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I need advice, I'm confused.

Derive the equations of motion, including gear ratios

Equation of motion including gear ratio

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System contains two lumped masses with angular velocity vectors $\boldsymbol{\omega}_1$, $\boldsymbol{\omega}_2$, angular positions vectors $\boldsymbol{\theta}_1$,$\boldsymbol{\theta}_2$, inertia matrices $\boldsymbol{J}_1,\boldsymbol{J}_2$, these masses are connected through a flexible shaft with stiffness matrix $\boldsymbol{c}_1$. $\boldsymbol{d}_1=\boldsymbol{0}$.

Generalized coordinates:

$q=\begin{bmatrix}\boldsymbol{\omega}_1\\\boldsymbol{\omega}_2\\\boldsymbol{\theta}_1\\\boldsymbol{\theta}_2\end{bmatrix}$

Kinetic energy of the system:

$T=\frac{1}{2}\boldsymbol{\omega}_1^T\boldsymbol{J}_1\boldsymbol{\omega}_1+\frac{1}{2}\boldsymbol{\omega}_2^T\boldsymbol{J}_2\boldsymbol{\omega}_2$

I have time-dependent transfer matrix $A(t)$ of the angular velocity vector $\boldsymbol{\omega}_1$ into the angular velocity vector $\boldsymbol{\omega}_2$, the structure of this matrix is known:

$\boldsymbol{\omega}_1=\boldsymbol{A}(t)\boldsymbol{\omega}_2$

To take into account the gear ratio between the angular generalized coordinates $\boldsymbol{\theta}_1$,$\boldsymbol{\theta}_2$, there must be a matrix $\boldsymbol{B}(t)$, i.e.

$\boldsymbol{\theta}_1=\boldsymbol{B}(t)\boldsymbol{\theta}_2$

But matrix $\boldsymbol{B}(t)$ is unknown.

By integrating the vector of the angular velocity of the first mass, one can obtain the vector of the angular position of the first mass, and then:

$\boldsymbol{\theta}_2=\int \boldsymbol{A}^{-1}(t) \boldsymbol{\omega}_2dt=\boldsymbol{A}^{-1}(t)\boldsymbol{\theta}_2-\int (\frac{d}{dt}\boldsymbol{A}^{-1}(t))\boldsymbol{\theta}_2dt$

Than, potential energy of the system:

$V=\boldsymbol{c}_1\cdot(\boldsymbol{i}_1(\boldsymbol{v}\boldsymbol{v}^T)\boldsymbol{i}_1+\boldsymbol{i}_2(\boldsymbol{v}\boldsymbol{v}^T)\boldsymbol{i}_2+\boldsymbol{i}_3(\boldsymbol{v}\boldsymbol{v}^T)\boldsymbol{i}_3)\cdot\begin{bmatrix}1\\1\\1\end{bmatrix}$

where $\boldsymbol{v}=(\boldsymbol{\theta}_1-(\boldsymbol{A}^{-1}(t)\boldsymbol{\theta}_2-\int (\frac{d}{dt}\boldsymbol{A}^{-1}(t))\boldsymbol{\theta}_2dt))$

$i_1=\begin{bmatrix}1 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}$

$i_2=\begin{bmatrix}0 & 0 & 0\\0 & 1 & 0\\0 & 0 & 0\end{bmatrix}$

$i_3=\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 1\end{bmatrix}$

Lagrangian $L=T-V$

Problem: Matrix $\boldsymbol{B}(t)$ is unknown. Lagrangian contains the integral of an unknown function $\int (\frac{d}{dt}\boldsymbol{A}^{-1}(t))\boldsymbol{\theta}_2dt$. How to derive the equation of motion $\frac{d}{d(\boldsymbol{\omega}_1,\boldsymbol{\omega}_2,\boldsymbol{\theta}_1,\boldsymbol{\theta}_2)}L$?

