2 motor drive VS 4 motor drive

[cgmv123]

Originally Posted by JamesCH95

I think you meant to say "the less per-motor/controller current draw you will potentially use" because with 4 cims (4x40A breakers) it is possible to draw twice as much total current as 2 cims (2x40A breakers).

That's right. Last year's rules prohibited splitting the output of a speed controller to 2 separate motors. What we did was split the PWM signal to 2 jaguars, each hooked up to their own 40-amp breaker, for a total of 4 jaguars and 4 CIM's. It worked just fine.

[DavisC]

So what you're saying is that with the torque divided upon the 4 motors vs 2; has less potential to break the victors (because of the current load); but has more potential to drain the battery faster (again because 4 motors are able to draw twice as much current as 2 motors).

[Marc S.]

When accelerating you will draw less current with 2 drive motors, but at speed 4 will draw less then 2 overal. This is presuming that both senatios are using the same gear ratio. Accounting for the powers loss (with a lower gear ratio) will lower the draw. Check it out with the jvn calc.

[AdamHeard]

Originally Posted by Marc S.

When accelerating you will draw less current with 2 drive motors, but at speed 4 will draw less then 2 overal. This is presuming that both senatios are using the same gear ratio. Accounting for the powers loss (with a lower gear ratio) will lower the draw. Check it out with the jvn calc.

Even at the same ratio, this really depends on the ratio. If you're at the point where two motors just isn't enough power to accelerate quickly, you'll end up drawing more current (when averaged over time) total.

At speed with the same ratios and friction, your net current draw should be the same, you'll just be going slightly faster.

[JamesCH95]

Originally Posted by Marc S.

When accelerating you will draw less current with 2 drive motors, but at speed 4 will draw less then 2 overal. This is presuming that both senatios are using the same gear ratio. Accounting for the powers loss (with a lower gear ratio) will lower the draw. Check it out with the jvn calc.

It's worth noting that the robots total coulombic consumption (number of electrons taken from the battery, i.e. energy) to reach top speed will be essentially the same with 4 cims as it will with 2 cims.

[AdamHeard]

Originally Posted by JamesCH95

It's worth noting that the robots total coulombic consumption (number of electrons taken from the battery, i.e. energy) to reach top speed will be essentially the same with 4 cims as it will with 2 cims.

Not likely. The motors are far less efficient operating near stall, which the 2-motor setup will spend much more time in.

The magnitude of this difference will depend greatly on the gear ratio.

[JamesCH95]

Originally Posted by AdamHeard

Not likely. The motors are far less efficient operating near stall, which the 2-motor setup will spend much more time in.

The magnitude of this difference will depend greatly on the gear ratio.

I was assuming that the 4-motor drive would accelerate about twice as quickly as a 2-motor drive, spending about half the time in the 'near stall' regime but drawing twice the current over that time, resulting in the same number of A*s (electrons) being used.

Is this a poor assumption to make?

[AdamHeard]

Originally Posted by JamesCH95

I was assuming that the 4-motor drive would accelerate about twice as quickly as a 2-motor drive, spending about half the time in the 'near stall' regime but drawing twice the current over that time, resulting in the same number of A*s (electrons) being used.

Is this a poor assumption to make?

I modeled a drivetrain with a 10:1 ratio, 3-stages of reduction at 90% efficiency, 150 lbs, 4" wheels and 2/4 CIMs. Timestep of .01 seconds, calculated colloumb use per timestep by multiplying timestep by current draw. This was done by modifying JVN's 2004 design calculator, under drive modeling (so that anyone can repeat these results).

With 2 Cims Total
-5.4 Amps per CIM at steady state (fully accellerated).
-1.41 seconds to reach 9 fps (9 fps was picked as a value just below it's steady state speed).
-45.4 C used per side in this 1.41 seconds, 90.8 Total.

With 4 Cims Total
- 8.1 Amp per side at steady state (4.05 per CIM). This difference is likely from the motors not having zero current draw at free speed.
-.58 seconds to reach 9 fps.
- 43 C used per side in this .58 seconds.

I was a bit surprised by these results, the difference in energy use exist, but isn't huge. I imagine this difference would be larger the higher the gearing.

I'm curious how well this data matches the real world.

[Ether]

JVN's 2011 calculator has a "Speed Loss Constant" of around 81%.

Is that not used in the 2004 calculator?

[AdamHeard]

Originally Posted by Ether

JVN's 2011 calculator has a "Speed Loss Constant" of around 81%.

Is that not used in the 2004 calculator?

I used .73, representing the 3 stages of gearing at 90% each.

[Ether]

Originally Posted by AdamHeard

I used .73, representing the 3 stages of gearing at 90% each.

