Friction, traction, torque - oh my...
 

After a week of robot camp, a couple of months searching around FIRST sites, and a few weeks at competitions, I still haven't found anyone who can link this all together for me: what is the link between friction, traction, and torque?

First off, torque. Motors have a neat, pretty little 4-line chart saying "if my motor is moving at this rpm it's generating this torque which is so and so efficient pulling this much current at 12 volts". I understand the individual bits and pieces of it, and how they all connect, but what I don't understand is how you use it. Do you calculate the torque of your machine and then gear to that? Do you calculate the speed of your machine and then gear to that? How do you get to "peak torque" on a robot?

Related to that is how on earth do you link torque to moving your robot? Granted, the little 4-line charts mentioned earlier will tell you your torque at certain conditions, and I assume you scale up torque through gearing and mechanical advantage systems (arms, etc). However, how is torque related to moving your drive train and your available pushing power? Where do friction and traction fit into these calculations?

On the note of friction and traction - I've heard a lot of teams pulling the "f = uN" stunt recently. That doesn't work. Tell me if this logic is correct - carpet is not a surface. It's a conglomeration of threads strung together, and so you can't accurately test any sort of friction. Also, if you have things like file cards (which have next to no traction on any other surface) they dig into the threads and latch on, and don't really count as traction...

Now, static friction I can see coming into account from a wheel turning. However, how do you calculate this on a robot? Do you disengage your motors and pull your robot across a carpet on a spring scale?

Any answers to these questions would be greatly appreciated, as well as the random comment of "hey, that question has nothing to do with it, what are you talking about?"

Thanks!

Somebody's signature says "Build first, ask questions later" I completely agree with this. Sure you could calculate all that stuff and come up with the winning combination but why bother. Just make something that looks right and it will probably work fine. If not, modify it. More speed, bigger tires. More torque, smaller tires. More traction, wider tires. It's far easier to just make it and fix problems rather than calculate it all first. There are real-world factors and problems that no amount of calculations could ever forsee.

I hate calculations.

Hope this helped but I know it probably didn't.

To see how much friction our bot had, we did exactly what you said. We had two guys stand on the carpet and small pulley system and a spring scale to yank the robot. Not _exact_ results but it gives you the basic idea whether your robot is going to do much moving.

actually, on the traction idea, smaller (As in thinner) tires are better, i think. when i lived upstate, my dad always put snow tires on in the winter. they were much thinner than the regular tires. this might not be true on carpet though, following gui's line of thought on how the carpet really isn't a surface...

That's thanks to the weight of the car. Smaller tires-allowed more force on a smaller surface area = better grabbing, thanks to crushing the snow below = allowed snow tires a good, hard non-slippery surface to grab on to. I'm assuming that he had some chains on thoes wheels too? This same concept applys to tha' chains! (I think this theory is right, but can't guarantee it! :D)

And the carpet....That is a different story. I would have to agree w/ Gui, and say it isn't a surface. I think of it as really tough sand. It is constantly moving, thanks to it being made up of tons of tiny fibers! For this, the more surface area you have, the better you are! (The more you have your robot weight spread out and less force that 1 fiber has to take and no bending back!! like in most Jeep off-roading, or any off-roading for that fact, have huge tires.) And That is where you get traction from. Your robot pushing off these little fibers and propelling it's self. Torque is how much "pushing power" you have in your robot, but you loose speed to gain torque. And friction. With out this, we all would be stuck. (no pun intended) That is what enables our wheels/tanks to drive on this ground. With out any traction, it won't move! (literally)

I know i probaly made you more confused. If you have some Q's, just e-mail me. Redwingvksm@msn.com

lol, no chains (it wasn't that far north). but i do see how the two situations are completely different.

For better traction, my team sawed grooves into the tires to give it better traction. The results were visible. We just taped 3 hack-saw blades together and clamped down the tire and sawed away. (We being I...heh...I got the fun job of doing it on 8 tires :))

OK - I guess it's time.......
 

I just caught your post - If you can wait a little while I'll clean up the whole answer and put it here. Anyone who saw the national finals is aware of the power, speed, traction, etc. of our little robot - thanks to 4 years of development of the drive train. Gary Jones - our drive train "Guru" - and I have been slowly putting together a paper that explains this and we would like to share it with the FIRST community. It's very detailed, based on theory, test data (including the pull test platform at nationals last year), and competition results that correlate pretty well. I guess your question was what I needed to get off TDC and actually finish it.

A few teasers till then:

It's all about power, not torque.
If you stall the motors before you slip the treads, you WILL burn them out.
Minimize friction losses in the drive train to minimize power loss.
The best traction we found is VERY expensive, but second best is pretty close and pretty cheap.
Believe it or not, f really does = uN (more or less) and carpet is indeed a surface.

Re: Friction, traction, torque - oh my...
 

