Drive Train Calcs - Not trusting my results

[Nemo]

I want to be able to believe the numbers I get when I do drive train calculations. But I think I'm doing something wrong.

Let’s say I’m using 6” wheels and I want to optimize a robot’s gear ratio to travel 15 ft distances as quickly as possible.

What I try:
Plug the numbers into a spreadsheet (JesseK, JVN, etc) and figure out what ratio minimizes the travel time. I also plug them into a spreadsheet that I made that basically traces JVN's first drive train spreadsheet (my sheet attached below).

Acceleration Spreadsheet.xlsx

What I get:
I find that a ratio of about 7:1, providing a theoretical top speed of about 20 ft/s, gets me there in 1.26 seconds from a stop. This is in a range of gear ratios that get me the approximate shortest sprint time. I plug those numbers into JesseK’s spreadsheet and get a similar time. According to the spreadsheet I made (mostly copying JVN’s original spreadsheet), the robot gets up to around 17 ft/s by the time it crosses the 15 foot line. And according to the numbers in my spreadsheet, it’s traction limited, because the gearbox can provide more force than my maximum tractive force in the first few instants of acceleration.

Those results seem wrong. What gives? I assume that a 20 ft/s drive with a CoF=1.1 is not traction limited, so what did I do wrong? Based on what I just calculated, my design decision would be to completely throw out those numbers and design something that’s perhaps 12-13 ft/s, because based on experience that seems like a nice speed that’s not too fast or too slow. I’d like to figure out what’s going on with my numbers so I don’t have to ignore the math.

Turning
I understand that in the end, the need to turn will constrain me into choosing a slower gear ratio than what I’d choose if I’m only considering sprint time over a certain distance. That means I can forget these problems to some extent. But I still want to understand what’s going on. If I’m designing for a strafe wheel (we may consider this depending on the game), I don’t care about the same sort of turning math anymore. Then I’ll be back to a case when I just want to know what ratio is the fastest over a certain distance. I want to know how the slower acceleration balances out with a faster top speed over a certain distance, and how hard we can push the motors without frying them or tripping breakers.

Speaking of tripping breakers,

Current Constraints?
I am basically ignoring the 40 amp limit, because the motor is going to spike initially anyway and the breakers can handle temporary overages. How should I decide exactly how much I dare to push those current limits? According to my spreadsheet, I’m pulling more than 40 amps for the first 0.70 seconds if I floor it from a stop, including a span of 0.29 seconds over 80 amps. That’s fine, right? The snap action breakers can handle (if I’m looking at the right spec sheet) a 200% overload for approx. 1.5-3.9 seconds, or 300% overload for 0.5-1.1 seconds. How much do I push that? Are there any good rules of thumb for figuring out how much you can push an RS-550 or FP motor?

Inefficiencies

Right now I have a 0.85 gearbox efficiency and a 0.93 carpet to wheel efficiency built into the spreadsheet. What other terms eat up power and slow down the robot? I lifted that 0.93 directly from JVN’s spreadsheet, and I have no idea if it’s anywhere close to a realistic number.

Thanks in advance for any guidance you may decide to offer.

[JesseK]

When gearing for 15 ft specifically, the "Time to Travel 15ft, vs. Gearing" curve is mostly flat from a ~5.5:1 ratio all the way to a ~9:1 ratio (assuming a 148-lb robot). Then consider that a robot on the field won't go in a straight line too many times during a match, particularly in elimination matches where there will be plenty of defense to get around.

The same negligibility applies when doing comparative 'drag race' graphs. A robot geared 5.5:1 won't gain any worthwhile distance on a 9:1-geared robot over 15 feet.

I'll poke around in your spreadsheet now. [edit -- seems fine]

[Kevin Sevcik]

One of the big things missing from most of those drive train calculators is electrical resistance. I'm pretty sure most of them assume a constant voltage power supply with superconducting wires. In reality, your battery and your wires have non-zero resistance which limits the maximum current available to your motor to something less than stall current. Sometimes significantly if you're doing a bad enough job of wiring. See this thread for the glorious details and possibly a better drivetrain model:
paper: Battery Voltage in Robot Drivetrain Simulation and Modeling

[JesseK]

Getting a precise measurement for a specified system is all well and good, yet when considering the end results of controllable decisions, the resistance of the battery & wiring doesn't necessarily effect two outcomes of the same FRC robot differently. Thus can we assume that, given gearing choice A and gearing choice B, the wiring/battery resistance is the same and doesn't need to be modeled?

Though taking the battery resistance into account would provide better results on modeling systems like rotary arms as they relate to desired performance metrics (current spikes & time of rotation). Thanks for pointing it out Kevin, I hadn't noticed that spreadsheet.

[apalrd]

Originally Posted by JesseK

Getting a precise measurement for a specified system is all well and good, yet when considering the end results of controllable decisions, the resistance of the battery & wiring doesn't necessarily effect two outcomes of the same FRC robot differently. Thus can we assume that, given gearing choice A and gearing choice B, the wiring/battery resistance is the same and doesn't need to be modeled?

