# Torque Required To Move a Vehicle

1. Oct 15, 2015

### John Vorberger

Hello,

I apologize if this is in the wrong sub-forum, as I am new to this site.

Anyways, I am trying to figure out how much torque is needed to move a vehicle that is about 500 pounds in weight. That is including the driver of the vehicle. For a little more information, the vehicle would have tracks that are powered by a chain and sprocket type setup. The drive sprocket would be 15 inches in diameter. I would be powering the vehicle with two hydraulic motors, and at the low pressure/high flow setting, the motors would produce roughly 20 ft-lbs of energy per motor, since I would be using two motors, one for each track, that would equate to 40 ft-lbs. Do you think the 40 ft-lbs would be enough to move the vehicle? I am talking about mostly flat ground, maybe a very minor incline.

Thank you,

John

2. Oct 15, 2015

### BvU

Hello John, welcome to PF !

Sounds like mechanical engineering alright !
Converting 40 ft . lbs at the motors to forward acceleration involves several steps:
torque from motor to track driving axle with sprocket wheel
(the radius axle to ground is determining there, if it is bigger than the sprochet wheel radius)​
track driving axle torque to propulsion force wrt ground
acceleration = propulsion force / total mass​
My return questions: 20 ft - lbs of energy, you say. Would that be per hour ? Or is it 20 ft - lbs of torque ?
Any idea of the friction losses ? Because when moving at a steady speed, that is all the propulsion force hass to compensate for.
What acceleration would you stil find acceptable on horizontal ground ?
(to be incremented with $g\sin\theta$ where $\theta$ is the maximum angle wrt horizontal you want to be able to climb)

3. Oct 15, 2015

### John Vorberger

Sorry for the confusion. I meant to type 20 ft-lbs of torque per motor, so 40 ft-lbs of torque total. I think an acceptable acceleration would be to get to 10 miles per hour within 3-4 seconds.

4. Oct 15, 2015

### BvU

I'm no good at all with this kind of units, but 40 ft-lbs sounds like what you apply when on a bicycle ?
Anyway, if the sprocket wheel diameter is also the distance (that's what I see here) from axle to ground, you have
propulsion force = torque * radius
acceleration = propulsion force / total mass
but you need to allocate a lot of force to overcome the friction.

Check my calculating: 10 mph/4s $\approx$ 4 ft/s2
40 ft-lbs/(15"/2) = 64 lb ?
64 lb/500 lbs = 64/500 * 32 ft/s2 ?? so about 4 ft/s2

5. Oct 15, 2015

### John Vorberger

I am a little confused about your calculations. I am sure they are correct but I am just confused were the (15"/2) comes from? Like I said I'm sure your calculations are sound I am just not very experienced when it comes to calculations like that.

Also, I just used ft-lbs because that is what the engine manufacturer used to describe torque.

6. Oct 15, 2015

### SteamKing

Staff Emeritus
If your chain sprocket is 15" in diameter, then the pull on the chain produced by applying a torque to the sprocket will be the torque divided by the radius of the sprocket, which is half the diameter. You also must convert the radius to feet to have the units match with the units for torque (ft-lbs).

7. Oct 15, 2015

### John Vorberger

Thanks for clearing that up. The problem with the hydraulic motor, which is the thing that is producing these torque numbers, is that it has two separate modes. The first mode is 13 gallons per minute (GPM) at 600-900 PSI. The second stage is 3.4 GPM at 3000 PSI. I think when I first accelerate, it might kick into the second stage, but shouldn't the PSI required to keep it rolling dramatically reduce once it is moving at around 7 MPH?

Thanks,

John