I was looking at a question on aviation.stackexchange and an interesting answer I found included that even today, we have no physical way to get around the fact that drag of a body is roughly the square of velocity. Theoretically is there any way to get around this?

  • 2
    $\begingroup$ What does it mean to "get around" it? $\endgroup$
    – Steeven
    Feb 18, 2018 at 1:09
  • $\begingroup$ If you can make the surface porous and blow air out of the pores, you can lower the drag. $\endgroup$ Feb 18, 2018 at 1:47
  • $\begingroup$ To echo Steeven's comment I don't know what you mean by getting around the $v^2$ law. Are you asking for drag reduction as Chester points out? A simplistic picture would be to represent drag $D$ as Taylor series in $v$: $D=a_1v+a_2v^2+a_3v^3+\ldots$ so the question really becomes "which term is dominant at what speed $v$?" $\endgroup$
    – Deep
    Feb 18, 2018 at 4:40
  • $\begingroup$ Yes that was what my question was aimed at sorry about the confusion. $\endgroup$
    – Jihyun
    Mar 4, 2018 at 22:02

2 Answers 2


It comes from the Navier-Stokes equation.

More specifically, it's the ram pressure term which goes as $ -\nabla (\frac{1}{2}\rho u^2)$. This gives a force per unit mass, since the LHS o the N-S equation is usually $\rho \frac{\partial \mathbf{u}}{\partial t}$.
The ram pressure is also referred to inertial terms.

Physically, in order to move at speed $u$, you need to "move away" the air in front of you, i.e. you need to provide it with kinetic energy $\frac{1}{2}m u^2$.



Let's consider the total derivative of the fluid velocity $\mathbf{v}$ in the presence of a pressure gradient $\nabla P$ and an external field $\phi$, which could be gravity:

Let's use tensor notation:

$$ \frac{\mathrm{d} (\rho v_i)}{\mathrm{d} t} = \rho \frac{\mathrm{d} v_i}{\mathrm{d} t} + v_i \frac{\mathrm{d}\rho}{\mathrm{d} t}. $$ The equation of motion is $$ \frac{\mathrm{d}(\rho\mathbf{v})}{\mathrm{d} t} = -\nabla P - \rho \nabla \phi,$$ and the total derivative $\mathrm{d}_t = \partial_t + v_i \partial_i$.

Hence we get:

$$ \frac{\partial \rho v_i}{\partial t} = -\rho v_j \partial_j v_i - \partial_i P -\rho \partial_i \phi - v_i \partial(\rho v_j) \\ = -\partial_j(\rho v_j v_i + P\delta_{ij}) -\rho \delta_i \phi$$ so that $$ \frac{\partial \rho v_i}{\partial t} = -\partial_j T_{ij} - \rho \partial_i \phi.$$

$T_{ij}$ is the stress energy tensor, the first term of which is the ram pressure due to the bulk motion of a fluid, and the second term is the pressure introduced by thermal effect (e.g. denser regions).


Inviscid drag due to lift (induced drag) of a wing or similar lifting body follows a different trend. It is roughly proportional to the inverse of the velocity squared, so it actually decreases with speed.

The total drag of the wing includes both viscous and inviscid components. The inviscid component dominates at very low speeds, while the viscous component dominates at high speeds. As a result, the drag-versus-speed curve is roughly parabolic, with a minimum at some value of speed greater than zero.


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