0
$\begingroup$

What physics is involved with wing flex when in air?

A simplified answer will suffice.

I can understand when on the ground that a heavier wing makes it bend more down.

My feeble understanding: When in air isn't it only positive G that makes it bend up, and weight of wing don't really matter, instead the weight of the body matters, yes? If that is the case the amount of normal force don't really matter, correct?

Improve this question
$\endgroup$
4
  • $\begingroup$ "the amount of normal force don't really matter..." Which normal force might you be referring to in this sentence? $\endgroup$
    – BowlOfRed
    Feb 28, 2020 at 22:01
  • $\begingroup$ Normal force. Its equal to lift at AoA of 0. At AoA of 90 its equal to drag. $\endgroup$
    – Invariant
    Feb 28, 2020 at 22:04
  • $\begingroup$ Yes, weight distributed well outboard on the wingspan decreases the wing-bending moment in actual flight, unlike the case when the aircraft is parked on the ground. $\endgroup$
    – quiet flyer
    Feb 28, 2020 at 22:38
  • $\begingroup$ Some info in this post $\endgroup$
    – Koyovis
    Feb 29, 2020 at 0:16

1 Answer 1

Reset to default
1
$\begingroup$

Here's a simplified, almost trivial answer.

Call "normal force" the component of the net force on the wing that is perpendicular to the wing chord. The net force: gravity (in whatever direction), plus any force due to air moving past it (in whatever direction(s)), plus any force due to mounted ordnance, or mounted thrust-vectored engines, or air-show wing walkers, or giant magnets, or whatever.

The greater that force, the more the wing bends.

The stiffer the wing (spar), the less the wing bends.

The more that force is nearer the tip than the root, the more the wing bends.

The wing just acts like a (complicated) spring, roughly obeying Hooke's law.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ It is important to mention that gravity means just gravity acting on the wing itself. If you add the weight of the fuselage, you'll obviously get zero for level flight, but the point is to get the force between the two. $\endgroup$
    – Jan Hudec
    Feb 29, 2020 at 23:28
  • $\begingroup$ Yes. "Net force on the wing," not on anything else. $\endgroup$
    – Camille Goudeseune
    Mar 1, 2020 at 3:46
  • $\begingroup$ Although I wrote a simplified answer will suffice, I maybe didn't mean that simple. But its okay, I found a good answer elsewhere. I will mark this as answer at least until somebody answers with a little more details, perhaps some formulas. $\endgroup$
    – Invariant
    Mar 4, 2020 at 12:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .