There are alot of questions in here, so if there’s a guide I can read to do this then maybe that would be simpler. Let’s put together a hypothetical situation for an experimental drive train. The goal of the drive train is to have almost the same pushing and acceleration power as a 4CIM drive train, but it’s ideal case is to only use 3 CIMS, combined with two window motors to disallow backdrive. Let’s also assume we can mate each motor to their respective transmissions (3 CIMS to one toughbox and a window motor to a stackerbox). The window motor in consideration is the Nippon-Denso window motor, yet this will all have to be reconfigured for whatever comes in a given KOP.

So how do I calculate the equivalent torque and overall current draw?

The free speed of the output shaft on a AM toughbox with two CIMs is 5500*(14/50)(14/50)=431rpm. How do I calculate the equivalent stall torque? It would seem that for two CIMs it is (2.222)(50/14)(50/14). Is this correct?

To match the toughbox output shaft free speed mentioned above, each window motor would have to go through a reversed stackerbox, then to a 16:28 sprocket expansion (is that the right word?). It’s an exact match – 92*(50/14)*(21/16) = 431rpm. Do I follow the above procedure to calculate the equivilant torque for two motors?

Now to add the torques together for 3 CIMs and 2 window motors – is it a straight addition? We can ignore efficiency losses for now.

I understand that each individual motor draws a current proportional to its torque contribution to the load, so assuming I can calculate the total current at stall I believe I can calculate each individual motor current. To calculate the equivalent stall current however, do I need to add the stall current together from the specs tabs for the motors or is there something else?

There is a new design I am prototyping in VEX right now, and it seems to work ok. Yet I know that there is much more to consider in FRC than plastic gears and milliamp motors will allow for. If I can figure things out, I’ll release the idea this Fall. Thanks in advance!

Before people start diving into the math/physics of this, I’d like to insert a reminder:
The internal gearing of the Nippon-Denso window motor is PLASTIC. (Nylon I believe.) I have witnessed shredding of the internal gears in this motor, even at torques less than the stall torque. Carefully consider the amount of stress you intend to put this motor under.

drive voltage in Volts
resistance in Ohms
torque constant in Nm per Ampere (which is coincidentally equal to the motor’s “speed” constant in V per radian/second which can be derived from the unloaded speed)

Be careful with “drive voltage”, you’ll need to account for losses in the wiring/drive/battery/connectors.

To find the combined torque of two motors, do I calculate the equivalent winding resistance, then back track to the stall torque?

For two identical motors this makes sense since their combined equivilant resistance is half of the original, so the torque is twice the original (which matches up with JVN’s super motor specs). Yet some of the other super motor specs don’t match up when I approximate the stall torques using the same method. I reverse-engineered each individual motor’s winding resistance using unit conversions and a rearranged version of your equation. I’ll let this sit for a while and think on it.

Excellent question - do motors in parallel add their torques?

In the simplest case when both are perfectly stalled, their torques should indeed add.

Once the motors start turning, however, one will inevitably start to dominate and things will go askew (possibly also due to gearbox efficiencies). At that point, you’ll need to carefully model the rotor and load inertias, gearbox efficiencies and differences between the motors and drive electronics.

You are on the right track. With completely stalled motors, the electrical model collapses to a pair of resistors in parallel. As Russ said, dynamics make the model harder. However, I don’t think it is nearly as hard as he makes it out to be.

Tomorrow I’ll try to put forth a better response, but here is the quick and dirty:

Assume your transmission perfectly lock-steps each motor with each other, so their speed is identical (or is at least a constant multiple). You can now sweep voltage and speed to get current and torque.

There are plenty of complex ways to look at this. I know I’ve played with many of them. When I started learning how to do this stuff, I used some fairly complex models, however these really aren’t needed.

For our applications, a simple model is MORE than sufficient. In fact, I frequently use approximations and get well within the ballpark for our type of systems.

I always tend to think in terms of an equivalent “super-motor” which represents the two motors geared together. In this case, you can assume the system will spin at some fixed speed (the super-motor represents the two motors and all the gearing in between them.) Based on this speed, each motor will see some percentage of the applied loading. This percentage is a function of the relationship of the speeds the motors are running at. (If both motors are identical and are geared the same, it makes sense that they will each see 1/2 the overall loading.)

To see these relationships in action you can take a peak at the attached (old-school) spreadsheet I used to use for motor combiner design. Digging through this should help you. (Your mileage may vary, this sheet is a few years old.)

One thing to think about:
Some of the best advice I ever got on this subject was “There is no voodoo involved, the motor torques add together, sometimes one motor may drag a little depending on how good your speed matching is, but if they are in the ballpark they will both contribute. Match them at free-speed and go about your business.”

Hope this helps. Feel free to email me if you have any specific questions. This sort of thing has been a hobby of mine for a while. (Teaching it to others has turned into a hobby of it’s own.)

That’s one heckuva spreadsheet John. I’ve played with numbers and setups and gear ratios for a couple of hours now. I’ve taken the ‘super motor’ over to the 1 and 2-speed drive train tabs and played with numbers there until I had a satisfactory quantity of torque values which I could take and play with back over in the super motor calculator.

Conclusion: when geared for 12 fps (bot weight = ~148lbs), if this experimental drive train even approaches stall, those window motors are toast haha. The bot would have to be less than ~85 lbs overall in order for the motors to be within safe tolerances to survive a full competition season. Even if the models are 20% off (which is the range of error I gave this experiment), I at least have a bigger picture.

Thanks alot for the help guys. I’m on vacation for the next four days so I’ll have plenty of time to think of reworks and time to calculate the minimum torque needed by a worm gear setup for the drive train to perform its key maneuvers. I’ll get a sketch up some time next week and a CAD later in the fall.