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78

vertical distance from the tephigram and also by using the theory of Box 1.5. Verify the value of θe at the surface using eq. 1.96b. Estimate the values of CIN and CAPE. 1.17 A second mode of convection: sea breeze, inertial oscillation and geostrophic balance. We have defined “convection” as being the result of (conditional) hydrostatic instability, which suggests that systematic vertical motion of air is not possible when the atmosphere is hydrostatically stable, i.e. when potential temperature increases with height. However, eq. 1.9 demonstrates that an air parcel can accelerate vertically, even when it does not possess buoyancy. This is why the climate-dynamicist, Victor Starr (page 26-27 of Starr, 1968) 35, distinguished two modes of convection. The first mode of convection, often called Rayleigh-Benard convection, is buoyancy driven, which means that it occurs when the driving force is along the vertical, as defined by the direction of gravity. The second mode of convection is brought about by horizontal contrasts in diabatic heating. In the second type of convection there need not necessarily be any vertical or hydrostatic instability due to potentially colder air being found above potentially warmer air. Here the potentially colder air tends to underrun the potentially warmer air and spread from the colder source to the warmer source. The potentially warmer air is displaced upwards, while maintaining a stable vertical stratification. An example of the second type of “convection” is the sea breeze circulation. On summer days with clear (cloudless) and calm weather, the temperature of the air in the boundaty layer (the lowest kilometre) over land increases more rapidly than the temperature of the air over sea. This is because Solar radiation heats a deeper layer of water than of soil, and also because a much larger portion of the Solar radiative energy is “used” to evaporate water over the oceans than over land. Therefore, assuming hydrostatic conditions, the vertical pressure gradients are larger over sea than over land. If we assume that the pressure at sea level (at the surface) is initially the same over land and over sea, we see that a horizontal pressure gradient perpendicular to the coast arises at upper levels, with higher pressure over land. This pressure gradient will drive air from land to sea (middle diagram in figure 1.30). The associated mass divergence (convergence) over land (over sea) will lead to a sea-level pressure decrease over land (increase over sea), which will subsequently initiate the sea breeze at the earth’s surface and so a closed sea-breeze circulation.

FIGURE 1.30 Three vertical cross-sections of the lowest 2 or 3 km of the atmosphere, schematically illustratingdifferentstagesofthedevelopmentofthepressuregradientsandsea-breezecirculation.Theleft-handdiagramshowstheinitialconditions,themiddlediagramshowstheinitiationofthecirculationandtheleft-handdiagramshowsthefinalsituationwithfully-developedseabreezecirculation(source:Tijm,A.B.C.,andA.J.vanDelden,1999:Theroleofsoundwavesinsea-breezeinitiation.Q.J.R.Meteorol.Soc.125,197-218). 35V.P.Starr,1968:PhysicsofNegativeViscosityPhenomena.MacGraw-Hill,256pp.

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FIGURE1.31.The“classical”analysisbyvanBemmelen.Isolinesofthespeed(inm/s)ofthenorth-southcomponentofthewind velocity above Batavia (nowDjakarta, Java) are obtained from hourlymean values evaluated from pibalobservationsfromMaytoNovember1909to1915.Windfromthesea(seabreeze)ishatched.Thereisapronouncedreturn currentwith amaximum shortly before sunset and in height of 2 km (VanBemmelen,W., 1922: Land-undseebriseinBatavia.Beitr.Phys.frei.Atmos.,10,169-177). Figure 1.31 shows observations of this circulation, over Djakarta, at the north coast of the island of Java, which lies very close to the equator. Farther away from the equator, the effect of the Coriolis force will become evident after a few hours, inducing a clockwise turning of the wind (in the northern hemisphere). This interesting process can be studied with a very simple model and illustrated with observations of the wind at a coastal station in The Netherlands. Let us adopt an idealised model configuration in which the coast lies at x=0, i.e. parallel to the y-axis. Land lies at x>0 and sea at x<0. The sea-level pressure gradient, perpendicular to the coast, on a summer day is approximated (“parametrised”) by

€

∂p∂x

= Acos Ωt +ϕ( ) . (1.103)

Here A is an amplitude, which might lie in the order of 0.1-1 hPa per 100 km (10-3 Pa m-1) (see figure 1.32), Ω is Earth’s angular velocity (=π/12 hr-1=7.2792×10-5 s-1) and

€

ϕisa phase angle. If we assume that the minimum pressure gradient occurs at t=12 hrs (midday), then

