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View Full Version : Torque/Bandsaw question again

05-03-2007, 08:57 AM
Bout got the bandsaw restoration complete, working on power options now (still want to convert to metal speed). It's a given I need about 20:1 or greater speed reduction. Forrest says the motor torque drops as speed drops so I need a huge motor or do something different. Darrin said a right-angle drive would work and someone on the PM site used a gear motor. I found a 1/4hp, 60rpm output gearmotor with 150 in/lbs torque that I can scrounge from work. If I use this motor with a 4" pulley driving the 8" axle pulley with a 14" tire I should get around 110 sfm, pretty much where I need to be.

My question is: will this motor have the required drive torque or will it stall in say 1/4" steel plate? 150 in/lbs is what most servo mill table feeds have which seems pretty beefy. I'm confused on the hp/torque/speed issue. If a regular induction motor hp/torque drops with speed, what happens with a 1725 rpm gear motor with a 28:1 built-in reduction to provide 60 rpm output shaft speed? Does a gear motor torque multiply with the pulley ratio advantage? Seems like I recall someone mentioned that with slow speed gear motors you don't need much hp to generate torque. How much is enough for cutting steel?

Can someone enlighten me on the mechanical aspects?

Scatterplot
05-03-2007, 09:48 AM
Not sure how much speed/torque/whatever you need, but I can help you a little on the mechanics. When Forrest said the torque drops with speed, that's talking about no gearing. Motors will have sort of an ideal run speed (maybe a range) and operating there they will produce the most torque they can. With gearing you can turn speed into torque and vice-versa. Imagine running a 2000 rpm motor with just enough electricity to turn it at 60 rpm, barely any torque. Now if you give it enough juice to make the motor shaft spin at 2000 rpm and Gear it to 60, then you'll have craploads of torque.

DR
05-03-2007, 10:03 AM
Consider this......have you ever seen a 14" metal cutting band saw with only a 1/4hp motor?

My 14" Powermatic wood/metal saw came with a 3/4hp motor.

Evan
05-03-2007, 10:06 AM
Horsepower is the product of rpm and torque. When you gear down a motor, rpm reduces and torque is multiplied so hp stays the same at the output shaft. As scatterplot said, Forrest was talking about slowing a motor via electronics, not gearing. Gears are torque multiplers when used as a step down ratio.

I'm using a 3450 rpm motor to drive my shaper but I need only around 200 rpm. I have a 15 to 1 worm reduction drive from an old concrete skimmer that will work perfectly.

05-03-2007, 11:33 AM
OK, then answer me this:

0.25hp x (33000 ft-lb/min per hp) = 8250 ft-lb/min which is motor power

if Power=Speed x Torque, then at 1725 rpm before reduction the torque is:

8250 ft-lb/min / 1725rpm = 4.78 ft-lb (times 12"/ft gives 57 in-lb)

after 28:1 internal reduction to 60rpm the torque must be:

8250 ft-lb/min /60rpm = 137.5 ft-lb (1650 in-lb)

Where/how would the 150 in-lb stated torque be derived from? Based on the power equation I see where torque has to go up/down if speed changes to produce the rated hp at the output shaft. This should mean if it's a stated 1/4hp motor then I get 1/4hp output at the shaft. This also means something doesn't add up: 150 in-lb stated torque is a far cry off of 57 and 1650 as calculated. Is there some motor efficiency/power factor adjustment that comes into play? It's a 110VAC/60Hz single phase motor. Further observation says that torque should increase further when I use the 4"/8" pulley ratio to bring the axle shaft speed to 30rpm, correct? But what torque is multiplied, 150 in-lb stated or 1650 in-lb that I calculated?

Evan
05-03-2007, 03:31 PM
Weak gears that break if the stated torque is exceeded? Designed in slip clutch?

05-03-2007, 05:54 PM
"Forrest says the motor torque drops as speed drops..." No, I didn't. Don't put the wrong words in my mouth. I said something like: the motor power decreases when the motor shaft RPM is electronically reduced while the motor torque stays roughly constant and proportional to the motor full load Amps. There a world of difference between the two statements. The first is utter BS such as you see in the pages of a car magazine and the second is a accurately stated equivalency using physics terms correctly in context. You will never see car mag level physics in my statements.

You cannot expect adequate metal cutting performance by slowing down a 1 HP three phase induction motor with a VFD or an equivalent DC motor and drive. If you want adequate metal cutting performance you'll need about 20 to 1 mechanical reduction with the accompanying and proportionate increase in tractive effort on the band.

Learn the physical concepts relating force to work to power, accelleration, etc. Ignorance of them in the machine shop is akin to running a race with one foot in a bucket.