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Basically I'm wondering if this is correct. Which essentially says that you need a torque to get the nose of the craft to turn and that this is provided by the rear tail surfaces.

After trying to simulate airplane flight in a game engine, I'm thinking that link is correct. Although airplane flight is rarely described like that.

If you just apply force in the direction of lift (and thrust), then during a roll, the horizontal component of lift causes the plane to get a velocity with a horizontal component. But the plane does not turn. The plane slips to the side while going forward. The horizontal component of lift does not stay perpendicular to the velocity vector so it doesn't move in a circle.

Imagine also a block sliding forward on ice (constant v) with a small thruster attached to its side (turned off). Now turn on the side thruster (pointed through cm). Does the block start turning in a circle or does it just get a horizontal component to it's velocity? Doesn't it need a torque, just like the plane to keep the thrust perpendicular to the velocity vector?

So am I out of my mind here or is this correct?

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the link looks correct but I'm troubled by the lack of discussion about the angular momentum in the plane's engines, which I believe is relevant (though possibly only as a second-order effect). –  Emilio Pisanty May 14 '12 at 15:22
    
Attach a tail fin to the block, and it will turn. On a car, the rear wheels serve the same function. –  Mike Dunlavey May 17 '12 at 2:57

5 Answers 5

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You're correct, the ice block will not turn automatically. It will require a torque. In aviation this is basically what is called coordinating a turn. With an airplane, if the pilot does not provide the necessary coordinating torque via rudder/elevator inputs, the torque will be generated automatically via the weathervane effect, which tends to align the fuselage with the velocity. So, you can effectively turn just by banking, if you're not anal about keeping your altitude and airspeed constant. In a helicopter, it's a different story, as the helicopter doesn't have much of a weathervane effect at least at low speeds. With a helicopter at low airspeed if you just bank without coordinating rudder then you just slide sideways.

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There were lot's of good responses to the posted question, but I'm choosing this as the answer because it's short and simple, even if less technical then some other posts. –  Fraggle May 15 '12 at 16:45

As a student pilot with 100 hours and 300 landings, not to mention interest in engineering and physics, I love questions like this.

My first recommendation is get a copy of Stick and Rudder. It's a delightful book, used by pilots for over half a century to understand the basics of flight.

My second recommendation is take an introductory flying lesson. It costs around $200 and is the most fun per buck you can have honestly.

Now to try to answer. (Almost any question like this can be understood with a simple balsa wood glider or a paper airplane.)

First to understand is that in straight flight at constant speed, for a normal airplane with single main wing and a tail, the wing pushes up and the tail pushes down. The wing supports not only the downward weight of the airplane but also the downward force on the tail. The plane is nose heavy, and if the tail were suddenly chopped off, the plane would immediately go down. Since the tail is some distance behind the wing, its downward lift creates a moment that holds the nose up. The lifts, up and down, and thus the moment, are proportional to speed squared. If for some reason the plane slows down, the moment decreases, the nose drops, and the plane speeds up. If it speeds up, the moment increases, the nose rises, and the plane slows down. enter image description here

This is the first thing you need to understand about airplanes - how the nose-heaviness and the downward lift on the tail cause its speed to be stable. The exact speed it seeks depends on the angle of the tail surfaces. The more they are canted upward, the slower the plane flies, and downward means faster. (In case you're wondering how the wing gets enough lift when it is flying slower, it's because the whole plane tilts up, giving it more angle of attack.) I know you are asking about turning, but this is the first thing you need to understand.

Second, how do you make it climb or descend? That's what the throttle is for. Just like in a car, you need more power to go up a hill, and less to go down. In a plane, if you increase power, it speeds up, raises the nose, settles back to its natural speed, and keeps climbing. If you decrease power, it slows down, drops the nose, and then settles to its natural speed, but on a descending slope.

OK, you were asking about turns. The way you turn a plane is by putting it into a bank. (With the ailerons, by monentarily turning the yoke left, and then re-centering it.) Suppose you bank it 30 degrees to the left. That tilts your lift vector, so that half of it is accelerating you to the left, and .866 of it is pushing up against gravity. To keep gravity from winning, you apply back pressure on the yoke, creating more lift. A turn has much in common with a climb, unless you want to descend at the same time. You also need to give it some more power, to maintain speed through the turn.

Here's a 60-degree bank: enter image description here where you have to pull 2Gs of lift in order to maintain constant altitude, and add quite a bit of power. That's a pretty stressful maneuver, and you can't do it at slow speed because if you pull that hard it will stall. That shows you the similarity between a turn and a climb.

You were asking why doesn't the plane just accelerate sideways rather than turn? If it did, then pretty soon it would be feeling this sideways wind on it. Every airplane is a weather-vane. It turns into the wind it feels. So, roundabout, that's the answer to your question.

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To roll, the ailerons (moveable surfaces on the trailing edges of the wings, near the tips) move in opposite directions, increasing the effective camber of one wing tip & reducing that of the other. This creates a spanwise difference in the lift distribution.

If no other input is commanded by the pilot, the nose will drop; assuming you started in straight & level unaccelerated flight (SLUF), as the plane rolls the component of the lift vector opposing the weight of the plane drops (to zero when in a 90$^{\circ}$ banks).

In order to maintain level flight, the pitch angle must be increased; this also changes the thrust vector. In order to maintain unaccelerated flight (specifically to not lose speed), thrust must be increased (as drag increases with increasing pitch angle).

The differential lift distribution on the wing tips also creates a differential drag distribution; for most conventional airplane configurations, this comes in the form of adverse yaw (i.e., the nose tends to track away from the direction of bank). This is countered by application of balanced rudder force.

