So I’ve been trying to put together a mini sheet for drive train calculations so that my team can enter in a specific gear ratio with some amount of motors and some sized wheels and etc, and it will spit out a speed after “x” amount of time. I know that there are some great design calculators out there but they don’t do exactly what I am looking for, and I’d like to actually understand what is going on in the calculations.
Now for my questions: (Note: some of these terms and calculations are via a mentor so correct me if I am wrong or if you need more explanation to help me out.)
Would I be correct in saying that the force(labeled as thrust?) of ONE singular wheel in Newtons is equal to my torque after a gear reduction(Newton-meters) divided by my effective wheel radius(meters)?
{Thrust=Torque/Effective wheel radius}
How would you go about finding the Back EMF of a motor(CIM)? Drag(Nm)? Resistance(ohm)?
There are some complicated theories and math involved but I know it can be done. ANY input helps!
You can get the other 50% by giving the equation for how much of that force actually transfers to the carpet to propel the robot forwards. For a singular wheel, you need to know both that wheel’s coefficient of friction, and how much of the robot’s weight is on that wheel (which will depend on how many wheels are on the robot among other things).
Yes, if there is only one wheel on that gearbox, and the wheel has enough traction force (coefficient of friction times the weight on that wheel); this is known as a torque-limited condition. If the applied thrust exceeds the traction, the wheels will slip and the effective thrust will be reduced to the traction force; this is known as the traction-limited condition. Having multiple wheels does not increase the available thrust assuming all are geared and sized the same; the thrust is distributed over the wheels, usually in proportion to the weight carried by each wheel as it accelerates.
Let’s start with the last. Look at the stall current. In the stall state, the back-emf is zero, so the DC current draw is simply due to the resistance. V=IR or R=V/I. For a CIM, the stall current is about 132A, so the resistance is 12V / 132 A = 0.91Ω
Next, the back-emf. Look at the free speed and current. Using the resistance just calculated, calculate the effective voltage across the motor, V=IR. Th remaining voltage (typically 12V - I R) is the back-emf at the free speed. The back emf is usually modeled as directly proportional to motor speed. Again, the CIM in free run at 12V applied draws about 2.7A and spins about 5330 rpm. The back-emf is is 12V - 2.7A * .091Ω = 12V - .025V = 11.975V. The back-emf is generally 11.975V / 5330 rpm = 2.25 mV/rpm.
Drag is robot-dependent, and there are at least two major schools of thought on how to model it. Until a year ago, the drag was typically modeled as primarily viscous (proportional to speed). Spreadsheet defaults typically assigned a value of drag which makes the top speed of the robot about 80% of the free speed of the wheels. In the past year, much of the drag has come to be understood as a constant force, independent of speed. If you have built a robot, you can roughly measure the constant drag force by finding the minimum voltage which you can give to your motors which causes the robot to move, and working through the torque and force calculations.
A paper with the evidence for the constant drag theory can be found in my signature. We also found a fairly significant frictional force proportional to acceleration, probably tooth to tooth friction in the gearbox. This can be really easily included in a drive calculator by scaling the stall torque down by a constant, which we found to be .5 on the one robot we tested.
So now that I understand the basics with one wheel on one gearbox let me take it a step further. Assume I have a 6WD tank drive robot with a center drop. On each side is a 2 CIM gearbox that will be powering all three wheels on a side(Only two are in contact with the ground at a time). When doing my calculations for that side should I double my torque output because both motors are driving the same gear? This would also double my thrust output(limited to my maximum tractive force for the wheel).
Another thing that I don’t quite understand is should I add my thrust together for each side? Or should it be the same throughout? Say my thrust per side is 150 Newtons, would I say my robot has 150 Newtons of force at this point or 300 Newtons?
So, you don’t just double the torque output because you either add a wheel or add a motor. It’s close, but not quite there.
Let’s walk through the pathway a little bit.
You put power into a gearbox, with a given speed and torque. If you double the motors, you double the power–but they’re running at the same speed, so in effect you double the torque. But then you run through the gearbox, dividing speed and multiplying torque with every reduction. Knock the final number from the gearbox down to about 80% of whatever you get to account for inefficiency. So you’re doing about 80% of 2x the input.
If you were running two identical CIM-gearbox setups you’d double the torque.
Now, assume that the torque is transmitted equally to all the wheels on the robot, including the ones not on the ground (we’ll consider it safe to ignore those for now–in reality there’s a touch of inertia to deal with). Each wheel gets the torque from its gearbox, times the wheel size, for it’s applied force, BUT then you add in the coefficient of friction and the fact that it’s now one of 4 wheels on the floor.
This would increase your thrust output for that side.
Now, on to the second part. You’re correct, BUT remember that now we’re talking vectors. If you assume that force causing the robot to move forwards is positive, and force causing the robot to move backwards is negative, you can add the thrusts… but if they aren’t equal, the resulting direction becomes a nice curve! (Case in point: 150N + -150N will result in a zero forward direction, but your robot will be moving. It’ll be spinning in place.)
Assuming you’re driving them the same, adding a motor to a gearbox increases the total torque coming out of the gearbox - two motors generate twice as much torque and three motors three times as much torque - and therefore that much more TOTAL available force for a given wheel size.
Adding wheels does NOT significantly increase the force/torque. The available force/torque is distributed among the wheels. This distribution is NOT equal if the wheels carry different weights. Unless you’re using a quite unusual system, all of the wheels are chained or belted or geared together so that they all rotate at the same SPEED. A wheel which is off the carpet will require very little torque to reach any given speed, but one which is on the carpet will utilize a lot of torque. The torque is approximately distributed among the wheels in proportion to the amount of weight each is carrying (assuming all are the same size, material, and gearing).
Assuming both sides are pushing the robot the same direction, the thrust/torque from the two gearboxes add to each other in accelerating the robot.
Overall, adding motors at a given gearing/wheel size DOES add roughly linearly to the amount of power/acceleration available; adding wheels is more about stability (robot not falling over and dealing with uneven surfaces) and adds little to acceleration at a given speed.
Note that I have only covered the mechanical considerations so far. Because of electrical considerations, 6 CIMs do not actually give you 3 times as much thrust as 2 CIMs. At some point in here, the limitations of your battery come into play - there are a number of threads on this topic; search on 6 CIMs, miniCIMs, or 775pros - or 8 775pros for examples if you want to get in those weeds.
