I know that an increase in velocity (TAS) reduces the AOA and that makes sense. But how just simply increasing the blade rotation (RPM) results in higher AOA?

  • $\begingroup$ Following through on the thought, increasing rpm will (eventually) speed the plane up too, so the forward vector grows as well. There for, the increase in prop AOA may only be temporary. Interesting though. $\endgroup$ Jan 21, 2020 at 2:47

1 Answer 1


enter image description here
Prop blades from side view at two RPMs

Vector diagrams help with visualizing prop blades. Above the vertical vector lines are two RPMs, the horizontal vector lines are the forward velocity (the same for both), and the connecting orange vector lines are the resultant airflows. Note the AOA for the same blade pitch and how it is affected by the airflow at the two RPMs.

Note: that's also why slow/windmilling RPMs result in high drag (negative AOA; airflow hitting top side), and why dead engines are feathered if that's an option.

  • $\begingroup$ @T.J.Crowder In vector diagrams, a longer vector represents more of whatever's being measured. In this case, it would be the speed of the blade. In the second diagram, the vector is about twice as long, so it would be representing about twice the speed. $\endgroup$ Jan 20, 2020 at 15:20
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    $\begingroup$ @T.J.Crowder: The diagram is a side view of the propeller, so from that view the prop is moving down (or up on the other side). The faster it's moving down, the higher the RPM. Let me know if that clears it up. $\endgroup$
    – user14897
    Jan 20, 2020 at 15:46
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    $\begingroup$ @T.J.Crowder: You'll note in the right diagram the AOA is bigger, i.e. air hitting underside more, if the engine can keep that fast RPM, it's actually more lift (thrust). The higher the AOA [and airspeed] (unless stalled), the more the lift, like your hand out the car window, the more tilted, the bigger the up force, than if your hand was parallel to the ground. $\endgroup$
    – user14897
    Jan 20, 2020 at 16:02
  • $\begingroup$ @ymb1 - Ah! Got it! $\endgroup$ Jan 20, 2020 at 16:07

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