chapter 10
Copyright © 2011, 2015 by Roland Stull. Meteorology for Scientists and Engineers, 3rd Ed.
Dynamics
Contents We use winds to power our wind turbines, push our sailboats, cool our Winds and Weather Maps 290 houses, and dry our laundry on the Height Contours on Isobaric Surfaces 290 10 clothesline. But winds can also be destructive — in Plotting Winds 291 hurricanes, thunderstorms, or mountain downslope Newton’s Second Law of Motion 292 windstorms. We design our bridges and skyscrap- Lagrangian Momentum Budget 292 ers to withstand wind gusts. Airplane flights are Eulerian Momentum Budget 293 planned to compensate for headwinds and cross- Horizontal Forces 294 winds. Advection 294 Winds are driven by forces acting on air. But Pressure-Gradient Force 295 these forces can be altered by heat and moisture car- Centrifugal Force 296 ried by the air, resulting in a complex interplay we Coriolis Force 297 call weather. The relationship between forces and Turbulent-Drag Force 300 Summary of Forces 301 winds is called atmospheric dynamics. Newto- nian physics describes atmospheric dynamics well. Equations of Horizontal Motion 301 Pressure, drag, and advection are atmospheric Horizontal Winds 302 forces that act in the horizontal. Other forces, called Geostrophic Wind 302 apparent forces, are caused by the Earth’s rotation Gradient Wind 304 (Coriolis force) and by turning of the wind around a Boundary-Layer Wind 307 Boundary-Layer Gradient (BLG) Wind 309 curve (centrifugal and centripetal forces). Cyclostrophic Wind 311 These different forces are present in different Inertial Wind 312 amounts at different places and times, causing large Antitriptic Wind 312 variability in the winds. For example, Fig. 10.1 shows Summary of Horizontal Winds 313 changing wind speed and direction around a low- Horizontal Motion 314 pressure center. In this chapter we explore forces, Equations of Motion — Revisited 314 winds, and the dynamics that link them. Scales of Horizontal Motion 315 Vertical Forces and Motion 315 Mass Conservation 317 Continuity Equation 317 Incompressible Continuity Equation 318 Boundary-Layer Pumping 319 Kinematics 320 Measuring Winds 321 Summary 322 Threads 322 Exercises 322 Numerical Problems 322 Understanding & Critical Evaluation 324 Web-Enhanced Questions 326 Synthesis Questions 327
Z 1 L1B “Meteorology for Scientists and Engineers, 3rd Edi- Y tion” by Roland Stull is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. To view a copy of the license, visit Figure 10.1 http://creativecommons.org/licenses/by-nc-sa/4.0/ . This work is Sketch of sea-level pressure (thin lines are isobars) & the result- available at http://www.eos.ubc.ca/books/Practical_Meteorology/ . ing near-surface winds (arrows). “L” is low pressure center. 289 290 CHAPTER 10 Dynamics
Winds and Weather Maps
B Height Contours on Isobaric Surfaces 1 Pressure-gradient force is the most important force because it is the only one that can drive winds L1B in the horizontal. Other horizontal forces can alter
[ LN an existing wind, but cannot create a wind from calm air. All the forces, including pressure-gradient force, are explained in the next sections. However, to understand the pressure gradient, we must first understand pressure and its atmospheric variation. Y We can create weather maps showing values of the pressures measured at different horizontal locations all at the same altitude, such as at mean- sea-level (MSL). Such a map is called a constant- height map. However, one of the peculiarities of meteorology is that we can also create maps on oth- C [ er surfaces, such as on a surface connecting points of equal pressure. This is called an isobaric map. L1B Both types of maps are used extensively in meteo- rology, so you should learn how they are related. Z In Cartesian coordinates (x, y, z), height z is geo- metric distance above some reference level, such - as the ground or sea level. Sometimes we use geopotential height H in place of z, giving a coor- LN dinate set of (x, y, H) (see Chapter 1). However, an Y alternative coordinate system can use pressure P as the vertical coordinate, because pressure decreases monotonically with increasing height. Pressure coordinates consist of (x, y, P). A monotonic variable is one that changes only D [ LNTVSGBDF E 1 L1BTVSGBDF in one direction. For example, it increases or is constant, but never decreases. Or it decreases or is 1 L1B [ LN constant, but never increases. Pressure in the atmo- Z Z sphere always decreases with increasing height. A surface connecting points of equal pressure is an isobaric surface. In low-pressure regions, this surface is closer to the ground than in high-pressure regions (Fig. 10.2b). Thus, this surface curves up and down through the atmosphere. Although isobaric Y Y surfaces are not flat, we draw them as flat weather maps on the computer screen or paper (Fig. 10.2d). Low pressures on a constant-height map cor- respond to low heights on a constant-pressure Figure 10.2 surface. High pressures on a constant height map (a) Vertical slice through atmosphere, showing pressure values correspond to high heights on a constant pressure (kPa). Thick dashed line is the 70 kPa isobar. Thin straight line surface. Similarly, regions on a constant-height map is the 3 km height contour. The location of lowest pressure on that have tight packing (close spacing) of isobars the height contour corresponds to the location of lowest height correspond to regions on isobaric maps that have of the isobar. (b) 3-D sketch of the same 70 kPa isobaric sur- tight packing of height contours, both of which are face (shaded), and 3 km height surface (white). (c) Pressures regions of strong pressure gradients that can drive intersected by the 3 km constant height surface. (d) Heights crossed by the 70 kPa surface. The low-pressure center (L) in (c) strong winds. This one-to-one correspondence of matches the low-height center in (d). both types of maps (Figs. 10.2c & d) makes it easier for you to use them interchangeably. R. STULL • meteorology for scientists and engineers 291
It is impossible for two different isobaric surfaces FOCUS • Why use isobaric maps? to cross each other. Also, these surfaces never fold back on themselves, because pressure decreases Constant pressure charts are used for five reasons. monotonically with height. However, they can in- First, the old radiosonde (consisting of weather sen- tersect the ground, such as frequently happens in sors hanging from a free helium balloon that rises mountainous regions. into the upper troposphere and lower stratosphere) Isobaric charts will be used extensively in the re- measured pressure instead of altitude, so it was easier mainder of this book when describing upper-air fea- to plot their measurements of temperature, humid- tures (mostly for historical reasons; see Focus box). ity and wind on an isobaric surface. For this reason, Fig. 10.3 is a sample weather map showing height upper-air charts (i.e., showing weather above the contours of the 30 kPa isobaric surface. ground) traditionally have been drawn on isobaric maps. Second, aircraft altimeters are really pressure Plotting Winds gauges. Aircraft assigned by air-traffic control to a Symbols on weather maps are like musical notes specific “altitude” above 18,000 feet MSL will actually in a score — they are a shorthand notation that con- fly along an isobaric surface. Many weather observa- cisely expresses information. For winds, the symbol tions and forecasts are motivated by aviation needs. is an arrow with feathers (or barbs and pennants). Third, pressure is a measure of mass in the air, so The tip of the arrow is plotted over the observation every point on an isobaric map has the same number (weather-station) location, and the arrow shaft is of air molecules above it. aligned so that the arrow points toward where the Fourth, an advantage of using equations of motion wind is going. The number and size of the feath- in pressure coordinates is that you do not need to con- ers indicates the wind speed (Table 10-1, copied from sider density, which is not routinely observed. Table 9-9). Fig. 10.3 illustrates wind barbs. Fifth, some weather forecast models use pressure coordinate systems in the vertical. However, more and more routine upper-air sound- ings around the world are being made with modern Table 10-1. Interpretation of wind barbs. GPS (Global Positioning System, satellite Symbol Wind Speed Description triangulation method) sondes that can measure geo- calm two concentric circles metric height as well as pressure. Also, some of the modern weather forecast models do not use pressure 1 - 2 speed units shaft with no barbs as the vertical coordinate. In the future, we might see 5 speed units a half barb (half line) the large government weather data centers starting to produce upper-air weather maps on constant height 10 speed units each full barb (full line) surfaces. 50 speed units each pennant (triangle) • The total speed is the sum of all barbs and pennants. For example, indicates a wind from the west at 65$ speed 75 units. Arrow tip is at the observation location. - +VMZ • CAUTION: Different organizations use different 1 L1B LN speed units, such as knots, m/s, miles/h, km/h, etc. [ LN Look for a legend to explain the units. When in doubt, / assume knots — the WMO standard. For unit conver- sion, a good approximation is 1 m/s ≈ 2 knots.
Solved Example Draw wind barb symbol for winds from the: (a) northwest at 115 knots; (b) northeast at 30 knots. / ) Solution 8 8 8 (a) 115 knots = 2 pennants + 1 full barb + 1 half barb. (b) 30 knots = 3 full barbs Figure 10.3 B C Weather map for a 30 kPa constant pressure surface over central Check: Consistent with Table 10-1. N. America. Solid contours show height z (km) of this surface above mean sea level (MSL). Hence, this is called a “30 kPa Discussion: Feathers (barbs & pennants) should be height chart”. Wind barbs (Table 10-1) show wind observations on the side of the shaft that would be towards low in knots (2 knots ≈ 1 m/s). The relative maxima and minima pressure if the wind were geostrophic. heights are labeled as H (high heights) and L (low heights). 292 CHAPTER 10 Dynamics
Newton’s Second Law of Motion
Lagrangian Momentum Budget FOCUS • In Newton’s Own Words Forces, winds, and acceleration are vectors pos- sessing both magnitude and direction. Newton’s Newton’s laws of motion, in his own words, are second law of motion (often shortened to New- given below. Actually, his original words were Latin, → → ton’s second law) states that a vector force F = m·a the language of natural philosophy (science) at that acting on an object such as an air parcel of mass m time (1687). Here is the translation from Newton’s → → Philosophiæ Naturalis Principia Mathematica (“Math- causes it to accelerateF = m · a in the same direction as the ematical Principles of Natural Philosophy”): applied force:
“Law I. Every body perseveres in its state of be- → → ing at rest or of moving uniformly straight forward, F = m·a •(10.1) except inasmuch as it is compelled by impressed forc- es to change its state. Acceleration is defined as the rate of change of ve- “Law II. Change in motion is proportional to the locity with time t: motive force impressed and takes place following the straight line along which that force is impressed. → → ∆V “Law III. To any action, there is always a con- a = (10.2) trary, equal reaction; in other words, the actions of ∆t two bodies each upon the other are always equal and opposite in direction. → where V is the vector wind velocity. “Corollary 1. A body under the joint action of Combining eqs. (10.1) and (10.2) give forces traverses the diagonal of a parallelogram in the same time as it describes the sides under their sepa- → rate actions.” → ∆V F = m· (10.3a) ∆t
If mass is constant, then the equation can be written as:
→ → ∆(m·) V Solved Example F = (10.3b) A 1500 kg car accelerates from 0 to 60 mph (0 to 96.6 ∆t km/h) in 9 seconds, heading south. (a) Find its aver- age acceleration. (b) What magnitude and direction of But mass times velocity equals momentum. Thus, force acted on it to make it accelerate? eq. (10.3b) describes the change in momentum with time following the air parcel; namely, it is the Solution Lagrangian momentum budget. → → Rearranging eq. (10.3a) gives a forecast equation Given: V initial = 0, V final = 60 mph = 27 m/s for the wind velocity: tinitial = 0, tfinal = 9 s. Direction is south. m = 1500 kg. → → → → → → FFind: = m ·a = ? m·s–2 , F = m? N·a ∆V Fnet (10.4) = → ∆t m → ∆V (a) Use eq. (10.2): a = ∆t The subscript “net” implies that there might be = (27 – 0 m/s) / (9 – 0 s) = 3 m·s–2 to the south many forces acting on the air parcel, and we need to
→ consider the vector sum of all forces in eq. (10.4), as → –2 (b) Use eq. (10.1): F = m(1500·a kg) · ( 3 m·s ) given by Newton’s Corollary 1 (see Focus box). = 4500 N to the south → → (From Appendix A, recall that 1 N = 1 kg·m·s–2) For situations where F net=/ mm· =a 0, eq. (10.4) tells us that the flow will maintain constant velocity due to → Check: Units OK. Physics OK. inertia. Namely, ∆ V /∆t = 0 implies that Discussion: If the car falls off a cliff, gravity would → → accelerate it at g = –9.8 m·s–2 , where the negative sign V = constant (not that V = 0). denotes the downward direction. R. STULL • meteorology for scientists and engineers 293
Eulerian Momentum Budget Solved Example For wind forecasts over a fixed location such as –4 –2 If Fx net/m = 1x10 m·s acts on air initially at rest, a town or lake, use an Eulerian reference frame. then what is the final wind speed after 10 minutes? Define a local Cartesian coordinate system with x increasing toward the local East, y toward the local Solution –4 –2 North, and z up (see Chapter 1). Given: U(0) = 0 , Fx net/m = 1x10 m·s , ∆t = 600 s Wind components in a Cartesian framework can Find: U(∆t) = ? m/s. Assume: V=0. be found by rewriting the vector equation (10.4) as three separate scalar equations: one for the west-to- Use eq. (10.6a): U(t+∆t) = U(t) + ∆t · (Fx net/m) –4 –2 east wind component (U), one for the south-to-north = 0 + (600s)·(1x10 m·s ) = 0.06 m/s. component (V), and one for the vertical component (W): Check: Units OK. Physics OK. ∆U Fx net Discussion: Not very fast, but over many hours it = •(10.5a) ∆t m becomes large. Positive U means it is toward the east.
∆V Fy net = •(10.5b) ∆t m ON DOING SCIENCE • Be Creative
∆W Fz net Isaac Newton grew up on his mother’s farm at = •(10.5c) ∆t m Woolsthorpe, and built model windmills, clocks, and sundials. His grades in school were OK. He got into where subscripts (x, y, z) indicate the component fights with his classmates, and carved his name in his of the net vector force toward the (east, north, up), desk. His schoolmaster saw a spark of talent in Isaac, respectively. Relationships between the horizontal and suggested to his mother that Isaac should go to “speed and direction” method of representing a vec- college, because he would never be a good farmer. tor wind versus the “U and V component” method At age 18, Isaac Newton went to Cambridge Uni- were given in Chapter 1. versity in England in 1661, and after working at odd jobs to pay his way, finally graduated with a B.A. in Recall that ∆U/∆t = [U(t+∆t) – U(t)]/∆t. Thus, we 1665. Later that year the plague hit, killing 10% of the can rewrite eqs. (10.5) as forecast equations: London population in three months. For fear that the plague would spread, Cambridge Univ. was closed Fx net U() t + ∆t= U()t + ·∆t •(10.6a) until 1667. Isaac and the other students went home. m He continued his scientific studies in seclusion at his mother’s farmhouse during the 18 months that Fy net V() t + ∆t= V()t + ·∆t •(10.6b) school was closed. During this period, at age 23 to m 24, he laid the groundwork for many of his major dis- coveries. This included the laws of motion, the study Fz net W() t + ∆t= W()t + ·∆t •(10.6c) of optics, the invention of the reflecting telescope, the m explanation for the orbits of planets, the understand- ing of gravity, and the co-invention of calculus. These equations are often called the equations Often the most creative science, music, literature, of motion. Together with the continuity equation and art are done by young men and women who have (later in this chapter), the equation of state (ideal gas not been biased and (mis)directed by studying the law from Chapter 1), and the energy budget equations works of others too much. Such knowledge of past in the Heat and Moisture chapters, they describe the work will often subconsciously steer one’s research in dynamic and thermodynamic state of the air. the directions that others have already taken, which To forecast winds [U(t+∆t), V(t+∆t), W(t+∆t)] at unfortunately discourages novel ideas. some future time, we must know the winds now Be creative, and learn from your mistakes. So what [U(t), V(t), W(t)], and the forces acting on the air. if you “re-invent the wheel”, and “discover” some- Mathematically, this is known as an initial-value thing that was already discovered a century ago. The problem, because we must know the initial winds freedom to make personal discoveries and mistakes and the knowledge you gain by doing so allows you to forecast the future winds. Even numerical weath- to be much more creative than if you had just read er forecast models (see the NWP chapter) must start about the end result in a journal. with an analysis of current weather observations. So we have a paradox. I wrote this book to help Average horizontal winds are often 100 times you learn the meteorological advances made by oth- stronger than vertical winds, except in thunder- ers, but I discourage you from studying the works storms and near mountains. We will focus on hori- of others. As a scientist or engineer, you must make zontal forces and winds first, and return to vertical your own decision about the best balance of these two winds later in this chapter. philosophies that will guide your future work. 294 CHAPTER 10 Dynamics
Horizontal Forces
To solve the equations of motion for horizontal Science Graffito winds in an Eulerian framework, we need to know “If I have been able to see further than others, it the horizontal forces acting on the air. The net was because I stood on the shoulders of giants.” “force per unit mass” consists of contributions from – Sir Isaac Newton. advection (AD), pressure-gradient force (PG), Coriolis force (CF), and turbulent drag (TD). In addition, imbalances between forces sometimes bal- ance centrifugal force (CN):
Fxnet Fx ADF x PG FxCNF xCF FxTD = + + + + (10.7a) m m m m m m
Fy net Fy ADF y PG FyCNF yCF FyTD = + + + + (10.7b) m m m m m m
Units of force per mass are N/kg, which is identical to units of m·s–2 (see Appendix A). We will use these latter units.
