Mathematical Analysis of a two degrees of freedom differential system

[TechnicBrickPower]

Hi All,

Just made a video that presents an analysis of a two degrees of freedom differential gearing system that I have been basing some previous designs on. I use this to explain why the "Slow reverse Car" mechanism did not work as expected. Only for maths fans - don't watch otherwise ;)

 

 

[TeamThrifty]

Maximum respect. To be able to take a relatively simple lego component and explain it in this way... whether everyone can follow the maths or not ..should be used as an educational tool. Its so much more engaging than just cold, soulless maths. Excellent.

(i may need to watch it again though as i lost the plot part way through!!  )

[piterx]

i wish there were more people doing the stuff you do dude!
this is so interesting

[Aleh]

It appears ti be very serious despite I understood almost nothing 

[GroundskeeperWillie]

[TechnicBrickPower]

22 hours ago, TeamThrifty said:

Maximum respect. To be able to take a relatively simple lego component and explain it in this way... whether everyone can follow the maths or not ..should be used as an educational tool. Its so much more engaging than just cold, soulless maths. Excellent.

(i may need to watch it again though as i lost the plot part way through!!  )

 

22 hours ago, piterx said:

i wish there were more people doing the stuff you do dude!
this is so interesting

Thanks guys - great feedback - I thought no one would be interested when I posted this. Hope to do more in the future.

[m00se]

Thanks a lot!

A while ago I tried to make such a mechanism that has a different speed depending on the direction, and I couldn't get it to work. Now at least I understand better why.

[lcvisser]

Interesting video, but as you've found out, differentials don't work like that when you drive the outer gear. Without considering the external torques (e.g. you holding the output axle), the problem is ill-defined.

Basically, the differential is a three-"port" system, where every "port" is a pair of rotational velocity and torque:

          ______|
(w1, T1) |  __  | (w2, T2)
---------|-|  |-|---------
         |______|
                |
                 (w0, T0)

(excuse me the poor ASCII drawing).

Power = rotational velocity x torque, or P  = wT, and since a differential is a passive component, the sum of all three products must be zero, i.e. all power that goes in must go out somehow:  0 = P0 + P1 + P2 = w0 x T0 + w1 x T1 + w2 x T2

If you choose to drive w1 and w2, then we know that w0 will be the average of the two. This is a consequence of the mechanics: the third internal gear is fixed to the housing and thus constraint. This means that w0 is an output variable and w1 and w2 are input variables: w0 = (w1 + w2)/2

In matrix form, you'd have 

                 [w1]
w0 = [ 0.5  0.5 ][  ]
                 [w2]

If you have studied linear algebra, you know this relation cannot be inverted: there's infinitely many solutions for w1 and w2 that satisfy this equation. That's exactly what you found out when you drew the two lines in your graph and noted that there's not a unique value for x or y.

The complement of the above equation is

[T1]   [0.5]
[  ] = [   ] T0
[T2]   [0.5]

Which should be familiar also: both shafts of the differential get halve the torque in nominal conditions. To properly analyse your mechanism you need to consider the torques that act on the input and output as well, then the problem becomes well-defined and you can solve the equations uniquely.

 @ukbajadave showed how playing around with internal torques can give you some interesting results.

(I'm still trying to find time to do the math on that)

Edited July 25, 2020 by Ludo Visser

[Ivorrr]

More videos like that please! Simple and logical

[arieben]

I've been watching your previous videos over and over again because I've been needing to use differentials to get odd ratios for my GBC module, so I understood this one instantly! 

Thank you for making this great and unusual content. Love this stuff.

[TechnicBrickPower]

6 hours ago, arieben said:

I've been watching your previous videos over and over again because I've been needing to use differentials to get odd ratios for my GBC module, so I understood this one instantly! 

Thank you for making this great and unusual content. Love this stuff.

Hi @arieben, thank you and am really happy you found my content on differentials useful - it's good to hear when someone can use it for their own projects. I find it very satisfying to use math to discover solutions you'd never find by just trial and error.

9 hours ago, Ivorrr said:

More videos like that please! Simple and logical

Will do my best ;)

[TechnicBrickPower]

18 hours ago, Ludo Visser said:

Interesting video, but as you've found out, differentials don't work like that when you drive the outer gear. Without considering the external torques (e.g. you holding the output axle), the problem is ill-defined.

Basically, the differential is a three-"port" system, where every "port" is a pair of rotational velocity and torque:

          ______|
(w1, T1) |  __  | (w2, T2)
---------|-|  |-|---------
         |______|
                |
                 (w0, T0)

(excuse me the poor ASCII drawing).

Power = rotational velocity x torque, or P  = wT, and since a differential is a passive component, the sum of all three products must be zero, i.e. all power that goes in must go out somehow:  0 = P0 + P1 + P2 = w0 x T0 + w1 x T1 + w2 x T2

If you choose to drive w1 and w2, then we know that w0 will be the average of the two. This is a consequence of the mechanics: the third internal gear is fixed to the housing and thus constraint. This means that w0 is an output variable and w1 and w2 are input variables: w0 = (w1 + w2)/2

In matrix form, you'd have 

                 [w1]
w0 = [ 0.5  0.5 ][  ]
                 [w2]

If you have studied linear algebra, you know this relation cannot be inverted: there's infinitely many solutions for w1 and w2 that satisfy this equation. That's exactly what you found out when you drew the two lines in your graph and noted that there's not a unique value for x or y.

The complement of the above equation is

[T1]   [0.5]
[  ] = [   ] T0
[T2]   [0.5]

Which should be familiar also: both shafts of the differential get halve the torque in nominal conditions. To properly analyse your mechanism you need to consider the torques that act on the input and output as well, then the problem becomes well-defined and you can solve the equations uniquely.

 @ukbajadave showed how playing around with internal torques can give you some interesting results.

(I'm still trying to find time to do the math on that)

Hi Ludo. Thanks so much for your detailed answer on the mathematics of the torques and forces involved - that is one area I'd like to study more. Also very interesting link you sent about the differential that won't go backwards - I have discovered the same in some cases and found it quite frustration - I don't fully understand what is behind this problem!

[lmdesigner42]

Another great video on differentials! The O/I ratio makes me think of transfer functions lol.