Explain inertia matching for selection of actuator.

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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

$\quad J_{M}=\frac{J_{L}}{N^{2}}$

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