Advection Not only can wind blow air of different tempera- ture or humidity into a region, but it can also blow air of different specific momentum (i.e., momen- tum per unit mass). Recall that momentum is de- fined as mass times velocity, hence specific momen- tum equals the velocity (i.e., the wind) by definition. Thus, the wind can blow different winds into a re- gion. Namely, winds can change due to advection, in an Eulerian framework. B C D This is illustrated in Fig. 10.4a. Consider a mass of Z 7 air (grey box) with slow U wind (5 m/s) in the north and faster U wind (10 m/s) in the south. Thus, U Y decreases toward the north, giving ∆U/∆y = negative. NT This whole air mass is advected toward the north 6 over a fixed weather station “O” by a mean wind V( 7 = positive). At the time sketched in Fig. 10.4b, a west 0 0 0 wind of 5 m/s is measured at “O”. Later, at the time NT NT of Fig. 10.4c, the west wind has increased to 10 m/s at the weather station. The rate of increase of U at 7 6 “O” is larger for faster advection (V), and is larger if ∆U/∆y is more negative. Thus, ∆U/∆t = –V · ∆U/∆y for this example. The NT NT advection term on the RHS causes an acceleration of U wind on the LHS, and thus acts like a force per 6 unit mass: ∆U/∆t = Fx AD/m = –V · ∆U/∆y . Advection is not usually considered a force in the NT traditional Lagrangian sense, but you must always include it when momentum budget equations are Figure 10.4 rewritten in Eulerian frameworks. You have seen Illustration of V advection of U wind. “O” is a fixed weather similar advection terms in the Eulerian heat and station. Grey box is an air mass containing a gradient of U moisture budget equations earlier in this book. wind. Initial state (a) and later states (b and c). R. STULL • meteorology for scientists and engineers 295
In general, the components of advective force per Solved Example unit mass are: Vancouver (British Columbia, Canada) is roughly Fx AD ∆U ∆U ∆U 250 km north of Seattle (Washington, USA). The winds = −U··− V − W· •(10.8a) m ∆x ∆y ∆z (U, V) are (8, 3) m/s in Vancouver and (5, 5) m/s in Se- attle. Find the advective force per unit mass. Fy AD ∆V ∆V ∆V = −U··− V − W· •(10.8b) Solution m ∆x ∆y ∆z Given: (U, V) = (8, 3) m/s in Vancouver, where ∆U/∆x is the gradient of U-wind in the x-di- (U, V) = (5, 5) m/s in Seattle rection, and the other gradients are defined similar- ∆y = 250 km, ∆x = not relevant (unknown) –2 –2 ly. Advection needs a gradient (i.e., change across a Find: Fx AD/m =? m·s , Fy AD/m =? m·s distance). Without a change of wind with distance, momentum advection cannot cause accelerations Use the definition of a gradient: ∆U/∆y = (8 – 5 m/s)/250,000 m = 1.2x10–5 s–1 Vertical advection of horizontal wind (–W·∆U/∆z ∆U/∆x = 0 (unknown in this problem) in eq. 10.8a, and –W·∆V/∆z in eq. 10.8b) also exists. ∆V/∆y = (3 – 5 m/s)/250,000 m = –0.8x10–5 s–1 But W is often very small outside of thunderstorms, ∆V/∆x = 0 (unknown) so we neglect vertical advection here. Average U = (8 + 5 m/s)/2 = 6.5 m/s Average V = (3 + 5 m/s)/2 = 4 m/s Pressure-Gradient Force Use eq. (10.8a): Pressure-gradient force always acts perpendicu- –5 –1 Fx AD/m = –(6.5m/s)·0 – (4m/s)·(1.2x10 s ) larly to the isobars (or height contours) on a weather = –4.8x10–5 m·s–2 map, from high to low pressure (or heights). This Use eq. (10.8b): –5 –1 force exists regardless of the wind speed, and does Fy AD/m = –(6.5m/s)·0 – (4m/s)·(–0.8x10 s ) not depend on the wind speed. It starts the horizon- = 3.2x10–5 m·s–2 tal winds and can accelerate, decelerate, or change the direction of existing winds. On a weather map, Check: Units OK. Physics OK. more closely spaced isobars (i.e., more closely Discussion: The U winds are slower in Seattle than packed, with smaller distance ∆d between them) in- Vancouver, but are being blown toward Vancouver by dicate a greater pressure-gradient force (Fig. 10.5). the southerly flow. Thus, advection is decreasing the The components of pressure-gradient force are: U-wind, hence, the negative sign. The V-wind is faster in Seattle, and these faster winds are being blown to- Fx PG 1 ∆P ward Vancouver, causing a positive acceleration there. = − · •(10.9a) m ρ ∆x
Fy PG 1 ∆P •(10.9b) = − · Solved Example m ρ ∆y Milwaukee is 100 km east of Madison, Wisconsin, USA. The sea-level pressure at Milwaukee is 100.1 kPa and at Madison is 100 kPa. What is the pressure gradi- ent force per mass? Assume ρ = 1.2 kg·m–3. OPSUI L1B Solution 1 L1B Define: Cartesian coord. with x = 0 at Madison. L1B Given: P = 100.1 kPa at x = 100 km, P = 100.0 kPa at x = 0 km. ρ = 1.2 kg·m–3. ' –2 1( Find: Fx PG/m =? m·s åZ Use eq. (10.9a): 1 åE åY Fy PG 1 (,101 000 − 100,)000 Pa = − · m (.1 1kg·)m−3 (4000,)000 − 0 m
1 å1 –4 –2 TPVUI = –8.33x10 m·s . –1 –2 XFTU FBTU where 1 Pa = 1 kg·m ·s was used (Appendix A).
Check: Units OK. Physics OK. Figure 10.5 Discussion: The negative answer implies that the Pressure gradient force (heavy line) is perpendicular to isobars force is in the negative x-direction; that is, from Mil- (medium lines) from high to low pressure. H and L indicate waukee toward Madison. regions of high and low pressure, respectively. 296 CHAPTER 10 Dynamics
where ρ is air density, and ∆P is the change of pres- Solved Example sure across distance ∆x or ∆y. The negative sign If the height of the 50 kPa pressure surface decreas- es by 10 m northward across a distance of 500 km, what makes the force act from high toward low pressure. is the pressure-gradient force? The magnitude of pressure-gradient force is
F 1 ∆ P (10.10) Solution PG = · Given: ∆z = –10 m, ∆y = 500 km, |g|= 9.8 m·s–2 . m ρ ∆ d –2 Find: FPG/m = ? m·s where ∆d is the distance between isobars. Use eqs. (10.11a & b): The hydrostatic equation (1.25) can be used to –2 Fx PG/m = 0 m·s , because ∆z/∆x = 0. Thus, FPG/m convert the pressure-gradient terms from height = Fy PG/m. to pressure coordinates. On isobaric surfaces, the Fy PG ∆z m −10m pressure-gradient terms become: = − g ·.= − 9 8 · m ∆y s2 500, 000m –2 Fx PG ∆z FPG/m = 0.000196 m·s = − g · m ∆x (10.11a) Check: Units OK. Physics OK. Sign OK. Discussion: For our example here, height decreases Fy PG ∆z = − g · toward the north, thus a hypothetical ball would roll m ∆y (10.11b) downhill toward the north. A northward force is in the positive y direction, which explains the positive where ∆z/∆x and ∆z/∆y are the slopes of the isobaric sign of the answer. surfaces (i.e., change of height with distance), and |g| = 9.8 m·s–2 is the magnitude of gravitational ac- celeration. If you could place a hypothetical ball on the isobaric surface plotted in Fig. 10.2b, the direc- tion that it would roll downhill is the direction of the pressure-gradient force, and the magnitude is Table 10-2. Sign s for centrifugal-force equations.
For flow around a F ∆ z (10.12) Hemisphere PG = g · Low High m ∆ d Northern +1 –1 where ∆d is distance between height contours. Southern –1 +1 Pressure-gradient force is the ONLY force that can drive the horizontal winds in the atmosphere. The other forces, such as Coriolis, drag, centrifugal, and even advection, disappear for zero wind speed. Hence, these other forces can change the direction and speed of an existing wind, but they cannot cre- Solved Example ate a wind out of calm conditions. On the back side of a low pressure center in the northern hemisphere, winds are from the north at 10 m/s at distance 250 km from the low center. Find the Centrifugal Force centrifugal force. Newton’s laws of motion state that an object tends to move in a straight line unless acted upon by Solution '$/ 3 5 a force in a different direction. Such a force, called Given: R = 2.5x10 m, centripetal force, causes the object to change direc- V = – 10 m/s 7 tion and bend its trajectory. Centripetal force is Find: F /m = ? m·s–2. x CN the sum or the imbalance of other forces. Use eq. (10.13a), with s = +1 from Table 10-2. Centrifugal force is an apparent force that is opposite to centripetal force and pulls outward from FxCN (·−5m/s) ( 5m/s) = –4x10–4 m·s–2. the center of the turn. The components of centrifu- = −1· 5 m 5× 10 gal force are FxCN VM· Check: Units OK. Physics OK. Sketch OK. = +s· •(10.13a) Discussion: The negative sign indicates a force to- m R ward the west, which is indeed outward from the cen- Fy CN UM· ter of the circle. = −s· •(10.13b) m R R. STULL • meteorology for scientists and engineers 297
OPSUI *G/PSUIFSO)FNJTQIFSF where s is a sign coefficient given in Table 10-2, M = ( U2 + V2 )1/2 is wind speed (always positive), XJOE XJOE and R is the radius of curvature. The sign depends '$' on whether air is circulating around a high or low ' pressure center, and whether it is in the Northern or $' Southern Hemisphere. The total magnitude is: TPVUI *G4PVUIFSO)FNJTQIFSF F M2 XFTU FBTU CN = (10.14) m R Figure 10.6 Coriolis force (dark lines). Coriolis Force Coriolis force is an apparent force caused by the Solved Example rotation of the Earth. It acts perpendicular to the a) Plot Coriolis parameter vs. latitude. b) Find Coriolis force at Norman, OK, USA, for a wind of U = 10 m/s. wind direction, to the right in the N. Hemisphere, and to the left in the Southern (Fig. 10.6). Solution To understand Coriolis force, we need to quan- Given: U = 10 m/s, ϕ = 35.2°N at Norman. tify the rotation rate of the Earth. The Earth rotates Find: Plot f vs ϕ. Also: F /m = ? m·s–2. one full revolution (2π radians) during a sidereal c y CF day (i.e., relative to the fixed stars,P is a bit less –1 sidereal a) Find fc (s ) vs. ϕ(°) using eq. (10.16). For example: than 24 h, see Appendix B), giving an angular rota- –4 –1 –5 –1 fc = (1.458x10 s )·sin(35.2°) = 8.4x10 s . tion rate of
Ω = 2·π / Psidereal •(10.15)
= 0.729 211 6 x 10–4 radians/s
The units for Ω are often abbreviated as s–1. Using this rotation rate, a Coriolis parameter is defined -BUJUVEF m as fc = 2·Ω·sin() φ •(10.16) m
–4 –1 m where 2·Ω = 1.458423x10 s , and ϕ is latitude. m m m m m This parameter is constant at any fixed location. At $PSJPMJT1BSBNFUFS GD T mid-latitudes, the magnitude is on the order of fc = 1x10–4 s–1. b) Coriolis force in the y-direction (eq. 10.17b) is: –5 –1 –4 –2 In the N. Hemisphere, the Coriolis force is: Fy CF/m = –(8.4x10 s )·(10m/s) = – 8.4x10 m·s .
