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Newton’s second law for a parcel of air in an inertial coordinate system (a coordinate system in which the coordinate axes do not change direction and are not accelerated) ma F In the case of more than one force: dVi m Forces acting on an object dt dVi m Forces acting on an object dt Let’s first take a look at the term in the yellow box… dVi An inertial coordinate system and the meaning of the vector Vi and dt y Earth not rotating x z (toward you) Inertial coordinate system: y x z (toward you) y x z (toward you) Vi is with respect to a distant star dVi is with respect to a distant star dt Earth-based (non-inertial) “spherical” coordinate system (favored by meteorologists) x = positive toward east y = positive toward north z = positive toward zenith dx u wind component toward east dt dy v wind component toward north dt dz w wind component toward zenith dt = longitude = latitude dx a cos d dy ad Earth-based (non-inertial) “spherical” coordinate system (favored by meteorologists) x = positive toward east y = positive toward north z = positive toward zenith dx d a cos dt dt dy d a dt dt dz w dt V ui vj wk Where unit vectors are a function of position on the earth (not Cartesian) What is the relationship between Vi ui vj wk dVi dt and dV dt Since the unit vectors are not constant… dVi du dv dw di dj dk i j k u v w dt dt dt dt dt dt dt We need to determine what these are di Let’s start with dt d i i i i i u v w dt t x y z d i i i i i u v w dt t x y z i At a point (constant x,y,z) none of 0 the unit vectors change with time so t i As one moves north or south the i 0 direction experiences no change so y As one moves up or down the i i direction experiences no change so z 0 So: di i u dt x di i u dt x From figure on right looking down at north pole at latitude : i i x a cos i 1 x a cos This gives us the magnitude, but not the direction. Note that i is pointed toward the center of the earth at the original point Looking at the figure at the right: We see that i has components in the j and k directions The unit vector describing the direction of i has two components: i j sin k cos di i j u sin k u cos u tan u u j k So: u dt x a cos a a REMEMBER THIS SLIDE? Vi ui vj wk Since the unit vectors are not constant… dVi du dv dw u tan u dj dk i j k u j k v w dt dt dt dt a a dt dt We still need to determine what these are Let’s do dj next dt dj j j j j u v w dt t x y z dj j j j j u v w dt t x y z j does not change with time or elevation, but does change in the x and y directions dj j j u v dt x y Lets first figure out j x Look at light gray triangle in (a): a cos sin or a tan Dark gray triangle is shown in a different view in (b) a x tan j j So: j tan x a j Direction of is the -x direction Now lets figure out j y y a j j j 1 y a a Direction of j is the k j 1 k y a So: direction u tan v dj i k dt a a REMEMBER THIS SLIDE? Vi ui vj wk Since the unit vectors are not constant… dVi du dv dw u tan u u tan v dk i j k u j k v i k w dt dt dt dt a a a a dt We still need to determine what this is Let’s do dk next dt dk k k k k u v w dt t x y z dk k k k k u v w dt t x y z k does not change with time or elevation, but does change in the x and y directions dk k k u v dt x y dk k k u v dt x y Let’s do k first x x a k k Direction of k is the positive i direction k 1 i x a dk k k u v dt x y k Let’s do next y y a k k Direction of k is the positive i direction k 1 j dk u v i j y a dt a a k k REMEMBER THIS SLIDE? Vi ui vj wk Since the unit vectors are not constant… dVi du dv dw u tan u u tan v u v i j k u j k v i k w i dt dt dt dt a a a a a a or dV du uv tan uw dv u 2 tan vw dw u 2 v 2 i k j dt a a dt a a dt a dt dV dV uv tan uw u 2 tan vw u 2 v 2 i k j dt dt a a a a a j What are the correction terms and what do they mean? Consider third equation: dw u 2 v 2 forces dt a Air moving in a straight line at a constant speed initially southward. u = 0, v = -10 m/s, w = 0 Air will accelerate upward! The earth curves away from the path of the air parcel. Newton’s second law in an inertial coordinate system dV m i Forces acting on an object dt