€

ϕ=0. The horizontal wind at the coast is governed by eq. 7 and 8 of box 1.2, which are repeated here:

€

dudt

= −1ρ∂p∂x

+uv tanφ

a+ fv , (1.104a)

€

dvdt

= −1ρ∂p∂y

−u2 tanφ

a− fu , (1.104b)

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80

FIGURE 1.32: Surface pressure change (tenths of hPa) over Europe between 1000 and 1300 UTC in June (fromSimpson,J.E.,1994:SeaBreezeandlocalwind.CambridgeUniversityPress). where the Coriolis parameter (also defined in eq. 9, Box 1.2) is

€

f ≡ 2Ωsinφ . For the moment, we neglect frictional forces. Let us assume that the pressure-gradient, parallel to the coast, ∂p/∂y=0. For simplicity, we neglect the “tan φ“ –terms in eqs. 1.104a,b. We apply these equations to a point exactly at the coast. Substituting eq. 1.103 into eqs. 1.104a and expanding the total derivative (eq. 1.5), eqs. 1.104a,b become

€

∂u∂t

+ u ∂u∂x

= fv − Aρcos Ωt +ϕ( ) , (1.105a)

€

∂v∂t

+ u ∂v∂x

= − fu , (1.105b)

Here we have, in fact, neglected all derivatives with respect to y and assumed that w=0 at the coast. The non-linear advection terms pose a serious problem to the general solution of these equations, but at the coast (x=0) it is reasonable to assume that

€

∂u∂x

= 0 and ∂v∂x

= 0 ,

so that eqs. 105a,b become linear:

€

∂u∂t

= fv − Aρcos Ωt +ϕ( ) , (1.106a)

€

∂v∂t

= − fu , (1.106b)

This set of equations can be simplified easily to

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81

€

∂2u∂t2

+ f 2u =AΩρsin Ωt +ϕ( ) , (1.107)

which has the following solution:

€

u = C1 sin ft( ) +C2 cos ft( ) +AΩ

ρ f 2 −Ω2( )sin Ωt +ϕ( ) . (1.108a)

The constants C1 and C2 are determined by the initial conditions (see below). The solution of eq. 1.107 consists of oscillations with two different frequencies, one, which is associated with the thermal forcing, and the other, which is associated with the natural (“eigen”-) response of the physical system. The “natural” frequency is isolated by putting A=0. We thus find that the Coriolis parameter, f, represents the natural frequency of the system. The associated oscillation is called an inertial oscillation. The Coriolis parameter is the inertial frequency. During one inertial oscillation the wind vector turns clockwise (if f>0), completing a full circle (a so-called inertial circle) in one inertial period, 2π/f. The solution (eq. 1.108a) for u is easily supplemented with the solution for v. Assuming that u=v=0 at t=0, and assuming ϕ=0, yields C2=0. Substituting v=0 at t=0 in eq. 1.106a and using 1.108a yields

€

v = C1cos ft( ) +fA

ρ f 2 −Ω2( )cos Ωt( ) (1.108b)

with

€

C1 = −Af

ρ f 2 −Ω2( )

Figure 1.33 shows the solution, in terms of wind direction, for a latitude of 52.5°N. Wind direction is defined as the direction from which the wind is blowing. A direction equal to 0° (or 360°) corresponds to a wind blowing from the north, parallel to the coast, and 270° corresponds to a wind blowing from sea to land, perpendicular to the coast (the “sea breeze”). The wind direction changes at a rate, which varies slightly with time. Between t=12 hours and t=30 hours the wind vector turns clockwise through a full circle, i.e. in 18 hours. The following two circles are completed in 12.5 hours and in 16 hours, respectively. The average of 18, 12.5 and 16 hours is 15.5 hours, which is slightly larger than the inertial period, 2π/f, at 52.5°N, which is 15.1 hours. Figure 1.34 shows hourly observations of the wind direction at a coastal station in The Netherlands, (Ijmuiden, Figure 1.35), during two days in early May 1976. The latitude of IJmuiden is 52.47°N. The wind at Ijmuiden is measured at a height of 10 m at the end of a pier, at a distance of about 1 km from the coast. The surface temperature of the North Sea in early May typically is about 10°C. During this period in 1976 summer came very early, with temperatures tipping 30°C at distances of about 50 km from the coast (Figure 1.36). At IJmuiden the wind blows from the sea during the day and from the land during the night (Figure 1.34), but hardly never really perpendicular to the coast (either 300° or 120°). In accordance with the analytical solution (figure 1.33), the wind vector changes direction at an approximate constant rate. After t=41 hours the wind vector turns clockwise through a full circle within 20 hours. The turning of the wind, therefore, appears somewhat slower than according to the analytical solution. In reality, the forcing, i.e. the time-varying pressure gradient, which has a period 24 hours, determines the wind

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direction more strongly than is predicted by the analytical theory. By introducing the effect friction, we may be able to reconcile these differences (Problem 1.15).

