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STATISTICAL CONSULTANT/ANALYST/TUTOR/CIVIL ENGINEER /MATHEMATICIAN/SUBJECT MATTER EXPERT

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- 1. TARUNGEHLOT Conversion from Rectangular to polar coordinates and gradient wind1. Derivation of horizontal equation of motion in polar coordinates Many problems in Meteorology and Oceanography have circular symmetrieswhich make them much easier to deal with in a polar coordinate system. As we learnedin our discussion of vectors, the results of a physical problem are independent of ourchoice of coordinate system. The coordinate system choice is a matter of convenience inthe calculation of the problem (in the case of rectangular versus polar coordinates – it ismainly a matter of how much trigonometry we wish to endure). It also allows us toexamine matters from a different point of view, from which we can gain better insightinto the nature of the physical problem. For example, an extremely simple but surprisingly accurate representation ofatmospheric flow is the Geostrophic wind where the Coriolis force is balanced by the 1pressure gradient force in the horizontal plane. f u h P . The resulting flow isaxi-symmetric (no dependence on ) and, by examining this result in polar coordinates,we can reduce two differential equations to just one. If we examine the more general Du h 1equation of horizontal motion in polar coordinates, f u h h p , we can Dtcome to a better physical understanding on what it means to neglect the advection term inthe Geostrophic approximation. To start the process of relating rectangular and polar coordinates, consider a radialvector in the Cartesian coordinate plane as in figure 1. (A radial vector or ray is aspecific position vector which always starts at the origin or pole) 1
- 2. ^ j ^ ^ x r xi y j y ^ i Figure 1 – Representation of a radial vector in the Cartesian plane.From figure 1 we can see thatx r cos (1)y r sinUsing equations (1) and figure 1 we can also see that 2 2r x y (2) 1 y tan x Equations (2) describe the two parameters of the polar CS which can be used todefine any point in the 2-D plane. Since we are interested in polar vectors, we need toderive the basis vectors in the new polar coordinate system. We wish to express anyvector in terms of a radial component from the origin and its angular distance, , from ^the x axis. The radial basis vector, r , is simply a unit vector in the same direction as the ^radial vector r shown in figure 1. Using the requirement that r 1 along with ^ ^equations (1) we can relate the radial unit vector to our rectangular unit vectors i and j : ^ ^^ ^ ^ r cos i sin j ^ ^ r xi y jr cos i sin j (3) r x 2 y 2 r ^ ^ ^Notice that r is a function of the azimuthal angle, r r . This will be important later inthe discussions. 2
- 3. ^ ^ To find the azimuthal unit vector, with respect to i and j , we simply require it ^ ^to be orthogonal to r . Use of simple trigonometric identities allow us to find :^ ^ ^ ^ ^ cos 90 i sin 90 j sin i cos j (4) ^ ^ ^Notice that is also a function of the azimuthal angle, . ^ ^ Equation (3) and (4) are two simultaneous equations with unknowns i and j .With some algebra we can show that^ ^ ^i cos r sin (5)^ ^ ^j sin r cos ^ ^Figure 2 shows the unit vectors r and in the x-y plane. ^ j ^ ^ ^ r xi y j ^ r r r ^ iFigure 2. Graph showing the radial and azimuthal unit vectors, Note, as is the case with ^ ^all unit vectors, r 1 and 1 We now have the tools to relate the polar components and associated basis vectorsto their rectangular counterparts. Since the equations of motion deal with temporal andspatial variations of the velocity vector, we also need to derive the variation of our polarcoordinates in equation (2) with respect to our rectangular coordinates so that we can thendefine the gradient or advection operator in polar coordinates. For starters, we can see bytaking the derivative of equations (3) and (4) with respect to that 3
- 4. ^ ^ ^ ^dr sin i cos jd (6) ^ ^ ^ ^d cos i sin j rdFinally, we need to relate the variation of our polar components with respect to their r rrectangular counterparts: , , and . x y x y r r To find and , simply take the derivative of equation (2) with respect to x x yand y: 1 1 r 2 2 1 2 2 x x y 2 x y 2 2x cos x x 2 r (7) 1 1 r 2 2 1 2 2 y x y 2 x y 2 2y sin y y 2 r Rather than use the tedious tangent function or its inverse in equation (2) to find and , we can use the orthogonality of the x and y coordinates and equations (1) as x yfollows: y r sin 0 sin r cos x x x x r (8) x r cos 0 cos r sin y y y y rGradient operator We can now find the gradient in terms of polar coordinates and their associatedbasis vectors by using equations (5), (7) and (8). First use the chain rule to expressvariations with respect to x and y as variations with respect to r and : ^ ^ ^ ^ r r h i j i j x y x r x y r yNow substitute equation (5) to obtain the polar basis vectors 4
- 5. ^ ^ ^ ^ sin cos h cos r sin cos sin r cos sin r r r r ^ ^ 1 h r (9) r rSpatial and velocity components in polar coordinates: Equation (3) already established the relationship of a radial vector in terms of the ^polar components and unit vectors as r r r . It is instructive to verify this relationshipby directly converting spatial rectangular coordinates and unit vectors into polarcoordinates as defined above. x from Eq (5) y from Eq (5) ^ ^ ^ ^ ^ ^r xi y j r cos( ) cos r sin r sin( ) sin r cos ˆ r co s( ) sin ( )) ˆ ˆ 2 2 2 2r ( r co s ( ) r sin ( )) r ( r sin ( ) co s( ) r r (co s ( ) sin ( )) ^r rr (10)Now, to find the associated velocity components in polar form, take the time derivative of ^equation (10) and use equation (6) for the variation of the unit vector r with respect to ^ ^ ^ ^ ^ dr d dr dr d dr du rr r r r r dt dt dt d dt dt dtWe now have the velocity vector in the form of polar components and unit vectors.We will define the associated velocity components of the velocity vector as follows: drur dt du r dtso ^ ^u ur r u (11)Advection derivative: From equations (9) and (11) and the use of the dot product, the conversion of theadvective derivative from rectangular to polar coordinates is straightforward: 5