OK, I've not looked at the 2004 version.

The 2011 version has a "Speed Loss Constant" and a "Drivetrain Efficiency". (see attachment)

The "Drivetrain Efficiency" affects the Wheel Stall Torque but not the Max Speed.

The "Speed Loss Constant" affects the Max Speed but not the Wheel Stall Torque.


With a 10:1 reduction and 4" wheels, the CIM's free speed of 5310 rpm corresponds to a vehicle speed of 9.27 fps. The purpose of the "Speed Loss Constant" in the 2011 version appears to be to allow the user to de-rate the vehicle max speed calculation to better match reality.

[JamesCH95]

Originally Posted by AdamHeard

I modeled a drivetrain with a 10:1 ratio, 3-stages of reduction at 90% efficiency, 150 lbs, 4" wheels and 2/4 CIMs. Timestep of .01 seconds, calculated colloumb use per timestep by multiplying timestep by current draw. This was done by modifying JVN's 2004 design calculator, under drive modeling (so that anyone can repeat these results).

With 2 Cims Total
-5.4 Amps per CIM at steady state (fully accellerated).
-1.41 seconds to reach 9 fps (9 fps was picked as a value just below it's steady state speed).
-45.4 C used per side in this 1.41 seconds, 90.8 Total.

With 4 Cims Total
- 8.1 Amp per side at steady state (4.05 per CIM). This difference is likely from the motors not having zero current draw at free speed.
-.58 seconds to reach 9 fps.
- 43 C used per side in this .58 seconds.

I was a bit surprised by these results, the difference in energy use exist, but isn't huge. I imagine this difference would be larger the higher the gearing.

I'm curious how well this data matches the real world.

Is there an inefficiency associated with the CIM's output gear to the first stage gear? That is the only difference in efficiency between the two that I can think of at 7:15 in the morning

I am surprised at how much longer it took the 2-CIM drive train to reach 9ft/s, though it makes sense when one factors in drive-train loss.

[FrankJ]

Hard question to answer in a vacuum. 4 motors will give you almost twice the power. The three questions are do you need it? can you use it? Would a limited resource be better used some where else? The answer to those questions is why we have have the competition.

[martin417]

An important consideration here that everyone seems to be ignoring is the loss that occurs due to internal resistance. The internal resistance that the manufacturer reports for the battery is .011 ohms. Based on my experience with these batteries, this number is very low. As a battery ages, the internal resistance gets higher. In addition, there are other resistive losses such as the battery connection terminals, the Anderson connector, the PD connections, the PD itself, the breaker connections, the breaker itself, the wires, the motor controller the.... But you get the point. All this can add up up to a significant number. What this means is that when you go from a two motor drive to a four motor drive, you do NOT double the available torque. The current draw goes up, but doesn't double because the voltage goes down. In 2010, we had a three motor per side drive (2 CIMs and one FP). After the season, we did some testing and found that the current draw caused the voltage to drop so far that we had more available torque without the FP motor than with it. We also tried adding another battery (two in parallel) to reduce the drop in voltage. The difference was incredible. I almost think two batteries with a two motor drivetrain would out-perform one battery and a four motor drivetrain. (so far, not legal in competition)

We know for certain that we can't draw full current because even with our three per side drivetrain, we never tripped a 120 amp breaker. With all those motors stalled, we should been pulling ~750 amps. We definitely stalled the motors on more than one occasion, but never tripped a breaker (the battery voltage dropped down below 7 volts at times though)

If you want to run some numbers, decide on a number for total resistance for the system, .035 to .045 ohms is not a bad guess. Multiply this number by the current draw to get the voltage drop. Subtract that number from the starting battery voltage (~12.6V for a charged battery) to get the voltage available to the motor.

Unfortunately, the motor curves are all shown at 12V so it is difficult to estimate the real torque available when the voltage drops to 8V or so.

Try some calculations, then run some experiments. Tell us what you found.

[JesseK]

I don't mean to toot my own horn per-se, but there is a tool available that shows the tradeoffs for 4-vs-2 motors with respect to top speed and current draw.

Attached are screenshots of how I'd design a 2-motor drive and the respective 4-motor equivalent (all else equal). It uses the 2011 KOP transmission with 6" skyway wheels. The exact #'s are always subject to verification (efficiency measurements, exact motor specs, weight, etc), yet it does show a meaningful difference between the two.

Note that the difference in max speed will be noticeable if the robot ever has to traverse the whole field. Other than that, if the 'zones of play' this year are small, 2 motors might not be that big of a deal if you avoid pushing matches.

edit -- I'd use a 12:34 sprocket ratio instead of a 12:28. Single-speed geared for ~10 ft/s.