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Educating the Debate – Part I - Oct. 9, 2013, 6:50 a.m.

That would be great Keith! I have always wanted to see this type of analysis. Look forward to your future articles.

Educating the Debate – Part I - Oct. 8, 2013, 2:48 p.m.

The kinetic energy calculations are for just the moving wheels themselves, in isolation, if I understood correctly. To understand the impact on the rider (that is, what the rider might actually notice), would it be better to look at the delta in kinetic energy for the entire system (that is, including the rider and the rest of the bike)?

If we assume a 180 lb rider on a bike which weighs about 26 lbs not counting the wheels (206 lbs / 2.2 = about 94 kg), then the kinetic energy of the rest of the system is 0.5 * 94 kg * (10 m/s)^2 = 4700 J.

26: Ek = 4700 + 694 = 5394 J (-0.7% change from 27.5)
27.5: Ek = 4700 + 730 = 5430 J
29: Ek = 4700 + 779 = 5479 J (+0.9% change from 27.5)

Another way to look at it is that 36 J delta between 26″ and 27.5″ wheels at 10 m/s is the Ek for a mass of 0.72 kg, or 1.6 lbs. So if you put a 750 mL water bottle on your bike, you would notice that about as much as the difference in the wheel change for accelerating and decelerating, based on the kinetic energy view of things. This is not taking into account the angle of attack or any other factor.

I am an electrical guy, not mechanical, so if I messed something up in the math please feel free to slag me mercilessly.

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