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shredder review

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riverstone

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larryccf
A bit late for a reply but you control motor speed by altering the frequency of the power. You guys are on 110v @ 60 hertz and we are on 230v @50 hertz. We have a moving target system at our pistol range and we set the speed initially by pulley size and fine tune it by using a controller on the frequency. We are running at 47 hertz and the speed is spot on. There is no notable difference in power/torque, only speed. Reducing speed by using pulleys or gears increases the torque of the motor. I had a question asked of me last week by a guy who imports reloading gear (Dillon) from the states and one of his customers was having trouble using a motorised case trimmer. The unit was 110v and the customer was running through a transformer from 230v - 110v. A transformer converts voltage but the cycles stay the same.So he was running on 110v 50 hertz. The guy had lost 17.5% of his motor speed. I don't how they got round it but a new small motor would be a lot cheaper than a controller or import the 230v unit from Dillon.
 

larryccf

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THANKS - i've ordered the pulleys & bearings - the new ratio should bring my rollers down to ~180 rpm. After i've built V.2, i'll see if it needs further tuning.
 

Jitterbugdude

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Just to clear a few things up... Reducing speed by using pulleys will NOT increase the torque at the motor... it will lower it.
Reducing the frequency to lower the rpm in this case will work but don't try that on anything like a refrigerator, computer or TV.
 

larryccf

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Just to clear a few things up... Reducing speed by using pulleys will NOT increase the torque at the motor... it will lower it.
......

Seriously? - i'm not an engineer, mechanical or electrical, so i'm not asking this argumentivally, but not sure how reducing speed mechanically would affect the torque output of the elec motor. I can understand some additional loss in efficiency of power being delivered to the roller head because of the additional resistance of the added bearings and the additional pulley belt. If you can explain that in simple "fred flinstone" speak, which is my literacy level, i'd appreciate
 

ChinaVoodoo

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I'm not sure what you mean either. On a bicycle, it's easier to go up hill when you have it on a slow gear. On fast gears you can power out and stop due to lack of torque on a steep hill.
 

Jitterbugdude

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I'll try.. it gets complicated quick without formulas.

Torque for a spinning motor is: Torque = Speed x Force (Force is constant). So if you have a 1 ft-lb motor running at 1725 rpm (common in the U.S.) and reduce the rpm you would, according to the definition of Torque (for a spinning motor) reduce the torque.

And for a belted system
Take a 1hp motor with 1 ft-lb of torque and spinning at 1725 rpm. Put a 1 inch pulley onto the motor and belt it up to another pulley that is 14 inches. The shaft of the 14 inch pulley will now spin at 246 rpm. So going back to Torque = Speed x Force we have now significantly reduced our speed at the shaft that the second pulley is attached to, thus lowering the Torque at the 2nd pulley.

Conversely, if you reversed the situation and put a smaller pulley after the motor the smaller pulley would be running faster, hence increasing the torque.

Hopes this helps. A little hard to describe without all the math.

For a direct drive system: You have a motor spinning at 1725 rpm and use a 10:1 gear reducer coupled directly to the motor and a tobacco shredder. The shredder will now spin at 1725 rpm and will see an increase of 10x the torque of the motor. A gear reducer works something like this: for a 10:1 reducer lets say the gear on your motor has 100 teeth. That means the gear on your shredder would have to have 10. So if we arbitrarily say the motor gear is 10 inches the shredder gear would have to be much smaller since it would only have 10 teeth. In other words the smaller gear now has to spin faster to keep up with the bigger gear. we've increased the speed of the smaller gear (Torque = Speed x Force) so, we've increased Torque at the shredder.
 
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ChinaVoodoo

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I usually take for granted that you're right, because you usually are. But I think you've got it backwards. Force is a constant, and by increasing diameter, you are decreasing angular speed, but also increasing the moment/length of the lever arm, and since torque is force x distance (@90 degrees), you have more torque.

Also:
Screenshot_20170301-093653.jpg
 

Jitterbugdude

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But I think you've got it backwards.

You are absolutely correct.. just seeing if anyone was paying attention..:rolleyes:

I think a better way to describe torque when used with a geared system is: The Torque at the destination Gear (call it the shredder torque) = the Motor Torque(MT) x Gear Ratio. So, we have: MT (1 ft-lb) x Gear Ratio (10:1) = 10 ft-lb torque at the shredder.

The Torque calculation for a non spinning application is different. T= Force x Distance applied at an angle of Sine Theta. So the longer the distance the more torque as long as it is applied 90 degrees since the sine of 90=1.

