# Transmitting power to the outer wheels of a drivetrain

So, I am writing a drivetrain presentation for my team’s new recruits and I have a pretty dumb question .

Why is power being transmitted to the outer wheels of a drivetrain rather than just powering the middle ones and having the outer ones just rotate freely on a shaft (lets say there are no obstacles to pass and the field is flat).

I’m sure there’s a pretty obvious answer, I’m just looking for the right definition for my presentation.

If there were no obstacles to pass, and the field is flat, AND there are no other robots, or game pieces you have to move around, then it would work. But real life isn’t like that, and often, much of the weight of the robot gets transferred to the outboard wheels. This leaves very little weight on the center wheels, and they can’t move the robot if they are spinning relative to the floor. So, you have to power the outboard wheels.

If you want to see how it works in real life, take the belts off and play a match, see how you do.,

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Jim’s explanation also highlights why in a typical 6 wheel set up the center wheels are dropped 1/8" or so. The robot has a tendency to tip under acceleration. Powering all wheels allows 4 wheels to be powered (rear and center) during that tip.

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Thanks for the quick replay!

Just a follow up question, does this explanation means that the number of wheels effect the power and speed of my robot?

Jim’s got the right answer. It’s the same reasoning as to why all-wheel-drive cars and trucks do better on uneven ground than two-wheel-drive vehicles.

To your followup - it’s complex.

The ultimate goal is to maximize the amount of power pulled from the battery, and drive it into the ground via the wheels. Changing the number of wheels changes the distribution how that power gets driven into the ground. More wheels means more mechanical losses and more weight, but more surface area contact with the ground (which translates to higher force prior to slip, which is wasted energy). There’s more too that I’m sure I’m forgetting…

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The number of wheels does not impact speed, but maximum acceleration. As others have said the force transferred before wheel slip. This may not be a factor depending on wheels, robot mass, center of gravity, motors, and gearing. When accelerating the least pressure is on the middle wheel in a 6 wheel drop center. The higher your center of gravity the more pronounced the impact is.
Increasing the wheels beyond 6 makes turning more difficult due to scrubbing (sideways resistance). A well balanced 6 wheel drop center turns easily. Often fields have at least a small platform to climb. With only the center wheel driving it does not take much to lift the center wheel off the ground. Climbing the lowest level hab would take a running start to avoid being stranded.

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From 254 blog post.

Sometimes the middle wheels don’t touch the ground.

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There’s another answer that I haven’t seen yet. So I’ll give it.

It’s because you actually lose pushing force if you have unpowered wheels on the ground. And you’re pretty much guaranteed–with a 6WD–to have unpowered on the ground.

Here’s the physics behind that.

First, we assume that Traction is mu* Fn, for any given wheel. Fn is the normal force of the robot, divided by the wheels in contact with the floor; mu is the frictional force between the wheel and the carpet. (Also include any manipulators–but that’s not important here unless you’re driving a manipulator into the carpet.)*

Next, we assume that for a 6WD robot with a dropped center wheel, no more than 4 wheels are on the floor at any given instant, and possibly only 2. Also assume that the robot weight is 150 lb.

Now, a few case studies.

1. 2WD, 2 wheels on the ground
in this case, each wheel has 75mu lbf of traction. 150mu total.
2. 6WD, 2 wheels on the ground. See case 1, but the outer wheels spin with the inner ones.
3. 2WD, 4 wheels on the ground.
Each wheel now has 37.5mu lbf of traction. But, with only 2 wheels driving… 75mu lbf total. The rest? Well… that’s taken up by the two non-powered wheels… uh-oh.
4. 6WD, 4 wheels on the ground.
Each wheel has 37.5mu lbf of traction… and there’s 4 wheels on the floor. 150mu lbf total traction.

When 4 hits 3, 3 is likely to lose any resulting pushing match just on available pushing force. Oh, and did someone mention that the robot may or may not actually be on the drive wheels?

It’s a couple pounds of belt or chain that can have a pretty large effect on robot performance. Between ability to get unstuck and the theoretical traction boost, as well as a reduction in drivetrain drag and/or scrub…

*For the pedantic/detail-oriented–not all wheels will have the SAME normal force due to things like CG placement at the moment we’re observing, how tipped up the robot is… stuff like that. But for an overall approximation, it’s a reasonable assumption.

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^^ This is the real answer. Assuming your drive train can provide the torque, your maximum acceleration is equal to (one gravity) times (the fraction of your robot weight carried by driven wheels) times (the coefficient of friction of those wheels with the floor/surface). Drive all the wheels so that middle term is at the maximum value of 1.

Added: You can’t control the first term (gravity) except by selecting events at lower altitudes, and that won’t help you any because the other robots experience the same gravity. Your control over CoF is also rather limited, and selecting a higher CoF often results in drastically shorter wheel life. Most FRC

games require good acceleration in order to be competitive. The bottom line is that for most* FRC
games, there is no good reason to put undriven wheels bearing significant weight on the carpet.

* Exceptions: I do not consider “follower” wheels designed to only carry a few pounds of weight but deliver good odometry as “carrying significant weight”. The 2015 game, “Recycle Rush”, was the only game I can think of since 2012 (likely longer) where intentionally limiting your acceleration was a viable strategy.

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