Blade Loading Coefficient

The thrust coefficient divides thrust by the whole disc, and the disc is about ninety percent empty air. Divide instead by the blade area — that is, divide $C_T$ by the solidity $\sigma$ — and you get the number that says how hard the blades themselves are working. $C_T/\sigma$ is called the blade loading coefficient, and it is the single most useful figure in rotorcraft performance.

It is proportional to the mean lift coefficient of the blade sections. The usual approximation, which comes from integrating a uniform lift coefficient against the $r^2$ variation of dynamic pressure along a blade, is that the mean $C_L$ is about six times $C_T/\sigma$. A rotor at $C_T/\sigma = 0.09$ is therefore running its blades at a mean lift coefficient near 0.54, which for an aerofoil that stalls around 1.2 to 1.4 sounds like plenty of room — until you remember that a mean is a mean. In forward flight the retreating blade sees far less airflow than the average and must take far more angle to carry its share, so the retreating side is working well above the mean while the advancing side is working well below it. That asymmetry is what puts the practical ceiling at roughly 0.12 to 0.14 rather than anywhere near the sectional stall value.

Working values run about 0.07 to 0.09 in normal flight. The number matters because it is the common currency of everything that eats into a helicopter's margin. Gross weight raises it. Altitude raises it, because thinner air means less $\rho$ underneath. Load factor in a turn or a pull-up raises it in direct proportion — a 2 g manoeuvre doubles it. They all draw on the same allowance, which is why a heavy machine, high and hot, has almost no manoeuvre margin left even though nothing about the rotor has changed.

Approaching the ceiling has a distinctive signature, and it is worth knowing. Retreating blade stall arrives as a rise in vibration at once per revolution, then a pitch-up and a roll toward the retreating side, and the correct response is to reduce collective and slow down — not to pull, which raises the blade loading further. Increasing rotor speed, if there is any available, also helps by raising the tip speed and lowering $C_T$.

A warning that follows straight from the previous page. Because $C_T/\sigma$ is built from $C_T$, it inherits the convention trap entirely. A thrust coefficient taken from a source that includes the factor of one half will produce a blade loading exactly twice what it should be, which will make a perfectly healthy rotor look as though it is in stall. Check the convention before you draw a conclusion, and check the solidity definition too — geometric or thrust-weighted — because that moves the answer by several percent as well.