Re: Motors - another concept for FRC
 

2337 did just that this year. I will try to convince someone from that team to post the experience.

Re: Motors - another concept for FRC
 

2815 did the same thing this year for their drivetrain. Not so much for weight savings as the fact that all four CIMs were tied up in our conveyor and shooter drum system. No issues all season (nor in its several off-season demos) driving on a slick floor, though we did keep cans of compressed air in the pits. I don't think we'd do it like this in a carpet game, but it worked great for the task at hand.

Re: Motors - another concept for FRC
 

i'd take the great weight for if just to get the Reliability of a CIM
fp's are vented, have really bad bushings, and spin at 15,000rpm it's not a good combo for what we do to them

Re: Motors - another concept for FRC
 

Sorry for dragging this up again -- I want to (try to) clarify what I said in an earlier post, using some data.The 12 Volt CIM and FP motors available from AndyMark have different parameters, consistent with their different sizes, weights, and costs:

CIM: 2.22 N-m stall torque, 115 Ampere stall current, 1280 g mass, $28 price
FP: 0.45 N-m stall torque, 70 Ampere stall current, 272 g mass, $13 price

To get about the same torque that is available from a CIM, others have pointed out that the FP can be combined with a 3.67:1 gearbox that has 409 g mass and $98 price.

Rated voltage (V), stall torque (ST), and stall current (SC) can be conveniently combined to give a figure-of-merit that is often called the 'motor constant' and is defined by:

kM = ST / sqrt (V*SC), which has units of torque over the square root of power.

When we use Newton-meter as the torque unit, kM is expressed in Newton-meter per root Watt. Its physical significance lies in the following simple expression for Joule (i.e., I^2 * R) losses in the motor for a given torque:

Pj = (torque / kM)^2

As an example, let's calculate Joule losses in the CIM and FP motors when each is used to drive 35 lbf traction load on a 6" diameter wheel through a 12.75:1 Toughbox. (Note that this loading, while extreme, has very likely been seen in many FRC situations over the past few competition seasons. Maybe not so often in 2009, when traction on the playing surface was limited.)

The specified load corresponds to 105 lbf-in torque at the wheel, and to 8.2 lbf-in torque at the Toughbox input. (Note: for simplicity I am neglecting mechanical losses in gearboxes; this would make conclusions regarding absolute rates of motor heating somewhat optimistic, but will not impact conclusions regarding relative rates of motor heating.)

So in this example the CIM torque is 0.931 Newton-meter (42% of its stall torque) and the FP torque is 0.931 / 3.67 = 0.254 Newton-meter (56% of its stall torque).

Now let's calculate kM for each motor:

CIM: kM = 2.22 / sqrt(12*115) = 0.0598 Newton-meter per root Watt
FP: kM = 0.45 / sqrt(12*70) = 0.0155 Newton-meter per root Watt

From these figures we calculate the Joule losses in each motor at their respective torque loads:

CIM: Pj = (0.931 / 0.0598)^2 = 242 Watt
FP: Pj = (0.254 / 0.0155)^2 = 267 Watt

These losses will cause the motors to heat up. How fast will the each motor's temperature rise? This brings us to Richard's second rule:

Rate of temperature rise is roughly proportional to the ratio of the motor's power losses to its mass. (Note the proportionality will not be the same for motors made of dissimilar materials; however, relatively conventional motors like the ones used in this example are made mostly of steel and copper, so the approximation is valid.) The loss-to-mass ratios in this example are:

CIM: 242 / 1.28 = 189 Watt/kg
FP: 267 / 0.272 = 979 Watt/kg

So the FP will exhibit about 979/189 = 5.17 times the rate of temperature rise. Put another way, if the CIM can take a 35 lbf wheel traction load for 30 seconds before overheating, then the FP can take the same load for about 6 seconds.