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| | #1 |
| Senior Member Join Date: Nov 2006 Location: DFW
Posts: 791
| I was wondering what method(s) you use to explain to a student pilot how stall speed will change with weight and bank angle? I can see how I will confuse the student by saying the aircraft will stall at the same AoA (flaps up) regardless of airspeed, weight, power setting, bank angle but the stall speed will be changing. Now when referring to CAoA, and how it is published as the CAoA at 1G, when you increase the load factor, such as in a turn, the stall speed will increase but the CAoA will remain the same. Maybe it is me but I just see it hard for the student to relate stall speed changing in these various scenarios with angle of attack always remaining the same and then bringing in load factor and after saying "The airplane will always stall at the same CAoA at 1G, but higher load factors will produce a higher stall speed but the CAoA will remain the same." I am trying to get at that CAoA and stall speed seem totally independent of each other and then it is like I am trying to relate them in the sentence. Also, for weight, does anyone have a quick and easy method of describing how increased weight will increase stall speed? The point that I came to is that a heavier a/c will have to have an increased AoA to stay straight and level flight compared to the same a/c that is lighter. Therefore as the AoA is increased at the same rate on both a/c, the heavier a/c will stall first because it already had the increased AoA over the lighter a/c which will stall the heavier a/c first and the lighter a/c may have X amount of degree's left before it reaches its CAoA which will equal more time for the a/c to lose its airspeed. Any help is appreciated; if I didn't confuse you ![]() |
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| | #2 | |
| Old Skool Join Date: Sep 2006 Location: Memphis, TN
Posts: 2,224
| Quote:
__________________ Core Concepts of Flight If an error is corrected whenever it is recognized as such, the path of error is the path of truth --Hans Reichenback | |
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| | #3 |
| Senior Member | I begin with a layman's definition of a stall: As angle of attack is increased on an airfoil or a wing, (and everything else remains the same), lift is increased. But this increase only works up to a point. Eventually you get to the point where any more increase in angle of attack will actually cause a decrease, rather than an increase in lift. This maximum angle of attack is known as the critical angle of attack. If you exceed the critical angle of attack, instead of getting more lift, you will produce less. This drop in the amount of lift produced is frequently quite drastic. This drop in lift, caused by exceeding the critical angle of attack is known as a stall. From there I move to the lift equation, and explain that lift can be increased by the pilot by increasing any of the factors of the lift equation. Typically, though the only ones that he has control over are velocity (V^2) and angle of attack (which is a key factor in CsubL), since mother nature controls density for all practical purposes, and the size and shape of the wing is pretty much fixed if you don't count certain devices like flaps. From there I look at the different scenarios that you mention from the point of view of exactly how much lift does the airplane need to make, and how does it make that lift. Obviously the pilot only has 2 "tools" to "make" lift: angle of attack and velocity. And I go from there.
__________________ Dude, what are you trying to do? Land the airplane or adjust the field elevation? |
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| | #4 |
| Senior Member | This isn't entirely true. It is absolutely true for unaccelerated maneuvers, (straight and level flight, constant speed climbs, constant speed descents) and it's also true for level turns (which are accelerated maneuvers). But it wouldn't be true in all acclerated maneuvers. Like a loop, for example.
__________________ Dude, what are you trying to do? Land the airplane or adjust the field elevation? |
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| | #5 | |
| Old Skool Join Date: Sep 2006 Location: Memphis, TN
Posts: 2,224
| Quote:
Our discussion assumes an equilibrium has been established. An aircraft in a loop isn't in equilibrium, but it's in the process of establishing one. Turn loose of the yoke in a loop and the aircraft will eventually assume a lift = weight situation.
__________________ Core Concepts of Flight If an error is corrected whenever it is recognized as such, the path of error is the path of truth --Hans Reichenback | |
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| | #6 | |
| Junior Member Join Date: Aug 2006 Location: FL
Posts: 115
| Quote:
Now, when an airplane climbs straight up like an F-16. I know it's all thrust. but doesn't airspeed change without change in AOA there? I'm not trying to say you're wrong. but I think I can learn alot by "trying" to prove one of your core concepts wrong! this way I'm learning something either way ![]() | |
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| | #7 | |
| Old Skool Join Date: Sep 2006 Location: Memphis, TN
Posts: 2,224
| Quote:
<<think I can learn alot by "trying" to prove one of your core concepts wrong!>> Go for it. About seven of them have some "yeah, but" possibilities.
__________________ Core Concepts of Flight If an error is corrected whenever it is recognized as such, the path of error is the path of truth --Hans Reichenback | |
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| | #8 | |
| Junior Member Join Date: Aug 2006 Location: FL
Posts: 115
| Quote:
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| | #9 |
| Old Skool Join Date: Sep 2006 Location: Memphis, TN
Posts: 2,224
| If we assume a non-cambered airfoil, then zero AOA should approximate zero lift coefficient. We'll assume your target is a zero lift coefficient. I'd say this airplane would end up in a vertical nose dive, with drag supporting the entire weight of the airplane.
__________________ Core Concepts of Flight If an error is corrected whenever it is recognized as such, the path of error is the path of truth --Hans Reichenback |
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