Friction, traction, torque - oh my...

[Gary Dillard]

I'm thinking I'll work on it this weekend and try to finish it next week after I get Gary J. to review my part of it.
I can't make MiM - I'll be in Boston that weekend

[Paul Copioli]

Gary,

You missed the 217 pull. We were a write in above 121 with 196lbs (if memory serves). We tested with the exact methods you guys did and found our mu was between 1.5 and 1.6 (several different tests). The brecoflex is good stuff, but we used the neoprene padding from McMaster Carr (and small parts). The neoprene pad is much less expensive than the Brecoflex stuff.

-Paul


P.S. - We are also looking forward to your white paper. We have derived several equations for the motion of a robot operating under dc motors and hope our equations match yours. (For a level of sanity check).

[Gary Dillard]

Sorry Paul - I don't see your team on our data sheet. My record keeping was less than adequate for sure - hopefully we'll do better next year (and maybe we won't get shut down for blocking the aisle).

We love the BrekoFlex but you have to decide if you're willing and able to budget it. If we hadn't bought spare belts we'd be dead - we actually broke one during qualifying rounds - but they ain't cheap (I think about $250 apiece). Ouch! Certainly depending on the game and strategy anything with reasonable coefficient of friction will get you there, and there are a alot of cheaper alternatives.

[Paul Copioli]

Hey Gary,

One of my students gently remined me that we didn't do your test. We did the pull off that was outside (I think by the technoticks). I pretty much screwed up every bit of information that I put in my last post. I apologize for being a doofus.

-Paul

[Gary Dillard]

's OK. We'll look for you next time

[Lloyd Burns]

Gui:
Suppose we have a motor with these specs:
....no load speed at 12 V is 9600 rpm, drawing 1.2 A.
....stall torque 2 N*m Stall current is 37.2 A.

Suppose we want to use wheel of 10 cm (= 0.1 m, ~4") radius, and the box is propelled at 3.14 m/s (about 10 feet per second).

Suppose we want maximum power out of the motor to do this.
(This may or may not be a good asumption! You may decide that you want max power at some other speed.)

-+-+-+-+-+-+-+-+-

Max power is found close to the midpoint of the speed range, giving about half the stall torque, at about half the stall current. Electrically and mathematically, the product of voltage (=k1 * rpm, graph \) times current (= k2 * torque, graph /) reaches a maximum for speeds and torques in the mid-range. Let's conjure with 4800 rpm, or 80 rps.

The driven wheel is 0.1 m radius: to go at 3.14 m/s, it must turn at 3.14 m/s / 2piR = 3.14 / 2 * 3.14 * 0.1 = 1/.2 = 5 rps or 300 rpm.

Our speed ratio must be 4800:300 = 16:1. Since ratios of 8:1 were supposed (on this board) to be the highest practicable for efficiency, let's use two drops at 4:1 each. It was also mentioned that every gear interface loses something, lets assume 25% for our spur gears.

The first drop finds the driven gear going 4800/4 = 1200 rpm, but the torque available is not 4 x 1 N*m, but due to losses, only (4 - 25%of 4) * 1 N*m, = 3 N*m. The next gear interface turns the output shaft at 1200/4 = 300 rpm, and the torque available is 9 N*m.

The wheel pushes along at 3.14 m/s, but the torque available is 9 N*m * 0.1 m = 90 N.

The torque could also be figured as 16 * 1 N*m (1-0.25)(1-0.25).

If the friction your box encounters is less than the available force, 90 N here, you're home free. If there is more friction than available force, it ain't gonna work: either revise your speed expectations and change the ratios for enough torque, or choose a more powerful motor.

To measure your Force requirements, disconnect the gears and pull the robot with a force scale. To read the force required to turn the gears, reconnect them, and repeat the experiment. Don't forget that the force is multiplied at the upper end of th echain, nearer the motor.

Power efficiency:

At this speed, the motor is drawing ((37.2 - 1.2)/2) + 1.2 A, or 19.2 A at 12 V for a power input of 230.4 W or 230.4 W/746 kW/HP = 0.308 H.P.

The 25% losses in two gear interfaces leave us with about 60% of the power we put in, as an approximation.

The wheel moves at 3.14 m/s, with a force of 90 N. Power out = Work out/time = Force x Displacement/time = 3.14 m/s x 90 N = 202.6 W, about 60% of what we put in.

Fine Print:

If we decide that max power should occur at a lower speed, we'll need to gear it down farther, and the torque available will be more, but the top speed will be lower. If the max power floor speed is to be greater, then we won't gear it down as much, and maybe, whem we need to accelerate from a standstill (1..2..3..go!) there will not be enough torque: then the motor will stall, the current will go way up, the motor temperature rises, and "pop" goes the breaker.