Newton’s second law in a spherical coordinate system dV uv tan uw m i a a dt u 2 tan vw u 2 v 2 k Forces j a a a Now let’s look at the right side of the equation! Forces and the governing equations Three Fundamental Forces in the lower atmosphere Pressure Gradient Force Gravity Friction Two Apparent Forces in the lower atmosphere due to the rotation of the earth Centrifugal Force Coriolis Force 1. Pressure gradient force: Directed from high pressure toward low pressure. Simple illustration of a pressure gradient acting on a wall (left) and a molecule (right) A wall with higher pressure on its left side (indicated by greater molecular density on the left side) An air molecule with higher air pressure on its left (indicated by greater molecular density on the left side) Volume of the fluid element V xyz Mass of the fluid element m xyz = density (g m-3) Pressure at center of the fluid element p0 px0 , y0 , z0 Definition of pressure: Pressure is equal to the force applied per unit area Find Pressure at A and B Use Taylor expansion: p x p A p0 higher order terms x 2 p x pB p0 higher order terms x 2 Find Force at B and A Force = Pressure Area p x FA p A Area A p0 yz x 2 p x FB pB AreaB p0 yz x 2 Find net x-direction force per unit mass p x p x Fx FA FB p0 yz p0 yz x 2 x 2 p Fx xyz x Fx 1 p m x Fx 1 p m x In a similar way: 1 p m y Fy Fz 1 p m z The vector form of the Pressure Gradient Force can therefore be expressed as F 1 p 1 p 1 p k j i m x y z F 1 p m 2. Gravity: Directed toward center of earth Newton’s law of universal gravitation GMm r Fg 2 r r G = 6.67310-11 N m2 kg-2 M = mass of Earth m = mass of fluid element r = vector from Earth center r = magnitude of r vector Newton’s law of universal gravitation… GMm r Fg 2 r r …expressed as a force per unit mass Fg GM 2 m r r r Over the depth of the troposphere the change in the force of gravity is insignificant and we can approximate r as the earth radius, a Fg GM 2 m a r r 3. Friction: Directed opposite the flow Friction is manifested as: A drag force in a very thin layer (a few mm) near the surface Turbulent mixing of blobs of faster and slower air at altitudes above the surface. Although largely irrelevant to the atmosphere, we will consider the drag force first and then make a simple analogy to formulate friction for the rest of the atmosphere moving Not moving Expect that drag force on upper plate will be proportional to: Speed of the plate (u0) Area of the plate (A) And inversely proportional to the depth of the fluid (l) F Au0 l Where is the proportionality constant called the dynamic viscosity coefficient F Au0 l Next let’s define a “Shearing Stress” , as the Force/Unit Area on the plate Let’s also assume the lower plate is moving at an arbitrary speed so that the speed difference is u Let’s also assume that the distance between the plates is arbitrary and that that difference is z Then: u zx z and u zx z as z 0 zx u z Force per unit area acting in x direction due to shear in z direction Now let’s consider how that force acts on a fluid element of volume xyz Illustration of x component of shearing stress (F/A) on a fluid element Use Taylor expansion Stress acting across the upper boundary on the fluid below it zx z zx z 2 Stress acting across the lower boundary on the fluid below it zx z zx z 2 Using Newton’s third law: Stress across the lower boundary on the fluid above it must be: zx z zx z 2 To find the net stress, we want to sum the forces that act on fluid within the fluid element zx z zx z zx x y x y xyz zx zx z 2 z 2 z Dividing by the mass of the fluid element we have the viscous force per unit mass arising from the vertical shear in the x direction on the fluid element Fzx 1 zx 1 u m z z z Fzx 1 u m z z Assuming is constant, this can be written Fzx 2u 2u 2 2 m z z where is called the kinematic viscosity coefficient We have figured out one component of the frictional force per unit mass: Fzx m Considering how shear can act in each direction to create a frictional force in each direction, there are nine components Fxx Fxy Fxz Fyx Fyy Fyz Fzx Fzy Fzz , , , , , , , , m m m m m m m m m 2u 2u 2u Frx 2 2 2 m z x y 2v 2v 2v 2 2 2 m x y z 2w 2w 2w Frz 2 2 2 m y z x Fry Unfortunately, all of the preceding theory applies to the air in a layer a few mm thick above the earth’s surface For blobs of air mixing, an analogous “eddy viscosity coefficient”, K, is defined that is calculated based on the average length that an eddy can travel before mixing out its momentum. 