Stabizer Pressure

[John Leidle]

   Trying to understand the amount of pressure force on the stab during turns.  Wonder if I use a conventional stab setup with it 1/2" above the  wing CL or built the stab inline with the wing  CL would pressure be  greater either way ?
  John L.

[Brent Williams]

   Trying to understand the amount of pressure force on the stab during turns.  Wonder if I use a conventional stab setup with it 1/2" above the  wing CL or built the stab inline with the wing  CL would pressure be  greater either way ?
  John L.

 

"Pressure,"  meaning what exactly? 
This can lead to other important questions that I hope others can comment on.

Effort to use, ie, hinge moment?
Effectiveness?
Ability to stabilize after the turn?
Size of tail for CG toleration?
Location of tail in "clean or dirty" air? Down wash?
Aspect ratio and pressure/effort differences?

[John Leidle]

  During the turn a certain amount of pressure is put on the stab surfaces to  stabilize the aft section similar  to what the elevators are doing.

[Tim Wescott]

I can tell you how to calculated, it, but it's complerkated (and very likely wrong, because I'm not an aerodynemicist).

It's easiest for a flapless plane -- just look at some center of pressure plots, and you quickly realize that the CP of a symmetrical wing stays pretty darned close to 25%.  So in a turn, it's as if you're concentrating the entire weight (not mass) of the airplane at the MAC of the wing.

For a stunter going 25 meters/second (55 mph) on a circle of radius 20 meters (that works out to about 64 feet handle to canopy), the acceleration in a loop is 8g, give or take a bit, plus the acceleration due to gravity.  So at the bottom of the loop it's 9g.

So a 4 pound stunter weighs 36 pounds at the bottom of the loop.  If you assume a 12-inch chord and a CG at 20% of the wing chord, then you're putting a moment of 22 inch-pounds onto the plane that has to be overcome by the stab.  If MAC of the stab is 22 inches back from the MAC of the wing, then the stab + elevator needs to carry one pound of force (wheelward, in an inside loop) to maintain the circular motion (note that I don't know how much of this is carried by just the stab -- I'm already above my pay grade here, and that's more so).

The further forward the CG is, the more load the stab has to carry -- but if I did my ciphering right, it doesn't look like the numbers are too bad.

[Brett Buck]

If MAC of the stab is 22 inches back from the MAC of the wing, then the stab + elevator needs to carry one pound of force (wheelward, in an inside loop) to maintain the circular motion (note that I don't know how much of this is carried by just the stab -- I'm already above my pay grade here, and that's more so).

The further forward the CG is, the more load the stab has to carry -- but if I did my ciphering right, it doesn't look like the numbers are too bad.

   The static case is relatively small. Starting/stopping the corner is where the real load comes. I have some notes somewhere.

    Brett


 p.s. I get something between 6 and 11 lbs, depending on how hard you think you hit the corner. Not trivial, which is why you occasionally see stabs fold.

[Howard Rush]

Trying to understand the amount of pressure force on the stab during turns.  Wonder if I use a conventional stab setup with it 1/2" above the  wing CL or built the stab inline with the wing  CL would pressure be  greater either way ?

"No" is the best answer to your question.  The dynamic pressure (the pressure from the airstream) is reduced a little in the wake of a wing, but when you're flying stunt, sometimes the stab is in the wing wake, sometimes the air comes over the top of the wing and blows on the stab, sometimes it comes up from under the wing.  Depending on where you put the stab, you might see a difference in response turning one way or the other, but that can be compensated for by screwing the pushrod in or out of the elevator link.  Any actual reduction in pressure from the wake can be compensated for by adjusting the control horn length.  The prop blowing on the tail probably makes the dynamic pressure at the tail greater than that of the free stream.  Frank Williams thinks that could explain why low aspect ratio tails are good: more of the stab is in the prop blast. 

Horizontal tails do two things: 1) they provide a force up or down as needed for the size loop you're turning, and 2) they provide stability, which is the rate of change of pitching moment with angle of attack or with pitch rate.  Both these are proportional to dynamic pressure.  They are also proportional to tail size, so if you're worried about the tail being in the wake of the wing in one position or the other, just make it a little bigger.

[Brett Buck]

Wonder if I use a conventional stab setup with it 1/2" above the  wing CL or built the stab inline with the wing  CL would pressure be  greater either way ?

