# Are the lift / drag coefficients considered dimensionless?

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. Commented Jul 5, 2021 at 7:00
• Not considered, but defined! Related: Why do we use dimensionless expressions in flight mechanics and aerodynamics? Commented Jul 5, 2021 at 10:47
• @Sanchises, yes, but only when they are in radians ;)
– Zeus
Commented Jul 6, 2021 at 0:46
• If it helps, the Mach number is also dimensionless. And very obviously not a constant. Commented Jul 6, 2021 at 4:32
• 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 $N \cdot m$. Commented Jul 6, 2021 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$$^3$$]
• A [m$$^2$$]
Dimension of $$\rho V^2 \cdot A$$ = $$\frac{kg}{m^3} \cdot \frac{m^2}{s^2} \cdot m^2 = kg \cdot m / sec^2 = N$$