# When rearranging the lift force equation to solve for lift coefficient instead of lift force, why is the numerator often given as F not 2F?

the standard description of the Lift force affecting the lift surface is:

$$F={\dfrac{1}{2}}\rho v^{2} SC_{L}$$ the basic rearrangement gives the following definition of the lift coefficient: $$C_{L}={\dfrac{2F}{\rho v^{2}S}}$$

why is the above expression met so rarely? with the standard skipping the 2 multiplier? $$C_{L}={\dfrac{F}{\rho v^{2}S}}$$

I've checked 30 scientific papers, those measuring the lift\drag coefficients.

• 28 papers have had F
• 2 papers have had 2F
• Could it be related to some assumptions about S, say S in one case is for one wing and in the other case, for the total wingspan ? It really depends on the context. But to me there should definitely be a 2 factor Commented Apr 29, 2018 at 16:00
• Are you sure that there’s no 0.5 in the denominator ? Can you give a reference to confirm? Commented Apr 29, 2018 at 18:16

The definition of the coefficients differs between countries/regions. The factor ½ is used in most of Europe and in Russia, because it is part of the dynamic pressure equation (in metric units). In the USA working with USA units, the 2 is incorporated in the coefficient.

As for instance mentioned in Principles Of Helicopter Aerodynamics, section 2.5 by J. Gordon Leishman, Cambridge University Press:

It is important to note that the US customary definition of the thrust, torque and power coefficients is different to that used in some parts of the world (mainly in Britain, most of Europe, and Russia), where a factor of one half is used in the denominator..

• Not contradicting anybody, just note the(1/2) factor in the lift force comes directly from Bernoulli equation Commented Jun 19, 2019 at 11:37
• Really? Anderson's Fundamentals of Aerodynamics, which I consider quite American, uses the "normal" (to me as a European) definition of q = 1/2 rho V^2 and C_L = L / (q S), which implies C_L = 2 L / (rho V^2 S). Commented Mar 24, 2022 at 15:20
• @Raketenolli Does Anderson use metric units? Commented Mar 25, 2022 at 6:27
• @Koyovis both, but he stresses the fact that he uses a consistent set of units. I just saw your edit, how is 1/2 part of the dynamic pressure equation in metric units - but (I guess that's what you mean) not in US customary units? Commented Mar 25, 2022 at 13:28
• Doesn't matter whether units are metric or US. Apparently, OP's noted discrepancy is due to a typo. The lift coefficient is dimensionless, of course. And 1/2 in the denominator translates to 2 in the numerator. Evidently, somebody forgot the 2 in setting up the press galley and missed the error in proof. Commented Mar 26, 2022 at 1:05

The total aerodynamic force acting on a body is usually thought of as having two components, lift and drag.

By definition, the component of force parallel to the oncoming flow is called drag

and the component perpendicular to the oncoming flow is called lift

It is convenient to express the induced drag as an expression having similarities with that of the lift, thus by taking the ratio of lift to drag the « 2 multiplier » disappears, but considering solely the lift coefficient the expression you are giving is correct.