Are the lift / drag coefficients considered dimensionless?

Question

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.

For what it's worth, angles are typically considered dimensionless, too. — Sanchises, Jul 05 '21 at 07:00
Not considered, but defined! Related: Why do we use dimensionless expressions in flight mechanics and aerodynamics? — Peter Kämpf, Jul 05 '21 at 10:47
If it helps, the Mach number is also dimensionless. And very obviously not a constant. — Jason, Jul 06 '21 at 04:32
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 $N \cdot m$. — Peter Cordes, Jul 06 '21 at 07:21

Koyovis · Accepted Answer · 2021-07-05T10:29:43.687

11

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

edited Jul 05 '21 at 10:29

answered Jul 05 '21 at 05:29

Koyovis

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same explanation holds for CD, and very similar for CM, and Ch (hinge moment coefficient), and other aerodynamic coefficients of forces and moments. – Gürkan Çetin Jul 05 '21 at 16:45

Are the lift / drag coefficients considered dimensionless?

1 Answers1