90° Bank Turns
90° Bank Turns
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Depends on the acft. type. Presuming you mean 180/360 deg turn, to maintain alt. at 90 deg bank, you need alot of HP and a real strong airplane. There are alot of books on aerodynamics, if you need to write a report, do your own research.
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Since no lift in the vertical plane is produced by the wings at 90 degrees bank, either the fuselage and vertical stabilizer must hold the airplane up, or it will descend. Aerobatic airplanes with relatively large vertical stabs and rudders and a plethora of power can maintain level 90 degree bank turns for a limited time.
Also, don't forget that "down" is sideways with respect to the airplane, so everything has to work with 0 "vertical" G and 1 lateral G.
Also, don't forget that "down" is sideways with respect to the airplane, so everything has to work with 0 "vertical" G and 1 lateral G.
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Surely your aircraft will be happy to turn for as long as you want at a 90-degree bank angle; as far as it knows, you're just pulling up. Therefore if you have a +6g airframe, you can pull as hard as that, and so on.
Problem is, you'll be descending at the same time (no lift in the vertical plane, see?). So, start the turn at 50 feet and you'll die quickly; start it at height and the effect of the sideways movement will be yaw in the direction of the ground. If you're pulling truly as hard as possible, you'll be just above the accelerated stall, so it will be an incipient spin.
I could bullsh1t for England, me.
Problem is, you'll be descending at the same time (no lift in the vertical plane, see?). So, start the turn at 50 feet and you'll die quickly; start it at height and the effect of the sideways movement will be yaw in the direction of the ground. If you're pulling truly as hard as possible, you'll be just above the accelerated stall, so it will be an incipient spin.
I could bullsh1t for England, me.
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the angle of bank depends on the TAS and the radius of turn.
at 90 degrees of bank the load factor becomes infinite as the lift force is completely horizontal and there is no vertical lift component produced.
at 90 degrees of bank the load factor becomes infinite as the lift force is completely horizontal and there is no vertical lift component produced.
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the angle of bank depends on the TAS and the radius of turn
The "radius" of the turn will depend and the angle of bank, the "g" pulled during the turn and the KIAS/TAS being flown. The "rate" of the turn will also be effected by the same factors
Also remember that the stall speed can be effected. Calculated as such--the stall speed x the square root of the load factor. The aircraft will always stall at the same KIAS at 1 G and level.
ie 79kts (normal stall speed clean) 4 G square root is 2, therefore stall speed is now 158Kias.
Turning at 90 AOB you will have to use the fuselage to generate lift either that or pull infinate "G" which is impossible due to power and stall speeds. Not to mention cannot build it
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The maximum level (ie, no sink rate) turn you can do with 6G Wing Loading is 80.4 Degrees.
The turn rate will be dependent on the velocity of the aircraft.
With a 9G Wing Loading, a 83.7 Degree Level Turn.
I suspect at exactly 90 Degrees, most airframes will have a tendency to yaw towards the ground and would be a handfull to fly....
As for bank angle in a sustained turn. Its dependent on the wing loading and wing loading only.... The rate of turn is dependent on the wing loading, bank angle and airspeed....
Example a:
Say a 1 ton (1000KG) airplane is flying at 450 knots and has a maximum sustianable wing loading of 6G and wants to perform a sustained turn..
Figures:
Speed, 450 knots = 225 m/s
Max G = 6G
Maximum sustain bank angle 90 - (sine X) ^1 x 1/6 = 80.40
F=(Mv^2)/r => 22300r=1000(50625) => r =2270m i.e. implies turn rate of 6 degrees per second.
At 225 knots (ie 112.5 m/s) radius will be = 567 m turn rate will be 11.7 degres persecond
For a 9 G turn at 150 knots(75 m/s) turn radius will be 200 m, turn rate will be 22.5 degrees...
For a aerobatic plane such as a extra 300 at 75 knots doing a 9 G turn level turn , turn radius will be 50 m..turn rate will be 43 degress per second......Things are starting to get high here now....hehe
The turn rate will be dependent on the velocity of the aircraft.
With a 9G Wing Loading, a 83.7 Degree Level Turn.
I suspect at exactly 90 Degrees, most airframes will have a tendency to yaw towards the ground and would be a handfull to fly....
As for bank angle in a sustained turn. Its dependent on the wing loading and wing loading only.... The rate of turn is dependent on the wing loading, bank angle and airspeed....
Example a:
Say a 1 ton (1000KG) airplane is flying at 450 knots and has a maximum sustianable wing loading of 6G and wants to perform a sustained turn..
