View Single Post
  #8   Spotlight this post!  
Unread 23-01-2017, 19:15
Ether's Avatar
Ether Ether is offline
systems engineer (retired)
no team
 
Join Date: Nov 2009
Rookie Year: 1969
Location: US
Posts: 8,128
Ether has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond reputeEther has a reputation beyond repute
Re: Velocity Control - Battery Compensation Term

Quote:
Originally Posted by Brian Selle View Post
This ^ with a 0.8-0.9 for gearbox efficiency will get you close. For this example, 65% command: 10,000 rpm * 0.65 * 0.8 = 5200 rpm free speed. 5200/2500 yields approx 2:1 gear ratio.
Consider the following thought experiment:

Take a CIM whose free speed is 5310 rpm at 12 volts and connect it to a good quality properly assembled and lubed 2:1 gearbox whose output shaft is connected to nothing (no load).

Using the same computation you did above, the CIM's speed would now be 5310*0.8 = 4248 rpm.

Now go to the motor curves for CIM and you'll see that CIM is drawing 29 amps.

Do you believe that? If not, where is the error?


Reply With Quote