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if I had two objects, scaled perfectly to each other so that one is $5$ times the size of another while keeping the shape the same, would their coefficients of drag, $C_d$, be the same in the formula $D = \frac12 \rho C_d S V^2$?

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Not necessarily.

If we take the fluid dynamical system under consideration to be only a function of fluid density $\rho$, some characteristic speed $V$, fluid viscosity $\mu$, and some characteristic length $R$, then it has to be the case from dimensional analysis that the drag coefficient is only a function of the Reynolds number: $$C_d = f(Re)$$ Since altering the characteristic length of the object changes the Reynolds number, you'd expect the drag coefficient to change as the size of the object changes.

That being said, at high enough Reynolds numbers, the drag coefficient "paradoxically" becomes independent of the Reynolds number (and by extension of everything), so size doesn't affect the drag coefficient in that scenario. You should certainly expect a dependence at low Reynolds numbers, where the drag coefficient is asymptotically inversely proportional to the Reynolds number.

For a more detailed discussion on the mathematical derivations of why this is so, feel free to read my notes on lift/drag here.

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  • $\begingroup$ Before you point people to your pages, it might be opportune to understand what a boundary layer is! $\endgroup$ Feb 4, 2020 at 23:09

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