EDIT (thanks to @Cesareo):

Clear["Derivative"];

ClearAll["Global`*"];

Remove[c, J, A];

pars = {Subscript[J, 1] = 1, Subscript[J, 2] = 1, c = 10, A[t] = 2};

s = NDSolve[{M'[t] + M[t] == -Subscript[\[Theta], 1]'[t] + 1, 
    Subscript[J, 1] D[D[Subscript[\[Theta], 1][t], t], t] + 
      D[\[Lambda][t], t] + 
      c (Subscript[\[Theta], 1][t] - Subscript[\[Theta], 2][t]) == 
     M[t], Subscript[J, 2] D[D[Subscript[\[Theta], 2][t], t], t] - 
      D[\[Lambda][t], t] A[t] - \[Lambda][t] D[A[t], t] - 
      c (Subscript[\[Theta], 1][t] - Subscript[\[Theta], 2][t]) == 0, 
    D[D[Subscript[\[Theta], 1][t], t], t] - 
      A[t] D[D[Subscript[\[Theta], 2][t], t], t] - 
      D[A[t], t] D[Subscript[\[Theta], 2][t], t] == 0, M[0] == 0, 
    Subscript[\[Theta], 1][0] == 0, Subscript[\[Theta], 2][0] == 0, 
    Subscript[\[Theta], 1]'[0] == 0.1, 
    Subscript[\[Theta], 2]'[0] == 1/2 0.1, \[Lambda][0] == 
     0}, {Subscript[\[Theta], 1], Subscript[\[Theta], 
    2], \[Lambda]}, {t, 100}];

Plot[{Evaluate[Subscript[\[Theta], 1]'[t] /. s], 
   Evaluate[Subscript[\[Theta], 2]'[t] /. s]}, {t, 0, 100}, 
  PlotRange -> Full];

Plot[{Evaluate[\[Lambda][t] /. s]}, {t, 0, 100}, PlotRange -> Full]

Solve[Subscript[\[Theta], 1]'[t] - A[t] Subscript[\[Theta], 2]'[t] == 
  0, Subscript[\[Theta], 2]'[t]] 

EDIT №2: $$ \cases{ J_1\dot\omega_1 +\dot\lambda+c(\theta_1-A\theta_2) = T_1\\ J_2\dot\omega_2-\dot\lambda A-\lambda\dot A-c(\theta_1-A\theta_2)= T_2\\ \dot\omega_1-A\dot\omega_2-\dot A\omega_2 = 0 } $$

EDIT №3:

Clear["Derivative"]

ClearAll["Global`*"]

Remove[c, J, A]

pars = {Subscript[J, 1] = 1, Subscript[J, 2] = 2, c = 1000, 
   A[t] = 2 + 0.1 Sin[0.1 t]};

s = NDSolve[{M'[t] + M[t] == -Subscript[\[Theta], 1]'[t] + 1, 
    Subscript[J, 1] D[D[Subscript[\[Theta], 1][t], t], t] + 
      D[\[Lambda][t], t] - 
      c (Subscript[\[Theta], 1][
          t] - (A[t] Subscript[\[Theta], 2][t] - 
           Subscript[\[CapitalTheta], 2][t])) == M[t], 
    Subscript[J, 2] D[D[Subscript[\[Theta], 2][t], t], t] - 
      D[\[Lambda][t], t] A[t] - \[Lambda][t] D[A[t], t] - 
      c (Subscript[\[Theta], 1][
          t] - (A[t] Subscript[\[Theta], 2][t] - 
           Subscript[\[CapitalTheta], 2][t])) == 0, 
    D[D[Subscript[\[Theta], 1][t], t], t] - 
      A[t] D[D[Subscript[\[Theta], 2][t], t], t] - 
      D[A[t], t] D[Subscript[\[Theta], 2][t], t] == 0, 
    Subscript[\[CapitalTheta], 2]'[t] == 
     D[A[t], t] Subscript[\[Theta], 2][t], M[0] == 0, 
    Subscript[\[Theta], 1][0] == 0, Subscript[\[Theta], 2][0] == 0, 
    Subscript[\[Theta], 1]'[0] == 0, 
    Subscript[\[Theta], 2]'[0] == 0, \[Lambda][0] == 0, 
    Subscript[\[CapitalTheta], 2][0] == 0}, {Subscript[\[Theta], 1], 
    Subscript[\[Theta], 2], \[Lambda], Subscript[\[CapitalTheta], 
    2]}, {t, 500}];

Plot[{Evaluate[Subscript[\[Theta], 1]'[t] /. s], 
   Evaluate[Subscript[\[Theta], 2]'[t] /. s]}, {t, 0, 500}, 
  PlotRange -> Full];