Check: Units OK. Physics OK. FxCF = f· V •(10.17a) Discussion: The – sign means force is north to south. m c
Fy CF •(10.17b) FOCUS • Coriolis Force in 3-D = − fc · U m Eqs. (10.17) give only the dominant components of Coriolis force. There are other smaller-magnitude Thus, there is no Coriolis force when there is no Coriolis terms (labeled small below) that are usually neglected. The full Coriolis force in 3-dimensions is: wind. Coriolis force cannot cause the wind to blow; FxCF it can only change its direction. =f·· V − 2Ωcos( φ)·W (10.17c) The magnitude of Coriolis force is: m c [small because often W< Fy CF | FCF /m | ≈ 2 · Ω ·|sin(ϕ)·M| (10.18a) = − fc · U (10.17d) or m | FCF /m | ≈ | fc · M | (10.18b) FzCF = 2Ω·cos()φ ·U (10.17e) nd m as is shown in the 2 Focus box on Coriolis force. [small relative to other vertical forces] 298 CHAPTER 10 Dynamics FOCUS • What is Coriolis Force? FOCUS • Coriolis Force (continuation) In 1835, Gaspar Gustave Coriolis used kinetic-en- /1 ' ergy conservation to explain the apparent force that () '$/7 now bears his name. The following clarification was provided by Anders Persson in 1998 and 2006. ' 3 $/ Background Coriolis force can be interpreted as the difference '$/) between two other forces: centrifugal force and gravi- FBSUI tational force. '( As discussed previously, centrifugal force is 2 G &2 FCN/m = (Mtan) /R , where Mtan is the tangential velocity of an object moving along a curved path Figure 10.c. Forces. '(7 with radius of curvature R (see Fig. 10.a). This force increases if the object moves faster, or if the radius Now that “up” and “down” are identified, we becomes smaller. The symbol X marks the center of can split centrifugal and actual gravitational forces rotation of the object, and the small black circle indi- into local horizontal (subscript H) and vertical (sub- cates the object. script V) components (Fig. 10.c). Because FCN is al- ways parallel to the equator (EQ), trigonometry gives '$/ F ≈ F ·sin(ϕ), where ϕ is latitude. CNH CN .UBO 3 Objects at Rest .UBO The Earth rotates counterclockwise when viewed from above the north pole (NP). During time interval ∆t, any single meridian (a longitude line, such as la- 3 beled with distance R in Fig. 10.d) will rotate by angle 9 Ω ∆t, where Ω is the angular velocity of Earth (= 360°/ sidereal day). Suppose an object (the dark circle) is at rest on this meridian. Then during the same time interval ∆t, it Figure 10.a. Centrifugal force, F . CN will move as shown by the grey arrow, at speed Mtan = Ω·R. Because this movement follows a parallel (latitude line), and parallels encircle the Earth’s axis, Because the Earth is plastic (i.e., deformable), cen- the stationary object is turning around a circle. trifugal force due to the Earth’s rotation and gravi- This creates centrifugal force. F have shaped the surface into an '$/) tational force G The horizontal component ellipsoid, not a sphere. This is exaggerated in Fig. FCNH balances FGH, 10.b. The vector sum of F and F is the .UBO G CN effec- giving zero net , F . This effective gravity acts per- 8p3 tive gravity EG apparent horizontal '() pendicular to the local surface, and defines the direc- force on the object 8påU tion we call . Thus, a stationary object feels down (Fig. 10.d). /1 9 no net force (the downward force is balanced by the 3 Earth holding it up). FBSUI Figure 10.d. /PSUI Object at rest. 1PMF /1 &2 '$/ Figure 10.b. Objects Moving East or West North-south Next, we can ask what happens if the object moves ' slice through ( eastward with velocity M relative to the Earth’s sur- Earth. ' face (shown by the thin white arrow in Fig. 10.e). The &( Earth is rotating as before, as indicated by the thin &RVBUPS &2 meridian lines in the figure. Thus, the total move- ment of the object is faster than before, as shown by &BSUI the grey arrow. This implies greater total centrifugal force, which results in a greater horizontal compo- nent FCNH. However, the gravity component FGH is unchanged. (continues in 4PVUI (continues in next column) 1PMF 41 next column) R. STULL • meteorology for scientists and engineers 299 FOCUS • Coriolis Force (continuation) FOCUS • Coriolis Force (continuation) '$/) gravitational component FGH, which hasn’t changed ' $' . ' very much relative to the other changes. However, . $/) the east-west component is acting to the right of the relative object motion M, and is identified as Coriolis '$' force: FCF = FCNH-ew. ' 8påU () '() Similarly, a southward moving object has a larg- 9 /1 9 er radius of curvature, giving a Coriolis force to the /1 3 3 right. In fact, an object moving in any arbitrary direc- FBSUI FBSUI tion has Coriolis force acting to the right in the North- ern Hemisphere. &2 &2 Figure 10.e. Figure 10.f. Magnitude of Coriolis Force Object moving east. Object moving west. For an object at rest (Figs. 10.c & d): Thus, those two forces (FCNH & FGH) do NOT bal- FGH = FCNH ≡ FCNHR (C1) ance. The difference between them is a net force to the right of the relative motion M. This force differ- where subscript R denotes “rest”. At rest: ence is called Coriolis force, FCF , and is indicated by the thick white arrow in Fig. 10.e. Mtan rest = Ω · R (C2) Similarly, for an object moving westward (thin white arrow in Fig. 10.f), the net tangential velocity For an eastward moving object (Fig. 10.e), Coriolis (grey arrow) is slower, giving an imbalance between force is defined as: FCNH & FGH that acts to the right of M. This is identi- fied as Coriolis force, as shown with the thick white FCF ≡ FCNH – FGH (definition) arrow. = FCNH – FCNHR (from eq. C1) Objects Moving North or South For a northward moving object, the rotation of the = sin(ϕ) · [FCN – FCNR] (from Fig. 10.c) Earth (dashed thick grey line) and the relative mo- tion of the object (M, thin white arrow) combine to Divide by mass m, and plug in the definition for cen- cause a path shown with the solid thick grey line (Fig. trifugal force as velocity squared divided by radius: 10.g). This has a smaller radius of curvature (R) about 2 2 a center of rotation (X) that is NOT on the North Pole FCF / m = sin(ϕ) · [ (Mtan) /R – (Mtan rest) /R ] (O). The smaller radius causes a greater horizontal component of centrifugal force (FCNH), which points Use Mtan = Mtan rest + M, along with eq. (C2): outward from X. 2 2 FCF / m = sin(ϕ) · [ (Ω·R+M) /R – (Ω·R) /R ] '$/) ' $/)OT = sin(ϕ) · [(2·Ω·M) + (M2/R)] The last term is small & can be neglected compared to the first term. Thus, the magnitude of Coriolis force ' . $/)FX is: '$' FCF /m ≈ 2·Ω·sin(ϕ) · M (10.18) 3 ≡ fc · M (from eq. 10.16) /1 This answer is found for motion in any direction. FBSUI 9 $FOUFSPG '() 3PUBUJPO &2 BEYOND ALGEBRA • Apparent Forces Figure 10.g. Object moving north. In vector form, centrifugal force/mass for an object at We can conceptually divide this horizontal cen- rest on Earth is –Ω × (Ω × r), and Coriolis force/mass trifugal force into a north-south component (F ) CNH-ns is –2Ω × V , where vector Ω points along the Earth’s and an east-west component (F ). The south- CNH-ew axis toward the north pole, r points from the Earth’s ward component F balances the horizontal CNH-ns center to the object, V is the object’s velocity relative (continues in next column) to Earth, and × is the vector cross product. 300 CHAPTER 10 Dynamics Turbulent-Drag Force XJOE GSFFBUNPTQIFSF At the Earth’s surface the air experiences drag against the ground. This turbulent-drag force in- creases with wind speed, and is always in a direc- IFJHIU tion opposite to the wind direction. Namely, drag CPVOEBSZ slows the wind (Fig. 10.7). XJOE MBZFS Only the boundary layer (see the Atmospheric [J Boundary Layer chapter) experiences this drag. It '5% is not felt by the air in the remainder of the tropo- sphere (except for deep vigorous thunderstorms or . for mountain-wave drag, see the Local Winds chap- ter). The drag force acting on a boundary layer of Figure 10.7 depth zi is: Turbulent drag force FTD opposes the wind in the boundary layer. FxTD U = −wT · •(10.19a) m zi FyTD V = −wT · •(10.19b) m zi Solved Example where w is a turbulent transport velocity. Find the drag force per unit mass on a wind of U T = 10 m/s, V = 0 for a: (a) statically neutral boundary The total magnitude of turbulent drag force is layer over smooth ground; and (b) statically unstable boundary layer with w = 45 m/s. The boundary layer B FTD M is 1 km thick. = wT · (10.20) m zi Solution Given: U = M = 10 m/s, zi = 1000 m, and is opposite to the wind direction. –3 CD = 2x10 , wB = 45 m/s. During windy conditions of near neutral static –2 Find: Fx TD/m = ? m·s . stability, turbulence is generated primarily by the wind shear (change of wind speed or direction (a) Combine eqs. (10.21) and (10.19a): with height). This turbulence transports frictional 2 FxTD U ()15m/s information upward from the ground to the air at = −CMD ·· = −(.0 02)· m zi 1500m rate: = –2x10–4 m·s–2. wT = CD · M (10.21) (b) Combine eqs. (10.22) and (10.19a): FxTD U where wind speed M is always positive, and CD is = −bDB·· w m z a dimensionless drag coefficient in the range of i –3 –2 ()15m/s 2x10 over smooth surfaces to 2x10 over rough = −(.0 00183)·()50m/s · or forested surfaces (see the Atmospheric Boundary 1500m Layer chapter). It is similar to the bulk heat transfer –4 –2 = –8.24x10 m·s . coefficient that was discussed in the Heat chapter. During statically unstable conditions of light Check: Units OK. Physics OK. winds and strong surface heating (e.g., daytime), Discussion: The negative sign means that the drag buoyant thermals transport the frictional informa- force is toward the west, which is opposite the wind direction. Both mechanical and buoyant turbulence tion upward at rate: are equally effective at transporting frictional infor- mation to the air. wT = bD · wB (10.22) where wB is the buoyancy velocity scale (always –3 positive, see the Heat chapter), and bD = 1.83x10 is dimensionless. R. STULL • meteorology for scientists and engineers 301 Summary of Forces Table 10-3. Summary of forces. Name of Magnitude Horiz. (H) Remarks (“item” is in col- Item Direction Force (N/kg) or Vert. (V) umn 1; H & V in col. 5) F hydrostatic equilibrium 1 gravity down G = g = 9.8 m·s–2 V m when items 1 & 2V balance pressure from high to low FPG 1 ∆ P the only force that can drive 2 = · V & H gradient pressure m ρ ∆ d horizontal winds 90° to right (left) geostrophic wind when Coriolis of wind in North- FCF 3 = 2·Ω· sin() φ ·M H* 2H and 3 balance (explained (apparent) ern (Southern) m later in horiz. wind section) Hemisphere boundary-layer wind when turbulent FTD M 4 opposite to wind = w · H* 2H, 3 and 4 balance (explained drag m T z i in horiz. wind section) 2 centripetal = opposite of centrifugal away from center F M 5 CN = H* centrifugal. Gradient wind (apparent) of curvature m R when 2H, 3 and 5 balance advection FAD ∆(UVor ) neither creates nor destroys 6 (any) = M · V & H (apparent) m ∆ d momentum; just moves it *Horizontal is the direction we will focus on. However, Coriolis force has a small vertical component for zonal winds. Turbulent drag can exist in the vertical for rising or sinking air, but has completely different form than the boundary-layer drag given above. Cen- trifugal force can exist in the vertical for vortices with horizontal axes. Note: units N/kg = m·s–2. Equations of Horizontal Motion 10x10–4 m·s–2 (which is equivalent units of N/kg, see Appendix A for review). For some situations, Combining the forces from eqs. (10.7, 10.8, 10.9, some of the terms are small enough to be neglect- 10.17, and 10.19) into Newton’s Second Law of Mo- ed compared to the others. For example, above the tion (eq. 10.5) gives simplified equations of horizon- boundary layer the turbulent-drag term is near zero. tal motion: Near the equator, Coriolis force is near zero. At the •(10.23a) center of high or low pressure regions, the pressure ∆U ∆U ∆U 1 ∆P U gradient is near zero. = −U −V − · +fc · V −w T · There are other physical processes that have been ∆t ∆x ∆y ρ ∆x z i neglected in the simplified equations just presented. Molecular friction is significant in the bottom few •(10.23b) millimeters of the ground. Mountain-wave drag ∆V ∆V ∆V 1 ∆P V can be large in mountainous regions (see the Local = −U −V − · −f · U −w · ∆t ∆x ∆y ρ ∆y c T z Winds chapter). Cumulus clouds can cause turbu- i lent convective mixing above the boundary layer } } } } } (see the Atmospheric Boundary Layer and Thunder- storm chapters). Vertical advection of the horizontal horizontal pressure turbulent tendency Coriolis advection gradient drag wind has been neglected because it often is small. Mean vertical motions (e.g., large-scale subsidence) These are the forecast equations for wind. Centrifu- will be examined later in this chapter. gal force is not included because it is the opposite of In the section that follows, the equations of mo- centripetal force, which is the net imbalance of the tion are simplified for some special cases, to yield other forces already included above. theoretical winds in the horizontal. Where appro- As was shown in the solved examples, each of priate, the forces and winds will also be given in iso- the terms can be of similar magnitude: 1x10–4 to baric coordinates. 302 CHAPTER 10 Dynamics Table 10-4. Names of idealized steady-state horizon- tal winds, and the forces that govern them. Horizontal Winds ∆U 1 ∆P U VM· 0 = −U − · +fc· V −w T · +s When air accelerates to create wind, forces such ∆x ρ ∆x z R i as Coriolis and drag change too, because they de- pressure turbulent centri- Coriolis Forces: gradient drag fugal pend on the wind speed. This, in turn, changes the acceleration via eqs. (10.23), so there is a feedback Wind Name process. This feedback continues until the forces Geostrophic • • finally balance each other. At that point, there is no Gradient • • • net force, and no further acceleration. Boundary Layer • • • This final condition is called steady state: BL Gradient • • • • Cyclostrophic • • ∆U ∆V Inertial • • = 0, = 0 •(10.24) ∆t ∆t Antitriptic • • In steady-state, wind speeds do not change with 1 L1B - time, but are not necessarily zero. Only the accel- eration is zero. Under certain idealized conditions, some of the '1( forces in the equations of motion are small enough L1B to be neglected. For these situations, theoretical steady-state winds can be found based on only the ( remaining larger-magnitude forces. These theoreti- cal winds are given special names, as listed in Table L1B 10-4. These winds are examined next in more detail. The real winds under these special conditions are often close to the theoretical winds. Z '$' L1B Geostrophic Wind ) Y The geostrophic wind (Ug , Vg) is a theoretical wind that results from a steady-state balance be- Figure 10.8 tween pressure-gradient force and Coriolis force Idealized weather map, showing geostrophic wind (G, grey ar- (Fig. 10.8). After setting the other forces to zero, eqs. row) caused by a balance between two forces (black arrows): (10.23) become: pressure-gradient force (FPG) and Coriolis force (FCF). P is pressure, with isobars plotted as thin black lines. L and H are 1 ∆P 0 = − · + f · V (10.25a) low and high-pressure regions. The small sphere represents an ρ ∆x c air parcel. 1 ∆P 0 = − · − f · U (10.25b) Solved Example ρ ∆y c Pressure increases 1 kPa eastward across a dis- tance of 500 km. What is the geostrophic wind speed, 3 –4 –1 Solving these equations for U and V, and then de- given ρ = 1 kg/m and fc = 10 s ? finingU ≡ Ug and V ≡ Vg , gives: Solution 3 –4 –1 Given: ∆P =1 kPa, ∆x =500 km, ρ =1 kg/m , fc =10 s . 1 ∆P Find: G = ? m/s Ug = − · •(10.26a) ρ· fc ∆y Ug = 0, thus G = Vg. Use eq. (10.26b): 1 ∆P −1 ()−2kPa Vg = + · •(10.26b) Ug = · = 20 m/s ρ· f ∆x (.1 2kg/m3)·(.1 1× 10−4s − 1) (800km) c Check: Units OK. Physics OK. –4 –1 where fc = (1.4584x10 s )·sin(latitude) is the Coriolis Discussion: “Kilo” in the numerator & denominator parameter, ρ is air density, and ∆P/∆x and ∆P/∆y are cancel. Given that ∆P/∆x = 0.002 kPa/km, the answer the horizontal pressure gradients. agrees with Fig. 10.10. The wind is toward the north. R. STULL • meteorology for scientists and engineers 303 OPSUI å1åE L1BLN ( NT TPVUI XFTU FBTU Figure 10.9 Geostrophic winds (grey arrows) are faster (longer arrows) where isobars (thin lines) are closer together. (For N. Hemis.) -BUJUVEF In regions of straight isobars above the top of the boundary layer and away from the equator, the ac- tual winds are approximately geostrophic. These ( NT winds blow parallel to the isobars or height con- tours, with low pressure to the left in the Northern å[åE NLN Hemisphere (Fig. 10.9). The wind is faster in regions where the isobars are closer together (i.e., where the Figure 10.10 isobars are tightly packed) and at lower latitudes Geostrophic wind speed G vs. latitude and height gradient (Fig. 10.10). (∆z/∆d) on a constant pressure surface. Top scale is pressure gradient at sea level. The total geostrophic wind speed G is: GU=2 + V 2 (10.27) g g Solved Example If height increases by 100 m eastward across a dis- If ∆d is the distance between two isobars (in the di- tance of 500 km, then what is the geostrophic wind rection of greatest pressure change; namely, perpen- speed, given f = 10–4 s–1 ? dicular to the isobars), then the magnitude of the c geostrophic wind is: Solution Given: ∆z = 100 m, ∆x = 500 km, f = 10–4 s–1 . 