   I think it makes no difference. The load comes from trying to start and stop rotation, depending on how heavy and how big your airplane is, you require a lot of torque to accelerate it to the rate necessary for a corner in a very short time (since the corner only takes about 1/4 second, figure at most .1 seconds to get it up to the maximum pitch rate, less if you are competitive) it takes a lot of torque to do it. The static loads of sustaining a corner is minimal, Tim did a decent guess for a round loop, although it's probably overstated.

    Brett

[Howard Rush]

   I think it makes no difference. The load comes from trying to start and stop rotation, depending on how heavy and how big your airplane is, you require a lot of torque to accelerate it to the rate necessary for a corner in a very short time (since the corner only takes about 1/4 second, figure at most .1 seconds to get it up to the maximum pitch rate, less if you are competitive) it takes a lot of torque to do it. The static loads of sustaining a corner is minimal, Tim did a decent guess for a round loop, although it's probably overstated.

    Brett

I can quantify that stuff when I get my dad gum DataTUT.

[Brett Buck]

I can quantify that stuff when I get my dad gum DataTUT.

   Which presumably includes some sort of MEMS gyro?

     Brett

[Howard Rush]

If I had a MEMS gyro I could do a loop kill.

[Brett Buck]

If I had a MEMS gyro I could do a loop kill.

     Or you could just use a real engine like an adult.

    Maybe we are being too clever. The most relevant unknown variable in computing the lift required from the tail is the angular acceleration, which would presumably be determined by taking the back-difference of the rate from a gyro with sufficiently high sample rate (say 500 Hz).

      Brett

[John Leidle]

   Thanks everyone.
  John L.

[Igor Burger]

I can tell you how to calculated, it, but it's complerkated (and very likely wrong, because I'm not an aerodynemicist).

It's easiest for a flapless plane -- just look at some center of pressure plots, and you quickly realize that the CP of a symmetrical wing stays pretty darned close to 25%.  So in a turn, it's as if you're concentrating the entire weight (not mass) of the airplane at the MAC of the wing.

For a stunter going 25 meters/second (55 mph) on a circle of radius 20 meters (that works out to about 64 feet handle to canopy), the acceleration in a loop is 8g, give or take a bit, plus the acceleration due to gravity.  So at the bottom of the loop it's 9g.

So a 4 pound stunter weighs 36 pounds at the bottom of the loop.  If you assume a 12-inch chord and a CG at 20% of the wing chord, then you're putting a moment of 22 inch-pounds onto the plane that has to be overcome by the stab.  If MAC of the stab is 22 inches back from the MAC of the wing, then the stab + elevator needs to carry one pound of force (wheelward, in an inside loop) to maintain the circular motion (note that I don't know how much of this is carried by just the stab -- I'm already above my pay grade here, and that's more so).

The further forward the CG is, the more load the stab has to carry -- but if I did my ciphering right, it doesn't look like the numbers are too bad.

It is more complex than that, it needs at least CG moment to aerodynamic center as you described, it needs also wing pitching moment minus tail pitching moment, and both of them must be calculated in round stream (what makes wing moment higher and tail moment lower) and plus moment of rotating prop. That all makes for my model and minimal projected corner radius (3.5m) in my program force aproximately 18 N  ... that is IN steady radius force.

For acceleration like Brett wrote I estimated something like 1.5N. I comes from experiment with model supported in CG (very close to AC, but not exactly) with weight on tail, after releasing I got aproximatel the same angular acceleration like gyroscope measured in real flight.

So looks like force for keeping in circular path is 10x higher than acceleration to the corner. However force doable on instantly deflected elevator in straight flight is really hig as Brett wrote, but in reality is not that strong, because hand does not deflect it instanty and until it is deflected, it has already some angular speed and thus elevator cannot make so high force in circular flow. It also means they do not act in the same time and so cannot be added.

[Brett Buck]

It is more complex than that, it needs at least CG moment to aerodynamic center as you described, it needs also wing pitching moment minus tail pitching moment, and both of them must be calculated in round stream (what makes wing moment higher and tail moment lower) and plus moment of rotating prop. That all makes for my model and minimal projected corner radius (3.5m) in my program force aproximately 18 N  ... that is IN steady radius force.

For acceleration like Brett wrote I estimated something like 1.5N. I comes from experiment with model supported in CG (very close to AC, but not exactly) with weight on tail, after releasing I got aproximatel the same angular acceleration like gyroscope measured in real flight.

