Are six wheels better than four?

[MrForbes]

Originally Posted by Hachiban VIII

Unpowered idlers of any type are a horrible waste of normal force.

In most cases, yes. However, this is a special case, in that the wheel does not need to support any of the robot's weight, it is being used only for instrumentation. If the speed sensing wheel is attached to a hinged arm, and carries only it's own weight, then it should not affect robot performance noticeably.

[Ragnarokae]

Driven (motor is powering the wheel) wheels will provide forward force.
Undriven (not powered by motor) wheels will distribute the weight more.

Traction depends on pounds per square inch of wheel that is in contact with the ground.

For example: you have 4 wheels, one in each corner. Robot = 120lb. Assuming that the robot is perfectly balanced, each wheel is carrying 30lb.

Example 2: 6 wheels, three on each side (left/right) with a 120lb robot gives 20lb per wheel.


More drive wheels still decrease weight/area ratio, but make up for it (possibly) with the added power.

[=Martin=Taylor=]

Originally Posted by squirrel

In most cases, yes. However, this is a special case, in that the wheel does not need to support any of the robot's weight, it is being used only for instrumentation. If the speed sensing wheel is attached to a hinged arm, and carries only it's own weight, then it should not affect robot performance noticeably.

Fat lot of good it will do as an encoder if it isn't pushing on the floor with significant force. If it slips, it won't be sensing anything meaningful.

[MrForbes]

I suppose someone who wants to use an extra wheel for instrumentation could calculate how much force it needs to contact the floor with, to prevent it slipping under maximum robot acceleration conditions (it would slip if it takes more torque to overcome it's own inertia, than will be transfered to it by the frictional force of it's contact with the arena). I expect it won't take (much) more than it's own weight. I haven't done the calculations....I'll leave that for the students

[RyanCahoon]

Originally Posted by squirrel

I suppose someone who wants to use an extra wheel for instrumentation could calculate how much force it needs to contact the floor with, to prevent it slipping under maximum robot acceleration conditions (it would slip if it takes more torque to overcome it's own inertia, than will be transfered to it by the frictional force of it's contact with the arena). I expect it won't take (much) more than it's own weight.

The point being, any normal force used to creating friction between the instrumentation wheel and the ground is normal force that's not able to be used to create friction between the drive wheels and the ground.

On a different note:

Our team is still considering different drivetrain options, but we've nearly decided on a six-wheel, dropped center wheel method if we follow a differential drive design, as the ability of the robot to rock slightly will allow the robot to not be as affected by transverse friction when trying to turn. We've tested a 4-wheel differential drive and while accelerating was not as difficult as one might think, turning was not as responsive as we would like. We're waiting for two more rover wheels to test out the 6-wheel drive.

--Ryan

[shannieliz]

my thinking is that it will just give you more speed.
meaning, you may go sliding more on the crater floor.
now i have no clue if this is true or not, this is just my opinion from what i've heard.

[MrForbes]

Originally Posted by RyanCahoon

The point being, any normal force used to creating friction between the instrumentation wheel and the ground is normal force that's not able to be used to create friction between the drive wheels and the ground.

Yup. Making this decision might involve comparing the control advantge to be gained by implementing a good traction control algorithm, vs. the loss in tractive force by putting a small part of the robot's weight on the unpowered sensor wheels.

[Molten]

squirrel: The only thing I want to know is why you need normal force to create the traction control algorithm? You can do it without such a reading. All you really need is to do a little bit of math (calculus at most) and you should be able to come up with it as a simple formula. A little programming should allow you to use this formula for the traction control.

[Rick TYler]

Originally Posted by Hachiban VIII

Unpowered idlers of any type are a horrible waste of normal force.

I would normally agree with this, but in this game there is so little friction available you might be talking about the difference between nil and .95(nil). I think comparing the rotational rate of an idler wheel compared to that of the powered wheel might be useful in managing wheelspin.

[MrForbes]

Originally Posted by Molten

squirrel: The only thing I want to know is why you need normal force to create the traction control algorithm? You can do it without such a reading. All you really need is to do a little bit of math (calculus at most) and you should be able to come up with it as a simple formula. A little programming should allow you to use this formula for the traction control.

There are different ways to implement traction control. Having a sensor wheel is one of them. What type of system you want to use is up to you..maybe the sensor wheel type could work better or be easier to program, or eliminate the need for other sensors. If you can do it a different way, great!

[rl1806]

If you had a six-wheel drive with a lowered center wheel, wouldn’t the weight of the trailer only allow the center and rear wheel touch the ground? If the center and rear wheel are the only ones touching the ground, the front wheel would be doing nothing to help drive the robot.

[Ferdi17]

Another thing to think about is that the trasverse coefficent of friction is greater than the inline so not only dose it matter how many wheels but if they are turned.

[Deathnmasses]

It would not give you more traction. In fact it would do better to have less wheels so that you could get more weight onto your motorized wheels.

[wireties]

I think encoders in the drive train are good enough to see slippage in the direction you are trying to go (just look for unnatural acceleration of a single wheel). Detecting slippage normal to the direction of travel is much tougher. I'd try a gyro before an optical mouse kind of thing. This is a tough surface to get an optical mouse to work well.

A couple of posts talk about the increase in traction if the floor is not smooth. This might be true but it is not because there is more surface area or a change in the coefficient. The reason is because there is now something to push against (bump/s in the material). There is an additional force term in the direction of travel.

[DMetalKong]

Originally Posted by squirrel

I suppose someone who wants to use an extra wheel for instrumentation could calculate how much force it needs to contact the floor with, to prevent it slipping under maximum robot acceleration conditions (it would slip if it takes more torque to overcome it's own inertia, than will be transfered to it by the frictional force of it's contact with the arena). I expect it won't take (much) more than it's own weight. I haven't done the calculations....I'll leave that for the students

According to this link, you shouldn't have problems with instrumentation slippage done in this fashion, due to the fact that you can use a higher COF wheel.