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Question: How many in-lbs or ft-lbs of torque are required to crush barley using a mill such as the Barley Crusher?

Motivation: I'm interested in motorizing my Barley Crusher. There are endless blogs entries and such on how to do this but the parts used often vary based on industrial salvage or the cheap motor of the week. Selecting equipment for this type of project requires a minimum torque specification I don't have, hence the question.

No guessing please; I'm looking for a definitive answer for the community.

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+1 I'm interested in the answer as well. The last time I researched this, I came up with wildly different answers. –  JoeFish Dec 11 '12 at 19:24
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It may be hard to get a definitive answer - the torque varies according to the type of grain, and the roller gap. –  mdma Dec 11 '12 at 20:51
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I was told there would be no math on the homebrewing exam :( –  Graham Dec 27 '12 at 17:24
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3 Answers 3

up vote 3 down vote accepted

I've built 2 motorized grain mills. The first used a 600w (input power) 3/8" drill. It worked well enough with a loose crush (0.040") but I ended up frying the drill when trying to crush at 0.035", so this is a good candidate to gauge the minimum torque.

A 600W drill produces about 380W mechanical output. I was using this on just under half power - the drill used PWM to reduce the motor speed. The rotational speed, was 240rpm, or 4 rps. This gives a torque of

 T = 190 / 2 x 3.141 x 4
   = 190 / 25.128
   = 7.5 Nm

That's about 66 lb.in.

However, unless you can get a motor with a rpm that matches the sweet spot for milling grain, 120-260 rpm, then you'll have to use gears, sheaves, PWM, variac or some other speed controller to regulate the rotational speed of the mill. All of these have an affect on the torque - most motors spin too quickly and so you'll need to slow them down, which increases the torque if using mechanical speed control. This makes the torque of the motor less of a pressing issue. This leads to the typical advice found in the forums that a 1/2 hp motor is more than sufficient when used with speed reducing transmission.

My second mill used a 340W washing machine motor, with a 1:15 sheave. This offered more than enough torque - the limiting factor was the belt slipping, which happens on the rare occaision I get a rock in there, or when I inadvertently drop the feeler gauge in the grain chute!

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Mdma, I'm a little confused. If torque is force x displacement vector, don't we need to consider the radius of the object, not just the number of rotations per second? –  Keith Hoffman Dec 18 '12 at 5:07
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I used power = torque x 2pi x rotational speed - the displacement vector is radius r which cancels out. see en.wikipedia.org/wiki/Torque#Derivation for the derivation. –  mdma Dec 20 '12 at 20:22
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If you don't want speculative answers based on what might be other users real experiences, I suggest contacting the Barley Crusher people directly.

So few homebrewers are being that specific about the torque in the way they use the mill that I think you'd have to cruise many forums to find the 1% of the brewing populace that has already done the homework and found the precise amount of torque required.

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Brewchez, I haven't found the Barley Crusher people to be very responsive to previous questions. And as mdma says, I may not have posed a question with an especially definitive answer. –  Keith Hoffman Dec 13 '12 at 3:16
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I've accepted mdma's answer but also want to throw out the number my local homebrewing community suggested: 50 in-lbs. This is very in-line with mdma's better documented answer. Happy automations!

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