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-   -   Best Form Of Six Wheel Drive? (http://www.chiefdelphi.com/forums/showthread.php?t=56092)

kramarczyk 26-03-2007 13:29

Re: Best Form Of Six Wheel Drive?
 
Quote:

Originally Posted by M. Krass (Post 604254)
I've been thinking a lot lately about why some teams use a high traction wheel as the center pair in a 6WD. There may be some benefit -- again with respect to maneuverability -- but such an arrangement does not offer increased pushing force over 6WD with three pairs of identical wheels, so I'm not convinced that accomodating the differing wheels is worth the trouble.

Should have posted this earlier...

The center axle in a 6WD system carries a disproportionate amount of weight on it. In a uniformly distributed frame the center axle carries about 62.5% of the weight vs. 18.75% for the outer axles. Since traction is proportional to weight, increasing the friction coefficient at the center axles has the biggest bang for the buck in terms of pushing force while minimizing the reduction in turning due to high skidding losses.

Assuming a uniformly distributed 120lb bot with 6 skyways (0.7 CoF) then the pushing force is 84 lbs. Swap the center wheels to a ridgetop material (1.3 CoF I think) and the pushing force jumps to 129 lbs (120*.625*1.3+120*.1875*.7*2), an improvement of over 50%. In reality most bots are not uniformly distributed, but instead biased to the center which makes this change even more significant.

Vikesrock 26-03-2007 13:49

Re: Best Form Of Six Wheel Drive?
 
Quote:

Originally Posted by kramarczyk (Post 605182)
Should have posted this earlier...

The center axle in a 6WD system carries a disproportionate amount of weight on it. In a uniformly distributed frame the center axle carries about 62.5% of the weight vs. 18.75% for the outer axles. Since traction is proportional to weight, increasing the friction coefficient at the center axles has the biggest bang for the buck in terms of pushing force while minimizing the reduction in turning due to high skidding losses.

Assuming a uniformly distributed 120lb bot with 6 skyways (0.7 CoF) then the pushing force is 84 lbs. Swap the center wheels to a ridgetop material (1.3 CoF I think) and the pushing force jumps to 129 lbs (120*.625*1.3+120*.1875*.7*2), an improvement of over 50%. In reality most bots are not uniformly distributed, but instead biased to the center which makes this change even more significant.

Also wouldn't changing the center wheels to omnis allow other robots to translate you sideways in addition to being able to spin you around the center axis? Then they wouldn't even have to hit you on the corner, any part of either side of your robot perpendicular to the drive direction would do.

Madison 26-03-2007 13:50

Re: Best Form Of Six Wheel Drive?
 
Quote:

Originally Posted by kramarczyk (Post 605182)
Should have posted this earlier...

The center axle in a 6WD system carries a disproportionate amount of weight on it. In a uniformly distributed frame the center axle carries about 62.5% of the weight vs. 18.75% for the outer axles. Since traction is proportional to weight, increasing the friction coefficient at the center axles has the biggest bang for the buck in terms of pushing force while minimizing the reduction in turning due to high skidding losses.

Assuming a uniformly distributed 120lb bot with 6 skyways (0.7 CoF) then the pushing force is 84 lbs. Swap the center wheels to a ridgetop material (1.3 CoF I think) and the pushing force jumps to 129 lbs (120*.625*1.3+120*.1875*.7*2), an improvement of over 50%. In reality most bots are not uniformly distributed, but instead biased to the center which makes this change even more significant.

This makes sense, I think. I had neglected to consider that the weight would be distributed unevenly on 'typical' 6WD arrangements since they don't use all six wheels at a time and I presume that's where the 62.5% figure comes from.

Perhaps I'm conceptualizing something incorrectly or misunderstanding your math, but wouldn't it seem that the most torque the outer wheels could transmit before slipping in the scenario above is 120*.1875*.7 (for the pair), or ~15.75 lbs? Once the output torque of your gearbox exceeds that value, those wheels no longer transmit torque to the carpet. In such a case, the torque transmitted to the carpet then becomes only that of the center wheel and its higher coefficient of friction -- 120*.625*1.3 or ~97.5 lbs. This is still an improvement over a .7 coefficient of friction at all points of contact, but it's not as significant as 50%.