Carpet is indeed a surface but F=uN is not the correct equation for dealing with rolling friction. Take drag racers, for example. In a physics lecture, they will explain to you that if you have a box of uniform substance that has more surface area on one side and less on another, it will take it the same time to slide down a hill on either side. Why do drag racers have big tires then? Surface area does matter when it comes to wheels.

~Tom Fairchild~

P.S. Carpet is a surface but things like the file cards just have very high u's. If you experiment with it, you can get fairly consistant static friction and kenetic friction measurements. Once again though, that doesn't work for rolling friction.

Friction, Traction and Torque
 

Friction, Traction, & Torque

Friction comes in many forms: static, sliding, and rollling are but a few. Sometimes we call an energy loss 'friction' rather tham explain exactly what it is, because the loss acts just like a friction. Remember that friction is a human construct to make it easier to explain Nature - it probably doesn't work that way at all. The 1940's cartoon joke about "aerodynamically, a bee cannot fly - but the bees never read the book, and they just go on flying" really says that the jokester's aerodynamics are woefully inadequate for the study of bee flight.

Static friction is sometimes thought of as the microscopic mountains of roughness on one surface fitting into the microscopic valleys of roughness on another surface, locking the two together. Often we get more friction by roughing up the surfaces - but not for the shiny surface of leather belts turning smooth drive pulleys on lathes and other factory machines in the days when electric motors were really costly (the days of the steam engine in Dean's 'house').

Sliding friction could be envisioned as the two surfaces bouncing along, the peaks of their microscopic mountains banging into one another as the peaks of the moving surfaces fly over the valleys. We could then evision putting something (like oil) between the surfaces to float the surfaces away from one another to lower the friction.

Static friction, by definition only occuring when the one surface does not move relative to the other, is often more than the friction when the same surfaces slide (move past each other). A rolling wheel's point of contact is stationary on the winter road, until the brakes siezes it, and prevents turning: then it slides over the road. A bookcase, pushed to one side by little brother to re-arrange the furniture, resists movement (static friction with the floor is high). Little brother's shoes slip on the floor: the friction he can muster is not equal to the friction between the case and the floor. Big Brothe comes along, and with the extra weight on his shoes helping F=uN to make more friction, his shoes do not slip (static). The bookcase resists (static), then with a bit of a bang, starts to slide (!). Big brother now doesn't need to push as hard to keep the case sliding.

Rolling friction is a funny kind of friction: It is due to a bike wheel of a railway wheel having to become flat under the load it carries. Since it would be round if it carried no load (bike on a rack), this means it had to change shape (deform). The energy to change shape had to come from somewhere. It is easy to calculate the energy loss if you conjure it as a friction; hard, if you try to examine the dynamic plastic deformation of the rail wheel.

Torque is a force acting in a circle, or turning, often within a shaft. A given amount of torque is constant whether it exerts a large force tangentially, close to the centre of rotation, or a small force farther out. (T = F(tangential) x radius) Hold a weight at the end of your arm, and think of the arm being 'torqued' around your shoulder. The weight is pulled down, at right angles to your arm's length, that is, tangential to the circle your hand goes in around your shoulder. More weight requires more pull from your muscles; different muscles if somebody tries to lift the weight and you resist. If you notice, this example uses forces (gravity, muscles) which are not torques, but because their lines of action do not not go through the pivot point (shoulder) they generate torques to 'spin' the arm one way or the other. A motor continuously generates a tangential pull around the armature, a short distance away from the axis of the armature's (rotor's) turning: what you'd observe in the shaft is a turning force, a torque.

If you turn a shaft with the torque, and put a wheel on the shaft, and the wheel has friction with a surface, the torque of the shaft will turn the wheel (or not) and a force of the wheel pushing the surface (and vice versa) will be observed. A small diameter wheel exerts a large force at its circumference, a larger wheel (or gear) exerts a smaller force as the radius increases. (F = T / radius)

Traction depends on the friction between the surfaces. The push depends on the torque delivered to the levers that push the one surface past the other. If your robot has adequate static friction, and adequate torque, it will move in the desired direction. If it has insuffienct friction with the floor, it will slide.

-=-=-=-=-=-=-=-=-=-=-=-=

BTW, on a macroscopic level, sand is a surface (try jumping of a cliff onto a beach) and so is the carpet. Each can be examined microscopically or treated as a surface.

Don't confuse reality with human tricks to make things easier to work with. The physics of Aristotle asked why a thrown javelin keeps moving: they figured all motion that wasn't the object returning to its 'natural' place (where its nature demanded it should be) was "Violent" motion, which, by definition, required a force to continue. The javelin was not in "Natural" motion, so a force must exist to keep it going horizontally. They invented a force to keep it going (the air molecules, closing in around the end of the javelin, to avoid a vacuum [the 'void'] banged into the end of the javelin, propelling it forward. Following this idea, Boeing should make the aft end of their fuselages come not to a point, but a nice flat circle.

