Re: Fastest drivetrain in Aerial assist?
 

Of course, driver training is extremely important. That's why I suggested a software limit for speed at first and later.
Personally, I think the game doesn't matter as much as driver training. The drivetrain should be what the driver can handle; pretty much every game in FRC in the past several years has required speed.

If you used PWM control to the motor, wouldn't it technically accelerate faster with a software limited top speed? If you don't stall the CIMs, that is. I feel like that doesn't work out, but can anybody explain why?

From what I've seen, acceleration can probably be mitigated as a problem with a 3 + 2/3 cim per side gearbox. (heheheh)
Does anybody know what the maximum speed it is physically possible to go on an FRC robot? Assuming you want to drive for 10 seconds before the main breaker blows with a 100lb robot.

Re: Fastest drivetrain in Aerial assist?
 

There was an interesting white paper a while back about how acceleration in FRC robots is affected almost entirely by weight and number of motors, and not much by gear ratio. I'll see if I can dig it up.

Re: Fastest drivetrain in Aerial assist?
 

Empirically higher top speed = less torque which in turn = less acceleration because T = I*M*A/R (Moment of Inertia, Mass, Acceleration, Radius)

If torque decreases with the others staying the same, then acceleration must decrease. That's why adding motors increases acceleration, because it adds torque to the drive.

A small piece I found on HP, I havn't checked the math but it seems logical

Re: Fastest drivetrain in Aerial assist?
 

Circuit breakers don't limit current. The 40A circuit breakers used by the PDB won't trip for a few seconds at ~100 amps.

Re: Fastest drivetrain in Aerial assist?
 

Complete BS psuedo math just to set an upper bound based on energy...

Assume 200 amps of draw for 10 seconds, this drops the battery to about 11V. That's 22 kJ of energy.

Assume the motors + drive average about 40% efficiency from electrical power to kinetic energy of the robot, that's 8.8 kJ.

Assume 68 kg vehicle, and E = .5mv^2, then velocity is 52 feet per second.

You'd require many shifting stages with instant engagement, a friction-less environment (in terms of air drag, etc..), and a much longer field though ;)

Re: Fastest drivetrain in Aerial assist?
 

The concept here is correct, but T=I*A/R. The formula that really shows the torque-velocity relationship is Power=Torque*angular velocity, where if torque is increased, then angular velocity must decrease assuming a constant power.

Re: Fastest drivetrain in Aerial assist?
 

I'm pretty sure the fastest robot that I saw this year was 1986. I wonder if Aaron can give us a number?

Re: Fastest drivetrain in Aerial assist?
 

Not sure who was the fastest out there but 368 was certainly one of the quickest...and nimble too!

Re: Fastest drivetrain in Aerial assist?
 

11 tooth pinion direct driving a 70 tooth gear on a 4 inch blue nitrile wheel. 6 CIMs, no shifter.

(5,310 RPM CIM Free Speed) x (11/70 gear reduction) x (4 inch x Pi wheel circumference) x (1/12 inches to feet conversion) x (1/60 per minute to per second conversion) = 14.56 ft/sec theoretical unloaded speed.

We never measured the actual speed.

Not the fastest out there, but fast enough for our strategies. Having a great driver helps. ;)

Re: Fastest drivetrain in Aerial assist?
 

Interesting. I was more wondering about friction and stuff like that though, because it is (technically) possible to go faster than that using more motors/ power.
But 52fps... my god.

Re: Fastest drivetrain in Aerial assist?
 

Okay, I'll play this game.

d = maximum linear distance robot can travel
μ = coefficient of static friction with carpet
g = acceleration due to gravity

The robot somehow manages to provide a constant force of μgm in the forward direction. It's mass is m, so the robot has a constant acceleration of a = μg. Assuming the robot starts at rest and x(0) = 0, then v(t) = μgt, and x(t) = (μg/2)*t^2. At x(t) = d, the velocity is the largest. Solving x(t) = d for t gives t = sqrt(2d/μg), so vmax = sqrt(2dμg).

Plugging in d = 60 feet, μ= 1.5, and g = 32ft/s^2 gives vmax = 76ft/s. Then either the robot or the field breaks depending on the quality of your bumpers. ;)

I suppose the robot doesn't necessarily have to drive in a straight line though. If the robot spun in a circle for the whole match, then vmax would be μg*(match length). Using μ= 1.5, and g = 32ft/s^2 again, and match length = 150 s, vmax = 7200 ft/s. Although they would probably not continue the match after the sonic boom.