Does Smeaton's coefficient have a modern accepted value or it is dependent of the air density?

Question

1. Does Smeaton's coefficient, k, have a modern value or it is dependent of the air density?

2. Why is the accepted value of k so high?

In various texts about the Wright brothers (see 1 and 2) one can read about Smeaton's coefficient that troubled them a lot and that they finally discovered the parameter had a much lower value reaching the conclusion $k = 0.0033 lbf/ft^2/mph^2 = 0.79 kg/m^3$ (instead of $k = 0.005$), a fact also noticed by others before them.

1. "the Wright brothers calculated a new average value of 0.0033. Modern aerodynamicists have confirmed this figure to be accurate within a few percent." Source: Correcting Smeaton's Coefficient

2. $L = k \cdot S \cdot V^2 \cdot C_L$

   $L$ = lift in pounds

   $k$ = coefficient of air pressure (Smeaton coefficient)

   $S$ = total area of lifting surface in square feet

   $V$ = velocity (headwind plus ground speed) in miles per hour

   $C_L$ = coefficient of lift (varies with wing shape)"

Source: The Wiki page of the Wright brothers

However, knowing that the modern formula for lift is $$L = 0.5 \cdot \rho \cdot S \cdot V^2 \cdot C_L$$ Where $\rho$ = the air density.

It appears that $k = 0.5 \cdot \rho$ and so it does not have a standard average value. Also a $k = 0.0033 lbf/ft^2/mph^2 = 0.79 kg/m^3$ leads to a $\rho = k/0.5 = 1.58 kg/m^3$ that corresponds to a sea level air temperature well below -25 C, which is unusual.

If the two relations for lift are correct, the Smeaton's coefficient can not be 0.0033 but closer to 0.0025 a value corresponding to a standard air density at $20 ^\circ C$ close to $1.2 kg/m^3$.

Answer 1

There is no modern equivalent to the Smeaton coefficient, therefore it has no "modern" value.

The Smeaton coefficient is supposed to represent a "coefficient of air pressure" (and thus indirectly density), but it is a false constant: There is no single value because air pressure is not a constant value, but rather a variable (depending on altitude, temperature, etc.).

As you noted lift is calculated by a different formula today, using actual density of the fluid (air) you're moving through in place of the Smeaton coefficient.

$ L = Cl \cdot \frac{(A \cdot \rho)}{2} \cdot V^2$

You can calculate the Smeaton coefficient at any given location/altitude with simple algebra if you know the air density: Solve the modern lift equation, plug the values into the "old" Wright Brothers era equation, and solve for $k$.

The density-based calculation also has the benefit of working in other fluids besides air. (The same equation can give you hydrodynamic lift for boats and submarines by plugging in the density of water - itself a dynamic value depending on temperature, salinity, etc.)

Answer 2

Wright brothers used the following 1900s equation:

$$ L = k \cdot V^2 \cdot S \cdot C_L $$

$k$ = pressure force (drag) on a one foot square flat plate moving at one mile per hour through the air.

$$ D = k \cdot V^2 \cdot S \cdot C_D \tag{1} $$

Modern drag equation is as follows: $$ D = \frac{1}{2} \cdot \rho \cdot V^2 \cdot S \cdot C_D \tag{2} $$ Therefore, by equating $(1)$ and $(2)$, we get:

\begin{align} k &= \frac{1}{2} \cdot \rho \\\\ & = \frac{1}{2} \cdot 1.225 \\\\ & = 0.6125 \left[\frac{kg}{m^3}\right] \end{align}

Value of k was given in units of lb force/m.p.hr sq/ft sq

1 lb force/m.p.hr sq /ft sq = 0.45359237 Kg × 9.8066 m/sec sq /(0.44704 m/sec) sq / 0.0929 m sq =239.5939 kg/m3

Therefore k = 0.6125/239.5939 = 0.00256 lbf/m.p.hr sq./ft sq.

Thus modern value of k is 0.00256 if we take air density as 1.225 kg/m3