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To clarify my question further I'll use a practical example.

Here is the simplest paper plane design I could think of - it is a rectangular thick drawing paper with a weight attached at the middle of one of the long sides. I used a paper clip holding a metal key.

enter image description here

Then I perform this simple experiment: I hold the sheet of paper at the "tail" with approximately 45° nose down attitude and simply let it go - that is, there is no initial velocity. If the weight has the proper mass, the sheet of paper glides gracefully forwards (in fact I was quite surprised that such a crude design behaves so well).

What bothers me is that I can not explain to myself why the plane has such behavior. Here's the way I try to analyze it (which should be wrong since it does not stand to the facts). The plane is drawn as seen from the side:

enter image description here

Since the wing is of very simple rectangular shape the center of pressure should be in its center. On the other hand by placing the weight at the nose we have moved the center of mass a little forward (which is said to be necessary for longitudinal stability of an airplane).

At moment t=0s when I let go of the plane there are no other forces acting on it except gravity. The plane gains a little velocity downwards and immediately the pressure forces begin to act. I've drawn the total pressure force and for clarity decomposed it into lift and drag components (the magnitudes and directions are only illustrative). The lift component explains why the plane starts to gain forward velocity.

If the forces act in this way then the pressure force should create a torque around the center of mass and rotate the plane counterclockwise (from the perspective of the drawing). But in reality the exact opposite happens - the plane rotates clockwise and continues to gain forward speed. We know that in airplanes the tail is used to produce torque which cancels the torque of the main wing but there is nothing here to cancel it. In fact the torque should act until the angle of attack is nullified and only the drag component remains. That is how a dart or an arrow will fall if dropped in the same way! The difference between dart and the plane above is that the center of mass of the former is much further away from the center of pressure. But still I don't understand where is that clockwise torque coming from and how it gets cancelled later when the plane gains enough forward velocity and achieves pitch stability.

EDIT/ADDON: Thanks everyone for the answers.

Since the bending of the paper really adds much complexity to the analysis I made the plane from firm cardboard as you can see from the following footage:

Anim

The wing is also square now. Since it is considerably heavier than before it does not glide so far but still the effect is the same.

Now with the wing curvature out of the picture, the explanation should lie in what Peter suggests: that the center of lift is not at the center of the wing but 1/4 closer to the nose. Something like this:

enter image description here

Why the lift center is not at the center but at 1/4th chord is quite unclear to me but I'll try and read about it.

Then the last thing that I would like to know would be how the wing flies in more or less stable angle of attack (or at least in some oscillating mode). The explanation would be easy if this 1/4th rule is not exactly solid and the point of pressure can move forwards and backwards in function of the angle of attack and/or velocity. Then if it moves very close to the c.m the moment will tend to zero. If it moves behind the c.m the torque will be ccw, the nose will drop and then the pitch oscillations could be explained. But is this really the case? Or the physics are much more complex for such simplification?

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very professional graphics! –  innisfree May 7 at 20:30
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Beautiful question! –  Floris May 7 at 20:43
    
Could you please elaborate on this part? "I hold the sheet of paper at the "tail" with approximately 45° nose down attitude..." If your drawing of the initial conditions is accurate, are you holding the paper flat at that angle? (it would have to be resting on something, like your hand) –  Xeren Narcy May 8 at 0:54
    
You're seeing "decalage" in action. Here's some basic stuff to read. –  Mike Dunlavey May 8 at 18:56
    
@XerenNarcy see the edited question –  Cleric May 8 at 18:57

4 Answers 4

up vote 6 down vote accepted

This isn't an exact description but it should give some insight.

I think it's better to tackle this problem in stages. First what happens if we drop a horizontal paper? Well, it's edges curve up like this:

Now what happens with 45° paper? Its edges curve up again. Pressure drag is a very sensitive function of angle of attack. It's maximum for a plate perpendicular to the flow drops to zero for a plate parallel to the flow (at which point friction drag dominates). It drops rapidly (most empirical relations predict something between $cos(\theta)$ and $cos^2(\theta)$). So center of pressure is shifted to the left as you see in the picture (red arrows represent pressure vectors).

After gaining some horizontal speed the relative airspeed vector changes so that there's even less pressure on the right edge (shifting the center of pressure even more) and it drops faster than the left edge while the left edge curve up more until it stalls and the situation is reversed. It means that it is unstable, like the second plane bellow:

Now for a positive lift airfoil to be stable its center of mass should be closer to the leading edge than the center of pressure. That's what the weight does. It keeps angle of attack of leading edge low so that it does not stall.

