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

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Discussion Overview

The discussion revolves around the question of how fast one needs to cycle in order to simulate the inflation of a bicycle tire, particularly when dealing with a flat tire. Participants explore the forces involved, including centripetal force and the mass of the tire in contact with the ground, while considering the complexities of tire deformation and the dynamics of cycling.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant calculates the force exerted on the tire based on the combined mass of the bike and rider, suggesting that this force must counterbalance the centripetal force for the tire to behave as if it were inflated.
  • Another participant mentions the importance of considering the mass of the tire patch in contact with the ground, implying that this mass is crucial for the calculations.
  • There is a suggestion that a small portion of the tire can be treated as a point mass for the purpose of calculations, but this is met with skepticism regarding the appropriateness of this simplification.
  • One participant argues that a point mass may not be suitable and suggests considering a finite mass, such as a sector of the tire, to account for the stiffness of the tire case and its contribution to the forces involved.
  • Concerns are raised about the role of aerodynamics in the overall forces when calculating the necessary speed for simulating tire inflation.

Areas of Agreement / Disagreement

Participants express differing views on the appropriateness of modeling the tire as a point mass versus a finite mass, indicating a lack of consensus on the best approach to the problem. The discussion remains unresolved regarding the exact calculations and assumptions needed.

Contextual Notes

Participants note the complexity of the problem, including the need to define the mass of the tire patch and the effects of tire deformation, which are not fully resolved in the discussion.

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?
 
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The mass of the tire patch with ground contact.
 
A.T. said:
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
 
Sam VdA said:
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.
 
Sam VdA said:
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.
 

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