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I have researched some about dihedral effect due to a question that I had, and I discovered that my question is essentially already answered, here and here, collectively.

This has led me to a new question because the information talks about slips.

If an airplane is flown in perfect coordination during a turn, does the dihedral effect still work to level the wings?

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    $\begingroup$ Related: How does the dihedral angle work?. It shows why the flow must be asymmetrical for the dihedral angle to work. $\endgroup$
    – mins
    Commented Jun 15, 2016 at 18:51

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No.

The dihedral effect depends on a flow asymmetry between left and right wing. Coordinated flight means that both wings have the same flow conditions, and the only asymmetry between both is caused by control deflections.

You do have a flow asymmetry during a turn (coordinated or not) from the speed difference between the inner and the outer wing which grows with the turn rate and wing span. This does indeed cause more lift on the outer wing, but due to the difference in dynamic pressure, not due to dihedral. Since dihedral causes a side force component of lift, you can argue that the outer wing creates more side force which pushes the plane into the turn. But the pilot will compensate the difference in dynamic pressure with the ailerons to stop any rolling motion once the desired turn rate is stabilized. This will also equal the lift on both wings.

Another asymmetry is caused by the location of the vertical tail aft of the center of gravity. Essentially, the tail swings around during the turn, which causes a sideslip condition at the vertical tail. The result is a side force which grows with turn rate and tail lever arm, and due to the high location of most vertical tails also causes a small rolling moment. But again the wing's dihedral has nothing to do with that.

Dihedral only comes into play if you build up a sideslip, and then the turn is not coordinated any longer. @Zeus is right with his point that dihedral stabilizes the aircraft and makes it easier to keep the turn a coordinated one. Now we can split hairs all day if flying in turbulence makes a perfectly coordinated turn impossible. When it comes to dynamic stability, a moderate dihedral effect is highly recommended.

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No but yes.

In a perfect turn (or level flight), everything will be symmetrical (except for the minor effects that Peter mentioned). So seemingly no effect.

However, the whole concept of stability requires that when a disturbance occurs, an opposing force or moment is generated. Even if you think that you are making a perfect turn, this perfection comes largely from inherent stability of the airplane, which automatically compensates small disturbances that are always happening. This includes the 'dihedral effect'.

Try to make a perfect turn in a severely unstable aircraft, say, with strong anhedral (perhaps by programming it in a simulator), and you'll see what I mean.

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  • $\begingroup$ Of course, my question has a hypothetical element for the purpose of isolating the specific phenomenon. $\endgroup$ Commented Jun 15, 2016 at 15:47
  • $\begingroup$ Uh, no: The disturbance will awaken the dihedral effect, but this will both cause a rolling motion and a yawing moment, so this will not just reduce the disturbance, but will awaken a lot of additional effects which together will make the aircraft roll and yaw in what is called a dutch roll. $\endgroup$ Commented Jun 15, 2016 at 16:33
  • $\begingroup$ @PeterKämpf, of course, there will be many effects (this is why I said 'includes'), but if we start talking about dynamic stability, it will be yet another topic. My point was just that in the real world, 'perfect' conditions don't exist without some sort of stability, which in turn involves the effects in question. It's worth keeping this in mind. $\endgroup$
    – Zeus
    Commented Jun 16, 2016 at 0:12
  • $\begingroup$ Your point about flying with strong anhedral is very valid. +1 $\endgroup$ Commented Jun 16, 2016 at 8:24
  • $\begingroup$ Would dihedral have the opposite effect in a skid then? $\endgroup$ Commented Jun 17, 2016 at 4:49

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