Drag force on an upright cylinder: which cross-section to use?

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avannoord
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Hey guys,

I'm trying to prove to a friend something but I couldn't find a proof online. Imagine wind coming in from the side and hitting an upright cylinder. We're trying to find the force exerted, which then requires which kinda of cross section to use. I think the best way to calculate it is by using the a rectangle with the dimensions 2*r*h. since all the wind is not hitting the curved surface perpendicular, it's not correct to use a curved surface model, which would be (2Pi*r*h)/2. I know this is basic physics but sometimes it's the basic you need to know down well. What do you think? Thank you for the help.
 
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This is really a problem that is solved using fluid mechanics. Typically, the drag coefficient on the cylinder is regarded as a function of the Reynolds number for the flow. The relationship between the drag coefficient and the Reynolds number has been studied and quantified experimentally, and can also be obtained mathemtaically using computational fluid dynamics (CFD) for both laminar and turbulent flow.
 
avannoord said:
Hey guys,

I'm trying to prove to a friend something but I couldn't find a proof online. Imagine wind coming in from the side and hitting an upright cylinder. We're trying to find the force exerted, which then requires which kinda of cross section to use. I think the best way to calculate it is by using the a rectangle with the dimensions 2*r*h. since all the wind is not hitting the curved surface perpendicular, it's not correct to use a curved surface model, which would be (2Pi*r*h)/2. I know this is basic physics but sometimes it's the basic you need to know down well. What do you think? Thank you for the help.

The drag exerted on a cylinder in crossflow is not a trivial problem- in fact, it's the subject of Stokes' paradox when the flow is slow (low Reynolds number). In this case, basic physics is not sufficient. I have a few references in my office, if I remember tomorrow morning I'll add them to this post.