10 ATMOSPHERIC FORCES & WINDS

Transcription

1 Copyright 2017 by oland Stull. Practical Meteorology: An Algebra-based Survey of Atospheric Science. v1.02b 10 ATMOSPHEIC FOCES & WINDS Contents Winds and Weather Maps Heights on Constant-Pressure Surfaces Plotting Winds Newton s 2 nd Law Lagrangian Eulerian Horizontal Forces Advection of Horizontal Moentu Horizontal Pressure-Gradient Force Centrifugal Force Coriolis Force Turbulent-Drag Force Suary of Forces Equations of Horizontal Motion Horizontal Winds Geostrophic Wind 302 INFO Approach to Geostrophy Gradient Wind 304 INFO The ossby Nuber Atospheric-Boundary-Layer Wind ABL Gradient (ABLG) Wind Cyclostrophic Wind Inertial Wind Antitriptic Wind Suary of Horizontal Winds Horizontal Motion Equations of Motion Again Scales of Horizontal Motion Vertical Forces and Motion Conservation of Air Mass Continuity Equation Incopressible Idealization Boundary-Layer Puping Kineatics Measuring Winds eview Hoework Eercises Broaden Knowledge & Coprehension Apply Evaluate & Analyze Synthesize 327 Practical Meteorology: An Algebra-based Survey of Atospheric Science by oland Stull is licensed under a Creative Coons Attribution-NonCoercial-ShareAlike 4.0 International License. View this license at This work is available at Winds power our wind turbines, push our sailboats, cool our houses, and dry our laundry. But winds can also be destructive in hurricanes, thunderstors, or ountain downslope windstors. We design our bridges and skyscrapers to withstand wind gusts. Airplane flights are planned to copensate for headwinds and crosswinds. Winds are driven by forces acting on air. But these forces can be altered by heat and oisture carried by the air, resulting in a cople interplay we call weather. Newton s laws of otion describe how forces cause winds a topic called dynaics. Many forces such as pressure-gradient, advection, and frictional drag can act in all directions. Inertia creates an apparent centrifugal force, caused when centripetal force (an ibalance of other forces) akes wind change direction. Local gravity acts ostly in the vertical. But a local horizontal coponent of gravity due to Earth s non-spherical shape, cobined with the contribution to centrifugal force due to Earth s rotation, results in a net force called Coriolis force. These different forces are present in different aounts at different places and ties, causing large variability in the winds. For eaple, Fig shows changing wind speed and direction around a lowpressure center. In this chapter we eplore forces, winds, and the dynaics that link the. north 100 south west P = 98.5 kpa Figure 10.1 Winds (arrows) around a low-pressure center (L) in the N. Heisphere. Green lines are isobars of sea-level pressure (P) east

2 290 CHAPTE 10 ATMOSPHEIC FOCES & WINDS (a) 7 z (k) (b) south north south z 5 k west west nor th west kpa (c) z = 5 k surface P=60 kpa L east L north south (d) P = 50 kpa surface west Figure 10.2 Sketch of the siilarity of (c) pressures drawn on a constant height surface to (d) heights drawn on a constant pressure surface. At any one height such as z = 5 k (shown by the thin dotted line in the vertical cross section of Fig. a), pressures at one location on the ap ight differ fro pressure at other locations. In this eaple, a pressure of 50 kpa is located idway between the east and west liits of the doain at 5 k altitude. Pressure always decreases with increasing height z, as sketched in the vertical cross section of (a). Thus, at the other locations on the cross section, the 50 kpa pressure will be found at higher altitudes, as sketched by the thick dashed line. This thick dashed line and the corresponding thin dotted straight line are copied into the 3-D view of the sae scenario is sketched in Fig. b. L indicates the cyclone center, having low pressure and low heights L east east P = 50 kpa z = 5 k z=6.0 k east WINDS AND WEATHE MAPS Heights on Constant-Pressure Surfaces Pressure-gradient force is the ost iportant force because it is the only one that can drive winds in the horizontal. Other horizontal forces can alter an eisting wind, but cannot create a wind fro cal air. All the forces, including pressure-gradient force, are eplained in the net sections. However, to understand the pressure gradient, we ust first understand pressure and its atospheric variation. We can create weather aps showing values of the pressures easured at different horizontal locations all at the sae altitude, such as at eansea-level (MSL). Such a ap is called a constantheight ap. However, one of the peculiarities of eteorology is that we can also create aps on other surfaces, such as on a surface connecting points of equal pressure. This is called an isobaric ap. Both types of aps are used etensively in eteorology, so you should learn how they are related. In Cartesian coordinates (, y, z), z is height above soe reference level, such as the ground or sea level. Soeties we use geopotential height H in place of z, giving a coordinate set of (, y, H) (see Chapter 1). Can we use pressure as an alternative vertical coordinate instead of z? The answer is yes, because pressure changes onotonically with altitude. The word onotonic eans that the value of the dependent variable changes in only one direction (never decreases, or never increases) as the value of the independent variable increases. Because P never increases with increasing z, it is indeed onotonic, allowing us to define pressure coordinates (, y, P). An isobaric surface is a conceptual curved surface that connects points of equal pressure, such as the shaded surface in Fig. 10.2b. The surface is higher above sea level in high-pressure regions, and lower in low-pressure regions. Hence the height contour lines for an isobaric surface are good surrogates for pressures lines (isobars) on a constant height ap. Contours on an isobaric ap are analogous to elevation contours on a topographic ap; naely, the ap itself is flat, but the contours indicate the height of the actual surface above sea level. High pressures on a constant height ap correspond to high heights of an isobaric ap. Siilarly, regions on a constant-height ap that have tight packing (close spacing) of isobars correspond to regions on isobaric aps that have tight packing of height contours, both of which are regions of strong pressure gradients that can drive strong winds. This one-to-one correspondence of both types of aps (Figs. 10.2c & d) akes it easier for you to use the interchangeably.

