Selecting a Gear Drive or Gear Motor

[franklineng]

I had recently fabricated a set of welding turning idlers for our weld shop using a set of 10 diameter wheels, 28 inches center to center. We are using it to turn 50" diameter tanks while welding. The tanks will be over 1500#, welding at a maximum of 45 ipm. I want to motorize the unit using a gear drive and electric motor or a gear motor? I've calculated the torque based on Tangential Force x (50/2). The tangential force given from the equation:

(Iy)x(A)/R=Tangential Force
where Iy is the mass moment of inertia of the cylinder.
A is angular acceleration (rads/sec^2) - uniformly accelerated from rest. If I expect the tank to be at final angular velocity in 20 seconds from rest.
R= 1/2 diameter of the tank.

Is this a correct assumption? If so, how would I go about selecting the gear drive and electric motor combination?

 

[dimjim]

Remember to use at least 1.5 times the running
torque for the approximate starting torque.

 

[tygerdawg]

peak torque requirement will be
(1) torque to overcome friction at start
(2) torque to overcome minute indentations of one cylinder on another (acts like friction)
(3) torque to accelerate rotationally the different masses like the gearmotor, couplings, chains/belts, cylinders, sprockets, everything that rotates
(4) everything else you can think of, then try to quantify it

Peak Torque will typically be more than running torque, more than torque to decelerate.

Search the net for "Smart Motion Cheat Sheet" PDF file for a collection of useful formulas. Also get gearmotor manufacturer catalogs & engineering guides. Once you start to narrow down the speed/horsepower/torque selections of gearmotors, you will probably have to go through a second calculation iteration for number of starts per hour of the motor. More starts == more motor current == more heat == bigger motor required to survive.

Then your third iteration may involve selection of the type of drive, either straight current on/off or a variable frequency drive for smooth ramping.

TygerDawg

 

[franklineng]

Thank you both for the information. I did pickup one of the available smart motion cheat sheets. Some of the information is unavailable for larger AC motor applications such as Load Inertia of the motor (at least in the catalogues I have in front of me).

 

[franklineng]

I'm going to give a bit more information here:

Known quantities:

Cylinder weight: 1504 lbs.
Outside Diameter: 50.5 inches
Inside Diameter: 50.29 inches
Time to accelerate approximately 15 seconds (T)
Full speed rpm: 0.283 rpm (calculated from below)
- welding speed 45 ipm, perimeter of cylinder 158 inches
- 45/158= 0.283 rpm.
Angular Acceleration: w=(Pi*n/30)/T)= 0.0296 rads/sec.
Diameter of Drive Wheel (10 inches) or 0.83 ft

Mass Moment of Inertia of Cylinder:
= (W/8*386)*(do^2 + di^2)
= 2473.87 lb-in.-sec^2

Torque (lb*ft)= [(2437.87*386)/(144)]*0.0296
= 13.063 lb-ft

Tangential Force Driven(lbs):
= 13.063/(50.5/2*12)= 6.210 lbs

Value seems small.

Torque at Drive wheel:
= (6.210 lbs)*(0.83/2 ft)= 2.557 lb-ft

Does this appear correct so far? Where do I go from here?

HP= (Torque lb-ft)*(rpm)/5250
= (2.557)*(1750)/5250
= 0.85 HP if the electric motor is rated for 1750 rpm.

I have not accounted for any gear reduction.

Your thoughts or input?

 

[dimjim]

What is the resistance or drag tourque
of the two idlers? You will have to
include that.

 

[MikeHalloran]

RPM of drive wheel = 0.283 rpm * 50.5" / 10" =~ 1.5 rpm

Buy a reducer, or a reducer and a chain drive, to give you that output speed, or a little more. Chain drive gives you the opportunity to drive multiple rollers with one motor.

Chances are you'll be able to accelerate the big cylinder much faster than you estimated, even with a small motor. Plug that speed (1.5 rpm) into your HP equation to see how small a motor you need in theory. In practice, buy 1/3 or 1/2 HP or whatever comes attached to the reducer. That will give you torque margin for future workpieces that are not symmetrical.






Mike Halloran
Pembroke Pines, FL, USA

 

[franklineng]

RPM of drive wheel = 0.283 rpm * 50.5" / 10" =~ 1.5 rpm

HP=(2.557)*(1.5)/(5252)=0.0007303 HP (1/1369 HP ?)

Not much horse power required at minumum, but like dimjim mentioned, whats the resistance of the two idlers? The 10" diameter idlers had bearings and little torque resistance.