On the other hand, Newton and Gallileo saw that a moving object will remain moving in a straight line unless a force prevents this. They had to invent something to slow things, and turn things. They invented what we call friction. We use rubber tires for more of it, and oil for less of it. We use the friction idea because it works better than the violent motion option. Being pragmatists, we don't care if it's true, if it works.

Maybe you will create the next construct - maybe we'll all talk of demons grabbing things, and of oil offered as a sacrifice to make the demons go away.

Sorry this is so long, I hope it helps.

Lloyd, you did a pretty good job of summarizing the "rubber hitting the road" end of it - I was leaning toward the same description of friction; if I plagiarize any of your post in our paper I'll be sure to give you credit.

On another note, has anybody seen the teaching unit from FIRST Place on motor power? Goes through pretty good detail on correlation between torque, speed, efficiency, current, etc. Just curious - I wrote it and haven't gotten any feedback from anyone who has used it.

Thanks to everyone that responded so far, I'm learning a lot (everything from rolling friction to why people's robots didn't work as they intended... ;-) )

Lloyd, can you walk through a physics problem combining all of these concepts into one? This is what I think it would look like:

First, find your place on the motor torque/efficiency/speed chart. Then, re-calculate the torque by multiplying it times the gear ratios in your drive train. Then, convert the radius of your wheel into whatever units of length your torque is in (ft-lbs, oz-in, n-m) and divide torque by the covnerted length. Now you have the torque of your wheel.

Now lets see if we can cheat with the units a little bit... you would then proceed to say (if you used the metric newton-meter) "For every meter I move, I can push with such and such newtons of force". You would then take that and subtract the losses to rolling friction, and would get a pretty good estimate of the pushing force of your robot.

To see if you can push another robot, you would take the force you have available for pushing and compare to either their "static" friction, if they are standing still and not pushing their motors, or you would do the same calculations for their drive train and compare the pushing forces - the robots would move in the direction of whoever has the greater pushing force, and the force of them moving in that direction would be equal to the difference of forces in their drive trains.

Yes, no, maybe?

I did take a year of physics, but right now I'm flying by ear.

Gary, I look forward to your White Paper !!!! Question, What value of Mu did you use and how did you obtain it? I have always just taken a swag of what I figured our wheel mat'l to the carpet would be.
And a HELLO!!! from Kyle to you!!

Thx
Dennis

OK - time to spill the beans a little more.....

Dennis - we measured 1.7 with the green BrekoFlex belts - under load on the robot by pulling against a spring scale (the traction pit we brought to Disney); 200# pull with a 115# robot. The belts were slipping on the carpet since we had plenty of motor power. We measured the same on team 144's robot with the same belts at stall (they were underpowered). We had measured 1.8-1.9 by just dragging the belts across the carpet before we built the robot.
Here's the big surprise from the pit results: Second best was pneumatic tires! 2 teams (457 and 121) pulled 180# or a u of about 1.4.

Gui - for your physics problem you need to start at the other end.

Say, hypothetically, you wanted to push rolling goals full of balls and other robots around. You want to maximize your pushing power (we'll address speed later), and you can assume the other team is doing the same. Only 2 things impact whether you will push or be pushed: friction factor between you and the carpet, and downward force. Take the friction factor times your weight plus any downward force from lifting up on the goal, and thats the most you can push.

Now, ignoring internal drive train friction losses for a moment, the maximum pushing force times the wheel radius divided by the total gear ratio through to the motor equals the torque that the motor will see at any point after the belt or wheel starts slipping (if you have multiple motors, divide by the number of motors). That tells you where you are on the motor curve. In design it works the other way around; you pick the point on the motor curve where you want to operate (generally either max efficiency or max power) and determine your gear ratio from the same equation.

Nothing about speed yet, right? That's a fallout of your chosen design point and the losses in the system. If you have a higher gear ratio than necessary you will drive slower but push the same (remember, the pushing force is limited at the carpet, not by the motor) - the tread will slip first. The motor operating curve times the gear ratio times the wheel radius gives you the speed.

So why a 2 speed gearbox? We were (and usually are) current limited. We have one gear ratio for speed (under no load) and a different gear ratio for pushing. Actually it turned out this year with the Chips that we had plenty of power to run in high gear most of the time. Finding the optimimum gear ratios is a lot of field testing, but the method above gets you in the ballpark.

Ooops - lunch is over; gotta get back to work. Hope this helps. Stay tuned for the full version.

I've been looking forward to that white parper, too, Gary <crack the whip - crack the whip> :D

...could it be ready by MiM?
Eric

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 :(

Gary, You missed 217
 

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).

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.

Ooops
 

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

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

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

Alright, let me get this straight, once again.
 

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?

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)?

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 ?)

Mu can be infinite (in theory)
 

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

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

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?

"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 !

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

Max Power???
 

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

Whoops! Originally we designed around max power. Then we discovered that we only had 30 amps to work with and designed around that number.

Andrew
Team 356