If the friction with the floor that you obtain with your wheels is greater than the force available, locked rotors will be common. Many FIRST teams try to make the wheels break free before the stall is reached.

We could go at this from desired torque output, as well. (Ex., How fast will the output shaft turn if we need 270 N of force on the edge of a wheel 5 cm in diameter ?) (No, I don't want to answer this one :-) !)


HTH

[Gui Cavalcanti]

Alright, tell me if the steps for this are off, or if I have it right, or what.

To determine the gear ratios for your drive train by solving for torque:

- Your maximum pushing power is going to be the friction coefficient of your materials on whatever surface FIRST decides on multiplied by the weight of your robot and whatever you are lifting. You want to increase this number a little bit in order to spin your wheels/treads without burning out motors and breakers.

- Shoot for a peak torque/power/efficiency to gear from on the motors charts, making sure it's within current limits and the like. Divide the torque of your robot overall by the radius of your wheels (drive pulleys for belts). Compare the torque from your motors to the torque from your robot with the wheel figured to get the ratio you're shooting for in the gearing. Remember that gears lose efficiency (10% I believe for every spur gear stage). Figure out the gears you need.

- Now that you have your maximum pushing power, you determine the torque needed to simply move the drive train around. To do this, you buy all the gears and assembly your drive train and pull it around with a spring scale (*NOTE* What happens if you have a worm gear in there...?) to determine the torque necessary to drive your robot. Once you find that out, you plug that torque into your equations and figure out what torque the motor is outputting, grab the speed at that torque, and plug the speed into the equation. Out pops your robot's ground speed with that type of drive train.

*gasp* Ok then. That seems easy enough. Now, my question is, how do you make a drive train based on speed? Don't you need to know the minimum torque to move your robot around, but you don't get that until you put gears on it? (Enter catch-22) How did other teams deal with this?

[Jon Lawton]

All this talk about having high friction coefficents... how can the u in F=uN be greater than 1? Isn't this why people lifted up the goals (increasing N increases F)?

[Lloyd Burns]

Gui:

You get it. I'd think you might even look at decreasing the friction coefficient ('mu') if you can stall your motors. Changing the gear ratios is also a possibility.

The process is indeed recursive, trying out and changing things and trying again. This is another advantage of the older teams: after a few times through, you have a fair idea of what the values will be :-). Much of engineering, involves approximating, whereby you fill in the equations with guesses, work it out, try something, measure the results, and plug in better guesses, which give better answers.

How much is your robot going to weigh ? Aren't you usually just a little over or a bit light ? Or do you calculate the weight of your completely CADded model before you build ? Even then, do you ever get it exactly right ?

Even flying VFR, you plot a course, and fly the heading, then you check where you are, and correct the heading, as you fly.

Jon:

"Mu > 1" merely means that the surfaces have more frictional force along the floor, say, than the weight (normal force) that is pushing the surfaces together. Mu = Fparallel/Fnormal. One text book had a pair at mu = 1.05. Lifting the goals is all about increasing Fnormal, while experimenting with different tire surfaces is about increasing 'mu', both in the interest of increasing Fparallel (Fhorizontal ?)

[Paul Copioli]

The quantitative (testing) way to determine Mu is by putting the material you want to test on an inclined surface you wnt to test it against. In our csae, this is the carpet. Keep lifting the incline until your material starts to slip. You can calculate Mu by taking the tangent of the angle from horizontal. Mu = tan(theta).

Note that the tan(90deg) = infinite. Note that it does not depend on the weight. Only the frictional force depends on weight Ff=Fn*Mu. Here is the derivation:

(1) Definitions:
Mu = coeficient of static friction
W = weight of material (block, for example)
Theta = Angle above horizontal
Fn = Force Normal to the incline
Ff = Frictional force parallel to the incline

(2) Ff = Fn*Mu (definition for Mu)
(3) From geometry sin(Theta)=Ff/W & cos(Theta)=Fn/W
(4) Substitute (2) for Ff and Sin(Theta)=Fn*Mu/W
(5) tan(Theta) is defined as sin(Theta)/cos(Theta) so:

Tan(Theta) = (Fn*Mu/W) / (Fn/W)

(6) Simplify and cancel terms ... Tan(Theta) = Mu

-Paul

[Andrew]

I've got to stick my $0.02 into this very interesting discussion!

