P Factor?

Student pilots, and their patient and brave instructors, are well acquainted with the empirical evidence that small single-engine prop-driven airplanes have a demonstrated left-turning tendency in certain phases of flight, particularly when flying slowly under high engine power, such as when climbing steeply or in the takeoff roll.

The general wisdom is that what we see as a left-turning tendency is, in fact, the combined effects from multiple, independent sources. These effects are well-described in various places, a particularly good one is here: http://www.av8n.com/how/htm/yaw.html#8-5

There is some disagreement, though, about one of the sources of left-turning tendency, known as "p-factor". The FAA describes P factor by pointing out that when the airplane is flying with a nose-high attitude, then when the prop blade is descending (on the right side of the airplane, on US-built engines, as viewed from the cockpit), it is also moving forward, as our nose-high attitude has the prop spinning in a plane that's tilted up towards the sky a bit, relative to the horizon. And when the prop blade is ascending up the left-side, it's moving backwards slightly. The FAA talks of the blade on the right side as getting a bigger "bite" of the air than the left-side, and thus generating more "lift" (which translates to "thrust"), resulting in a "pull" of the nose to the left.

So is this a correct characterization? Do we really get a bigger pull from the right side of the prop? If so, is it because of a greater angle of attack seen at the prop blades? Then again, isn't the descending blade also heading slightly into the relative wind, while the ascending blade is heading slightly away from it, leading to a higher airspeed seen at the prop blade, thus generating greater lift?

So let's take a shot at calculating the effect, maybe if we creep up on it step by step, it won't see us coming!

Imagining the relative wind of an airfoil that's spinning while moving forward while tilted upwards with the blades twisted to a chosen pitch, all at the same time, can stretch our brains a bit. So, for the most part, we'll try to leave our prop pitch to a few choice places near the ends of our arguments, and focus instead of the relative wind as the prop blade's airfoil sees it. Once we know the speed and direction of the relative wind, it's fairly simple to compare that to the blade's pitch to determine angle of attack.

For this experiment, it may help to forget about the shape of the prop blade initially, as that has no impact on relative wind, and just throws in an additional detail to make everything harder to visualize. Just think of it our prop blade as a wire, or a very thin, round, metal "dowel", being spun through the air. We're going to choose a particular spot out along this conceptually shapeless prop, one which we'll call, affectionately, "spot".

Note that the following text uses the term "plane" in the geometric sense (e.g. the "plane" in which the prop spins), whereas "airplane" refers to the thing we fly in.

Sit and spin

First, let's imagine a propeller spinning in a plane that's perfectly straight "up and down", i.e. a plane that's perpendicular to the surface of the earth. We might imagine the airplane being in this attitude at the beginning of the takeoff roll, if the airplane's attitude on the ground had the nose just right so that the prop's plane was perfectly vertical. Let's also imagine that there is no relative wind, e.g. the airplane is just starting its takeoff roll, but we haven't released the brakes yet, and there's no wind. (Note that this is an idealized model, as a prop will necessarily be pulling some air through itself).

We'll use feet-per-second (f/s) for our units of velocity. If our prop is spinning at E (for engine) rpm, and we imagine a spot on our prop that's 'r' (for radius) feet from the hub, then that spot on the prop is traveling in a circle at:

v = (2 * π * r) * (E / 60)

The "2 * π * r" is just the equation for the circumference of a circle, and the (E / 60) just converts our rpm into "revolutions per second".

As an example, let's place "spot" at a point 2 ft from the center of the prop hub, and spin the prop at 2500 rpm. We get:

v = (2 * π * 2ft) * (2500 rpm / 60) = 523 f/s

So our little spot on the prop is moving in a circle at 523 feet per second, which is about 300kts. As the relative wind seen by the prop is purely a result of the movement of the prop itself, this is the airspeed seen by the prop blade, and the angle of attack depends solely on the prop pitch at that point on the blade:

aoa = prop pitch

Let's get moving...

For our next trick, we're going to keep the prop in a plane perfectly perpendicular to the earth, as before, but now we're going to add in a relative "headwind". We can imagine ourselves still in that same attitude on the ground, but now rolling forward at some speed. Or we can imagine flying through the air in an attitude with the nose perfectly on the horizon. We can even be sitting still on the ramp but facing into a stiff breeze.

Either way, let's imagine we're now seeing a headwind of A (for "airspeed") feet per second. What is the relative wind that our little spot on the hub sees now?

Well, our little spot on the hub is now moving in three dimensions. It has circular motion in the "plane" of the spinning prop, and it has "forward" motion through the oncoming "headwind" air. The net motion of these two effects, summed together, has our little spot on the prop tracing out something like a corkscrew pattern through the air. Let's consider the motion of our spot at two instants of time, when the blade is at the 9 o'oclock and 3 o'clock positions.

