gearbox calculator

[1683cadder]

I'm designing a three motor gearbox right now (2 Cim and 1 minicim), so I designed an excel chart to help me figure out the total gearing I need for the robot and some other specs. I am totally lost. I based my calculations from the information given here (http://www.instructables.com/id/Unde...arbox-Design/). What am I doing wrong? Is there other variables that I need to consider?

[Tom Line]

I think your first step should be to look at JVN's calculation sheet. You can find it in the media section under papers. It already has something similar to what you are attempting to do. You can compare your calculations to his and see where you're going wrong. Start with 2 motors, then add a third.

[1683cadder]

Is the math right?

[ablatner]

JVN's is right. Lots of people use it.

[1683cadder]

oh lol, I finally got how to use JVN. But the question still remains. If you are using two different motors in a gearbox, then how do you factor that into the calculations?

[apalrd]

Originally Posted by 1683cadder

oh lol, I finally got how to use JVN. But the question still remains. If you are using two different motors in a gearbox, then how do you factor that into the calculations?

There are a couple ways to do this.

I use JVN's Old design calculator for this, since it was written in the olde days before the CIM was the king of drivetrain motors and people mixed all kinds of things to drive. It's available here. I also have a version Here which includes battery voltage modeling, useful for some advanced simulation but not necessary for the basics.

To calculate the contribution of a motor, it's easiest to start with the speed and calculate the torque. Since the speed of each motor is the same (or tied through a fixed gear ratio), we can use it to calculate the torque contribution of each motor. You don't have to do this at free speed.

The JVN Old Calculator method (Acceptable for basic modeling, not quite ideal) basically creates a single 'super-motor' which you specify a free speed and gear ratio for each input motor and it calculates the sum torque for use as an assumed single motor. This is good enough when you match to free speeds. You can match free speeds and get moderately good results. IF free speeds are matched perfectly, then each motor will always be running at the same % operating point, and produce it's share of torque. For example, say we have a motor (Motor A) which produces 4 units of torque at 5 units speed. Another motor (Motor B) produces 1 unit of torque at 10 units speed. We gear motor B 2:1 relative to motor A such that they are spinning at the same speed when mixed, and the torque output at the mixed point will force motor A to produce 2x the power and torque of motor B, while consuming 2x the current.

It's also possible to design for non free speed matching. In that case, you could force the 'weaker' motor to produce more than it's share of power at your design speed. This can be useful sometimes.

[Ether]

Originally Posted by apalrd

For example, say we have a motor (Motor A) which produces 4 units of torque at 5 units speed. Another motor (Motor B) produces 1 unit of torque at 10 units speed. We gear motor B 2:1 relative to motor A such that they are spinning at the same speed when mixed, and the torque output at the mixed point will force motor A to produce 2x the power and torque of motor B, while consuming 2x the current.

That last part -- "while consuming 2x the current" -- is generally not true for different motors.

Generally speaking, if you gear Motor B to produce half as much mechanical power as Motor A at some operating point, it will not be drawing half as much current at that operating point.

Likewise, if you gear Motor B to draw half as much current at some operating point, it will not be producing half as much mechanical power at that operating point.

[apalrd]

Originally Posted by Ether

That last part -- "while consuming 2x the current" -- is generally not true for different motors.

Yes and no.

Assuming the motors were both perfectly efficient (or identical efficiencies), then what I said would be true.

For the CIM and MiniCIM, this is roughly true (the miniCIM will contribute it's share relative to it's power potential), as they are nearly identical motor designs.

[Ether]

Originally Posted by Ether

That last part -- "while consuming 2x the current" -- is generally not true for different motors.

Originally Posted by apalrd

Yes and no.

Assuming the motors were both perfectly efficient (or identical efficiencies), then what I said would be true.

For the CIM and MiniCIM, this is roughly true (the miniCIM will contribute it's share relative to it's power potential), as they are nearly identical motor designs.

For the sake of students who may be reading I need to make a couple observations:

When I wrote "generally not true for different motors" it was not intended to include "nearly identical motor designs". I assumed that would be inferred but perhaps it should have been stated explicitly.

If two different motors have identical efficiencies at the chosen operating point, that would be a special case... not general. In general, the efficiencies of different motors will not be identical for an operating point selected based on current or power.

Your posts are usually very practical. I'm sure you did not expect your readers to infer you were discussing perfectly efficient motors.

[1683cadder]

Originally Posted by apalrd

. In that case, you could force the 'weaker' motor to produce more than it's share of power at your design speed. This can be useful sometimes.

Very helpful, but please elaborate on this point.

[apalrd]

Originally Posted by 1683cadder

Very helpful, but please elaborate on this point.

brushed DC motor torque curves are linear, such that peak torque is produced at stall and 0 torque is produced at free speed. Current is proportional to torque, so at peak torque the motor will consume peak current, and at 0 torque the motor will consume free current. In this example, I'm going to use ideal motor specs as listed on the mfg's website. All speeds are rpm, torque in N*m, current in a.

