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How can I get the y force of an airplane in ground effect when using image theory? is the y force zero because of the mirror cancelling it out? bare in mind I'm talking about a case where there is a sideslip angle.

Many thanks

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Ground effect happens because the air cannot flow around the airplane as it would in free flight. The impenetrable ground gets in the way, and using the mirror image in CFD is done only to create an opposing flow field so the direction of flow in the ground plane (= plane of symmetry for the mirror image) will always be horizontal.

In order to create lift, the airplane is adding a vertical speed component to the air and ground effect is the consequence of a restricted vertical movement for this air. In order to create a side force, air would need to have a horizontal speed component added, orthogonal to the direction of the initial flow speed. Since there is nothing which restricts this horizontal movement, the flow pattern will not really differ from free flight, and so should the forces.

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The mirror image won't cancel it out.

To a first order, I would assume it is unaffected. I haven't ever run a detailed analysis to see if there was an effect, but I would be surprised to see one that wasn't very small and configuration dependent. I.e. I doubt a general statement can be made about trends and magnitude.

Any aero tool that can model a ground effect should be able to answer this question for you with a little experimentation.

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  • $\begingroup$ Thank you for your reply. Does that mean (side force in ground effect - side force out of ground effect) = 0 (or approx 0)? $\endgroup$
    – MarcoD
    May 4, 2023 at 19:26
  • $\begingroup$ I would think so, yes. $\endgroup$ May 4, 2023 at 20:03

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