- #1

SevenToFive

- 56

- 3

The other is a conveyor to travel 28 feet in 20 seconds with 2" diameter rollers. IF I did this correctly I divided 28ft by 20 seconds to get 1.4ft/sec, which would give me 160rpm.

Thanks for checking my work.

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- Thread starter SevenToFive
- Start date

- #1

SevenToFive

- 56

- 3

The other is a conveyor to travel 28 feet in 20 seconds with 2" diameter rollers. IF I did this correctly I divided 28ft by 20 seconds to get 1.4ft/sec, which would give me 160rpm.

Thanks for checking my work.

- #2

Baluncore

Science Advisor

- 12,339

- 6,410

A 2” diam roller has 6.283” circumference. So roller will be at 12.733 RPM.

How you get a 12:1 gearbox ratio I do not know. It depends on the motor RPM.

A 1500 RPM motor will need a reduction of 1500 / 12.733 = 117.8 : 1 ratio.

28 feet in 20 sec = 84 ft per min = 1008” per min.

2" diameter roller has 6.283” circumference. So roller will be at 160.5 RPM.

A 1500 RPM motor will need a reduction of 1500 / 160.5 = 9.35 : 1 ratio.

- #3

Asymptotic

- 782

- 528

How is power transferred to the 2" rollers?

The other is a conveyor to travel 28 feet in 20 seconds with 2" diameter rollers. IF I did this correctly I divided 28ft by 20 seconds to get 1.4ft/sec, which would give me 160rpm.

Thanks for checking my work.

For instance, are they connected to a main drive shaft equipped with pulleys via heavy "rubber bands"?

If so, is there a chain or belted reduction between line shaft and gearbox?

- #4

SevenToFive

- 56

- 3

I was using 1750 rpm for a motor speed. Thanks again for the help.

A 2” diam roller has 6.283” circumference. So roller will be at 12.733 RPM.

How you get a 12:1 gearbox ratio I do not know. It depends on the motor RPM.

A 1500 RPM motor will need a reduction of 1500 / 12.733 = 117.8 : 1 ratio.

28 feet in 20 sec = 84 ft per min = 1008” per min.

2" diameter roller has 6.283” circumference. So roller will be at 160.5 RPM.

A 1500 RPM motor will need a reduction of 1500 / 160.5 = 9.35 : 1 ratio.

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