Need Help Calculating what Motor and Gearbox I need (building myself a Trommel)

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To build a trommel for sieving soil, a motor with a power rating of about 0.5 HP is recommended based on the weight of the soil and the required drum rotation speed of 20 to 25 RPM. A reduction ratio of approximately 75 is necessary, which can be achieved through a combination of pulleys and possibly a gearbox. Calculations indicate that lifting 20 kg of soil 0.4 meters at 25 RPM requires around 300 watts of power, accounting for additional friction. A 1 HP motor would be suitable and provide a safety margin for performance.
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Hello

I recently got into veg gardening. I have a lot of good soil but it needs to be sieved to remove rocks and other crap from it. I decided to build myself a Trommel based on this article http://www.popularmechanics.com/home/how-to-plans/...
But I decided to make a more robust version of it, everything else stayed the same.

I need to calculate what One Phase Motor I need to purchase and if any gearbox (peed reducer) I need. Below is the measurements of the trommel that I think are important for this calculation.

The Drum Weight is 52KG (up to 70KG with Soil inside, that’s about 2 x 12 Liter buckets full)
The length of the drum is 50” (127cm)
The outside diameter is 30” (76cm)
Inside diameter is 28 ½” (72.5cm) the difference of the inside and outside diameter is mainly due to the channels that the casters and belt run on.
The channel for the belt is 20mm wide.
I would need the drum to spin at about 20 to 25rpm.

Going by this information can someone tell me how to calculate what type of motor I would need, what HP and what would be the best way to reduce the rpm?

Thanks,
P
 
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Paul Glavey said:
I would need the drum to spin at about 20 to 25rpm.
On a 50 Hz supply, a low cost induction motor runs at just below 1500 RPM. 1500/20 = 75 reduction ratio.
At 60 Hz it will be just below 1800 RPM. 1800/25 needs a 72 ratio.

Drum diameter is 30” so motor pulley would need to be about 30”/73 = 0.4” which is impossibly small. So you will need another stage of reduction. If the motor pulley used was 3” diameter the extra reduction stage would need to be about 3”/0.4” = 7.3 which could be done with a gearbox or an intermediate shaft and an extra belt.

The compost is being continuously lifted up as the drum rotates, then falls back. The mass of compost rolling in the trommel and the hight it falls from will determine the HP required.
 
Hello Baluncore

I was thinking if I used a 1500rpm motor and with a 2" pulley (A) connected to it and a belt from that going to a 8" pulley (B) on an axial with another 2" pulley (C) and the second 2" pulley connected via a belt to the drum (D) which is 30", then I would end up with 25rpm. But what HP motor should I start with, if the weight of the soil in the drum is 18kg and the height it falls is 28.5"?

Thanks for the help,
P
 
Last edited:
Paul Glavey said:
But what HP motor should I start with, if the weight of the soil in the drum is 18kg and the height it falls is 28.5"?
It does not fall from the top. It falls from half height or less as the drum rotates. That is 20 kg of soil being lifted 0.4 metre, 25 times per minute.
PE = m·g·h = 20 * 9.8 * 0.4 = 78.4 joule per tumble. There will be about 6 tumbles per rev.
78.4 joule * 6 tumbles * 25 RPM / 60 sec = 200 watt. Belt friction will be something like 100 watt. Total power = 300 watt.
300 watt = 0.4 HP
So I expect a 1/2 HP motor will be just sufficient.
 
Baluncore said:
It does not fall from the top. It falls from half height or less as the drum rotates. That is 20 kg of soil being lifted 0.4 metre, 25 times per minute.
PE = m·g·h = 20 * 9.8 * 0.4 = 78.4 joule per tumble. There will be about 6 tumbles per rev.
78.4 joule * 6 tumbles * 25 RPM / 60 sec = 200 watt. Belt friction will be something like 100 watt. Total power = 300 watt.
300 watt = 0.4 HP
So I expect a 1/2 HP motor will be just sufficient.

Wasn't able to figure out what your math was there for a few seconds but I understand it now. I was trying to calculate the needed HP differently using Power = (Force x Distance) /Time which was giving me 112.32 HP. Then I divided that by 60 because 1500/25 is 60. That was telling me I needed a 1.8 HP motor to start with.

Thanks for putting me straight. One final query, there would be no harm using a 1.5 HP Motor like this one http://www.clarkedistributors.ie/index.php?route=product/product&path=95_98&product_id=710 or should I stick to a 1/2HP one?

P
 
That 1 HP motor should do the job.
 
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