3 . STULL PACTICAL METEOOLOGY 291 Isobaric surfaces can intersect the ground, but two different isobaric surfaces can never intersect because it is ipossible to have two different pressures at the sae point. Due to the sooth onotonic decrease of pressure with height, isobaric surfaces cannot have folds or creases. We will use isobaric charts for ost of the upper-air weather aps in this book when describing upper-air features (ostly for historical reasons; see INFO bo). Fig is a saple weather ap showing height contours of the 50 kpa isobaric surface Plotting Winds Sybols on weather aps are like usical notes in a score they are a shorthand notation that concisely epresses inforation. For winds, the sybol is an arrow with feathers (or barbs and pennants). The tip of the arrow is plotted over the observation (weather-station) location, and the arrow shaft is aligned so that the arrow points toward where the wind is going. The nuber and size of the feathers indicates the wind speed (Table 10-1, copied fro Table 9-9). Fig illustrates wind barbs. Table Interpretation of wind barbs. Sybol Wind Speed Description cal two concentric circles 1-2 speed units shaft with no barbs 5 speed units a half barb (half line) 10 speed units each full barb (full line) 50 speed units each pennant (triangle) The total speed is the su of all barbs and pennants. For eaple, indicates a wind fro the west at speed 75 units. Arrow tip is at the observation location. CAUTION: Different organizations use different speed units, such as knots, s 1, iles h 1, k h 1, etc. Look for a legend to eplain the units. When in doubt, assue knots the WMO standard. For unit conversion, a good approiation is 1 s 1 2 knots. INFO Why use isobaric aps? There are five reasons for using isobaric charts. 1) During the last century, the radiosonde (a weather sensor hanging fro a free heliu balloon that rises into the upper troposphere and lower stratosphere) could not easure its geoetric height, so instead it reported teperature and huidity as a function of pressure. For this reason, upper-air charts (i.e., aps showing weather above the ground) traditionally have been drawn on isobaric aps. 2) Aircraft altieters are really pressure gauges. Aircraft assigned by air-traffic control to a specific altitude above 18,000 feet MSL will actually fly along an isobaric surface. Many weather observations and forecasts are otivated by aviation needs. 3) Air pressure is created by the weight of air olecules. Thus, every point on an isobaric ap has the sae ass of air olecules above it. 4) An advantage of using equations of otion in pressure coordinates is that you do not need to consider density, which is not routinely observed. 5) Soe nuerical weather prediction odels use reference pressures that vary hydrostatically with altitude. Ites (1) and (5) are less iportant these days, because odern radiosondes use GPS (Global Positioning Syste) to deterine their (, y, z) position. So they report all eteorological variables (including pressure) as a function of z. Also, soe of the odern weather forecast odels do not use pressure as the vertical coordinate. Perhaps future weather analyses and nuerical predictions will be shown on constant-height aps L W 5.8 H 60 N N 40 W N 60 W Saple Application Draw wind barb sybol for winds fro the: (a) northwest at 115 knots; (b) northeast at 30 knots. (a) 115 knots = 2 pennants + 1 full barb + 1 half barb. (b) 30 knots = 3 full barbs Check: Consistent with Table Eposition: Feathers (barbs & pennants) should be on the side of the shaft that would be towards low pressure if the wind were geostrophic. (a) (b) W H 5.9 Figure 10.3 Winds (1 knot 0.5 s 1 ) and heights (k) on the 50 kpa isobaric surface. The relative aia and inia are labeled as H (high heights) and L (low heights). Table 10-1 eplains winds. 40 N H z = 5.9 k 50 kpa Heights (k) & Winds (knots) 12 UTC, 20 Sep N

4 292 CHAPTE 10 ATMOSPHEIC FOCES & WINDS INFO Newton s Laws of Motion Isaac Newton s published his laws in Latin, the language of natural philosophy (science) at the tie (1687). Here is the translation fro Newton s Philosophiæ Naturalis Principia Matheatica ( Matheatical Principles of Natural Philosophy ): Law I. Every body perseveres in its state of being at rest or of oving uniforly straight forward, ecept inasuch as it is copelled by ipressed forces to change its state. Law II. Change in otion is proportional to the otive force ipressed and takes place following the straight line along which that force is ipressed. Law III. To any action, there is always a contrary, equal reaction; in other words, the actions of two bodies each upon the other are always equal and opposite in direction. Corollary 1. A body under the joint action of forces traverses the diagonal of a parallelogra in the sae tie as it describes the sides under their separate actions. Saple Application If a 1200 kg car accelerates fro 0 to 100 k h 1 in 7 s, heading north, then: (a) What is its average acceleration? (b) What vector force caused this acceleration? Given: V initial = 0, V final = 100 k h 1 = 27.8 s 1 t initial = 0, t final = 7 s. Direction is north. = 1200 kg. Find: (a) a =? s 2, (b) F =? N (a) Apply eq. (10.2): = V a t = ( s 1 ) / (7 0 s) = 3.97 s 2 to the north (b) Apply eq. (10.1): F = (1200 kg) ( 3.97 s 2 ) = 4766 N to the north where 1 N = 1 kg s 2 (see Appendi A). Check: Physics and units are reasonable. Eposition: My sall car can accelerate fro 0 to 100 k in 20 seconds, if I a lucky. Greater acceleration consues ore fuel, so to save fuel and oney, you should accelerate ore slowly when you drive NEWTON S 2 ND LAW Lagrangian For a Lagrangian fraework (where the coordinate syste follows the oving object), Newton s Second Law of Motion is F = a (10.1) where F is a force vector, is ass of the object, and a is the acceleration vector of the object. Naely, the object accelerates in the direction of the applied force. Acceleration is the velocity V change during a short tie interval t: = V a t Plugging eq. (10.2) into (10.1) gives: F = V t (10.2) (10.3a) ecall that oentu is defined as V. Thus, if the object s ass is constant, you can rewrite Newton s 2 nd Law as Lagrangian oentu budget: = ( V F ) (10.3b) t Naely, this equation allows you to forecast the rate of change of the object s oentu. If the object is a collection of air olecules oving together as an air parcel, then eq. (10.3a) allows you to forecast the oveent of the air (i.e., the wind). Often any forces act siultaneously on an air parcel, so we should rewrite eq. (10.3a) in ters of the net force: V Fnet = (10.4) t where F net is the vector su of all applied forces, as given by Newton s Corollary 1 (see the INFO bo). For situations where F net / = 0, eq. (10.4) tells us that the flow will aintain constant velocity due to inertia. Naely, V / t = 0 iplies that V = constant (not necessarily that V = 0).

