Differential drag on a cup anemometer

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    Differential Drag
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Discussion Overview

The discussion revolves around the calculation of differential drag on a four cup anemometer, specifically focusing on how changes in cup shape and size affect the drag coefficient and, consequently, the anemometer's rotation. The conversation includes theoretical considerations and mathematical reasoning related to drag coefficients and their implications for different configurations of the cups.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant inquires about the calculation of differential drag on a specific size cup at a constant wind velocity.
  • Another participant suggests that the drag coefficient of open and closed cups is necessary, along with the normal drag equation.
  • A participant expresses gratitude for a provided link that helped them start their calculations.
  • One participant questions whether the differential drag coefficient remains the same when altering the shape of the cup, specifically when modifying it to a quarter sphere with a flat plate.
  • Another participant provides specific drag coefficient values for open and closed cups, indicating that the force can be calculated using the area and the difference in drag coefficients.
  • A participant asks whether the drag coefficient for a half cup differs from that of a whole cup due to shape changes, speculating on the magnitude of any potential change.

Areas of Agreement / Disagreement

Participants express varying views on how changes in cup shape affect the drag coefficient, indicating that there is no consensus on whether the drag coefficient remains constant or changes significantly with shape modifications.

Contextual Notes

Participants discuss the implications of shape on drag coefficients without resolving the specific mathematical relationships or assumptions involved in these calculations.

cocoon95
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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.?
 
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thanks for the link. It got me started in the right direction.
 
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!
 
An open cup is around CD of 1.42, closed side is only 0.38.
http://www.windpower.org/en/tour/wtrb/drag.htm , the area is the same - so the force is just the area * the difference in Cd
 
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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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