-A 775 pro, in a Versaplanetary with a 9:1 stage, with a 40 tooth gear on the output shaft
-A second 775 pro, in a Versaplanetary with a 3:1 stage, with a 40 tooth gear on the output shaft
-A third 40 tooth gear, linking together the output of both planetary gearboxes, with a final output shaft in it.
Some interesting questions:
What would be the output free speed? My guess is about 1/6th the free speed of 775 pro
What would be the stall torque? My guess is 12 times that of a single 775 pro.
For a given load - say, for example, lifting an object from a spool on the output shaft - what would the current draws be on each motor. My guess is that the second motor would draw three times as much current as the first, but I could be easily convinced that they would be equal.
Unless the 775pro in the 3:1 is given about 1/3 output speed in the program, which at that point is probably defeating the purpose of the gearbox.
Running this gearbox with the 9:1 at full speed and the 3:1 at 1/3 speed would give the speed of 1/9 of the motor
Torque is correct (ignoring the fact that the motors kinda fight each other): 12 times the stall torque of the gearbox (4/3 of a motor’s worth of power times the 9:1 reduction)
I WOULD NOT RECOMMEND RUNNING THIS GEARBOX
Both as 9:1s would be much better: it would have the same speed, more torque, and would run into less of an issue with one motor handling most of the load
What is the purpose of this gearbox? Could you achieve the results you want with a different gear ratio? I would not be sure that this would work so it might be wise to try a different gear setup/ratio. If this design did work, I think that the load would not be evenly distributed between the two motors.
If you are just curious what would happen, my guess would be load would be different on each motor and one motor would have to almost have to help the other motor along, thus making a inefficient use of the motors. I also think that this would have an effect on motor longevity and heat from the motor.
Can’t comment on speed without doing some math first. The 775Pro in the 9:1 would begin to resist movement. However I can tell you that your stall torque would be limited to only 3 times a single 775 unless there was a greater reduction later down the line. Otherwise the 775Pro with the 9:1 gearbox would start to back-drive it’s counterpart in the 3:1 gearbox.
Edit: Did some quick math. A 775 with a 9:1 reduction would have a free speed of 2000 RPM, and one with a 3:1 would result in 6000 RPM. I believe it’s safe to say that the 9:1 has greater torque, and therefore the 3:1 wouldn’t be able to speed it up easily, so the speed would be somewhere in the middle, with the balance weighing towards the 9:1’s free speed by what I’m assuming is two-thirds. This would give you a free speed of 3333.3333 RPM, but I’m not completely certain on this math.
This gearbox gives you the worst of both speed and torque and could cause irreparable damage to your motors, especially fan-cooled 775Pros.
This was more of a thought excercise than something I would actually do. However, i’d imagine that teams that mix CIMs, Mini CIMs, and 775s in a drive gear box must deal with a similar issue, albeit on a less extreme scale.
And I should have clarified, the intent of the thought excercise would be to put the same voltage into each motor.
The gearbox would behave almost exactly like the 9:1 gearbox would alone, but with some performance loss due to having to fight the second motor. Most of the energy put into the 3:1 gearmotor would be lost as it could not really force the 9:1 gearmotor to spin faster (3x more torque in that setup, after all). (Wait, do I have this backwards?)
Everyone’s comments about matching free speed and how it would sort of work if the 3:1 was software limited to 1/3 speed are not really correct. There is nothing magical about matching free speed vs some other design or operating speed, which depends on gearbox load. The key is that each motor is contributing something as long as it is not forced to spin above free speed, and if it is forced into that speed then that motor is actively resisting the motion in the system.
I’ll note what mismatched gearboxes are typically used for - if you’re using a mixed gearbox (ex. a 775pro with a CIM), you find that they have vastly different free speeds. You could gear them 1:1, but that results in inefficiencies. If you’re clever, you set the relative gearing across each motor such that they provide the same speed at the expected load.
There’s a thread somewhere where Paul C and Joe Johnson derive the equations for speed/torque of combined gearboxes of mismatched motors, if that’s what you were hoping to gain out of the thought experiment.
Yeah, my science project this year was a planetary gear infinitely variable transmission, so I’ve spent a lot of time the last few weeks doing gearbox calculations. That does not make me an expert, so please, correct me if I’m wrong.
My calculations for free speed were done by taking the difference between the two free speeds, and dividing it by the difference in strengths between the two gearboxes. (6000-2000)/(9/3) I just added this to the slower speed, as the slower gearbox should be difficult for the faster one to overdrive. However, due to the torque curve in each being linear, the 3:1 would still have significant power output over the free speed of the 9:1. The more I think about it, the more I think I may be wrong, and I’ll have to look into overdriving motors.
You really just need to solve a system of equations where the torque needed to overpower the 9:1 is equal to the torque provided by the 3:1 at the same output speed.
If this isn’t the math you were asking for, I’d love to help.
From the thread I linked earlier, credits to Joe Johnson:
Say you have one motor, with a stall torque of T_s1 and a free speed of W_f1. Likewise, the other motor has a stall torque of T_s2 and W_f2.
The combined stall torque is simply T_s1 + T_s2.
The combined free speed is (T_s1 + T_s2) * (K1 * K2) / (K1 + K2) where K1 = W_f1 / T_s1 and K2 = W_f2 / T_s2.
Let’s take the scenario above, where the two motors are the same but one has an extra 3:1 reduction. This gives us the constraints that T_s1 * 3 = T_s2 and W_f1 / 3 = W_f2.
If we want everything in terms of motor 1 (the motor with the less reduction), we can substitute our new equations in:
So, if you had a 3:1 775Pro and a 9:1 775Pro geared 1:1 to each other, your free speed would be about 40% of the free speed of the 3:1 775Pro. 3 * 10/4 = 7.5, so you would have the same free speed and stall torque as a dual 775Pro gearbox geared at 7.5:1.
Note this is JUST free speed and stall torque. This does NOT say anything about power, efficiency, current draw, etc. It does not say anything about speed or torque at other points on the curve.
Btw Paul Copioli has a calculator that, among other things, gives combined motor specs for gearboxes with multiple motors of different types with different gear ratios.
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