5 . STULL PACTICAL METEOOLOGY 293 In Chapter 1 we defined the (U, V, W) wind coponents in the (, y, z) coordinate directions (positive toward the East, North, and up). Thus, we can split eq. (10.4) into separate scalar (i.e., non-vector) equations for each wind coponent: U t = Fnet V t = Fynet W t = Fznet (10.5a) (10.5b) (10.5c) where F net is the su of the -coponent of all the applied forces, and siilar for F y net and F z net. Fro the definition of = final initial, you can epand U/ t to be [U(t+ t) U(t)]/ t. With siilar epansions for V/ t and W/ t, eq. (10.5) becoes Fnet Ut ( + t) = U() t + t (10.6a) Fynet Vt ( + t) = V() t + t (10.6b) Fznet Wt ( + t) = W() t + t (10.6c) These are forecast equations for the wind, and are known as the equations of otion. The Nuerical Weather Prediction (NWP) chapter shows how the equations of otion are cobined with budget equations for heat, oisture, and ass to forecast the weather Eulerian While Newton s 2nd Law defines the fundaental dynaics, we cannot use it very easily because it requires a coordinate syste that oves with the air. Instead, we want to apply it to a fied location (i.e., an Eulerian fraework), such as over your house. The only change needed is to include a new ter called advection along with the other forces, when coputing the net force F net in each direction. All these forces are eplained in the net section. But knowing the forces, we need additional inforation to use eqs. (10.6) we need the initial winds [U(t), V(t), W(t)] to use for the first ters on the right side of eqs. (10.6). Hence, to ake nuerical weather forecasts, we ust first observe the current weather and create an analysis of it. This corresponds to an initial-value proble in atheatics. Average horizontal winds are often 100 ties stronger than vertical winds, ecept in thunderstors and near ountains. We will focus on horizontal forces and winds first. Saple Application Initially still air is acted on by force F y net / = s 2. Find the final wind speed after 30 inutes. Given: V(0) = 0, F y net / = s 2, t = 1800 s Find: V( t) =? s 1. Assue: U = W = 0. Apply eq. (10.6b): V(t+ t) = V(t) + t (F y net /) = 0 + (1800s) (510 4 s 2 ) = 0.9 s 1. Check: Physics and units are reasonable. Eposition: This wind toward the north (i.e., fro 180 ) is slow. But continued forcing over ore tie could ake it faster. A SCIENTIFIC PESPECTIVE Creativity As a child at Woolsthorpe, his other s far in England, Isaac Newton built clocks, sundials, and odel windills. He was an average student, but his schoolaster thought Isaac had potential, and recoended that he attend university. Isaac started Cabridge University in He was 18 years old, and needed to work at odd jobs to pay for his schooling. Just before the plague hit in 1665, he graduated with a B.A. But the plague was spreading quickly, and within 3 onths had killed 10% of London residents. So Cabridge University was closed for 18 onths, and all the students were sent hoe. While isolated at his other s far, he continued his scientific studies independently. This included uch of the foundation work on the laws of otion, including the co-invention of calculus and the eplanation of gravitational force. To test his laws of otion, he built his own telescope to study the otion of planets. But while trying to iprove his telescope, he ade significant advances in optics, and invented the reflecting telescope. He was years old. It is often the young woen and en who are ost creative in the sciences as well as the arts. Enhancing this creativity is the fact that these young people have not yet been overly swayed (perhaps isguided) in their thinking by the works of others. Thus, they are free to eperient and ake their own istakes and discoveries. You have an opportunity to be creative. Be wary of building on the works of others, because subconsciously you will be steered in their sae direction of thought. Instead, I encourage you to be brave, and eplore novel, radical ideas. This recoendation ay see paradoical. You are reading y book describing the eteorological advances of others, yet I discourage you fro reading about such advances. You ust decide on the best balance for you.

6 294 CHAPTE 10 ATMOSPHEIC FOCES & WINDS HOIZONTAL FOCES Five forces contribute to net horizontal accelerations that control horizontal winds: pressure-gradient force (PG), advection (AD), centrifugal force (CN), Coriolis force (CF), and turbulent drag (TD): (10.7a) Fnet FAD FPG FCN FCF FTD = (10.7b) Fynet FyAD FyPG FyCN FyCF FyTD = Centrifugal force is an apparent force that allows us to include inertial effects for winds that ove in a curved line. Coriolis force, eplained in detail later, includes the gravitational and copound centrifugal forces on a non-spherical Earth. In the equations above, force per unit ass has units of N kg 1. These units are equivalent to units of acceleration ( s 2, see Appendi A), which we will use here. (a) y O V 5 /s U 10 /s (b) V O 5 /s U 10 /s 5 /s O 10 /s Figure 10.4 Illustration of V advection of U wind. O is a fied weather station. Grey bo is an air ass containing a gradient of U wind. Initial state (a) and later states (b and c). (c) V U Advection of Horizontal Moentu Advection is not a true force. Yet it can cause a change of wind speed at a fied location in Eulerian coordinates, so we will treat it like a force here. The wind oving past a point can carry specific oentu (i.e., oentu per unit ass). ecall that oentu is defined as ass ties velocity, hence specific oentu equals the velocity (i.e., the wind) by definition. Thus, the wind can ove (advect) different winds to your fied location. This is illustrated in Fig. 10.4a. Consider a ass of air (grey bo) with slow U wind (5 s 1 ) in the north and faster U wind (10 s 1 ) in the south. Thus, U decreases toward the north, giving U/ y = negative. This whole air ass is advected toward the north over a fied weather station O by a south wind (V = positive). At the later tie sketched in Fig. 10.4b, a west wind of 5 s 1 is easured at O. Even later, at the tie of Fig. 10.4c, the west wind has increased to 10 s 1 at the weather station. The rate of increase of U at O is larger for faster advection (V), and when U/ y is ore negative. Thus, U/ t = V U/ y for this eaple. The advection ter on the HS causes an acceleration of U wind on the LHS, and thus acts like a force per unit ass: U/ t = F AD / = V U/ y. You ust always include advection when oentu-budget equations are written in Eulerian fraeworks. This is siilar to the advection ters in the oisture- and heat-budget Eulerian equations that were in earlier chapters.