MrForbes 26-03-2007 13:57

Re: Best Form Of Six Wheel Drive?
 
Quote:

Originally Posted by M. Krass (Post 605200)
Once the output torque of your gearbox exceeds that value, those wheels no longer transmit torque to the carpet.

just because they're slipping, does not mean that they are not transmitting force. The coefficient of sliding friction is usally less than that of static friction, but is still much higher than zero.

mathking 26-03-2007 14:05

Re: Best Form Of Six Wheel Drive?
 
One other thing to consider is how your weight distribution characteristics change when you are pushing hard against something like another robot. When happens to a lot of well balanced robot is a tendency for the front to raise slightly, causing the center of mass to shift backward. This can be a quite pronounced shift if your robot rocks, even a small amount.

Such changes can make your back wheels (relative to the object being pushed) as or more important than center front wheels. We found this out the hard way last year. (Fortunately we were able to correct at the Championships.)

Madison 26-03-2007 14:07

Re: Best Form Of Six Wheel Drive?
 
Quote:

Originally Posted by squirrel (Post 605204)
just because they're slipping, does not mean that they are not transmitting force. The coefficient of sliding friction is usally less than that of static friction, but is still much higher than zero.

Yes, this is also true. Laziness always gets me into trouble. :)

Does anyone know the coefficient of friction for the KoP AndyMark wheels whilst they're sliding? I'm interested now in seeing how large of a benefit there is to varying your wheel types, though almost nothing will get me to abandon "rock-less" 6WD. :p

For 2008, I'm trying to stress efficiency of design and making the best use of resources in our drive rather than trying to win by overwhelming force. We've ended up running this season with a single CIM per side at 17.5:1 with the AndyMark KoP wheels and AndyMark omniwheels and have had very little trouble making our way around the field. It's been an eye-opener and a refreshing change from the prevailing mindset that seems to indicate that you won't survive a match without the biggest, meanest tread you can find and nineteen motors powering your wheels.

MrForbes 26-03-2007 14:29

Re: Best Form Of Six Wheel Drive?
 
That's an interesting way of looking at it.

My oldest son Steve (a freshman in engineering school now) went with us the AZ regional, and noticed that 60 was doing very well with a very light robot. It does have a powerful 6wd (rocking) drivetrain, but didn't have all the weight that we normally associate with strong pushing bots. He's trying to convince 1726 to try a few matches without ramps or bumpers, to see if the added agility of a lightweight bot will be more advantageous than the pushing power of the full weight version. The difference would be about 25 lbs.

I was thinking about the rocking thing some more...in this years game, it looks to me like rocking is not much of a concern at all, as long as you are hanging tubes by dropping or hooking them onto the spider foot. And with the rack moving around sideways as much as it does, if you are scoring some other way, vertical rocking would be a very small concern.

kramarczyk 26-03-2007 14:43

Re: Best Form Of Six Wheel Drive?
 
Andy his whole line of wheels in one place http://www.chiefdelphi.com/forums/sh...ad.php?t=50284 'course the KOP wheels were still a surprise to come, but based on the numbers he arrived at I think that using a sliding friction value thats ~90% of the should be reasonable. Sooooo...... If they are 1.0 static then ~.9 for the KOP wheels

Updating my math for the sliding friction...

with skyways...
120*.7*.9 = 75.6 lbs
swap center skyways to traction...
120*.625*1.3*.9+120*.1875*.7*2*.9 = 116.1 lbs

As far as rock from a dropped center wheel goes, with our 1/8 drop across a ~30" wheel base the max rock is ~0.5 degrees. We can see it on concrete, but not carpet... it basically is rockless. C'mon, drop that wheel I know you want too. Everybody's doing it. :cool:

BTW... does anybody have a good metrics for describing a drivetrain?
Pushes around 170 lbs (static)
15.62 fps High (analytic)
6.10 fps Low (analytic)
turning moment of 2400+ lbf*in (Is this a good metric for turning? Is current draw better?)

lukevanoort 26-03-2007 15:11

Re: Best Form Of Six Wheel Drive?
 