Figures:
Speed, 450 knots = 225 m/s
Max G = 6G
Maximum sustain bank angle 90 - (sine X) ^1 x 1/6 = 80.40
F=(Mv^2)/r => 22300r=1000(50625) => r =2270m i.e. implies turn rate of 6 degrees per second.
At 225 knots (ie 112.5 m/s) radius will be = 567 m turn rate will be 11.7 degres persecond
For a 9 G turn at 150 knots(75 m/s) turn radius will be 200 m, turn rate will be 22.5 degrees...
For a aerobatic plane such as a extra 300 at 75 knots doing a 9 G turn level turn , turn radius will be 50 m..turn rate will be 43 degress per second......Things are starting to get high here now....hehe
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ah, don't ya just love aerodynamic questions....
saudipc-9,
-I understand where your coming from, but I think you'll find it is really true, just not put in the way us pilot chappies normally think. I was simply trying to answer the original question.
I'm quite aware that in combat terms "rate kills". Aircraft g and airspeed. Rate and radius.
I was just putting it in a way the aerodynamics guys seem to favour.
tan bank angle equals tas squared over gR.
jetdriverwannabe
your assertion here is flawed. Remember, at 90 degrees of bank the load factor becomes infinite, so the situation you describe is impossible in the real world. A 90 degree bank turn in level flight requires infinite g, so it's not "most" airframes as you assert, but all, although it doesn't necessarily mean they're a handful to fly. It's just that in aerodynamic terms, a level turn at 90 degrees of bank is impossible. Simple as that.
Intruder is correct in saying that you can be in "knife-edge" flight, ie 90 degrees of bank and still level, in a high performance aerobatic aircraft. But you are using the rudder and fuselage for lift in this case, so in proper aerodynamic terms it does not constitute a turn, as the wings are not loaded.
saudipc-9,
Ahh not really true.
I'm quite aware that in combat terms "rate kills". Aircraft g and airspeed. Rate and radius.
I was just putting it in a way the aerodynamics guys seem to favour.
tan bank angle equals tas squared over gR.jetdriverwannabe
I suspect at exactly 90 Degrees, most airframes will have a tendency to yaw towards the ground and would be a handfull to fly....
Intruder is correct in saying that you can be in "knife-edge" flight, ie 90 degrees of bank and still level, in a high performance aerobatic aircraft. But you are using the rudder and fuselage for lift in this case, so in proper aerodynamic terms it does not constitute a turn, as the wings are not loaded.
Last edited by Maximum; 5th May 2003 at 08:21.
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Yeah, I should reword that to. At a bank angle of 90 degrees the airframe concerned will be accelerating towards the earth at 9.8ms^2 (probably less due to air resisitance) as there is no vertical component to lift to conteract gravity.
Last edited by JetDriverWannabe; 5th May 2003 at 11:47.
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Maximum,
So what you really mean to say is "the angle of bank required is dependant on the TAS and radius of turn required"
The aerodynamics guys are putting the cart before the horse by saying it the other way.
So what you really mean to say is "the angle of bank required is dependant on the TAS and radius of turn required"
The aerodynamics guys are putting the cart before the horse by saying it the other way.
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"Intruder is correct in saying that you can be in "knife-edge" flight, ie 90 degrees of bank and still level, in a high performance aerobatic aircraft. But you are using the rudder and fuselage for lift in this case, so in proper aerodynamic terms it does not constitute a turn, as the wings are not loaded."
Maximum:
In knife-edge flight as you describe, you can additionally use the elevator to load the wings and cause the airplane to turn.
All the equations and assertions already provided by others are likely valid for COORDINATED turns in LEVEL, CONSTANT-SPEED flight. However, other modes of flight do exist, and are used for various purposes.
Maximum:
In knife-edge flight as you describe, you can additionally use the elevator to load the wings and cause the airplane to turn.
All the equations and assertions already provided by others are likely valid for COORDINATED turns in LEVEL, CONSTANT-SPEED flight. However, other modes of flight do exist, and are used for various purposes.
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saudi-pc-
well, what I meant to say is exactly what I said! However, I agree with you absolutely that in the real world the way you've described it is quite rightly the way it happens.
Intruder
...erm, I know, I've even used some of them.
Pienkie, how about some feedback seeing as you're the one who got us all into this?
So what you really mean to say is
Intruder
All the equations and assertions already provided by others are likely valid for COORDINATED turns in LEVEL, CONSTANT-SPEED flight. However, other modes of flight do exist, and are used for various purposes.
Pienkie, how about some feedback seeing as you're the one who got us all into this?
