I'm aware of the viscous drag formulae to calculate the drag force acting on a moving sphere inside a viscous fluid. But how do you improve the formulae to get the force acting on a cube with side length a?
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1$\begingroup$ I am not sure a cube can be described similar to a sphere. The problem may be that there is always a scale on which the corners and edges of the cube cause turbulence, which is not the case for a sphere in a viscous liquid. That's just a hunch, and I am looking forward to a real answer by somebody who knows. $\endgroup$– CuriousOneCommented May 2, 2015 at 4:07
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$\begingroup$ I get the feeling if I equal the volume of the cube to a sphere with equal volume and hence obtain it's radius i will be able to put it directly into the stroke's equation. But as u mentioned if the corners cause a turbulance that is not possible! $\endgroup$– slhulkCommented May 2, 2015 at 5:01
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$\begingroup$ If you look at the drag coefficient for very high Reynolds numbers (fully developed turbulence), then the cube (frontal flow) has a drag coefficient of 1.05, while a rough sphere is 0.47 or so. A smooth sphere is much smaller (cited as low as 0.1!)... so it seems really dependent on how the turbulence forms around the body. I am still trying to find a publication that has a diagram with the drag coefficient as a function of Reynolds number. You have really triggered my interest in how flow develops around a cube. $\endgroup$– CuriousOneCommented May 2, 2015 at 6:02
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$\begingroup$ Let me know if you find one. :) $\endgroup$– slhulkCommented May 2, 2015 at 6:12
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1$\begingroup$ Will certainly do. $\endgroup$– CuriousOneCommented May 2, 2015 at 6:27
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