Design > Engineering board
Stabizer Pressure
Brett Buck:
--- Quote from: Howard Rush on April 19, 2020, 09:55:46 PM ---If I had a MEMS gyro I could do a loop kill.
--- End quote ---
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:
--- Quote from: Tim Wescott on April 19, 2020, 03:54:26 PM ---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.
--- End quote ---
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:
--- Quote from: Igor Burger on April 20, 2020, 12:58:15 AM ---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.
--- End quote ---
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?
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