6WD vs 8WD

[JamesCH95]

Originally Posted by Lil' Lavery

Before we get carried away, let's examine this statement.

Let's start with the simple equation F=ma. Based on this, when you reduce your weight (and consequently mass) and keep your force the same, your acceleration should increase (which agrees with the bolded statement). However, when examined further, we find that the force does not always remain constant in the FRC world when your mass changes.
The force we're speaking about in this application is the force exerted by the drive wheels. This is limited by friction between the wheel surface and the playing surface. The frictional force is calculated as F_f=μN, where μ is the coefficient of friction and N is the normal force. Subbing this back into the first equation, we now have:
μN=ma
In most situations, the normal force is going to be a function of the mass, usually just N=mg when on a flat surface. Putting this back into the previous equation we have:
μmg=ma
Which simplifies to:
μg=a
In other words, in friction limited drivetrains (ie drive systems that have enough power to "spin out" their wheels), acceleration is not governed by the mass of the robot, as the mass plays into both sides of the equation.

Good point, and well worded.

However, consider that the driving force (i.e. output torque of the motors) decreases with speed. At some speed the motors will lack the torque required to spin the wheels, barring extreme cases. At that point, the situation will become the F=ma situation we all know and love and the lighter robot will be able to accelerate faster.

Another interesting tidbit is that the wheels' coefficient of friction may not be constant, but rather vary with contact pressure. Assuming that the coefficient of friction is constant the heavy and light robots will accelerate at the same rate in a friction-limited case. However, if the tread/playing surface is sensitive to contact pressure then the lighter robot will have the advantage, all else being equal.

Also note that a drive-train does not require power to break traction, but rather requires torque, specifically torque on the drive wheels.

[Lil' Lavery]

Originally Posted by JamesCH95

Good point, and well worded.

However, consider that the driving force (i.e. output torque of the motors) decreases with speed. At some speed the motors will lack the torque required to spin the wheels, barring extreme cases. At that point, the situation will become the F=ma situation we all know and love and the lighter robot will be able to accelerate faster.

Another interesting tidbit is that the wheels' coefficient of friction may not be constant, but rather vary with contact pressure. Assuming that the coefficient of friction is constant the heavy and light robots will accelerate at the same rate in a friction-limited case. However, if the tread/playing surface is sensitive to contact pressure then the lighter robot will have the advantage, all else being equal.

Also note that a drive-train does not require power to break traction, but rather requires torque, specifically torque on the drive wheels.

Good points (especially the last one, I have to be more careful about my technical diction around here ).

[KrazyCarl92]

Originally Posted by Lil' Lavery

The frictional force is calculated as F_f=μN, where μ is the coefficient of friction and N is the normal force. Subbing this back into the first equation, we now have:
μN=ma

The equation is actually F_f <= μN, which brings us back to the question of traction limitted drive trains. Because 67 used traction wheels that would be driven and in contact with the ground at all times, I highly doubt that they would be fraction limited. Let's assume that the coefficient of friction of their wheels is .8 (conservative estimate), then their force of friction would be .8 x 90 <= 72 lbs. That means that they would have up to 72 pounds of force moving them forward. This lead to the acceleration of a 90lbs robot up to:
mass = 40.82 kg
force = 320.27 N
acelleration = 320.27N / 40.82 kg = 7.846 m/s2 = 25.74 ft/s2

Given that no robots can get up to 25.74 ft/s in a single second (or at all for that matter), the acceleration is clearly limited by gearing/motors, not the traction of the wheels. So in this case 67 would in fact accelerate about at about 4/3 the rate of the 120 robots that they compete against.

[Lil' Lavery]

Originally Posted by KrazyCarl92

The equation is actually F_f <= μN, which brings us back to the question of traction limitted drive trains. Because 67 used traction wheels that would be driven and in contact with the ground at all times, I highly doubt that they would be fraction limited. Let's assume that the coefficient of friction of their wheels is .8 (conservative estimate), then their force of friction would be .8 x 90 <= 72 lbs. That means that they would have up to 72 pounds of force moving them forward. This lead to the acceleration of a 90lbs robot up to:
mass = 40.82 kg
force = 320.27 N
acelleration = 320.27N / 40.82 kg = 7.846 m/s2 = 25.74 ft/s2

Given that no robots can get up to 25.74 ft/s in a single second (or at all for that matter), the acceleration is clearly limited by gearing/motors, not the traction of the wheels. So in this case 67 would in fact accelerate about at about 4/3 the rate of the 120 robots that they compete against.