Summary: in an unaccelerated level turn, the side force on the vertical tail due to rudder deflection keeps the nose tangential to the circle, while the horizontal component of lift due to bank angle provides the centripetal force.

Emilio is correct that engine torque (gyroscopic coupling) can be an issue; it is typically negligible at low rotation rates.

EDIT (not enough rep yet to comment on other's answers): yes, the lift vector 'naturally' stays normal to the velocity vector by definition; it's the component of the aerodynamic force that is normal to velocity, while the component parallel to velocity is labeled 'drag'.

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See, here's the interesting part, why does a deviation from SLUF result in a drop of the nose as opposed to a loss of altitude. We do get a bit of that from your answer, particularly with mention of the vertical tail and the differential drag distribution (although the latter seems to work against the desired turn). –  Alan Rominger May 14 '12 at 16:19
    
For a given configuration of the control surfaces, there is a flight condition (airspeed & angle of attack) toward which the static & dynamic stability of the aircraft will drive; this is called its 'trim condition'. If the plane were to drop without pitching, the angle of attack would increase, decreasing the downforce from the horizontal stabilizer, at which point the nose-down pitching moment from the wing producing positive lift would be unbalanced until the aircraft returns to trim. –  Ghillie Dhu May 14 '12 at 16:26

I like this question, and I find the link to be of great utility, but I'll still go over several issues I have with its explanation.

First, we need to establish the completely plausible situation of a balanced plane operating normally, then being slanted to the left or right, and the fact that it will shimmy in that respective direction and downward. This requires only a first-level force diagram. The gravity vector is the same magnitude but rotated relative to the plane and the aerodynamic forces are unchanged relative to the plane. There are several mentions of force moments in the link, but I...

  1. Don't find them useful or necessary for the explanation of aircraft turning and
  2. Don't see any reason to predict a moment imbalance in the aircraft's vertical plane

Consider that normally the wings exert a lift force upward (relative to the aircraft) and the horizontal tail exerts a lift force downward. If we assume that we do not change the air velocity relative to the aircraft then changing the gravity vector (at CM) has no ability to create a net force moment in any direction. Both lift and gravity act at the CM, and one has to move for us to obtain a moment (like they do in buoyancy problems).

In the simplistic view of a point force balance, steering the plane is not possible at all. The link presents an attempt to resolve this conflict.

in order to avoid losing altitude, the pilot has to compensate with a bit of back-pressure on the yoke

I disagree with this. The pilot has to put back pressure on the yoke in order to bank, but the idea that he/she does so to compensate for losing altitude shouldn't hold up to scrutiny. Consider the fundamental perception of movement - velocity with no acceleration or reference frame is not perceptible. The pilot would have to literally look at the altimeter and pull the yoke in order to compensate. Provided he/she did so, the turn would actually happen, yes.

It is much more perceptible that you start pointing toward the ground as opposed to moving toward the ground. Again, if you're losing altitude, you might notice eventually, but seeing the horizon move creates... a very instinctive correction by the pilot.

Here I will get into the superior resolution.

For instance, when an airplane is banked and moves sideways through the air (as in the parallel move discussed above), the relative wind will hit the airplane from the side it is banked towards.

If the plane is moving right/left and downward (as a result of the acceleration from the point force diagram) then the tail presents a large area directly to the wind. This predicts a force moment exclusively in the aircraft's horizontal plane, which will point the aircraft down and right/left. The pilot will then pull the yoke back (possibly imperceptibly) to compensate for the horizon's movement. From a controls system analysis, the pilot's primary responsibility is likely to be maintaining the up/down orientation. As any other control input changes this, the up/down orientation control will automatically respond very vigilantly.

This is my preferred explanation, and I should disclaim that I have very little experience directly with flying planes and my writing is strictly academic physics oriented. With my current hypothesis, we have one major prediction to note:

  • Turning left/right is a 2nd order effect from aircraft banking

The alternative to this would be that the orientation of the aircraft left/right directly imparts a moment (which you should remember I disagree with). It is interesting to consider applying this to experience. How fast does does the turn start once the aircraft is oriented in the banked position? Does turning inherently result in a loss of altitude? My explanation predicts "slow" and "yes".


One more point was brought up about the moment of the engines/propellers. Unless you had counter-rotating dual (or more, even number) engines, this adds an asymmetrical component. It becomes easier to turn left or right, depending on the direction of spin. I know WWII era dogfighters were acutely aware of this fact in many situations. I think it's obvious one could use this fact to their advantage/disadvantage.

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I believe that the banked turn could be done like this:

  1. Tilt one wing's surfaces up and the other's down to get the plane to bank. The lift force is now slightly off to the side. You are now sliding to the side, as mentioned in the link you posted.
  2. Get the plane to pitch "upwards" from its own point of view. The force at this point is off to the side and slightly backwards. I don't know if this would be done with the wings or the tail.
  3. Undo step 2. You are on a different heading, but still banked and still sliding.
  4. Undo step 1. You are on that new heading and no longer banked.

For the block-on-ice portion: you describe: any thruster that applies force along a line that passes through the center of mass will not cause the object to rotate. You can add a horizontal component of acceleration, as you suggested, but the block won't spin and you won't be able to get it to move in a circle with just that thruster.

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Yes, this is essentially what the link posted describes. But it is rarely discussed. I'm wondering though if the plane's lift vector "naturally" stays perpendicular to both it's velocity vector and it's wing plane, without adjusting other control surfaces. This would help explain how a paper airplane turns. So maybe I should revise the question a bit. –  Fraggle May 14 '12 at 16:00

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