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Two-mass rotational system has the following form and is represented in following structural diagram.

enter image description here

where $\tau_e$, $\omega_1$ and $J_m$ - motor torque, angular velocity and moment of inertia

$\tau_s$, $\tau_s$, $\omega_2$ and $J_d$ - shaft torque, load torque, angular velocity and load moment of inertia;

$K_{md}$ - shaft stiffness

Problem: how to include a gear ratio $N=\frac{\omega_1}{\omega_2}$ in equation of motion and in in a block diagram respectively?

enter image description here

I wrote down the Lagrangian for each mass in terms of the gear ratio, but in the usual case, I just compiled a system of differential equations according to the Lagrange equation, but I cannot understand how now using the TWO Lagrangian to get one block diagram with a gear ratio.

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    $\begingroup$ What have you considered so far? This is not a free homework or exam question solution site - people like to see original effort on here. $\endgroup$
    – Solar Mike
    Jun 27 at 14:59
  • $\begingroup$ @SolarMike These are my own activities. Not an exam or homework. I am mastering simulink. I just need a hint. $\endgroup$
    – dtn
    Jun 27 at 15:01
  • $\begingroup$ First comes the motor, then the gearbox, then the shaft with the stiffness factor, then the load. $\endgroup$
    – dtn
    Jun 27 at 15:06
  • $\begingroup$ @AJN Let's consider two cases. ibb.co/ckJhm0g Let's take $N = 2$. 1.angular velocity $\omega_2$ is equal to the $w_{1_{ref}}$, angular velocity of the shaft $\omega_1$ increase by N times. 2. angular velocity $\omega_2$ is in N times less than $w_{1_{ref}}$. $\endgroup$
    – dtn
    Jun 27 at 15:26
  • $\begingroup$ @AJN Is the second variant more correct than the first? $\endgroup$
    – dtn
    Jun 27 at 15:26
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Assuming that the gear box is on the left end of the shaft (i.e. no flexible shaft between motor and gearbox).

  1. The angular velocity on the left end of the gear box is $\omega_1$.
  2. The angular velocity of the shaft side of the gear box is assumed as $\omega_1' = \frac{\omega_1}{N}$.
  3. The angular velocity on the right end of the shaft is $\omega_2$. So the torque on the shaft is $\pm K_m (\frac{\phi_1}{N} - \phi_2)$. (sign to be checked).
  4. Because of the way I described the gearbox, $\omega_1' < \omega_1 $. so the torque on the shaft when acting on the motor through the gearbox is $\frac{1}{N}$. This can be seen in the below derivation.
  5. Since I have assumed that shaft is directly connected to the load, the torque in the shaft is made available 1:1. This can also be seen in below derivation.

(Below derivation to be verified independently by OP) $$ L = \frac{J_m \omega_1^2}{2} + \frac{J_d \omega_2^2}{2} + \frac{Km (\frac{1}{N} \phi_1 - \phi_2)^2}{2} $$

$$ \frac{d}{dt} \frac{\partial L}{\partial \omega_1} = \frac{d}{dt} J_m \omega_1 = J_m \frac{d \omega_1}{dt} $$

$$ \frac{\partial L}{\partial \phi_1} = \frac{K_m}{\color{red}{N}} (\frac{1}{N} \phi_1 - \phi_2) $$

Similarly for the other body also (exercise left to you).

$$ \frac{\partial L}{\partial \phi_2} = -K_m (\frac{1}{N} \phi_1 - \phi_2) $$

I have not considered the input torque. It can be added to this result.

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  • $\begingroup$ Hello, I have a serious problem. I invite you to take a look at the topic. mathematica.stackexchange.com/questions/250791/… $\endgroup$
    – dtn
    Jul 10 at 7:44
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    $\begingroup$ I am not a member of that community and don't understand how the problem is related to that software; But regarding : "... it is impossible to establish a direct analytical connection between the angles ... θ1,θ2,θ3 with ... ϕ,Θ,σ and ... wi, and therefore, it is impossible to differentiate the ...". As an analogy, AFAIK it is not possible to find analytic relation between the coefficients of a polynomial and its roots, but the rate of change of the roots with respect to the coefficients can be calculated (used in root locus technique). @dtn $\endgroup$
    – AJN
    Jul 10 at 13:28
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    $\begingroup$ 1, 2. implicit function theorem, 3. $\endgroup$
    – AJN
    Jul 10 at 13:33
  • $\begingroup$ Wow, with such a tool I am not familiar. I will try to learn it. What a pity that I do not have much time as I would like. $\endgroup$
    – dtn
    Jul 10 at 13:35
  • $\begingroup$ What I meant to say is that, a closed form(?) expression is not required to differentiate between expressions. (BTW, I will be deleting my comments here in a while since they are to related to this question). $\endgroup$
    – AJN
    Jul 10 at 13:37

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