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Author Topic: Effect of wind on maneuvers  (Read 3704 times)

Offline Chuck_Smith

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Re: Effect of wind on maneuvers
« Reply #50 on: November 23, 2019, 05:41:47 AM »
But as I think about this thread I must have had some old synapse fire off - maybe there is a fundamental advantage to electric power for stunt speed control.

On electric, the back emf reduces the torque as the rpm increases. That's a big deal and we can use it to our advantage.

Chuck
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Online Brett Buck

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Re: Effect of wind on maneuvers
« Reply #51 on: November 23, 2019, 12:22:43 PM »
But as I think about this thread I must have had some old synapse fire off - maybe there is a fundamental advantage to electric power for stunt speed control.

On electric, the back emf reduces the torque as the rpm increases. That's a big deal and we can use it to our advantage.

   There's a lot more left to come on this thread. But as an aside, it's perfectly clear to me that even just having a governor (with whatever response it has) that keeps the engine RPM up (or quickly recovers it),  regardless of load in the corner,  is *far better* at controlling short-term airspeed variations that any combination of high pitch/low rpm, that otherwise allows the RPM to drop and stay dropped when the speed is depressed. That's why electric system with pitches we would consider high, like 6", seem to have better short-term airspeed response than anything we have come up with pitch <4".

   I also strongly suspect that this is the a large part of the phenomenon behind the far-forward CGs on electric - when the airspeed is better-controlled, you can tolerate more wing loading, but you also have more control effectiveness. The same sort of thing happens with IC engines, it's just happening to a much greater degree. It's no accident that Paul Walker's 40VF setup, with the teeny prop and lots of engine response, also wound up with the CG so far aft, and that adding diameter and reducing the pitch  - which should cause less speed depression - also permits you to handle the CG further forward.

     Accelerometer feedback systems like Igor's add a completely new factor - everything to date operated almost entirely on applied load, which is caused almost entirely by airspeed variation causing load variation, and the engine/prop response to it. Adding the accelerometer makes it additional responsive to the *inertial* acceleration and velocity, not just the apparent airspeed. This comes a lot closer to addressing the real issue, since we also care about the speed relative to an inertial reference - the ground.

     Brett

 

Offline phil c

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Re: Effect of wind on maneuvers
« Reply #52 on: November 24, 2019, 07:36:20 PM »
A plane with a fair wind- 15mph+ doesn't need any prop to fly, albeit mostly on the downwind side.  Even light winds add significant energy at various point in a maneuver where the wing is at an angle to the wind.   The wind can speed up loops and eights particularly.  You can blow the wings off a plane doing large round loops downwind  unless you run out of elevator to keep the loops round at the bottom.

It shows up more in places such as Texas and the other plains states where it' often windy with steady winds.

I can see where a lower pitch prop would help reduce the effect because it has less load in the first place.
phil Cartier

Online Brett Buck

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Re: Effect of wind on maneuvers
« Reply #53 on: November 24, 2019, 07:38:29 PM »
I can see where a lower pitch prop would help reduce the effect because it has less load in the first place.

    It's actually *more* load, in that you have to overcome much more parasitic drag at 12,000 than you do at 8000. The shaft HP required for a low-pitch prop is *much* higher.

    Brett

Online Brett Buck

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Prop angle of attack at various speeds
« Reply #54 on: December 01, 2019, 03:38:42 PM »
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 Vrel= sqrt(402 +1492) = 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.html

and the blade AoA changes cyclicly during the rotation. This is what causes P-factor torques, because one blade has a different AoA than another.
   







« Last Edit: December 01, 2019, 04:45:34 PM by Brett Buck »

Offline RandySmith

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Re: Effect of wind on maneuvers
« Reply #55 on: December 02, 2019, 08:04:55 PM »
Hello,
A single propeller stunt plane has a tendency to speed up in maneuvers when it is windy. 

The plane with the counter-rotating propellers does not show this behavior or shows very little of it.

Why?

Thank you,
M

There is a difference, example same plane, 2 blade prop, speeds up a little more with 2 blade than it does, with a 3 blade, or a  4 blade, or a conterrotaing 4 blade,  i have seen this personally.  however there is so much more to it.  the  same plane  with several different 2 blade props , ones can wind up more than others, Like the wide blade 2 blade  will wind up more than the higher aspect ratio 2 blade,  Same  with  the 3 or 4 blade.  Under cambered  props can wind up  much more,
I had a magnum  with a  conterrotaing  prop, and it was at it best for  not winding up with that setup. worse  with the  2 blade setup.
don't read  everything into this though, using the number of blades

Randy

Offline Peter Germann

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Re: Effect of wind on maneuvers
« Reply #56 on: December 03, 2019, 02:47:48 AM »
When recently talking to a very successful electric pylon racer he mentioned a very significant brake effect when when diving. As he states the effect being much more when using a geared-down motor, as opposed to an outrunner, could it be the reverse-driven gearbox influences braking on counter-rotating electrics?
Peter Germann


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