# Why is the drag force proportional to $v^2$ and defined with a factor of $1/2$?

$$Drag = \frac{1}{2}C_d \rho Av^2$$

I understand that the strength of the drag depends on the density of the fluid the body passes through, the reference area of the body, the drag coefficient, and the velocity of the object.

I don't, however, understand the 1/2 and the $$v^2$$ in the equation.

• I think this is a very interesting question. For example why don't we just make $C_D$ half of what we currently use and remove the 1/2 altogether. – JMac Jun 21 '19 at 1:59
• The Factor $\frac{1}{2} \rho\,v^2$ is obtain form the stagnation (dynamic) pressure – Eli Jun 21 '19 at 8:13
• Keep in mind that this drag rule is not universal. It applies for Reynolds numbers from about 1000 up (even then in some regimes the $v^2$ dependence is not exact, though this is usually expressed by varying the drag coefficient rather than by changing the form of the expression). – dmckee --- ex-moderator kitten Jun 21 '19 at 14:27

In short, the squared speed $$v^2$$ appears in the equation because when moving faster, you increase both

• how much momentum ($$p=mv$$) that is transferred to the air molecules (they must be moved away faster) and
• how much air that must be moved away (because you sweep through more air per second).

Increasing the speed means increasing both of these factors that both make it tougher to fall. Thus, speed appears "twice", so to say.

The half $$\frac 12$$ that also appears in the equation, is - as others also point out - due to the drag coefficient $$C_d$$ being neatly written as $$C_d=\frac{D}{Aq}$$, where $$q=\frac12\rho v^2$$ is the dynamic pressure, an important aerodynamic property.

Sure, you could have included the half in the drag coefficient to simplify the drag formula. But you would simultaneously complicate the relationship $$C_d=\frac{D}{Aq}$$. You could also ask, why there is a half in $$K=\frac12 m v^2$$. Why isn't that half just included in the mass $$m$$? Well, because then many other relationships that include $$m$$ would become more complicated.

The drag force is doing negative work on the object that it is decelerating. By the work/kinetic-energy theorem, the work done is equal to the change in kinetic energy that the object experiences. Since kinetic energy is defined as $$E_k = 1/2 mv^2$$, you can expect the "1/2" and the $$v^2$$ terms to show up in the equation.

• Ah, I haven't learned the kinetic energy formula yet. – Austin Gae Jun 21 '19 at 1:57
• It still does leave an interesting question of why we keep the 1/2 and don't just account for it with the drag coefficient. – JMac Jun 21 '19 at 2:00
• Maybe and maybe not. I would rather keep the 1/2 in the formula so I could associate it with a kinetic energy term. When I taught high school physics, there were a few instances when kinetic energy was involved, and a 1/2 showed up in the denominator of the answer. From a math viewpoint, you don't want fractions in the denominator, but I told students to leave it there so people could recognize that kinetic energy was involved. – David White Jun 21 '19 at 2:10
• @JMac i think it is the same reason why kinetic energy is 1/2 mv^2 and not just mv^2, as mentioned in the answer. – QuIcKmAtHs Jun 21 '19 at 2:22
• @QuIcKmAtHs m and v have many other uses, so the 1/2 relationship there is needed for consistency everywhere. $C_D$ is just a constant to fit different geometries to the drag equation. – JMac Jun 21 '19 at 3:31

Adding just a bit to the previous answer, I believe that the drag coefficient definition is based on the dynamic pressure term in Bernoulli's equation, $$\frac{1}{2}\rho v^2$$. Thus dependence on velocity squared is expected, and is often observed. However, fluid flow is complicated, and $$C_d$$ determined empirically in many cases varies with flow velocity, and Reynolds number, depending on boundary layer transition from laminar to turbulent flow, form drag, wake, etc.