2u 2u 2u Frx K 2 2 2 m z x y In general, vertical wind shear is much stronger than horizontal shear for synoptic scale flows so this reduces to: Frx 2u K 2 m z 2v K 2 m z Fry Frz 0 m Force balance in the atmosphere for a non-rotating earth expressed in spherical coordinates 2 2 2 2 dV 1 GM r V uv tan uw u tan vw u v F p 2 K 2 i k j dt a a a a a m a r z Acceleration Correction for spherical Coordinate system gravity friction pressure gradient force Same statement expressed as 3 orthogonal scalar equations (East-West equation) (North-South equation) (Up-Down equation) du uv tan uw 1 p 2u K 2 dt a a x z dv u 2 tan vw 1 p 2v K 2 dt a a y z dw u 2 v 2 1 p GM 2 dt a z a Pressure gradient Gravity Acceleration Force Correction terms for earth Friction based coordinate system Apparent forces on a rotating earth – the Centrifugal Force We will derive the force terms twice First from a more conceptual approach The second time from a more formal vector approach Apparent forces on a rotating earth – the Centrifugal Force Consider an object at rest on the rotating earth Actual speed due to rotation of earth: 60 N: Speed = 232 m/s 40 N: Speed = 356 m/s Equator Speed = 465 m/s A ball whirling on a string experiences a CENTRIPETAL acceleration toward the axis of rotation The acceleration is equal to r where is the angular rotation rate 2 A person on the ball whirling on a string experiences a CENTRIFUGAL acceleration toward the axis of rotation The acceleration is equal to r where is the angular rotation rate 2 Vectors not to true size! Standing still on the rotating earth Outward acceleration away from Earth’s axis is 2 r where is the Earth’s angular rotation rate 2 86156.09s Where 86156.09 s is the time it takes for the Earth to rotate once with respect to a fixed star (a sidereal day) g 9.81 ms2 2 2 R s 1 6378100 m 0.034 ms 2 86156.09 2 Earth has distorted its shape into an oblate spheroid in response to this force. The distortion is such that the combined gravitational force and centrifugal force (g*) act exactly perpendicular to the Earth’s (flat) surface. GM r 2 * r 2 gk 2 R g R a r r We combine the centrifugal force with gravity and forget about it!! Apparent forces on a rotating earth – the Coriolis Force In the absence of a twisting force called a torque, air in motion across the earth must conserve its angular momentum (MVR) where M is the parcel mass, V is velocity about the axis of rotation and R is the distance from the axis of rotation. A simple interpretation of the Coriolis effect: Air moving across the earth’s surface will try to come to equilibrium at a latitude/altitude where its angular momentum equals that of the earth so that the parcel has no relative motion. Motion on a rotating earth – the Coriolis Force Consider air moving: Eastward: The air has greater angular momentum than the earth beneath it. It will experience an “outward” centrifugal acceleration equatorward Westward: The air has less angular momentum Than the earth beneath it. It will experience an “inward” centripetal acceleration poleward. Poleward: The air will progressively move over points on the earth with less angular momentum. Air will accelerate eastward relative to the earth below it. Equatorward: The air will progressively move over points on the earth with greater angular momentum. Air will accelerate westward relative to the earth below it. Suppose an object is moving eastward on the rotating earth at speed u The total centrifugal force it will experience will be its angular velocity squared distance to axis of rotation 2 u R u R 2 CEN R R 2u 2 R R R 2 Centrifugal force (combined with gravity) 2 2 10 ms 1 10 3 ms 2 86156 s Magnitude of 2nd term on RHS 102 m 2 s 2 105 ms 2 6378100 m Magnitude of 3rd term on RHS Suppose