# How does blade solidity ratio relate to thrust/power/torque of a propeller?

for quite some time I was using ABBOTT formula for static thrust estimation on a two-bladed propellers:

$$T=6.8\times10^{-5}\times D^{3}\times p\times RPM^{2}$$

$T$ is Static Thrust (N); $D$ is propeller diameter (m), $p$ is propeller pitch (m); RPM (1/s)

I was, however, unable to find any formula which would include number of propeller blades.

Standard thrust estimation methods all seem to rely upon the "propeller disk area" which is always assumed for a two-bladed propellers.

The closest I was able to get is something called "blade solidity ratio", or "rotor solidity":

$$\sigma = (B c)/2\pi R$$

where $B$ is number of propeller blades, $c$ is chord of each blade, $R$ is radius of the rotor.

The question is - how does blade solidity relate to thrust/power/torque?

## 2 Answers

You'll find this info in helicopter performance design literature. The power to drive the rotor (and a propellor) can be sub-divided into three parts: useful power, induced power, and profile power. The solidity ratio shows up in the profile power part.

Like in aerodynamics, the power/thrust equations are often made dimensionless and evaluated in coefficients. From Helicopter Test and Evaluation by Cooke and Fitzpatrick, section 2.4:

$C_P = C_T \left( \frac{V_C + v_i}{V_T} \right) + \frac{s \cdot C_D}{8}$

with $s$ being solidity ratio. So solidity ratio shows up in the profile drag power portion, which makes sense.

Each propeller blade is a wing in itself, and like a wing carries the weight of the plane, the propeller blade carries its fraction of the total thrust of the propeller. The more blades, the lower the fraction of each blade.

Low disc loadings are associated with two- or three-bladed propellers. Those can be found on GA aircraft and older, slow designs like pre-WW II aircraft. With turboprops and near-transsonic designs, more blades are needed to distribute the aerodynamic loads and to reduce the lift coefficient especially at the tips.

There is no strict formula, but in general a higher disk loading is associated with a higher solidity ratio.