The physics of pushing

[bardd]

I'm trying to decide on a good gear ratio for our drive system, but I'm kind of stuck on one thing - I can't figure out how to calculate the force the robot could inflict on another robot while pushing (not momentarily). I alraedy have the torqe of each wheel, I'm just missing the next step.
Would the number of wheels make a difference?

Thanks in advance,
Bar #2212

[AlexH]

from personal experience gearing for about 5fps and using any high grip wheel (plaction, colson, traction ect) will give you a very strong pushy bot. you still won't have enough torque to break traction with the carpet though.

[bardd]

Originally Posted by AlexH

from personal experience gearing for about 5fps and using any high grip wheel (plaction, colson, traction ect) will give you a very strong pushy bot. you still won't have enough torque to break traction with the carpet though.

I actually want to do the math... And to know how to. The thing is I already have the built robot and all I want to do is decide whether I want to chaneg the gear ratio or not. Besides, I don't want to have a very strong pushy robot, I want a combination of strength and speed that would be the best for the robot out of several options I have.

[Ether]

Originally Posted by bardd

I actually want to do the math... And to know how to.

For each drive motor, multiply the motor's torque by the total gear ratio (gearbox and sprockets etc) and divide by the wheel diameter radius. Then add those up. Watch your units. Post your results here and we'll check it for you.

[bardd]

Originally Posted by Ether

For each drive motor, multiply the motor's torque by the total gear ratio (gearbox and sprockets etc) and divide by the wheel diameter. Then add those up. Watch your units. Post your results here and we'll check it for you.

I got that step... What I want to know is how to calculate the force the robot can push with using the torqe

[Adam.garcia]

Originally Posted by bardd

I'm trying to decide on a good gear ratio for our drive system, but I'm kind of stuck on one thing - I can't figure out how to calculate the force the robot could inflict on another robot while pushing (not momentarily). I alraedy have the torqe of each wheel, I'm just missing the next step.
Would the number of wheels make a difference?

If you'd look at JVN's Mechanical Design Calculator, you can see the effects of gear ratio, # of wheels, # of motors, etc. on the effective pushing power of a robot. It allows for easy "prototyping", as the numbers are as simple to change as the push of a button.

I would defer the mathematical explanation to somebody of more knowledge, but for starters you can look at the formulas that he uses in this spreadsheet. JVN's Design Calculator may be found here:

http://www.chiefdelphi.com/media/papers/2059

[EricVanWyk]

Draw a "free body diagram" with all the forces you think are relevant and post it here. In your post, tell us all the assumptions that you have made. From there we can create some simple equations and walk through it. The math is actually quite easy - the hard part is finding all the things that affect it.

[Ether]

Originally Posted by bardd

I got that step... What I want to know is how to calculate the force the robot can push with using the torqe

Assuming you are not traction limited, that is the force the robot can push.

Or perhaps I am not understanding what you want.

[Craig Roys]

Pushing force is basically your static friction force between your wheels and the ground. The max static friction force is equal to your [Normal Force (the weight of the robot on a level surface)]x[coefficient of static friction].

The max weight of a robot including bumpers and battery is around 150 lbs.

Assuming a coefficient of static friction between the wheels and carpet of ~1.5, you can create about 225 lbs of pushing force before your wheels start to slip. Once they start to slip, that number will drop.

Keep in mind that you also can only draw 40 amps per motor before you start popping breakers. The trick is to get your gearing such that your wheels will begin to slip just before you get to 40 amps to avoid popping your breakers. Once the wheels slip, the resistance force goes down and your current draw should decrease.

That's an over simplified look at it, but it's a place to start. There are many other factors such as type of drive-train, number of wheels, type of tread, weight distribution of robot, etc. All can affect your pushing power.

[JamesCH95]

You're looking for:

Torque applied to wheels (in*lbs)/Radius of Wheel (in) = Pushing Force (lbs)

Assuming you're not slipping.

I'd also ding about 5% per gearing stage you have (gears sets and sprocket/chain sets). Easiest way is to multiply your end result by 0.95^(number of gear reductions). You can apply more realistic numbers with some minor research.

[DonRotolo]

OK, maybe this will help:

If each wheel gets 20 ft-lbs of torque, and the wheel is 6" in diameter, the force acting on the carpet* is 20 / 0.25 = 80 Lbf. (20 is the torque in ft-lbs, 0.25 is the radius of the wheel in ft, and Lbf is "pounds force" (different from pounds weight)).

*Theoretically. Remove maybe 5% for losses (as suggested above), you get 76 Lbf.

Four wheels, 76 * 4 = 304 pounds of force.

Assumes you don't break traction with the carpet.

[Ether]

Originally Posted by DonRotolo

OK, maybe this will help:

If each wheel gets 20 ft-lbs of torque, and the wheel is 6" in diameter, the force acting on the carpet* is 20 / 0.25 = 80 Lbf. (20 is the torque in ft-lbs, 0.25 is the radius of the wheel in ft, and Lbf is "pounds force" (different from pounds weight)).

*Theoretically. Remove maybe 5% for losses (as suggested above), you get 76 Lbf.

Four wheels, 76 * 4 = 304 pounds of force.

Assumes you don't break traction with the carpet.

Be aware that if you are powering 2 wheels with 1 motor, each wheel gets half the torque. Similarly, if you are powering 3 wheels with 2 motors, each wheel gets 1/3 the torque produced by adding the torques of the 2 motors together (and factoring that torque up by the gear ratio, of course). That's why I worded my post in terms of drive motors instead of wheels.

[JamesCH95]

Originally Posted by DonRotolo

OK, maybe this will help:

If each wheel gets 20 ft-lbs of torque, and the wheel is 6" in diameter, the force acting on the carpet* is 20 / 0.25 = 80 Lbf. (20 is the torque in ft-lbs, 0.25 is the radius of the wheel in ft, and Lbf is "pounds force" (different from pounds weight)).

*Theoretically. Remove maybe 5% for losses (as suggested above), you get 76 Lbf.

Four wheels, 76 * 4 = 304 pounds of force.

Assumes you don't break traction with the carpet.

Bardd: note that this would require a wheel tread with a coefficient of friction of around 2.0, much higher that commonly used tread materials. When you finish your calculation you should double-check it with other potentially limiting factors such as traction and component strength.

[IKE]

There is also a lot of good info in this thread on designing for pushing from the power electronics aspect:
http://www.chiefdelphi.com/forums/sh...hlight=40+amps

Just because the motors can provide the power doesn't mean the electronics can provide the power (for very long).

[Ken Streeter]

Originally Posted by Ether

Assuming you are not traction limited, that is the force the robot can push.

Or perhaps I am not understanding what you want.

Ether is right that the torque delivered by the wheels (assuming no wheel slip) is the maximum pushing force that can be provided by the robot.

However, in my experience (at least with 1519 robots), we haven't been able to actually get that much pushing force due to a different limiting factor. However, the limiting factor isn't the wheels slipping (in recent years we have used very grippy wheels) but rather the robot starting to flip itself over. (See photo below for an example.) Once teams design robots with sufficient torque and traction, the robot's own weight distribution and geometry is the next limiting factor.

For robots with high torque capabilities and very grippy wheels, "lift up" seems to be the limiting factor. Figuring this out is something that we've never actually done before on our team, but would probably be good to do. There's little point in having more torque or traction in a pushing contest than the amount that starts to result in the robot being lifted up, as that torque (or traction) can't be effectively used.

(By the way, I was hoping that this thread was taking these other factors into account...)

Example of the problem is shown in the photo below from this old CD thread: http://www.chiefdelphi.com/media/photos/35102