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?


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 '20 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 '20 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 '20 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 '20 at 14:05
  • $\begingroup$ I specified the problem and tried to make the questions consistent and not confusing. $\endgroup$ – dtn Jan 13 '20 at 16:15

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