Blade Angle and Tip Mach on a Turboprop
A twin turboprop with 3.93 m diameter propellers cruises at FL250, TAS 280 kt, propeller speed 1 200 rpm. The AFM quotes a propeller efficiency of 0.84 at this condition. For a long-range cruise trial the crew is asked to consider 1 000 rpm at the same TAS and altitude, to reduce tip Mach number and cabin noise. OAT at FL250 is −35 °C. On the subsequent take-off at sea level the aircraft is rotated at VR with maximum take-off power, and the pilot notes a strong yaw to the left needing substantial right rudder; the yaw increases noticeably as the nose is raised. The first officer suggests that a small rudder trim bias will solve the problem for all take-offs.
1. Compute the advance ratio and the blade angle at 75% radius at 1 200 rpm and at 1 000 rpm for 280 kt TAS. What must the constant-speed unit do, and what limits it?
Read the worked answer
280 kt = 144 m/s; R = 1.965 m, r75 = 1.474 m.
At 1 200 rpm (n = 20 rev/s): J = V/(nD) = 144/(20 × 3.93) = 1.83 ωr75 = 2π × 20 × 1.474 = 185 m/s β75 = arctan(V/ωr75) = arctan(144/185) = arctan 0.777 ≈ 37.9°
At 1 000 rpm (n = 16.67 rev/s): J = 144/(16.67 × 3.93) = 2.20 ωr75 = 2π × 16.67 × 1.474 = 154 m/s β75 = arctan(144/154) = arctan 0.933 ≈ 43.0°
The blade element sees the relative airflow at the helix angle φ = arctan(V/ωr) and produces lift at α = β − φ. When RPM falls, the rotational component shrinks, φ rises, and the blade angle must rise by about 5° to hold the same α and therefore the same thrust. That is exactly the function of the constant-speed unit: it increases pitch as RPM tries to rise and decreases it as RPM falls, holding the selected RPM.
The limits are the fine-pitch (ground) stop and the coarse-pitch or feathering stop, plus the torque limit. If the unit runs out of coarse pitch the blade angle of attack falls, thrust and efficiency drop, and the propeller may run away towards the fine stop - the opposite of what was intended.
2. Compute the helical tip Mach number at both settings using −35 °C at FL250 and comment on the trade-off the crew is being asked to accept.
Read the worked answer
a = sqrt(γRT) with T = 273 − 35 = 238 K: a = sqrt(1.4 × 287 × 238) ≈ 309 m/s.
1 200 rpm: ωR = 2π × 20 × 1.965 = 247 m/s Helical tip speed = sqrt(144² + 247²) = sqrt(20 700 + 61 000) ≈ 286 m/s → Mtip ≈ 0.93
1 000 rpm: ωR = 2π × 16.67 × 1.965 = 206 m/s Helical tip speed = sqrt(144² + 206²) = sqrt(20 700 + 42 400) ≈ 251 m/s → Mtip ≈ 0.81
The reduction removes about 0.12 from tip Mach. Above roughly M0.9 the tip sections begin to shock, with a sharp rise in profile drag and noise and a fall in efficiency; at M0.81 the tips are comfortably subcritical. That is the gain.
The cost is a 5° increase in blade angle and, for the same power, a higher torque at lower RPM, which the gearbox and the engine torque limit may not accept at high power. Efficiency also depends on where the design point sits: with J rising from 1.83 to 2.20 the propeller moves along its efficiency curve and may pass the peak. The judgement is not 'lower RPM is better' but 'compare the AFM efficiency and torque charts at the actual TAS, power and altitude, and stay clear of the pitch stops and the torque limit'.
3. Explain the yaw on rotation, why it increases as the nose is raised, and evaluate the proposal to fix it with rudder trim.
Read the worked answer
This is the asymmetric blade effect, or P-factor. When the propeller axis is at a positive angle of attack to the relative airflow, the descending blade on one side meets the air at a larger angle of attack and a higher dynamic pressure than the ascending blade, so it produces more thrust. For a clockwise propeller seen from the cockpit the descending blade is on the right, so the extra thrust is on the right and the yaw is to the left. The effect is largest at high power, low forward speed and a high nose-up attitude - precisely VR at maximum take-off power.
Raising the nose increases the disc's angle of attack, so the asymmetry grows: that is why the yaw increased with pitch attitude. Three other effects add to it simultaneously: torque reaction (roll to the left for a clockwise propeller), slipstream swirl striking the fin on one side, and, on a twin, the different moment arms of the two engines - which is why the critical engine is the one whose P-factor arm is longer.
Rudder trim is not an acceptable fix. The asymmetry varies continuously with power, speed and attitude, while a trim tab is a fixed setting; trimming for the take-off would leave a large residual yaw at climb and cruise, and could mask a genuine directional control problem. The correct management is procedural: apply the rudder needed to hold the centreline, rotate smoothly at VR, and respect VMC. If directional control cannot be held with full rudder before VMC, the take-off is rejected.