Chapter 3

Space, Time, and Motion

(1) Wind Observations

• Vectors have both magnitude and direction.

• Wind is a vector quantity.

• The components of wind can be expressed in the Cartesian coordinates: x, y, z.

• Wind Instruments:– Anemometer and wind vane– Aerovane– Sonic anemometer

• Anemometer and wind vane.

• Aerovane

• Sonic Anemometers

(2) Plotting Data

Each element has a standard location and format.

Station V N dd ff TT TdTdTd PPPPP a ppp ww W Cl Cm Ch

Cordoba,

Argentina

11/2 8 25 08 17 16 1021.2 3 017 58 6 7 2 8

o

Station Id.

Air Temp.

(C)

Dew Point Temp.

(C)

Height (meters)

Wind Direction (degrees)

Wind Speed (knots)

McGrath, Alaska. 70231

-30

-34

5450

100

27

O

(3) Derivatives in Time and Space

• If a graph is drawn with the values of a quantity on the vertical and time on the horizontal, then drawing a line tangent to the graph line and determining its value (slope) will determine the derivative of that quantity with respect to time.

• The tangent technique measures the derivative - how the quantity changes with respect to time.

• Similarly, if the quantity was drawn on a graph with a spatial direction; e.g., the x-, y-, or z-direction. Then, the slope of the tangent to the line at the location of interest determines the derivative of the quantity with respect to that direction.

• Actually, these derivatives are estimations since it is very difficult to draw an accurate line tangent to the parameter line, and we are looking at the change over a rather large time interval or large x-direction interval.

• Calculus considers the denominator value as approaching zero.

• Secondly, these derivatives should be considered partial derivatives because temperature changes in all three directions as well as time.

• If we are considering only the change in the x-direction, we are assuming that there are no changes in the other directions or time.

(4) Advection

• Advection (in meteorology) is the rate of change of some property of the atmosphere by the horizontal movement of air.

• The rate of change is a derivative (with respect to time).

• If at 2:00 pm the carbon monoxide concentration was 80 ppb at a location 30 nm upstream from Savannah, Georgia, and the wind were blowing at 15 knots toward Savannah, when would the concentration at Savannah reach 80 ppb with no sources or sinks.

• The 80 ppb air must travel 30 nm and it is moving at 15 knots. Divide 30 nm by 15 knots (nm/hr) and you get a travel time of 2 hours. 2:00 pm + 2 hours = 4:00 pm.

• Suppose the concentration at Savannah at 2:00 pm were 60 ppb. How rapidly will the carbon monoxide concentration change.

• The rate of change = (change in concentration)/(change in time)

• The change in concentration of carbon monoxide can be written as:

• The subscript is there to indicate we are only considering the change at Savannah which is not moving. The x-direction is toward Savannah.

€

Δ CO[ ]( ) x

• Thus, the rate of change = =

• The wind speed (magnitude of the horizontal wind velocity vector in the direction of interest) can be written as:

• Then,

€

(80 ppb - 60 ppb)

(2 hours)=

20ppb

2 hrs=

10ppb

hr= 3 x 10−3 ppb/sec

€

rv =

Δx

Δt

€

Δt =Δxr v

€

Δ CO[ ] x

Δt

• We can then write the rate of change as:

• We can get the rate of change with time (derivative with respect to time) from the rate of change with distance (derivative with respect to distance) if we know the velocity of the wind.

• Essentially, it is simply:

€

Δ CO[ ]( ) x

Δt=

r v

Δ CO[ ]( ) x

Δx

€

Δ CO[ ]

Δt=

Δx

Δt⋅

Δ CO[ ]

Δx=

r v ⋅

Δ CO[ ]

Δx

(5) The One-Dimensional Vector Equation

• If we consider the change in time and the change in x as approaching zero, we have the instantaneous rate of change at a point (e.g., Savannah).

• Writing in calculus form (partial differential equation since we are only considering the change in the direction of the wind field (our x-direction), we have:

€

∂ CO[ ]

∂t

⎛

⎝ ⎜

⎞

⎠ ⎟x

= −u∂ CO[ ]

∂x

⎛

⎝ ⎜

⎞

⎠ ⎟t

• This is a general advection equation (along the x-coordinate - the west to east direction). (We are using “u” the component of the wind along the x-coordinate

• One can write such an equation for the advection along the y-coordinate, or z-coordinate.

Consider the following analysis of [CO].

To get the rate of change of [CO] (partial derivative of [CO] with respect to time) at Savannah, we need the partial derivative of [CO] with respect to x and the average wind speed.

Notice that concentrations at Savannah are less than they are upwind, so the rate of change of [CO] over time should be positive.

Just as on a graph, to get the change in concentration of [CO], pick two points on either side of the point of interest and get the difference between those values. In this case, 50-70= -20.

Now, divide by the distance between those points, 30nm. This will be the slope of the graph line at Savannah which will be the rate of change of [CO] with respect to distance at Savannah.

