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Lots of Wheels and F = u x N

Posted by Ernie P at 1/25/2001 9:41 AM EST


Engineer on team #548, RoboStangs, from Northville High School and Robert Bosch.



Hi all you gearheads out there,
I'm an RF electronics guy ... so all this traction stuff is just another exercise in vector analysis to me. My old physics classes told me that F = u x N, or propulsion force equals the coefficient of friction times the force normal to the surface.

So, why does everyone want to add 356,215,492 wheels to the robot???? When you add another wheel, the force is distributed so your N is reduced by 50% .... net result is no increase in propulsion force. I can see for real trucks that they need to distribute the load on roads and may be happier on soft surfaces, but why are more wheels better for us???

Your words of wisdom to enlighten my meager knowledge will be accepted as a ray of light in my void of ignorant darkness.
Thanx, Ernie P


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Re: Lots of Wheels and F = u x N

Posted by Matt Berube at 1/25/2001 11:26 AM EST


Engineer on team #49, Delphi Knights, from Buena Vista High School and Delphi Automotive.


In Reply to: Lots of Wheels and F = u x N
Posted by Ernie P on 1/25/2001 9:41 AM EST:



Your classical physics model is right for a hard whel on a hard surface.

More wheel surface does seem to give better traction however.

I think what happens with these robots is that wider wheels or more wheels get more traction because the wheel treads are actually "hooking into" the carpet.

IMHO

Matt B.
T49
"that OTHER Delphi team"

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More comments on this please!

Posted by Eric Reed at 1/25/2001 2:23 PM EST


Coach on team #481, NASA Ames / De Anza High School, from De Anza High School and It could be you!.


In Reply to: Re: Lots of Wheels and F = u x N
Posted by Matt Berube on 1/25/2001 11:26 AM EST:



I imagine I'm not the only one who is terribly interested in this topic. My physics book says that regardless of the material, surface area has nothing to do with friction. But it makes sense to me that if there is some kind of perpendicular force (does this make sense?), such as texture in the wheel vs texture in the carpet, that surface area might make a difference.

Please, more teams post your opinion, experience!

Eric - 481


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Re: Lots of Wheels and F = u x N

Posted by Michael "Special K" Krass at 1/25/2001 3:24 PM EST


Other on team #271, Mechanical Marauders, from Bay Shore High School and Verizon.


In Reply to: Lots of Wheels and F = u x N
Posted by Ernie P on 1/25/2001 9:41 AM EST:



>

What you've written above is the formula for the force of friction alone. The total propulsion force, as you call it, is actually the sum of the forces in the horizontal direction. That's made of several components.

Force of Wheels On Carpet = a vague definition for the forces at work, but it can and will get ridiculously complicated.
Force of friction = m * g * u, where m is the mass of the robot, g is (9.8 m/s^2), and u is mu, the coefficient of friction.

Treating the entire robot as a point object (which just makes life easier, even though it's not entirely accurate), we see that:

the force of the wheels on the carpet - the force of friction = propulsion force

>

The normal force is affected by the addition of wheels only in that you're increasing the mass of the robot, and consequently, increasing friction. You're also adding additional friction to the system along the new bearing surfaces that additional wheels provide.

However, the additional surface area provided by additional wheels can be advantageous in that in increases tractive effort, can sometimes just look really cool.

It's just a matter of weighing the pros and cons. Additional wheels can mean more traction, in more places. That's good.

But, it can also make a robot a lot harder to turn, increase friction a bit.

That's why a tank tread design seems to be so popular. It allows for a large surface area to be in contact with the floor, with increased tractive benefits that far outweigh the added friction (in most cases . . . ask me about that some other time). But, with a curved or pointed bottom surface, it also facilitates fast rotation at a single point, rather than along the 3' length of the robot.

Anyway, I hope my ramblings are a bit useful. Really, I do, because I really dislike physics and wouldn't want my efforts to be for naught.

~ Michael ~

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Re: Lots of Wheels and F = u x N

Posted by Ernie P at 1/25/2001 5:18 PM EST


Engineer on team #548, RoboStangs, from Northville High School and Robert Bosch.


In Reply to: Re: Lots of Wheels and F = u x N
Posted by Michael on 1/25/2001 3:24 PM EST:



Michael,
Thank you for the info. I'm with ya up to the tank tread part, then I lose the theoretical interpretation.

