No, I was not thinking of a stall, but of a steep bank. No nose-up attitude, no turbulence over the airfoil - but no (or at least, not enough) vector component in the upward direction, either.
I also doubt the 70[sup]o[/sup] figure which would result in 2.7 g’s. That’s a pretty heavy load for people who might not be in the best physical condition. In almost all cases the pilot warns, and certainly should warn, the passengers of any unusual maneuvers.
One time we were taking off on Southern Airlines from Dothan, AL. The pilot came on and told us there was a thunderstorm not too far off the end of the runway and we would be making a rather steep turn to avoid any possibility of turbulance from it. The bank was probaly 30-40[sup]o[/sup] and when you are close to the ground that seems awfully steep.
There is a particular instrument departure from Balitimore that has a noise abatement procedure, or used to have at any rate. The plane would take off, enter the clouds climbing steeply with engines going all out. Then, all of sudden the engines were throttled back to what seems like idle, but wasn’t, and nose was pitched sharply down to level or nearly so. It was almost enough to cause heart failure. Pilots on the flights I was on never warned us about it and I think they should have.
To some extend, bith Broomstick and I are right, but we’re getting into some pretty esoteric differences between the types of turns and the way the lift vector of an airfoil is resolved.
For those interested in the nitty-gritty of this, I recommend this page which has the diagrams and math that are not easily reproduced in a Straight Dope post. For those not interested in the details, a brief quote:
The nose may not look elevated, but at some point you have to increase the angle of attack relative to flight in order to achieve greater lift - and you have to achieve lift in excess of what you need for level flight or else you don’t have the lift to pull you around the turn. Unless you are in a grossly overpowered airplane flying on pure thrust - but while an F-16 might fall into that category your passenger airline does not. Yes, they have lots of thrust. They also have a lot of weight, too.
Yes, I am strictly a “practical flight” person when it comes to aerodynamics - my understanding comes from actually flying, not from mathematical studies. Maybe I’m wrong on this (in which case one of the Big Iron guys can correct me) but my understanding of airliners is that they don’t fly solely on thrust, they really do need the aerodynamic lift generated by the wings, they turn in a conventional manner, and that includes increased angle of attack when in a steep bank. In which case, as the bank steepens, they eventually approach and enter a stall when the the lift required exceeds the lift that can be generated by wing+engines. In something less than maximum bank you might have a choice between increased engine power vs. higher angle of attack in order to generate the turn vector of lift, but that point will quickly be exceeded, long before you reach a steep bank (usually definied as over 45 degrees, but perhaps where passenger are concerned it may be anything over 30)
Sounds like it’s in the ballpark to me - I don’t really have a direct way of measuring it precisely.
In level flight the turning radius is proportional to the square of the velocity for the same acceleration. 200/20 = 5 and 5[sup]2[/sup] = 25
It might help to think of it this way.
A 5000# airplane must go 80 MPH to overcome gravity and fly.
A 5000# airplane in a coordinated turn with a 60 degree bank is at 2 G’s. That is 10,000#'s. His airspeed needed to remain aloft using the same wing will be much higher.
In an coordinated turn with a 60 degree bank and held level, you will slow down unless you add power. So, if you keep the aircraft at the same altitude and power setting but keep increasing the turn, the weight of the airplane and loss of airspeed will soon cross the point where the speed is sufficient to prevent stalling.
Power changes, climbing, descending, turbulence, or abrupt control movements throw a lot of variables into it.
If you are turning correctly, you will stall before you side slip into the turn.
As the bank angle increases, back pressure (on the controls) must increase to provide the increased lift needed to cater for the fact the lift vector is inclined. At some point–around 65º on most aircraft–the angle of attack required to maintain height reaches the stalling AoA and the aircraft stalls. If you were to turn without increasing lift, only then would you slip into the turn (the B52 airshow practice crash is a good example).
Of course, some aircraft have enough power and fuselage keel surface to be able to maintain straight and level flight with a 90º bank angle. The fuselage becomes the primary lifting surface with the engine power providing a thrust vector sufficiently inclined upwards to suplement the lift from the fuselage. That’s a specific aerobatic manoeuvre though.
Correct, however, all but the very worst of pilots make their turns coordinated, meaning that for any given bank angle, the aircrafts rudder is controlled in a way that ensures that the g forces are felt straight down through the pilots body. When flown in the normal way, g-forces are directly related to angle of bank in a level turn. In a descending turn, the g-forces will be less.
I’m not sure what it is here that is controllable. Military pilots can maintain tight turns at high angle of banks because they are fit and are wearing g-suits that allow them to pull up to 9 gs. Additionally they are flying aircraft that have the power to maintain these bank angles at a speed that gives some margin above the stall. They most certainly do have to worry about those pesky g-forces.
Any turn in an airliner will not be at such an angle to warrant any concern about side slipping into the ground.