Yes, but what's the point of accurately calculating the ground effect? This effect is small because it only concerns the induced power, which is less than one third compared to the friction power of the blades plus the parasitic power of the airframe.
A very simple solution for calculating the ground effect of a gyrocopter, whose wake is always almost parallel to the ground, is to replace the ground by an identical, mirrored rotor. So, his marginal circulations cancel out the vertical components at ground level of the first, as would the true ground.
So all we just need to do is quantify the difference in induced velocity at the center of the real rotor with and without the image rotor, knowing that circulation gives a tangential velocity inversely proportional to tips distance
For example, if the real rotor flies at R above the ground, then the edge of the mirrored rotor is 1.56 R from the center of the first, and the circulation it adds is inclined at 50° to the vertical. This gives a vertical induced speed 0.6 times that of the mirrorless rotor.
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