Continuing on with the examination of the way the thrust varies with speed, the next thing that is required is to understand why the thrust goes down as the velocity goes up. For all intents and purpose, each section of the blade, along the diameter is inclined at some angle to the axis of rotation (what you measure as "pitch") and the rotation of the engine provides some local velocity. For example, a 6" pitch prop at 8500 RPM looks like this:

The velocity at any point along the diameter is the angular velocity, omega, usually expressed in radians/second, times the radius from the center of rotation to the you are interested in (feet, in this case, to get the result in feet/second). To get from RPM to radians/second, you divide by 60 seconds/minute and multiply by 2pi radians/revolution.

At 8500 RPM omega = ~ 889 radians/second, so

r = 2" v=~149 feet/second

4" ~298

6" ~447

At 11000 RPM, omega = ~1152 rad/second

2" ~192

4" ~384

6" ~576

Looking at what is happening at each point, the prop is being driven "in plane" at the velocity omega*r at each point r. It is an airfoil, it has an angle of attack, that to first approximation, at zero forward velocity, is just the alpha angle of the blade at that point. So:

So, 8500 RPM, 6" of pitch, at zero velocity on the ground and various points on the prop see:

r = 2" V = 149 ft/sec AoA = 25.5 degrees

4" 298 13.4

6" 447 9.05 Of course, this is an airfoil with a substantial AoA and a high velocity, so it creates lift. I would note that for this prop, the 2" station and probably the 4" station are *stalled*, since most airfoils stall at around 11-12 degrees. More on that later, but for now, it is interesting to see what happens as the airplane is released, and begins to accelerate. At that point, the various points on the prop see the vector sum of the prop tangential velocity as shown above, and the forward velocity of the airplane. For our current purposes, assume that the airflow is parallel to the engine shaft*, so the forward velocity is perpendicular to the prop rotational motion. Graphically, it looks like this for 40 feet/second (about half level flight speed):

The way it works is that you add the 40 feet/second in the "vertical" direction to the 149 feet/second at r=2" in the horizontal direction, to get a net velocity of V

_{rel}= sqrt(40

^{2} +149

^{2}) = 154 fps, at an angle of arctangent(40/149) = ~15 degrees from the horizontal. The blade sees an AoA = 25.5-15, or ~10.5 degrees. Of course the angle made by 40 FPS goes down towards the tips as the tangential velocity increases - but so does the blade alpha angle. That's *why* it goes down towards the tips instead of being a constant angle. Note that all of them are now out of the stall region.

Note that this is the *primary* effect causing speed stability - the AoA of the blades goes down with forward velocity, lower AoA = less lift = less thrust. It's just a rotating wing. This, and the fact that different parts of the blade needed different alpha angles was one of the breakthroughs the Wright Brothers made. Note that we are coming close to being able to approximate the thrust analytically, at least well enough to start seeing qualitative effects of various pitch/rpm combinations. Which is what Howard and I were referring to early in this thread.

Of some interest, the pitch/rpm combination I picked was not quite arbitrary, although it was rounded off. See what happens when you get to 80 FPS with this combination:

For all intents and purposes, and considering round-off errors, the AoA is effectively *0* or slightly negative for all positions. This is not a complete accident, it is an inevitable consequence of the definition of pitch and the way we measure it. 6" of pitch is 0.5 feet, 8500 rpm is ~141 revolutions/second. 141*0.5 = 70.5 feet/second, not far off of 80 feet/second.

If the theory behind measuring the pitch (the angle of the back of the blade) was in some way valid, you would reach 0 AoA at 70.5 fps, and the prop would not be "slipping" at all. But, it's not really like that, because the blade is cambered and has lift even at zero angle of attack. But it is why, if you take the marked pitch and somehow figure out the in-flight RPM, the airplane seems to be going "much faster" than it is supposed to. It has to be generating significant thrust in level flight (maybe 2-ish lbs of thrust for a typical 40-60-sized stunt plane), because that's what is overcoming the drag,

More later still...

Brett

*When the forward velocity of the airplane *is not* perpendicular to the plane of prop rotation, the triangles for the vector sums are not right triangles, they follow the law of sines:

https://www.mathsisfun.com/algebra/trig-sine-law.htmland the blade AoA changes cyclicly during the rotation. This is what causes P-factor torques, because one blade has a different AoA than another.