Plot[{Evaluate[
    Subscript[\[Theta], 1][t]/Subscript[\[Theta], 2][t] /. s], 
   A[t]}, {t, 0, 500}, PlotRange -> Full];
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  • $\begingroup$ Are you sure that the stiffness is zero? Because, wouldn't that essentially imply no shaft? Instead, maybe you meant infinite stiffness, such that the shaft acts as a rigid connection? $\endgroup$ Oct 9, 2021 at 11:11
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    $\begingroup$ Hint. Use Lagrange multipliers to include the restrictions. $\endgroup$
    – Cesareo
    Oct 9, 2021 at 11:40
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    $\begingroup$ As far as I remember, the finite stiffness introduces an extra torque, $T_{12} = c_1(\theta_1-\theta_2)$. Then you have $J_1\dot \omega_1 = T_1-T_{12}$ and $J_2\dot \omega_2 = T_2+T_{12}$. $\endgroup$
    – Arastas
    Oct 11, 2021 at 15:56
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    $\begingroup$ @dtn You are right, $\omega_1(t) = A(t)\omega_2(t)$ for all $t$ does not generally imply $\theta_1(t) = A(t)\theta_2(t)$ as $\theta_1$ and $\theta_2$ are solutions of an ode. But is does not change the dynamics, i.e., it is the same ode. $\endgroup$
    – Arastas
    Oct 14, 2021 at 10:09
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    $\begingroup$ @dtn Then you probably need to write the solution of your system. It is a linear time-varying system, and its solution is given in the terms of the fundamental matrix. It is not obvious if you can compute it without explicitly solving the ode for the given $A(t)$. $\endgroup$
    – Arastas
    Oct 14, 2021 at 10:23

2 Answers 2

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Having in consideration the document about non holonomic constraints from a reputable source, here (item 6 - constraint linearity on $\omega_1,\omega_2$) we can dare to develop a Lagrangian model as follows

$$ L(\omega,\theta,\lambda) = \frac 12\omega_1^TJ_1\omega_1+\frac 12\omega_2^TJ_2\omega_2-\frac 12(\theta_1-\theta_2)^T c(\theta_1-\theta_2)+\lambda(\omega_1-A\omega_2) $$

thus obtaining the movement equations

$$ \cases{ J_1\dot\omega_1 +\dot\lambda+c(\theta_1-\theta_2) = T_1\\ J_2\dot\omega_2-\dot\lambda A-\lambda\dot A-c(\theta_1-\theta_2)= T_2\\ \dot\omega_1-A\dot\omega_2-\dot A\omega_2 = 0\\ \theta_1 = \int_0^t A(\tau)\dot\theta_2(\tau) d\tau } $$

Here $\omega_i = \dot\theta_i$

NOTE

This ODE system is equivalent to

$$ \cases{ J_1\dot\omega_1+\dot\lambda+c(\Theta_1-\theta_2)=T_1\\ J_2\dot\omega_2-\dot\lambda A- \lambda\dot A-c(\Theta_1-\theta_2)=T_2\\ \dot\omega_1-A\dot\omega_2-\dot A\omega_2=0\\ \dot\Theta_1 = A\dot\theta_2\\ \dot\theta_2 = \omega_2 } $$

EDIT

Included a MATHEMATICA script to simulate a very simple system

Clear[A, J1, J2, c]
sols = Solve[{J1 dw1 + dlambda + c (Theta1 - theta2) == T1, 
              J2 dw2 - dlambda A - lambda dA - c (Theta1 -theta2) == T2, 
              dw1 - A dw2 - dA w2 == 0}, {dw1, dw2,dlambda}][[1]] // FullSimplify;
odes0 = sols /. Rule -> Equal;
odest = odes0 /. {w1 -> w1[t], w2 -> w2[t], lambda -> lambda[t], Theta1 -> Theta1[t], theta1 -> theta1[t], theta2 -> theta2[t], A -> A[t], dA -> A'[t], dlambda -> lambda'[t], dw1 -> w1'[t], dw2 -> w2'[t]};

tmax = 10;
A[t_] := 2 + 0.1 Sin[0.1 t]
J1 = 1;
J2 = 2;
c = 500;
T1 = 1;
T2 = 2;
odestot = Join[odest, {theta1'[t] == w1[t], theta2'[t] == w2[t], Theta1'[t] == A[t] w2[t]}];
cinits = {theta1[0] == 0, theta2[0] == 0, w1[0] == 0, w2[0] == 0, lambda[0] == 0, Theta1[0] == 0};
solode = NDSolve[Join[odestot, cinits], {w1, w2, theta1, theta2, Theta1, lambda}, {t, 0, tmax}];