So with the prop blade at 3 o'clock (from the perspective of the pilot looking out from the cockpit, i.e. with the prop blade going down), our little spot has downwards vertical motion as it did before:

(2 * π * r) * (E / 60)

We'll rewrite this more succintly as:

πrE / 30

As our prop blade is moving downwards, we're seeing this much relative wind coming upwards at our spot.

But "spot" is also moving forward at A f/s. So to determine the total relative wind seen by our spot, we have to add together these two vectors which are oriented at a right angle to each other, and determine both the magnitude and direction of the resultant.

The magnitude is calculated as:

sqrt((πrE / 30)^2 + A^2)

Our spot is moving downwards and slightly forwards. So it is seeing a relative wind that's now tilted slightly "forwards" from the straight down direction. We'll call this angle the relative wind makes relative to "straight down" α, we can calculate it like so:

α = arctan(A / (πrE / 30))

Let's plug in some numbers again. We pick a new number of 200 feet per second as the speed at which we're moving forward (roughly 120 kts), and use our old prop numbers from above (2500 rpm, spot 2 ft from hub). Plugging these in we get:

v = 560 f/s
α = 20.8°

So our friend "spot" is seeing a relative wind that's about 20.8 degrees up from "straight down". If we imagine ourselves sitting on top of the prop at the 3 o'clock position, we were previously feeling the relative wind coming straight up at us from below, hitting the bottoms of our feet, but now we're feeling it coming from a direction 20.8 degrees forward of that, still coming up from below, but sort of hitting us in our shins, now.

So what effect resulted from adding in our 120kt headwind? Well, the airspeed seen by "spot" has increased from 523f/s to 560f/s, and it's now shifted forward 20.8 degrees, from the "straight up" (as prop goes down) direction. Whereas we were previously seeing an angle of attack equal to the blade pitch, the 120kts headwind has reduced our angle of attack by 20.8 degrees, so we now have:

aoa = prop pitch - 20.8°

(I found it surprising that the loss of angle of attack is as high as 20+ degrees, but I've read that constant-speed props vary in blade pitch by over 30 degrees. It's certainly illuminating as to the compromises inherent in a fixed-pitch prop).

So what's doing over at the 9 o'clock position? Well, our prop is still spinning in a plane that's perfectly perpendicular to our oncoming 120kt headwind, so "spot" sees the same effects on the upward swing as he sees on the downward swing, with the only difference being that at the 9 o'clock position, up and down are swapped, so the relative wind now comes from an angle shifted forward 21 degrees from "coming straight down". If we imagine ourselves sitting on top of the prop at the 9 o'clock position, we were previously feeling the relative wind coming straight down onto the tops of our heads, but now we're feeling it coming from a direction 21 degrees forward, sort of hitting us in the fore-head now. As far as airspeed and angle of attack are concerned, the blades see everything the same on both sides of the prop (i.e. at both 3 o'clock and 9 o'clock positions).

Here comes the pitch...

Well, so far, we haven't managed to demonstrate any P-factor, everything's been symmetrical so far, but our luck is about to run out. Let's continue our perfectly horizontal direction of flight, but let's add in a nose-up pitch attitude, so that the plane in which our prop is spinning is now tilted up a little bit. The "top" of the prop plane moves back, the bottom moves forward, as the prop tilts up a few degrees to better face the sky). We'll tilt our prop backwards ρ degrees (cause it kinda looks like a 'p' for propeller).

How fast is spot moving now, and in what direction?

Well, spot's motion from the spinning action of the prop is no longer perpendicular to the airplane's motion through the air. Since our vector arithmetic is easier when we're adding up components along perpendicular axes, we're going to take the motion resulting solely from "prop spin" and break it up into vertical and horizontal components.

First, let's check out the motion due to prop spin alone, at the 3 o'clock position. As before, our motion due to prop spin is

πrE / 30

but this is now slanted ρ degrees from the vertical. That yields a vertical (i.e. straight downwards) component of:

cos(ρ) * πrE / 30

and a horizontal (i.e. straight forwards) component of:

sin(ρ) * πrE / 30

(As a sanity check, let's note that when ρ = 0, we have cos(ρ) = 1, and sin(ρ) = 0, so the horizontal component is zero, and the vertical component is πrE / 30, and everything is back as we had it above, before we tilted our prop, as we'd expect.)

