2 Motors is Faster?!

Okay, so I think we all saw the demo where Dean and Dave race a 2 motor robot against a 4 motor robot and they reach the finish at the same time.

Having taken my fair share of physics I do not for the life of me understand how this is possible.

More force should equal greater acceleration, for the same mass. Correct?
They will reach the same speed, but the 4 motor bot will accelerate faster.

So did they reach the end at the same time simply because the 2-motor bot’s lightness made up for its lack of power? :confused:

After several transmission/motor swaps in '08 our team pretty much convinced ourselves that 4 motors was ALWAYS the way to go.

If torque is greater than traction/FoF, then the wheels will skid reducing your ability to apply your forward force to the ground.

2 motors have less torque than a 4 motor setup, so the wheels skidded less

They accelerated at the same rate because acceleration is a function of mass and force. If both bots were maximum weight, the mass and the force of friction provided by the wheels would be the same. Since the motors provided more force than the friction of the wheels, either robot could only apply, at most, the frictional force of the wheels. This is different than in previous years where the frictional force of the wheels was greater than the output possible with the motors, a situation where the motor output was the limiting factor (and more motors = more force, assuming identical motors). The difference is that this year the limiting factor is friction, while previously it has been motor power.

I see… That makes sense…

Two FP’s for drive and four CIMS driving a beastly manipulator… :slight_smile:

I predict this ending in a LOT of magic smoke. At least use the AM planetary so you can switch back to CIMs when you smoke your drive train…

4 motors can create more force, but the low friction wheels will spin before that additional force can be used to accelerate the robot, The acceleration is limited by the wheels, not the motors. I think that the 2 motor robot accelerated faster because of the gear ratios used.

Depending how it is set up, I think this could work fine. The best idea would be a software traction control system, so the FPs are only ever doing as much work as is required. If they aren’t spinning too fast, and only doing as much work as the wheels allow, I don’t think they would have too much trouble. Then again, it’s nice to keep the cg low, so having the heavy motors in the bottom would probably be preferable. Note that I’ve never used the Fisher Prices, so I’m just guessing. If someone who knows a bit more than me could comment, that would be helpful.

High speed and low torque is the way to go. Try using some encoders in a special way as to create your own ABS. Might just work out better than you plan. Try it out.

I think it wouldn’t be too difficult to figure out if the robot is slipping. If you have encoders and an accelerometer you could compare the measured acceleration to the expected acceleraion at the measured output speed. If the measured acceleration is significantly less than what is expected, you would simply reduce the power to the motors. Does this sound like a good system (I haven’t messed with any of the sensors in the kit, but this is what I assume would be the simplest solution).

I punched the numbers into the JVN calc and with a reduction of 48:1 (big reduction) I got a good speed of around 8 fps with a current draw of 13 Amps per Motor. Perhaps the smallest amp draw I’ve ever seen in a drive.

Going from CoFs of 1.4 to 0.1 makes a huge difference :slight_smile:

48:1 is a bit of reduction, but if you use the AM planetary adapters then you could use the Toughbox and get a ratio close to that, though I suspect you would want to get gears for the toughbox for less reduction, maybe the 8:1 set, to eliminate the need for a funky sprocket ratio.

I see no reason to use four CIM motors in a drive train this year.

There isn’t really a way in which a team would be torque-limited this year (unless you use the Mabuchi motor); you will be entirely traction-limited. Using four CIMs on the drive is just wasting hundreds of Watts of power.

Another thing answering your question;
In this game traction is everything. And you want the most traction.
So with the two wheel bases, they both are relitively the same weight (I know weight is added for extra motors and wheels, but I’m talking on a very general level).
So with four wheels, if the weight was 100, each wheel carried 25 pounds (ok, so I know this is a little off, and I don’t have enough physics background to properly explain this, but once I’m done, this should make sense), but on the two wheel, each wheel carried 50 pounds, which created more friction which gave it more… power (if thats the right word…)

EDIT: No one really explained that to you, and thats why the two-wheeled went faster (or as my mentors explained)…

Not necessarily, that’s with the assumption that your team feels traction is most important. Even if Dean or Woodie said it doesn’t make it true either. There’s no defining basis that says that teams want most traction possible in this situation.

Ok, well, in my mind, on the ride from kick-off to shop, that was the most important thing to me, because I couldn’t think of anything else to worry about without knowing the rules/restrictions. And so thats where that comes from.

I should have specified that that was an “In My Opinion” thing.

Notice the coefficients of friction published in the manual. The static coefficient (where the wheel is NOT skidding) is .6 the coefficient of dyanamic friction (where the wheel IS skidding) is .5

If a four motor drive and a two motor drive are both given the same input signal the four motor drive will generate twice as much torque. This will make the wheels skid (spin freely, do a burnout, whatever you want to call it) sooner than with a two motor drive. At this point you have shifted from static to dyanamic friction.

As soon as that happens, your friction, and thus your accelleration, drops by almost 20%.

It is quite possible for four motors to be equally as fast as two on this surface, but since the determining factor is traction, teams will have to build their robots to avoid spinning their wheels in order to achieve maximum accelleration and pushing power.


Actually, I think they were 0.05 and 0.06. They’re really easy to slide…

This was posted in another thread, but here are the equations:

General force equation (assuming constant mass and somewhat constant accel which is good enough for now):

F= mass * accel

The wheel to floor interaction is governed by this equation:

F = mass * g * COF (Coefficient of friction)

Set the equations equal to each other and solve for acceleration:

accel = g * COF

So your maximum acceleration is entirely based on the planet you are located and your coefficient of friction. Since the GDC couldn’t control where we play the game, they simulated this by mandating a certain COF for all.

Assuming the COF is actually 0.05, then our maximum accel is 32.2 f/s^2 * 0.05, or 1.61 feet / sec^2

Anyone with more acceleration than that will slip. This is not about 1 motor or 2, but about how we manage acceleration.

For those of you that think this doesn’t make sense, just think of this:

While it is true you can get more pushing force from a higher weight, it is also true that you get less acceleration due to the increased mass. This trade off is the center of drive base design for this year’s game. If you can control acceleration, then you will control the game play.

4 motors give out too much torque, which when there is low traction just spins the wheels. If you have ever watched a drag race where a car didn’t heat up the tires well, they go, but also just spin the tires down the track. The trick for this year is finding the right amount of torque i think.

Not quite true. You can gear the output of 4 CIMS so that your wheels won’t slip (Read Pauls post above). That being said, I don’t believe you’d gain much by doing so.

@ Paul Copioli,

Technically I believe that the wheel to floor interaction is governed by
F = CoF * downward force (which usually is gravity * mass, but can have other forces interacting with it as well)