Gear Motor Torque Requirements 1

[IClnYourAir]

I need a sanity check here.

I'm in the process of designing a "rough handling table" The table consist of a rigid base mounted to the floor. There are two parallel shafts, supported by pillow blocks, mounted to the base. Each shaft has two cam lobes on it, one towards each end of the shaft. Each shaft has a sprocket mounted on the end. The sprockets of the two shafts are connected with a roller chain that is also connected to a gear motor.

A flat "table" rests on the cam lobes. The cam lobes are designed and positioned such that they raise and lower the table, with a total stroke of 1". The shafts are required to operate at 200 rpm. The combined weight of the table and the load on the table is roughly 500 lbs.

My question is: What size gear motor is required and what should the torque output on the gearbox be?

Based on my calculations:

work = force * distance = 500lbs * 1in = 500in.lb = 41.67ft.lb

200 rpm = 3.333 revs/sec, thus 0.3 secs/rev

I want to do 41.67 ft.lb of work in 0.3 secs = 138.9 ft.lb/s

knowing that 1 hp = 550 ft.lb/s, this translates to 0.25hp

Of course, I have to assume friction between the cam lobes and the rollers on the bottom of the table against which they ride. Also have to include drives losses, motor efficiency, etc..

Assuming 50% efficiency, this equates to 0.50hp.

Do I simply specify a 0.50hp motor, with a gearbox that will give me an output of nearly 200 rpm, and then use the sprockets to attain the 200 rpm requirement? Do I need to be concerned about the output torque of the gearbox? Do these calculations make sense?

I look forward to hearing your comments.

 

[haggis]

IClnYourAir

I dont want to make things complicated but:

When designing something like you described, actual practice and theory are two very different things. Your HP calculations are fine in theory if you are lifting the 500 lbs in a vertical linear motion at a steady rate of speed. The torque at the shaft where the cam is, is your primary consideration to achieve the 200 cycles per minute by accelerating the 500 lb weight from rest to a distance of 1” in 0.3 seconds.

Lets work back from there. You can substitute your dimensions for the ones I’ll use. We’ll say you have a 2” diameter shaft and your cam lobe is 1” for the required lift. This puts your cam lobe 2” from the shaft centre.

To arrive at the force required to be exerted by the cam lobe we have to find out the rate of acceleration:

So by applying the following:

V2 = V1 + at
where:
V1 = initial velocity in inches/second
V2 = final velocity in inches/second
T = time in seconds
a = acceleration inches/second/second

V2 = V1 + at
a = 0 + V2 / t
a = 0 + 3.33 / 0.3 = 11.1 inches/second/second.

We arrive at the rate of acceleration and apply the following to find the force required:

F = ma
where:
F = force
m = mass
a = acceleration

F = ma
= 500 * 11.1 = 5550 lbs.force


This would require a torque of 11100 lbs.in (925 lbs.ft) at the shaft. We’ll also say your motor output speed is 1750 RPM so a speed reduction of 8.75:1 is needed for your final 200 RPM. Since torque is directly related to the speed reduction, the torque at the motor shaft would be:

925 / 8.75 = 105.7 lbs.ft of torque.


Motor HP is calculated as follows:

HP = torque * RPM / 5252 = 925 * 1750 / 5252 = 35.2 HP

Having this said, this is where practicality becomes important. From the brief description you gave of your machine, every component is going to take a pounding. Shafts will have to be big enough to handle the torsion and resist bending. The bigger the shaft, the farther the cam lobe gets from the centre. The farther the distance from the centre, the greater amount of torque is required and so on right back to the motor. Pillow block bearings are not designed to consistently take impact loads. To say the least, it works in theory but will probably self destruct in very short order.

Have you given any thought to moving the load up and down in a linear fashion by use of an air cylinder(s) and linear bearings as guides? The cylinder(s) can be sized to accommodate the force required and the movement in the given time cycle of 200” strokes/min.

To size the cylinder, you know that it has to stroke 1” every 0.3 seconds to do the work, requires a force of 5550 lbs. and we’ll assume that you have plant air at 80 psi.

(square root (5550 / 80 / pi)) * 2 = cylinder bore.

The added benefit of using a pneumatic or hydraulic system is that it can be adjusted easily by use of pressure regulators, speed control valve etc.

Just as a footnote, you might also consider what this machine will be sitting on. In our plant we have a “shaker machine” consisting of four pads with a cylinder under each where we place a vehicle on the pads to simulate various road conditions like gravel roads and paved roads with potholes etc. We had to pour a seismic mass (large concrete foundation) to dissipate the vibrations to prevent it from eventually driving itself into the
ground and also dampen the vibrations for the surrounding area.

Haggis.