Here’s my 2nd gearbox design. It’s geared at 13.81 fps in high, and 5.48 fps in low, but can be adjusted by changing the CIM pinion gear. All gears, shafts, and plates(except the CIM gears) are aluminum, the 54 tooth gear is lightened, and the 56 tooth gear needs to have a dog pattern added. The 40 tooth gear comes from vex with the dog pattern and hex bearing included. The gearbox is set up to use the shifting cylinder mount and coupling/bearing from the ball shifter/WCP DS gearbox. Without motors or pneumatic cylinder, SW predicts the weight to be 1.63 lbs. The gearbox also has the room to add a banebots rs-775 motor, which would have the 89% of the power a 3 CIM gearbox, but I can’t find anywhere to buy .6 module gears.
My biggest concern is the shifting dog, which would need to have a diameter that is about .03" less, and I’m not sure what consequences of this would be. One of our previous dogs split on the really thin walls around the threaded hole, but I don’t think this will be the failure point for the WCP dogs, which are much thicker.
I’m not sure adding an rs775 to the gearbox is a good idea. I’m pretty sure the rpm curve is different for the rs775 as compared to the cim (whereas I think the miniCIM is meant to have a very similar/the same curve). This could mean that the motors are trying to drive each other to some degree instead of all working together to drive the wheel.
Of course, I could also just be completely wrong here … I’d love someone to confirm/deny what I’m saying with a better explanation.
It is possible to add a motor to a gearbox that has a different curve as long as you take into account the specifications of the motor you are adding. Team like 254 and 118 have both done this in the past, and our team did it in 2003. The motors will only fight each other if one is forced to travel faster than its free speed by another. Because drive gearboxes will operate at both really slow and really fast speeds, it makes sense to match their free speeds, so that their is no condition where one motor forces another motor to go faster than its free speed.
If you are interested, here’s an example of a single CIM and one rs-775 gearbox, with free speeds not matched, then matched. To get these, you need to find what the rpm of each motor will be, use the motor data to calculate the torque at the shaft of each motor, then you need to find the torque at the output, which is the CIM motor shaft in all these examples.
Free Speed Stall Torque Stall Current Free Current (Amp) Power (W)
CIM 5310 2.43 133.00 2.70 337.81
RS-775 (12V) 13000 0.78 86.67 1.80 266.60
CIM torque (Nm) vs. rpm
R/5310 + τ/2.43 = 1
τ/2.43 = 1 - R/5310
τ = 2.43 - R*0.000457627
RS-775 torque (Nm) vs. rpm
R/13000 + τ/0.78 = 1
τ/0.78 = 1 - R/13000
τ = 0.78 - R*0.00006
with same gear ratio, both at 12V, at 6000 rpm (this is possible)
τ775 = .78 - 6000(.00006) = 0.42
τCIM = 2.43 - 6000(0.000457627) = −0.315762 The CIM is providing resistance, and reducing availible torque
Total torque = 0.104 Nm
geared so that free speeds match at 5310 (2.45:1), at 12v, at 3000 rpm (nothing higher than 5310 is possible) The rs-775 will be geared down.
rpm of the 775 will be (2.45 X 3000) = 7350
τ775 = .78 - 7350(.00006) = .339 this is at the shaft of the motor
at the CIM, it will be 2.45 times more = .83055
τCIM = 2.43 - 3000(0.000457627) = 1.057119
Total torque = 1.888 Nm
It makes sense because when free speeds are matched, no motor ever has to push the other past its free speed, so the torque contributed by each motor will always be positive, and the CIM will always contribute more. The ratio between the contributed torque of the CIM and the contributed torque of the rs-775 will always be equal to the ratio of their power, if the free speed is matched. The ratio is about 79:100 for the rs-775 to the CIM.