Differential drag on a cup anemometer

[cocoon95]

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.?

 

[mgb_phys]

You need the drag coefficient of an open and closed cup
and then the normal drag equation http://en.wikipedia.org/wiki/Drag_coefficient

 

[cocoon95]

thanks for the link. It got me started in the right direction.

 

[cocoon95]

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!

 

[mgb_phys]

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

 

[cocoon95]

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.