7 . STULL PACTICAL METEOOLOGY 295 For advection, the horizontal force coponents are FAD U U V U W U = y (10.8a) z FyAD U V V V W V = y (10.8b) z ecall that a gradient is defined as change across a distance, such as V/ y. With no gradient, the wind cannot cause accelerations. Vertical advection of horizontal wind ( W U/ z in eq. 10.8a, and W V/ z in eq. 10.8b) is often very sall outside of thunderstors Horizontal Pressure-Gradient Force In regions where the pressure changes with distance (i.e., a pressure gradient), there is a force fro high to low pressure. On weather aps, this force is at right angles to the height contours or isobars, directly fro high heights or high pressures to low. Greater gradients (shown by a tighter packing of isobars; i.e., saller spacing d between isobars on weather aps) cause greater pressure-gradient force (Fig. 10.5). Pressure-gradient force is independent of wind speed, and thus can act on winds of any speed (including cal) and direction. For pressure-gradient force, the horizontal coponents are: FPG 1 P = ρ (10.9a) FyPG 1 P = ρ y (10.9b) where P is the pressure change across a distance of either or y, and ρ is the density of air. y 100 kpa 101 kpa P P+ P Figure 10.5 The dark arrow shows the direction of pressure-gradient force F PG fro high (H) to low (L) pressure. This force is perpendicular to the isobars (solid curved green lines). L y P = 102 kpa d F PG H Saple Application Minneapolis (MN, USA) is about 400 k north of Des Moines (IA, USA). In Minneapolis the wind coponents (U, V) are (6, 4) s 1, while in Des Moines they are (2, 10) s 1. What is the value of the advective force per ass? Given: (U, V) = (6, 4) s 1 in Minneapolis, (U, V) = (2, 10) s 1 in Des Moines y = 400 k, = is not relevant Find: F AD / =? s 2, F y AD / =? s 2 Use the definition of a gradient: U/ y = (6 2 s 1 )/400,000 = s 1 U/ = not relevant, U/ z = not relevant, V/ y = (4 10 s 1 )/400,000 = s 1 V/ = not relevant, V/ z = not relevant Average U = (6 + 2 s 1 )/2 = 4 s 1 Average V = ( s 1 )/2 = 7 s 1 Use eq. (10.8a): F AD / = (7 s 1 ) ( s 1 ) = s 2 Use eq. (10.8b): F y AD / = (7 s 1 ) ( s 1 ) = s 2 Check: Physics and units are reasonable. Eposition: The slower U winds fro Des Moines are being blown by positive V winds toward Minneapolis, causing the U wind speed to decrease at Minneapolis. But the V winds are increasing there because of the faster winds in Des Moines oving northward. Saple Application Minneapolis (MN, USA) is about 400 k north of Des Moines (IA, USA). In (Minneapolis, Des Moines) the pressure is (101, 100) kpa. Find the pressure-gradient force per unit ass? Let ρ = 1.1 kg 3. Given: P =101 = 400 k (north of Des Moines). P =100 = 0 k at Des Moines. ρ = 1.1 kg 3. Find: F y PG / =? s 2 Apply eq. (10.9b): FyPG = (. 11 ) (,, ) Pa 3 kg ( 400, 000 0) = s 2. Hint, fro Appendi A: 1 Pa = 1 kg 1 s 2. Check: Physics and units are reasonable. Eposition: The force is fro high pressure in the north to low pressure in the south. This direction is indicated by the negative sign of the answer; naely, the force points in the negative y direction.

8 296 CHAPTE 10 ATMOSPHEIC FOCES & WINDS Saple Application If the height of the 50 kpa pressure surface decreases by 10 northward across a distance of 500 k, what is the pressure-gradient force? Given: z = 10, y = 500 k, g = 9.8 s 2. Find: F PG / =? s 2 Use eqs. (10.11a & b): F PG / = 0 s 2, because z/ = 0. Thus, F PG / = F y PG /. FyPG z = g y = s 500, 000 F PG / = s 2 Check: Physics, units & sign are reasonable. Eposition: For our eaple here, height decreases toward the north, thus a hypothetical ball would roll downhill toward the north. A northward force is in the positive y direction, which eplains the positive sign of the answer. Table To apply centrifugal force to separate Cartesian coordinates, a (+/ ) sign factor s is required. Heisphere For winds encircling a Low Pressure Center High Pressure Center Southern 1 +1 Northern +1 1 Saple Application 500 k east of a high-pressure center is a north wind of 5 s 1. Assue N. Heisphere. What is the centrifugal force? Given: = 510 5, U = 0, V = 5 s 1 Find: F CN / =? s 2. H Apply eq. (10.13a). In Table 10-2 find s = 1. FCN 5 5 = 1 ( /s) ( /s) = s 2. F CN Check: Physics and units OK. Agrees with sketch. Eposition: To aintain a turn around the high-pressure center, other forces (the su of which is the centripetal force) are required to pull toward the center. V If pressure increases toward one direction, then the force is in the opposite direction (fro high to low P); hence, the negative sign in these ters. Pressure-gradient-force agnitude is F PG P = 1 ρ d (10.10) where d is the distance between isobars. Eqs. (10.9) can be rewritten using the hydrostatic eq. (1.25) to give the pressure gradient coponents as a function of spacing between height contours on an isobaric surface: FPG z = g (10.11a) FyPG z = g y (10.11b) for a gravitational acceleration agnitude of g = 9.8 s 2. z is the height change in the or y directions; hence, it is the slope of the isobaric surface. Etending this analogy of slope, if you conceptually place a ball on the isobaric surface, it will roll downhill (which is the pressure-gradient force direction). The agnitude of pressure-gradient force is F PG z = g d (10.12) where d is distance between height contours. The one force that akes winds blow in the horizontal is pressure-gradient force. All the other forces are a function of wind speed, hence they can only change the speed or direction of a wind that already eists. The only force that can start winds blowing fro zero (cal) is pressure-gradient force Centrifugal Force Inertia akes an air parcel try to ove in a straight line. To get its path to turn requires a force in a different direction. This force, which pulls toward the inside of the turn, is called centripetal force. Centripetal force is the result of a net ibalance of (i.e., the nonzero vector su of) other forces. For atheatical convenience, we can define an apparent force, called centrifugal force, that is opposite to centripetal force. Naely, it points outward fro the center of rotation. Centrifugal-force coponents are: FCN s VM =+ (10.13a) FyCN s UM = (10.13b)

