Re: 4" vs. 6" + 8" Diameter Wheels
 

Wouldn't this vary somewhat significantly based on the material of the wheel as well as any lightening? I'm sure we could calculate worst case scenarios (solid wheel of aluminum billet) but I'm not sure it would actually be terribly useful. Perhaps assign real materials to the AM plaction/performance wheels and use your CAD program of choice to computer the moments?

Re: 4" vs. 6" + 8" Diameter Wheels
 

Roughly speaking, it will increase with the square of the radius. (I say roughly, because the different sizes aren't scaled versions of each other; some things, like hub size and rim thickness, tend to be driven to a large degree by other design constraints.)

Re: 4" vs. 6" + 8" Diameter Wheels
 

Yes, of course. That's why I said "comparable" in my post. Compare a 6" wheel that you would make for your robot to a 4" wheel that you would make for the same robot.


Exactly. That's why I said "comparable" :-)

Re: 4" vs. 6" + 8" Diameter Wheels
 

The major reasons have been stated, but the biggest for me is the redundancy in gear ratio of having a big wheel. Gearing down then using a big wheel is redundant. The wheel size is part of your gear ratio. If you have a 2 stage reduction gearing down 12:1 with an 8 inch wheel, you could achieve the same ratio with one stage 6:1 and a 4 inch wheel. The weight savings add up, lighter wheel, less reduction, therefore fewer gears etc...You will also get better acceleration (relative to the gear ratio) with a smaller wheel because, in a general comparison, smaller wheels have a smaller moment of inertia.

(Big wheels do have some advantages in tackling rough terrain, but we don't see that too much in FIRST).

Re: 4" vs. 6" + 8" Diameter Wheels
 

Worst case scenario, solid wheel made of 6061 Aluminum 1 inch thick. (I picked a type of Aluminum)

I used (m*r^2)2 for the moment of inertia.

Once you start lightening it could get different since the amount of material you have to leave on the edge of the wheel doesn't scale with size so we would start treating it as a tick walled tube + however many support spokes you have. This starts getting a little more complicated but it is still doable. I am attaching an excel spreadsheet that (unless I did something stupid) should compute a rough estimate of the rim and spoke style wheels.

Edit: I had originally forgotten the width of the spoke so I was finding the area instead of volume, this has been fixed.

Re: 4" vs. 6" + 8" Diameter Wheels
 

Less reduction != easier to produce.

Re: 4" vs. 6" + 8" Diameter Wheels
 

Yes, useful exercise for students.

The numbers above can't be used, of course, to answer the original engineering question, to wit: all other things being equal, how much acceleration advantage does a 4" wheel vehicle have over a 6" wheel vehicle (where both the 4" and 6" wheels have been reasonably individually optimized for the vehicle).

Does anybody have CAD models (from this or past years) of 4" vs 6" drive train options which included enough detail to calc the mass and moment of the wheels?

Re: 4" vs. 6" + 8" Diameter Wheels
 

We don't need a CAD model to do a rough estimate. We could assume 140 pounds (close to full size robot + battery + bumpers). We could assume a somewhat reasonable gear ratio (4 CIMS geared down 18:1 with 6 inch wheels and 12:1 with 4 inch wheels). Theoretically these are geared to move at the same speed, the problem is we don't have estimates on the exact weight savings with a smaller gear ratio and smaller wheels (someone else might be able to help estimate the weight savings).

Re: 4" vs. 6" + 8" Diameter Wheels
 

For us we use the smallest practical wheel we can get away with. Why?

1. Less gear ratio required for same pushing force at the wheel.

2. Move the wheels out further front and back.

3. Less linear feet of tread to deal with (less chance of losing tread)


This really adds up to less weight overall. Smaller wheels = less weight. Less gear ratio usually = less weight.

For us, 4" seems to be the sweet spot.

Re: 4" vs. 6" + 8" Diameter Wheels
 

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. :ahh:

Re: 4" vs. 6" + 8" Diameter Wheels
 

Of course.

Yes, that is the problem.

Re: 4" vs. 6" + 8" Diameter Wheels
 

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.

Re: 4" vs. 6" + 8" Diameter Wheels
 

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.

Re: 4" vs. 6" + 8" Diameter Wheels
 

Our 4x2" wheels this season are .8 lb-in^2 and our 1" ones are .467 lb-in^2.

Modifying the 1" wide wheel to a 6" diameter and changing no other factors (face thickness, spoke width, etc...) results in 1.999 lb-in^2. The design appears to weak for such a diameter though, and would likely need more material, increasing that number. These are both very light compared to the available COTS wheels.

Re: 4" vs. 6" + 8" Diameter Wheels
 

No, you were quite clear.


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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.