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Old 8th Jul 2009, 13:05
  #3290 (permalink)  
ClippedCub
 
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Just to review, 3 1/2 other things can possibly replace the sideslip.
1) When close to the critical AOA, dropping an aileron (say in rapid response to the start of roll-off on one side) effectively increases the AOA in the region of the aileron and could stall that region.
2)When close to the critical AOA, having a some upward gust applied to the outer wing section on one side, causing it to exceed critical AOA.
3)When close to the critical AOA, making a smaller radius turn so that the inside wingtip is flying more slowly that the outer wingtip and effectively causing a greater AOA on the inside wingtip (same vertical vector amount, less horizontal). More an issue with long-winged aircraft.

and 1/2) The typical spin entry we're all aware of, when close to the AOA kicking in rudder for some reason. With the A330 you can substitute differential engine thrust for the rudder. While it may or not slip sideways overall, that's not the problem. It's the yawing so that the inside wingtip flies more slowly or even backwards relative to the other tip. Due to vertical vs horizontal flight vectors of the tips, the inside tip goes critical and the other away from critical.
Your first three points are confusing roll with sideslip. Coordinated roll will not invoke a spin by itself, the conversion of aoa to yaw angle with roll off creates the effect. Your first three points explain roll-off.

Your 1/2 point is what is taught in ppl straight wing trainers. While this applies to the A330, it's effect is minor compared to the effect of unsweeping the upwind wing and increasing sweep on the downward wing. This sweep effect due to sideslip is easy to visualize. Draw an A330 planform on paper with a vector coming from 20 degrees off the nose. Tilt the paper till the vector is vertical. Observe that upwind wing gets longer and is unswept. The downwind wing gets shorter with increased sweep.

Also understand that CLmax changes with the cosine of the sweep.

Hope this helps. And thanks for demonstrating that stall and sideslip are the two components needed for spin entry.

Edit: The 1/2 point also explains roll-off. So all 3.5 points don't replace sideslip, but do explain how a roll-off can occur which does induce sideslip once roll-off occurs.

Last edited by ClippedCub; 8th Jul 2009 at 13:22.
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