FIGURE1.33.Winddirection,accordingtotheanalyticalsolution(eq.1.108a,b),forlatitude=52.5°N,A=0.001Pam-1,ϕ=0andρ=1.16 kg m-3,assumingthatu=v=0att=0.Awinddirectionequalto270°impliesaseabreezeperpendiculartothecoast.Notethatthewinddirectiondoesnotfollowthesamedailycycleeachday!

FIGURE1.34.Hourlytemperature(blacksquares)andwinddirection(redcircles)atapointonthecoastoftheTheNetherlands (IJmuiden, see figure1.35)onMay7and8,1976(hour=20corresponds to20UTConMay6). Winddirectionisdefinedinthetext.Thehorizontaldashedlinesindicatethedirectionsperpendiculartothecoast,whilethehorizontalsolidlinesindicatethedirectionsparalleltothecoast.Thewindhasalandwardcomponentbetween15UTCand20UTCon7Mayandbetween8UTCand20UTCon8May.

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FIGURE 1.35. Map of Holland (on the right) and the North Sea (on the left), showing the positions of measuringstations.Awinddirectionof360°(or0°)correspondstowindblowingfromthenorth(indicatedbytheletter“N”).

FIGURE1.36.Hourlytemperature(blacksquares)andwinddirection(redcircles)atDeBilt(figure1.34)onMay7and8,1976.Horizontaldashedlineindicatesthedirectionperpendiculartothecoast;Horizontalsolidlineindicatesthedirectionparalleltothecoast.Thenortherlydirectionisindicatedinfigure1.35.Awinddirectionof270°impliesthatthewindisblowingeastward.ThecoastlinearoundIJmuidenisrotatedclockwisebyabout30°. At a latitude of 30° one inertial circle of the wind vector is completed in 24 hours. At this latitude we, therefore, expect a resonance between the thermal forcing, due to the land-sea temperature contrast, and the natural response of the atmosphere (the inertial oscillation). Indeed, according to eq. 1.108, u beomes very large when (

€

f 2 −Ω2) approaches zero. This is a manifestation of resonance. This occurs when sin φ approaches ±1//2, i.e near a latitude of ±30°.

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84

In reality resonance does not seem to occur at these latitudes. One reason for this is that cooling at night is in general not sufficient to reverse the sea-level pressure gradient perpendicular to the coast. Instead, the pressure gradient perpendicular to the coast tends to attain a component of constant sign, which we will denote here by B, such that eq. 1.103 becomes

€

∂p∂x

= Acos Ωt +ϕ( ) + B . (1.109)

This is observed, for example, over the Iberian Peninsula (figure 1.32) and also over Turkey, where “thermal lows” are observed nearly permanently over land in summer. In order to incorporate this effect we use eq. 1.109 instead of eq. 1.103 and rewrite eqs. 1.106a,b as

€

∂u∂t

= fv − Aρcos Ωt +ϕ( ) − B

ρ , (1.110a)

€

∂v∂t

= − fu , (1.110b)

If B is constant and A=0, this set of equations has the steady state solution:

€

u = 0; v =Bρf

≡ vg , (1.111)

This steady state represents the state of geostrophic balance, where

€

vg is the geostrophic wind component parallel to the coast. We can now rewrite eq. 1.110a as

€

∂u∂t

= f v − vg( ) − Aρ cos Ωt +ϕ( ) , (1.112a)

The full wind is equivalent to the sum of the geostrophic wind and so-called ageostrophic wind. The ageostrophic wind is identified with the sea breeze. The sea breeze, therefore, has a component parallel to the coast, which is equal to

€

va ≡ v − vg( ).The wind component perpendicular to the coast may also be composed of a geotrophic component,

€

ug , and an ageostrophic component,

€

ua (the sea breeze). During fair weather situations in The Netherlands a large-scale high-pressure system (an

anticyclone) is typically observed to the north, centred over the North Sea and Scandinavia. Therefore, the pressure gradient has a component parallel to the coast. The large-scale geostrophic wind, associated with this pressure gradient, is from the east (from land). In order to incorporate this effect into our the theory, eq. 1.110b should be written as

€

∂v∂t

= − f u − ug( ) . (1.112b)

The geostrophic wind component perpendicular to the coast is

€

ug = −1ρf

∂p∂y. (1.113)

These definitions are used in the following problems, which represent an attempt to apply the simplified theory of this section to the situation at the west coast of The Netherlands on May 6 to May 8, 1976.

