2-Speed vs 1-Speed drivetrain

[waialua359]

The ideal situation for us several years ago was to go with a 2 speed transmission, and use code to do "auto-shifting" depending on the speed of the robot.
It proved short lived when we started having roll pin issues with our shifter. Although there are now fixes to this issue, I still dont think that shifting often in every match is a reliably good thing to do.

I say run as quick as your drivers can handle comfortably and give yourself the option to switch into a ~5fps low gear range in pushing situations. There are many more options available that you can buy compared to just a couple of years ago. Our robot this season accelerated much better vs. our 2013 robot.
In 2015, we plan to run faster (add additional motors also), with the intent that our 5th year driver will be able to handle it and he wants to.

[Chris Fultz]

Here is some data on 3 gear ratios and acceleration / top speed tests.

http://www.chiefdelphi.com/media/papers/2911

[Abhishek R]

So this year we ran a two-speed 4 CIM drive with auto-shifting, geared about ~11 in low and 16 or 17 in high, on 4 inch wheels. With the current auto shifting setup, would it be beneficial to make the low gear lower to accelerate faster?

Originally Posted by waialua359

I say run as quick as your drivers can handle comfortably and give yourself the option to switch into a ~5fps low gear range in pushing situations. There are many more options available that you can buy compared to just a couple of years ago. Our robot this season accelerated much better vs. our 2013 robot.
In 2015, we plan to run faster (add additional motors also), with the intent that our 5th year driver will be able to handle it and he wants to.

So what was your 2014 setup like? I didn't see it to much in the pits, but it was definitely one of the fastest robots I saw this year.

So would an ideal system (barring weight and other limiting factors) be a 6 CIM two-gear drive, the low being only used for pushing, and high gear for everything else? I think this is what 254 used this year.

[colin340]

boy this is a mess

2 speed= Lets push stuff!!. yaya robot

1 speed 6 cim (13-14fps) = We don't have time to push, bum us and watch us run. (see 610/talk to Mr.Lim he is the MAN)

[sdcantrell56]

Everyone should try datalogging acceleration if they have encoders on their drivetrain, particularly with 2 speed transmissions. Depending on the ratios, there really is no difference in acceleration.

[Monochron]

Originally Posted by DampRobot

No, the opposite. Gearing down below 7 fps or so doesn't add any pushing power to your drivetrain, as you're traction limited. Basically, if you push with more than a certain force, your wheels will slip no matter what. If you're pushing with 6 CIMs at 5 fps, your wheels will likely slip, and your main breaker will likely be OK (for a while, at least). On the other hand, if you're pushing in an 18 fps gear, your wheels won't slip, your motors will stall, and you will likely trip your main breaker quite quickly.

Quick aside question, am I right in assuming that the wheels slipping is the only thing that prevents the breaker from tripping at low fps?

For instance, if a weird situation occurred where a gearing of 5 fps did NOT slip the wheels in a pushing match, the breaker would still trip in about the same amount of time as if the gearing was 16 fps correct? Basically, the CIMs draw the same amount of current at stall regardless of the gearing so any stalling of a 6 CIM drive is just as dangerous.

Am I right in that assessment?

[Chris is me]

Originally Posted by Monochron

Quick aside question, am I right in assuming that the wheels slipping is the only thing that prevents the breaker from tripping at low fps?

For instance, if a weird situation occurred where a gearing of 5 fps did NOT slip the wheels in a pushing match, the breaker would still trip in about the same amount of time as if the gearing was 16 fps correct? Basically, the CIMs draw the same amount of current at stall regardless of the gearing so any stalling of a 6 CIM drive is just as dangerous.

Am I right in that assessment?

If the motors are stalled, they will draw their stall current.

That said, there's pretty much no traction material available that won't slip drive wheels with 6 CIMS geared for 5 fps.

Drive trains are "traction limited" (wheels slip) for the most part. The amount of current drawn per motor while the wheels are slipping depends on your gear ratio. Going well above 40A per motor can trip the resetting breakers. Going well above 200A per system risks tripping your main breaker a bit too quickly.

[themccannman]

Originally Posted by Monochron

Quick aside question, am I right in assuming that the wheels slipping is the only thing that prevents the breaker from tripping at low fps?

For instance, if a weird situation occurred where a gearing of 5 fps did NOT slip the wheels in a pushing match, the breaker would still trip in about the same amount of time as if the gearing was 16 fps correct? Basically, the CIMs draw the same amount of current at stall regardless of the gearing so any stalling of a 6 CIM drive is just as dangerous.

Am I right in that assessment?

Yes, stall current is stall current. Ideally your low speed gear ratio in your 2 speed drivetrain should be as fast as possible while still being traction limited. That gives you a higher speed in low gear without sacrificing any pushing power.

[Michael Hill]

Originally Posted by themccannman

Yes, stall current is stall current. Ideally your low speed gear ratio in your 2 speed drivetrain should be as fast as possible while still being traction limited. That gives you a higher speed in low gear without sacrificing any pushing power.

Well....stall current is kinda stall current. The stall current per motor for 4 motors is slightly different than the stall current for 6 motors. Stall current for a 6 CIM system is about 75 Amps/motor and the stall current for a 4 CIM system is about 87.5 Amps/motor. This is because when you add more CIMs, the motor voltage will decrease. This voltage can drop to about 7.5-8 volts when stalling. The stall current cited in the CIM motor documentation (~131 Amps) is assuming the voltage is a constant 12 volts (not what we see in robots). It is correct that friction is what limits the stall current. Also, note that there's a difference between the static friction and kinetic coefficient of friction. In fact, you can even have just as much pushing power with a 4 CIM drive as a 6 CIM drive (if geared properly).

