There is an ordinary linear three-mass system.

enter image description here

If we write its Lagrangian, we get the following equation.

enter image description here where $W_k$ and $W_n$ - kinetic and potential energy.

To find the moment of rotation of the first mass, we must differentiate lagrangian first by the angle of rotation of the first mass $\phi_1$, then find the rate of change in time of the Lagrangian derivative with respect to speed $\omega_1$.

Lagrange equation

Suppose we want to find the moment of the first mass $J_1$ (we assume that on the shaft of the drive motor).

$\frac{\partial L}{\partial q} = \frac{\partial L}{\partial \phi_1} = -c_{12} (\phi_1 - \phi_2)$

$\frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = \frac{d}{dt}\frac{\partial L}{\partial \omega_1} = J_1 \frac{\partial \omega_1}{\partial t}$

Which ultimately gives us a dynamic equation:

$M - c_{12} (\phi_1 - \phi_2) = J_1 \frac{\partial \omega_1}{\partial t}$

My questions are as follows:

  1. Where did the masses of $J_2$ and $J_3$ go? Because we are looking for a derivative and speeds of this masses $\omega_2$ and $\omega_3$ do not explicitly depend on $\omega_1$, no matter how massive they are, they do not include into the equation of motion and do not affect the torque M.

  2. Do I understand correctly that taking into account the influence of the moments of inertia of the remaining masses $J_2$ and $J_3$ is possible only by bringing the moments of inertia to the motor shaft $J_1$ (by correcting the coefficients of the square of the gear ration).

  3. Is it possible to do without this operation and include the gear ratio in the model?


1 Answer 1


Welcome to Robotics, Andrew Sol. I think I'm a little confused with your question as it appears to be about independent rotational masses connected by rotational springs, but then later you're asking about gear ratios.

This is a succinct as I can think to put it, and again I may have misunderstood the question so please feel free to comment on this answer or edit your question to clarify:

The equations of motion govern the dynamics of the system. If you have several bodies that are rigidly coupled, as in a gearbox, then you would (I would, at least) modify the moments of inertia to account for the gearbox ratio and then treat the entire gear train as one system.

Consider this: If $\omega_2$ could be calculated by $\omega_1$ and a gear ratio, then what dynamics do you need to solve?

  • $\begingroup$ Lagrangian formula contains components of kinetic energy. We differentiate the Lagrangian by velocity $\omega_1$. In this case, the components in $\omega_2$ and $\omega_3$ will "leave", because they do not depend on $\omega_1$, which means that when finding the derivative they can be removed., i.e. no matter how massive they are (i.e. $J_2$ and $J_3$), they have no effect on the moment of the first mass, in this case ... $\endgroup$
    – dtn
    Jan 13, 2020 at 4:55
  • $\begingroup$ Do I understand correctly that they can only be taken into account when correcting the moment of inertia of the first mass? Can you do without this adjustment? I need to get a multi-mass system taking into account stiffness, and then convert it into a state space. I can not consider the system as absolutely rigid. $\endgroup$
    – dtn
    Jan 13, 2020 at 4:59
  • $\begingroup$ Pay attention to how the structural diagram of a two-mass system looks like, taking into account the gear ratio. Figure 5. Servo controller structure for two-mass model Design and servo control of a leak tightness machine working based on hydrostatic pressure aging method. $\endgroup$
    – dtn
    Jan 13, 2020 at 5:25
  • 1
    $\begingroup$ @AndrewSol - The thing that's confusing me about your question is that there's no gearbox in your diagram and no description anywhere of where a gearbox would be or what it's connecting, but then your questions all seem to be about gear ratios. If the motion of masses 2 and 3 don't depend on the motion of mass 1 then they're not coupled and are thus two separate systems. I think it would help if you could please edit your question to post a diagram that better reflects the system you're asking about. $\endgroup$
    – Chuck
    Jan 13, 2020 at 14:05
  • $\begingroup$ I specified the problem and tried to make the questions consistent and not confusing. $\endgroup$
    – dtn
    Jan 13, 2020 at 16:15

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