1 ∆ P c G = · •(10.28) Find: G = ? m/s ρ· fc ∆ d Ug = 0, thus G = Vg. Use eq. (10.29b): Above sea level, weather maps are often on g -2 ∆z 9. 8ms 50m = 19.6 m/s isobaric surfaces (constant pressure charts). The Vg = + · = · f ∆x 0.00009s-1 200,, 000m geostrophic wind as a function of horizontal dis- c tances between height (z) contours on a constant- pressure chart is: Check: Units OK. Physics OK. Discussion: This is nearly the same answer as be- g ∆z fore. ∆z of 100 m on an isobaric surface corresponds Ug = − · •(10.29a) fc ∆y to ∆P of roughly 1 kPa on a constant height surface. A hypothetical ball rolling downhill would start moving g toward the west, but would be deflected northward by ∆z •(10.29b) Vg = + · Coriolis force in the Northern Hemisphere. fc ∆x where |g| = 9.8 m·s–2 is gravitational acceleration FOCUS • Approach to Geostrophy magnitude, and fc is the Coriolis parameter. The corresponding magnitude of geostrophic wind on How does an air parcel, starting from rest, ap- an isobaric chart is: proach the final steady-state geostrophic wind speed G sketched in Fig. 10.8? g ∆ z G = · •(10.29c) Start with the equations of horizontal motion (10.23), fc ∆ d and ignore all terms except the tendency, pressure- gradient force, and Coriolis force. Use the definition of continues on next page 304 CHAPTER 10 Dynamics If the geopotential Φ = |g|·z is substituted in FOCUS • Appr. to Geostrophy (continued) eqs. (10.29), the resulting geostrophic winds are: geostrophic wind (eqs. 10.26) to write the resulting simplified equations as: 1 ∆ Φ Ug = − · (10.30a) ∆U / ∆ t = − fc ·() V g − V fc ∆ y ∆V / ∆ t =fc ·() U g − U 1 ∆ Φ (10.30b) Next, rewrite these as forecast equations: Vg = · fc ∆ x UUnew= old − ∆t · fc·(VV g − old ) VVnew= old + ∆t · fc ·(Ug − U new ) Gradient Wind Start with initial conditions (Uold, Vold) = (0, 0), and then iteratively solve the equations on a spreadsheet Around a high or low pressure center, the steady- to forecast the wind. state wind follows the curved isobars, with low pres- For example, consider the conditions given in the sure to the left in the Northern Hemisphere. Around previous solved example, where we would anticipate lows, the wind is slower than geostrophic, called the wind should approach (Ug, Vg) = (0, 20) m/s. The subgeostrophic, regardless of the hemisphere. actual evolution of winds (U, V) and air parcel posi- Around highs, the steady-state wind is faster than tion (X, Y) are shown in Figs. below. geostrophic, or supergeostrophic. The curved steady-state wind is called the gradient wind. I The gradient wind occurs because of an imbal- I ance (Figs. 10.11 & 10.12) between pressure-gradient 7 NT I (FPG) and Coriolis forces (FCF); namely, the net force (F ) is not zero. This net force is called centripe- 6 7 net H H tal force, and is what causes the wind to continu- NT ally change direction as it goes around a circle. By I describing this change in direction as causing an I apparent force (centrifugal), we can find the steady- U state gradient wind: I m 1 ∆P VM· 0 = − · +f · V +s· 6 NT ρ ∆x c R (10.31a) Surprisingly, the winds never reach geostrophic equilibrium, but instead rotate around the geostrophic wind. This is called an inertial oscillation, with 1 ∆P UM· 0 = − · −f · U −s· (10.31b) period of P = 2·π/fc. Twice this period is called a pen- ρ ∆y c R dulum day. For our solved example, P = 17.45 h. The net result in the figure below is that the } } } wind indeed moves at the pressure Coriolis centrifugal geostrophic speed of 20 m/ I gradient s to the north (≈ 1250 km in 17.45 h), but along the way it I staggers west and east with Because the gradient wind is for flow around a an additional ageostrophic I circle, we can frame the governing equations in ra- (non-geostrophic) part. dial coordinates: Inertial oscillations are I : LN sometimes observed at night I I I in the boundary layer, but 1 ∆P M2 ··=f M + tan (10.32) rarely higher in the atmo- ρ ∆R c tan R sphere. Why not? (1) The I U ageostrophic component of wind (wind from the East where R is radial distance from the center of the I ( NT in this example) moves air circle, fc is the Coriolis parameter, ρ is air density, mass, and changes the pres- I I ∆P/∆R is the radial pressure gradient, and Mtan is sure gradient. (2) Friction the magnitude of the tangential velocity; namely, m damps the oscillation to- the gradient wind. The signs of the terms in this ward a steady wind. 9 LN equation are for flow around a low. R. STULL • meteorology for scientists and engineers 305 - L1B L1B ) L1B1 1 L1B L1B ' $' L1B '1( M Q F B SD UI B P SQ G L1B BJ BJS G QB UIP SDF 'OFU QB . M 'OFU UBO ( L1B .UBO ( L1B ' $' L1B '1( Z Z Y Y ) - Figure 10.11 Figure 10.12 Forces (dark arrows) that cause the gradient wind (solid grey Forces (dark arrows) that cause the gradient wind (solid grey arrow, Mtan) to be slower than geostrophic (hollow grey arrow) arrow, Mtan) to be faster than geostrophic (hollow grey arrow) when circling around a low-pressure center (called a cyclone in for an air parcel (grey sphere) circling around a high-pressure the N. Hem.). The short white arrow with black outline shows center (called an anticyclone in the N. Hemisphere). centripetal force (the imbalance between the other two forces). Centripetal force pulls the air parcel (grey sphere) inward to force the wind direction to change as needed for the wind to turn along a circular path. OPSUI 1 L1B Define U ≡ U and V ≡ V as the components tan tan - ( of gradient wind, with a total gradient wind speed of .UBO 2 2 1/2 Mtan = [Utan + Vtan ] . One solution to eq. (10.32) is 2 Mtan (10.33) MGtan = ± fc · R where Mtan is the gradient-wind speed, and where ) the negative sign is used for flow around low pres- TPVUI sure centers, and the positive sign for highs. This XFTU FBTU solution demonstrates that wind is “slow around Figure 10.13 lows” (Fig. 10.13), meaning slower than geostrophic Geostrophic wind G and gradient wind Mtan around a low pres- G. However, the solution is implicit because the de- sure center in the Northern Hemisphere. sired wind Mtan is on both sides of the equal sign. Solved Example What radius of curvature causes the gradient wind to equal the geostrophic wind? Solution Given: Mtan = G Find: R = ? km 2 Use eq. (10.33), with Mtan = G: G = G ± G /( fc·R) This is a valid equality G = G only when the last term in eq. (10.33) approaches zero; i.e., in the limit of R = ∞ . Check: Eq. (10.33) still balances in this limit. Discussion: Infinite radius of curvature is a straight line, which (in the absence of any other forces such as turbulent drag) is the condition for geostrophic wind. 306 CHAPTER 10 Dynamics Solving the quadratic equation (10.33) for the cy- Solved Example If the geostrophic wind around a low is 10 m/s, then clonic flow (around a low-pressure center) yields –4 –1 the gradient wind M : what is the gradient wind speed, given fc = 10 s tan and a radius of curvature of 500 km? Also, what is the curvature Rossby number? 4·G M= 0.· 5 f ··R −1+ 1 + •(10.34a) tan c f· R Solution c –4 –1 Given: G = 10 m/s, R = 500 km, fc = 10 s Find: Mtan = ? m/s, Roc = ? (dimensionless) For anticyclonic flow (around a high-pressure cen- ter): Use eq. (10.34a) 4·G −4 −1 •(10.34b) Mtan = 0.· 5 ()10 s ·(500000m)· Mtan = 0.· 5 fc ··R 1− 1 − fc · R 4·( 10m/) s −1 + 1 + (()10−4 s −1 ·(500000m) A “curvature” Rossby number (Roc) can be de- = 8.54 m/s fined that uses the radius of curvature R( ) as the rel- Use eq. (10.35): evant length scale: G ()10m/s Ro = Roc = −4 −1 5 = 0.2 c (10.35) ()10 s·(5× 10 m) fc · R Check: Units OK. Physics OK. [Neither R nor Roc are the “Rossby deformation ra- Discussion: The small Rossby number indicates that dius” (see eq. 10.70, and the Airmasses, Fronts, and the flow is in geostrophic balance. The gradient wind Extratropical Cyclones chapters).] Small values of is indeed slower than geostrophic around this low. the Rossby number indicate flow that is nearly in geostrophic balance. The cyclonic gradient wind (around a low) is: G 1/ 2 (10.36a) Mtan =·· −1+( 1 + 4 Roc ) 2·Roc and anticyclonic gradient wind (around a high) is: G 1/ 2 (10.36b) Mtan =··1 − ( 1 − 4 Roc ) 2·Roc where G is the geostrophic wind. For high-pressure centers, steady-state physical ) (non-imaginary) solutions exist only for Roc ≤ 1/4 . - Thus, around anticyclones, isobars cannot be both 1 L1B closely spaced and sharply curved. In other words, the pressure cannot decrease rapidly away from high centers. There is no analogous restriction on 3 LN cyclones, because any value of Roc is possible. Thus, pressure gradients and winds must be gentle in Figure 10.14 highs, but can be vigorous near low centers (Figs. Variation of surface pressure across an anticyclone (H) and cy- 10.14 and 10.15). clone (L), showing that the curve can have steep pressure gra- By combining the definition of the Rossby num- dients near the low and a cusp at the low, but not at the high. ber with that for geostrophic wind, and setting Roc = Arbitrary center pressures of 103 kPa and 97 kPa were chosen to 1/4, we find that the maximum horizontal variations illustrate the anticyclone and cyclone, respectively. of pressure P or height z near anticyclones are: 2 2 z= zc − ( f c ·/ R ) (8· g ) •(10.37a) 2 2 •(10.37b) PP=c − (ρ··f c R )/8 R. STULL • meteorology for scientists and engineers 307 'BTU ( i4MPXFS - BSPVOE-PXTu .UBO 5VSCVMFOU%SBH .#-( - - 8JOE 4QFFE . 5VSCVMFOU%SBH UBO ) i'BTUFS .#-( BSPVOE)JHITu 4MPX ) ( -PX )JHI 1SFTTVSF$FOUFS Figure 10.15 Figure 10.16 Sea-level pressure (contoured every 2 mb= 0.2 kPa), for 00 UTC Relative magnitudes of different wind speeds around low- and on 24 Nov 1998. Notice the tight packing of isobars around high-pressure centers. G = geostrophic wind, Mtan = gradient lows, but looser spacing near high-pressure centers. wind speed, MBLG = boundary-layer gradient wind speed. G is smaller in highs than in lows, because it is not physically pos- sible to have strong pressure gradients to drive strong steady- state winds at high centers. where zc and Pc are the reference height or pressure at the center of the anticyclone, fc is the Coriolis pa- rameter, |g| is gravitational acceleration magnitude, ρ is air density, and R is distance from the center of the anticyclone (Fig. 10.14). Figs. 10.14 and 10.15 show that pressure gradi- ents, and thus the geostrophic wind, can be large near low centers. However, pressure gradients, and thus the geostrophic wind, must be small near high centers. This difference in geostrophic wind speed G between lows and highs is sketched in Fig. 10.16. The slowdown of gradient wind Mtan (relative to geostrophic) around lows, and the speedup of gra- 1 L1B - dient wind (relative to geostrophic) around highs is also plotted in Fig. 10.16. The net result is that gradi- ' ent winds, and even boundary-layer gradient winds 1( .#- MBLG (described later in this chapter), are usually L1B stronger (in an absolute sense) around lows than highs. For this reason, low-pressure centers are of- B ( ten windy. '5% L1B Boundary-Layer Wind Turbulent drag in the boundary layer slows the '$' Z wind below the geostrophic value, and turns the L1B wind to point at a small angle (α) across the isobars toward low pressure (Fig. 10.17). For flow along Y straight isobars, the steady-state equations of mo- ) tion become: 1 ∆P U Figure 10.17 0 = − · +fc · V −w T · (10.38a) Balance of forces (black arrows) creating a boundary-layer wind ρ ∆x zi (MBL, solid grey arrow) that is slower than geostrophic (G, hol- low grey arrow). The grey sphere represents an air parcel. Thin 1 ∆P V black lines are isobars. L and H are low and high-pressure cen- 0 = − · −fc · U −w T · (10.38b) ρ ∆y zi ters. 308 CHAPTER 10 Dynamics Solved Example Namely, the only forces acting for this special case Find the boundary-layer winds given Ug = 10 m/s, –4 –1 are pressure gradient, Coriolis, and turbulent drag Vg = 0, zi = 1 km, CD = 0.002, and fc = 10 s . Also, what angle do the winds cross the isobars? This is a (Fig. 10.17). statically neutral boundary layer. DefineU BL = U and VBL = V as components of the boundary-layer wind. An implicit solution of eqs. Solution (10.38) is: wTA· V BL Given: (see above) UUABL= g − (10.39a) Find: UBL =? m/s, VBL =? m/s, MBL =? m/s, α = ? ° fc· z i 2 2 1/2 Use eqs. (10.41): with G = (Ug + Vg ) = 10 m/s wTA· U BL 0.003 VVABL= g + (10.39b) a = = 0. 02 s/m fc· z i ()10−4 s −1 ·(1500 m) where (U , V ) are geostrophic wind components, f Check: a·G = (0.02s/m)·(10 m/s) =0.2 ( is < 1. Good.) g g c is Coriolis parameter, z is ABL depth, and w is the U =[1–0.35·(0.02s/m)·(10m/s)]·(10m/s) ≈ 9.3 m/s i T BL turbulent transport velocity. VBL=[1–0.5·(0.02s/m)·(10m/s)]· (0.02s/m)·(10m/s)·(10m/s) ≈ 1.8 m/s Eqs. (10.39) can be solved by iteration (guess VBL and plug into the first equation, solve for UBL and MU=2 + V 2 =13..42 += 3 89.47 2 m/s ABLABL ABL plug into second equation, solve for VBL and repeat The geostrophic wind is parallel to the isobars. the process). The magnitude of the boundary-layer The angle between BL wind and geostrophic is wind is: –1 V U –1 11° 2 2 1/2 α =tan ( BL/ BL) = tan (1.8/9.3) = . MBL = [U BL + V BL ] (10.40) Check: Units OK. Physics OK. If the boundary layer is statically neutral with Discussion: The boundary-layer wind speed is in- strong wind shear, then wT = CD·MBL, where CD is deed slower than geostrophic (9.47 vs. 10 m/s), but only the drag coefficient (see eq. 10.21). An approximate slightly slower because the drag coefficient for this explicit solution for the wind at most altitudes in the example was very small. Also, it crosses the isobars boundary layer is: slightly toward low pressure. (The geostrophic wind toward the east means low pressure is to the north. ) •(10.41a) UABL ≈(.1 − 0 35··a UUg )· g −(.1 − 0 5··a Vg )·a·· Vg G •(10.41b) Solved Example VABL ≈(.1 − 0 5··a Ug )· a··GUg +(.1 − 0 35··a Vg )·Vg Find the boundary layer winds given Ug = 10 m/s, –3 a G a C f z G Vg = 0, zi = 1 km, wB = 45 m/s, bD = 1.83x10 , and fc for · < 1 , where = D/( c· i) and is the = 10–4 s–1. Also, at what angle does the wind cross the geostrophic wind speed. If this condition is not met, isobars? This is a statically unstable boundary layer. or if no reasonable solution can be found using eqs. (10.41), then use the iterative approach described in Solution the next section, but with the centrifugal terms set to Given: (see above, for a convective BL) zero. Eqs. (10.41) do not apply to the surface layer Find: UBL = ? m/s, VBL = ? m/s, MBL = ? m/s, α = ?° (bottom 10% of the neutral boundary layer). For a statically unstable boundary layer with Use eqs. (10.42): light winds, use wT = bD·wB (see eq. 10.22). An exact (.1 83× 10−3)·() 50m/s explicit solution for the winds at most altitudes in c1 = = 0.824 (dimensionless) ()10−4 s −1 ·(1500 m) the boundary layer is: 2 c2 = 1/[1+(0.8242) ] = 0.60 (dimensionless) U= c ·[U − c ·]V UBL = 0.6·[(10m/s) – 0 ] = 6.0 m/s ABL2 g 1 g •(10.42a) VBL = 0.6·[0 + (0.824)·(10m/s)] = 4.94 m/s Use eq. (10.40): V= c ·[V + c ·]U •(10.42b) 2 2 1/2 ABL2 g 1 g MBL = [U BL + V BL ] = 7.77 m/s –1 –1 α =tan (VBL/UBL) = tan (4.94/6.0) = 39.5° bDB· w 1 where c1 = , and c2 = 2 . Check: Units OK. Physics OK. fc· z i []1 + c1 Discussion: As before, the boundary layer winds are Again, this solution does not apply to the sur- subgeostrophic, and cross the isobars toward low pres- face layer (bottom 5% of convective boundary lay- sure. Turbulent drag has similar effects, regardless of er). See the “Forces” section earlier in this chapter whether the turbulence is generated mechanically by for definitions of the factors in c . wind shear, or by buoyant rising thermals. 1 R. STULL • meteorology for scientists and engineers 309 Thus, for both neutral and unstable static stabili- ties, boundary-layer winds (for most mid-boundary- layer altitudes) cross the isobars at a small angle (α) toward low pressure. This cross-isobaric flow oc- curs for both straight and curved isobars. Boundary-Layer Gradient (BLG) Wind In regions of curved isobars at the bottom of - L1B M L1B1 cyclones and anticyclones, drag force exists in the F D S boundary layer (BL) in addition to pressure gradi- B Q S ent and Coriolis force. The imbalance (Fnet) of these J L1B B forces creates a centripetal force that causes the air G P I to spiral in towards low-pressure centers (Fig. 10.18) ' U 1( B and spiral out from high-pressure centers. Fig. 10.1 Q .