looking for pulley/RPM formulation

[huntersridge]

I plan on build a 4' x 6' fur/hide tumbler for breaking leather and am wanting a rough idea of the final RPM of my tumbler. I plan on using a 2hp elctric motor with a speed of 1800 rpm. The plans I have call for a 3 inch pulley on the motor going to a 8 inch pulley on a 16 inch mandrel with a 3 inch pulley on the other end (of the mandrel or "jackshaft"). Running from the 3" pulley I will have a 173" v-belt (A) going around a 4' (yes 4 foot) wooden pulley on the end of the 6 foot diameter drum. I am wanting between 16 and 18 RPM for my final drive. How can I calculate this before I build it for a rough idea?


Thank you,
Jim

 

[hydtools]

By ratio of pulley diameters.

1800rpm * (3/8) * (3/48) = 50.6 rpm

Ted

 

[dimjim]

Isn't the rpm based on the motor shaft size
rather than the pulley size?

 

[hydtools]

No.

The 3" pulley on the motor drives the 8" pulley on the jackshaft. 3" pulley on the jackshaft output drives the 48" wheel on the drum. The pulley on the motor turns at the motor speed regardless of the motor shaft size. Unless the motor is a synchronous motor, the motor shaft speed is probably 1750 rpm and not 1800 rpm.

Ted

 

[huntersridge]

The 1800 RPM is what the motor RPM is on the tag. I am trying to get this final drive down do 16-18 RPM. Thank you for the replies.

Jim

 

[electricpete]

Sounds like hydtools nailed it.

The basic principle is the linear speed of the belt is approx the same at each pulley -> rpm1*D1 = rpm2*D2. rpm2=rpm1*D1/D2.

Start with input speed. Then for each belt, multiply by ratio of drivinG pulley diameter to driveN pulley diameter.

Two belts. One has driving pulley 3 and driven pulley 8. The other has driving pulley 3 and driven pulley 48.

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[huntersridge]

The plans that I purchased for this project call for using a Dayton 1-2 hp motor with a 3inch pulley on the motor going to a 8 inch pulley on a jackshaft with a 3inch pulley on the opposite end of the jackshaft. This then ties a 173inch v-belt to a 48inch wood pulley on the end of the foot drum. The author says that this set-up gives around 16-18 RPM as the final RPM on the drum. He also states that if it does not come out to this RPM that you may have to make some pulley changes. Again, I am trying to get this right in the "ballpark" before I start building.


Jim

 

[hydtools]

Sounds like you need to make a pulley change.

One change would be change the 8" to 18".

Then 1800 * (3/18) * (3/48) = 18.75rpm.

Or to 20", then 1800 * (3/20) * (3/48) = 16.88

Ted

 

[huntersridge]

Ted,

Thank you!

The 20" pulley is going to be the tough part. My jackshaft would have to be raised by just over 10" in order for a 20" pulley to spin on the shaft. By keeping the 3" pulley on the motor and using a 2" on the opposite end of the jackshaft, what would I be looking at as far as the larger pulley?

 

[hydtools]

If you just change the jackshaft 3" pulley to 2", you final speed will be 28.1 rpm. To get 18 rpm the large wheel will then have to be at least 75".

One way or another you are looking for a 100 to 1 change is speed. 1800rpm input/18rpm output = 100

Ted

 

[hydtools]

Can you add another intermediate shaft?
Add one like the one you have with a 3" and 8" pulley would give you an ouput speed of 18.1rpm

If you add one with a 3" and 7" pulley you would get an output speed of 15.8rpm.

Ted

 

[sreid]

Replacing 3" pulleys with 2" pulleys would get you there (if 2" pulleys can be found that fit the motor and jack shafts).