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PROBLEM 1.13. Analysis of the sea-level pressure and wind from observations Observations are performed at randomly spaced points (figure 1.35), while theoretical analysis and/or numerical models usually require data on a regular grid of points. This requires interpolation of the observations. The most simple interpolation method is called “piecewise linear interpolation”. Suppose that we want to interpolate observations to a point with horizontal coordinates, (x0, y0), e.g. IJmuiden (figure 1.35). The horizontal coordinates of three measuring points in the vicinity of this point are given by (xi, yi), with i=1, 2, 3. The value of a variable (for instance the sea level pressure) at one of these points can be approximated as follows.

€

pi = p0 + (xi − x0)∂p∂x

+ (yi − y0)∂p∂y

, (1.114)

The desired value of the pressure at (x0, y0) is p0. Applying eq. 1.115 to the three measuring points yields three equations with three unknowns, p0, ∂p/∂x and ∂p/∂y, which can be written concisely as

€

p1p2p3

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

=

1 x1 − x0( ) y1 − y0( )1 x2 − x0( ) y2 − y0( )1 x3 − x0( ) y3 − y0( )

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

p0∂p /∂x∂p /∂y

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ .

This system of equations can be written shortly as

€

p1p2p3

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

= Mp0

∂p /∂x∂p /∂y

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ which implies that

€

p0∂p /∂x∂p /∂y

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

= M−1p1p2p3

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

,

To find, p0, ∂p/∂x and ∂p/∂y w need to invert the matrix,

€

M =

1 x1 − x0( ) y1 − y0( )1 x2 − x0( ) y2 − y0( )1 x3 − x0( ) y3 − y0( )

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ .

In this problem the point i=0, with coordinates, (x0, y0), corresponds to IJmuiden. Observations from three measuring sites are used to estimate the sea level pressure gradient. These measuring sites and their locations are listed in table 1.3. Because we need to estimate only the pressure gradient, IJmuiden is both the reference point and one of the measuring sites. Station Latitude [°] Longitude [°] IJmuiden 52.47 4.57 Schiphol 52.32 4.78 Valkenburg 52.18 4.42 Table1.3.Coordinatesofmeasuringstations(seefigure1.34).IJmuiden,ValkenburgandSchipholformatriangle Compute the sea level pressure-gradient at IJmuiden for 7 and 8 May 1976 from the measurements at IJmuiden, Schiphol and Valkenburg. Note that, since the locations of stations are given in degrees latitude and longitude in table 1.3, it might be quicker to work in these units. The measurements can be downloaded from http://www.knmi.nl/nederland-nu/klimatologie/uurgegevens. The distance from IJmuiden to Valkenburg is only about 0.3 degree in latitude. You may therefore assume that the latitude-longitude grid is Cartesian, where 1 degree in longitude corresponds to 67737 m and 1 degree in latitude corresponds to 111195 m. The coast at IJmuiden is rotated clockwise by 30° with respect to the south to north direction. Decompose the sea level pressure gradient perpendicular to the coast at IJmuiden into (1) a constant component, (2) a component associated with a linear trend and (3) a sinusoidal time-varying component, as is done in eq. 1.109 (see Box 1.6). Is there a residual fourth component? The first

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and second components are associated with both the slow change of the location and intensity of the large-scale pressure distribution, which might partially be thermally induced. The third component is associated with the diurnal thermal forcing. Estimate the value of A from the data. PROBLEM 1.14. A linear numerical model of the sea breeze at the coast Construct a numerical model of the sea breeze at the coast based on eqs. 1.112a,b. Approximate the time derivative with a finite difference scheme (e.g. the semi-implicit Euler method; problem 1.6). Check the accuracy of your numerical scheme by comparing the numerical solution with the analytical solution (eq.1.108) for idealised initial conditions (i.e. u=v=0 at t=0 and no geostrophic wind). PROBLEM 1.15. Simulation of the wind at IJmuiden and verification with observations Simulate the wind at IJmuiden on 7 and 8 May 1976 with the model, which was constructed in problem 1.14. Estimate the values of A and ϕ from the analysis (problem 1.13) of the observed average daily cycle of the pressure gradient at IJmuiden. Compare the model outcome with the measurements at IJmuiden (figure 1.34). The simulation can be improved by assuming a realistic geostrophic wind as well as by adding a term, representing friction, to the model equations. Eqs. 1.112a,b become