#-( shows a sketch of the BL gradient winds in the N. L1B Hemisphere, and the associated isobars. ' . Assume steady state, neglect advection, and re- OFU B UBO ( quire an imbalance of forces equal to the centrifugal '5% force, to reduce the equations of motion to: L1B '$' 1 ∆P U VM· 0 = − · +fc · V −w T · +s · (10.43a) Z ρ ∆x zi R Y 1 ∆P V UM· ) 0 = − · −f · U −w · −s · (10.43b) ρ ∆y c T z R i } } } } Figure 10.18 Imbalance of forces (black arrows) yield a net centripetal force pressure turbulent Coriolis centrifugal (Fnet) that causes the boundary layer gradient wind (MBLG, sol- gradient drag id grey arrow) to be slower than both the gradient wind (Mtan) and geostrophic wind (G). The resulting air-parcel path crosses the isobars at a small angle α toward low pressure. Without actually solving these equations, we can anticipate the following from our previous under- standing of geostrophic, gradient, and boundary- layer winds. Boundary-layer gradient (BLG) wind speed is slower than the gradient wind speed due to the drag. The BLG winds flow counterclockwise around lows in the N. Hemisphere, and clockwise around highs. Instead of blowing parallel to the curved isobars like the gradient wind, BLG winds cross the isobars at a small angle (α, tens of degrees) toward low pressure (Fig. 10.19). The turbulent drag term is different for highs and lows. Winds are strong around lows and skies are often overcast, hence the transport velocity is best - represented by the statically neutral parameteriza- tion: 3 2 2 1/2 . wT = CD · M = CD · [ U + V ] (10.21 again) Z #-( 7 which adds even more nonlinearity to eqs. (10.43). 6 In highs, winds are light and skies are clear, sug- Y gesting that transport velocity should be given by Figure 10.19 the statically unstable parameterization during day- Tangential (U) and radial (V) components of the BLG wind in time by: the N. Hemisphere, for that one vector south of the Low. 310 CHAPTER 10 Dynamics w = b · w (10.22 again) Solved Example T D B Find the BLG winds south of a low pressure center in the N. Hemisphere, given G = 20 m/s at radius R = where wB is not a function of wind speed. For stati- cally stable conditions at night in fair weather, steady 500 km from the low center, zi = 1 km, CD = 0.01, and fc = 10–4 s–1. Also, find the speed. state is unlikely, so eqs. (10.43) are invalid. While the equations of motion for geostrophic Solution and gradient winds were simple enough to allow an Given: (see above) analytical solution, and we could devise an approxi- Find: UBLG = ? m/s, VBLG = ? m/s, MBLG = ? m/s mate analytical solution for the BL wind, we are not so lucky with the BLG wind. The set of coupled Use eqs. (1.1) and (10.44) in a spreadsheet. Set the time equations (10.43) are nonlinear and nasty to solve. step to ∆t = 1 h. Choose U = 0 and V = 0 as the first Nonetheless, we can numerically iterate towards guess. Make 20 iterations (see the Focus box). the answer by including the tendency term on the LHS of each equation. For example, LHS = ∆U/∆t = [U(t+∆t) – U(t)]/∆t. Also, rewrite eqs. (10.43) in cylin- drical coordinates, letting U be the tangential com- ponent, and V be the radial component (Fig. 10.19). Use geostrophic wind G as a surrogate for the radial pressure gradient. For BLG winds around a low in the N. Hemi- sphere (i.e., s = +1), equations (10.43) can be rewrit- ten as the following set of coupled equations, which is valid day or night: M = ( U2 + V2 )1/2 (1.1 again) (10.44a) CMD ··U VM· U() t +∆t = U + ∆t·· fc V − + s zi R (10.44b) U = 8.51 m/s, V = 8.09 m/s, M = 11.74 m/s, CMD··V UM· BLG BLG BLG V() t +∆t = V + ∆t·· fc ()GU − − − s where (UBLG, VBLG) are (tangential, radial) parts. zi R where (U, V) represent (tangential, radial) parts for Check: Units OK. Physics OK. Check that the com- the wind vector south of the low center. puted wind approaches the: (1) geostrophic wind as R approaches ∞, with CD = 0; (2) gradient wind when CD = 0; (3) boundary-layer wind when R approaches ∞. I performed these checks using a modified spread- FOCUS • Solution by Iteration sheet that relaxed the results using a weighted aver- Equations (10.44) are difficult to solve analytically, age of new and previous winds, and found: (1) for but we can iterate as an alternative way to solve for geostrophic: UBLG = G = 20 m/s, VBLG = 0; (2) for gradi- the BLG wind components. To use this approach: ent: UBLG = 15.31 m/s, VBLG = 0; and (3) for BL: UBLG = 7.81 m/s, VBLG = 9.76 m/s. This BL solution is the exact (1) Make a first guess that U( , V) = (0, 0). solution; it is better than approximate eqs. (10.41). (2) Plug in these values everywhere that U, U(t), Discussion: The solution converges toward the BLG V, or V(t) appears in eqs. (1.1) and (10.44). wind, as if an air (3) Solve eqs. (10.44) for the new values of parcel started from [U(t+∆t), V(t+∆t)]. rest and began accel- (4) Repeat steps 2 & 3, but using the new winds. erating. The Fig. at (5) Continue until the [U(t+∆t), V(t+∆t)] wind right used ∆t = 1000 components converge to steady values, which .#-( s (not 1 h). Starting 7 NT by definition are the U( BLG, VBLG) components at zero wind speed, that we want. each dot shows the wind forecast for the Iterative approaches are tedious when done on a hand next time step in the calculator, so use a spreadsheet or computer program iteration. instead, such as was done in the solved example. 6 NT R. STULL • meteorology for scientists and engineers 311 Similar equations can be derived for convective boundary layers in high-pressure regions. You can use the iterative method described in the Focus box to solve equations (10.44). In the solved example, notice that the spreadsheet iterations do not proceed directly to the final solution, but spiral toward it. This spiral is called a damped inertial oscillation. Equations (10.44) are also valid for unsteady (time varying) solutions, such as at night. At night when drag is weak, the winds may never reach steady state, and may continue as undamped or weakly-damped inertial oscillations (see a previous Focus Box on Approach to Geostrophy). Such oscillations can tem- Z porarily cause winds to be greater than geostrophic - L1B in regions of straight isobars, or greater than gradi- 1 ent in regions of curved isobars. This is one reason L1B Y for the supergeostrophic (faster than geostrophic) '1( nocturnal jet, which will be covered in the Atmo- L1B spheric Boundary Layer chapter. 'OFU M F SD Cyclostrophic Wind QB JS In intense vortices, strong winds rotate around a L1B GB IP QBU very tight circle. Winds in tornadoes are about 100 .DT m/s, and in water spouts are about 50 m/s. As the tornado strengthens and tangential winds increase, '$' centrifugal force increases much more rapidly than Coriolis force. Centrifugal force quickly becomes the dominant force that balances pressure-gradient Figure 10.20 force (Fig. 10.20). Thus, a steady-state rotating wind Around tornadoes, pressure gradient force is so strong that it is reached at much slower speeds than the gradient greatly exceeds all other forces. The net force (Fnet) pulls the air wind speed. around the tight circle at the cyclostrophic wind speed (Mcs). For steady state winds, the equations of motion reduce to: 1 ∆P VM· (10.45a) 0 = − · +s · ρ ∆x R Solved Example 1 ∆P UM· 0 = − · −s · (10.45b) A 10 m radius waterspout has a tangential velocity ρ ∆y R of 45 m/s. What is the radial pressure gradient? } } Solution pressure centrifugal gradient Given: Mcs = 45 m/s, R = 10 m. Find: ∆P/∆R = ? kPa/m. Because of the cylindrical nature of these flows 3 as they rotate around intense low-pressure cen- Assume cyclostrophic wind, and ρ = 1 kg/m . Rearrange eq. (10.46): ters, it is easier to write and solve the equations for 3 2 cyclostrophic wind Mcs in cylindrical form: ∆P ρ 2 ()1kg/m·( 45 m/s) =· Mcs = ∆RR 10m RP∆ (10.46) ∆P/∆R = 202.5 kg·m–1·s–2 / m = 0.2 kPa/m. M = · cs ρ ∆R Check: Units OK. Physics OK. where R is radial distance outward from the center Discussion: This is 2 kPa across the 10 m waterspout of rotation, ∆P/∆R is the local radial pressure gradi- radius, which is 1000 times greater than typical synop- ent, and Mcs is the tangential speed. tic-scale pressure gradients on weather maps. 312 CHAPTER 10 Dynamics Cyclostrophic winds never occur around high Z pressure centers, because the strong pressure gra- dients needed to drive such winds are not possible. 3 Around lows, cyclostrophic winds can turn either Y counterclockwise or clockwise in either hemisphere, because Coriolis force is not a factor. 'OFU '$' Inertial Wind QB UI P Steady-state inertial motion results from a bal- GB JSQ ance of Coriolis and centrifugal forces: BSDFM .J 2 Mi 0 =fc· M i + (10.47) Figure 10.21 R Imbalance of forces (F, black arrows) on an air parcel (grey ball), creating an anticyclonic inertial wind M , grey arrow). R is i where Mi is inertial wind speed, fc is the Coriolis pa- radius of curvature, and FCF is Coriolis force. rameter, and R is the radius of curvature. Since both of these forces depend on wind speed, the inertial wind cannot start itself from zero. It can occur only Solved Example after some additional force first causes the wind to For an inertial ocean current of 5 m/s, find the ra- blow, and then that extra force disappears. dius of curvature and time period to complete one cir- The inertial wind coasts around a circular path cuit. Assume a latitude where f = 10–4 s–1. c of radius R, Solution M R = − i Given: M = 5 m/s, f = 10–4 s–1. (10.48) i c fc Find: R = ? km, P = ? h Use eq. (10.48): R = –(5 m/s) / (10–4 s–1) = –50 km where the negative sign implies anticyclonic rotation Use P = 2π/fc = 62832 s = 17.45 h (Fig. 10.21). The time period needed for this iner- tial oscillation to complete one circuit is P = 2π/fc, Check: Units OK. Magnitudes OK. which is half of a pendulum day (see Approach to Discussion: The tracks of drifting buoys in the ocean Geostrophy Focus Box earlier in this chapter). are often cycloidal, which is the superposition of a Although rarely observed in the atmosphere, circular inertial oscillation and a mean current that inertial oscillations are frequently observed in the gradually moves the whole circle. ocean. This can occur where wind stress on the ocean surface creates an ocean current, and then after the wind dies the current coasts in an inertial - Z oscillation. 1 L1B Antitriptic Wind .B Y A steady-state antitriptic wind Ma could result from a balance of pressure-gradient force and tur- '1( L1B bulent drag: ( 1 ∆ P Ma (10.49) 0 = − · − wT · ρ ∆ d zi L1B where ∆P is the pressure change across a distance '5% ∆d perpendicular to the isobars, fc is the Coriolis pa- rameter, wT is the turbulent transport velocity, and zi ) is boundary-layer depth. This wind blows perpendicular to the isobars Figure 10.22 (Fig. 10.22), directly from high to low pressure: Balance of forces (F, black arrows) that create the antitriptic wind M (grey arrow). G is the theoretical geostrophic wind. z·· f G a M = i c (10.50) F is turbulent drag, and F is pressure-gradient force. a TD PG wT R. STULL • meteorology for scientists and engineers 313 Solved Example For free-convective boundary layers, w = b ·w is T D B In a 1 km thick convective boundary layer at a loca- not a function of wind speed, so Ma is proportional –4 –1 tion where fc = 10 s , the geostrophic wind is 5 m/s. to G. However, for windy forced-convection bound- The turbulent transport velocity is 0.02 m/s. Find the ary layers, wT = CD·Ma, so solving for Ma shows it to antitriptic wind speed. be proportional to the square root of G. This wind would be found in the boundary layer, Solution –4 –1 and occurs as an along-valley component of “long Given: G = 5 m/s, zi = 1000 m, fc = 10 s , gap” winds (see the Local Winds chapter). It is also wT = 0.02 m/s sometimes thought to be relevant for thunderstorm Find: Ma = ? m/s cold-air outflow and for steady sea breezes. How- ever, in most other situations, Coriolis force should Use eq. (10.50): –4 –1 not be neglected; thus, the boundary-layer wind and Ma = (1000m)·(10 s )·(5m/s) / (0.2 m/s) = 25 m/s BL Gradient winds are much better representations of nature than the antitriptic wind. Check: Magnitude seems too large. Units OK. Discussion: Eq. (10.50) can give winds of Ma > G for Summary of Horizontal Winds many convective conditions, for which case Coriolis Table 10-5 summarizes the idealized horizontal force would be expected to be large enough that it winds that were discussed earlier in this chapter. should not be neglected. Thus, antitriptic winds are On real weather maps such as Fig. 10.23, isobars unphysical. However, for forced-convective bound- ary layers where drag is proportional to wind speed or height contours have complex shapes. In some squared, reasonable solutions are possible. regions the height contours are straight (suggesting that actual winds should nearly equal geostrophic or boundary-layer winds), while in other regions the height contours are curved (suggesting gradient or boundary-layer gradient winds). Also, as air parcels move between straight and curved regions, they are sometimes not quite in equilibrium. Nonetheless, when studying weather maps you can quickly esti- mate the winds using the summary table. FOCUS • The Rossby Number The Rossby number (Ro) is a dimensionless ratio defined by M M Ro = or Ro = fc · L fc · R where M is wind speed, fc is the Coriolis parameter, L is a characteristic length scale, and R is radius of curvature. - In the equations of motion, suppose that advection - terms such as U·∆U/∆x are order of magnitude M2/L, and Coriolis terms are of order fc·M. Then the Rossby number is like the ratio of advection to Coriolis terms: (M2/L) / (f ·M) = M/(f ·L) = Ro. Or, we could consider [ LN c c ) the Rossby number as the ratio of centrifugal (order of 2 M /R) to Coriolis terms, yielding M/(fc·R) = Ro. Use the Rossby number as follows. If Ro < 1, then Coriolis force is a dominant force, and the flow tends to become geostrophic (or gradient, for curved flow). If Ro > 1, then the flow tends not to be geostrophic. For example, a midlatitude cyclone (low-pressure –4 –1 system) has approximately M = 10 m/s, fc = 10 s , Figure 10.23 and R = 1000 km, which gives Ro = 0.1 . Hence, mid- Upper-air 30 kPa height chart valid at 12 UTC on 9 July 99. latitude cyclones tend to adjust toward geostrophic Shading indicates isotachs in the jet stream, where light grey balance, because Ro < 1. In contrast, a tornado has denotes roughly 25 m/s or greater winds, and darker shading –4 –1 roughly M = 50 m/s, fc = 10 s , and R = 50 m, which is roughly 50 m/s or greater. Thick lines are height contours. gives Ro = 10,000, which is so much greater than one Faster winds occur where contours are packed. [Adapted from a that geostrophic balance is not relevant. US Navy Fleet Numerical Meteor. and Ocean. Ctr. chart.] 314 CHAPTER 10 Dynamics Table 10-5. Summary of horizontal winds**. Name of Item Forces Direction Magnitude Where Observed Wind faster where isobars parallel to straight are closer together. aloft in regions pressure-gradient, 1 geostrophic isobars with Low pres- where isobars are Coriolis 1 ∆ P sure to the wind’s left* G = · nearly straight ρ· fc ∆ d similar to geostrophic slower than wind, but following pressure-gradient, geostrophic around aloft in regions curved isobars. Clock- 2 gradient Coriolis, Lows, faster than where isobars are wise* around Highs, centrifugal geostrophic around curved counterclockwise* Highs around Lows. similar to geostrophic near the ground pressure-gradient, slower than boundary wind, but crosses in regions where 3 Coriolis, geostrophic (i.e., layer isobars at small angle isobars are nearly drag subgeostrophic) toward Low pressure straight pressure-gradient, similar to gradient boundary- near the ground in Coriolis, wind, but crosses slower than gradient 4 layer regions where iso- drag, isobars at small angle wind speed bars are curved gradient centrifugal toward Low pressure either clockwise or tornadoes, water- stronger for lower pressure-gradient, counterclockwise spouts (& sometimes 5 cyclostrophic pressure in the vortex centrifugal around strong vortices in the eye-wall of center of small diameter hurricanes) coasts at constant Coriolis, anticyclonic circular ocean-surface 6 inertial speed equal to its centrifugal rotation currents initial speed * For Northern Hemisphere. Direction is opposite in Southern Hemisphere. ** Antitriptic winds are unphysical; not listed here. To be accurate, an additional vertical advection Horizontal Motion term should be included in the right-hand side of each equation. Namely, –W·∆U/∆z in the forecast Equations of Motion — Revisited equation for U wind, and –W·∆V/∆z in the equation The geostrophic wind can be used as a surrogate for V wind. For example, fast jet-stream horizontal for the pressure-gradient force, based on the defini- winds aloft can be advected down toward the sur- tions in eqs. (10.26). With this substitution, we can face, causing fast, damaging surface winds. Simi- then group this term with the Coriolis term in the larly, slow boundary-layer horizontal winds can be equations of horizontal motion (10.23): advected upward to spread the effects of surface drag higher into the atmosphere. •(10.51a) A centrifugal term could also be added for winds ∆U ∆U ∆U U associated with curved isobars, which is an artifice = −U −V +f · V − V − w · to account for the continual changing of wind di- ∆t ∆x ∆y c() g T z i rection caused by an imbalance of the other forces (where the imbalance is the centripetal force). •(10.51b) The third term on the right is called the ∆V ∆V ∆V V geostrophic departure term. The wind difference = −U −V −f · U − U − w · c( g ) T is also called the ageostrophic wind (Uag, Vag): ∆t ∆x ∆y zi } } } } } Uag = U – Ug •(10.52a) horizontal pressure turbulent tendency Coriolis V = V – V •(10.52b) advection gradient drag ag g R. STULL • meteorology for scientists and engineers 315 Scales of Horizontal Motion 5BCMF4DBMFTPGIPSJ[POUBMNPUJPOJOUIFUSPQPTQIFSF In the atmosphere, motions of many scales are superimposed: from small turbulent eddies through 4J[F 4DBMF /BNF LN thunderstorms and cyclones to large planetary-scale NBDSPB QMBOFUBSZTDBMF circulations such as the jet stream. Scales of hori- LN NBDSPC TZOPQUJDTDBMF zontal motion are classified in Table 10-6. LN NFTPB Small atmospheric phenomena of horizontal LN NFTPC NFTPTDBMF dimension less than about 10 km are frequently LN NFTPH isotropic; namely, their vertical and horizontal di- LN NJDSPB CPVOEBSZMBZFSUVSCVMFODF mensions are roughly equal. Horizontally-larger N NJDSPC TVSGBDFMBZFSUVSCVMFODF phenomena are somewhat pancake-like, because the N NJDSPH vertical dimension is generally limited by the depth N JOFSUJBMTVCSBOHFUVSCVMFODF of the troposphere (about 11 km). NN NJDSPE Fig. 10.24 shows that time scales τ and horizontal NN NJDSPTDBMF GJOFTDBMFUVSCVMFODF length scales λ of many meteorological phenomena NN WJTDPVT EJTTJQBUJPOTVCSBOHF early follow a straight line on a log-log plot. This N NFBOGSFFQBUICFUXFFONPMFD implies that N NPMFDVMBS b NPMFDVMFTJ[FT τ/τo = (λ/λo) (10.53) /PUF%JTBHSFFNFOUBNPOHEJGGFSFOUPSHBOJ[BUJPOT where ≈ 10–3 h, ≈ 10–3 km, and b ≈ 7/8. For ex- 4ZOPQUJD".4LN 8.0LN τo λo .FTPTDBMF".4LN 8.0LN ample, microscale turbulence about 1 m in diameter .JDSPTDBMF".4LN 8.0DNLN might last about a 1 s. Boundary-layer thermals of XIFSF".4 "NFSJDBO.FUFPSPMPHJDBM4PDJFUZ diameter 1 km have circulation lifetimes of about 25 BOE8.0 8PSME.FUFPSPMPHJDBM0SHBOJ[BUJPO min. Thunderstorms of size 10 km might last a few hours. Cyclones of size 1000 km might last a week. In the next several chapters, we cover weather EFDBEF phenomena from largest to smallest horiz. scales: HMPCBM DJSDVMBUJPO • Chapter 11 Global Circulation (planetary) DMJNBUF ZFBS • Chapter 12 Airmasses and Fronts (synoptic) WBSJBUJPOT • Chapter 13 Extratropical Cyclones (synoptic) NPO NPOUI TPPO • Chapter 14 Thunderstorms (meso β) 3PTTCZ • Chapter 15 Thunderstorm Hazards (meso γ) XFFL XBWFT DZDMPOFT • Chapter 16 Hurricanes (meso α & β) GSPOUT • Chapter 17 Local Winds (meso β & γ) EBZ IVSSJDBOF MPDBM • Chapter 18 Atm. Boundary Layers (microscale) XJOET .