an object is moving eastward on the rotating earth at speed u The total centrifugal force it will experience will be its angular velocity squared distance to axis of rotation 2 u R u R 2 CEN R R 2u 2 R R R 2 ignore Centrifugal force (combined with gravity) Coriolis Force R COR 2u R Coriolis acceleration motion parallel to latitude circle j 2u sin k 2u cos dv 2u sin dt dw 2u cos dt Suppose an object is moving equatorward on the rotating earth at speed -v Apply principle of conservation of momentum 2 u 2 R R R R R Expand this expression 2 u 2 2 R R 2 R R R R R Since u and R are small, neglect terms with their product uR 2 2 2 R u R R 2RR R R 2 2 2 R u R R 2RR R R (From previous slide) 2 R u 2RR R R If we assume that R << R this expression can be approximated as: 2RR Ru or u 2R u 2R For a displacement in the –y direction, from Figure: u 2 y sin y2 sin Dividing by t and taking the limit as t 0 du dy 2 sin v 2 sin dt dt u 2R For a displacement in the z direction, from Figure: u 2 z cos z 2 cos Dividing by t and taking the limit as t 0 du dz 2 cos w2 cos dt dt Combining the results for east-west and north-south movement of air du v 2 sin w2 cos fv w2 cos dt dv 2u sin fu dt dw Where: 2 sin f 2u cos dt and f is the Coriolis parameter Motion on a rotating earth – the Coriolis Force In a reference frame on the earth the Coriolis effect appears as a force acting on an air parcel. The Coriolis Force Causes air to deviate to the right of its direction of motion in the Northern Hemisphere (and to the left in the Southern Hemisphere); Affects the direction an object will move across the earth’s surface, but has no effect on its speed; Is strongest for fast-moving objects and zero for Stationary objects; and Has no horizontal component at the equator and has a maximum horizontal component at the poles A formal way of deriving the Coriolis and Centrifugal Forces In our previous discussion of spherical coordinates, the coordinates were fixed in position. A B At Point A, for example, the x, y, and z axes always point toward distant stars. At point B, the axes point at different stars, but always in the same direction. When the earth rotates, the axes at point A find themselves at a later time at point B. The direction the axes at a specific location point is a function of time To account for the local movement of the coordinate system mathematically, we must consider the behavior of a vector in a rotating coordinate system Consider a vector in a stationary coordinate system A Ax i Ay j Az k where i , j , and k are fixed at local points The components of the vector in the rotating coordinate system are: A Ax i Ay j Az k The time derivative of the vector A Ax i Ay j Az k in a stationary coordinate system would be: dA dAx dAy dAz i j k dt dt dt dt The time derivative of the vector A Ax i Ay j Az k in the rotating coordinate system are: dA dAx dAy dAz di dj dk i j k Ax Ay Az dt dt dt dt dt dt dt dA dAx dAy dAz i j k dt dt dt dt Stationary Rotating dA dAx dAy dAz di dj dk i j k Ax Ay Az dt dt dt dt dt dt dt We can write the equation for the rotating frame of reference as: dA dA di dj dk Ax Ay Az dt dt dt dt dt dk di dj The derivatives , , and , represent the rate of dt dt dt change of the unit vectors i , j , and k , that arise because the coordinate system is rotating Let’s determine and expression for di dt Rotation vector, , points upward from north pole By similar triangles i i di The magnitude of is given by: dt lim i di d i i t 0 t dt dt di The direction of is given by: dt di i dt Let’s determine and expression for dj dt By similar triangles Rotation vector, , points upward from north pole j j dj The magnitude of is given by: dt lim j dj d j j t 0 t dt dt dj The direction of is given by: dt dj j dt Let’s determine and expression for dk dt Rotation vector, , points upward from north pole By similar trianglesk k dk The magnitude of is given by: dt lim k dk d k k t 0 t dt dt dk The direction of is given by: dt dk k dt Our previous equation was: dA dA di dj dk Ax Ay Az dt dt dt dt dt and we can write di dj Ax Ax i Ax i Ay Ay j Ay j dt dt dk Az Az k Az k dt then dA dA di