Optimal Wheelbase Length on Skid Steer Drivetrains

Is there a calculation or general rule that one can turn to so that you can best determine the maximum wheelbase length for skid steer drive systems? Such that your turning scrub doesn’t start to negatively affect your driving dynamics.

My understanding of why 6WD/8WD systems usually drop the center wheel(s) is to reduce the effective wheelbase length. How does one know the best length to use for a particular wheel diameter? Empirical testing? …or is this something that physics gives a specific answer for?

Here’s a white paper for the physics approach: White Paper Discuss: Drive Train Basics

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If you know the coefficient of friction of your wheels and the weight on them, you should have all you need to calculate the torque needed to turn, then from there the torque to the drive wheels needed to produce that turning torque. That being said, I’ve never heard of anyone explicitly trying to optimize for turning scrub - unless you have some really weird aspect ratio on your drivetrain, I’d say you’re bikeshedding.

You are likely correct. I just know we had issues on a 8 wheel skid steer drive we ran two years ago during the offseason. We didn’t drop the center wheels instead we had corner omni wheels. I think in the end… it was a combination of our wheel size (6") and the gear ratio not being high enough. In the end, I just wanted to understand the physics better.

Those pneumatic wheels have a ton of friction. We had the same issue, and our solution was to inflate them to the absolute maximum, and switch the front and back wheels to plaction with a slightly smaller diameter. That gave us enough rocker to fix the problem. We were shocked because even with a 3/16 drop they were all scrubbing.

We were running 6" white AM traction wheels in the center 4. Not pneumatic tires.

This reminds me of when my team used pneumatic wheels in 2016 for Stronghold. We actually wrapped duct tape around the wheel to reduce the friction. Surprisingly, it worked well for how crude it was, though we did have to change the duct tape periodically.

General rule of thumb, if your track width is wider than your wheel base is long, you should be able to turn. Give yourself some margin for error here.

As laid out in the white paper Joe Ross linked, a lot of this simply comes down to lever arms around your center of rotation. When turning, the distance between your wheels and your CoR on the “x axis” will be acting as a lever for the wheel rotation trying to turn your bot, while the distance in the “y axis” will be acting as a lever for the scrub forces resisting that turning. Assuming the coefficients of friction are the same for each motion of your wheel (they very obviously wouldn’t be for an Omni wheel), the bigger lever is going to win out.

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.

In reality, it comes down to driver preference. Less scrub usually means more responsive turning, at the cost of some oversteer.

A general trend I’ve noticed recently is to have very minimal scrub (corner omnis on a 6w, for example) with some software compensation for the oversteer.

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