I've been reading the FAA's Glider Flying Handbook, 2013 (FAA-H-8083-13A). In chapter 3 ("Aerodynamics of Flight"), the book is discussing stability. On page 3-12, it states:

Dihedral is the upward angle of the wings from a horizontal (front/rear view) axis of the plane. As a glider flies along and encounters turbulence, the dihedral provides positive lateral stability by providing more lift for the lower wing and reducing the lift on the raised wing. As one wing lowers, it becomes closer to perpendicular to the surface and level. Because it is closer to level and perpendicular to the weight force, the lift produced directly opposes the force of weight. This must be instantly compared to the higher and now more canted wing referenced to the force of weight. The higher wing's lift relative to the force of weight is now less because of the vector angle. This imbalance of lift causes the lower wing to rise as the higher descends until lift equalizes, resulting in level flight.

That doesn't sound right to me.

This paragraph says that there's more lift on the lowered wing and less lift on the raised wing. That's not true, is it? The amount of lift only depends on a wing's airspeed and angle of attack, not on the wing's bank angle.

The paragraph then explains that the lowered wing produces more upward lift than the raised wing. I think this is true, but it's not relevant, because upward lift isn't the only lift which contributes to rolling moment. Rolling moment depends on the total lift (and the direction of the lift).

All in all, I think that dihedral can't produce a stabilizing rolling moment in the way that the book says that it does. Any stabilizing effect must come from a difference in airspeed or angle of attack between the two wings.

Is the book's description of dihedral correct?

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    $\begingroup$ Possible duplicate of How does the dihedral angle work? $\endgroup$ – Pilothead Jul 11 '18 at 18:59
  • $\begingroup$ I agree with you. Dihedral effect requires sideslip. They seem to be describing what people call the "pendulum effect," which is due to the dihedral but is not the same thing. $\endgroup$ – TomMcW Jul 11 '18 at 19:19
  • $\begingroup$ @Pilothead I'm asking about this one particular explanation of the dihedral effect. Only one of the answers on that question addresses the FAA's explanation. That answer essentially says that the FAA's explanation is correct, but it's also downvoted, and there are comments which say that the answer (and therefore the FAA's explanation) is not correct. So, there is apparently no answer to that question which correctly answers my question. $\endgroup$ – Tanner Swett Jul 11 '18 at 19:23
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    $\begingroup$ @PeterKämpf I think I'll post up a question about it later so I can figure it out. We haven't had a question specific to the "pendulum" thing. $\endgroup$ – TomMcW Jul 11 '18 at 22:13
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    $\begingroup$ I don't think this is a duplicate. Although the question is about the same topic, this question asks about understanding a particular detail which isn't really mentioned in the other question or its answers. $\endgroup$ – Dan Hulme Jul 12 '18 at 8:33

No, the explanation is not correct.

Tanner, you are right when you say that lift does not depend on bank angle. Lift is caused by a pressure difference, and pressure can only act perpendicularly to a surface. Therefore, the lift on each wing and its lever arm to the center of gravity won't change with bank and no "correcting" rolling moment is created.

Instead, what does happen is a side force from the wing's bank angle which will accelerate the airplane sideways. This in turn will result in a sideslip condition, and only now will a dihedral effect show up: By changing the angle of attack on each side differently, the sideslip in combination with dihedral will create a correcting rolling moment.

As you correctly observe, a stabilising effect can only come from a difference in angle of attack between both wings, and that only occurs in a sideslip.


No the paragraph is not correct; the same is true of any "explanation" of dihedral that doesn't rely on the difference in angle-of-attack between the two wings generated by sideslip.

There is a shocking amount of garbage in the FAA's "Glider Flying Handbook". It is the most error-filled flight training publication I've ever seen. It doesn't even contain an accurate depiction of the L-D-W vector triangle to explain why the L/D ratio is the same as the glide ratio through still air. A good question for Aviation SE would be "What are all the errors in the FAA's "Glider Flying Handbook"?

