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#21
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Mounting a turn coordinator on the tail?
Ron Natalie wrote: Stubby wrote: The horizontal component of lift behaves like a string tied to a rock being swung around. No, it doesn't. The lift vector points in a direction (roughly) perpendicular to the wing. Nothing causes it to point to the a "center" other than the other aerodynamic surfaces . The lift vector(s) point perpendicular to the wing (s). With a dihedral angle in the two wings there will be horizontal components of the lift vectors and in level flight they will be equal and opposite, canceling. In a turn with the plane banked, it's easy to see how one lift vector will point entirely straight up while the other doubles its horizontal component. This pulls the plane toward the center of the turn. (It sure is easier to explain with a blackboard!) |
#22
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Mounting a turn coordinator on the tail?
Didn't anybody read "Stick and Rudder"?
To make an object travel on a curved path, you need a "centripetal" force. This isn't a "centrifugal force", which is a made-up construct that helps to understand what it feels like to be in a vehicle that's moving you in a circle, a centripetal force is a real force that's acting upon an object that pulls it towards the center of a circle. Now, if you're traveling in a circle, the force that's pulling you toward the center of the circle acts in a direction perpendicular to your direction of motion, and since your direction of motion is always changing (you're going around a circle, after all), the direction of that centripetal force is necessarily continually changing as well. If you bank the wings to the left, the lift vector will start dragging you to the left. You'll begin "slipping", drifting sideways through the air. If this was the end of it, you'd never end up traveling in a circle, you'd just keep drifting slightly sideways, you nose would keep pointing where it was pointing before, and your track over the ground would still be a straight line, but just slightly "diagonal", at an angle to what it was before. But when you slip, there's now a net component of drag acting sideways against the side of the plane. Since most of our surface area is behind the CG, when you blow really hard against the side of a plane, it tends to yaw into the wind. As the plane yaws to the left, the wings also turn to the left, and the direction that the lift vector is pointed in turns as well. Bingo, now you've got a real "centripetal force", one that continually changes direction to point at a 90 degree angle to your curving path. Now your ground track can form a circular path. You need that yaw to turn the wings to continually update the direction of lift, to create circular motion. Now you're a little closer to that rock on a string model. Alternately, if you're intentionally doing a slip, like if you're landing in a cross-wind, you could always apply opposite rudder. This acts counter to the "weather-vaning" moment, and prevents the plane from yawing, which prevents the wings from "rotating", preventing the lift vector from changing direction. By stopping the nose from changing direction, you stop the plane from moving in a circular path, and you just "slip" sideways up against the wind (possibly just to resist a cross-wind and maintain a straight path along the ground). -harry |
#23
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Mounting a turn coordinator on the tail?
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#24
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Mounting a turn coordinator on the tail?
... What's more, because the horizontal force is
acting through a point behind the CG, the plane would spin in the opposite direction to the force. And when the plane is flying straight and level, since the wing is behind the CG, shouldn't the wing, similarly, be continually pulling the tail up over the nose, leading the plane to tumble through space? Why doesn't it? Because while the wing is aft of the CG, the tail provides a downward force to balance it. Since the tail is so much further away from the CG than the wing is, it can leverage a relatively smaller force to balance out the torque, while allowing the much larger force at the wing to provide a net upwards force to support the plane's weight. When the plane is in a bank, yes, the wing is providing a horizontal force from behind the CG again. But nothing has changed from the straight and level case, there's still an upside-down wing back there at the tail, and it's still balancing out the torque produced by the wing being behind the CG. When the wing is banked, the tail is banked too. It's not easy to imagine this, but think of the space shuttle stationary in space, then having a sideways horizontal force applied behind the CG But imagine another sideways horizontal force applied in the opposite direction at the tail, exactly strong enough to balance the "torque". If your left force is applied aft of the CG, but fairly close to it (like where the wings are), and your counter-balancing right force is applied at the tail, much further from the CG, then a relatively small force at the tail will be enough to counter the much larger force at the wing, due to leverage. So our torques are equal, preventing any rotation, but there's still a net force to the left (the force up near the wing is much larger), so there will be a net translational force to the left. Your space ship will accellerate to the left without rotating. -harry |
#25
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Mounting a turn coordinator on the tail?
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#26
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Mounting a turn coordinator on the tail?
Now that's a VERY interesting way of thinking about it.
It's the proper and textbook way to think about it. difficult to keep the ball exactly centered in a 50-degree bank, so I started thinking about whether it's even possible to do this in a 60-degree bank. Of course it is. However, in a steep bank, the rate of turn is mostly controlled by the elevator. No. The rate of turn is controlled by load factor at a given airspeed. You can increase the rate of turn by banking more, or you can pull back on the yoke to increase the load factor (temporarily). In a 60 degree banked turn, you will get a load factor of 2 regardless of what you do with the elevator. If you don't increase the AOA during turn entry however, the aircraft will accelerate and the 60 degree banked turn will occur at a higher airspeed, in a rapid descent. |
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