As has already been mentioned, F = mu * N is a convenient mathematical fiction to capture some of the relevant behavior seen when two materials rub against each other.

It should really say, Fmax = mu * N, though. The "real force" of contact with the ground is that force which is required to keep the point of contact of the wheel moving at the same speed as the ground. This was referred to earlier as rolling friction.

When the required force exceeds Fmax, then the wheel slips.

Whenever you see an equation like y = a * x in physics, you know that someone is trying to capture the dominant behavior under most circumstances. In this case, Coulomb said, "The maximum force that two materials can exert depends on the normal force. Let's try a linear fit."

When we design our drive system, we don't worry about coefficient of friction. We design around the rolling contact force. If we need more traction, we can always put more material in contact with the ground (or cut treads into wheels...). If we need less traction, we can take some off (or put a more slippery material on the wheel). We treat the issues of desired drive system force and traction as separate design issues.

Andrew
Team 356

[Gui Cavalcanti]

Andrew -

What do you mean when you say you design around rolling contact force? Is there some formula to calculate this, and what exactly is it? (Rolling "friction" force, as mentioned above?)

Also, why don't you just take into account both things, or is that a bit more difficult?

[Lloyd Burns]

"Rolling Contact Friction" is the same concept as "Static Friction", since both are the friction between two surfaces in contact with each other and not in motion relative to one another.

Maximum Effective Drive Thrust = Maximum Friction (Traction). They cannot be solved at different times, they are solved simultaneously, in Newton's Third.

If your robot's thrust is less than the available friction, it is possibly inefficient (slow to accelerate), or is ineffective (run into a wall, and it'll stall). If your robot has more force with which to push than the maximum available friction, then any acceleration > mu*Fn/mass could spin your wheels (burning carpet?).

Uncanny how much Ike Newton knew about robots !

[Andrew]

Andrew -

What do you mean when you say you design around rolling contact force? Is there some formula to calculate this, and what exactly is it? (Rolling "friction" force, as mentioned above?)

Also, why don't you just take into account both things, or is that a bit more difficult?

As Lloyd said (one post down), rolling contact force is created by static friction. It will be less than mu*N. Once it reaches mu*N, you slip (the wheels start to burn the carpet) and you have a much lower force exerted by the wheels, since kinetic friction is less than static friction.

The formula which we use (and which many other people also use) for calculating the desired force of contact is:
F = (motor torque at desired operating point [usually max power])*(GR1*.9)*(GR2*.9)*...*(GRn*.9)/(wheel radius)

GR1...GRn are the gear ratios of our various stages in between the motor and the load. The .9 is there to account for efficiency. It could be lower! But, short of doing a bunch of experiments on the gear stages themselves, this should get us into a ball park for pushing force.

Why don't we account for both things at one?
I was taught that you should separate your requirements and figure out how to satisfy each independently.

We decide what speed/pushing power we want (this is a trade off unless you add in a transmission or more than two drive motors). We also decide what level of complexity we can handle in the design/fabrication based on available man power and work done in the off-season.

This year (and hopefully in the future), we had access to unlimited skyway wheels. Although our design choices were quantized around the available diameters (6", 8", 10", I think), we figured out how to "double up" the tires. So, we could get more friction by adding another tire to the back or less friction by removing tires, by cutting grooves circumferentially, or by plastering short lived teflon tape over part of the tire. We have been toying with the idea of making replaceable tires which fit over our wheels, perhaps using the Small Parts rubber and a mold... We'll have to see how next year's rules changes impact this problem.

Anyway, per Lloyd's comments, we adjust friction between too much slip and motor stall. I think there is also another plateau on the "too much friction" side where you may see bits in your motor/gear box getting broken (although we have fortunately not found this plateau). That probably could only occur if you add weight to your drive system and get back driven and have fresh fuses so that they don't pop until you see 100 amps.

I would be interested in hearing comments on when is too much friction a bad thing. I remember that Woodie commented that you want your tire to slip at maximum push. However, we have designed to blow our circuit breakers as the weakest component in our system. If someone has an idea, I'd like to hear it before we learn the hard way.
Andrew
Team 356

[Paul Copioli]

A lot of people have been talking about designing around max power of the motor. Be very, very careful!!! For example, the Chiaphua motor pulls 55 amps at max power. The 30 amp breakers will trip if you run at max power for long. We take a little more conservative approach and design all our drivetrains to be optimized around 28 amps. The max power ratings of motors can be deceiving: Look at the FP(this years), Chiaphua, and Drill motor curves only from 0 to 30 amps and see how similar they really are.

Food for thought.

-Paul