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Why does the cricket ball swing?

  1. Mar 28, 2010 #1
    Hello guys,

    65913.jpg

    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 :smile:
     
  2. jcsd
  3. Mar 28, 2010 #2

    sophiecentaur

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    This tends to go against intuition but high speed air (and this is the situation where there's turbulence) has a reduced pressure compared with still air. (The Bernouili effect is described all over the place to various levels - just Google it and it will find at least one hit that will make sense to your personal taste). Hence, the ball will experience a net sideways force in the direction, as shown, of the side where there's more turbulence.
    Motor vehicles experience drag, mainly due to the turbulence which is set up behind them as the air which they have pushed out of the way, starts to come together at the back. Drag is not, primarily, due to hitting the air head-on at the front.
     
  4. Mar 28, 2010 #3

    rcgldr

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    It's a continous flow situation, the turbulent boundary layer separates further back and while moving more downwards then the smooth boundary layer. The result as shown in your diagram is a net downwards diversion and acceleration of air, resulting in the ball exerting a downwards force onto the air, coexistant with the air exerting an upwards force onto the ball, which creates the lift in the swing direction.

    The pressure in a boundary layer (turbulent or laminar) is generally the same as the static pressure of the air outside the boundary layer, so any pressure differential is ultimately due to a pressure differential outside the boundary layers, not within them. In this case, it's the downwards diversion (acceleration) of the air as seen by the wake aft of the ball that corresponds to the presssure differential.

    Turbulent flow is generally a higher energy flow, it takes more drag to initiate turbulent flow, but a lot of that energy ends up as increase in speed of the air as it's spins in a series of tiny vortices, but again it's static pressure is the same as the static pressure of the air just outside the boundary layer. As shown in the diagram a turbulent boundary layer tends to be thinner than a laminar boundary layer, reducing "form drag" which is related to thickness of the boundary layer, but it's not a key factor in producing swing direction.
     
    Last edited: Mar 28, 2010
  5. Mar 28, 2010 #4
    Thanks a lot for all the answers :smile: Ok I so now I understand why turbulent air has lower pressure but I'm still bit confused.

    1. So when the air comes in front one layer becomes turbulent and one becomes laminar, so air pushes the ball to the turbulent side.
    Jeff has also mentioned that further down the turbulent layer pushes down on the air and according to Newton 3 rd law the ball
    moves up. My question is wouldn't the ball move up anyway because of Bernoulli why is Newton 3rd law involved in this?
    2. Also Jeff has mentioned the boundry layer and air outside is the same. Do you mean that if the turbulent layer has lower pressure
    air outside that has lower pressure as well? Also I just got another question, if I get a paper and blow on top of that. The air
    would move faster and it would lift but why is it that only air in the bottom pushes the paper, what about the air above the
    region I'm blowing can't that push down on the paper.

    Thanks again for all the help. Please excuse my poor knowledge on aerodynamics!!
     
    Last edited: Mar 28, 2010
  6. Mar 29, 2010 #5

    rcgldr

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    Turbulence of air doesn't translate into lower pressure. Turbulent or laminar air flow can both have low or high pressure.

    That's because the rough surface and/or ridges trips up the air flow in the boundary layer, so that it transitions into turbulent flow.

    That's the final outcome, but it's an indirect consequence of the turbulent boundary layer, which tends to remain attached to the back of the ball longer than the the laminar flow, resulting in a diversion of the wake flow.

    Bernoulli doesn't explain how pressure differentials are created and/or maintained, it explains how the air reacts once those pressure differentials exist. The root cause here is the fact that the turbulent flow exits in a more downwards direction than the laminar flow, resulting in a downwards diversion and acceleration of air in the wake (flow aft of the ball). This downwards acceleration corresponds with a force between the ball and air, the ball exerting a downwards force onto the air, coexistant with the air exerting an upwards force onto the ball. Whethere the ball actually moves up depends if the upward force from air is greater than the downwards force from gravity.

    The pressure of a boundary layer and the adjacent air just outside the boundary layer is the same. If there's a momentary difference in pressure, due to start up conditions, then the pressure differential results in a mixing of the air between the boundary layer and adjacent air until there the pressures equalize.

    The static ports that are flush mounted onto the side of a fuselage of an aircraft rely on this fact. These ports are small holes that "hide" inside the boundary layer. The holes are the ends of a pipe connected to a chamber inside the aircraft. Regardless of the airspeed of the aircraft (within reason, perhaps any air speed below mach 0.3), the pressure inside that chamber is essentially the same as the ambient air.

    That's a type of Coanada effect, not Bernoulli. If the paper isn't curved into a convex shape, and you don't blow below it then nothing significant happens. If the paper is curved, then beyond the peak point of the paper a void (very low pressure zone) would be created if the paper didn't rise or the air didn't descend in order to prevent the creation of a void. In order for the air to follow that curve, it has to accelarate (mostly centripetally) to follow that curve, and the air's momentum and viscosity result in a lowering of pressure while it accelerates to fill in what would otherwise become a void. The lower pressure above the paper and the ambient pressure below the paper result in an upwards force on the paper, causing it to rise. The lower pressure above the paper also draws in the surrounding air, but with the paper preventing upwards flow, the result is a net downwards acceleration of air.
     
  7. Mar 29, 2010 #6
    Thanks a lot for the answers Jeff. That does make sense :smile: However I still have some misunderstanding.

    1. All you said makes sense if I can stop thinking that turbulent has low pressure and laminar has high pressure. When the air first separates into turubulent and laminar, is the reason that turbulent doens't have low pressure because the air surrounding it fills it up?, same for laminar.
    So most of the time turbulent and laminar air are just like normal air because air surrounds it equalizes it? Is that right?
     
  8. Mar 29, 2010 #7

    rcgldr

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    This isn't an issue of laminar and turblent flows in general but instead the flows in the boundary layers, which are the flows very close to the surface of the ball. Wiki article on boundary layers:

    http://en.wikipedia.org/wiki/Boundary_layer

    Since the amount of air within a boundary layer is small compared to the amount of air outside the boundary layer, any change in pressure of the surrounding air is going to be a dominant factor and change the pressure within the boundary layer.

    In case you're wondering, the wiki article shows a thicker turbulent boundary layer on a flat wall in that example. If the wall was convex, curving downwards, then it turns out that a turbulent boundary layer would follow a convex wall better than a laminar boundary layer, and the resulting turbulent boundary layer would thinner in the case of a convex wall. The wiki article is showing a single boundary layer that transitions into turbulent flow rather than comparing a laminar boundary layer with a turbulent one. Your diagram from the first post shows the difference in a laminar boundary layer versus a turbulent boundary layer, although the thicknesses are greatly exaggerated.
     
  9. Mar 29, 2010 #8
    Oh right so the air surrouding the boundary layer normally equalizes the boundary layer for normal pressure. Ok I think I have some sufficient understanding now. Thanks a lot Jeff, I was thinking something entirely different before :smile:
     
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