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Old 05-02-2008, 03:55 PM   #4 (permalink)
aerohead
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Quote:
Originally Posted by Patrick View Post
Phil, thanks for that analogy. But I'm a bit confused. If the ball is moving in the direction of 3 o'clock, wouldn't the separation point be 5 seconds BEFORE midnight? Doesn't the air cling to the ball for a short distance past maximum circumference?

Also, since water is 800 times as dense as air, wouldn't the Reynolds numbers for the bowling ball and golf ball be very different? Or was this compensated for in the bowling ball analysis?

Thanks.
Hi Patrick! From the images,with the laminar boundary layer,flow does separate before midnight(90-degrees).If 3-O'clock denotes 0-degrees(the stagnation-point),flow separates at 80-degrees from there (clockwise).With the sand glued to the front of the bowling ball,or dimples with the golfball,a turbulent boundary layer is established immediately,which forestalls flow separation until the bell has sounded for Cinderella at 5-seconds after 12:00.When the flow breaks off,the wake is smaller and has a higher base pressure and consequently lower profile drag.For the golfball,drag is cut in half,and its range (for a club head velocity of 110-mph) is doubled,from 150-yards,to 300-yards.Drag reduction is so remarkable with the turbulent boundary layer,artificial roughening of the golfball is critical to it's performance even at the low Reynold's Number it experiences for it's size and velocity.I should have posted the velocity of the submerged ball.And you're correct that under water,the bowling ball it would see much higher Reynold's Number.
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