Physics Asked by Gnubie on January 27, 2021
gives examples of drag coefficients of assorted shapes from 0.045 to 1.28. I understand that $C_D=large{frac F{frac 12Av^2rho}}$.
Does a drag coefficient of exactly one correspond to any significant shape? Or is the value $C_D$ simply a result of the arbitrary choice of the definitions of the SI units?
I contrast this with the coefficient of friction, for which a value of unity means that the drag is equal to the object’s weight, implying that the object put on a ramp of increasing slope will begin to slide when the ramp angle reaches 45° if prevented from toppling.
Does a drag coefficient of exactly one correspond to any significant shape?
Cd = 1 and Cd = F / (1/2 * A * v^2 * ro) <=> F = (1/2) A * v^2 * ro That can occur with different combinations of: fluid, area, speed and drag. So it does not imply a particular shape.
Is Cd a result of the arbitrary choice of the definitions of the SI units?
Note that Cd has no units, therefore the value will not change provided that the system of units used is consistent (SI is a consistent system of units, but there are others). Note that the expression is non dimensional:
[] means dimensions, [l] are dimensions of length, [t] dimensions of time, [m] dimensions of mass and so on.
From the expression of Cd we have [Cd] = [F] / (1/2 * [A] * [v]^2 * [ro]) = [l][t]^-2 / ([l]^2 ([l].[t]^-1)^2 . [m].[l]^-3) <=> (after simplification) [Cd] = 1 (no units)
Conclusion: Cd depends only on the shape of the object and type of surface. Just like as it happens with the phenomena of friction, there is only need for a single experimental constant. However Cd = 1 has no particular special meaning in my view.
Recommendation: It might be worth to have a look on the so called "Buckingham Pi theorem" and the subject of dimensional analysis, for a deeper compression of this and other experimental phenomenas and constants.
Good luck for your studies.
Answered by Lord_David on January 27, 2021
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