€

∂u∂t

= f v − vg( ) − Aρ cos Ωt +ϕ( ) − λu , (1.115a)

€

∂v∂t

= − f u − ug( ) − λv , (1.115b)

where λ is a constant coefficient, called the “Rayleigh damping constant”. Implement a realistic geostrophic wind (estimated from the analysis of the pressure gradient in problem 1.13) and realistic initial conditions. Construct a Taylor diagram (Box 1.7) showing the performance of the model. Investigate the model solution as a function of the value of λ. The parameter, λ, can be used to “tune” the model, to fit the observations. Indicate the performance of the model for different values of λ in the Taylor diagram. Box 1.6 Spectral or Fourier analysis and the mesoscale range of scales A series of N consecutive measurements of a variable,

€

Un (n=0 to n=N-1), regardless of how complex in form (except that it must be continuous or without breaks, and there can only be one value of U for each value of t), can be represented as the real part of the sum of a series of complex exponential functions, or sine/cosine time (t) dependent wave-forms, as follows.

€

U t( ) = Re Ckk=0

N / 2∑ exp i 2πkt /T( ){ }

⎡

⎣ ⎢

⎤

⎦ ⎥ .

The amplitudes or spectral (“Fourier”) coefficients,

€

Ck , are complex, i.e.:

€

Ck = αk + iβk T is the total length in time of the record or series of measurements. The parameter k is the wave number, or frequency. The corresponding period is

€

Tk =Tk.

The amplitudes for each wave number, k, i.e. the spectral coefficients,αkand βk,for wave numbers k=0 to k=N/2, can be found by evaluating

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FIGURE 1 (Box 1.6): Spectrum of the kinetic energy associated with the horizontal wind measured at differentlocations.Eachcurve is fromadifferentstudy.Thedominantpeak in thewavelengthrange from3000to6000kmrepresents the troughs and ridges in the large scale pressure field atmid-latitudes. The small bump near 500 kmrepresents the fronts, while the peak at about 1 km represents the energy associated with cumulus convection.Source: J.S.A. Green, 1979: Topics in dynamicalmeteorology: 8. Trough-ridge systems as slantwise convection (1).Weather,34,2-10.

€

αk =2N

Unn=0

N−1∑ cos −2πkn

N⎛

⎝ ⎜

⎞

⎠ ⎟ ,

€

βk =2N

Unn=0

N−1∑ sin −2πnk

N⎛

⎝ ⎜

⎞

⎠ ⎟ and

€

α0 =1N

Unn=0

N−1∑ .

€

α0 represents the average. Note that

€

β0 = 0.Hence we have N+1 spectral (Fourier) coefficients. The amplitude of the kth harmonic is

€

Ak = αk2 + βk

2 . A summary of spectral analysis of several meteorological time series is shown in figure 1 (this box). The maximum in the spectrum at periods of several days is identified with the synoptic scale (Rossby waves, cyclones, anticyclones, etc.), while the maximum at periods of several minutes is identified with the micro-scale (e.g. turbulence). A minimum in the variance spectrum, i.e. a spectral gap, at periods ranging from half an hour to several hours is identified with the meso-scale. The existence of a spectral gap is not an expression of the fact that individual meso-scale circulation systems contain little energy and that they are therefore unimportant, rather it is an indication that these circulations are highly intermittent36. They occur only if certain conditions are fulfilled. For example, a sea breeze circulation (section 1.17) will occur only when the temperature difference between the sea and the land is large enough, and even then, other conditions must be met. Thunderstorms (section 1.30), hurricanes (problem 1.18), lee-troughs (figure 1.8) and sea-breeze circulations, all of which are examples of meso-scale circulations, which are characterized significant departures form hydrostatic- and/or geostrophic-balance, occur only in certain areas and in certain seasons. This is in contrast to the ever-present small scale turbulent eddies in the boundary layer as well as the continuous meandering of the isentropic tropopause (figure 1.53). What makes the range of time-scales between 1 hour and 1 day "intermediate" (mesoscale) and, therefore, defines the boundary between "slow" and "fast"? These lecture notes intend to provide physical and mathematical insight into the answer to this question. This insight is gained by studying the 36Ishida,H.,1990:Seasonalvariationsofspectraofwindspeedandtemperatureinthemesoscalefrequencyrange.Boundary-LayerMeteorol.,52,335-348.

