| Author |
Message |
David Anderson (168.215.176.209)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Thursday, October 23, 2003 - 2:20 pm: | |
Not a bus question, but the best minds in the world lurk on this board.
I need to know the size in HP of a standard 110 volt 1725rpm motor to turn a couple of pulleys on a gearbox. I'm not sure how to describe what I want, but I need to know what variables are needed to determine motor size, obviously torque and weight, I guess.
I guess a reasonable question would be: what factors would you need to know to give a qualified answer? I could then post them on another thread.
Pretty vague, but I'm not sure how to ask what I need.
Any engineers out there, please give me some insight, and I'll post right back.
David Anderson |
Peter E (Sdibaja) (200.76.240.28)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Thursday, October 23, 2003 - 2:48 pm: | |
I have no clue about how to answer your question, I think we need to know what you are trying to turn... with what you have told us so far the sky is the limit.
Peter E
yes the best mind in the world does lurk here... I check in most every day!
engineers? this is about buses, not trains...
sorry, I had to be a smart ass, I just got home in Mexico and feel GOOD! |
ED-NJ (67.85.224.113)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Thursday, October 23, 2003 - 3:52 pm: | |
Hi David,
How well a motor converts electrical energy into work-performing mechanical energy. For example, a 70% efficient motor drawing 400 watts produces 280 watts of useful energy (400 x .70 = 280) and loses 120 watts (to heat). |
Erich (216.132.61.162)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Thursday, October 23, 2003 - 4:24 pm: | |
You need to know how much torque it takes to move your whatever and you need to know how fast you want it to move.
HP= Torque (ft-lbs) X RPM/5252
Erich |
ED-NJ (67.85.224.113)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Thursday, October 23, 2003 - 4:49 pm: | |
Try this-http://shop.emotorstore.com/estore/TD_MotorEquations.asp? |
Gary McFarland (Gearheadgary) (209.128.79.46)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Thursday, October 23, 2003 - 5:08 pm: | |
Ooh that's a keeper.
Gary |
James Cotes (68.184.122.103)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Thursday, October 23, 2003 - 11:09 pm: | |
Uhh, what he said!
(sorry, I couldn't help myself) |
David Anderson (168.215.176.217)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Friday, October 24, 2003 - 12:38 am: | |
Thanks, guys.
The website link answered my question perfectly.
I knew I could find an answer here. Busnuts always come through.
David Anderson |
jmaxwell (66.81.63.42)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Saturday, October 25, 2003 - 1:54 am: | |
thnks ed |
Henry R. Bergman, Jr. (Henryofcj) (63.224.197.10)
Rating: N/A
Votes: 0 (Vote!) | | Posted on Saturday, October 25, 2003 - 3:46 pm: | |
My best quess would be to take the required running power times about 125% then add another 25% as a fudge factor.
This would let the motor start under load and also compensate for the efficiency loss normal in an electric motor??? |