$4 Although hurricanes are larger than thunderstorms, IPVS 5JNF4DBMF I we cover thunderstorms first because they are the UPSOBEP UIVOEFSTUPSNT building blocks of hurricanes. Similarly, midlati- UIFSNBMT tude cyclones often contain fronts, so fronts are cov- UVSCVMFODF QSPEVDUJPO ered before extratropical cyclones. NJOVUF UVSCVMFODF DBTDBEF &BSUI$JSDVNGFSFODF Vertical Forces and Motion )PSJ[POUBM4DBMF LN Forces acting in the vertical can cause or change Figure 10.24 vertical velocities, according to Newton’s Second Typical time and spatial scales of meteorological phenomena. Law. In an Eulerian framework, the vertical com- ponent of the equations of motion is: (10.54) ∆W ∆W ∆W ∆W 1 ∆P Fz TD Science Graffito = −U −V −W − −g − ∆t ∆x ∆y ∆z ρ ∆z m “The book of nature is written in the language of mathematics.” } } } } } – Galileo, as paraphrased by Alex Stone, 2005. Dis- pressure gra- turb. cover, 26, p77. tendency advection gradient vity drag 316 CHAPTER 10 Dynamics (a) where the left hand side is the vertical acceleration, and the right hand side lists the vertical forces per mass. (U, V, W) are the three Cartesian velocity components in the (x, y, z) directions, P is pressure, [ LN ρ is air density, t is time, Fz TD is the turbulent drag force in the vertical, and mass is m. Gravitational acceleration magnitude is |g| = 9.8 m·s–2. Coriolis force is negligible in the vertical (see the Focus box on Coriolis Force in 3-D, earlier in this chapter), and is 1 L1B not included in the equation above. Recall from Chapter 1 that our atmosphere has (b) an extremely large pressure gradient in the verti- cal, which is almost completely balanced by gravity (Fig. 10.25). Also, there is a large density gradient in the vertical. We can define these large terms as a mean background state or a reference state of [ LN the atmosphere. Use the overbar over variables to indicate their average background state. Define this background state such that it is exactly in hydro- static balance (see Chapter 1): ∆ P (10.55) = −ρ· g S LHN ∆ z Figure 10.25 Mean background state, showing variations with height z of (a) However, small deviations in density and pres- atmospheric pressurePP= +andP' (b) densityρ= ρ +(fromρ′ Chapter 1). sure from the background state can drive important non-hydrostatic vertical motions, such as in ther- mals and thunderstorms. To discern these effects, we must first remove the background state from the full vertical equation of motion. Focus on the pressure-gradient and gravity terms of eq. (10.54), Solved Example rewritten here as: An updraft of 8 m/s exists 2 km west of your loca- 1 ∆P tion, and there is a west wind of 5 m/s. At your loca- − − ρ g (10.56) tion there is zero vertical velocity, but the air is 3°C ρ ∆z warmer than the surrounding environment of 25°C. What is the initial vertical acceleration of the air over Split the total density ρ into a backgroundρ = ( ρ ) + ρ′ your location? and deviationρ= ρ + (ρ′ ) part: ρ= ρ + ρ′ . Do the same for pressure: PP= + P' . The terms above become: Solution Given: ∆θ = 3°C, Te = 273+25 = 298 K, U= 5m/s ∆W/∆x= (8m/s – 0) / (–2,000m – 0) 1 ∆P ∆ P′ − − −ρg − ρ′ g (10.57) Assume: V = 0 . Drag = 0 initially, given W = 0. (ρ+ ρ′) ∆z ∆ z Dry air, thus Tv = T. Find: ∆W/∆t = ? m·s–2 The first and third terms in square brackets in eq. ∆W ∆W θp− θ e Use eq. (10.59): = −U + · g (10.57) cancel out, due to hydrostatic balance (eq. ∆t ∆x Te 10.55) of the background state. = –(5 m/s)·(–8m·s–1/2,000m)+(3/298)·(9.8m·s–2) Next, a Boussinesq approximation is made = 0.020 + 0.099 = 0.119 m·s–2 that ρ′ g /( ρ + ρ′) ≈ ( ρ′/ ,ρ which)· g implies that density deviations are important in the gravity term, but Check: Units OK. Physics OK. negligible for all other terms. This is reasonable be- Discussion: Extrapolated over a minute, this initial cause ρ′ << ρ . In the Stability chapter, it was shown acceleration gives W = 7.1 m/s. However, this vertical that density deviations can be described by virtual velocity would not be achieved because as soon as the temperature T deviations (with a sign change be- velocity is nonzero, the drag term also becomes non- v cause low density corresponds to high tempera- zero and tends to slow the vertical acceleration. ture): R. STULL • meteorology for scientists and engineers 317 ρ′ θ′ θv p − θv e FOCUS • Eötvös Effect − ··g = v g = ·'g= g (10.58) ρ Tv Tv e When you move along a path at constant distance R above Earth’s center, gravitational acceleration ap- where subscript p denotes the air parcel, and sub- pears to change slightly due to your motion. The script e is for the environment, and g’ is called the measured gravity |gobs| = |g| – ar , where: reduced gravity. T in the denominator must be ve a = 2·Ω·cos(ϕ)·U + (U2 + V2)/R in Kelvin. r Plugging this back into eq. (10.54) gives: The first term is the vertical component of Coriolis force (eq. 10.17e in the Focus box on p.297), and the last ∆W ∆W ∆W ∆W term is centrifugal force as you follow the curvature = −U −V − W ∆t ∆x ∆y ∆z of the Earth. Thus, you feel lighter traveling east and •(10.59) heavier traveling west. This is the Eötvös effect. 1 ∆ P′ θv p − θv e Fz TD − + · g − ρ ∆ z T v e m 8 å8 7 å7 Terms from this equation will be used in the Lo- cal Winds and Thunderstorms chapters, to explain strong vertical velocities. Turbulent drag is the resistance of a vertically 6 6 å6 å[ moving air parcel against other surrounding (sta- tionary environmental) air. This is a completely dif- åZ ferent effect than drag against the Earth’s surface, 7 åY and is not described by the same drag equations. [ The nature of F is considered in the chapter Figure 10.26 z TD Inflow and outflow 8 Z on Air Pollution Dispersion, as it affects the rise of Y from a fixed volume. smoke-stack plumes. Air with no vertical movement relative to its environment has no drag. Solved Example Just before a tornado strikes your garage, the air density inside is 1 kg/m3. Winds of 100 m/s enter the Mass Conservation open garage from the west, but nothing leaves from the east. Also, your garage is temporarily intact, so the Barring any nuclear reactions, air molecules are other walls, floor, and ceiling prevent winds in those other directions. The east end of your garage is 8 m not converted into energy, and air mass is conserved. from the open west end. What is the density change in In an Eulerian framework, mass flowing into a fixed your garage during the first 1 s, neglecting advection? volume minus the mass flowing out gives the change of mass within the volume (Fig. 10.26). The equation Solution describing this mass balance is called the continu- Given: Udoor = 100 m/s, Uend = 0 m/s, ∆x = 8 m, ity equation. The name “continuity” is based on ρ = 1 kg/m3. the observation that gases such as air tend to spread Find: ∆ρ/∆t = ? kg·m3·s–1 . smoothly and evenly within a volume. Use eq. (10.60), with V = W = 0 because the other walls, roof, and floor prevent winds in those directions: Continuity Equation ∆ρ VV− kg ()−60 − 0 m/s Recall that mass within a unit volume is defined = −ρ N.. entr S..entr = − 1. 2 ∆t ∆ y 3 ()20 m as density, ρ. The mass budget is: tunnel m (10.60) ∆ρ/∆t = +12.5 kg·m3·s–1 . ∆ρ ∆ ρ ∆ρ∆ ρ ∆U ∆V ∆W = −U − V − W − ρ + + Check: Units OK. Physics OK. ∆t ∆x ∆y ∆z ∆x ∆y ∆z Discussion: Due to Bernoulli’s principle (Local Winds chapter), you won’t have to worry about this which is also called the continuity equation. The high density for long. The pressure inside your garage terms in curly braces { } describe advection. Eq. will increase so fast that it will blow out your walls and roof as if a bomb exploded. So don’t leave your ga- (10.60) can be rearranged (using calculus) to be: rage door (or your windows) open during a tornado. 318 CHAPTER 10 Dynamics ∆ρ ∆(ρU) ∆( ρV) ∆(ρW) Horizontal divergence (D) is defined as = − + + (10.61) ∆t ∆x ∆y ∆z ∆U ∆V D = + (10.63) ∆ x ∆ y where U, V, and W are the wind components in the x, Negative values of D correspond to convergence. y, and z directions, respectively, and t is time. Plugging this definition into eq. (10.62) shows that Be careful when calculating the wind gradients; vertical velocities increase with height where there calculate them as wind at location 2 minus wind at is convergence: ∆W location 1, divided by distance at location 2 minus = −D (10.64) distance at location 1. Do not accidently subtract 1 ∆ z from 2 in the numerator, and then subtract 2 from 1 in For situations such as circular isobars around the denominator, because it gives the wrong sign. a low pressure center, cylindrical coordinates are easier to use (Fig. 10.27). The continuity equation is then Incompressible Continuity Equation 2·V ∆W in = •(10.65a) Density at any fixed altitude changes only a little R ∆z with temperature and humidity for most non-violent where R is the radius of the cylinder, and ∆z is the weather conditions. Therefore, we can neglect mass cylinder depth. We assume that the radial inflow changes within the volume, compared to the inflow velocities V through the sides of the cylinder are and outflow. The air is said to be incompressible in equal everywhere. When V is positive (indicating when the density does not change (∆ρ ≈ 0). This ap- in horizontal inflow), then W∆ is also positive (indicat- proximation fails in strong thunderstorm updrafts ing vertical outflow). and tornadoes. If a cylinder of air is at the ground where W = 0 at For incompressible flow, the left hand side and the cylinder bottom, then W at the cylinder top is: the advection terms of eq. (10.60) are zero. This re- quires inflow to balance outflow: W = (2 · Vin · ∆z) / R (10.65b) ∆U ∆V ∆W 8 å8 + + = 0 •(10.62) ∆x ∆y ∆z This continuity equation is illustrated in Fig. 10.26 Figure 10.27 for an example of inflow and outflow across the faces Mass conserva- 3 7 7JO JO of a cube. The length of the grey arrows represents tion in cylindrical the strength of the wind components. Also, for this coordinates. example ∆x = ∆y = ∆z. å[ 7JO 7JO In the x-direction of this particular illustration, there is less wind (U) entering the cube than leaving 8 (U + ∆U). This is called divergence. For this case, ∆U is positive. In the z-direction (vertical), there is Solved Example more air entering (W) than leaving (W + ∆W), in Fig. At a radius of 400 km from a low center, boundary 10.26. This is called convergence, and ∆W is nega- layer winds have a 2 m/s component that crosses the tive. In the y-direction, the air entering (V) and leav- circular isobars inward. If the boundary layer is 1 km ing (V + ∆V) are equal, so there is no convergence or thick, estimate the average vertical velocity out of the divergence. For this example, ∆V is zero. top of the boundary layer. For this example shown in Fig. 10.26, the continu- ity equation is Solution Given: Vin = 2 m/s, R = 400 km, ∆z = zi = 1 km Find: W = ? m/s (positive) + (0) + (negative) = 0 ∆z 1km Use eq. (10.65b): WV= 2··in = 2·(3m/s)· so we anticipate that mass is conserved. In the real = 0.01 m/s R 500km atmosphere, the directions having convergence or divergence might differ from this example, but the Check: Units OK. Physics OK. sum must always equal zero. Namely convergence Discussion: For non-thunderstorm conditions, this in one or two directions must be balanced by diver- magnitude of about 1 cm/s is typical for vertical ve- gence in the other direction(s). locities in the atmosphere. Such small velocities allow use of the hydrostatic assumption (see Chapter 1). R. STULL • meteorology for scientists and engineers 319 Boundary-Layer Pumping 8#- Around low-pressure regions near the surface, turbulent drag causes horizontal inflow (conver- - gence) and slower tangential winds within the 7JO 7 boundary layer. There is no vertical air motion (W = [ JO [J 0) at the bottom of the boundary layer because of the Z ground. Thus, horizontal inflow in the boundary Y ( layer must be balanced by vertical outflow from the boundary-layer top. Figure 10.28 This mechanism for creating mean upward mo- Convergence and updrafts in a cyclone. tion is called Ekman pumping or boundary- layer pumping. The upward motion carries water vapor, which then condenses in the adiabatically cooled air, causing clouds and precipitation. Thus, G z (10.67) low-pressure regions generally have foul weather. i ≈ NBV Around high-pressure centers, turbulent drag causes horizontal outflow (divergence). This is bal- where N is the Brunt-Väisälä frequency (see the anced by downward motion called subsidence at BV Stability chapter) above the boundary layer. This al- the top of the boundary layer. Subsidence warms air lows b to be rewritten as: b = { 1 – 0.5·[C ·N /f ]}. adiabatically, thereby evaporating most clouds and D BV c The approximation for boundary-layer cross-isobar- causing fair weather in high-pressure regions. ic flow is valid only when C[ ·N /f ] < 1. In the section on the BLG wind, we definedV D BV c BLG The updraft velocity in eq. (10.66) depends on as the radial component of wind, which by defini- both the size and rotation speed of the cyclone. tion equals the inflow velocity V for eq. 10.65 (see in Vorticity is one measure of the combined effects Fig. 10.28). Unfortunately, we were unable to find an of rotation speed and size. analytical solution for V . However, there is an Geostrophic relative BLG vorticity can be defined as analytical solution available for VBL from an earlier section, which is slightly greater than VBLG. For the 2·G subsequent analysis, we will use Vin ≈ VBL, know- ζg = (10.68) ing that the result will be an upper limit on what R are likely slower inflow velocities in the real atmo- which is a measure of the rotation of the air. More sphere. details of vorticity will be covered in the Global Cir- Assuming that the winds are relatively strong culation chapter. around lows, and that cloudy conditions preclude Using the geostrophic vorticity in eq. (10.66) strong surface heating or cooling, we can assume yields an alternative expression for the vertical ve- that the boundary layer is statically neutral. For locity pumped out of the top of a cyclone: this situation, the inflow velocity across the isobars is given by eq. (10.41b). Combining this inflow velocity with the cylin- G CN· WC= ··ζ ·.1− 0 5 DBV (10.69) drical form of the continuity equation (10.65) allows BL D g fc fc us to calculate the vertical velocity WBL at the top of the boundary layer (Fig. 10.28): The terms outside of the brackets suggest that greater vertical velocities (and hence nastier storms) 2 2··b CD G occur for stronger geostrophic winds (i.e., tighter WBL = · •(10.66) fc R packing of isobars) over rougher terrain, and where there is greater curvature of the cyclonic flow. Also, lower latitudes give smaller Coriolis parameters, where R is the radius of curvature of the isobars which allow greater vertical velocity. around the low center, G is the geostrophic wind The term in square brackets shows that the fastest speed, f is the Coriolis parameter, and C ( ≈ 0.005 c D vertical velocity is expected in statically neutral flow over land) is the drag coefficient for a neutral bound- (where N = 0) above the boundary layer. Greater ary layer. BV static stabilities cause weaker vertical velocities. The factor b = { 1 – 0.5·[C ·G/(f ·z )] } is from D c i By utilizing the approximation for mixed-layer eq. (10.41b), where boundary-layer depth is z . For i depth (eq. 10.67), an simplicity, the statically neutral boundary-layer internal Rossby radius of can be approximated as depth within a cyclone can be approximated as deformation 320 CHAPTER 10 Dynamics Solved Example G zT λR ≈ · (10.70) A geostrophic wind of 10 m/s blows cyclonically fc zi around a low-center with radius of curvature of 1000 –1 km. The latitude is such that fc = 0.0001 s , and the drag coefficient is 0.005. The tropospheric lapse rate is where zi is the boundary-layer depth and zT is the standard above the BL. depth of the troposphere. The internal Rossby radi- What is the vertical velocity out of the top of the us of deformation is discussed in more detail in the boundary layer? Also, estimate the boundary-layer Global Circulation chapter, and an external Rossby depth, geostrophic relative vorticity, and internal radius is given in the Airmasses & Fronts chapter. Rossby radius. Thus, an alternative form for the vertical velocity equation can be written using the Rossby radius of Solution deformation: 6 –1 Given: G = 10 m/s, R = 10 m, fc = 0.0001 s , –1 CD = 0.005, NBV = 0.0113 s (from a previous solved example using the standard atmos.) zi λR Find: z = ? m, ζ = ? s–1, λ = ? km, W = ? m/s WCBL = D ··λR·· ζ g 1− 0.