dj dk Ax Ay Az dt dt dt dt dt dA dA Ax i Ay j Az k dt dt dA dA A dt dt For a vector A, the coordinate transformation from a non-rotating to a rotating system is given by: dA dA A dt dt Let’s apply this to two vectors: Let r be a position vector, perpendicular to the axis of rotation of the earth with magnitude equal to the distance from the surface of the Earth to the axis of rotation By definition: d a r dr r dt dt Va V r Also apply vector transform to V a d aVa dVa Va dt dt The absolute velocity of an object on a rotating earth is equal to its velocity relative to the rotating earth + the velocity of the Earth’s rotation The absolute acceleration of an object on a rotating earth is equal to its acceleration relative to the rotating earth + acceleration due to the Earth’s rotation d aVa dVa Va dt dt (1) Va V r Put (2) into (1) d aVa d V r V r dt dt d aVa dV dr V r dt dt dt d aVa dV V V r dt dt d aVa dV 2 V 2 r dt dt Coriolis Force Centrifugal Force (2) d aVa dV 2 2 V r dt dt i j k 2 V 0 2 cos 2 sin u v w d a ua du v 2 sin w2 cos dt dt d a va dv 2u sin dt dt d a wa dw 2u cos dt dt Equations of motion on non-rotating earth (East-West equation) (North-South equation) (Up-Down equation) du uv tan uw 1 p 2u K 2 dt a a x z dv u 2 tan vw 1 p 2v K 2 dt a a y z dw u 2 v 2 1 p GM 2 dt a z a Pressure gradient Gravity Acceleration Force Correction terms for earth Friction based coordinate system Equations of motion on rotating earth Answer We must add terms to account for the acceleration required for air to conserve its angular momentum. In scalar form the equations of motion for each direction become: (East-West equation) du uv tan uw 1 p 2u K 2 dt a a x z dv u 2 tan vw 1 p 2v K 2 (North-South equation) dt a a y z (Up-Down equation) dw u 2 v 2 dt a 1 p z 2v sin 2w cos 2u sin g 2u cos Pressure gradient Effective Force Gravity Correction terms for Friction Coriolis Force spherical coordinate system Acceleration The complete momentum equations on a spherical rotating earth du uv tan uw 1 p 2u K 2 dt a a x z 2v sin 2w cos dv u 2 tan vw 1 p 2v K 2 dt a a y z 2u sin dw u 2 v 2 dt a 1 p z g 2u cos FOR SYNOPTIC SCALE MOTIONS….. WHICH TERMS ARE LARGE AND IMPORTANT? WHICH TERMS ARE SMALL AND INSIGNIFICANT? See Scale Analyses: Table 3.1 P. 60, Table 3.2 P. 64 Scale Analysis of the horizontal momentum equations du uv tan uw 1 p 2u K 2 dt a a x z dv u 2 tan vw 1 p 2v K 2 dt a a y z U2 L U2 a 104 105 UW a p L 108 103 U 10 ms 1 W 0.01 ms1 U H 2 107 L 106 m p 1000 Pa 10 mb 3 2 2 10 m s 3 1 kg m 2v sin 2w cos 2u sin f 0U 103 f 0W 106 H 104 m K 1 m2 s 1 *numbers in yellow boxes apply to free atmosphere above boundary layer Scale Analysis of the vertical momentum equations dw u 2 v 2 dt a 1 p z g 2u cos p H g 101 101 U 10 ms 1 W 0.01 ms1 L 106 m UW L U2 a 107 105 p 105 Pa 1000 mb 5 2 2 10 m s 3 1 kg m f 0u 103 H 104 m g 1 0ms2 ABOVE THE BOUNDARY LAYER du uv tan uw 1 P 2u K 2 dt a a x z 2v sin 2w cos dv u 2 tan vw 1 P 2v K 2 dt a a y z 2u sin dw u 2 v 2 dt a 1 P 0 g 2u cos z WITHIN THE BOUNDARY LAYER du uv tan uw 1 P 2u K 2 dt a a x z 2v sin 2w cos dv u 2 tan vw 1 P 2v K 2 dt a a y z 2u sin dw u 2 v 2 dt a 1 P 0 g 2u cos z The Hidden Simplicity of Atmospheric Dynamics: du 1 P fv dt x dv 1 P fu dt y WITHIN THE BOUNDARY LAYER, ALL P u K 2 fv HORIZONTAL PARCEL ACCELERATIONS x z CAN BE UNDERSTOOD BY COMPARING THE MAGNITUDE AND DIRECTION OF THE P 2v K 2 fu PRESSURE GRADIENT, CORIOLIS AND y z FRICTIONAL FORCES 2 du 1 dt dv 1 dt 0 ABOVE THE BOUNDARY LAYER, ALL HORIZONTAL PARCEL ACCELERATIONS CAN BE UNDERSTOOD BY COMPARING THE MAGNITUDE AND DIRECTION OF THE PRESSURE GRADIENT AND CORIOLIS FORCES 1 P g z THE ATMOSPHERE IS IN HYDROSTATIC BALANCE – VERTICAL PGF BALANCES GRAVITY – ON SYNOPTIC SCALES Note the Total Derivative: d dx dy dz dt dt dt x y , z ,t dt y x, z ,t dt z x, y ,t dt t x, y , z Note that: u dx dy dz , v , w dt dt dt d u v w dt x y , z ,t y x , z ,t z x, y ,t t x , y , z d u v w t x , y , z dt z x , y ,t x y , z ,t y x , z ,t The rate of change of a property (temp) at a fixed point (x,y,z) = The rate of change of the property Following a parcel as it moves to the point (x,y,z) The