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Ok, so my boss is trying to make a car turntable. In essence, he has a two boards that sit atop a rotating ring. He wants to put two wheels at the end of each board (8 wheels total). He thinks that you can angle wheels properly (2d) such that they can be flat (axis is parallel to the floor) and the whole contraption move smoothly.

Take a look at this picture (Edit: copied below - Mark) alt text

Ok, so the question, will this design move smoothly? My guess is "no." I reason that if the wheels are unbanked then they will move in a straight line when driven, and the ring will realign them periodically to its circular path (probably causing a skip at the wheel as the stress is released). He thinks it will just work.. Please use technical terms with very easy layman definitions so I can reference wikipedia and texts and learn more about this.

Sorry, for the utter lack of useful physics terms.

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I'm a little unclear on what's supposed to happen. It sounds like the ring is on bottom (therefore presumably touching the ground), but also the ring is the thing that rotates. Is that right? Are the wheels attached to the boards or to the ring? –  Mark Eichenlaub Dec 15 '10 at 7:33
    
They're attached to the end of the boards, which are wielded onto the rotating ring. –  Evan Carroll Dec 15 '10 at 7:37
    
An important factor will be the ratio of the thickness of the wheels to the distance of the wheels from the center of the ring. Do you know roughly what that ratio is? Are the wheels gimbaled? –  Mark Eichenlaub Dec 15 '10 at 8:00
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2 Answers

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I think you are essentially correct that the device does not run smoothly, but the importance of this effect will depend on things like the thickness of the wheels and how well-built the table is.

I will try to answer the following question: "What is the minimum force we must pull with on the outside rim to make the table spin?"

For this calculation we'll pretend that the construction is essentially perfect - the wheels are super-smooth, perfectly aligned, etc.

If the turntable is spinning, any given wheel is moving in a circle. An ideal wheel contacts the ground beneath it with just the very bottom of the wheel. This bottom portion is a line segment whose width is the width of the wheel. Let's call that the "contact line" and make its width $d$ (I want to use $w$ later for the weight).

Suppose we painted the rim of one of the wheels. Then as it went around in the circle, it would paint the ground beneath it with a ring whose thickness is $d$. Let's call the radius of that ring $R$. This is the same as the distance from the center of the device to the wheel.

Resistance to turning arises when the bottom of the wheel slips, or moves relative to the ground. For a wheel going in a straight line, this slipping could ideally be zero - the bottom of the wheel would not be moving at all relative to the ground (while the top of the wheel would be moving at double speed to make up for it.) When the wheel moves in a circle, though, there must be some slipping because different parts of the wheel's contact line move different distances.

When we go around one full circle, we want to know the total amount of slipping that occurs. We'll say that the very center of the contact line doesn't slip at all. The outside and inside edges move in larger and smaller circles respectively, so their total slipping distance is $d\pi$ each (in opposite directions, but that doesn't matter much). The average slipping distance across the wheel is $d \pi/2$.

Next we want to know what force friction with the ground exerts on slipping wheels. This is proportional to the weight of the turntable, $w$, and to an unknown coefficient of friction with the ground $\mu$. $\mu$ might be around one half or so.

The total work done is the force $w\mu$ times the slipping distance $2 d \pi$, so the work needed to rotate the ring one full rotation is

$$W = 2w\mu d \pi$$

(The number of wheels doesn't matter, because if we add more wheels, the weight on each wheel goes down proportionately.)

This work is also equal to the force we exert on the ring times the distance around, which in this case is $2 \pi R$. This means we can set

$$2 \pi R F = 2 w \mu d \pi$$

and simplify to

$$F = w \mu \frac{d}{R}$$

The wider the wheels, the heavier the table, the higher the friction, and the smaller the table, the harder it will be to turn the table.

Now we want to know whether this matters. Is this effect comparable in size to the rolling resistance we expect with wheels anyway?

The Wikipedia article linked above says that a typical car tire has a rolling resistance of $0.01$. Perhaps a well-constructed turn table could achieve the same. The comparable statistic from the sliding of the wheels is $\mu d/R$. A reasonable ratio of the wheel thickness to the radius is $1/10$, so if $\mu \approx 1/2$ we get that the resistance due to slipping is around $0.05$, perhaps five times as big as the rolling resistance.

The punch line is this: if the cart is exquisitely well-constructed and has very thick wheels, then the fact that the wheels don't align perfectly will be the dominant cause of resistance. If the cart is poorly-constructed, with sticky wheels, or if the wheels are very thin, then the slightly-misaligned wheels don't matter much. In my best guess for an average scenario, the extra resistance added by slipping wheels is, to an order of magnitude, about five times the ordinary rolling resistance, and the spinning turntable will be significantly harder to spin than a simple cart constructed the same way would be to push in a straight line.

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Mentally, I picture it like this, the wheel however wide will be wider at the outer end than in the inner end. The wheels will also want to travel in a straight line. The automatic re-coursing of them to a different trajectory that more closely resembles the path tangent to the desired circle path will be provided by force gained from releasing stress in the wheel. This will likely result in a violent skipping, or the wheel physically giving way. –  Evan Carroll Dec 15 '10 at 15:47
    
I don't see why it needs to skip. The wheel does need to slide over the ground. But in that respect it's like pushing a box over the floor, and a box pushed over the floor doesn't skip. –  Mark Eichenlaub Dec 15 '10 at 16:58
    
Well, would it be fair to say that a box doesn't have the necessary traction to skip? My boss was talking about potentially resting up to a fourth of the weight of a car, on two fixed wheels that I believe are rubber. –  Evan Carroll Dec 15 '10 at 18:11
    
@Evan I'm not really sure about the answer - you're getting too much into the real world for me! The analysis I posted is the best I can do. Of course, when your boss builds the turntable we'll be able to find out. –  Mark Eichenlaub Dec 15 '10 at 19:06
    
ahaha, well said. –  Evan Carroll Dec 15 '10 at 19:10
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I'm going to give a sort of meta-answer to this question.

This sounds like a hokey design for a turntable. Having multiple bearings for the same rotation motion is begging for alignment problems.

Your best bet is probably to use a single large tapered-roller bearing in the base of ring, and make sure the "boards" are rigid enough to not cause any deflection loading when they're loaded.

For a 25,000 N pickup truck and a safety factor of 4, you're probably looking at around $150 for a bearing that will handle whatever car you plunk onto it, no wheels necessary.

If you decide to go the wheels route anyways (probably a bit cheaper), use casters, and you won't have to worry about wheel alignment.

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