Why isn't a tank tread, which has more contact points, just distributing the same robot wieght over more contact points, and resulting in no increase in "propulsion force"?? Is it because the individual points each sink into the carpet slighly, which provides a small force vector component which is no longer normal to the flat surface of the carpet??

Once again my lowly grey matter would be infinitely expanded by your words of superior erudition.
Thanx, Ernie P

: >

: What you've written above is the formula for the force of friction alone. The total propulsion force, as you call it, is actually the sum of the forces in the horizontal direction. That's made of several components.

: Force of Wheels On Carpet = a vague definition for the forces at work, but it can and will get ridiculously complicated.
: Force of friction = m * g * u, where m is the mass of the robot, g is (9.8 m/s^2), and u is mu, the coefficient of friction.

: Treating the entire robot as a point object (which just makes life easier, even though it's not entirely accurate), we see that:

: the force of the wheels on the carpet - the force of friction = propulsion force

: >

: The normal force is affected by the addition of wheels only in that you're increasing the mass of the robot, and consequently, increasing friction. You're also adding additional friction to the system along the new bearing surfaces that additional wheels provide.

: However, the additional surface area provided by additional wheels can be advantageous in that in increases tractive effort, can sometimes just look really cool.

: It's just a matter of weighing the pros and cons. Additional wheels can mean more traction, in more places. That's good.

: But, it can also make a robot a lot harder to turn, increase friction a bit.

: That's why a tank tread design seems to be so popular. It allows for a large surface area to be in contact with the floor, with increased tractive benefits that far outweigh the added friction (in most cases . . . ask me about that some other time). But, with a curved or pointed bottom surface, it also facilitates fast rotation at a single point, rather than along the 3' length of the robot.

: Anyway, I hope my ramblings are a bit useful. Really, I do, because I really dislike physics and wouldn't want my efforts to be for naught.

: ~ Michael ~


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F = u x N, At Most - the "normal" force

Posted by Dodd Stacy at 1/25/2001 10:27 PM EST


Engineer on team #95, Lebanon Robotics Team, from Lebanon High School and CRREL/CREARE.


In Reply to: Re: Lots of Wheels and F = u x N
Posted by Ernie P on 1/25/2001 5:18 PM EST:



Ernie,

F = u x N is the MAXIMUM force that can be developed between the robot and the floor, parallel to the floor, before the wheels slip. Hook a spring scale to your dead robot and drag it over the carpet, and that tells you what "u" (mu) is, using the entire weight (m x g) of your robot as "N."

Now, are you able to exploit this MAXIMUM possible horizontal force for propulsion? Depends. Is any of the robot's weight being supported by wheels which are not powered by the motors? If the answer is yes (eg: 2 wheel drive, 4 wheels on the ground), then the maximum PROPULSION force you can deliver to the ground is limited to F = u x Ndw, where "Ndw" is the weight carried by the drive wheels. Unpowered wheels that help support or balance the robot cannot develop propulsion force. Powering all the wheels that carry the robot's weight is how you guarantee that you produce as much propulsion force as possible. This is why people use 4 wheel drive.

(This is especially important when robots reach out and pick up a weight, such as lifting a goal off the ground. The robot's wheels now support the weight of the robot and the goal, and the wheels closer to the goal will bear a lot more weight than the wheels at the other end. (It might even tip over!) The robot has more total weight that it has to drive around or climb up the bridge with. Which wheels are powered? Which ones should be powered? If ALL the wheels are powered, we know that we aren't wasting any possibility of delivering the maximum PROPULSION force.)

OK, so why have 356,215,492 wheels (actually, I think 6 is the maximum I've seen suggested)? Dodd is saying that 4 powered wheels (3, actually) would support any kind of weight (till it tipped over - other physics) and not waste any available PROPULSION force. What's this 6 stuff? What's wrong with 4 wheel drive (or 3)?

Errr, sorry, gotta steer. Joe and others (us too!) like to turn the wheels, swerve, swoop, holonomically omni-navigate, etc. You can, too, next year (or later) if you're masochistic enough (or worse). For now, here's the deal. Imagine steering the bot as rotating without moving forward or backward. That's tank steering, skid steering. For reasons not important here, the bot tends to steer/rotate around its center of mass, which is probably near the middle of the bot.

