How fast do you need to cycle to simulate bicycle tire inflation

[Sam VdA]

My question comes from a moment when I was riding my bike. I had a flat tire. My question then was how fast do i need to cycle so that i don't ride on my rim's anymore. I tride calculating this with the force that i put on the tire (15kg+70kg= mass of the bike and I)*g= 833,85 N that devided by 2 becaus I ride on 2 tires.=416,925N this force needs to be opposite to the centripetal force and than i would have an 'inflated' tire .This means that Fc= m*ω²*r but in this formula the mass is a point and here the tire is a point so which formula do I need?

 

[A.T.]

The mass of the tire patch with ground contact.

 

[Sam VdA]

The mass of the tire patch with ground contact.

So I can see it as a Point mass? but using a small bit of the tire as the mass in my formula and not the whole tire

 

[A.T.]

So I can see it as a Point mass? but using a small bit of the tire as the mass in my formula and not the whole tire

Yes. How much of the tire depends on how much deformation you allow. But this is more complicated.

 

[sophiecentaur]

So I can see it as a Point mass? but using a small bit of the tire as the mass in my formula and not the whole tire

I don't think that a point mass is involved (in fact it doesn't work unless you consider some finite orbiting mass). You have a sector of the tyre that will contribute to the force against the road because there is some stiffness in the tyre case. If the mass to consider is, say 5% of the mass of the tyre case, it gives you something to work with. (You need a value for this mass) All you need to do is to equate the centripetal force on this mass, to the weight of (half or some fraction of) the bike and rider. It would at least give you some idea of the tangential velocity involved.

Not having done any sums, I suspect that the necessary speed would mean that aerodynamics would be more significant in the total vertical forces involved.