Is there ground effect flying below an infinite 2D plane as opposed to above one?
In computational fluid dynamics (CFD), ground effect can be simulated by flying two airplanes / wings in mirror formation (the lower one inverted).
That mirrored wing is only a mathematical trick. By distorting the flowfield in an identical, albeit inverted, way, the effects of the lower wing cancel out those of the upper wing in the plane of symmetry, and vice versa. This way, the flowfield on both sides looks exactly like one where the plane of symmetry cannot be crossed by the flow, just as the ground cannot inhale or expel air.
Now, with the more precise description: Yes, there is also a ground effect for the plane flying below a hypothetical ceiling. Of course, it would help if the vertical tail would not stick up, for minimum distance and maximum effect. Just as the ground stops the downward movement of the wake, a ceiling will prevent air from filling the space left by the downward movement of the wake, so the wake will stick to the ceiling. Call it the ceiling effect, if you will. In both cases the induced drag drops and the center of pressure moves a bit backwards. I wager to predict that the drag reduction over distance relative to wing chord is the same as in &quot;regular&quot; ground effect.
Proof: Fly a model helicopter indoors and get it up to the ceiling. You will notice that it will get stuck there and needs much less power to stay airborne. The rotor blades are like wings, and due to the rotor position on top of the craft, the ground effect is very noticeable. In order to get the helicopter unstuck you must reduce power a lot, such that the helicopter will drop like a stone.
In CFD, this can readily be proven by using the trick described above, only now with both wings / airplanes flying inverted to their original orientation for the ground effect calculation.