9 . STULL PACTICAL METEOOLOGY 297 where M = ( U 2 + V 2 ) 1/2 is wind speed (always positive), is radius of curvature, and s is a sign factor fro Table 10-2 as deterined by the heisphere (North or South) and synoptic pressure center (Low or High). Centrifugal force agnitude is proportional to wind speed squared: F CN = 2 M (10.14) Coriolis Force An object such as an air parcel that oves relative to the Earth eperiences a copound centrifugal force based on the cobined tangential velocities of the Earth s surface and the object. When cobined with the non-vertical coponent of gravity, the result is called Coriolis force (see the INFO bo on the net page). This force points 90 to the right of the wind direction in the Northern Heisphere (Fig. 10.6), and 90 to the left in the S. Heisphere. The Earth rotates one full revolution (2π radians) during a sidereal day (i.e., relative to the fied stars, P sidereal is a bit less than 24 h, see Appendi B), giving an angular rotation rate of Ω=2 π / P sidereal (10.15) = radians s 1 The units for Ω are often abbreviated as s 1. Using this rotation rate, define a Coriolis paraeter as: f c = 2 Ω sin( φ ) (10.16) where ϕ is latitude, and 2 Ω = s 1. Thus, the Coriolis paraeter depends only on latitude. Its agnitude is roughly s 1 at id-latitudes. The Coriolis force in the Northern Heisphere is: FCF = fc V (10.17a) FyCF = fc U (10.17b) In the Southern Heisphere the signs on the right side of eqs. (10.17) are opposite. Coriolis force is zero under cal conditions, and thus cannot create a wind. However, it can change the direction of an eisting wind. Coriolis force cannot do work, because it acts perpendicular to the object s otion. The agnitude of Coriolis force is: or F CF / 2 Ω sin(ϕ) M (10.18a) F CF / f c M (10.18b) Northern Heisphere Southern He. F CF wind Saple Application ( ) a) Plot Coriolis paraeter vs. latitude. b) Find F CF / at Paris, given a north wind of 15 s 1. : a) Given: ϕ = N at Paris. Find f c (s 1 ) vs. ϕ( ) using eq. (10.16). For eaple: f c = ( s 1 ) sin( ) = s 1. Latitude ( ) F CF wind Low Latitudes Figure 10.6 Coriolis force (F CF ) vs. latitude, wind-speed, and heisphere wind F CF F CF wind wind F CF F CF wind Coriolis Paraeter, f c (10 4 s 1 ) b) Given: V = 15 s 1. Find: F CF /=? s 2 Assue U = 0 because no info, thus F y CF /= 0. Apply eq. (10.17a): F CF / = ( s 1 ) ( 15 s 1 ) = s 2 Eposition: This Coriolis force points to the west. INFO Coriolis Force in 3-D High Latitudes wind F CF wind Eqs. (10.17) give only the doinant coponents of Coriolis force. There are other saller-agnitude Coriolis ters (labeled sall below) that are usually neglected. The full Coriolis force in 3-diensions is: FCF = fc V 2Ω cos( φ ) W (10.17c) [sall because often W<<V] FyCF = fc U (10.17d) FzCF = 2Ω cos( φ ) U (10.17e) [sall relative to other vertical forces] F CF

10 298 CHAPTE 10 ATMOSPHEIC FOCES & WINDS INFO On Coriolis Force Gaspar Gustave Coriolis eplained a copound centrifugal force on a rotating non-spherical planet such as Earth (Anders Persson: 1998, 2006, 2014). Basics On the rotating Earth an ibalance can occur between gravitational force and centrifugal force. For an object of ass oving at tangential speed M tan along a curved path having radius of curvature, centrifugal force was shown earlier in this chapter to be F CN / = (M tan ) 2 /. In Fig 10.a the object is represented by the black dot, and the center of rotation is indicated by the X. INFO On Coriolis Force (continuation) Earth Equator ϕ ais of rotation F GH F G V F CNV F GV F CN F CNH Figure 10.c. Horizontal & vertical force coponents. H X M tan F CN / = (M tan ) 2 / Split the vectors of true gravity into local vertical F GV and horizontal F GH coponents. Do the sae for the centrifugal force (F CNV, F CNH ) of Earth s rotation (Fig. 10.c). Total centrifugal force F CN is parallel to the equator (EQ). Thus, for an object at latitude ϕ, you can use trig to show F CNH F CN sin(ϕ). Figure 10.a. Basics of centrifugal force (F CN ). The Earth was ostly olten early in its foration. Although gravity tends to ake the Earth spherical, centrifugal force associated with Earth s rotation caused the Earth to bulge slightly at the equator. Thus, Earth s shape is an ellipsoid (Fig. 10.b). The cobination of gravity F G and centrifugal force F CN causes a net force that we feel as effective gravity F EG. Objects fall in the direction of effective gravity, and it is how we define the local vertical (V) direction. Perpendicular to vertical is the local horizontal (H) direction, along the ellipsoidal surface. An object initially at rest on this surface feels no net horizontal force. [Note: Ecept at the poles and equator, F G does not point eactly to Earth s center, due to gravitational pull of the equatorial bulge.] North Pole (NP) V F CN Objects at est with respect to Earth s Surface Looking down towards the north pole (NP), the Earth turns counterclockwise with angular velocity Ω = 360 /(sidereal day) (Fig. 10.d). Over a tie interval t, the aount of rotation is Ω t. Any object (black dot) at rest on the Earth s surface oves with the Earth at tangential speed M tan = Ω (grey arrow), where = o cos(ϕ) is the distance fro the ais of rotation. o = 6371 k is average Earth radius. But because the object is at rest, its horizontal coponent of centrifugal force F CNH associated with oveent following the curved latitude (called a parallel) is the sae as that for the Earth, as plotted in Fig. 10.c above. But this horizontal force is balanced by the horizontal coponent of gravity F GH, so the object feels no net horizontal force. Equator Earth F CNH F GH X NP Ω t M tan = Ω Any Parallel (const. latitude) Earth (with eaggerated oblateness) Equator (EQ) F G F EG H Figure 10.d. Looking down on the North Pole (NP), for an object at rest on Earth s surface. (SP) South Pole Figure 10.b. Earth cross section (eaggerated). (continues in net colun) Objects Moving East or West relative to Earth Suppose an object oves with velocity M due east relative to the Earth. This velocity (thin white arrow in Fig. 10.e) is relative to Earth s velocity, giving the object a faster total velocity (grey arrow), causing greater centrifugal force and greater F CNH. But F GH is constant. (continues in net colun)

