I'm building a ground effect vehicle, but I'm failing to stabilize the lift-off. At a certain speed, the lift force is equal to the weight force, so the vehicle lifts off. But this implements a change in angle of attack, so also an uncontrolled change in lift, as can be seen in this video.

I know that I can solve this with an elevator that can oppose the change in angle of attack. Now I'm searching for an equation to calculate the elevator angle to get a stabile and controllable lift off.

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    $\begingroup$ I think your problem lies in the CoM placement compared to where the CoL is, each time you get off the ground the vehicle pitches back dramatically and stalls (or that is just the drag of the tail section) $\endgroup$ Nov 9, 2014 at 15:00
  • $\begingroup$ I would approach this problem a little differently. Think about calculating the amount of lift that the elevator is producing given its angle and speed, and go from there. $\endgroup$
    – Keegan
    Nov 9, 2014 at 21:21

1 Answer 1


You need more than the correct elevator incidence. Natural longitudinal stability works by distributing lift over the flow direction such that forward surfaces create proportionally more lift than rear surfaces.

Now it is important to know that only pure supersonic lift will be evenly distributed over the chord length of a wing. In subsonic flow the center of lift is at 25% (and even less for wings of small aspect ratio), so you now need to place the center of gravity such that it will be ahead of the 25% chord line of the sum of all lifting surfaces. To distribute lift for this forward center of gravity location, the incidence of the elevator needs to be several degrees smaller that that of the wing. Adjust incidences (say 5° for the wing and 0° for the tail), and then trim the center of gravity by adding ballast or shifting masses to make the aircraft fly straight.

Since airfoil camber shifts the incidence of zero lift to more negative values, you can alternatively use a cambered airfoil for the wing (say, at now 2°) and use a symmetric airfoil for the elevator at again 0° incidence.

  • $\begingroup$ tl;dr, adjust center of mass forward $\endgroup$ Nov 9, 2014 at 18:34
  • $\begingroup$ @ratchetfreak: Right, but I hope they now understand why. $\endgroup$ Nov 9, 2014 at 20:59

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