# The Naked Scientists Forum

### Author Topic: Which are the main forces involved in a caterpillar track?  (Read 3630 times)

#### Lcorp7

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##### Which are the main forces involved in a caterpillar track?
« on: 11/10/2011 01:00:52 »
Hello,

I was wondering, If I was to build an UGV and instead of wheels I'd use tracks, How may I do the proper calculations of the forces involved in the system?

Cheers.
« Last Edit: 17/10/2011 00:56:41 by Lcorp7 »

#### CliffordK

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##### Re: Which are the main forces involved in a caterpillar track?
« Reply #1 on: 11/10/2011 12:55:16 »
Are you talking about a military "Unmanned Ground Vehicle"?

http://en.wikipedia.org/wiki/Unmanned_ground_vehicle

Whew, doesn't that remind one of the Terminator Movies?  We certainly need to concentrate on better diplomacy rather than better ways to kill people.

However, the term does also include robotics and civilian rovers, for example for use in explorers and nuclear facility maintenance.

It appears as if some of the vehicles are using rubber tracks rather than steel tracks.  That would simplify the forces within the tracks, although I suppose the bending forces would still be there to some extent.

So, what forces?

You have downward weight/mass.  Tracked vehicles actually have good mass distribution due to the very large are of the tracks on the ground, although many dozers are extremely heavy.  You can calculate weight per square inch by dividing the weight of the vehicle by the area of the tracks on the ground.  Compare that to a wheeled vehicle with a much smaller are in contact with the ground under the tires.

You would have a drive force that is essentially a sprocket on a chain, as well as looking at the friction of the tracks with the ground.

Most tracked vehicles use some kind of a skid-steer mechanism, although perhaps some snow cats would use more of a conventional steering mechanism.  I believe the skid-steer is accomplished in 2 methods.  Cletrac & Oliver crawlers use a typical differential like an automobile differential as the drive for the tracks, then applies brakes to slow down one track over the other.  Most other dozers use clutch plates to apply different forward driving forces to the two tracks.  No matter how you do it, the steering is accomplished by applying differential forces to the two drive sprockets (and perhaps reverse forces on the sprockets).  An electric drive could be done clutchless, by employing two independent motors.  You may still choose to add friction brakes.

Turning would also create torque in the system at the point the tracks contact the ground, and tend to pull the tracks off of the drive sprockets.

You would have some kind of a spring tension system within the tracks.  Multiple track rollers give better weight distribution.

Forces might be slightly different when considering uneven terrain and gulleys vs a static system on flat ground, but it is all essentially the same.

Is there any specific design question you are dealing with?

#### Lcorp7

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##### Re: Which are the main forces involved in a caterpillar track?
« Reply #2 on: 12/10/2011 02:19:15 »
Thank you for your answer, I guess that the very words has been misconceived, UGV doesn't intrinsically means 'military'. The problem I have to deal is basically how much mechanical power a given UGV needs to climb an inclined surface, ex... hill, ramp, or even stairs? Then after knowing the powers required, determine the kind of batteries I would use and also the kind of motors.

Cheers.

#### CliffordK

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##### Re: Which are the main forces involved in a caterpillar track?
« Reply #3 on: 12/10/2011 03:01:08 »
If it is a tracked vehicle, you will have some intrinsic losses in the driveline of your system.

If you read some of the Cletrac information, they will claim better power to the tracks than the equivalent Caterpillar or other brands due to the use of their differential, however, I'm not convinced the difference would be so great.

Otherwise, it is just a typical acceleration, mass, horsepower problem.

So, basically, you would have two problems.

What is the HP requirements to move the system on the level?

For the incline, you would essentially count the vertical climb.

HP = 550 ft lbs per second.

So, to get the climbing HP requirements, you would take the vertical climb speed in feet per second (or equivalent) * the weight of the vehicle in pounds.

You should be able to add the HP requirements on the flat plus the calculated HP requirements for climbing to get the total HP requirements.

Gearing may allow you to slow the rate of climb, and thus diminish the actual HP requirements.

There is a discussion about fuel tanks and mileage which will give some idea of system loss.
http://www.thenakedscientists.com/forum/index.php?topic=41263.0

As far as a battery electric vehicle, your batteries will be related to the volts and amps of the system.  Obviously the more batteries you add, the more weight, and intrinsic power losses in the system.

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##### Re: Which are the main forces involved in a caterpillar track?
« Reply #3 on: 12/10/2011 03:01:08 »