11 . STULL PACTICAL METEOOLOGY 299 INFO On Coriolis Force (continuation) INFO Coriolis Force (continuation) F CNH Earth F CF F GH M Ω t X NP Figure 10.e. Eastward oving object. F M CNH F CF X F GH NP Earth Figure 10.f. Westward oving object. Objects oving south have a Coriolis force to the right due to the larger radius of curvature. egardless of the direction of otion in the Northern Heisphere, Coriolis force acts 90 to the right of the object s otion relative to the Earth. When viewing the Southern Heisphere fro below the south pole, the Earth rotates clockwise, causing a Coriolis force that is 90 to the left of the relative otion vector. Coriolis-force Magnitude Derivation Fro Figs. 10.c & d, see that an object at rest (subscript ) has Horizontal force F CNH does NOT balance F GH. The thick green arrow (Fig. 10.e) shows that the force difference F CF is to the right relative to the object s otion M. F CF is called Coriolis force. The opposite ibalance of F CNH and F GH occurs for a westward-oving object (thin white arrow), because the object has slower net tangential velocity (grey arrow in Fig. 10.f). This ibalance, Coriolis force F CF (green arrow), is also to the right of the relative otion M. Northward-oving Objects When an object oves northward at relative speed M (thin white arrow in Fig. 10.g) while the Earth is rotating, the path traveled by the object (thick grey line) has a sall radius of curvature about point X that is displaced fro the North Pole. The saller radius causes larger centrifugal force F CNH pointing outward fro X. Coponent F CNH-ns of centrifugal force balances the unchanged horizontal gravitational force F GH. But there reains an unbalanced east-west coponent of centrifugal force F CNH-ew which is defined as Coriolis force F CF (green arrow). Again, it is to the right of the relative otion vector M of the object. F CNH F CNH-ns F GH = F CNH F CNH (C1) and M tan rest = Ω (C2) Fro Fig. 10.e, Coriolis force for an eastward-oving object is defined as F CF F CNH F GH Apply eq. (C1) to get or F CF = F CNH F CNH F CF = sin(ϕ) [F CN F CN ] (fro Fig. 10.c) Divide by ass, and plug in the definition for centrifugal force as velocity squared divided by radius: F CF / = sin(ϕ) [ (M tan ) 2 / (M tan rest ) 2 / ] Use M tan = M tan rest + M, along with eq. (C2): F CF / = sin(ϕ) [ (Ω +M) 2 / (Ω ) 2 / ] F CF / = sin(ϕ) [(2 Ω M) + (M 2 /)] The first ter is usually uch larger than the last, allowing the following approiation for Coriolis force per ass: F CF / 2 Ω sin(ϕ) M (10.18) F CNH-ew = F CF M Earth Define a Coriolis paraeter as f c 2 Ω sin(ϕ). Thus, F CF / f c M latitude line (paraallel) NP F GH X Center of otation Figure 10.g. Northward oving object. (continues in net colun) HIGHE MATH Apparent Forces In vector for, centrifugal force/ass for an object at rest on Earth is Ω (Ω r), and Coriolis force/ass is 2Ω V, where vector Ω points along the Earth s ais toward the north pole, r points fro the Earth s center to the object, V is the object s velocity relative to Earth, and is the vector cross product.

12 300 CHAPTE 10 ATMOSPHEIC FOCES & WINDS z z i 0 0 wind M F TD G free atosphere atospheric boundary layer (ABL) theral turbulence eddies caused by shear turbulence Figure 10.7 Wind speed M (curved black line with white highlights) is slower than geostrophic G (vertical dashed line) because of turbulent drag force F TD in the atospheric boundary layer Turbulent-Drag Force Surface eleents such as pebbles, blades of grass, crops, trees, and buildings partially block the wind, and disturb the air that flows around the. The cobined effect of these eleents over an area of ground is to cause resistance to air flow, thereby slowing the wind. This resistance is called drag. At the botto of the troposphere is a layer of air roughly 0.3 to 3 k thick called the atospheric boundary layer (ABL). The ABL is naed because it is at the botto boundary of the atosphere. Turbulence in the ABL ies the very-slow near-surface air with the faster air in the ABL, reducing the wind speed M throughout the entire ABL (Fig. 10.7). The net result is a drag force that is norally only felt by air in the ABL. For ABL depth z i the drag is: FTD = w U T zi FyTD = w V T zi (10.19a) (10.19b) where w T is called a turbulent transport velocity. The total agnitude of turbulent drag force is Saple Application What is the drag force per unit ass opposing a U = 15 s 1 wind (with V = 0) for a: (a) statically neutral ABL over a rough forest; & (b) statically unstable ABL having convection with w B = 50 s 1, given z i = 1.5 k. Given: U = M = 15 s 1, z i = 1500, C D = 210 2, w B = 50 s 1. Find: F TD / =? s 2. (a) Plugging eq. (10.21) into eq. (10.19a) gives: FTD = C M U D = (. ) ( ) 2 15/s 002 zi 1500 = s 2. (b) Plugging eq. (10.22) into eq. (10.19a) gives: FTD = b w U D B zi = (. ) ( ) ( 15/s /s ) 1500 = s 2. Check: Physics and units are reasonable. Eposition: Because the wind is positive (blowing toward the east) it requires that the drag be negative (pushing toward the west). Shear (echanical) turbulence and convective (theral/buoyant) turbulence can both cause drag by diluting the faster winds higher in the ABL with slower near-surface winds. FTD = w M T (10.20) zi and is always opposite to the wind direction. For statically unstable ABLs with light winds, where a war underlying surface causes therals of war buoyant air to rise (Fig. 10.7), this convective turbulence transports drag inforation upward at rate: w T = b D w B (10.22) where diensionless factor b D = The buoyancy velocity scale, w B, is of order 10 to 50 s 1, as is eplained in the Heat Budget chapter. For statically neutral conditions where strong winds M and wind shears (changes of wind direction and/or speed with height) create eddies and echanical turbulence near the ground (Fig. 10.7), the transport velocity is w T = C D M (10.21) where the drag coefficient C D is sall (210 3 diensionless) over sooth surfaces and is larger (210 2 ) over rougher surfaces such as forests. In fair weather, turbulent-drag force is felt only in the ABL. However, thunderstor turbulence can i slow near-surface air throughout the troposphere. Fast winds over ountains can create ountain-wave drag felt in the whole atosphere (see the egional Winds chapter).