Look at your poor 4 wheel drive bot. You're driving the right side wheels forward and your left side wheels backwards to turn/rotate left. The two front wheels are sliding directly to the left and the two rear wheels are sliding directly to the right. UNGH (how they DO that?) This is called converting battery power into carpet warmth.

Gee, wouldn't it be nice if almost all of the bot's weight were carried by just two wheels side by side with the robot's center of mass? These two wheels would just roll forward and backward when the robot rotates and wouldn't have to slip on the carpet at all. AND we would have almost maximum PROPULSION FORCE, because almost all of the weight is being carried on these two driven wheels.

But how to keep the 2 wheeled robot from falling on its face? Talk to Dean about the Ibot. For now, try putting extra wheels at the corners of the bot's footprint, just to lightly balance the bot if it wants to tip forward or backward. Imagine mounting these 4 wheels 1/8" higher than the middle two. When you turn/rotate, and these "corner" wheels need to skid sideways, they do so easily because there is almost no weight on them. So now we have a 2 wheel drive/6 wheeled platform that steers/rotates easily (because most of the weight is carried on the middle wheels) and has great PROPULSION force capability.

If the robot doesn't change its weight distribution during the match, you're done. If it has to go up a bridge/ramp incline and its weight shifts onto the rear wheels, or if it picks up a weight at one end, you better be driving the newly loaded wheels or you're giving up PROPULSION force capability. That's why 6 wheel drive.

All of these remarks relate to making sure that normal force (weight) is carried by POWERED wheels and that most of the weight is carried near the middle. Treads are powered along their full length, so they don't waste any PROPULSION force capability if the loading shifts forward or aft. If they have a bit of "rocker," then they are also easy to turn/rotate.

...TO BE CONTINUED (too ____ long already)

Dodd


: Once again my lowly grey matter would be infinitely expanded by your words of superior erudition.
: Thanx, Ernie P

: : >


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Re: F = u x N, At Most - "myewww"

Posted by Dodd Stacy at 1/25/2001 11:20 PM EST


Engineer on team #95, Lebanon Robotics Team, from Lebanon High School and CRREL/CREARE.


In Reply to: F = u x N, At Most - the "normal" force
Posted by Dodd Stacy on 1/25/2001 10:27 PM EST:



OK, Ernie. Wake up - here comes the second installment. Now we know why it's important to maximize N on the DRIVEN wheels (or "drive interface surfaces&quot to have the greatest PROPULSION force. Now, how do we maximize u (mu), and why DO those guys use tank treads?

Think of mu, the friction coefficient, as relating to the tendency of one surface to mechanically interlock with another surface on a microscopic scale. Teeth in teeth. We all know how to get great robot traction on a FIRST carpet. Cover your wheels/treads with those nifty little stiff wire brushes in the SPI catalog they use to clean gradoo out of files. The little wire needles sink down into the carpet, make a great mechanical interlock, and produce a marvelous mu.

There are lots of other ways to do it, but they all involve carrying the robot's weight onto the carpet over a small surface area of actual contact (needles, wires, spikes, etc) so that they sink in and mechanically interlock with the carpet. Terrific. We have a very high mu and all of the robot's weight on the drive wheels and we can push anything that's in the way. Look out!

Until the carpet says uncle. It has a certain shear strength (ability to withstand horizontal force, per square inch of surface), and then it lets go. In the end, it is the weak link, the limiting factor. How do we deliver more horizontal PROPULSION force if the carpet can only take so much force per square inch before it shreds? Apply the force over more square inches. If you want the ultimate in straight line traction without damaging the carpet, make a 36" long by 30" wide tank tread drive with each cleat of the track being a wire brush. (Just don't try to turn.)

By way of closure, the game is all about balancing the factors that limit performance. There is the torque/force that the drive system can deliver to the traction interface (motor and gearing). There is the friction coefficient that can be developed at the traction interface (surface micro geometry). There is the normal force that can be applied to the traction interface when the weight shifts as the robot goes about ALL of its operations (chassis layout and drive train). And, finally, there is what the carpet can sustain before it fails.

It makes no sense to develop enough traction, without damaging the carpet, to win an ox pull and then gear your motor so tall as to stall when climbing the bridge. Conversely, it's a lost cause to gear way down for pushing power and then run smooth hard drive wheels with no weight on them. I apologize for belaboring what is known and obvious to many. I'm hoping to dispel some of the confusion I've read here from both the question and the answer side.

