Bicycle Tire Question: Why Does Air Loss Affect Cycling Difficulty?

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Air loss in bicycle tires increases cycling difficulty primarily due to greater rolling resistance caused by tire deformation. When tires are under-inflated, they compress and create a torque that must be overcome to maintain motion. The normal force acting on the tire is redistributed, leading to inefficiencies in energy transfer. While some believe that increased contact area contributes to resistance, the main issue lies in how the tire's shape changes under load. Understanding these dynamics is crucial for optimizing cycling performance.
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bicycle tires question...

ok...heres a question that was given in the interview to a student who wanted to study physics at cabridge university.

why when there is air loss in the tires of a bicycle(pressure decrease)...the bicycle becomes(heavier)?,meaning...harder to cycle.
 
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There is a bunch of energy that goes into deforming the rubber of the tires and shifting the air around that doesn't get used if the tires are filled.
 
Originally posted by NateTG
There is a bunch of energy that goes into deforming the rubber of the tires and shifting the air around that doesn't get used if the tires are filled.

thats true...but its not by far the main reason of the huge difference in the effort u have to put into continue cycling.

pls try again!
 
I have always assumed that it was the greater area of contact between the the tire and the road which caused greater rolling resistance thus low pressure tires are MUCH harder to push down the road.

Is there more?
 
Originally posted by Integral
I have always assumed that it was the greater area of contact between the the tire and the road which caused greater rolling resistance thus low pressure tires are MUCH harder to push down the road.

Is there more?

the (point),(area...) of contact of the tire with the road is stationary...the tyre at that point does't move with respect to the road...so there is no point of discusing resistance due to friction.(so the greater area of contact is irrelevant in that respect).

TRUE there is greater resistance in the rolling of the (wheel) tire...but what causes it?

i have given u a strong hint now.
 
Originally posted by fotonios
TRUE there is greater resistance in the rolling of the (wheel) tire...but what causes it?
I would expect that the major effect of soft tires is this: the tire is squashed forward and bunched up---essentially pulled away from the rim. In turn, the rubber pulls against the rim, creating a torque on the wheel that must be overcome to keep riding.

Looked at in terms of external forces, the normal force (the ground pushing the tires up) has been redistributed. In a firm tire, the normal force is directly underneath the center of the wheel. In the soft tire case, the normal force is now squashing the rubber in front of center. Thus the normal force now exerts a torque on the wheel.
 
Originally posted by Doc Al
I would expect that the major effect of soft tires is this: the tire is squashed forward and bunched up---essentially pulled away from the rim. In turn, the rubber pulls against the rim, creating a torque on the wheel that must be overcome to keep riding.

Looked at in terms of external forces, the normal force (the ground pushing the tires up) has been redistributed. In a firm tire, the normal force is directly underneath the center of the wheel. In the soft tire case, the normal force is now squashing the rubber in front of center. Thus the normal force now exerts a torque on the wheel.

you got it!
good thinking!
and good and complete explanation...not just a dry answer.
you should apply for cabridge...:)
 
Last edited:
Ahhh! Beautiful! It makes very good sense now.

Thanks for completing a picture for me.
 
I am not fully convinced. Wouldn't the normal force also act (equally) on the other side of the axis of rotation, that is, at the back end of the footprint?
 
  • #10
Originally posted by Moose352
Wouldn't the normal force also act (equally) on the other side of the axis of rotation, that is, at the back end of the footprint?
No, since the "footprint" is not symmetric. The footprint is distorted towards the front.
 
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