13 . STULL PACTICAL METEOOLOGY Suary of Forces Table Suary of forces. Ite Nae of Force Direction 1 gravity down Magnitude (N kg 1 ) Horiz. (H) or Vert. (V) F G = g = 9.8 s 2 V earks ( ite is in colun 1; H & V in col. 5) hydrostatic equilibriu when ites 1 & 2V balance 2 pressure gradient fro high to low pressure F PG = z g d V & H the only force that can drive horizontal winds 3 4 Coriolis (copound) turbulent drag 90 to right (left) of wind in Northern (Southern) Heisphere opposite to wind F CF = 2 Ω sin( φ) M H* FTD = w M T H* zi geostrophic wind when 2H and 3 balance (eplained later in horiz. wind section) at. boundary-layer wind when 2H, 3 and 4 balance (eplained in horiz. wind section) 5 centrifugal (apparent) away fro center of curvature F CN = 2 M H* centripetal = opposite of centrifugal. Gradient wind when 2H, 3 and 5 balance 6 advection (apparent) (any) F AD = M U d V & H neither creates nor destroys oentu; just oves it *Horizontal is the direction we will focus on. However, Coriolis force has a sall vertical coponent for zonal winds. Turbulent drag can eist in the vertical for rising or sinking air, but has copletely different for than the boundary-layer drag given above. Centrifugal force can eist in the vertical for vortices with horizontal aes. Note: units N kg 1 = s EQUATIONS OF HOIZONTAL MOTION Cobining the forces fro eqs. (10.7, 10.8, 10.9, 10.17, and 10.19) into Newton s Second Law of Motion (eq. 10.5) gives siplified equations of horizontal otion: (10.23a) U = U U V U W U 1 P t y z + f V w U c T ρ z i (10.23b) V = U V V V W V 1 P t y z y f U w V c T ρ zi tendency advection pressure gradient Coriolis turbulent drag These are the forecast equations for wind. For special conditions where steady winds around a circle are anticipated, centrifugal force can be included. The ters on the right side of eqs. (10.23) can all be of order to s 2 (which is equivalent to units of N kg 1, see Appendi A for review). However, soe of the ters can be neglected under special conditions where the flow is less coplicated. For eaple, near-zero Coriolis force occurs near the equator. Near-zero turbulent drag eists above the ABL. Near-zero pressure gradient is at low- and high-pressure centers. Other situations are ore coplicated, for which additional ters should be added to the equations of horizontal otion. Within a few of the ground, olecular friction is large. Above ountains during windy conditions, ountain-wave drag is large. Above the ABL, cuulus clouds and thunderstors can create strong convective iing. For a few idealized situations where any ters in the equations of otion are sall, it is possible to solve those equations for the horizontal wind speeds. These theoretical winds are presented in the net section. Later in this chapter, equations to forecast vertical otion (W) will be presented.

14 302 CHAPTE 10 ATMOSPHEIC FOCES & WINDS Table Naes of idealized steady-state horizontal winds, and the forces that govern the. 0 = U U Forces: 1 P + + f V w U s VM c T ρ zi pressure gradient Coriolis turbulent drag centrifugal Wind Nae Geostrophic Gradient At.Bound. Layer ABL Gradient Cyclostrophic Inertial Antitriptic P = 99.2 kpa 99.6 kpa kpa kpa L H F PG F CF Figure 10.8 Idealized weather ap for the Northern Heisphere, showing geostrophic wind (G, grey arrow) caused by a balance between two forces (brown arrows): pressure-gradient force (F PG ) and Coriolis force (F CF ). P is pressure, with isobars plotted as thin green lines. L and H are low and high-pressure regions. The sall sphere represents an air parcel. Saple Application Find geostrophic wind coponents at a location where ρ = 1.2 kg 3 and f c = s 1. Pressure decreases by 2 kpa for each 800 k of distance north. Given: f c = s 1, P= 2 kpa, ρ=1.2 kg 3, y=800 k. Find: (U g, V g ) =? s 1 But P/ = 0 iplies V g = 0. For U g, use eq. (10.26a): Ug = (. 12 ) ( s ) ( kpa ) 3 =18.9 s 1 kg/ ( 800k) Check: Physics & units OK. Agrees with Fig Eposition: As the pressure gradient accelerates air northward, Coriolis force turns it toward the east. y G HOIZONTAL WINDS When air accelerates to create wind, forces that are a function of wind speed also change. As the winds continue to accelerate under the cobined action of all the changing forces, feedbacks often occur to eventually reach a final wind where the forces balance. With a zero net force, there is zero acceleration. Such a final, equilibriu, state is called steady state: U = V 0, t t = 0 (10.24) Caution: Steady state eans no further change to the non-zero winds. Do not assue the winds are zero. Under certain idealized conditions, soe of the forces in the equations of otion are sall enough to be neglected. For these situations, theoretical steady-state winds can be found based on only the reaining larger-agnitude forces. These theoretical winds are given special naes, as listed in Table These winds are eained net in ore detail. As we discuss each theoretical wind, we will learn where we can epect these in the real atosphere Geostrophic Wind For special conditions where the only forces are Coriolis and pressure-gradient (Fig. 10.8), the resulting steady-state wind is called the geostrophic wind, with coponents (U g, V g ). For this special case, the only ters reaining in, eqs. (10.23) are: 1 P 0 = + fc V (10.25a) ρ 1 P 0 = fc U ρ y (10.25b) Define U U g and V V g in the equations above, and then solve for these wind coponents: 1 P Ug = ρ fc y (10.26a) 1 P Vg =+ (10.26b) ρ fc where f c = ( s 1 ) sin(latitude) is the Coriolis paraeter, ρ is air density, and P/ and P/ y are the horizontal pressure gradients.