[Caleb Sykes]

Originally Posted by Michael Hill

Well....stall current is kinda stall current. The stall current per motor for 4 motors is slightly different than the stall current for 6 motors. Stall current for a 6 CIM system is about 75 Amps/motor and the stall current for a 4 CIM system is about 87.5 Amps/motor. This is because when you add more CIMs, the motor voltage will decrease. This voltage can drop to about 7.5-8 volts when stalling. The stall current cited in the CIM motor documentation (~131 Amps) is assuming the voltage is a constant 12 volts (not what we see in robots). It is correct that friction is what limits the stall current. Also, note that there's a difference between the static friction and kinetic coefficient of friction. In fact, you can even have just as much pushing power with a 4 CIM drive as a 6 CIM drive (if geared properly).

I am aware that the voltage can drop down to 8 volts for brief instants in time, but does it stay down this low for extended periods during a pushing match? I will have to remember to check the driver station data logs next time I have access to the driving laptop.

I had always thought that, for things like pushing matches, voltage briefly dropped down to ~8 volts but then jumped back up to some value around 12. The battery should be applying a relatively constant voltage, so I don't see why the voltage would remain notably lower for any extended period of time.

So yes, the stall current might be ~90 amps for short periods of time, but I am thinking that the stall current is much closer to 130 amps for most of the time in a pushing match.

[Orion.DeYoe]

Originally Posted by Abhishek R

I searched some old threads but couldn't find a clear answer.

What is it about a single speed drive train that allows it to accelerate faster than a two gear one? I don't see why it does, but it seems like there should be a technical reason that I'm missing as to why it does accelerate faster.

As most people have said, it's the gear ratio (or number of motors).
However there is one other thing that makes a bigger difference on acceleration than most realize. That would be the number of rotating components (gears) in the gearbox. Typically a good 1-speed gearbox will have 1 or two stages and have 3-5 gears (2 small pinions). Most shifters have 2-3 stages with 6-9 gears (3-4 small pinions). Not only do the motors have to get the robot moving they also have to get the gears spinning. And in the early stages of the gearbox they don't have the mechanical advantage of multiple gear reductions to increase torque. This is made worse when the gears are steel not aluminum.
Now whether or not that actually creates a noticeable difference, maybe not. However it certainly is something to keep in mind when designing a gearbox.

[JesseK]

There is a lot of talk about speed in this thread.

There is no talk (yet) about designing for speed and still being able to turn well. There are extra variables in play here would make the fastest robots worthless without special consideration:

• Weight distribution (want it slightly off-center from track center-of-area)
• Wheel Base (want it wider than track length)
• Track Length
• Wheel Tread Composition
• Drive Train Frame Rigidity (want it STIFF)
• Center of Mass (want it LOW)

[Oblarg]

Originally Posted by inkling16

So yes, the stall current might be ~90 amps for short periods of time, but I am thinking that the stall current is much closer to 130 amps for most of the time in a pushing match.

This is false.

This year we were having problems tripping our main breaker, and did a fair number of tests pushing against a wall. With a not-quite-full battery and the compressor running (i.e. pretty reasonable in-match conditions), we were only drawing ~50 amps per motor on a 6CIM drive when stalled.

[AlecS]

The voltage the battery outputs is proportional to the amperage being drawn, and the resistance of the circuit. This is also known as ohm's law.

V=IR

Although batteries are not always perfect voltage sources, you can use ohm's law to determine the voltage drop in your robot fairly accurately. If you were drawing 300 amps, and the total circuit resistance was .020 ohms, you would lose 6 volts. This does not fluctuate over time, it is directly related to the current being drawn. (ignoring temperature and other factors). Hopefully this clears things up.

[dellagd]

Originally Posted by AlecS

The voltage the battery outputs is proportional to the amperage being drawn, and the resistance of the circuit. This is also known as ohm's law.

V=IR

Although batteries are not always perfect voltage sources, you can use ohm's law to determine the voltage drop in your robot fairly accurately. If you were drawing 300 amps, and the total circuit resistance was .020 ohms, you would lose 6 volts. This does not fluctuate over time, it is directly related to the current being drawn. (ignoring temperature and other factors). Hopefully this clears things up.

I'm not an engineer (yet), but doesn't this calculation have to be related back to the capacity of the battery somehow? When drawing, say, 5 amps, intuition tells me some tiny lead-acid battery would have a much larger drop in voltage than a huge deep-cell marine battery would. Maybe I'm rusty with my basic electrical physics, but with V=IR, if you had a 300 amp draw with .02 ohms, that would mean that you must be drawing from a 6 volt source (or be in series such to draw 6 volts), not "losing" 6 volts. Our applications are of course done in parallel however on the robot, so series is not an option. If we know the current draw and resistance of the whole circuit like you said, then we solve for our voltage across the whole circuit. I'm thinking the theoretical voltage drop of the battery under load is another calc entirely. I'm simply raising a concern, not saying this it absolutely correct. I'd love to be educated otherwise if someone has a good explanation.



Here, they relate a voltage drop on current draw to a capacity fraction.