  • $\begingroup$ You will notice this was written early in my tenure here. As your "crusade" to smite all you do not agree with and handle by saying "no, you are 100% wrong" and calling handbooks "shocking garbage" continues, I will comment. What I was referring to (accompanied with vector diagrams) was (respect to the universal and unchanging CG vector) a difference in lift caused by the dihedral rolling relative to CG. Further more, if one can picture trying to bend the rope in a tug of war between lift and drag, what will happen? If have simply gone crazy, I would stop and evaluate why. $\endgroup$ – Robert DiGiovanni Oct 29 '18 at 17:22
  • $\begingroup$ Correction, lift and weight, tug of war. Try bending rope. It is not a pendulum, but it might be useful as a training aid. It's important for you to understand it, getting other people to is a little different, and yes, can be frustrating. $\endgroup$ – Robert DiGiovanni Oct 29 '18 at 17:33
  • $\begingroup$ This answer would be improved if the last sentence read "A good question for Aviation SE would be "What are all the errors or conspicuous omissions in the FAA's 'Glider Flying Handbook'"? $\endgroup$ – quiet flyer Oct 29 '18 at 17:37
  • $\begingroup$ OK, I do not know if this ASE has a committee to discuss modernizing or upgrading aviation information, but this is what I am sort of trying to do. Speaking the same language. Yes, there are a boatload of misconceptions out there and better understanding and training will certainly make it safer. My first pick would be "forward CG is always good!". We work towards these things. Sometimes it takes time. $\endgroup$ – Robert DiGiovanni Oct 29 '18 at 18:00
  • $\begingroup$ This answer would be improved by stating (re the FAA Glider Flying Handbook) "It also claims that it is dangerous to slip gliders with swept leading edges due to the resulting slip-roll coupling-- totally ignoring the much larger slip-roll coupling created by dihedral (e.g. any modern glass ship other than Fox, Swift, etc) and/or high wing placement (e.g. 2-33, 2-22) in a typical sailplane." $\endgroup$ – quiet flyer Oct 29 '18 at 21:30

A "dihedral" wing will also AERODYNAMICLY roll away from a slip AT ANY ROLL ANGLE. An " upward " pointing vertical stabilizer also contributes to this.

But don't forget those gravity force vectors, they are always there!

Think of "lift" as that which opposes gravity. Your vertical components. Roll one wing parallel to the ground. ALL of its force is "lift". The other wing, tilted or "canted" sideways, has some of its force as "lift" (vertical component) and some as sideways (horizontal component). Relative to CG, one side has more VERTICAL lift than the other, which rolls the plane back to level.

You will note dihedral stabilization, being GRAVITY driven, ceases to exist when the plane rolls 90 degrees, and is destabilizing (anhedral), when the plane rolls 180 degrees.

It is important to look at the whole picture. The AERODYNAMIC " rolling moments " of both wings are equal. They are producing the same amount of force perpendicular to their surface. Look at a plane as 2 mirror images right and left. A dihedral will roll back to level at any roll angle up to 90 degrees due to GRAVITATION FORCES.

Difference here in rudder induced slip, with strong change in relative wind, and a turbulence induced roll, where slip rate is initially very slow.

Dihedral and pendulum stabilization are gravity driven and are REAL. They become insignificant, but no less real, as aerodynamic forces predominate (faster).

Remember, this is a quote from a GLIDER handbook (slow).

  • $\begingroup$ Thanks for your answer. I have the same question for you that I had for Carlo's answer. It looks to me like you're only looking at the effects of the vertical component of lift, and not the horizontal component. Is that right? When you take the horizontal component of lift into account, is there still a net force making the low wing rise? $\endgroup$ – Tanner Swett Aug 8 '18 at 14:22
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    $\begingroup$ Hi Tanner: that's the beauty of breaking it down into components. Roll until one wing is parallel to the ground. All of its "lift" is straight up. Now look at your other wing, it's "lift" is more to the side. When you see this, imagine how $\endgroup$ – Robert DiGiovanni Aug 8 '18 at 19:19
  • $\begingroup$ how the forces are different RELATIVE TO GRAVITY. The plane slips to the side and rolls until the wing forces equalize. Bob $\endgroup$ – Robert DiGiovanni Aug 8 '18 at 19:26
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    $\begingroup$ Please do not create multiple answers to the same question. If you have to add anything to your existing answer, please edit it. $\endgroup$ – Federico Aug 10 '18 at 15:42
  • $\begingroup$ "You will note dihedral stabilization, being GRAVITY driven, ceases to exist when the plane rolls 90 degrees" -- not so $\endgroup$ – quiet flyer Oct 29 '18 at 16:28

Yes, it is correct. Consider an airplane with a positive 5° dihedral to its wings. If the lift the airfoils generate is perpendicular to its span line, this means that in level flight the lifting force parallel to the plane’s vertical axis is L*cos(5°) for both wings. If the airplane is rolled 30° to the left, then the lift generated by the left wing parallel to the force of gravity is L * cos(25°) whereas the lift generated by the right wing is L*cos(35°). This imbalance in lift causes a moment about the plane’s CG, causing it to roll back level and providing both a positive lateral static and dynamic stability.

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    $\begingroup$ It looks to me like you're only looking at the component of lift which is parallel to the force of gravity, and not the component which is perpendicular to the force of gravity. Is that right? When you add in the component which is perpendicular to gravity, is there still an overall rolling moment, or is the rolling moment canceled out to zero? $\endgroup$ – Tanner Swett Jul 11 '18 at 20:16

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