Question about torque through a differential. 4

[sean7457]

I've currently got a gearbox that is driven by two electric motors. Each motor is ran through the same differential such that if one motor were to fail, the other motor could still drive the system.

My question is: Will the differential act as a torque multiplier?

I tend to think of it like a traditional differential in a car, only backwards -- replace each tire with a motor, and use the driveshaft as the output for the system. Assume the gear ratio between the motor and diff are 1:1.

The argument I've been having is that one of my co-workers seems to think that in this setup, the diff will add a 2:1 gear reduction, and thus double the torque output. In other words, if both my motors are putting out 10 ft*lb of torque, and they are tied directly to the differential at 1:1 and the differential is tied directly to my output at 1:1, I will still have a total of 20 ft*lb of torque coming out of my gearbox.

I understand that if you lose one motor, the speed will be halved, but the total torque out should remain constant? The load would just shift from one motor to the other?

...I hope I haven't confused anybody. Thanks in advance for any and all help.

 

[sean7457]

Lionel,

If I understand you, you are saying that if one motor is stopped, the other motor will see no load change?

My current design is using two electric motors on a 24V power supply that is restricted to a maximum of 2 amps. So based on your example above, I would not have to size one motor to handle the total required output torque? In other words, each motor has no idea whether or not the other motor has stopped working?

To take it one step further, you are saying that I could essentially lock one side of the diff and use it as a 2:1 reduction?

 

[patprimmer]

The one remaining motor, IF IT RETAINS THE SAME SPEED will also require the same torque.

The output will retain the original torque but the speed will be halved so the power will be halved.

The 2:1 reduction in speed gives a 2:1 multiplication in torque at the output.

Removing drive from one motor halves the total input power. The change in effective drive ratio corrects for the loss in torque at the expense of speed or distance travelled. This of course presumes that it is accepted that power equals torque times speed.

Regards
Pat
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[LionelHutz]

Pat - It's OK, I'll confirm that you're not confused.

Compositepro - no, the torque load on the remaining motor does not double.

Sean - YES. To go back to your first post, your co-worker is correct.

The differential must always have the same torque on both input shafts and it always adds the torque on both input shafts together to the torque on the output side. Don't confuse the stopped motor as meaning there is not any input torque applied to that shaft. The stopped input shaft still has the same torque applied, it's just applied at 0rpm via a brake of some sort instead of being applied at a rotating speed via a motor. If you didn't have any input torque applied to one of the shafts then you'd get no output torque (both input shafts would just rotate opposite directions).

So, thinking about it like above, it's not technically a torque multiplier or torque doubler, it's a torque adder. Funny though, since T1 = T2 and Tout = T1 + T2 you always get Tout = 2 x T1 and so it always does doubles the torque of a single input shaft.

This is why I posted it's a HP adder. You don't remove any torque from the differential input shaft when you stop the motor, you just remove HP.

 

[sean7457]

Lionel,

OK, so in THIS case it would act as a 2:1 reduction.

If, for example, the motors were not identical (T1 does not equal T2), then the reduction ratio would depend on the ratio of the motor sizes, correct?

I think the confusion for me from the beginning was that I was being told a differential will ALWAYS add a 2:1 reduction regardless of input.

Also, you stated in your first post:

'This type of gearbox could be considered a constant torque unit which more or less adds the HP and speed of the motors together.'

So in your example of two 1500RPM/500ft*lb motors, shouldn't your differential be outputting 1000ft*lbs at 3000RPM when both motors are running? And then when one motor goes down, you'd have 1000ft*lbs at 1500RPM?

 

[Warpspeed]

I am just extremely grateful that I have never had to teach this stuff to students.

And then we have various strange compound configurations of epicyclic gear trains to think about......

 

[LionelHutz]

The gearbox and more or less are important parts of that comment. I was making a general comment about a gearbox that used a differential.

What is this 2:1 reduction? What speed you are comparing the output speed to?

The differential itself adds 1/2 the speed of each input to get the final output speed.

S(out) = 0.5 x S(input1) + 0.5 x S(input2)

If you use the sum of the input speeds then the differential is always a 2:1 reducer. If you have matched motors and compare the output speed to a single motor's input speed then it's a 1:1 when both motors are running and 2:1 when one motor is running.

And I thought some simple numbers in an example would make the relationships clear.

 

[sean7457]

First of all, thanks for all your help and taking the time to fully explain this. I really appreciate it.

What I meant by THIS case is that because both motors happen to be the same, the ratio ends up being 2:1. But this isn't necessarily the case for all situations.

For example (assuming I'm understanding what you've said up to this point):

If you have two different motors:
M1 = 100ft*lbs @ 500RPM
M2 = 20ft*lbs @ 500RPM

Then your output with both motors running = 120ft*lbs @ 500RPM. With only one motor running, this would change to 120ft*lbs @ 100RPM (a reduction of 5:1 -- 100/20)

You could also have varying input speeds as well:
M1 = 100ft*lbs @ 500RPM
M2 = 20ft*lbs @ 200RPM

Then your output with both motors running = 120ft*lbs @ 350RPM ((500/2)+(200/2)). If one motor fails in this scenario, then your output would change to 120ft*lbs @ 70RPM (a reduction of 5:1 -- 100/20).

I know this might be going slightly beyond the scope of my initial question, but I'm trying to fully understand the fundamentals here.

Thanks again!

 

[LionelHutz]

No, you apparently don't understand at all. The torque from both input shafts (motors) must always be equal and it always sums to 2x a single input shaft torque at the output. The input speeds added together is 2x the output speed (or inputs speeds added together divided by 2 equals the output speed).

In both examples
Output = 120ft-lbs both running means each motor = 60ft-lbs.
M1 stopped you can get 40ft-lbs out.
M2 stopped you can get 200ft-lbs out.

Example #1
If either motor stops then the output speed is 250rpm.

Example #2
If M1 stops then the output speed is 100rpm.
If M2 stops then the output speed is 250rpm.


Once again, here are the formula to follow.

T(out) = T(input1) + T(input2)
where T(input1) must be equal to T(input2)

S(out) = 0.5 x S(M1) + 0.5 x S(M2)

T is the torque and S is the speed.

I have no idea how to make this any clearer.

 

[Tmoose]

Are these Induction motors?

 

[sean7457]

Tmoose -- no, these are DC servo motors.

Lionel -- OK, torque must be equal on both inputs. That being true, how could you have a torque higher than the rated torque of a motor on one side? Wouldn't the whole system be reduced to the max output of 2X the smallest motor?

In other words, in both the above examples, you could only expect 40ft*lbs (2x20ft*lbs) at the output. The 80ft*lb difference between the two motors would essentially 'bleed off'? Just like in a car when you hit an ice patch on one tire with this type of differential.

 

[LionelHutz]

It appears you figured out the problem with your example.

 

[sean7457]

...it was definitely a long journey, but I think I finally got there.

Thanks for all the help!