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Differential drag on a cup anemometer

  1. Feb 18, 2009 #1
    If differential drag on a four cup anemometer causes its rotation, how does one calculate the differential drag on a specific size cup and with a specific wind velocity assuming the wind is constant and in the same horizontal plane as the four cups.?
     
  2. jcsd
  3. Feb 18, 2009 #2

    mgb_phys

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  4. Feb 19, 2009 #3
    thanks for the link. It got me started in the right direction.
     
  5. Feb 21, 2009 #4
    okay, The drag coefficeint is clear. If however I slice half of the cup off with a vertical cut and blank the end, so I really have a 1/4 sphere with a flat plate at the end of the cup, does the differential Cd remain the same for a relative wind perpendicular to the 1/4 cup vs the 1/4 sphere. Or does if change and only depend on the surface area of the 1/4 cup and 1/4 sphere?
    Remember that shape matters!
     
  6. Feb 21, 2009 #5

    mgb_phys

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    An open cup is around CD of 1.42, closed side is only 0.38.
    http://www.windpower.org/en/tour/wtrb/drag.htm [Broken], the area is the same - so the force is just the area * the difference in Cd
     
    Last edited by a moderator: May 4, 2017
  7. Feb 22, 2009 #6
    Knowing that the CD for an open cup is 1.42 and the closed side 0.38, is the CD for a 1/2 cup the same as a whole cup or does the CD change due to the different shape. If it changes would it be expected to change by a very small amount say < 0.05 or something larger like 0.10.
     
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