You raise valid points about what is limitting acceleration (and, as you and JamesCH95 alluded to, it's often elements of the motor/drivetrain itself), but you have a pretty significant flaw in your argument. 25.74 ft/s² isn't equal to 25.74 ft/s. Just because you're accelerating at rate x, doesn't mean your maximum speed is x.

Another point is that radial wheel [drivetrain] acceleration and linear robot acceleration are two different matters, though often linked together in FRC scenarios.

[Hawiian Cadder]

i think that 6wd is a more agile drivtrain, it turns a bit easier, and is also a bit faster accelerating, due to less stuff needing to be accelerated. however an 8wd is generally beefier, more precise. i think they tend to be a bit slower turning, but also harder to turn. i noticed this year that we had little difficulty spinning some top teams with 6 wheel tank. however any 8wd teams we attempted to defend were much more difficult to knock of course, notably 610 and 691. if it were a game like overdrive, or this year, where speed kills, 6wd would probably perform slightly better. if it were a game like 2010 or aim high, i might choose 8wd due to the way it is less likely to be pushed.

[Andrew Schreiber]

Originally Posted by Hawiian Cadder

i think that 6wd is a more agile drivtrain, it turns a bit easier, and is also a bit faster accelerating, due to less stuff needing to be accelerated. however an 8wd is generally beefier, more precise. i think they tend to be a bit slower turning, but also harder to turn. i noticed this year that we had little difficulty spinning some top teams with 6 wheel tank. however any 8wd teams we attempted to defend were much more difficult to knock of course, notably 610 and 691. if it were a game like overdrive, or this year, where speed kills, 6wd would probably perform slightly better. if it were a game like 2010 or aim high, i might choose 8wd due to the way it is less likely to be pushed.

Look carefully at how an 8wd with the center two wheels dropped is built. Specifically their wheel base. Then look at 6wd. Then apply that thought to this white paper. Does the math support your hypothesis? If not what other factors could be coming into play?

[apalrd]

Originally Posted by Grim Tuesday

I dont know if anyone saw 610's 2010 drive. It was a 6 wheel, but the middle wheels were mobile (I dont know the details), so that they could either move the wheel up or down to traverse bumps. It was amazing.

<image snipped>

We had a similar system.
The ride height was controlled by the claw, and is automated (more drop when turning, less when not). The driver must request a bump-cross, but when not bump-crossing, it handles ride height on its own.

From Danial Ernst's pictures from the State Championship:


The rear chassis (the part we attach the bot-bottoms to) is mobile, pivoting around the rear axle (omni wheels), and containing the drive motors (4 CIM + AM shifters), and most of the control system. The gearbox is chained directly to the middle and rear wheels, and the rear wheels are chained to the front wheels via two jack shafts per side (over the top above where the chassis floats). In software, the normal claw movement (including the bump-crossing and flat centered positions) is handled with the claw axis pot, and center dropping is handled by a string pot in the chassis which measures drop.

[IKE]

So, I was very inspired by some of the physics guys up above, and have a challenge:
Assume 16m dash (approx. 54 feet- robot length), and COF of 1.0. Also assume a 4 cim power-train at 300 Watts/CIM.
What weight is your bot so that you are just barely traction limited the moment you strike the other wall? Is this less than 4 cims & a battery?

[KrazyCarl92]

Originally Posted by IKE

So, I was very inspired by some of the physics guys up above, and have a challenge:
Assume 16m dash (approx. 54 feet- robot length), and COF of 1.0. Also assume a 4 cim power-train at 300 Watts/CIM.
What weight is your bot so that you are just barely traction limited the moment you strike the other wall? Is this less than 4 cims & a battery?

First of all THANK YOU for putting it in meters to spare me the conversions necessary to use watts and meters. However, there would be a missing variable in the necessary calculations if I am not mistaken, because we would need to know a time that it would take to travel said 16m. But I can solve for this with time as a variable anyway.

P_T = 4P_CIM
P = 1200 watts (~80 - 95 % efficient depending on number of gears, transmission, and other variables)

F_f = mgµ = m*9.8*1 = 9.8m

Work = Force * Distance = 9.8m * 16 = 156.8*m

time = Work / Power = 156.8*m / (between 960 and 1140) watts = between .138*m and .163*m seconds

so to solve for mass (in kg) it would work out to between 6.135 * t to 7.246 * t depending on the efficiency.

