Inertia matching

• Motor connected to load via coupling medium as shown below

• Let JM =Moment of inertia of motor

• Let J, Moment of inertia of load

• Then if inertia of these two component will get matched then energy dissipation in the motor will get decrease.

i.e. Efficiency of motor will get increases

• Coupling medium may be

a) Gear transmission

b) Belt and pulley transmission

c) Lead screw drive

• To discuss inertia matching cons connected to the load via gęar mechanism as shown in fig Np = No. of teeth present on motor pinion a motor

  N = No. of teeth present on gear

• The parameters related with the load are OL, T, JL.

• Load is connected to motor via coupling medium.

• This parameter will get reflected at the shaft of motor.

• These reflected parameter are indicated by 0L, T'L, J'L.

• The reflected parameters on the motor related with parameter of load depending upon gear ratio (N) as follow.

$$ \begin{aligned} &J_{L}^{\prime}=J_{\phi} / N^{2} \\ &T_{L}^{\prime}=T_{L} / N \\ &\theta_{L}^{\prime}=N \theta_{L} \end{aligned} $$

• We know that, energy dissipated in motor during time tc is given by

$$ W_{c}=\frac{R}{K_{\tau}^{2}}\left[\lambda \frac{J_{\tau}^{2} \theta^{2}}{t_{c}^{3}}+T_{L}^{2} t_{c}\right] $$

$$ W_{c}=\frac{R}{K_{T}^{2}}\left[\frac{\lambda \theta^{2}}{t_{c}^{3}}\right]\left[J_{T}^{2}+\frac{t_{c}^{\wedge} T_{L}^{2}}{\lambda \theta^{2}}\right] $$ - Similarly, energy dissipated in motor with Gear Transmission with reflected parameters during time tc is given by.

$$ W_{c}=\frac{R}{K_{T}^{2}}\left[\frac{\lambda \theta^{2}}{t_{c}^{3}}\right]\left[J_{T}^{2}+\frac{t_{c}^{\wedge} T_{L}^{2}}{\lambda \theta^{2}}\right] $$

• Where Jr is Motor total moment of inertia
• JJ = JM +J'i

$$ W_{c}=\frac{R}{K_{T}^{2}}\left[\frac{\lambda \theta_{L}^{12}}{t_{c}^{3}}\right]\left[\left(J_{M}+J_{L}^{\prime}\right)^{2}+\frac{t_{c}^{4} T_{L}^{12}}{\lambda \theta_{L}^{12}}\right] $$

$$ \begin{gathered} W_{c}=\frac{R}{K_{\tau}^{2}}\left[\frac{\lambda N^{2} \theta_{L}^{2}}{t_{c}^{3}}\right]\left[\left(J_{\mu}+\frac{\mathcal{P}_{L}}{N^{2}}\right)^{2}+\frac{t_{c}^{4} \frac{T_{L}^{2}}{N^{2}}}{\lambda N^{2} \theta_{L}^{2}}\right] \\ W_{c}=\frac{R}{K_{\tau}^{2}}\left[\frac{\lambda N^{2} \theta_{L}^{2}}{t_{c}^{3}}\right]\left[J_{L}^{2}\left(\frac{J_{\mu}}{J_{L}}+\frac{1}{N^{2}}\right)^{2}+\frac{t_{c}^{4} T_{L}^{2}}{\lambda N^{4} \theta_{L}^{2}}\right] \end{gathered} $$

$$ \begin{gathered} W_{c}=\frac{R}{K_{r}^{2}}\left[\frac{\lambda N^{2} \theta_{L}^{2}}{t_{c}^{3}}\right]\left[J_{L}^{2}\left(\frac{J_{M}}{J_{L}}+\frac{1}{N^{2}}\right)^{2}+\frac{t_{c} T_{L}^{2}}{\lambda N^{4} \theta_{L}^{2}}\right] \\ W_{c}=\frac{R}{K_{t}^{2}}\left[\frac{\lambda J_{L}^{2} \theta_{L}^{2}}{t_{c}^{3}}\right]\left[N^{2}\left(\frac{\left.J_{M}+\frac{1}{J_{L}}\right)^{2}}{N^{2}}+\frac{t_{c}^{4} T_{L}^{2}}{\lambda N^{2} \theta_{L}^{2} J_{L}^{2}}\right]\right. \end{gathered} $$ Let $\gamma=\left[\frac{t_{c}^{\iota} T_{L}^{2}}{\lambda \theta_{t}^{2} J_{L}^{2}}\right]$

