4" vs. 6" + 8" Diameter Wheels

[James Critchley]

The rotational inertia is of the form

K*M*R*R.

In the limit, a small wheel is a solid object and K = 1/2. Large wheels put more mass at the rim and begin to approximate a hoop, K = 1.0. Not only does the mass get larger, but the ratio of mass at a distance tends to do this too (K goes up).

The wheel must also get heavier because the stresses to perform similar maneuvers are higher.

Interestingly, radius has nothing to do with this discussion of acceleration and drops out of the equation.

T = I * alpha

The gear ratio must change to keep the same ground force and free speed, so force is constant not torque.

F * R = I * alpha

Pluggin in the inertia

F * R = K*M*R*R * alpha

Then relating rotational acceleration to linear acceleration

F * R = K*M*R*R * A / R

Then dividing through by radius gives

F = K*M * A

If M goes up faster (proportionally speaking) than K, then acceleration must go down. Per prior logic, K and M generally move upwards together when scaling the same "spoke" type design.

These effects should be in the noise compared to the reflected inertia of the motor (through all of those gears) and the associated losses. Also keep in mind that most teams use chain drives to keep the wheels on each side moving together. The inertia of the chain is more signifficant? And it has the oposite effect, causing large wheel drive trains to have a lower effective rotational inertia. I expect all of this to be in the noise... now I'm just waiting to be surprised by the results Ether's calculations.

[Ether]

Originally Posted by James Critchley

now I'm just waiting to be surprised by the results Ether's calculations

Well I wouldn't want to disappoint :-)

Looking at the wheel without the vehicle (as you did):

I*alpha is equal to the net torque on the wheel. If "tau" is the driving torque on the wheel and F is the floor reaction force responsible for the linear acceleration of the wheel, then a free-body analysis of the torques and forces on the wheel gives:

tau - F*R = I*alpha

tau - (M*A)*R = (K*M*R*R)*(A/R)

tau = = M*A*R*K + M*A*R

A = tau/(M*R*(K+1))



For the analysis of wheel plus vehicle, see attached PDF.

a = tau/(I/r +r*(Mw+Mv/4))

a is vehicle acceleration;
tau is driving torque on wheel from gearbox;
I is wheel moment;
r is wheel radius;
Mw is wheel mass;
Mv is vehicle mass.

[James Critchley]

Sorry Ether, I'm just not being clear.

The torque available "at the motor" and "by the motor" is the same for all wheel sizes. However, keeping the same top speed and low end torque requires a different gear box (this is a common assertion by others in this thread). So the torque applied TO THE WHEEL must go through a different gearbox, and will then be a different torque. Having so designed all gearbox-wheel combinations, at stall the force at the exterior rim of any wheel will actually be the same (no losses). So the applied torque as indicated is actually F_stall * R which was used correctly.

Then I'm spinning the wheel under no load... I don't mean to set a bad example, but there is no need for a free body diagram, just the applied load (analytical dynamics). My "linear acceleration" term is also somewhat abusive, but it relates to the same setup (e.g. rad/ss converted to ft/ss). Clearly if the vehicle weighs more it will also accelerate slower. That doesn't help prove the assertion that "the vehicle accelerates slower BECAUSE the moment of inertia is higher for larger wheels." This statement was one of several independent reasons to use smaller wheels. This simple setup really lets you isolate everything.

I've shown that radius drops out of the equation entirely and that only mass and mass distribution ratio "k" contribute. In the strictest sense, I've actually disproved this assertion. YAY ME!!! If the mass and mass distribution ratio "k" of the wheel remain constant (e.g. use a lighter material as the wheel gets larger) then the moment of inertia will actually go up (as it MUST with R squared) but there is absolutely no performance penalty!!! In fact the inertia can go up and you can increase performance using a lower mass or k value.

None of this is practical, the product of mass and k REALLY should go up in any reasonable manufacturing process. So I did not bother to argue the causality bit.

I worked out the equation because I suspected that something neat would happen to the radius, and it did.

[Ether]

Originally Posted by James Critchley

Sorry Ether, I'm just not being clear.

No, you were quite clear.