advection of the property from upstream to the point (x, y, z) Simplified u Momentum Equation expressed in terms of local time derivatives du 1 P fv dt x u 1 P u u u u v w fv x x y , z ,t y x, z ,t z x, y ,t t x, y , z u u 1 P u u fv u v w t x y z x The rate of change of west-east wind at a fixed point (x,y,z) PGF COR The advection of the west-east wind Component from upstream to the point (x, y, z) Simplified v Momentum Equation expressed in terms of local time derivatives dv 1 P fu dt y v 1 P v v v u v w fu y x y , z ,t y x , z ,t z x, y ,t t x, y , z v v 1 P v v fu u v w t y y z x The rate of change of south-north wind at a fixed point (x,y,z) PGF COR The advection of the south-north wind Component from upstream to the point (x, y, z) Pressure as a vertical coordinate Pressure decreases monotonically with altitude, and therefore can be used as a vertical coordinate instead of altitude We will find this coordinate system more convenient mathematically and so our task will be to recast the momentum equations in pressure coordinates The simplified horizontal momentum equations in height coordinates are given by: du 1 P fv dt x dv 1 P fu dt y or more succinctly: dV 1 p fk V dt In order to cast these equations in pressure coordinates, we need to convert the PGF term into an equivalent expression in isobaric coordinates…….. Consider the differential dp on a constant pressure surface Expanding… p p dx p dy p dz p x y , z z x, y y x , z dp p p (Subscripts indicate differentiation carried out holding subscripted variable constant) p p p 0 dx p dy p dz p x y , z z x , y y x, z Now expand dz as a function of x and y p z p p z 0 dx p dy p dx p dy p x y , z z x, y x y , p y x, z y x , p p z p p z 0 dx p dy p dx p dy p x y , z z x, y x y , p y x, z y x , p Rearrange: p p p z p z 0 dx p dy p y x, z z x, y y x, p x y , z z x, y x y , p For this statement to be true for all dx p and dy p , the statements in each square bracket must equal zero p p z x y , z z x , y x y , p p p z z x , y y x , p y x , z p p z x y , z z x , y x y , p Use hydrostatic equation p z g x y , z x y , p p p z z x , y y x , p y x , z P g z p z g y x , z y x , p Divide both sides by 1 p z g x y , z x y , p z 1 p g y x , z y x , p or 1 p x y , z x y , p where gz 1 p y x , z y x , p is called the geopotential height 1 p x y , z x y , p 1 1 p y x , z y x , p p y , z y , p We will now drop the use of subscripts and write the momentum equations in p coordinates Pressure coordinates In pressure coordinates, the horizontal momentum equations (above the boundary layer) become: du fv dt x dv fu dt y dV p fk V dt where , the geopotential height, is given by = gz and f = 2sin and the vertical momentum equation becomes Rd T p p (using the ideal gas law) The Total Derivative in pressure coordinates d dx dy dp dt dt dt x y , p ,t dt y x , p ,t dt p x , y ,t dt t x , p , z Note that: u dx dy , v dt dt Note that the vertical velocity in pressure coordinates has a different form: dp dt The term, , represents the rate of change of pressure following parcel motion as a parcel rises or descends in the atmosphere Note the Total Derivative: d dx dy dp dt dt dt x y , p ,t dt y x , p ,t dt p x , y ,t dt t x , p , z Note that: u dx dy , v dt dt d u v dt x y , p ,t y x , p ,t p x , y ,t t x , y , p d u v t x , y , p dt x y , p ,t y x, p ,t p x, y ,t The rate of change of a property (temp) at a fixed point (x,y,z) = The rate of change of the property Following a parcel as it moves to the point (x,y,z) The advection of the property from upstream to the point (x, y, z) Simplified u Momentum Equation in pressure coordinates expressed in terms of local time derivatives du fv dt x u u u u fv u v t x y p x The rate of change of west-east wind at a fixed point (x,y,z) PGF COR The advection of the west-east wind Component from upstream to the point (x, y, z) Simplified v Momentum Equation expressed in terms of local time derivatives dv fu dt y v v v v fu u v t y y p x The rate of change of south-north wind at a fixed point (x,y,z) PGF COR The advection of the south-north wind Component from upstream to the point (x, y, z)
