Since the lift coefficient, $C_L$, and the drag coefficient, $C_D$, are obtained by rescaling the full lift and drag by

$$\frac12 \rho A v^2$$

does that mean they're dimensionless numbers?

It's slightly confusing, because after rescaling, $C_L$ and $C_D$ depend on the angle of attack, $\alpha$, which somehow makes me think of $C_L$ and $C_D$ as dimensionful quantities.

  • 3
    $\begingroup$ For what it's worth, angles are typically considered dimensionless, too. $\endgroup$
    – Sanchises
    Jul 5 at 7:00
  • 4
    $\begingroup$ Not considered, but defined! Related: Why do we use dimensionless expressions in flight mechanics and aerodynamics? $\endgroup$ Jul 5 at 10:47
  • $\begingroup$ @Sanchises, yes, but only when they are in radians ;) $\endgroup$
    – Zeus
    Jul 6 at 0:46
  • $\begingroup$ If it helps, the Mach number is also dimensionless. And very obviously not a constant. $\endgroup$
    – Jason
    Jul 6 at 4:32
  • $\begingroup$ Fun fact: work and torque both have the same dimensions (Newton-meters), but are very different things: Work / energy is based on a dot product (force in the direction of motion), the other on a cross product (force at a distance from a pivot point). Unfortunately, the way we've constructed our mathematical formalism for doing physics, dimensional analysis doesn't help us sort out $N \times m$ from $N \cdot m$. $\endgroup$ Jul 6 at 7:21

Yes they are dimensionless numbers, which does not mean that they are constants. $C_L$ $C_D$ are variables. Dimensionless meaning: no physical unit. $$L = C_L \cdot \frac{1}{2} \rho V^2 \cdot A$$ with metric units:

  • L [N] = [kg*m/sec$^2$]
  • $\rho$ [kg/m$^3$]
  • V [m/sec]
  • A [m$^2$]

Dimension of $\rho V^2 \cdot A$ = $\frac{kg}{m^3} \cdot \frac{m^2}{s^2} \cdot m^2 = kg \cdot m / sec^2 = N$

  • 1
    $\begingroup$ same explanation holds for CD, and very similar for CM, and Ch (hinge moment coefficient), and other aerodynamic coefficients of forces and moments. $\endgroup$ Jul 5 at 16:45

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