- 6. ^ ^ ^ ^ 1 uu ur r u r ur (12) r r r rEquation (12) is valid for a scalar function, r , . For a vector however, we saw fromequations (6) that the polar unit vectors vary with respect to so we must take this intoconsideration in our results. For the advection of a flow field for example ^ ^ ^ ^ uu u ur ur r u ur r u r rLet us examine each term in the above equation. The first term is simplified to: ^ ^ ^ ^ ur uur ur r u r ur ur r r rThe second term is a bit more complicated due to the -dependence of the basis vectors.It simplifies to: ^ ^ ^ ^ ^ ^ 2u u ur u u u ur r u ur r u r r r r r r 2 ^ ^ ^ ^ u ur u u u ur u r r r r r r 2 ^ ^ u ur u u u u ur r r r r rCombining both terms we see that 2 ^ ^ ur u ur u u u u u uru u r ur ur r r r r r r Du 1The resulting components of the equation of motion, f u h h p, Dtin polar coordinates are: 2 ur u u 1 p ur ur fu t r r r r (13) u u u ur 1 p ur u fu r t r r r r 6
- 7. We can see equations (13) contain an advective operator just like in equation (12) whenoperating on a scalar. The above equations also contain an additional term that accountsfor the curvature of the problem. The additional term in the radial equation above issimply the centrifugal acceleration.2. Gradient Wind: Looking back at Figure 1, we notice that curvature is significant in the flow field.The continually changing direction of the fluid parcel as it travels around the lowpressure system will lead to a centrifugal acceleration and associated force to theequations of motion. By converting the original equations of motion, given by equations(1), to cylindrical polar coordinates we can derive a simple relationship for the axial flowaround a low or high pressure system that includes the effects of rotation.Recall the results of equation (13) and including conservation of mass 2 ur u u 1 p ur ur fu t r r r r (14) u u u ur 1 p ur u fu r t r r r r u v w 0 x y zWe now make the following assumptions:1) The flow is steady – therefore 0 t2) The flow is axi-symmetric – therefore 03) Assume the flow has no radial component (tangential): is u r 0Given the above assumptions, the radial component of the conservation of momentum is 2u 1 p fu (15) r o rAll terms of the axial component of conservation of momentum are eliminated and thevertical component of momentum conservation is the equation of hydrostatic balance.The axial flow in equation (15) is called the gradient wind and accounts for both theCoriolis force and pressure gradient force. Unlike a geostrophic flow however, there is 7
- 8. force imbalance in such a way that we obtain a steady state curved flow where the onlyacceleration is a continual change in direction caused by the centripetal acceleration. Since equation (15) is simply a quadratic equation for u , we can obtain a simpleanalytic solution for the gradient wind: 2 fr fr r pu (8) 2 2 o rThere are four possible physical situations for equation (8) depending on the sign of thesquare root and the sign of the pressure gradient. They are: 2 fr fr r p p1) Normal Cyclonic flow: u ;u 0 and 0 2 2 o r rSince the radial gradient of the pressure field is greater than 0, we are talking about apressure field that increases as one travels away from the origin. Both terms in the frradical are of the same sign and the sum is greater than implying that the axial flow is 2counter-clockwise or cyclonic. From a force relationship perspective, consider first ageostrophic balance. The straight flow is a balance between an inward directed pressuregradient and an outward directed Coriolis force. In the case of the gradient wind, there isa force imbalance where the pressure gradient force is larger than Coriolis force. Thisleads to net force directed inward and a resultant centripetal acceleration causing a curvedcyclonic flow. 2 fr fr r p p2) Normal anti-cyclonic flow : u ;u 0 and 0 2 2 o r r:The decreasing radial pressure gradient indicates a high pressure system. The root term fris positive but less than indicating the axial flow is clockwise. There is a force 2imbalance where the outward directed pressure gradient force is less than inward directedCoriolis force. This force imbalance leads to net force directed inward and a resultantcentripetal acceleration causing a curved anti-cyclonic flow. Notice that, since the radial pgradient of pressure is less than 0, there is a limit on how large can be to ensure the r 2 p o f rradical is real. The restriction is r 4 2 fr fr r p3): Anomalous anti-cyclonic flow about a high: u ;u 0 2 2 o r pand 0 r 8
- 9. This third type of flow can exist theoretically but is not really seen empirically. It has thesame force balance as example 2 but utilizes the negative root of the quadratic equation.Simple calculation shows that the required pressure gradients for the above flow to existas a gradient wind are extremely small. 2 fr fr r p4): Anomalous anti-cyclonic flow about a low: u ;u 0 and 2 2 o r p 0 rThis fourth type of flow can exist theoretically and occasionally is seen in nature (i.e., ananti-cyclonically rotating tornado). It has the same force balance as example 1 butrequires that the pressure gradient (which is a positive number) be large enough tosupport anti-cyclonic flow (positive u ) . 9

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