Now we need to add our relative wind from our forward motion back in. So in the horizontal direction, spot now sees:

sin(ρ) * πrE / 30 + A

and in the vertical direction, spot now sees:

cos(ρ) * πrE / 30

Let's take a deep breath, and determine the total airspeed that spot's seeing. Adding (vector-wise) our components together, we get:

v = sqrt((sin(ρ) * πrE / 30 + A) ^ 2 + (cos(ρ) * πrE / 30) ^ 2)

Woof! As another diversion slash sanity check, let's consider what the above equation yields when ρ = 90° , i.e. imagining the prop tilted all the way backwards until it's oriented horizontally like a helicopter. sin(90) = 1 and cos(90) = 0, so we get:

v = πrE / 30 + A ;

Note that this πrE / 30 is what we calculated as the speed that spot gets solely from prop spin, and A is our airspeed. This gives us the result we'd expect from the 3 o'clock blade position (moving forward) of a "conceptual model" of a helicopter with a perfectly horizontal blade spinning on top, yet with the helicopter (somehow) moving forward through the air at A f/s.

Getting back to our airplane, what's the direction of the relative wind as seen by spot? That's:

α = arctan(((sin(ρ) * πrE / 30) + A) / (cos(ρ) * πrE / 30))

So let's plug in some numbers again, and see if we've broken the space-time continuum yet. We'll choose a value for ρ of 10 degrees. Cranking it all through, we get:

v = 592 f/s
α = 29.4 °

As we recall, before we tilted our prop, spot was seeing a relative wind of 560 f/s from an angle 20.8 degrees forward of vertical. With the prop tilt, spot is now seeing 592 f/s from an angle 29.4 degrees forward of vertical. Of course, tilting the prop plane by 10 degrees also tilts the blade by 10 degrees, so while the wind is now coming from 29.4 degrees forward (as opposed to 20.8 before), the prop blades are meeting the air with a prop "attitude" that's now twisted an extra 10 degrees forward. The net effect is that whereas our initial 100kts headwind took away 20.8 degrees of angle of attack, with our prop tilt that drops down to 29.4 - 10 = 19.4 degrees of angle of attack "lost". In other words, the angle of attack seen by the prop in the neighborhood of spot while at the 3 o'clock position is:

aoa = prop pitch - (29.4 ° - ρ) = prop pitch - 19.4°

What's happening at 9 o'clock?

Let's now take a look at what spot's up to in the 9 o'clock position, where the prop blades are moving upwards, and slightly backwards.

That small horizontal component of spot's motion that's derived solely from "prop spin" is negative now that we're over at 9 o'oclock, as the prop "trails" somewhat at this position given our nose-up tilt of the prop plane. So adding this to our "headwind" yields spot's total horizontal motion:

-sin(ρ) * πrE / 30 + A

And our vertical motion is the same as before (i.e. as it was at 3 o'clock):

cos(ρ) * πrE / 30

Except it's now up, instead of down, of course.

Recalculating our total resultant airspeed and direction, we get:

v = sqrt((-sin(ρ) * πrE / 30 + A) ^ 2 + (cos(ρ) * πrE / 30) ^ 2)
α = arctan(((-sin(ρ) * πrE / 30) + A) / (cos(ρ) * πrE / 30))

Plugging in our same numbers from before, we get:

v = 527f/s
α = 11.9°

So "spot" is seeing a relative wind that's 11.9 degrees "forward of vertical". By tilting back the plane that the prop is spinning in, we've tilted the prop blades "backwards" by 10 degrees from the vertical, so the angle of attack seen by the prop blades at spot's location while in the 9 o'clock position is actually:

aoa = prop pitch - (11.9° + ρ) = prop pitch - 21.9°

Relatively long winded...

So given our sample numbers of 120kts forward airspeed, 2500 rpm, 10 degrees of nose-up pitch attitude, and a spot 2 feet out along the prop, we calculated that when compared to the upward-moving prop blade, the downward-moving prop blade had about 2.5° greater angle of attack (i.e. 21.9° - 19.4° = 2.5°) as well as about 65f/s greater airspeed, both of which would lead to greater lift on the right side.

How much of a difference is this? Well, the extra airspeed came out to about 12%. Since the lift equation states that lift increases with the square of airspeed, a 12% increase in airspeed is expected to produce about a 25% increase in lift force (1.12 * 1.12 = 1.25). Note, of course, that this value is only relevant for this particular set of parameters; in particular, it will vary at different positions along the prop's radius.

The 2.5° difference in angle of attack would also depend on the prop pitch at "spot's" position; without having knowledge of a particular prop's pitch at this position, I don't have a good way to determine this as a percentage, and thus no way to translate this into big vs. small difference in lift. So ultimately, I don't have enough "number stuff" at hand to compare the contribution to asymmetrical thrust that results from the differences in angle of attack, as opposed to differences in airspeed, as seen by the prop blades.

Note that this entire exercise is the fairly isolated effort of a guy who took a bunch of physics and engineering classes 20 years ago, and remembered some of it, but who is a software engineer by trade. I'm certainly interested in any insight or corrections.

Harry Mantakos / harry@meretrix.com