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Q & A: bikes: does weight matter?

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Most recent answer: 08/03/2013
My friend and i are having a debate about our bicycles and an upcoming 100-mile ride over flat terrain. I have an old 35 pound steel frame bike and he has a new 16 lb carbon fiber bike. Both bikes are in excellent mechanical shape (so essentially no difference in internal friction) and both of us are about the same size and weight (so aerodynamics should be about the same). He claims that since his bike is lighter, he is going to use substantially less energy over the 100-mile course. I disagree and say that on weight alone, we should both use about the same amount of energy. My analysis is that there are four forces at play that will cause the negative acceleration of the bike (once the bike has come to speed). 1) weight times road angle, 2) internal (mechanical) friction, 3) road / tire friction, and 4) wind resistance. Thus, the forces of negative acceleration would be: mg*sin[theta] + F[i] + mg*u[s] +c[i]*f(v) Assuming a flat course, theta = 0, so that term drops. The only other term dependent on mass is then the road /tire friction which I feel is relatively small for a bike with road tires. Also, I'm assuming that there is not a lot of accelerating / decelerating, which is probably true on a really long road course. So, am I missing anything, can you physics savants put this question to rest so that I continue to josh my friend about spending $3500 on a bicycle that provides little mechanical advantage!
- Andrew (age 29)
Bethlehem, PA USA

Your reasoning makes sense to me. Maybe there'll be a bit more accelerating and terrain unevenness than you're expecting, but you're right that it shouldn't be a big deal. Of course the next trip may be different, and then the lightweight bike may come in handy.  That's assuming that you don't need to bring locks, since a colleague assures me that there's a theorem that the sum of the weight of the bike and the lock it needs to protect against theft is a constant.

Mike W.


(published on 08/03/2013)

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