Quote:

Originally Posted by kramarczyk (Post 605237)
BTW... does anybody have a good metrics for describing a drivetrain?

For traction, I like to use the "V-Neun" unit of traction. One V-Neun is equivalent to the pushing force that year's kitbot's could produce, assuming it has enough torque to do so, and also assuming a weight of 133 lbs (robot + battery). So, in this years case, 1 V-Neun = 133lbs * .5 (for the amount of weight over the drive wheels) * 1.05 (I averaged AM's values for the kit wheel) = ~69.8 lbs of force. The reason I like this unit is that it gives some basis of comparison since, at least at competition, what matters more is how much traction you have relative to other robots, not the absolute numbers. So, instead of saying our robot has 139.7 lbs of pushing force, which leaves scouts trying to figure out if thats a lot/little/medium, I can say we have 2 V-Neuns, or twice the kitbot's pushing force, they can visualize that amount easier. Also, it takes into account yearly changes in the easiest to make drive system. For example our robot's 139.7 lbs of force would push a good number of teams in 2005, for example, but not so many number this year, since the base kitbot has more traction. (Amoung other factors, like the proliferation of IFI and AM traction wheels) This is reflected in the V-Neun number, this same robot has 3 V-Neuns of traction using the 2005 kitbot as the standard.

MrForbes 26-03-2007 15:19

Re: Best Form Of Six Wheel Drive?
 
there might be an error in your formula...the factor of 0.5 for weight distribution over the rear wheels ignores weight transfer when pushing. If you put a scale up against a wall, and have the robot push against that, you might discover the real number to be higher than you thought.

This weight transfer thing plays an important role in drag racing, cars that are set up right have the front wheels just skimming the track at launch, all the vehicle weight is on the rear wheels.

In robot pushing matches, a similar thing happens, but for a different reason. Draw a free body diagram of the robot, with the pushing point centered on the bumper zone, see what happens.

lukevanoort 26-03-2007 15:28

Re: Best Form Of Six Wheel Drive?
 
Quote:

Originally Posted by squirrel (Post 605264)
there might be an error in your formula...the factor of 0.5 for weight distribution over the rear wheels ignores weight transfer when pushing. If you put a scale up against a wall, and have the robot push against that, you might discover the real number to be higher than you thought.

This weight transfer thing plays an important role in drag racing, cars that are set up right have the front wheels just skimming the track at launch, all the vehicle weight is on the rear wheels.

In robot pushing matches, a similar thing happens, but for a different reason. Draw a free body diagram of the robot, with the pushing point centered on the bumper zone, see what happens.

How do you know a given team drives with the driven wheels in the rear? For example, our rookie year we used 2WD with the front wheels driven, then in '02 we drove the rear wheels. I just use a 50% weight distribution for two primary reasons 1) Its easier to calculate 2) The above issues with teams driving differently.

s_forbes 26-03-2007 15:28

Re: Best Form Of Six Wheel Drive?
 
Quote:

Originally Posted by kramarczyk (Post 605237)
As far as rock from a dropped center wheel goes, with our 1/8 drop across a ~30" wheel base the max rock is ~0.5 degrees. We can see it on concrete, but not carpet... it basically is rockless. C'mon, drop that wheel I know you want too. Everybody's doing it. :cool:

I'm going to have to agree with you on this statement. Dropping the center wheel of a drivetrain is the perfect fix for adding maneuverability to a six wheel drive robot, and the rocking that results is so small that it will have no effect on the performance of the bot. (I can't even see it on carpet)

As far as calculations go for drivetrains, I don't think we are able to make even close estimates to the amount of pushing force you can get out of a certain drivetrain. It depends on more than just the tread material; the geometry that you have your wheels set up in has an enormous effect on how much pushing power you get. Take team 842's drivetrain for example. They have four wheel drive, but because they have traction wheels set up the way they do they can get much more pushing force than a typical 6 wheel drive bot (240 lbs, actually). There are tradeoffs, though. Carl Hayden's bot can only push hard in one direction, and it can easily be spun by opponents.