$$ W_{c}=\frac{R}{K_{\tau}^{2}}\left[\frac{\lambda J_{L}^{2} \theta_{L}^{2}}{t_{c}^{3}}\right]\left[N^{2}\left(\frac{J_{M}}{J_{L}}+\frac{1}{N^{2}}\right)^{2}+\frac{\gamma}{N^{2}}\right] $$ As energy dissipation in motor depends upon gear ratio N. Differentiating above bracketed term w.r.t to N? & equate to it zero so as to get optimum gear ratio N, that minimizes power dissipation Wc in motor

$$ \frac{d}{d N^{2}}\left[N^{2}\left(\frac{J_{M}}{J_{L}}+\frac{1}{N^{2}}\right)^{2}+\frac{\gamma}{N^{2}}\right]=0 $$

$$ \frac{d}{d N^{2}}\left[N^{2}\left(\frac{J_{M}}{J_{L}}\right)^{2}+N^{2}\left(\frac{1}{N^{4}}\right)+N^{2}\left(2 \times \frac{J_{M}}{J_{L}} \times \frac{1}{N^{2}}\right)+\frac{\gamma}{N^{2}}\right]=0 $$

$$ \begin{gathered} \frac{d}{d N^{2}}\left[N^{2}\left(\frac{J_{M}}{J_{L}}\right)^{2}+N^{2}\left(\frac{1}{N^{4}}\right)+N^{2}\left(2 \times \frac{J_{M}}{J_{L}} \times \frac{1}{N^{2}}\right)+\frac{\gamma}{N^{2}}\right]=0 \\ \frac{d}{d N^{2}}\left[N^{2}\left(\frac{J_{M}}{J_{L}}\right)^{2}+\left(\frac{1}{N^{2}}\right)+\left(2 \times \frac{J_{M}}{J_{L}}\right)+\frac{\gamma}{N^{2}}\right]=0 \\ \frac{d}{d N^{2}}\left[N^{2}\left(\frac{J_{M}}{J_{L}}\right)^{2}+\left(\frac{1+\gamma}{N^{2}}\right)+\left(2 \times \frac{J_{M}}{J_{L}}\right)\right]=0 \\ \left(\frac{J_{M}}{J_{L}}\right)^{2}-\left(\frac{1+\gamma}{N_{0}^{4}}\right)+0=0 \end{gathered} $$

$$ \begin{array}{r} \left(\frac{J_{M}}{J_{L}}\right)^{2}-\left(\frac{1+\gamma}{N_{0}^{4}}\right)+0=0 \\ \left(\frac{J_{M}}{J_{L}}\right)^{2}=\left(\frac{1+\gamma}{N_{0}^{4}}\right) \\ \frac{J_{M}}{J_{L}}=\frac{\sqrt{1+\gamma}}{N_{0}^{2}} \end{array} $$

$$ \left(\frac{J_{M}}{J_{L}}\right)^{2}-\left(\frac{1+\gamma}{N_{\circ}^{4}}\right)+0=0 $$ $$ \left(\frac{J_{M}}{J_{L}}\right)^{2}=\left(\frac{1+\gamma}{N_{0}^{4}}\right) $$ If $T_{L}=0$ then $\gamma=0$ $$ \begin{aligned} \frac{J_{M}}{J_{L}}=\frac{1+\gamma}{N_{0}^{2}} \quad \Longrightarrow & \frac{J_{M}}{J_{L}}=\frac{1}{N_{0}^{2}} \\ N_{0} &=\sqrt{\frac{J_{L}}{J_{M}}} \end{aligned} $$ $J_{L}=J_{M} N^{2} \quad$ OR $\quad J_{M}=\frac{J_{L}}{N^{2}}$