How do the casters on a turntable support the weight of a motorcycle?

[D Hook]

Hi: New to this forum and have a quick question. Hope I'm posting in the right area.

First of all, I'm not an engineer, nor do I pretend to be one but I'm having to settle an argument between two people who think they are.

There is a ten foot diameter turntable constructed of a 2" steel frame and a 1 1/2" plywood top. On top of the turntable sits an 1100lb motorcycle. The table turns atop 12 casters rated at 600lb each spread out equa-distance around the perimeter under the table. In addition, there are 4 casters towards the center of the table, spaced equal-distance about 2 feet away from the pivot point. At some point, while the top is rotating, the load of the motorcycle is being directly above only two of the perimeter casters and all four of the interior casters. One person says to use casters that will pick up the load of the 3 foot x 10 foot wide path the motorcycle sits on, in addition to the load of the rest of the table. The other person says no, the load is equally spread out and the casters only need to be rated to carry the total load of the turntable and motorcycle, divided by 16 casters.

In other words, when the motorcycle rotates and is directly above those casters, are those casters carrying MORE of the load or the same load as the other 10?

Thanks for your time and hope this isn't too elementary. I think the first guy is correct but neither will budge unless they hear it from someone trained in the field.
Personally, I think the first guy is correct.

 

[Dense]

Unless the center of mass of the motorcycle is over the center of mass of the turntable, the area underneath it will be on the receiving end of greater force than a similar area on the opposite side. And so will the casters under that place. Think torque.

Simple experiment: Get a CD. Lay it flat in your hand. Press down with a finger on a spot halfway between the hole and the rim. Does your hand feel greater pressure where you're poking, or is the pressure applied evenly by the whole CD?

 

[D Hook]

Thanks! I'll pass on the reply.

If anyone else has any other input, love to hear it.

 

[Vixus]

It's a no-brainer if you think about it.

Pressure = Force/Area

Where Force is the weight of the bike (500 x 9.8) and the area is the area of tyre/bike touching the turntable. The pressure's not applied uniformly.

 

[LURCH]

Although I agree that the pressure is not the same throughout the turntable, I feer the explanation given may be a little too simple. If you try the CD experiment suggested above, you'll find that pressure is greater directly beneath your finger, and much less at the far edge of the disk. However, this much lesser pressure is not zero pressure. So the weight of the bike is not distributed evenly, but it is distributed some.

Now I hope somebody will check me on this, but I think this best way to model this is to think of the disk as a lever, with the edge nearest the bike as the fulcrum (and I'm assuming a ridgid turntable here). Downward pressure (the weight fo the bike) is trying to make the lever rotate, but the casters underneath prevent this motion. The amount of resistance required by each caster is determined by how close that caster is to the fulcrum. The casters at the edge of the disk farthest from the bike have the most leverage, and so they are not required to provide nearly as much support, but they still must provide some (support of the bike's weight, that is). So the casters underneath the bike must support the greatest share of its weight, but not all. Therefore, you could theoretically get away with placing casters directly under the bike that are not capable of supporting its entire weight, and they might still hold up.

Of course, if this model is correct, then the casters under the edge of the turntable closest to the bike are actually receiving more of its weight than those directly under the bike. That seems counterintuitive, but I think it's accurate.