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Unformatted text preview: 4.9 The Joukowski mapping: circles to ellipses A particularly useful application of the mapping idea concerns the flow around bodies. We have solved the problem of the flow around a cylinder. Thus if we can find a conformal mapping between the unit circle and any given shape, we have solve the flow problem around this shape. z R 2 c 2 d z = + b 2 Consider the (inverse) mapping z = + b 2 , (42) called the Joukowski mapping. Consider a circle of radius R , whose surface in the plane is described by the polar representation = R e i . Under the Joukowski mapping, z = R e i + b 2 R e i = parenleftbigg R + b 2 R parenrightbigg cos + i parenleftbigg R b 2 R parenrightbigg sin = c cos + id sin is an ellipse with axes length 2 c and 2 d . The semiaxes come out to be c = R + b 2 R , d = R b 2 R . Now consider the uniform flow past an ellipse. To model an arbitrary angle between the direction of the flow and the semimajor axis, we consider the flow around a cylinder that approaches the xaxis under an angle : w 1 ( ) = U parenleftbigg e i + R 2 e i parenrightbigg . In principle, one can find w ( z ) using the inverse of the Joukowski mapping = f ( z ) = 1 2 ( z + z 2 4 b 2 ) , so that w ( z ) = w 1 ( f ( z )). However, the resulting expressions are often not so useful. For example, to find the streamlines, it is much easier to find the streamlines of the w 1 ( ) in the plane, and then to transform them using (42). This is how the pictures were produced. If one wants to calculate the velocity, one uses u iv = d w d z = d w 1 d 1 d z/ d = U (e i R 2 e i / 2 ) (1 b 2 / 2 ) . On the cylinder, = R e i , so u iv = U (e i e i e 2 i ) (1 ( b 2 /R 2 )e 2 i ) = 2 iU sin( ) (e i ( b 2 /R 2 )e i ) . (43) This means there are stagnation points at = and = + . This point is where a streamline leaves the surface. In other words, this streamline (plotted in red) has the same value of the streamfunction then the surface of the ellipse. 4.10 Lift Now we want to fly! In principle, we know how to construct the flow around a wing of arbitrary shape, we only have to find the transformation, starting from a circle. We have seen already that the key ingredient is to have circulation around the wing. We have calculated the lift in the case of a cylindrical crosssection, but what is it for an arbitrary shape? 4.10.1 Blasius theorem C t = (cos , sin ) n = (sin , cos ) x y We now calculate the force on a fixed body in a stream, using complex notation. This is a close relative of the calculation done in section (3.6), leading to (33). Suppose a fixed rigid body, boundary C is in a steady flow, generating a potential w ( z ). As we saw in 4.4, dw dz = u iv = q e i , (44) where  u  = q is the speed of the flow, and is the angle the flow direction makes to the horizontal. So, according to Bernoulli (no gravity):horizontal....
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 Fall '11
 Eggers
 Unit Circle

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