€

50ppb − 70ppb

30nm=

−20ppb

30nm= −0.667ppb / nm

• The wind speed is everywhere 15 knots (nm/hr) so the average wind speed is 15 nm/hr.

• Then, the rate of change of [CO] with time is:

• Similarly, the equation can be set up for any spatially-varying atmospheric variable; such as, temperature:€

∂ CO[ ]

∂t

⎛

⎝ ⎜

⎞

⎠ ⎟x

= −u∂ CO[ ]

∂x

⎛

⎝ ⎜

⎞

⎠ ⎟t

= −15nm / hr ⋅ −0.667ppb / nm( ) = 10ppb / hr

€

∂T

∂t= −u

∂T

∂x

(6) Equations on the Brain

• To understand the equation and how it relates to the atmosphere:– Say it in words.– See if it makes sense if each variable or

derivative, in turn, is zero.– See if the signs make sense.– See it it makes sense if certain variables

get larger or smaller.– Make up a concrete example and work

through it.

• For the advection equation.– The equations states that the rate of

change of temperature with respect to time at a particular location is equal to the negative of the wind speed times the rate of variation of temperature in the direction toward which the wind is blowing.

€

∂T

∂t= −u

∂T

∂x

• For various terms set to zero.– If wind speed is zero, no air is being

transported, so the wind is not changing the temperature, so the advection is zero.

– If is zero, then the temperature

is uniform along the x-axis, so the

air blowing in is the same temperature as the air blowing out so advection is zero.

€

∂T

∂t= −0

∂T

∂x

€

∂T

∂t= −u ⋅0

€

∂T

∂x

• For this situation, remember, the equation relates to advection at that instant, at a particular location. Here the advection is zero because the change in temperature at that instant is zero. You would have to average u and ∂T/∂x over a much larger distance to get a non-zero value, not just locally.

• Does the sign make sense?– If this were analyzed

temperature, orient x to point

toward where the wind is

blowing, then u will always be positive.

– The sign of the temperature change with time then will be determined by the sign of the downwind variation with temperature.

If ∂T/∂x is positive, (T down - T up), then if you graphed T versus x, the slope would be positive.

•Warmer temperatures would be downwind.If ∂T/ ∂x is negative, then a slope of T vs x would have colder temperatures downwind (slope would be negative) as we have.

• Check the magnitude.– Suppose the average wind is 5 m/s toward

east and temperatures are colder upstream, warmer downstream.

– Then, if the wind were stronger, the change would occur faster and temperatures would drop faster - greater change in temperature with time - greater advection.

– If the temperature change with distance is small, the temperature change over time would be small.

• Check with numbers.– Assume a wind speed of 5 m/s and

temperatures are colder by 5oK over a distance of 100km upstream.

– Then,with a wind speed of 5 m/s, how long will it take the colder air to travel 100km?

– So, temperature should drop at a rate of 5oK every 2 x 104 seconds. This is:

€

100km

5 mm

=1x105 km

5 ms

= 2x104 sec

€

∂T

∂t=

−5o K

(2x104s)= −2.5x10

oKs

• Check equation with numbers.– Assume a wind speed of 5 m/s and

temperatures are colder by 5oK over a distance of 100km upstream.

– Then,– and

€

∂T

∂x= 5

oK100km = 5oK

100000m = 5.0 × 10−5 oKm

€

∂T

∂t= −u ⋅

∂T

∂x= −5 m

s ⋅50K

1x105 m

⎛

⎝ ⎜

⎞

⎠ ⎟= −25x10−5 oK

s = −2.5x10−4 oKs

(6) Space-Time Conversion

• In a constant negative wind field (wind blowing from the east toward the west at the same speed all about the point of interest, the advection equation becomes:

• Thus, a graph of temperature vs. time would have the same shape as a graph of temperature vs. location (x-coordinate).

• The only difference would be the scale determined by the constant wind speed.

• And, whatever changes in temperature occur at a given location (∂T/∂t) must correspond to variations in the upstream temperature pattern (∂T/∂x).

€

∂T

∂t= (constant)⋅

∂T

∂x

• The differences in the horizontal scale is simply due to the speed of the wind.

(8) Phase Speed• The examples have been using wind as the

cause of the advection, but the same concept can be applied to anything that is moving regardless of the cause.

• As long as you can tell how fast it is moving (e.g., a cold front), the speed can be used in the advection equation to determine the change over time.

• The time record of observed meteorological variables at a particular space (location) can be converted directly into a depiction of the horizontal structure of the phenomena.

• The front is moving at 20 knots and it passed North Platte two hours ago, so it should be 40 miles past North Platte, at location B.

• If temperature at North Platte two hours ago was 52oF and it is moving toward Dodge City at 20 knots, the 52oF air should arrive in Dodge City in 5 hours after being in North Platte, or in another 3 hours.

Questions:

• Do: 1, 2, 3, 4, 5