The drag force equation is \(R = \frac{1}{2}D\rho Av^{2}\), where \(v\) is the speed of an object, \(A\) is the cross-sectional area of the object, \(\rho\) is the density of the fluid, and \(D\) is the drag coefficient. To go about proving the formula, i did this

From experimentation, it is evident that the drag force acting on an object is related to the objects velocity somehow. Hence I analysed an object moving with constant velocity in a fluid, and analysed the system to be the fluid. The power lost by the object would be

\(P = \frac{W}{\Delta t}\)

now if we look at a disk of air which has not yet been touched by the object. Let its velocity be 0. once the object hits that disk of air of mass m, it does work on it, and changes its kinetic energy to \(\frac{1}{2}mv^{2}\)

hence \(P = \frac{W}{\Delta t} = \frac{\Delta K}{\Delta t} = \frac{mv^{2}}{2\Delta t}\)

\(\rho = \frac{m}{V}\Rightarrow m = \rho V\)

\(V = A\Delta r \Rightarrow m = \rho A\Delta r\), where \(\Delta r\) is the distance that disk of air travelled

\(v = \frac{\Delta r}{\Delta t} \Rightarrow \Delta r = v\Delta t\)

hence, \(m = \rho Av\Delta t\)

subbing this back into the equation for power lost, we get

\(P = \frac{\rho Av\Delta t v^{2}}{2\Delta t}\)

\(P = \frac{1}{2}\rho Av^{3}\)

Since object is travelling at constant velocity, air resistance is constant, so power loss is constant. hence instantaneous power is equivelant to our average power equation above.

equation for instantaneous power is \(P = \vec{F}\cdot\vec{v} = Fv\). Since the obect is in constant velocity, there is not net force so \(F = R\). hence \(P = Rv\)

sub this back into our last power equation: \(Rv = \frac{1}{2}\rho Av^{3}\)

\(R = \frac{1}{2}\rho Av^{2}\)

my question is... where is the drag coefficient gone?

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## Comments

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TopNewestWe know that the drag coefficient is given by the object's shape (if it's flat, spherical, a cone shape etc.) and friction against its surfaces. In your proof you don't take these things into consideration, so you won't get the drag coefficient in your equation.

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Is not the objects shape taken to consideration by the \(A\) in \(R = \frac{1}{2}D\rho Av^{2}\)

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Not really. \(A\) is only the cross section area of the object in the direction in which it is travelling. For example, a cylinder, a sphere and a cone (all with the same base radius) will have the same value for \(A\), but completely different values for \(D\).

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