Fastest drivetrain in Aerial assist?

[asid61]

Originally Posted by Dunngeon

ONLY if you have a low gear/two speed gearbox. Your acceleration will be impacted significantly with a top speed of 26FPS, which may mitigate the perceived higher top speed (not useful if you never get there). Personally, I would only design up to 24 theoretical, because beyond that number most drivers seem to struggle controlling the robot.

Top speed isn't the only thing to consider when choosing gear reduction. We factor in Driver Ability, Game Requirements (open field, field with obstacles) , and Robot factors (such as weight) when choosing the reduction.

This year we were geared for 18 FPS, and achieved 16 actual due to a robot that only weighed 107 pounds w/ battery and bumpers. We thought it was a good harmony between speed, and torque (we run single reduction). If we were to run a 2-speed, the actual would likely be bumped up to 18-20 FPS actual because we can now have a low gear (5-9 FPS) for pushing

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.

[Arpan]

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.

[Dunngeon]

Originally Posted by asid61

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.

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

Originally Posted by BBray_T1296

Fun fact: 6 CIMs at stall consume a total of roughly 9576 watts of electrical power, which is over 9000!


Seriously though, a robot with 6 CIMs under "normal load" output (excluding friction) a total of 2.6 Horsepower, while a 4 Cim drive does just 1.75. At exactly 40 Amps per motor the numbers are (theoretically) 3.86hp and 2.57hp, respectively.

And as everybody knows, power/weight ratio is the most important figure.

[cgmv123]

Originally Posted by Dunngeon

I *think* the reason is because we are power limited by our PDB and Talons to 40 Amps per motor. Thus all the data I have above for Amperage isn't true if the motors will be drawing more than 40 amps, which they are.

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

[AdamHeard]

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

[PingPongPerson]

Originally Posted by Dunngeon

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.

Put into HP

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.

[hrench]

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

[JB987]

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

[Alpha Beta]

Originally Posted by hrench

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

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.

[asid61]

Originally Posted by AdamHeard

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

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.

[Caleb Sykes]

Originally Posted by asid61

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.

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.