·5 CD · (10.71) i g R BL zT zT Assume: tropospheric depth zT = 11 km Use eq. (10.67): –1 zi ≈ G/NBV = (10 m/s)/(0.0113 s ) = 885 m Use eq. (10.68): Kinematics 2·( 15m/s) –5 –1 Kinematics is the study of patterns of motion, ζg = 5 = 2x10 s 5× 10 m without regard to the forces that cause them. We Use eq. (10.70): will focus on horizontal divergence, vorticity, and –1 ()15m/s 11km deformation. All have units of s . λR ≈ · = 1243 km (.0 0001s−1) 1.327km We have already encountered divergence, D, the spreading of air: First check that [CD·NBV/fc] < 1. [0.005·(0.0113s–1)/(0.0001s–1)] = 0.565 < 1. OK. ∆U ∆V D = + (10.72) ∆ x ∆ y Use eq. (10.71): WBL = (.1 327km) 0.·004 ·(1.)243 ×106 m·(6 × 10−5 s−1) Figure 10.29a shows an example of pure divergence. ()11km Its sign is positive for divergence, and negative for 1243km ·.1− 0 5·( 0.)004 · convergence (when the wind arrows point toward 11km a common point). = (0.01 m/s) · [0.718] = 0.0072 m/s Vorticity describes the rotation of air (Fig. 10.29b). The relative vorticity, ζr , is given by: Check: Units OK. Physics OK. Discussion: This vertical velocity of 7.2 mm/s is typ- ∆V ∆U ζ = − (10.73) ical of synoptic circulations. Although weak, it is suf- r ∆ x ∆ y ficient to lift air to cause condensation, releasing latent heat which allows stronger buoyant updrafts within The sign is positive for counterclockwise rotation (i.e., the clouds. cyclonic rotation in the N. Hemisphere), and nega- tive for clockwise rotation. Vorticity is discussed in greater detail in the Global Circulation chapter. Neither divergence nor vorticity vary with rotation of the axes — they are rotationally invariant. Two types of deformation are stretching defor- mation and shearing deformation (Figs. 10.29c & d). Stretching deformation, F1, is given by: ∆U ∆V (10.74) F = − 1 ∆ x ∆ y The axis along which air is being stretched (Fig. 10.29c) is called the axis of dilation (x axis in this R. STULL • meteorology for scientists and engineers 321 example), while the axis along which air is com- B %JWFSHFODF pressed is called the axis of contraction (y axis in this example). Shearing deformation, F2, is given by: ∆V ∆U F = + (10.75) 2 ∆ x ∆ y Z As you can see in Fig. 10.29d, shearing deformation is just a rotated version of stretching deformation. Y The total deformation, F, is: 1/ 2 FF 2 F 2 (10.76) = 1 + 2 C 7PSUJDJUZ Deformation often occurs along fronts. Most real flows exhibit combinations of divergence, vorticity, and deformation. Z Measuring Winds Y For weather stations at the Earth’s surface, wind direction can be measured with a wind vane mounted on a vertical axel. Fixed vanes and other D 4USFUDIJOH%FGPSNBUJPO E 4IFBSJOH%FGPSNBUJPO shapes can be used to measure wind speed, by us- ing strain gauges to measure the minute deforma- tions of the object when the wind hits it. The generic name for a wind-speed measuring device is an anemometer. A cup anemometer has conic- or hemispheric-shaped cups mounted on spokes that rotate about a vertical axel. A propellor anemometer has a propellor mounted on a hori- Z zontal axel that is attached to a wind vane so it al- ways points into the wind. For these anemometers, the rotation speed of the axel can be calibrated as a Y wind speed. Other ways to measure wind speed include a hot- wire or hot-film anemometer, where a fine metal Figure 10.29 wire is heated electrically, and the power needed to Kinematic flow-field definitions. Black arrows represent wind maintain the hot temperature against the cooling velocity. effect of the wind is a measure of wind speed. A pitot tube that points into the wind measures the dynamic pressure as the moving air stagnates in a dead-end tube. By comparing this dynamic pres- sure with the static pressure measured by a differ- ent sensor, the pressure difference can be related to wind speed. Sonic anemometers send pulses of sound back and forth across a short open path between two op- posing transmitters and receivers of sound. The speed of sound depends on both temperature and wind speed, so this sensor can measure both. Trac- ers such as smoke, humidity fluctuations, or clouds can be tracked photogramatically from the ground 322 CHAPTER 10 Dynamics or from remote sensors such as laser radars (lidars) equation), while the word dynamics describes how or satellites, and the wind speed then estimated forces cause winds (as given by Newton’s second from the change of position of the tracer between law). successive images. Measurements of wind vs. height can be made Threads with rawinsonde balloons (using a GPS receiver Forces cause winds, and winds flow over moun- in the sonde payload to track horizontal drift of the tains to form mountain waves and lenticular clouds balloons with time), dropsondes (like rawinsondes, (Chapter 17). Winds embody the global circula- only descending by parachute after being dropped tion (Chapter 11), which moves air masses around from aircraft), pilot balloons (carrying no payload, to create fronts (Chapter 12), extratropical cyclones but being tracked instead from the ground using (Chapter 13) and our global climate (Chapter 21). radar or theodolites), wind profilers, Doppler The equations of motion are solved every day on weather radar (see the Remote Sensing chapter), large computers to make the daily weather fore- and via anemometers mounted on aircraft. casts (Chapter 20). The interplay between buoyancy (Chapter 5) and dynamics creates phenomena such as thunderstorms, tornadoes (Chapters 14 and 15), and hurricanes (Chapter 16). Winds advect air pol- Summary lutants from place to place (Chapter 19). Not only do winds advect temperature (Chap- Pressure-gradient force can start winds moving, ter 3), but horizontal temperature variations cause and can change wind direction and speed. This horizontal pressure variations via the hypsometric force points from high to low pressure on a constant equation (Chapter 1), thereby driving the jet-stream altitude chart (such as at sea-level), or points from winds. Winds riding over colder air masses carry high to low heights on an isobaric chart (such as the water vapor (Chapter 4), some of which can con- 50 kPa chart). dense to make clouds (Chapter 6) and precipita- Once the air is moving, other forces such as tion (Chapter 7). Clouds imbedded in the air will turbulent drag or Coriolis force also act on the air. move horizontally with the air, and can be tracked Coriolis force is an apparent force that accounts for by satellite (Chapter 8) to estimate the wind speed our moving frame of reference on the rotating Earth. and direction. The complexity of the atmosphere is Turbulent drag is important only near the ground, in becoming apparent. the boundary layer. The relationship between forces and acceleration of the winds is given by Newton’s second law of motion. When all forces balance, the winds are steady. Exercises In regions of straight isobars above the boundary layer, pressure-gradient and Coriolis forces balance to cause winds that are geostrophic. Around highs Numerical Problems or lows, there are slight imbalances associated with N1. Plot the wind symbol for winds with the follow- centrifugal force, which causes steady-state gradient ing directions and speeds: winds. In the boundary layer, winds are slower than a. N at 5 kt b. NE at 35 kt c. E at 65 kt either geostrophic or gradient because of turbulent d. SE at 12 kt e. S at 48 kt f. SW at 105 kt drag. In tornadoes and waterspouts, centrifugal and g. W at 27 kt h. NW at 50 kt i. N at 125 kt pressure gradient forces nearly balance to create the intense cyclostrophic wind. In oceans, currents can N2. Find the acceleration of a 75 kg person when inertially flow in a circle. pushed by a force (N) of The two most important force balances at mid- a. 1 b. 2 c. 5 d. 10 e. 20 f. 50 latitudes are hydrostatic balance in the vertical, and g. 100 h. 200 i. 500 j. 1000 k. 2000 geostrophic balance in the horizontal. Outside of thunderstorms, winds are well de- N3. If the initial velocity of an object is zero, find the scribed by incompressible mass continuity. Thus, final velocity after 100 s for an applied net force per mechanisms that cause motion in one direction unit mass of: (horizontal or vertical) will also indirectly cause a. 5 N/kg b. 10 m·s–2 motions in the other direction as the air tries to c. 15 N/kg d. 20 m·s–2 maintain continuity, resulting in a circulation. Ki- e. 25 N/kg f. 30 m·s–2 nematics is the word that describes the behavior g. 35 N/kg h. 40 m·s–2 and effect of winds (such as given by the continuity i. 45 N/kg j. 50 m·s–2 R. STULL • meteorology for scientists and engineers 323 f. U = 5 m/s and V = 35 m/s N4. Find the advective “force” per unit mass given g. U = 25 m/s and V = –2 m/s the following wind components (m/s) and horizon- h. U = 0 m/s and V = 10 m/s tal distances (km): a. U=10, ∆U=5, ∆x=3 N10. For a statically unstable boundary layer, find b. U=6, ∆U=–10, ∆x=5 the turbulent drag force per unit mass, given a buoy- c. U=–8, ∆V=20, ∆x=10 ant velocity scale of 60 m/s and d. U=–4, ∆V=10, ∆x=–2 a. U = 5 m/s and V = 1 m/s e. V=3, ∆U=10, ∆y=10 b. U = –1 m/s and V = 3 m/s f. V=–5, ∆U=10, ∆y=4 c. U = 2 m/s and V = –4 m/s g. V=7, ∆V=–2, ∆y=–50 d. U = –2 m/s and V = –1 m/s h. V=–9, ∆V=–10, ∆y=–6 e. U = –4 m/s and V = 0 m/s f. U = 5 m/s and V = 3 m/s N5. What is the pressure-gradient force per unit g. U = 5 m/s and V = –2 m/s mass between Seattle, WA, and Corvallis, OR (about h. U = 0 m/s and V = 2 m/s 340 km south of Seattle), if sea-level pressure is 98.4 kPa in Seattle and has the following pressure (kPa) N11. Draw a northwest wind of 5 m/s in the S. Hemi- in Corvallis? sphere on a graph, and show the directions of forces a. 98.6 b. 98.8 c. 99.0 d. 99.2 e. 99.4 acting on it. Assume it is in the boundary layer. f. 99.6 g. 99.8 h. 100.0 i. 100.2 j. 100.4 a. pressure gradient b. Coriolis k. 100.6 l. 100.8 m. 101.0 n. 101.2 o. 101.4 c. centrifugal d. drag N6. Given U = –5 m/s and V = 10 m/s, find the com- N12. What is the geostrophic wind speed at a height 3 –4 –1 ponents of centrifugal force around a 800 km diam- where ρ = 0.7 kg/m , fc = 10 s , and the pressure eter: gradient ( kPa/100 km) magnitude is: a. low in the southern hemisphere a. 0.1 b. 0.2 c. 0.3 d. 0.4 e. 0.5 b. high in the northern hemisphere f. 0.6 g. 0.7 h. 0.8 i. 0.9 j. 1.0 c. high in the southern hemisphere k. 1.1 m. 1.2 n. 1.3 o. 1.4 p. 1.5 d. low in the northern hemisphere N13. At a latitude of 60°N, find the geostrophic N7. Calculate the Coriolis parameter for wind given a height gradient (m/km on a constant a. Munich, Germany pressure surface) of: b. Oslo, Norway a. 0.1 b. 0.2 c. 0.3 d. 0.4 e. 0.5 c. Madrid, Spain f. 0.6 g. 0.7 h. 0.8 i. 0.9 j. 1.0 d. Chicago, USA k. 1.1 m. 1.2 n. 1.3 o. 1.4 p. 1.5 e. Buenos Aires, Argentina f. Melbourne, Australia N14. If the geostrophic wind around a high is 10 g. Vancouver, Canada m/s, then what is the gradient wind speed, given fc h. Beijing, China = 10–4 s–1 and a radius of curvature of: i. Moscow, Russia a. 800 km b. 500 km c. 600 km j. Tokyo, Japan d. 1000 km e. 2000 km f. 1500 km k. (your town) g. 700 km h. 1200 km i. 900 km N8. For Chicago, find the Coriolis force per unit N15. Find the boundary layer winds given Ug = 5 –4 –1 mass in the N. Hemisphere for: m/s, Vg = 5 m/s, zi = 1500 m, and fc = 10 s . Also, a. U = 10 m/s b. V = 5 m/s c. U= 3 m/s what angle do the winds cross the isobars? This is a d. U = –10 m/s e. V = –5 m/s f. U = 8 m/s statically neutral boundary layer. Use CD = g. U = –3 m/s h. V = –8 m/s i. V = 40 m/s a. 0.002 b. 0.004 c. 0.006 d. 0.008 e. 0.010 f. 0.012 g. 0.014 h. 0.016 i. 0.018 j. 0.019 N9. For a neutral boundary layer, find the turbulent drag force per unit mass over a forest for N16. Same as previous problem, but for an unstable a. U = 5 m/s and V = 25 m/s boundary layer with wB (m/s) of: b. U = –10 m/s and V = 5 m/s a. 75 b. 100 c. 50 d. 200 e. 150 c. U = 5 m/s and V = –15 m/s f. 225 g. 125 h. 250 i. 175 j. 275 d. U = –5 m/s and V = –5 m/s e. U = –40 m/s and V = 5 m/s 324 CHAPTER 10 Dynamics N17(§). Recompute MBLG as in the solved example in (∆U/∆y , ∆V/∆y) in units of (m/s)/(500 km) as given the Boundary Layer Gradient Wind section, but with below: the following changes: a. (–5, –5) b. (–5, 0) c. (0, –5) d. (0, 0) e. (0, 5) a. G = 10 m/s b. zi = 2 km c. CD = 0.002 f. (5, 0) g. (5, 5) h. (–5, 5) i. (5, –5) –4 –1 d. R = 1000 km e. fc = 2x10 s f. G = 15 m/s g. zi = 1.5 km h. CD = 0.005 –4 –1 i. R = 1500 km j. fc = 1.5x10 s Understanding & Critical Evaluation U1. Compare eq. (10.1) with (1.24), and discuss. N18. Given a pressure gradient of 0.5 kPa/m, com- pute the cyclostrophic wind at the following radii U2. Can eqs. (10.6) be used to make a forecast if the (m): a. 10 b. 12 c. 14 d. 16 e. 18 initial conditions (i.e., the current winds) are not f. 20 g. 22 h. 24 i. 26 j. 28 k 30 known? Discuss. N19. For an inertial wind, find the radius of curva- U3. If all of the net forces (eq. 10.7) are zero, does ture (km) and the time period (h) needed to com- that mean that the wind speeds (eq. 10.5) are zero? plete one circuit, given f = 10–4 s–1 and an initial c Explain. wind speed (m/s) of: a. 1 b. 2 c. 3 d. 4 e. 6 f. 7 g. 8 h. 9 U4. Eqs. (10.8) suggest that winds can advect winds. i. 10 j. 11 k. 12 m. 13 n. 14 o. 15 How is that possible? N20. Find the antitriptic wind for the conditions of U5. In eqs. (10.8), why does advection depend on the exercise N15. gradient of winds (e.g., ∆U/∆x) across the Eulerian domain, rather than on just the value of wind that is N21. The boundary layer is 2 km thick. At a radius being blown into the domain? of 200 km from a low center, estimate the average vertical velocity through the top of the boundary U6. In Fig. 10.5, and on weather maps, what is the layer (BL), given an inward radial wind component relationship between packing of isobars (i.e., the (m/s) of: a. 2 b. 1.5 c. 1.2 d. 1.0 number of isobars that cross through a square cm of e. –0.5 f. –1 g. –2.5 h. 3 i. 0.8 j. 0.2 weather map) and the pressure gradient? N22. Estimate boundary layer depth within a cy- U7. In the N. Hemisphere, the pressure gradient clone given an isothermal environment above the points from high to low pressure. Which way does boundary layer, and a geostrophic wind (m/s) near it point in the S. Hemisphere? the surface of: a. 5 b. 10 c. 15 d. 20 e. 25 f. 30 g. 35 h. 40 i. 3 j. 8 k. 2 l. 1 U8. Eqs. (10.9) give the horizontal components of the pressure-gradient force. Combine those equations N23(§). For a 1 km thick boundary layer over a sur- vectorially to show that the vector direction of the face of drag coefficient 0.003, plot the vertical veloc- pressure-gradient force is indeed perpendicular to ity due to boundary-layer pumping as a function of the isobars and pointing toward lower pressure, for geostrophic wind speed, but for only wind speeds any arbitrary isobar direction such as shown in Fig. within the valid range for that eq. Plot separate 10.5. curves for the following radii (km) of curvature: (Assume a latitude of 45°, & standard atmosphere) U9. Eqs. (10.13) give the horizontal components a. 500 b. 1000 c. 2000 d. 3000 e. 4000 of centrifugal force. Combine those equations (Assume a latitude of 60°, & standard atmosphere) vectorially to show that the vector direction of cen- f. 500 g. 700 h. 900 i. 1500 j. 2500 trifugal force is indeed outward from the center of rotation, and is proportional to the square of the tan- N24. Estimate the value of internal Rossby radius