# Torque and Rolling Resistance

1. Jan 18, 2009

### tcitizen

Hello.

I'm trying to find the torque required of an electric vehicle to pull a 1000kg trolley on four Nylon Wheels on smooth concrete. The trolley fully supports its own weight. The vehicle has 2 rubber drive wheels of Radius( r = .125m). Now I'm not sure if am on the right track because i cant find much information on Rolling resistance.

Using Fr = Crr * N to find the rolling resistive force.
Where Fr = Rolling resistance force
Crr = coefficient of rolling resistance
N = Normal force

Rolling resistance of trolley
Assuming Crr = .004
N = 1000kg * 9.81 m/s^2
Ft= (.004*9810) = 39.24

rolling resistance of vehicle
Fv = (45kg * 9.81 m/S^2 * .08)
Fv = 35.316
Ftotal = Fv +Ft
Ftotal = 35.316 + 39.24 = 74.6N

So if I use the equation for toque
T = F * r
T = 74.6 * .125
T = 9.32 N.m

This should give me the minimum required torque to get the trolley and vehicle rolling on a smooth horizontal surface. Yes/No?

Also if my system were to reach maximum velocity of 5km/h (1.389m/s) how do I find the torque needed.

Regards
tcitizen

PS I've tried to include as much information as possible.

2. Jan 18, 2009

### Dr.D

Where does this come from: Fv = (45kg * 9.81 m/S^2 * .08)?
What is the 45kg? the .08?

3. Jan 18, 2009

### tcitizen

That is the equation for the rolling resistance of the vehicle.
Fv = Crr * N
Fv = rolling resistance of vehicle
0.08 = coeff of rolling resistance of vehicle
45kg = mass of vehicle

4. Jan 18, 2009

### Dr.D

Looks like what you have is OK, now that you have identified your additional calculations. Why do you doubt it?

If you are going forward at a speed V, then the power required (at steady state) is
P = Ftotal * V = Torque * omega_wheel

Solve this for the torque required after you get the drive wheel speed.

5. Jan 18, 2009

### Phrak

Given a cart and concrete floor at my disposal, I would load the cart and measure the force required to pull the cart. That's what I did. But depending upon application, I might be more concerned about the force required to overcome surface irregularities from a standing start or low speed.

Last edited: Jan 19, 2009
6. Jan 19, 2009

### nvn

tcitizen: Wouldn't you need to also include axle bearing friction as an additional term in your torque equation? The resistance due to axle bearing friction depends on bearing radius and bearing coefficient of friction. Also, to try to model the surface irregularities mentioned by Phrak, you could perhaps estimate the effective incline (slope angle) of the surface asperities, and work your problem on an inclined plane having that upward slope angle.

7. Jan 19, 2009

### Dr.D

And this can go one without end, this inclusion of additional effects. Besides the axle friction torque, we can include the friction in the transmission, the motor friction, etc. But that really goes beyond the original question which was the relation between the rolling resistance (only) and the torque required to overcome rolling resistance. This is a case of "problem creep."

8. Jan 19, 2009

### tcitizen

Yes, there are many many factors to consider, the loading case of 1000kg is the absolute maximum that could be applied in ideal situations, realistically loads of <500kg will be more commonly used.

I've got a query, What, if any, is the difference between friction force and traction force?
Because I would like to determine whether the drive wheels will slip in this situation.

Regards

9. Jan 19, 2009

### tcitizen

I think I just figured it out.

I just use the equation for static friction of the drive wheels and the surface.
Then use the torque equation to determine the maximum torque before slipping.

Fs = us * N

where Fs = static friction force
us = coeff of static friction
N = normal force

Fs = 1 * 45 *9.81
= 441.45

T = Fs * r
where T = torque
r = radius of wheel

T = 441.45 * .125
T= 55.18 n.m
therefore when 55.18 n.m of torque is applied wheels will slip.