The torque available "at the motor" and "by the motor" is the same for all wheel sizes.

check

keeping the same top speed and low end torque requires a different gear box

check

So the torque applied TO THE WHEEL must go through a different gearbox, and will then be a different torque.

check

Having so designed all gearbox-wheel combinations, at stall the force at the exterior rim of any wheel will actually be the same (no losses). So the applied torque as indicated is actually F_stall * R which was used correctly.

If a wheel of mass M_w and moment I and radius R is sitting on the floor and is free to accelerate, then the force F that it exerts on the floor when a torque tau is applied is not equal to tau/R.

It is equal to tau*M_w*R / (M_w*R² + I). That approximately equals tau/R only if I is negligible compared to M_w*R².

That wasn't made clear in your post.


If four of the above wheels are on a vehicle of mass M_v which is free to accelerate, then the force F with which each wheel pushes against the floor is given by

F = tau*R*(4*M_w+M_v) / (R²*(4*M_w+M_v)+4*I)

The above approximately equals tau/R only if I is negligible compared to R²*(M_w+M_v/4).


...


The acceleration is given by a = 4*tau*R / (R²*(4*M_w+M_v) + 4*I)

Letting I=K*M*R² this becomes:

a = 4*(tau/R) / (M_v+4*M_w*(K+1))

Compare the acceleration of two vehicles, one with wheels of mass M_w1, K= K₁, radius R₁, applied torque tau₁, and vehicle weight M_v1, and the other with wheels of mass M_w2, K= K₂, radius R₂=2*R₁, applied torque tau₂=2*tau₁, and vehicle weight M_v2. Then

a1/a2 = (M_v2+4*M_w2*(K₂+1)) / (M_v1+4*M_w1*(K₁+1))

... and R does not appear in the ratio, as you said.

[Gdeaver]

There is a difference in the contact patch. Also the carpeting is 3 dimensional. The wheels sink into the carpet. So how do the different size wheels affect the contact patch for traction and the ability to turn?

[Jared Russell]

Various teams have done experiments to measure CoF with different wheel configurations. The data that I've seen (it was given to me by another team, so I will let them post it if they choose to) was quite surprising - wheel diameters and widths can make a big difference! Definitely worth experimenting with.

[James Critchley]

Still not clear.

Originally Posted by James Critchley

Then I'm spinning the wheel under no load...

No load means no load. The robot is on jack stands. If radius drops out here, then it's not coming back as the model becomes more complicated.

The bottom line is that an increase in rotational inertia of large wheels alone is not a valid reason to choose smaller wheels. It is an increase in total mass of the robot or (EDIT) increase in rotational inertia attributed to (END EDIT) mass distribution (EDIT) and mass (END EDIT) of the wheels that contributes to any performance degradation. These and all of the other excellent reasons cited in this thread push for smaller wheels.

[Ether]

Originally Posted by James Critchley

It is an increase in total mass of the robot or mass distribution of the wheels that contributes to any performance degradation.

YMMV, but I think it's clearer to explain it the following way:

The equation for vehicle acceleration in the PDF file attached to this post can be re-arranged as follows:

a = (tau/r) / (I/r² + (1/4)*(M_v+4*M_w))

or, substituting K*M_w for I/r²

a = (tau/r) / (K*M_w + (1/4)*(M_v+4*M_w))


Since tau/r remains unchanged when you change the wheel radius (if you change the gearing accordingly), what remains is

a) the total vehicle mass M_v+4*M_w, and

b) the mass of the wheel times the mass distribution factor of the wheel (K*M_w)

Since neither (a) nor (b) is likely to remain the same when you change wheel size, there will be a change in vehicle acceleration.

The equation also gives a way to estimate the change.

[Paul Copioli]

All,

While the calculations in this thread appear correct (I haven't checked them against mine thoroughly, since mine are done a little differently), they really aren't the main reason for going to small wheels on our team.

It is all about the huge constraint placed on us in FRC for max robot weight. While the wheel weight effect on overall robot acceleration is low (if you do the math you will see), it is significant to help meet the weight restriction in FRC. The max weight restriction is really the driving force for us to go to smaller wheels. Second is the smaller packaging. Robot acceleration is not on our list of reasons.

Paul