Gearmotor Requirements 3

[motorquest]

Hello,

I'm building a motorized platform that will weight approx. 1500 lbs and will be used to move a load weighing ~32,000 lbs. I plan to use six 12" Dia. x 4" tread polyurethane on steel wheels. The surface is level, smooth concrete. I need to find the locked rotor torque necessary to overcome static friction in order to select the correct gearmotor.

According to my calculations I would need a force of 33,500 N (7520 lbs) to overcome friction and start the load moving (Coef. S-friction used = 1.0) without wheels. My question is, how do I determine the force required to start the load "rolling"? I've pulled every mechanical engineering and physics text that I can find but I'm overlooking something.

One engineer at a caster manufacturer told me that I could use 5% of the total load (33,500 #) as a guideline for selecting the gearmotor.

Am I anywhere near the mark on either count?

MotorQuest

 

[israelkk]

As I see it you need to calculate the friction torque at the wheel axes due to the (32000+1500)lbs load and then divide it by the wheel radius to get the pulling force.

Assuming the wheel axis is 2" diameter with a 0.1 coefficient of friction (if the wheels contain roller/ball bearings it may reduce to ~0.005). Then the friction torque in the wheel axes is: 33500lbs*0.1*1inch=3350lbf-inch

Divide this by the wheel radius=6inch and you will receive a pulling force of ~560lbf. This is how you can large man pulling a huge truck in tournaments. Be sure to take safety factors for unlevel surface, small bump etc.

 

[Barry1961]

With roller or ball bearings about the only difference in static and dynamic friction is the seal which will not be enough to consider in this application.

Five percent of you total load plus the force required to accelerate the total load should be about right if you are truly level with a smooth surface. If you have little bumps or pot holes the required force can go up very quickly.

The acceleration should be much more of a factor than any static friction with bearings.

Barry1961

 

[aolalde]

Hi MotorQuest

As you stated from basic dynamics for a horizontal plane:

Fr= mu*N

Fr = frictional force
mu = friction coef.
N = Vertical reaction equal to total weight.

Now if you apply a force equal to the friction force your body remains stand still !

You need extra force to break the equilibrium and accelerate the body ( 2nd Newton’s Law)

F = m * a or a = F/m

a= acceleration = (Vf-Vo)/t
m = mass
F = accelerating force = Ft-Fr
Ft = total force applied
t = time to reach the final speed.
Vf = final speed
Vo = initial speed, normally Vo = 0

Now it depends on the time required to accelerate and how the force will be applied by the motor. Traction on two wheels?, a winch type cable? , a chain?

Next, convert the lineal movement velocities and forces into rotational movement and reflected inertia. That will depend on the rotational speed of the motor shaft, and the diameters of the wheels or pulleys.

Neglecting slip, the linear speed ( V) becomes a rotational speed rpm =V*60/( 3.1416* D)
The total force (Ft) becomes a motor torque (T) T= Ft*D/2
The mass becomes a rotational inertia constant WK^2, reflected at the motor shaft.

After the mass is accelerated to the desired speed, the power to keep the body in motion consumption is proportional to the speed and frictional forces including the wind resistance.

 

[fwc]

What about the elastic deformation of the wheels, is it a significant factor in this application?

 

[motorquest]

fwc,

At this stage, I'm looking at using 18" all steel wheels. The load I need to move combined with the acceleration would probably severly damage the original polyurethane over steel wheel design I had planned to use.

 

[itsmoked]

motorquest; those 18" wheels would have very low friction! But they will add some non trivial inertia you will need to concider.