[thegolfer]

Hi All,

I am trying to work out how servo power ratings relate to their ability to actually move things around.

Theoretically, if I have the following parameters (excluding friction, losses and everything else):

Mass (m): 30 kg
Final velocity (v): 1 m/sec (60m/min)
Acceleration time (t): 0.5 sec

Then:

Acceleration (a) = v/t = 1/0.5 = 2 m/s²

Force (F) = ma = 30*2 = 60 N

Acceleration displacement (s) = 1/2*a*t² = 1/2*2*0.5² = 0.25 m

Therefore:

Work done (w) = Fs = 60*0.25 = 15 Joules

Power (p) = w/t = 15/0.5 = 30 Watts

In this instance, keeping in mind this system lives in a make believe no losses world and rejects a factor of safety, could a 30 W servo motor theoretically be used to accelerate this load under the above conditions? Further, if the motor was receiving 30VDC, theoretically would it only require 1 amp?

[JACKY2014]

Hi thegolfer,

The mechanical poweer transfrred to the load is ;
dwork= F*dx
Power = dwork/dt = F*dx/dt=F*v=m*a*v

dx is infinitesimal displacement
dwork is infenitisimal work change
V is instantaneous velocity

The power depends on the velocity
At final velocity the power is 30Kg*2m/sec^2 *1m/sec = 60 watt

i hope that answer your question

Hi All,

I am trying to work out how servo power ratings relate to their ability to actually move things around.

Theoretically, if I have the following parameters (excluding friction, losses and everything else):

Mass (m): 30 kg
Final velocity (v): 1 m/sec (60m/min)
Acceleration time (t): 0.5 sec

Then:

Acceleration (a) = v/t = 1/0.5 = 2 m/s²

Force (F) = ma = 30*2 = 60 N

Acceleration displacement (s) = 1/2*a*t² = 1/2*2*0.5² = 0.25 m

Therefore:

Work done (w) = Fs = 60*0.25 = 15 Joules

Power (p) = w/t = 15/0.5 = 30 Watts

In this instance, keeping in mind this system lives in a make believe no losses world and rejects a factor of safety, could a 30 W servo motor theoretically be used to accelerate this load under the above conditions? Further, if the motor was receiving 30VDC, theoretically would it only require 1 amp?

[thegolfer]

Hi Jacky,

Thank you, that helps me a great deal.

The thing that I found hard to understand with the definition p = F*v is that at the instant that v = final velocity, theoretically a = 0, F = 0 and therefore p = 0. I realise that this is just a mental hurdle, however, as the definition you pointed out (p = m*a*v) makes things easier to understand. Under constant acceleration, power increases from 0 to 60 W as velocity increases from 0 to 1 m/s, meaning that the peak power required during this acceleration phase is 60 W. Theoretically, once final velocity is reached and there are no losses, no force is required for the mass to continue travelling.

I'm going to go away and put together a more realistic model for the power requirements of a CNC router axis. I'll post it once I'm done for anyone that's interested.