Hi all,

I'm wondering what would be a (somewhat simple) way to reduce the RPM on a motor. the motor is a 1/2hp or 3/4hp, and I need to bring it down to around 500 RPM for a piece of machinery I have. If I could do this internally with the motor, great, but I'm thinking I'll need to make some sort of large flywheel and reduce it from there with a ratio orientated approach.

Thanks,

Aaron

2 things come to mind, check the plate on the motor , some AC motors have a high and low speed wiring setup.
the other is to get a VFD< (variable freguency Drive) and reduce the speed that way.


 

 

If all you want is a speed reduction, then the term you should search is 'jack-shaft' or even just 'pulley' rather than 'flywheel'. If you have a 1725 rpm motor (fairly typical in the US) then it is a matter of simple ratios, as you suggested: A 2 inch pulley on the motor and a 7 inch pulley on the driven end should do the trick. (Connected with a v-belt) If you have a higher speed motor, you may find it easier to add an intermediate shaft (Jack shaft) to get greater speed reduction. You should be able to buy a couple of pillow blocks (mounted bearings), a shaft, and the appropriate pulleys with relative ease and much less expensively than with a VFD.

 

Yes like fciron said- use a jack shaft, Here is a calculator that will help you determine the size of pulley that you need. It looks like if you take a 1.5 inch pulley on a 1725 rpm motor and run that to a 5 inch pulley you will get close to 500 rpm
http://gadi.agric.za/software/renting/pulley.php

You don't need a separate jack shaft if you can just do it with one set of pulleys, one on the motor and one on the machine. (KISS principle).

For more massive reductions---say 1725 to 60 you can find a gear reduction unit. I scrounge them off old machinery when I can as they are handy things to have around the shop...

Yes TP is right I was thinking about that after I posted, the ratio that I got was just a two pulley set up, no need for a separate shaft (duh)

Here at the factory where I work we use right angle (gear box) speed reducers. Grainger has pages of them with ratios of say 5:1 up to 60:1 in many cases. I don't know if that applies to what your doing with your specific situation but we sure do it a lot around here. They are in the $300-$500 range but rarely ever wear out. Just food for thought. Good luck with it. Spears.

Depends what your final application is, torque requirement etc, pulleys the simplest,reduction gearbox probably the best,I always thought a VFD would only work as a speed reducer on a 3 phase motor???

VFD's work on single or three phase. They will work with any induction motor where the rpm is dependent on line frequency. They can maintain constant torque but not horsepower. A gear or belt reduction can multiply the torque.

Could a simple rheostat fix the problem? Would it cause loss of performance, or a heat build up in the circuit?

Rheostats are fine for "universal" motors (aka AC-DC motors), but not good for shaded-pole induction motors. Induction motors have to have both the power AND the frequency changed - ergo we use the VFDs for induction motors. Rheostats can "work" for induction motors over a narrow range, but don't expect the motor to last as long as it should.

Also, by "rheostat" I'm assuming we mean "autotransformer" and not "potentiometer." Potentiometers (aka variable resistors) are seldom made big enough to handle anything but the smallest universal motors.

Thanks for giving me research for the next few days. I did not realize the difference. I was thinking of a dimmer switch, similar to ones that are installed in homes to control lighting and fans. I am still uncertain if it is actually a rheostat or an autotransformer. I have always heard them called rheostats.

I love learning and AC is one of my weak areas, so thanks for pointing this out. I did find a formula for figuring out the pulley size and rpms.


Drive Pulley diameter / driven pulley diameter * input RPM = Output RPM

 

I just got a 1 hp 3600 RPM ac motor that used to be part of a belt driven table saw. From these posts (if I understand correctly) something like a router speed control would not be my first choice to bring the RPMs down to th 1700 range. Is that correct?

 

thanks 

Ernest

Correct. Since the motor is already set up for belt drive, best solution to lowering the RPM's would be to change out the drive pulleys so that you end up at the final RPM you want.

DSW, 

 

I shall do just that. I saw one helve hammer video on YouTube (

)

 

where the guy makes the comment "P.S.S. Small pink bicycle wheels are a perfectly legitimate clutch."

Now this is chain driven but it is worth it for the comment. 

LOL. 

 

Ernest

 

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I feel I should give a plug to the place I got my motor. 

If any of you good folks need an AC motor for a power hammer or grinder and are near Huntsville you need to check out Midway Furniture & Appliance at , Madison, AL. Every time I go in there he has what I need. He has all sorts of appliance motors as well as bins full of sockets to replace ones that have been lost. He has everything from coins, Baseball cards, some power tools to tow bars for the car.  

"""Depends what your final application is, torque requirement etc, pulleys the simplest,reduction gearbox probably the best,I always thought a VFD would only work as a speed reducer on a 3 phase motor???"""

 

Agree with this, the torque requirement is key here.

hi, I have 3hp elect.  an air compresser   pump  with a  10.25in pully .  3hp turns 3400, pump needs to turn 700-900. what size pully on elect. motoer.

The motor pulley would need to be somewhere between 2" and 3". It's a ratio. Divide 3400 by the speed you want to get your ratio. Then divide 10.25 by the ratio for the small pulley. 800 would be midway and would take a 2.5" pulley on the motor.

Welcome aboard will 51. Have you read this yet? READ THIS FIRST  It will help you get the best out of the forum.

The Algebraic Expressions for Drive and Driven R.P.M. and Diameter.

1750 to 500 on a small motor, put a 2 inch pully on the motor and a 7 inch pulley on the tool, it will be really close.  This is a 3.5 to 1 ratio.  If you need to use different size pulleys, keep that ratio and you'll be okay.  Just remember physics really doesn't care about math.  So, if your motor pully gets too big, say 7 inch pully on motor to 25 inch pully on tool, the starter surge may not be enough to overcome the angular force it takes to start the motor.

2 X 1750 ÷ 500 = 7" Driven pulley.

On 3/5/2020 at 9:51 AM, Randy Griffin said:

The motor pulley would need to be somewhere between 2" and 3". It's a ratio. Divide 3400 by the speed you want to get your ratio. Then divide 10.25 by the ratio for the small pulley. 800 would be midway and would take a 2.5" pulley on the motor.

10.25 X 800  ÷ 3400 =2.41"   2.5" Off the shelf.

On 3/5/2020 at 8:21 AM, will 51 said:

hi, I have 3hp elect.  an air compresser   pump  with a  10.25in pully .  3hp turns 3400, pump needs to turn 700-900. what size pully on elect. motoer.

10.25 X 800  ÷ 3400 =2.41"   2.5" Off the shelf.