Of course, this all assumes that the robot would be constantly accelerating at the maximum possible value (no slipping/static friction) and that the robot would be able to keep getting faster and faster, but I'm pretty sure it would be impossible to gear 4 cims without a transmission such that they provided exactly maximum static frictional force and still could reach rotational speeds that would allow it to continue accelerating...would probably need really really small wheels, but not too sure about that one.

[h1n1is4pigs]

i personally like 6 wheel it is lighter and well tested. as a matter of fact we built a 6wd drivetrain this off season and it preforms excellent, very maneuverable and fast. video form testing this afternoon: http://www.youtube.com/watch?v=KwOr-Mpl8DA

[IKE]

Originally Posted by KrazyCarl92

First of all THANK YOU for putting it in meters to spare me the conversions necessary to use watts and meters. However, there would be a missing variable in the necessary calculations if I am not mistaken, because we would need to know a time that it would take to travel said 16m. But I can solve for this with time as a variable anyway.

P_T = 4P_CIM
P = 1200 watts (~80 - 95 % efficient depending on number of gears, transmission, and other variables)

F_f = mgµ = m*9.8*1 = 9.8m

Work = Force * Distance = 9.8m * 16 = 156.8*m

time = Work / Power = 156.8*m / (between 960 and 1140) watts = between .138*m and .163*m seconds

so to solve for mass (in kg) it would work out to between 6.135 * t to 7.246 * t depending on the efficiency.

Of course, this all assumes that the robot would be constantly accelerating at the maximum possible value (no slipping/static friction) and that the robot would be able to keep getting faster and faster, but I'm pretty sure it would be impossible to gear 4 cims without a transmission such that they provided exactly maximum static frictional force and still could reach rotational speeds that would allow it to continue accelerating...would probably need really really small wheels, but not too sure about that one.

Assuming constant acceleration at said traction limit, D=1/2*a*t^2.

[KrazyCarl92]

Now that I think about it, the size of the wheel really wouldn't matter...

Although I would usually love to sit down and work out the rest if this problem, at this point I feel like that would be selfish of me because I'm sure my team would rather have me well rested for kickoff which outweighs my own personal satisfaction of solving the problem

[548swimmer]

The final speed, as well as the acceleration are functions of the angular velocity and acceleration of the drive shaft (respectively), as well as the wheel radius.
Think about a wheel with radius = r and a driveshaft with some torque T. We then sum the torques about the center of the wheel and get Fa*r < Fsfmax*r (theta is ninety so it goes away sin(90)=1). Cancel the r, plug in uFn for Fsfmax, and you have your max Fa to not slip. Use T = F * rsin(theta) and you get the max torque from the driveshaft to be traction limited. From there it's just some basic ratios back to the torque output of the motor and that ratio is your desired gear ratio.
Now, this assumes that you have a perfectly efficient gearing system, and that your wheel has no mass and therefore no moment of inertia to slow the acceleration, but it gets you close.
For speed based, do the same basic thing only with rpm instead of torque.

[Chris is me]

@Adam: I misinterpreted that post as saying that in order to be traction limited you needed to have a free speed of 27 feet per second or something. My bad.

[Jared Russell]

Originally Posted by IKE

So, I was very inspired by some of the physics guys up above, and have a challenge:
Assume 16m dash (approx. 54 feet- robot length), and COF of 1.0. Also assume a 4 cim power-train at 300 Watts/CIM.
What weight is your bot so that you are just barely traction limited the moment you strike the other wall? Is this less than 4 cims & a battery?

1) How long does it take to go 16m when you are traction-limited the whole way?

Ffriction = 1.0 (CoF) * Wrobot (because Ff = mu * Fnormal)
a = Ffriction / Mrobot (because F=m*a, so a=F/m)
a = G (because Wrobot/Mrobot = G, gravity)

D = .5 * a * t^2
t = sqrt(16 m * 2 / G )
t = 1.81 seconds

2) How fast are you going just before you hit the wall?

v = a * t
v = G * 1.8 seconds
v = 17.7 m/s

3) At this ultimate speed, we are barely traction limited, so:

P = F * speed (1 Watt = 1 N * m/s)
F = Wrobot
speed = v = 17.7 m/s
1200W = Wrobot * 17.7 m/s

Wrobot = 67.8 N = 15.2 lbs

So yes, less than 4 CIMs and one battery.