You guys are making me work my brain too much.. I'm going go out and smoke my meer with some home grown Yenidje
 

larryccf

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only posting this as it may help someone else or give them an idea - I thought i'd check the rpm on the shredder and not having a digital counter at home, used my "Fred Flintstone" approach. I've got some video editing software that let's me view video clips by the second, ie i can scroll thru a 1 second segment in great detail (mainly used for when i'm combining two segments after removing a section) - taped a pc of paper to one of the spokes, and made a short video of the machine in operation. holy moly, I counted that spoke coming around 6 times in 1 second, which translates to 360 rpm. Using the pulley rpm calculator, 360 at the large pulley = 1954 at the motor

not sure why the difference from what the rpm calculator predicted but the motor has to be turning faster than it's spec'd on the mfgr's plate - i suspect it may be the incoming voltage to my house is 128 volts, while the motor sez 110V & 1470 rpm - maybe riverstone can confirm if higher voltage would generate increased motor speed.
 

Jitterbugdude

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One more try...

You guys are confusing how torque is applied. With a motor the torque comes from the voltage exciting the stator/commutator. It doesn't come from how much force is applied at a certain distance. In a static situation such as a bolt, you are applying the torque as T= Force x Distance applied at the sine of an angle.
No amount of angular anything is going to significantly impact the torque of a motor. If you have a pulley system the "belt load" will exert a lateral force on the bearing (making it wear out prematurely) but it will not impact the torque of the motor.

Torque for a spinning shaft, as in a motor shaft or the shaft at the other end of a pulley system is defined as: Torque = Speed x Force. It has nothing to do with distance or applying force at an angle.
 

ChinaVoodoo

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Using imperial, Torque= (HPx5252) / RPM

Therefore, changing the RPM mechanically, while not changing the power output, changes the torque.
For example, 1HP motor@ 1000rpm = (1x5252) / 1000 = 5.252ftlb
If you used pulleys to reduce the speed to 500RPM, it would be (1x5252) / 500 = 10.504ftlb

True, if all you do is put a bigger wheel on the end of the shaft, you're just putting excess strain on the mechanics of the motor and accomplishing nothing. The RPM has not changed.

However, if you take that quarter inch shaft, put a 2" wheel on it, then use that to run a belt which wraps around a 4" wheel, you will decrease the resulting rpm by half, because it takes two full turns of the original shaft to turn the second one once. This slower moving second wheel has twice the torque of the second. And yes, this is a factor of diameter, i.e.it is a lever.
 

Jitterbugdude

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No...No...No..

Torque = Speed x Force. Let's Solve for F and substitute back in.

Assumptions: Torque= 20 ft-lb, Speed= 1725 rpm so... F= .01159 ft-lb/rpm. So.. the shredder rpm is 200, sub back in.. T= 200 x .01159..T= 2.3 ft-lbs.

Mathematically its not quite that simple but you should be able to tell from the mathematical formulas I posted earlier that for a pulley system the torque at the shredder is less than the torque at the motor.

A geared system, like I said before is a "whole nother" ball game

And everyone... stop confusing applied torque to a static situation with rotational torque driven my electromagnetic means.
 

larryccf

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Jitterbugdude - i can't argue with you on engineering terms, but your position conflicts with one example in my shop - i have a drill press with 6 speed options, speeds are changed when i move the belt on the reduction pulleys

if i move the belt to the smallest diameter pulley driven directly by the motor, it gives me the slowest speed at the chuck but that drill press can then handle a 1" diameter drill bit going into metal. If i move the belt to the largest diameter pulley on the motor side, it will give me the fastest bit speed and start that 1" bit, but halfway in it will jam - it can't torque past the load it needs to continue drilling. That would seem to be exactly opposite of what you're stating.
 

Jitterbugdude

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Larryccf, I'm no expert on machine design but I'll give it a go.

When you drill a piece of metal, chips are formed and work their way up the flute. When drilling you have primary chip removal, secondary and tertiary chip removal. All three are important in order to execute a proper drilling technique ( has to do with the shear strain of the chip). I'm guessing here but by greatly increasing the speed you circumvent the proper chip removal sequence. At some point (by drilling at too fast a speed) you get chips stuck along the fluted area causing a jam. Ever notice a lot of drill presses will have a chart recommending what speed to drill a particular material? The people that sit around coming up with these exotic formulas take the speed as well as material in consideration as it relates to proper chip removal.
 

deluxestogie

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τ = (I * V * E *60) / (rpm * 2π)

T is torque
I is the current (amperage)
V is the voltage
E is the efficiency of power conversion (typically less than 50%) of that particular motor
60 (seconds per minute)
rpm is revolutions per minute
2pi radians = 360º

Aside from the brain ache, notice that torque (T) is equal to something divided by rpm. So for any given power input, T increases when the rpm decreases, and T decreases when the rpm increases.

Another way to consider the problem is that the force applied per degree of rotation is dependent on the radius of the motor pulley. A bigger motor pulley spreads that same applied power over more degrees of rotation. Various primary and secondary pulley ratios are effectively just changing the radius of the drive pulley. If the final rpm of the driven pulley is greater than the rpm of the motor, then the effective torque on the driven shaft is lower than that of the motor alone.

Automobile transmission analogy: the lower the gear, the greater the torque on the wheels, but the slower the speed of the wheels. You trade torque for speed.

Bob
 
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