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problem of adjustment to balance. The atmosphere is continuously brought out of balance by differential absorption of Solar radiation or by other non-adiabatic (i.e. diabatic) effects. Much of the motion that is observed in the atmosphere is a result of (re-)adjustment to hydrostatic and geostrophic balance. Observations of the sea breeze, discussed in section 1.17, reveal this process of geostrophic adjustment. PROBLEM 1 (BOX 1.6). Spectral analysis Perform a spectral analysis of a series of hourly pressure-, temperature-, and wind-measurements near the Earth’s surface at a weather station in the Netherlands (e.g. Schiphol or IJmuiden). Hourly measurements are available from 1951 until today. Plot the spectra, i.e. the amplitude as a function of the frequency or period. Interpret the spectra. The data can be downloaded from the following website: http://www.knmi.nl/nederland-nu/klimatologie-metingen-en-waarnemingen . Box 1.7 Measuring model performance: Taylor diagram On a Taylor diagram37 the correlation coefficient, the ratio of the standard deviations of two fields and the root-mean-square difference between two fields are indicated by a single point on a two-dimensional plot, thereby providing a short summary of the degree of correspondence of the two fields. The correlation coefficient, r, between two fields, A and B, is defined as

€

r =

1N

An − A ( )n=1

N∑ Bn − B ( )

σAσB

where

€

A and

€

B are the mean values and

€

σA and

€

σB are the standard deviations of A and B, respectively. The standard deviation is defined as

€

σA ≡1N

An − A ( )2n=1

N∑ .

The maximum value of r is 1 when

€

An − A ( )= a

€

Bn − B ( ) for all n, where a is a positive constant. If a =1, the fields are identical. The correlation coefficient is not a measure of the relative amplitudes of variation, as determined by their variances. To further quantify differences in two fields, the root-mean-square (RMS) difference, E, is introduced as,

€

E ≡1N

An − Bn( )2n=1

N∑ .

E can be resolved into two component: the overall biass,

€

E ≡ A − B , and the centred pattern RMS:

€

E '≡ 1N

An − A ( ) − Bn − B ( )[ ]2n=1

N∑ .

37Taylor,K.E.,2001:Summarizingmultipleaspectsofmodelperformanceinasinglediagram.J.Geophys.Res.,106(D7),7183-7192.

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FIGURE 1 (Box 1.7). “Taylor diagram”, for displaying pattern statistics. The radial distance from the origin is proportional to the standard deviation of a pattern (blue and red line). The centred RMS difference between the test and reference field is proportional to their distance apart (in the same units as the standard deviation). The correlation between the two fields is given by the azimuthal position of the test field. The full mean square difference is

€

E 2 = E 2 + E '2 . The statistical parameters r, E’, σA and σB, which are all useful in comparisons of patterns, are related by

€

E '2 ≡σA2 +σB

2 − 2σAσBr . This relation can be interpreted in terms of the law of cosines,

€

c2 ≡ a2 + b2 − 2abcosφ , where a, b and c are the lengths of the sides of a triangle and φ is the angle opposite to side c. We can now construct a so-called Taylor diagram that quantifies the statistical similarity between a reference field, A, which usually represents the observations, or reanalysis of observations, and a test field, B, which represents the model simulation. The statistical similarity is represented by a triangle in this diagram with sides of lengths E’, σA and σB, where cos-1 r represents the angle opposite the side of length E’. The side with length σA coincides with the abscissa (horizontal axis) of the Taylor diagram. A Taylor diagram, which quantifies the performance of n simulations, compared to the reanalysis, should show n+1 points and n hypothetical triangles. The influence of the differences in degree of heating of the earth’s surface can be recognised on a global scale (figure 1.37). In general, the sea-level pressure over the continents is lower in summer than in winter, while the reverse is the case over the oceans. This implies that the so-called “sea-breeze-effect” penetrates inland over thousands of �kilometers! Figure 1.32 indeed indicates that diurnal pressure oscillations at the Earth’s surface, induced by the land-sea thermal contrast, are observed everywhere inland over the European continent in the summer. Inertial oscillations are manifest in more phenomena than only the sea breeze. The low-level jet, which forms around sunset when the Earth’s surface cools and a stable stratification develops