Equivalent Inertia - Geared System

[CWAnthony]

Hi guys,

I've just been working through some info related to the titled 'equivalent inertia of a geared system' found in the following links (thanks to David Grieve for posting them, whoever you are)

It's probably all simple stuff to you guys, I understand most of it except I am confused by the inclusion of efficiency at the end.

In the explanation page here -

http://www.tech.plym.ac.uk/sme/mech226/gearsys/gearaccel.htm,

the efficiency constant (eta, weird n) is used to divide the second term, which makes sense. If the efficiency of a system is less than 1 (less than 100%), the equivalent inertia term is more, representing a greater loss of torque in the system, less output torque compared to the input torque. Makes sense to me.

But in the worked example of a gear hoist (found here -

http://www.tech.plym.ac.uk/sme/mech226/gearsys/hoistcalc.jpg),

eta is used to multiply this term, giving less equivalent inertia loss if the system is less than 100% efficient. This doesn't seem right?

Inertia at 100% efficiency is 4.47kgm^2, but at 80% efficiency is 3.876kgm^2. Surely it should be the other way round? Which is correct

Forgive if this seems trivial, thanks for your time

 

[Engdoitbetter]

The example is worked in the opposite way compared with the theoretical part.

In the theory, the inertia and moments are reduced to the driving shaft, so the inertia of the user shaft is multiplied by n^2 and divided by efficiency.
In the example, all the quantities are reduced to the user shaft, so the inertia of the driving and intermediate shafts are divided by n^2 (which I remind you is less than 1 in this case) and multiplied by efficiency.

This is clear if you think of the physical meaning of this operations. n^2 takes into account the different rotational velocity of the shafts, while eta is a way to describe the loss of energy/power in the transmission.

Hope it helps.

Stefano