So looks like force for keeping in circular path is 10x higher than acceleration to the corner. However force doable on instantly deflected elevator in straight flight is really hig as Brett wrote, but in reality is not that strong, because hand does not deflect it instanty and until it is deflected, it has already some angular speed and thus elevator cannot make so high force in circular flow. It also means they do not act in the same time and so cannot be added.

   I want to see Howard's gyro data, that will tell us how hard it accelerates in and out of the corner. I have indirect data - tails being deflected by load - that suggests it's in the range of at least several pounds.

     Brett

[L0U CRANE]

Brett, Howard and Igor-

Sorry it has taken so long, but...

Except for Igor's post late in the initial discussion, I think one idea has been overlooked:

Am I oversimplifying to remind us that the "load" on the horizontal tail surface relates more to the forces at the Aero CPs of the wing and tail than to the total 'g' on the model?  The tail load is what is required to raise the wing AoA to the lift coefficient  needed for the turn, not necessarily that the tail's lift load proportionally equals the load on the wing.

The simple graphic image is that of a shovel. The wing is the scoop end of the shovel; the tail is the handgrip end. The wing bears the major weight - the handgrip end merely 'aims' where the scoop is pointed and holds it there. Other than  the centrifugal (centripetal?) load on its structure what needs the tail support in a square corner?

Of course the force to hold the wing to the required AoA is greater than in rounds or level flight, but the lever arm to the tail CP plays in this picture, too. No?

[Chuck_Smith]

You need to know the moment coefficient of the wing, which changes with flap deflection. Next, use Max Munk's equations to determine the lift and moment created by the fuselage. Determine the inertia tensor for the aircraft. Now, the more difficult part - the apparent mass effects of realigning the flow field around the aircraft. This is non-trivial for large "excursions" as we use in CLPA. With all this in hand determine how fast you want the aircraft to rotate around the cg in he pitch axis. Plug and chug simple Newton in rotational coordinates and you'll get the horizontal tail force required for the rate of AoA change you require. 

Notice when you do the plug and chug that the aircraft's center of pressure is ( hopefully ) behind the cg. That means that lift is fighting the turn at this point. [OK, you may now pick up the pieces of your exploded head]

But here's the tricky part to remember: The airplane doesn't just make a turn. It's a TWO STEP process. First you rotate the aircraft to change the AoA, then the aircraft turns. Lift will help the turn- but it will oppose the rotation. Long story short, for fast maneuvers like our squares  or triangles the rate at which we change our angle of attack is important. For round maneuvers not so much. If you're flying at 40 mph and need to make a sharp radius for a square or triangle - I maintain that the initial rotation required to achieve AoA is the dominate consideration to achieve the desired flight path.

The above is simplified, but not that much. It's just that getting all the flow effects can be challenging. Hyper-maneuverability is a fairly new branch on the aerodynamic tree. The important takeaway here is that the derivative of the pitch rate plays an important role. It's not just the pitch rate, but how fast we can change it that matters.

Mind you, with a CLPA airplane the position vector isn't a constant in any dimension, but we won't go there in the undergraduate class.



Chuck

[Ken Culbertson]

And then the wind changed direction.....

I love this hobby!

Ken

[Howard Rush]

With all this in hand determine how fast you want the aircraft to rotate around the cg in he pitch axis.

Mine rotates at 300 degrees per second.  More might be nice.

Notice when you do the plug and chug that the aircraft's center of pressure is ( hopefully ) behind the cg.

not counting the contribution of the tail (That’s probably obvious from what you said, but it took me awhile to see.)

I maintain that the initial rotation required to achieve AoA is the dominate consideration to achieve the desired flight path.

Numbers, por favor

I can guess how the angle of the flow at the tail varies with stuff, but not how the force on an elevator and stabilizer vary with alpha at weird alphas.  See Igor’s amusing stabilator comments for further entertainment.

I think John’s original question was about stability, but my flimsy looking fuselage takeapart fittings and my weight-saving plywood deletion have me worried about load.  I asked Alexa, my conversion person, what 18 Newtons is in pounds.  She said, “Newtons cannot be converted to pounds sterling.  Newtons measure force; pounds sterling measure money.”

[Chuck_Smith]

Hi Howard,

Remember, it's a two step process. The tail force required in a constant radius turn (neglecting gravity and the fact that our loop described on the surface of a sphere) is pretty easy to quantify: You take the moment coefficient for the required angle of attack and then use statics to sum the moments to zero.
So lets say we're doing an inside loop. The wing has a nose-down pitching moment that is a function of airspeed and angle of attack. Choose any spot in the universe located along the fuselage's longitudinal axis and sum all the moments and solve for the tail force required.