Update That's why I said this isn't exact description. When an object is placed into a flowing fluid, some of it will flow over the object, some of it under. This is a wind tunnel test of a flat plate.

The dividing line is termed stagnation streamline and it's terminated at stagnation point on the object. The fluid will be stationary at this point. As you see in the picture for low angle of attack it's very close to the leading edge.

According to Bernoulli's principle, in a laminar flow pressure will be greatest where the fluid has no velocity (i.e stagnation point) because at that point all of the dynamic pressure of the fluid will be exerted at the object. So at the lower surface pressure will be greatest at stagnation point, but what about upper surface? Velocity distribution and consequently pressure distribution is not uniform at the upper surface but they're also in favor of leading edge. This figure belongs to the Aerodynamics book by Theodore von Karman (and it explains in chapter IV how they arrive at this):

Don't look at the left one, that's for supersonic. So center of pressure is closer to the leading edge for subsonic speeds at low angle of attack. But for very high angle of attack stagnation point is close to the middle and pressure distribution is uniform at the other side so center of pressure is at the middle. It's merely an empirical observation that for low angles of attack center of pressure is about quarter of the chord length behind leading edge.

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Thanks for the nice graphics! I've edited the question to make the wing rigid. I wonder if a similar pressure diagram can be drawn for the unbendable wing to explain the 1/4 chord rule? –  Cleric May 8 at 18:56
    
@Cleric Well not as nice yours! I've updated the answer –  Azad May 9 at 6:17

The first factor in stability is pitch inertia. Even though it is just a sheet of paper, its moment of inertia around the pitch axis is quite high, so any pitch motion is slow.

Next is aerodynamic pitch damping: As the paper starts to pitch, local forces at the edges will produce a stabilizing moment.

That the paper will slowly pitch up is due to the location of the center of lift, which will act at a quarter of the chord. The precise location is slightly ahead of the quarter chord, the more so, the more slender the paper is (ratio of span to chord length). This creates a pitch-up moment, and since the paper itself is rather floppy, will bend the whole paper into a graceful curve. This again helps to stabilize the paper, because a negatively cambered wing is naturally stable.

Your paperclip weight is needed to shift the center of gravity forward to one quarter of the paper airplane's chord. This is where the resulting lift force will act, so by shifting the cg forward, you minimize pitch moments. You could as well fold the first half of the sheet of paper into a tight roll of paper - this will produce the same effect as the paperclip. Once the pitch angle increases, so does lift and angle of attack. Beyond an angle of attack of maybe 10° or 15°, airflow on the top will separate, which will shift the center of lift backwards. Also, lift stops to increase with further angle of attack increases. Now the lift creates a pitch-down moment, which helps to regain the attitude at which the flow is still partially attached. Especially when using stiffer cardboard, this stabilizing effect of flying with a partially stalled wing is easy to reach once the center of gravity is right.

In folded paper airplanes, you will notice that a slightly bent-up trailing edge will help to stabilize the paper airplane, whereas folding it down will make sure it goes into a dive quickly. By folding the trailing edge at least partially up, you will create a local area of lower lift (or actual downforce), which will see a proportionally stronger increase of lift when the whole paper airplane pitches up. This results in a pitch-down moment, and vice versa. This is all what is needed to give a paper airplane positive stability.

A simple sheet will be too floppy to benefit from the stabilizing effect of a bent-up trailing edge, though.

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Thank you! Could you please take a look at the edited question? –  Cleric May 8 at 19:00
    
Thank you, Peter! I've selected the answer of Azad because it's more graphical. Otherwise yours is also most informative and gave me the fact of the 1/4 chord. I greatly appreciate your effort! –  Cleric May 9 at 8:43
    
@Cleric: If you want why the pitch-up (what you call "clockwise") moment dies down read the 4th paragraph: Beginning flow separation will shift the center of pressure backwards. If you want to know why there is a pitch-up moment, please read this and this. –  Peter Kämpf May 9 at 11:05

Cool! I just had to try it. A penny may / may not have been just the right mass, and I used a slightly heavier than average grade of paper. But it was sort of stable.

When I drop it, the front corners bend upward. This increases the drag at the front end. This seems to be the source of the bend you show. The nose comes up and the plane slows.

Sometimes it is stable. Sometimes the nose goes too far up, it stalls, and then the nose drops. If you drop it down some stairs, it has more time to develop unstable behavior.

