I don’t know that anyone has expanded the equations from this document to a 6+ wheel drive drop center, so let me briefly do that. For this purpose, let’s assume that the robot’s COM is over its geometrical center.

The purpose of a drop center is that only 4 wheels should be on the ground at any time. We can split these 4-wheel segments into two categories: those where the COM is inside the area (e.g. the center of an 8WD) and those where it is on the edge or outside (e.g. a 6WD).

For the first type, the analysis is the same as with a standard 4WD. But since the drivetrain is split into multiple sections, each section is significantly wider than it is long. Since the COM is over the we can use eqn(5), which can be rearranged to \frac{L_{TW}}{L_{LB}} > \frac{\mu_y}{\mu_x}. Assuming \mu_x \approx \mu_y, so \frac{\mu_y}{\mu_x} \approx 1, \frac{L_{TW}}{L_{LB}} > \frac{\mu_y}{\mu_x} is satisfied by L_{TW} > L_{WB}, a drivetrain section that is wider than it is long. So the robot can always turn on these drivetrain sections.

For the second type, when the robot’s CoM is over its geometrical center (between the two center wheels) the equivalent 4 wheel analysis would have L_{CY}=(k+\tfrac{1}{2}) L_{WB}, where k is the number of additional drivetrain segments between the one being analyzed and the robot’s center (k \geq 0). Eqn(4) can therefore be simplified to \frac{\mu_x L_{TW}}{\mu_y L_{WB}} > -4k(k+1). If k \neq 0, since \mu_x, \mu_y, L_{TW}, L_{WB}, and k are all positive, this equation is always true. If k=0, we end up with \mu_x L_{TW} > 0, which is also always true. Therefore the robot can always turn on these sections as well.