pic: Gearbox
 

Re: pic: Gearbox
 

No, your gearbox is not geared to X fps.. It's got a reduction. With a certain sized wheel it would drive a robot X fps.

What's the reductions?

What size wheel is it designed for?

Re: pic: Gearbox
 

3.25 Inch VersaWheel

First Stage - 11 to 54
Low Stage - 36 to 40
High Stage - 20 to 56

Re: pic: Gearbox
 

Sorry. I meant to post the picture with the 3.25" versa wheel DT so people could see how it mounts.

It is designed for a 3" to a 3.25" wheel. It can work with the 3" colson, and can go faster if the CIM gear is swapped to an 12, 13, or 14 tooth gear.

The overall reductions are 5.45:1 and 13.75:1, which are a little fast for 2 CIMs driving a 3.25" wheel. I plan to add an rs-775 to the gearbox.

Re: pic: Gearbox
 

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.

Re: pic: Gearbox
 

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.


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.