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I can't seem to understand this sentence from the FAA PHAK about why propellers are "twisted."

"If the blades had the same geometric pitch throughout their lengths, at cruise speed the portions near the hub could have negative angles of attack while the propeller tips would be stalled"

From what I understand, propeller blades are twisted to have a higher blade angle/pitch angle/AOA at the hub because the central part moves more slowly than the ends. Slower means less "airspeed" means it needs higher AOA/pitch angle to generate the same lift as the tip of the propeller. That makes sense to me.

On the image below, higher airspeeds (like cruise), would decrease AOA of a propeller since the relative wind would be closer to the chord line. If the airspeed is fast enough I see how there could be negative lift at the hub.

At the tips of the propeller though, why would there be a stall? Stalling means the "wing"/propeller reached it's stalling AOA, so that means AOA at the tip must be large. The only way to really increase AOA of a propeller is to increase airflow due to propeller spinning relative to airflow of plane (airspeed). Even so, if the plane is moving forward, why would the tips of the propeller be past critical AOA and stall? Are they saying that at cruise, the propeller is spinning so fast at the tip that despite the increased airflow from airspeed, the AOA increases so much that it would stall? How does this happen unless the plane was moving backwards?

Propeller AOA

I'd appreciate any input!

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    $\begingroup$ Have you taken into account that the "rotational velocity" linearly increases from root to tip? It is 0 at the hub and $\omega R$ at the tip. Btw, I'd have called it "velocity due to rotation" $\endgroup$
    – sophit
    Apr 18 at 7:36

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The small magnitude of velocity at the root does not dictate whether the section there is stalled or not.

The local angle of attack determines whether the flow is stalled. The magnitude of velocity at the root helps determine the local angle of attack.

The FAA document is trying to represent a standard diagram from the theory of the blade element method -- but they've added so much color fade and un-related stuff that it actually makes it more difficult to understand in my opinion.

Here is another example of the same diagram -- search for 'blade element theory' to find more.

enter image description here

Unfortunately, this image is rotated and flipped relative to the FAA one, so I've manipulated it to be in the same orientation here.... This makes the labels read like a mirror, but hopefully you can use this to convince yourself that it is depicting the same thing as the FAA version, then we can work with the one where the labels are readable.

enter image description here

It is helpful to talk about velocity around a propeller in three components. Instead of x,y,z components, we will use cylindrical components. Think of a cylinder whose axis is aligned with the propeller axis and the direction of flight. The circular outside of the cylinder corresponds with the tips of the spinning propellers. We will call the three components of velocity:

Axial -- along the cylinder axis, parallel to the axis of rotation of the prop, aligned with the direction of flight.

Radial -- along the cylinder radius, from the root of the blade towards the tip of the blade.

Tangential -- tangent to the motion of the blade, this direction follows a curved path, like the grooves on a record.

This diagram is drawn in the Axial-Tangential plane. The radial direction would be coming out of the paper and is not depicted.

The FAA diagram also ignores the $w_a$ and $w_t$ contributions (in cyan). We can ignore them for now too.

The forward velocity of the aircraft (airspeed) is $V_\infty$. This is the primary axial component of the velocity.

The rotational speed of the propeller is $\omega$. It gets multiplied by the radius $r$ to give the tangential component of velocity $u=\omega r$.

Near the center ($r$ is small), the tangential component is small. Near the tip, ($r$ is large), the tangential component is large.

Together, $V_\infty$ and $u$ add together through vector addition to form $w$, the velocity each blade section 'sees'.

Although every blade section sees the same $V_\infty$, the variation of $r$ from root to tip means that the magnitude and angle that $w$ makes with the plane of rotation varies from very steep at the root (where $r$ is small) to very shallow at the tip (where $r$ is large). This angle is labeled $\phi$ (it is not given a label in the FAA diagram).

The geometric blade pitch (measured relative to the plane of rotation) is $\beta$.

If we subtract the local wind angle $\phi$ from the blade pitch $\beta$, we get the local angle of attack $\alpha=\beta-\phi$.

So now we can tell that the local angle of attack of a blade section depends on the flight condition ($V_\infty$ and $\omega$) and also on the blade twist $\beta$.

We know that $\phi$ will be large at the root and small at the tip.

If $\alpha$ gets too big, the blade will stall.

The FAA mentioned an un-twisted blade (like a ceiling fan blade). In such a blade, the twist will be constant $\beta=Constant$. Lets assume $10^\circ$. In that case, considering the variation of $\phi$, we are forced to have $\alpha=10-\phi$. At the root, $\phi$ is large, so $\alpha$ might go negative. At the tip, $phi$ is small, so $\alpha$ might be large enough to stall the airfoil.

One idea for designing a propeller might be to seek to achieve constant $\alpha$ (real props are more sophisticated than this, but this is good to the first order). So if you want to achieve constant $\alpha$ (say $3^\circ$), and we know $\alpha=\beta-\phi$, then we will need $\beta$ to follow $\phi$ in behavior -- large at the root, small at the tip.

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  • $\begingroup$ Thank you! That is very clearly put. So is what I'm missing this: FAA document is trying to make a point, and I am taking it too literally. I was wondering if there were some forces that I missed or weren't discussed to cause this stall at the propeller tips. In an un-twisted blade, the propeller tip could stall. If the aircraft is not moving backwards, even if r is very large, ϕ would be very small but still non-negative. So unless the twist (β) is so large that it's past the critical angle of attack where the stall happens to begin with, the tip wouldn't stall? $\endgroup$ Apr 18 at 19:11
  • $\begingroup$ That is correct. If you work with a single value of twist, $\beta$ is probably quoted at 75% of the blade radius. That value would be small. However, at a blade root, $\beta$ will certainly be higher than the stall angle of attack. I believe the FAA document is trying to make a point -- without twist, propeller performance will suffer either at the root or tip (or both). By twisting the blade, you can try to get the most out of the entire blade. $\endgroup$ Apr 18 at 19:24

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