scilearner Posted March 28, 2010 Posted March 28, 2010 Hello guys, So how and why does a cricket ball swing in the conventional mode? Let us see what happens to the flow over a cricket ball released with the seam angled (Fig. 1). Between about 30 and 70 mph, the laminar boundary layer along the bottom surface separates at about the apex of the ball. However, the boundary layer along the top surface is tripped by the seam into a turbulent state and its separation is therefore delayed. This asymmetry results in a pressure differential (lower pressure over the top) and hence side force which makes the ball swing in the same direction that the seam is pointing (upwards). This is from an article I recently read. I have not done any aerodynamics but I'll be so greatful if anyone can explain what this guy means. What I'm thinking is if turbulent air delays the separation of air and the smooth air separates wouldn't the ball swing to the smooth side first and then to the turbulent side when that separates. Also why does turbulent air have lower pressure, in a tube if a fluid is turbulent wouldn't it exert higher pressure on the walls. Thanks guys
rogerxd45 Posted March 28, 2010 Posted March 28, 2010 just a guess but i think the turbulent air might take up more space which would reduce the pressure on that side making the ball move to the lower pressure area in a tube the volume is fixed so as the turbulent air trys to expand and cant it increases the pressure on the walls (as you said) but without the walls its free to expand which causes the lower pressure
Mr Skeptic Posted March 31, 2010 Posted March 31, 2010 If you throw a spinball, you can use Bernoulli's principle: there is some air rotating with the ball, and on one side the air flows faster and on the other side lower, causing a high pressure area that causes the ball to curve. If you throw a ball with almost no spin, irregularities on the ball's surface can change the direction slightly and in a random direction.
J.C.MacSwell Posted April 1, 2010 Posted April 1, 2010 Hello guys, This is from an article I recently read. I have not done any aerodynamics but I'll be so greatful if anyone can explain what this guy means. What I'm thinking is if turbulent air delays the separation of air and the smooth air separates wouldn't the ball swing to the smooth side first and then to the turbulent side when that separates. Also why does turbulent air have lower pressure, in a tube if a fluid is turbulent wouldn't it exert higher pressure on the walls. Thanks guys The turbulence from the seam on the upper surface (as drawn) makes the flow more robust on that surface of the ball, mixing it with the stagnant air behind the ball and sweeping it away. This allows the flow to travel further around the ball, with the air exiting more downward than it otherwise would. In contrast, the lack of turbulence on the underside allows the wake air to remain longer, forcing the flow to separate earlier and exit more parallel to the flow (flow relative to the ball). The net result is an upward lift on the ball since the air is deflected downward. Probably not enough to make the ball rise, but more likely drop less than expected. Thrown with the seam on the forward right quarter there would be a significant curve in that direction.
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