Sorry for dragging this up again -- I want to (try to) clarify what I said in an earlier post, using some data.
Quote:
Originally Posted by Richard
Richard's first rule of motor selection: you pay for torque.
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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:
Quote:
Originally Posted by Richard
Richard's second rule: choose a motor that can take the heat.
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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.