gential velocity. of deformation at latitude 60°N for a tropospheric depth of 11 km and geostrophic wind speed of 15 U10. An air parcel at rest (relative to the Earth) near m/s. Assume a boundary-layer depth of the equator experiences greater tangential veloc- a. 500 m b. 1 km c. 2 km d. 750 m e. 250 m ity due to the Earth’s rotation than do air parcels at f. 1.5 km g. 2.5 km h. 3 km i. 3.5 km j. 100 m higher latitudes. Yet the Coriolis parameter is zero at the equator. Why? N25. Given ∆U/∆x = ∆V/∆x = (5 m/s) / (500 km), find the divergence, vorticity, and total deformation for R. STULL • meteorology for scientists and engineers 325 U11. Eqs. (10.17) give the horizontal components of the Coriolis force. Combine those equations U22. a. Is there any limit on the strength of the pres- vectorially to show that the vector direction of the sure gradient that can occur just outside of the cen- Coriolis force is indeed 90° to the right of the wind ter of cyclones? (Hint: consider Fig. 10.14) direction, for any arbitrary wind direction such as b. What controls this limit? the two shown in Fig. 10.6. c. What max winds are possible around cy- clones? U12. For the subset of eqs. (10.1 - 10.17) defined by your instructor, rewrite them for the S. Hemi- U23. Calculate the geostrophic and gradient winds, sphere. as appropriate, at a number of locations using the height contours plotted in Fig. 10.3. Compare them U13. Eqs. (10.19) give the horizontal components of to the observed winds and comment. the turbulent drag force. a. Combine those equations vectorially to show U24. What is implicit about the implicit solution (eq. that the vector direction of the drag force is indeed 10.39) for the boundary-layer wind? opposite to the wind direction, for any arbitrary wind direction. U25. How accurate are the approximate boundary b. Show that the magnitude of the drag force is layer wind solutions (eq. 10.41)? Under what condi- proportional to the square of the wind speed, M, for tions are the approximate solutions least accurate? statically neutral conditions. (Hint: compare with an iterative solution to the im- plicit equations 10.39). U14. Compare the values of the turbulent transport velocity during windy (statically neutral) and con- U26. Why is an exact explicit solution possible for vective (statically unstable) conditions. Discuss. the steady-state winds in the unstable boundary layer, but not for the neutral boundary layer? U15. Plug eqs. (10.26) into (10.27) to find the vector speed and direction (see Chapter 1) of the geostrophic U27. Verify that eqs. (10.42) are indeed exact solu- wind as a function of the vector pressure gradient. tions to (10.39) or (10.38). U16. Re-derive the geostrophic wind eqs. (10.26) for U28(§). a. Recreate on a spreadsheet the solved ex- the S. Hemisphere. ample for the boundary-layer gradient winds. b. Recreate the checks of that equation for the U17. Derive eqs. (10.29) from eqs. (10.26). special cases where it reduces to the geostrophic wind, gradient wind, and boundary-layer wind. U18. The geostrophic wind approaches infinity as c. Compare the results from (b) against the re- the equator is approached (see Fig. 10.10), yet the spective analytical solutions (which you must com- winds in the real atmosphere are not infinite there. pute yourself). Why? d. Can the analytical solutions for the gradient wind and the (neutral) boundary layer wind be com- U19. Verify that eq. (10.33) is indeed a solution to the bined to yield an approximate analytical solution to gradient wind eqs. (10.31). the BLG winds. If so, what are the limitations, and the magnitude of the errors. (Hint: Try substitut- U20. Verify that eqs. (10.34) are solutions to eq. ing Mtan in place of G in the equations for boundary (10.33). layer wind.) U21. Imagine an idealized weather map that had U29. Photocopy Fig. 10.13, and enhance the copy a single high pressure center next to a single low by drawing additional vectors for the boundary- pressure center. Further, suppose that if you were layer wind and the BLG wind. Make these vectors to draw a line through those two centers, that the be the appropriate length and direction relative to pressure variation along that line would be the same the geostrophic and gradient winds that are already as Fig. 10.14. Given that information, and assuming plotted. circular cyclones and anticyclones, draw isobars on the weather map at ±0.5 kPa increments, starting at U30. Verify that the cyclostrophic winds are indeed 100 kPa. Comment on the packing of (i.e., how close- a solution to the governing equations (10.45). ly spaced are) isobars near the centers of the cyclone and anticyclone. 326 CHAPTER 10 Dynamics U31. Re-derive the equations for cyclostrophic wind, b. Same as (a), but do this for several days in a but in terms of height gradient on a constant pres- row. Then plot a graph of how the pressure gradient sure surface, instead of pressure gradient along a changes with time over your location. constant height surface. [Hint: Compare eqs. (10.29) to (10.26).] W4. Search the web for 50 kPa (500 mb) weather maps that cover part of the S. Hemisphere. Com- U32. What aspects of the Approach to Geostrophy Fo- pute the geostrophic and gradient wind speeds and cus Box are relevant to the inertial wind? Discuss. directions in regions of straight and curved isobars, respectively, and compare with upper-air observa- U33. a. Derive eq. (10.66) based on geometry and tions. Discuss the differences between these winds mass continuity (total inflow = total outflow). and corresponding winds in the N. Hemisphere. b. For horizontal winds, we know that an in- creased drag coefficient will reduce wind speed. W5. Find a current 50 kPa (= 500 mb) weather map Why in eq. (10.66) does an increased drag coefficient (or other map as specified by your instructor) that cause increased vertical wind speed? has height contours already drawn on it. Look for c. When considering that factor b in eq. (10.66) is a region in the map where the isobars are nearly a negative function of the drag coefficient, does your straight. answer to part (b) change? a. Compute the geostrophic wind speed compo- nents, and the total geostrophic wind speed vector U34. The paragraph after eq. (10.69) gives a physical (speed and direction). interpretation of the equation. Show how that inter- b. If upper-air (rawinsonde) observations are pretation was reached, by examining each term in available near your area, compare the measured the equation and discussing its impact on W. winds with the geostrophic value computed from part (a). U35. Given eq. (10.70), determine the physical mean- c. If there are multiple regions of nearly straight ing of the internal Rossby radius of deformation by isobars at different latitudes in your weather map, how it affects the various terms in (10.71). see how the observed winds in these regions vary with latitude, and compare with the expected lati- U36. Regarding horizontal balances of forces, if tudinal variation of geostrophic wind (as from Fig. only Coriolis and turbulent-drag forces were active, 10.10). speculate on the nature of the resulting wind. W6. a. Same as W5, except look for a region on the U37. Modify eqs. (10.66) through (10.71) if necessary, map where the isobars are curving in a cyclonic to describe the boundary-layer pumping around (counterclockwise in the N. Hemisphere) direc- highs (anticyclones) rather than around lows (cy- tion when following along with the wind direction. clones). Discuss the significance of your equations. Compute the gradient wind for this case. b. Same as (a) but for anticyclone (clockwise) U38. Rewrite the total deformation as a function of turning winds. divergence and vorticity. Discuss. c. Compare the measured winds from rawinsondes to the theoretical winds from (a) and (b). Confirm that the gradient winds (both theo- Web-Enhanced Questions retical and observed) are slower than geostrophic around lows, and faster around highs. W1. Search the web for historical discussions of Isaac Newton, and summarize what you find. What W7. a. Find a sea-level weather map from the web did you find most unusual or interesting? that has isobars drawn on it. Look for a day or a re- gion where there are neighboring regions of strong W2. Search the web for “Coriolis Force”, to find sites high and low pressure. Print this map, and draw a that show animations of the movement of objects in straight line connecting the centers of the high and a rotating coordinate system. Tabulate a list of the the low. Extend this line well past the centers of best sites. the high and low. Using the isobars that cross your drawn line, find how pressure varies with distance W3. a. Search the web for a current sea-level weather between the high and low, and plot the results simi- map that covers your area, and which includes iso- lar to Fig. 10.14. bars. Measure the distance (km) between isobars, b. If you set the location of the center of the and compute the pressure gradient force. high as the origin of your coordinate system, check R. STULL • meteorology for scientists and engineers 327 whether the shape of the pressure curve agrees with b. Do the same as (a), but using a sea-level weath- eq. (10.37b). Confirm that the pressure variation er map showing the isobars. Also comment on the across a low-pressure center can have a cusp, while effect of boundary layer drag. that across a high center cannot. W8. Search the web for a weather map at 50 kPa (500 Synthesis Questions mb) or any other altitude above the boundary layer, S1. Suppose Newton’s second law of motion was not that includes the equator. Find a region of relatively a function of mass. How would the motion of bul- straight isobars near the equator, and compute the lets, cannon balls, and air parcels be different, if at geostrophic wind speed based on the plotted pres- all? sure gradient. Compare this theoretical wind with observed upper-air winds for the same altitude. S2. What if Newton’s second law of motion stated Why don’t the observations agree with the theory? that velocity was proportional to force/mass. How would weather and climate be different, if at all? W9. Find a current sea-level weather map from the web, that shows both wind direction (such as the S3. What if wind could not advect itself. How would wind symbol on surface station observations), and the weather and climate be different, if at all? isobars. a. For a region of the map where the isobars are S4. Suppose that pressure-gradient force was along relatively straight, and hopefully over non-moun- isobars, rather than perpendicular to them. De- tainous and non-coastal terrain, confirm that the scribe how winds would be different, how weather observed boundary-layer winds indeed cross the maps would be different, and how this might affect isobars at some small angle from high to low pres- the weather and climate, if at all. sure. What is the average angle? b. For a region where the isobars are curved S5. There is some debate in the literature that our around a high or low, confirm that the winds spiral understanding of Coriolis force might be incorrect. in towards the center of the low, and out from the We think that Coriolis force is an apparent force. center of a high. Can an apparent force change the momentum and c. Around either the high or low, estimate the kinetic energy of air parcels? If so, would this vio- average value of the component of wind that repre- late Newton’s laws when viewed from a non-rotat- sents inflow or outflow from the low or high. Use ing frame? Discuss. (Hint: look for a series of de- that value of V (or V ) in eq. (10.65) to compute in out scriptive articles by Anders Persson that appeared the vertical velocity at the top of the boundary lay- in Weather magazine starting in year 2000.) er. d. Based on the inward or outward component of S6. Fig. 10.7 shows wind shear across the top of the velocity from (c), estimate the drag force acting on boundary layer, where subgeostrophic winds in the the air. If this calculation is for flow around a low boundary layer change to geostrophic winds above where the winds are relatively fast, find the value of the boundary layer. The shear can exist because a the drag coefficient C for statically neutral condi- D strongly statically stable layer of air caps the bound- tions. ary layer. If such a stable capping inversion did not exist, how might the wind profile be different over W10. a. If it is hurricane season, search the web for a the depth of the troposphere? weather map that is just above the top of the bound- ary layer (85 kPa or 850 mb might be good enough), S7. Suppose that turbulent drag force acted 90° to but which is well below the altitude of the 60 kPa the right of the wind direction in the boundary layer. pressure. Look for a map (either observed or fore- Discuss how the boundary-layer winds would work cast) that has the height contours around the hurri- around highs and lows, and in regions of straight cane on this pressure surface (85 kPa). Find the loca- isobars. How would the weather and climate be dif- tion just outside of the eye where the height contours ferent, if at all? are packed closest together, and calculate the pres- sure gradient there. Then use that pressure gradi- S8. Suppose that the boundary-layer drag force did ent to compute the cyclostrophic wind, and compare not increase with velocity (in the case of an unstable your theoretical value with the observed upper-air boundary layer) or with velocity squared (in the case values at that height. of a neutral boundary layer), but was constant re- gardless of wind speed. How would the boundary layer winds, weather, and climate change, if at all? 328 CHAPTER 10 Dynamics S18. Suppose the Earth did not rotate. How would S9. What if there were no Coriolis force? How would the winds, weather, and climate be different, if at winds, weather, and climate be different, if at all? all? S10. On our present world where we perceive a S19. Why is incompressibility such a good approxi- Coriolis force, are there situations where it is pos- mation for the real atmosphere? How does the at- sible for wind to blow directly high to low pressure, mosphere react to density changes, that might help rather than more-or-less parallel to the isobars? De- ensure little density change? (Hint: consider the first scribe such scenarios. solved example in the continuity equation section.) S11. What if geostrophic winds could turn around S20. Extend the discussion of the Focus box on high or low pressure systems without feeling cen- Coriolis Force by deriving the magnitude of the tripetal or centrifugal force. How would the weath- Coriolis force for an object moving er and climate be different, if at all? a. westward b. southward S12. Suppose that cusps in pressure were allowed at S21. For zonal (east-west) winds, there is also a verti- high-pressure centers as well as at low pressure cen- cal component of Coriolis force. Using your own di- ters (see Fig. 10.14) so that strong pressure gradients agrams similar to those in the Focus box on Coriolis could exist near the center of both types of pressure Force, show why it can form. Estimate its magni- centers. Describe how the winds, weather, and cli- tude, and compare the magnitude of this force to mate might be different, if at all? other typical forces in the vertical. Show why a ver- tical component of Coriolis force does not exist for S13. Suppose that the Earth’s surface were friction- meridional (north-south) winds. less. How would the weather and climate be differ- ent, if at all? S22. What if a cyclostrophic-like wind also felt drag near the ground? This describes conditions at the S14. Suppose that the tropopause acted like a rigid bottom of tornadoes. Write the equations of mo- lid on the troposphere, and that the air at the top of tion for this situation, and solve them for the tan- the troposphere felt frictional drag against this rigid gential and radial wind components. Check that lid. How would the winds, weather, and climate be your results are reasonable compared with the pure different, if at all? cyclostrophic winds. How would the resulting winds affect the total circulation in a tornado? As S15. What if the Earth were a flat spinning disk in- discoverer of these winds, name them after your- stead of a spinning sphere. How would the weather self. and climate be different, if at all? S23. The time duration of many weather phenom- S16. Suppose the Earth’s rotation were twice as ena are related to their spatial scales, as shown by fast as now. How would the weather and climate eq. (10.53) and Fig. 10.24. Why do most weather change, if at all? phenomena lie near the same diagonal line on a log- log plot? Why are there not additional phenomena S17. Suppose that the axis of the Earth’s rotation that fill out the relatively empty upper and lower were along a radial line drawn from the sun (i.e., in triangles in the figure? Can the distribution of time the plane of the ecliptic), rather than being more or and space scales in Fig. 10.24 be used to some ad- less perpendicular to the ecliptic plane. How would vantage? the weather and climate be different, if at all?