The voltage drop across a wire (which we can assume is the same resistance, the resistance is a constant) is dependent on current. A different gear ratio would pull a different current over time as it accelerates. So modeling that is good.

The original JVN calc had several battery voltage steps, which gets 'close enough'. The point is that a battery will drop down to at least 8v or lower during launch, the time it stays low is determined by the gearing and current draw. Essentially, torque = acceleration, a faster gear ratio gets you less total motor torque to the wheels, so you will stay at high torque output longer. Torque is proportional to current, also.


It would be hard to use the battery simulation with mechanisms because the current draw is (comparatively) so low that drivetrain motions would screw up your nice voltage simulation. It could be done, but this would be more an RT software kind of challenge (battery voltage compensation for control loops) then a simulation one. Just assuming you are running under a lower voltage and being pleasantly surprised when the voltage is higher is usually good for mechanisms.

[Nemo]

I never even thought of the battery voltage drop. I'll have to find that math. I'll add voltage to the spreadsheet. I might have to come back and ask for help on how to estimate that voltage drop, though. It seems like that should help me get a more realistic estimate of how fast the gearing can be while still being traction limited.

Related to post 2, I did notice how flat that curve is for 10-15 ft sprint times. I don't really understand why anybody would want to be on the fast side of that sweet spot. What is a good reason for gearing it to 17 ft/s when 12 ft/s gets you there just as quickly?

[Michael Hill]

Originally Posted by Nemo

I never even thought of the battery voltage drop. I'll have to find that math. I'll add voltage to the spreadsheet. I might have to come back and ask for help on how to estimate that voltage drop, though. It seems like that should help me get a more realistic estimate of how fast the gearing can be while still being traction limited.

Related to post 2, I did notice how flat that curve is for 10-15 ft sprint times. I don't really understand why anybody would want to be on the fast side of that sweet spot. What is a good reason for gearing it to 17 ft/s when 12 ft/s gets you there just as quickly?

Battery voltage drop shouldn't be too difficult. I imagine you take your current you're running at and multiply it by your battery's internal resistance, and you end up with the drop in voltage. Say, for example, if you're pulling 150 Amps and your battery's internal resistance is 20 mOhms, you'll have a voltage drop of 3 Volts, so instead of supplying 13V (which seems to be the typical charge at the start of a match), you're really supplying only 10V.

Don't "quote" me on that, but I think that's how the math would work out. Anyone with more authority on the subject care to tear my analysis apart?

[Michael Hill]

btw, my assumption for current is way high, but you get the idea. It comes out to a nice round number :-)

[Kevin Sevcik]

Originally Posted by Michael Hill

btw, my assumption for current is way high, but you get the idea. It comes out to a nice round number :-)

Your current assumption is actually too low. A stalled CIM will draw 100A+ at 12V, assuming no resistance. So a four CIM drivetrain will be trying to suck down 400A or so if you floor it from a dead stop. The only reason it won't be able to is all the added resistance from the battery, wires, PDB, speed controller, etc.

[Michael Hill]

Originally Posted by Kevin Sevcik

Your current assumption is actually too low. A stalled CIM will draw 100A+ at 12V, assuming no resistance. So a four CIM drivetrain will be trying to suck down 400A or so if you floor it from a dead stop. The only reason it won't be able to is all the added resistance from the battery, wires, PDB, speed controller, etc.

Well, there you have it. I remember seeing our Battery drop to at least 8.5 Volts, but we had some bad batteries (internal resistance was 35 mOhms), so I guess we were drawing ~240A.

[apalrd]

In all of the data logs I took on our 2012 practice robot, a straight-line accel always dropped the battery to 8v on launch. The time under voltage was highly dependent on the gearing and battery state of charge, but the low-voltage peak was always around 8v or lower.

At T0 where v=0, current in a 4-CIM drivetrain would be around 500a or so.

[Ether]

Originally Posted by apalrd

In all of the data logs I took on our 2012 practice robot, a straight-line accel always dropped the battery to 8v on launch. The time under voltage was highly dependent on the gearing and battery state of charge, but the low-voltage peak was always around 8v or lower.

At T0 where v=0, current in a 4-CIM drivetrain would be around 500a or so.

It would be if there were 12 volts at the motors. But not with 8 volts.

[IKE]

Originally Posted by apalrd

...

It would be hard to use the battery simulation with mechanisms because the current draw is (comparatively) so low that drivetrain motions would screw up your nice voltage simulation. It could be done, but this would be more an RT software kind of challenge (battery voltage compensation for control loops) then a simulation one. Just assuming you are running under a lower voltage and being pleasantly surprised when the voltage is higher is usually good for mechanisms.

That being said, it is often prudent to do your analysis for a mechanism at a lower than 12V Voltage to see how it may operate. This is especially true for end-game mechanisms as the batteries are often getting tired towards the end of matches.