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At the moment I am struggling with the problem that I do have an idea how to calculate this force by using the blade element theory and going over a whole propeller blade in order to calculate the torque inside the blade. I need this torque in order to calculate the force necessary for changing my pitch inflight. Is the blade element theory the correct way to calculate this?

Forces on a Propeller Blade

Source: Nelson, Wilbur C.: Airplane Propeller Principles (John Wiley & Sons, 1944).

Also I do lack information regarding the airfoil data. Is there a good estimate on how high the coefficient of torque will be? Is there a good airfoil database for propellers online?

The assumption that cm=-1/4*ca does not work, as profiles on propellers are not even plates.

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  • $\begingroup$ depends on the propeller, on the speed you're flying, on the arm available to apply the force (same force, but twice the arm gives twice the torque) $\endgroup$ – Federico Jan 11 '17 at 9:18
  • $\begingroup$ Edited question to clarify Just to get it correct: I am not looking for the torque my engine has to deliver to get the propeller spinning, but I am searching for the torque my blades generate, in order to calculate the force for an electric servo/hydraulic mechanism I need to change my pitch in flight. As this is a variable pitch propeller, the speed I am flying should be of lower importance, since my angle of attack (and thus my coefficient of torque) as well as the rotational speed are way higher. I do hope for a general method, but in my application case it will be a plane with Ma<0.3 $\endgroup$ – Lumis Jan 11 '17 at 18:21
  • $\begingroup$ Per this post, the X-Plane flight simulator uses blade element theory on the prop disk, which gives the drag and therefore torque. $\endgroup$ – fooot Jan 11 '17 at 19:05
  • $\begingroup$ This doesn't answer the question, but this link provides great info on how a Constant Speed Propeller and it's Governor work. I don't know how much engine oil pressure is needed to change the pitch, or what the formula might be for 2,3,4 blae propellors.<boldmethod.com/learn-to-fly/aircraft-systems/…> $\endgroup$ – CrossRoads Apr 1 '18 at 15:28
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In a well-designed propeller the hinge axis is very close to the line connecting the centers of pressure, so the aerodynamic pitch moment is rather small. The main force is friction; note that the centrifugal pull of the blades at the hub will put a considerable load on the blade bearings, so friction goes up with rotation speed. Regular propellers have plain bearings; if you know the mass of the blade, the rotation speed and the coefficient of friction, you can calculate the breakaway torque of one blade.

Since propeller blades have some (but not much) camber, the center of pressure moves forward with increasing angle of attack. This will drive the blade away from a medium angle of attack; in other words, the aerodynamic moment is destabilizing the propeller pitch.

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  • $\begingroup$ Paging @PeterKämpf to the chat room, paging Peter Kämpf to the white courtesy chat room! $\endgroup$ – FreeMan Jan 11 '17 at 21:57
  • $\begingroup$ Did I get something wrong with aerodynamics? With increasing angle of attack the coefficient of torque (cm) should decrease, not increase (assuming we are in the linear area), as you stated. Thus I should get a stabilizing moment for higher angles of attack and not a destabilizing one. $\endgroup$ – Lumis Jan 12 '17 at 0:12
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    $\begingroup$ @Lumis: That entirely depends on the axis around which you measure the torque. If you use the quarter chord, the coefficient should ideally be constant. If you place the axis of rotation at roughly one third of chord (what propeller designers normally do), the coefficient has a positive (instable) gradient over AoA. $\endgroup$ – Peter Kämpf Jan 12 '17 at 13:36
  • $\begingroup$ Is there some kind of literature or source with examples on the axis of rotation? Maybe even something I could use for citation? $\endgroup$ – Lumis Apr 22 '17 at 23:19
  • $\begingroup$ @Lumis: The root of a variable pitch propeller blade is cylindrical, and the center of this cylinder is the axis of rotation. Maybe google.com/patents/US2496169 will do. $\endgroup$ – Peter Kämpf Apr 23 '17 at 5:10

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