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Imagine we have a plane that is already moving at a speed $v_{plane}$. At a certain time $t=0$, a motor starts moving a propeller whose blades consist on symmetrical airfoils with $90^\circ$ of pitching (that is, their chords are perfectly parallel to the fuselage of the plane). Under which conditions (if any) would this propeller produce thrust?

Intuitively, one may think that, with $90^\circ$ of pitching, this will just be impossible. Note however that we start from the condition that the airplane is already moving.

Imagine that the rotating speed of the propeller was $\omega_{prop}$. If we analysed the forces generated by a blade in a cross section at radius $R$, they should look like the following (blue for the velocity of air and red for forces produced by the airfoil):

Forces

Since the plane is moving and the propeller is rotating, from the perspective of the blade the incoming air will have a nonzero apparent angle of attack. This will produce lift and, if $\frac{c_L}{c_D}$ is sufficiently large, the direction of the resulting force vector ($F_{TOTAL}$ in the diagram) will point slightly to the direction of movement, hence generating thrust.

Is this correct? If so, would a propeller with long blades spinning sufficiently fast generate thrust under these constraints?

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The horizontal component of the total aerodynamic force $F_{TOTAL}$ would indeed be a thrust.

Anyway the much bigger vertical component would be the force that had to be won by the torque of the engine. This design wouldn't be efficient: a lot of torque would be needed to produce just a small thrust. This is why real propellers have the airfoils almost perpendicular to your design and rotate with a speed much higher than the speed of flight.

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  • $\begingroup$ But the AOA would be negative, producing reverse thrust, right? $\endgroup$
    – Michael Hall
    Jul 7 at 18:10
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    $\begingroup$ For how it is drawn in the picture, the AoA is positive since the blade is moving forward ($V_{plane}$) and downward ($\omega_{prop}$). On the opposite side $\omega_{prop}$ would be the other way around so $F_{total}$ would point downward and slightly forward $\rightarrow$ still a lot of torque and few thrust. $\endgroup$
    – sophit
    Jul 7 at 18:46
  • $\begingroup$ Disregard, I was 90 degrees off in my thinking, presuming flat pitch... $\endgroup$
    – Michael Hall
    Jul 8 at 1:30
  • $\begingroup$ @MichaelHall Yes, for blades set at 0 deg pitch (i.e perpendicular to the airflow) the thrust is reversed. This is how thrust breaking is accomplished with the G-1. $\endgroup$
    – Thomas Perry
    Jul 8 at 3:37

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