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If you see it nosing up and stalling, that means move the weight further toward the front. It needs to be forward of the center of lift, wherever that is (but not too far forward). –  Mike Dunlavey May 11 at 17:24
    
Thanks. In this case, it was already at the front edge. So adding another paperclip might do. The flexing of the forward corners of the paper moves the center of lift back by giving it a somewhat triangular shape. So even stiffer paper might help. –  mmesser314 May 12 at 0:53

I'm still seeing explanations being pulled out of the air, so to speak. I'm a 100-hour student pilot, and let me tell you what every pilot must absolutely learn about stability in the pitch axis, because if you get it wrong just once, you void the warranty on the plane and on yourself. You don't need a PhD in physics to get the point.

Let's start with a simple ordinary small airplane like a Cessna 172 (details here), like I learned on. You must be aware of the center of gravity (CG), and it must be forward of the center of lift of the main load-bearing wing. It has a range, with a forward limit and an aft limit. You must be able to do a simple calculation based on the weight of fuel, the weight of passengers, and luggage, and their position forward or aft. You're basically adding up torques due to weight and moment-arm, and this tells you if your CG is within limits.

Why is the CG forward of the center of lift? Because that means the tail has to push down. (It is actually an inverted wing.) This creates a pitch-wise torque between the main wing and the tail, and the strength of that torque is proportional to speed squared.

Why do you need that? For stability of speed and pitch. If for some reason the plane hits an updraft and pitches up, what happens? Since it is going up, it loses speed, which lessens the pitch-wise torque, which allows the heavy nose to pull down, so it returns to its original pitch and speed. Same if it hits a downdraft - it speeds up, more nose-up torque, and it returns to original pitch and speed.

You may have heard of "trimming for a desired speed". That's just a little wheel in the cockpit that effectively adjusts the torque by adjusting the tail. If you set it for more torque, the plane slows down, less-torque - plane speeds up.

What happens if you move the CG too far aft, by putting lead bricks in the baggage? You no longer have that stability in pitch. You have to concentrate really hard to keep it level. The slightest finger pressure on the yoke (the stick that the pilot hangs onto), and you are zooming up and stalling, or zooming down and overspeeding, and oscillating chaotically between the two. If you're not careful, you can get nose up and slow and actually start a backslide, that you can't get out of, all the way to the ground.

Of course, if the CG it too far forward, you get different problems, that aren't too hard to figure out.

Fun?

Same goes for any simple "aircraft" like your paper board. There needs to be a speed-dependent torque causing it to pitch up, and a forward weight pulling the nose down. Even if the board is flat, its center of lift is forward of the center, so there is a torque. Just because a "flying wing" doesn't seem to have a tail, its trailing edge has enough of an upward tilt to provide that torque. Some planes have a canard instead of a tail, but it's the same idea - faster speed = greater nose-up torque, and the CG is always forward of the center of lift.

Exception: Military jet fighters. They prefer instability so they can maneuver very quickly. They have computers manage the controls, because a human could never do it. They look down on plain-old planes, calling them "God-fearing" aircraft :)

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Dear Mike - I added a link showing the Cessna 172 (so readers can know what it is like - I had no idea) and also a phrase explaining "yoke" which I hadn't heard before (change if you can think of a better one) Great explanation overall. I had NO idea whatsoever the tail was an inverted wing but it makes wonderful sense given your answer. I also understand military planes were the first to use quaternions in the mathematical model for the plane that the control system uses to "visualize" the plane's attitude and fiber gyros for the basic measurement to remove any possibility of gimbal lock. –  WetSavannaAnimal aka Rod Vance May 10 at 1:33
    
@Wet: Thanks for edits, and yes, quaternions are basic "rocket science". –  Mike Dunlavey May 10 at 15:19
    
Stability of airplane with tail has been easier for me to understand. But for single wing is not exactly the same. You say "There needs to be a speed-dependent torque". The torque magnitude certainly is speed-dependent but the question is when it goes to zero (which means pitch is stable)? One way is if the pressure force acts in line with CG-CoF axis, but we have no lift in this case - only drag along the wing. The other way is when CG and CoF coincide (moment=0 since the lever arm=0). Since we do have lift it turns out that CoF must move around CG in dependence of AoA. –  Cleric May 11 at 7:42
    
@Cleric: The speed-dependent torque due to differential lift doesn't go to zero. The weight in the nose produces its own torque that counteracts the speed-dependent torque, so it's stable when the total torque is zero. –  Mike Dunlavey May 11 at 12:22
    
But the weight can not produce a torque! All torques are created around the CG and since the weight acts through the CG (lever arm=0) there is no torque. The heavier nose moves the CG forward - it does not create "second" CG which rotates around the "first". If you hold the plane horizontally and drop it in vacuum no rotation will occur no matter how heavy the nose is - all particles accelerate the same. –  Cleric May 11 at 14:03

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