After pondering this for a while, I've been thinking that an ideal set up may be what 703 has done with their drivetrain this year, with a large amount of wheels set up in a large arc. Does anyone know how much their robot can push?

MrForbes 26-03-2007 15:41

Re: Best Form Of Six Wheel Drive?
 
Quote:

Originally Posted by lukevanoort (Post 605272)
How do you know a given team drives with the driven wheels in the rear? For example, our rookie year we used 2WD with the front wheels driven, then in '02 we drove the rear wheels.

If you are just looking for a quick calculation that has little to do with how hard a robot can actually push, fine....but as you should be well aware, the estimate will be way off compared to real life. We know all about how powerful FWD robots are (not), having done that rookie thing also. :)

teenmisfit 26-03-2007 16:17

Re: Best Form Of Six Wheel Drive?
 
After 3 years of playing with the idea we finally got what team 501 found to be the best possible drivetrain for us. 4 close grouped 8in by 2in wide high traction wheels at the back and 2 8in powered omnis at the front. It makes the robots center of turning at the back but this helps us in this years game. The layout is powered with 4 small cims and 2 2speed transmissions. This design has already helped us win a regional as well as part of the GM Motors Industrial Motors award this year.

Otaku 27-03-2007 01:13

Re: Best Form Of Six Wheel Drive?
 
Quote:

Originally Posted by kramarczyk (Post 605182)
Should have posted this earlier...

The center axle in a 6WD system carries a disproportionate amount of weight on it. In a uniformly distributed frame the center axle carries about 62.5% of the weight vs. 18.75% for the outer axles. Since traction is proportional to weight, increasing the friction coefficient at the center axles has the biggest bang for the buck in terms of pushing force while minimizing the reduction in turning due to high skidding losses.

Assuming a uniformly distributed 120lb bot with 6 skyways (0.7 CoF) then the pushing force is 84 lbs. Swap the center wheels to a ridgetop material (1.3 CoF I think) and the pushing force jumps to 129 lbs (120*.625*1.3+120*.1875*.7*2), an improvement of over 50%. In reality most bots are not uniformly distributed, but instead biased to the center which makes this change even more significant.

Hmm. I've not studied these sort of things (I want to though), but that seems to make sense, except I would imagine that traction is much more than the weight pushing down on top of the wheel, although that would have a noticeable effect. You also have to keep in mind the material used as tread (Say, a plastic/perspex "tread" would be much less efficient than normal traction tread) and the driving surface matters too (such as carpet, hardwood floor, concrete, cobblestone, sand, etc.).

How then, would you calculate the traction of a robot whose weight is mostly on their right side, and the rest is evenly distributed front to back? It's a 4-wheel bot, though. (I'm asking because 675 is using 4 CIM's/4 56mm Gearboxes @ 12:1 and we can push another bot AND the rack with normal traction tread on carpet, without losing traction)

Quote:

Originally Posted by Vikesrock (Post 605199)
Also wouldn't changing the center wheels to omnis allow other robots to translate you sideways in addition to being able to spin you around the center axis? Then they wouldn't even have to hit you on the corner, any part of either side of your robot perpendicular to the drive direction would do.

Assuming that the chassis is not a rocker chassis and also that the other 4 wheels on the bot are normal traction (which would be quite unintelligent, omnis in the center but tractions on the corners... wtf?), then maybe. If the chassis was a rocker, then yes, and probably with ease.

The optimal omniwheel/traction wheel combination is omnis at the corners, tractions in the center. Rocker chassis or not, it offers the turning radius of omnis, with the traction of, well, tractions.

I say if you're going to try to make an interesting drive system, go with mecanum. I think that a bot that moves in all directions is pretty cool.

Quote:

Originally Posted by squirrel (Post 605204)
just because they're slipping, does not mean that they are not transmitting force. The coefficient of sliding friction is usally less than that of static friction, but is still much higher than zero.

Hmm, good point.

Anywho, if you break traction in a pushing fight and you're sliding around, it doesn't matter the torque you're transmitting to the carpet, it just matters that it's below the number you want. ;)


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