This example isn't as obvious with the CIM and MiniCIM, so I'm going to use the CIM and RS-550 from BaneBots. Here are the graphs of their torque curves. CIM is blue, RS550 is green.
http://www.chiefdelphi.com/forums/at...d=13711310 01

Now we have to decide what ratio to mix them at. If we mix them at the ideal ratio to match free speeds (it happens to be ~3.63:1 reduction on the RS550 to match CIM speed), we can simply solve for the torque at a given speed and add them to get the yellow line. If we don't like the yellow line, we can mix at whatever speed we want. Note that the line is still linear, so we can simply adjust for the slope in the gearing. For example, the blue line is 3:1 RS550:CIM, the pink line is 2.7:1 RS550:CIM, the white line is 4:1, and the red line is 5:1.
http://www.chiefdelphi.com/forums/at...d=13711310 15

The real reason to do this is for current. Current is related to torque, so we can graph current like we did for torque (vs speed), assuming constant voltage. I skipped ahead and mixed them according to free speed, and summed them. In this graph, pink is CIM current, sloped blue is speed-adjusted RS550 current, red is sum current, light blue is 120a, yellow is 80a, green is 60a, flat blue is 40a. Based on the flat current lines, the CIM is at ~60a when the RS550 is at ~40a.
http://www.chiefdelphi.com/forums/at...d=13711310 27

If we make this graph again with 2.7:1 mix ratio, we end up with (RS550 current changed to white, other colors the same). Now the two motors hit ~40a at the same (adjusted) speed. Once the output speed overruns the slowest motor it will become a generator and produce current, loading the motor overrunning it. That is why the sum current intersects RS5550 current at CIM free speed then goes below it, as the CIM graph continues under the X axis as a negative current.
http://www.chiefdelphi.com/forums/at...d=13711310 34

So now the RS-550 is running at a lower operating point, and will draw more current than it's ideally mixed current (it's share).

Why would you do this? If you know the motor can handle more than it's share, you can match such that you get more low-end torque for accel, and more evenly distribute current through the electrical system (40a breakers handling different motors see closer to same current), which can improve accel performance. However, the motor consuming more than it's share will produce more than it's share of heat also, so you have to verify that the motor can handle the additional heat (for a CIM + MiniCIM it's probably OK, worry about the RS550 mixed 2.7:1).

It's really all about where you want to optimize the current and heat, since you can adjust for torque in the gearing later. You could mix these two motors at 2.7:1 so the currents are more evenly distributed, or mix at 4:1 so the RS550 has to handle less heat.

[Dillon Carey]

As I was reading through this thread I have been wondering. On a single speed 4 or 6 cim drive, would the acceleration benifit from having 1 of the cims in each gearbox geared differently? In the excel calculator it makes some significant improvemants, however I am not sure the program was meant to handle this situation.

My main worry would be when the lower geard cim goes past its free speed rpm, would it hurt the motor or hinder the drivetrains ability?

Thanks for the awesome explanations so far.

[apalrd]

The JVN excel calculator does not deal with this situation well.

The JVNcalc determines the 'supermotor' specs by:
-For speed, you input the design speed. There is no calculation of this.
-For torque, you input the gear ratios for each motor, and it adds the adjusted torques.
-For current, it adds them up.

This works fine if the adjusted free speed of all of the motors is close to the design speed, but if you deviate from this, then the free speed number is wrong, so you would incorrectly get more torque than you should.

Looking at the second link in my last post, moving from the yellow line to the red line will cause an increase in stall torque (which the simulator uses as the torque number) but a decrease in free speed - Since these torques are the sum of a CIM and adjusted RS-550, the CIM will overrun the RS-550 before 4500rpm, and at 4500rpm the negative torque produced by the RS550 and positive torque produced by the CIM total zero and you have your new free speed of the combined assembly (~4500 for that example). So while you would gain stall torque, you would lose free speed and when you adjusted your gearing appropriately, the result would be roughly the same. The JVN calc does not calculate free speed, so it does not show this. If you open the circuit on the motor that is being overrun at the point at which it starts being overrun (so instead of acting as a generator it does nothing), you end up with a piecewise torque curve with a higher free speed, but the accel simulator on the next two pages of the spreadsheet won't simulate the piecewise torque curve correctly either.

Increasing motor load also increases electrical load. There's a point at which adding more motors hurts more than it helps, I believe 6 motors is right around that point (so adding more than 6 is bad, adding 6 is good), but the benefits taper off after 4 (the improvement from 2 to 4 motors is huge, the improvement from 4 to 6 is not nearly as big).

[Dillon Carey]

Originally Posted by apalrd

If you open the circuit on the motor that is being overrun at the point at which it starts being overrun (so instead of acting as a generator it does nothing), you end up with a piecewise torque curve with a higher free speed, but the accel simulator on the next two pages of the spreadsheet won't simulate the piecewise torque curve correctly either.

This is something that I have been contemplating for a while.

I haven't figured out how to get this to work in both directions, but if you had a 3 cim gearbox and one of the cims was geared significanlty lower, lets say 3 times lower, and was connected using a one way locking bearing would there be any significant gain in acceleration? Is there any reason for doing this instead of a 2 speed gearbox?

Thanks again

[apalrd]

You would gain low-end torque at the expense of torque above ~1350rpm in that scenario.

I guess it depends on where you want to optimize torque. For driveability you would want more midrange torque, which three motors geared the same will provide more of than your solution.

If you shift at the right time, you could get a much nicer torque curve through accel and driving.

Illustration:
green is torque curve with 2.2:1 reduction (ball shifter shift spread) with 3 CIM input
dark blue is your proposed torque curve (2 CIM 1:1 plus 1 CIM 3:1 when torque output from that CIM is >0)
yellow is torque curve with 1:1 reduction (3 CIM 1:1)
http://www.chiefdelphi.com/forums/at...d=13714870 18

Mechanically shifting ends up switching between the green and yellow curves at the intersection (ideally), which will ALWAYS lead to more output torque than the blue curve.

Mechanically shifting is the answer.