Plot[Evaluate[{w1[t], w2[t]} /. solode], {t, 0, tmax}]
Plot[Evaluate[{theta1[t], theta2[t]} /. solode], {t, 0, tmax}]
Plot[Evaluate[{Theta1[t]} /. solode], {t, 0, tmax}]
Plot[Evaluate[{lambda[t]} /. solode], {t, 0, tmax}]
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  • 1
    $\begingroup$ 1. It is easy to include the gear ratio, 2. The derivative of $f(x) = 0$ is $f'(x) = 0$. 3. The ODE's reduction to differential form should be accompanied by coherence in the initial conditions to integrate conveniently. $\endgroup$
    – Cesareo
    Oct 18, 2021 at 15:20
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    $\begingroup$ Good idea. Included as a NOTE to the answer. $\endgroup$
    – Cesareo
    Oct 18, 2021 at 17:48
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    $\begingroup$ Note that $\dot\theta_1 = A\dot\theta_2$ then $\theta_1 = \int_0^t A(\tau)\dot\theta_2(\tau)d\tau$ $\endgroup$
    – Cesareo
    Oct 18, 2021 at 18:09
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    $\begingroup$ Yes. Of course. This formulation is a matricial one. $\endgroup$
    – Cesareo
    Oct 27, 2021 at 5:23
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    $\begingroup$ If you want to follow a motion law $\omega_z$ then we have a control problem. Now we will produce inputs $T_1,T_2$ such that $\omega_2$ follows $\omega_z$. This can be done within an optimal control formulation. $\endgroup$
    – Cesareo
    Nov 12, 2021 at 8:33
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Ok, following the discussion in the comments, here are some equations. Note that I have no idea what is the dynamics of your gearbox, so I assume something simple and approximative.

We have four (vector) states: $\theta_1$, $\omega_1$, $\theta_2$, and $\omega_2$. What we know for sure is that $\dot{\theta}_1(t) = \omega_1(t)$ and $\dot{\theta}_2(t) = \omega_2(t)$. Suppose that the torque $T_1$ acts on $\dot{\omega}_1$, e.g., it is a motor, and the torque $T_2$ acts on $\dot{\omega}_2$, e.g., it is the load torque or a friction.

Now we need to model the gearbox. Define the internal variable $\delta$; the dynamics of $\delta$ is assumed to be $\dot{\delta}(t) = \omega_1(t) - A(t)\omega_2(t)$, where $A(t)$ is the time-varying gear ratio matrix. Then the torque due to the stiffness is $T_{12}(t) = c \delta(t)$, where $c$ is the constant finite stiffness.

Finally, the equations are $$ \begin{aligned} \dot{\theta}_1(t) &= \omega_1(t),\\ \dot{\theta}_2(t) &= \omega_2(t),\\ J_1(t)\dot{\omega}_1 &= T_1(t) - c\delta(t), \\ J_2(t)\dot{\omega}_2 &= T_2(t) + c\delta(t), \\ \dot{\delta}(t) &= \omega_1(t) - A(t)\omega_2(t), \end{aligned} $$ where $J_1$ and $J_2$ are the time-varying inertias.

Note that for the constant gear ratio $A(t)\equiv A$ we can eliminate the dynamics of $\delta$ replacing $\delta(t) = \theta_1(t) - A\theta_2(t)$.

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  • $\begingroup$ Thank you for your answer! He cleared up some things I didn't understand in my head. Now do I know what to do? I need to compose a Lagrangian that will give me such equations of motion. $\endgroup$
    – dtn
    Oct 14, 2021 at 17:37
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    $\begingroup$ Note that $\omega_1(t)-A(t)\omega_2(t) = 0$ $\endgroup$
    – Cesareo
    Oct 18, 2021 at 10:46
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    $\begingroup$ @Cesareo Why should it be so for a finite stiffness? For example, if the stiffness is zero, you have $\omega_2=0$. $\endgroup$
    – Arastas
    Oct 19, 2021 at 18:20
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    $\begingroup$ @Arastas $\omega_1-A(t)\omega_2 = 0$ is a restriction. It always is verified. $\endgroup$
    – Cesareo
    Oct 19, 2021 at 18:42
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    $\begingroup$ @Cesareo If you think about a very flexible joint, at some moment of time even the rotation directions are opposite. $\endgroup$
    – Arastas
    Oct 19, 2021 at 18:44

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