But as we know, if we want to make a round loop instead of a longhand lowercase "l" then we need to be constantly changing pitch to adjust for the changing airspeed, gravity vector relative to the plane and bank angle.

Numbers  - I love numbers- are interesting. Let's take a stunt ship moving around at say, 65 fps. If I want to make a 10' radius, 90° turn at that speed it needs to happen in  about 300 milliseconds. But before that can happen there needs to be a pilot reaction, then the aircraft needs to rotate to change the AoA and then start the turn. 300ms is forever in CLPA, and the initial rotation needs to happen very, very quickly.  And since an airplane has mass, we know this will take time and be a second-order response ( hopefully with non-imaginary roots ).


BTW, 300 ms is glacial time in controline combat, right?

[Chuck_Smith]

Howard,

I'll wanted to also add that since we're long past 2D, non-viscous, infinite wing theory, There's actually a 3rd thing and it causes yet another delay, when the flaps move it takes times for the shed vortex to move aft to the tail and have the now greater wing downwash whomp the tail. Figure around .01 seconds (maybe about 8" of flight path) , give or take. So the moment to change the angular acceleration required to increase the AoA is a stepwise function and takes a non-trivial portion of the intended flight path. Pretty cool, eh? Luckily, mass hides most of this or we'd never be able to fly the danged things.

This discussion is getting pretty esoteric, but hopefully it demonstrates that the concept of the elevator simply steering the nose around is incorrect.

Chuck

[Howard Rush]

The tail force required in a constant radius turn (neglecting gravity and the fact that our loop described on the surface of a sphere) is pretty easy to quantify: ...

Here's a quantification: https://stunthanger.com/smf/engineering-board/some-forces/

I made a video recording of control deflections of Paul Walker doing a round loop.  He does quite round loops.  Control deflection is all over the place.  I'd post the video, but it's a zillion MB, and I never learned how to make it smaller.

[Chuck_Smith]

Cool Howard. As a point of reference, when I do a loop in a full scale aircraft, to make it nice and round I make a 3g pullup but at the top of the loop, I'm basically zero g. If I get it right the pullout ends at 3g and my original entry airspeed, so the elevator is always changing during the maneuver. A loop in a controline plane is a lot more complicated since it involves all six degrees of freedom.

Now for the final blow to the "steering fins" version of the elevator, let's address the horizontal tail's aspect ratio.  We know the aspect ratio is key to its  span-wise lift distribution along the surface since the AR determines the amount of upwash coming into the surface. The lower the aspect ratio of the tail, all other things being equal, the more it has to deflect the elevator to get create the same change in lift as one of equal area with a higher AR.

OK, now for a visualization that everyone can relate too regarding the shed vortex creating a warped flow field and all that jazz and how the 2D, non-viscous, infinite wing theory is almost completely useless on a real 3D airplane:

I give to you - the Flite Streak. That tiny little elevator attached to a much larger stabilizer doesn't inspire much confidence, but man, it sure as heck turns that little airplane - and quickly. (Notice too that modern combat ships have evolved to tiny elevators on extremely short tails, compared to the long booms and planks we used back in the day.)

Granted, there's a lot of flow visualization required to grok the whole deal of why that little strip of an elevator is so danged powerful, but hopefully it's enough empirical evidence to show that something a lot more complicated is going on than first meets the eye. 

Chuck

Aside: it also demonstrates why a "wicker bill" on a racecar wing has such a dramatic effect even though it's chord is negligible compared to the wing it's attached to. I can attach a .300" protrusion onto the TE of an Indy car's car rear wing and the effect is strong enough it will require a change in spring rate to keep from bottoming out at top speed with the added downforce it created.

[Howard Rush]

(Notice too that modern combat ships have evolved to tiny elevators on extremely short tails, compared to the long booms and planks we used back in the day.)

Just looking at them, you'd think they'd have no maneuvering stability.  They don't.

[Chuck_Smith]

Just looking at them, you'd think they'd have no maneuvering stability.  They don't.

Howard, back in our day combat was a lot more savage. We flew like falcons taking prey on the wing. Today they look more like a swarm of gnats

Chuck

[Ken Culbertson]

 

Howard, back in our day combat was a lot more savage. We flew like falcons taking prey on the wing. Today they look more like a swarm of gnats

Chuck

It is even hard to tell if there are even any tactics in play other than turn and burn.   I used to love combat.

Ken