15 . STULL PACTICAL METEOOLOGY 303 y d d z/ d ( / 100 k) wind H wind wind wind wind wind P = 26.0 kpa Figure 10.9 Isobars (green lines) that are ore closely spaced (i.e., tightly packed) cause stronger geostrophic winds (arrows), for N. Heisphere. 100 G (/s) latitude eal winds are nearly geostrophic at locations where isobars or height contours are relatively straight, for altitudes above the atospheric boundary layer. Geostrophic winds are fast where isobars are packed closer together. The geostrophic wind direction is parallel to the height contours or isobars. In the N. (S.) heisphere the wind direction is such that low pressure is to the wind s left (right), see Fig The agnitude G of the geostrophic wind is: 2 2 G= Ug + Vg (10.27) If d is the distance between two isobars (in the direction of greatest pressure change; naely, perpendicular to the isobars), then the agnitude (Fig ) of the geostrophic wind is: P G = 1 ρ fc d (10.28) Above sea level, weather aps are often on isobaric surfaces (constant pressure charts), fro which the geostrophic wind (Fig ) can be found fro the height gradient (change of height of the isobaric surface with horizontal distance): g z Ug = (10.29a) fc y g z Vg =+ fc (10.29b) where the Coriolis paraeter is f c, and gravitational acceleration is g = 9.8 s 2. The corresponding agnitude of geostrophic wind on an isobaric chart is: g z G = fc d (10.29c) P/ d at sea level (kpa / 100 k) Figure Variation of geostrophic wind speed (G) with horizontal pressure gradient ( P/ d) at sea level. Top scale is height gradient of any isobaric surface. Saple Application Find the geostrophic wind for a height increase of 50 per 200 k of distance toward the east. Assue, f c = s 1. Given: = 200 k, z = 50, f c = s 1. Find: G =? s 1 No north-south height gradient, thus U g = 0. Apply eq. (10.29b) and set G = V g : g z Vg =+ fc = s s 200, 000 =27.2 s 1 Check: Physics & units OK. Agrees with Fig Eposition: If height increases towards the east, then you can iagine that a ball placed on such a surface would roll downhill toward the west, but would turn to its right (toward the north) due to Coriolis force. INFO Approach to Geostrophy How does an air parcel, starting fro rest, approach the final steady-state geostrophic wind speed G sketched in Fig. 10.8? Start with the equations of horizontal otion (10.23), and ignore all ters ecept the tendency, pressuregradient force, and Coriolis force. Use the definition of continues on net page

16 304 CHAPTE 10 ATMOSPHEIC FOCES & WINDS INFO Appr. to Geostrophy (continuation) geostrophic wind (eqs ) to write the resulting siplified equations as: U/ t = fc ( Vg V) V/ t = fc ( Ug U) Net, rewrite these as forecast equations: Unew = Uold tfc ( Vg Vold ) Vnew = Vold + tfc ( Ug Unew ) Start with initial conditions (U old, V old ) = (0, 0), and then iteratively solve the equations on a spreadsheet to forecast the wind. For eaple, suppose P = 1 kpa, f c = 10 4 s 1, = 500 k, ρ = 1 kg 3, where we would anticipate the wind should approach (U g, V g ) = (0, 20) s 1. The actual evolution of winds (U, V) and air parcel position (X, Y) are shown in Figs. below. V (/s) h (U g, V g )= (0, 20 /s) 9 h 12 h 3 h t = 15 h 0 0 h U (/s) Surprisingly, the winds never reach geostrophic equilibriu, but instead rotate around the geostrophic wind. This is called an inertial oscillation, with period of 2 π/f c. For our case, the period is h. Twice this period is called a pendulu day. The net result in the figure below is that the wind indeed oves at the geostrophic 2500 speed of 20 s 1 to the north ( 1250 k in h h), but along the way it staggers west and east with an h additional ageostrophic 24 h (non-geostrophic) part Inertial oscillations are 21 h soeties observed at night Y (k) 15 h in the atospheric boundary 12 h 1000 layer, but rarely higher in the atosphere. Why not? (1) 9 h = t The ageostrophic coponent 500 of wind (wind fro the East 6 h in this eaple) oves air ass, and changes the pressure gradient. (2) Friction 3 h 0 0 h daps the oscillation toward a steady wind. X (k) h G = 20 /s If the geopotential Φ = g z is substituted in eqs. (10.29), the resulting geostrophic winds are: Ug = 1 Φ fc y Vg = 1 Φ fc (10.30a) (10.30b) Gradient Wind If there is no turbulent drag, then winds tend to blow parallel to isobar lines or height-contour lines even if those lines are curved. However, if the lines curve around a low-pressure center (in either heisphere), then the wind speeds are subgeostrophic (i.e., slower than the theoretical geostrophic wind speed). For lines curving around high-pressure centers, wind speeds are supergeostrophic (faster than theoretical geostrophic winds). These theoretical winds following curved isobars or height contours are known as gradient winds. Gradient winds differ fro geostrophic winds because Coriolis force F CF and pressure-gradient force F PG do not balance, resulting in a non-zero net force F net. This net force is called centripetal force, and is what causes the wind to continually change direction as it goes around a circle (Figs & 10.12). By describing this change in direction as causing an apparent force (centrifugal), we can find the equations that define a steady-state gradient wind: 1 P 0 = ρ + f V + s V M c (10.31a) 1 P 0 = f U s U M c (10.31b) ρ y pressure gradient Coriolis centrifugal Because the gradient wind is for flow around a circle, we can frae the governing equations in radial coordinates, such as for flow around a low: 2 1 ρ P tan tan = f M + M c (10.32) where is radial distance fro the center of the circle, f c is the Coriolis paraeter, ρ is air density, P/ is the radial pressure gradient, and M tan is the agnitude of the tangential velocity; naely, the gradient wind.