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