Dodd


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Beginnings of a white paper (eom)

Posted by Joe Ross at 1/25/2001 11:23 PM EST


Engineer on team #330, Beach Bot, from Hope Chapel Academy and NASA/JPL , J&F Machine, and Raytheon.


In Reply to: Re: F = u x N, At Most - "myewww"
Posted by Dodd Stacy on 1/25/2001 11:20 PM EST:



Thanks!

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Author! Author!

Posted by Joe Johnson at 1/26/2001 8:34 AM EST


Engineer on team #47, Chief Delphi, from Pontiac Central High School and Delphi Automotive Systems.


In Reply to: Beginnings of a white paper (eom)
Posted by Joe Ross on 1/25/2001 11:23 PM EST:



As the crowd often yells on opening night, I think I just heard the FIRST online community calling for Dodd to come out and take a bow.

Well done.

Will you make a whitepaper of it? Put in a few simple sketches and viola! you have a best seller on you hands.

Seriously, I hope you consider it. It would be a shame to let the information you just put together molder in the forum archives. It deserves a better fate.

Think about it.

Joe J.


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You the man Dodd!!!! You the Maaaan!!!

Posted by Ernie P at 1/26/2001 10:21 AM EST


Engineer on team #548, RoboStangs, from Northville High School and Robert Bosch.


In Reply to: Re: F = u x N, At Most - "myewww"
Posted by Dodd Stacy on 1/25/2001 11:20 PM EST:



Hey Dodd,
Wow .... thank you for your response. The weak link being "in the carpet" was the hole in my understanding of the traction system. By distributing the weight to multiple wheels we don't change the propulsion force much ... But, we do increase our ability to transfer that force without damaging the surface. This makes complete sense.
Thanx, Ernie P

p.s. Good luck on your white paper!!!!!

: OK, Ernie. Wake up - here comes the second installment. Now we know why it's important to maximize N on the DRIVEN wheels (or "drive interface surfaces&quot to have the greatest PROPULSION force. Now, how do we maximize u (mu), and why DO those guys use tank treads?

: Think of mu, the friction coefficient, as relating to the tendency of one surface to mechanically interlock with another surface on a microscopic scale. Teeth in teeth. We all know how to get great robot traction on a FIRST carpet. Cover your wheels/treads with those nifty little stiff wire brushes in the SPI catalog they use to clean gradoo out of files. The little wire needles sink down into the carpet, make a great mechanical interlock, and produce a marvelous mu.

: There are lots of other ways to do it, but they all involve carrying the robot's weight onto the carpet over a small surface area of actual contact (needles, wires, spikes, etc) so that they sink in and mechanically interlock with the carpet. Terrific. We have a very high mu and all of the robot's weight on the drive wheels and we can push anything that's in the way. Look out!

: Until the carpet says uncle. It has a certain shear strength (ability to withstand horizontal force, per square inch of surface), and then it lets go. In the end, it is the weak link, the limiting factor. How do we deliver more horizontal PROPULSION force if the carpet can only take so much force per square inch before it shreds? Apply the force over more square inches. If you want the ultimate in straight line traction without damaging the carpet, make a 36" long by 30" wide tank tread drive with each cleat of the track being a wire brush. (Just don't try to turn.)

: By way of closure, the game is all about balancing the factors that limit performance. There is the torque/force that the drive system can deliver to the traction interface (motor and gearing). There is the friction coefficient that can be developed at the traction interface (surface micro geometry). There is the normal force that can be applied to the traction interface when the weight shifts as the robot goes about ALL of its operations (chassis layout and drive train). And, finally, there is what the carpet can sustain before it fails.

: It makes no sense to develop enough traction, without damaging the carpet, to win an ox pull and then gear your motor so tall as to stall when climbing the bridge. Conversely, it's a lost cause to gear way down for pushing power and then run smooth hard drive wheels with no weight on them. I apologize for belaboring what is known and obvious to many. I'm hoping to dispel some of the confusion I've read here from both the question and the answer side.

: Dodd


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as usual, you are on the money

Posted by Ken Patton at 1/26/2001 7:24 PM EST


Engineer on team #65, The Huskie Brigade, from Pontiac Northern High School and GM Powertrain.


