p0
Constant pressure p0+Δp
Δx
We have, for instance, ∂p/∂x on a constant z surface in our derivation of the momentum equation.((p0+Δp)-p0)/Δx
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Getting pressure gradient in pressure coordinates
x
z
Constant pressure p0
Constant pressure p0+Δp
Δz
We can also calculate how pressure changes on on a z surface as we hold x constant.
(p0-(p0 +Δp))/Δz
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Getting pressure gradient in pressure coordinates
x
z
Constant pressure p0
Constant pressure p0+Δp
Δz
Which we project onto the x direction by how much z changes with x on the pressure surface. Δz/Δx
(p0-(p0 + Δp))/Δz
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Getting pressure gradient in pressure coordinates
x
z
Constant pressure p0
Constant pressure p0+Δp
ΔzΔx
xzg
xp
xzg
xp
xz
zp
xp
1
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xxzg
xp
xzg
xp
1
1
Implicit that this is on a constant z surface
Implicit that this is on a constant p surface
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Horizontal pressure gradient force in pressure coordinate is the gradient of geopotential
pz
pz
yyp
xxp
1
1
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Our momentum equation(height (z) coordinates)
jikuu fufvgpdtd
)(1 2
fuyp
dtdv
fvxp
dtdu
zz
zz
)1()(
)1()(
Horizontal momentum equations (u, v), no viscosity
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Our horizontal momentum equation(pressure coordinate)
p
pp
pp
fDtD
fuydt
d
fxdt
du
uku
)()v(
v)()(
Assume no viscosity
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Next time
• The material derivative.
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Weather
• National Weather Service– http://www.nws.noaa.gov/– Model forecasts:
http://www.hpc.ncep.noaa.gov/basicwx/day0-7loop.html
• Weather Underground– http://www.wunderground.com/cgi-bin/findweather/getForecast?
query=ann+arbor– Model forecasts: http://
www.wunderground.com/modelmaps/maps.asp?model=NAM&domain=US
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Derivation of Coriolis Force (conclusion)
return to body of lecture
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Displace parcel up (1)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
Let’s imagine we move our parcel of air up (or down). What happens? Δz
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Displace parcel up (2)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
We get some change ΔR
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Displace parcel up (3)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
We get some change ΔR
zR )cos(
For our upward displacement
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Displace parcel up (4)(Conservation of angular momentum)
auww
dtdu
)cos(2
Remember this is ONLY FOR a VERTICAL displacement.
Do the same form of derivation
return to body of lecture
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Displace parcel east (1)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
Let’s imagine we move our parcel of air east (or west). What happens? Δx
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Displace parcel east (2)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
Well, there is no change of ΔR.
But remember
)(2RuRL
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Displace parcel east (3)(Conservation of angular momentum)
• So, we have changed u (=dx/dt). Hence again we have a question of conservation of angular momentum.
• We will think about this as an excess (or deficit) of centrifugal force relative to that from the Earth alone.
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Displace parcel east (4)(Conservation of angular momentum)
This excess FORCE is defined as
RR
RR
2
2
22
2
)(
Ru
Ru
RuF lcentrifuga
excess
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Displace parcel east (5)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
Vector with component in north-south and vertical direction
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Displace parcel east (6)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
For the Coriolis component magnitude is 2Ωu. For the curvature (or metric) term the magnitude is u2/(acos())
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Displace parcel east (7)(Conservation of angular momentum)
These forces in their appropriate component directions are
auu
dtdw
auu
dtd
2
2
)cos(2
)tan()sin(2v
return to body of lecture