In Reply to: F = u x N, At Most - the "normal" force
Posted by Dodd Stacy on 1/25/2001 10:27 PM EST:



excerpt from Dodd's excellent post:

: Gee, wouldn't it be nice if almost all of the bot's weight were carried by just two wheels side by side with the robot's center of mass? These two wheels would just roll forward and backward when the robot rotates and wouldn't have to slip on the carpet at all. AND we would have almost maximum PROPULSION FORCE, because almost all of the weight is being carried on these two driven wheels.

: But how to keep the 2 wheeled robot from falling on its face? Talk to Dean about the Ibot.

END excerpt


Or, come see the Huskie Brigade's robot (Team #65) in the pits at Ypsi, Grand Rapids, or Nationals. )


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2 Legged Huskie Dog?

Posted by Dodd Stacy at 1/26/2001 9:52 PM EST


Engineer on team #95, Lebanon Robotics Team, from Lebanon High School and CRREL/CREARE.


In Reply to: as usual, you are on the money
Posted by Ken Patton on 1/26/2001 7:24 PM EST:




: : But how to keep the 2 wheeled robot from falling on its face? Talk to Dean about the Ibot.


: Or, come see the Huskie Brigade's robot (Team #65) in the pits at Ypsi, Grand Rapids, or Nationals. )


I can hardly wait to see what you guys are doing this year. I love this part.

Dodd


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Maybe Ken is pulling OUR leg!

Posted by Joe Johnson at 1/26/2001 10:02 PM EST


Engineer on team #47, Chief Delphi, from Pontiac Central High School and Delphi Automotive Systems.


In Reply to: 2 Legged Huskie Dog?
Posted by Dodd Stacy on 1/26/2001 9:52 PM EST:



I am parsing Ken's sentences pretty closely.

It is clear that he INTENDS for us to think his robot
is going to balance on two wheels ala the I-bot.

But... I know two things: 1) This is a VERY difficult
task that I seriously doubt a computer with a 25msec
maximum sample rate can accomplish reliably. 2) Ken is
a very clever person that is not only a great engineer
but has a fine command of the language as well (he
didn't win the Woodie Flowers Award for whistling Dixie ;-)

I put these two facts together to believe that the
Huskie Brigade (our cross town rivals / friends) are up
to something sneaky, but not an I-bot like drive.

Rest assured, I will be in the pits checking them out.

Joe J.



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Re: Maybe Ken is pulling OUR leg!

Posted by Dodd Stacy at 1/26/2001 10:10 PM EST


Engineer on team #95, Lebanon Robotics Team, from Lebanon High School and CRREL/CREARE.


In Reply to: Maybe Ken is pulling OUR leg!
Posted by Joe Johnson on 1/26/2001 10:02 PM EST:



: I am parsing Ken's sentences pretty closely.

: It is clear that he INTENDS for us to think his robot
: is going to balance on two wheels ala the I-bot.

: But... I know two things: 1) This is a VERY difficult
: task that I seriously doubt a computer with a 25msec
: maximum sample rate can accomplish reliably. 2) Ken is
: a very clever person that is not only a great engineer
: but has a fine command of the language as well (he
: didn't win the Woodie Flowers Award for whistling Dixie ;-)

: I put these two facts together to believe that the
: Huskie Brigade (our cross town rivals / friends) are up
: to something sneaky, but not an I-bot like drive.

: Rest assured, I will be in the pits checking them out.

: Joe J.

Funny, I had that same sense. I've got it! 2 wheels midships and dog sled runners just kissing the carpet fore and aft. But wait, how will they make it onto the bridge, or off? Like I said, I can't wait to see.

Dodd





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Unread 06-23-2002, 10:37 PM
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Re: Maybe Ken is pulling OUR leg!

Posted by Joe Ross at 1/27/2001 2:56 AM EST


Engineer on team #330, Beach Bot, from Hope Chapel Academy and NASA/JPL , J&F Machine, and Raytheon.


In Reply to: Re: Maybe Ken is pulling OUR leg!
Posted by Dodd Stacy on 1/26/2001 10:10 PM EST:



: Funny, I had that same sense. I've got it! 2 wheels midships and dog sled runners just kissing the carpet fore and aft. But wait, how will they make it onto the bridge, or off? Like I said, I can't wait to see.

I can think of a few ways to do it. I posted earlier about my team's experience using this type of drive. With a few modifications, I believe that one would be able to get up the ramp.

Joe Ross
Beach Bot, Team 330


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