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Most of the Friction In A Bicycle Chain
Someone on sci.engr.mech said a chain sprocket system was over 95% efficient at
transmitting power -- better than a belt drive. I'm guessing most of the friction is between the pins and the links as they wrap around the sprocket. The friction would increase directly with rpm and with tension in the chain -- in other words, the % friction would remain constant over all power ranges. 5% might not sound like much but . . . Bret Cahill |
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Most of the Friction In A Bicycle Chain
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Most of the Friction In A Bicycle Chain
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Most of the Friction In A Bicycle Chain
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Most of the Friction In A Bicycle Chain
Dan Musicant wrote:
A lot of that friction is not between the pins and links but between the links and the cogs. Actually, the links shouldn't be slipping over the cogs so there should be either almost no friction or 100% friction depending upon how you look at it. Either way, the connection between the chain and the cog isn't what's eating your energy. There's friction in the links. There's also friction in the pulleys, hubs, bottom bracket and pedals. There's also rolling resistance and air resistance eating away at efficiency. --Bill Davidson -- Please remove ".nospam" from my address for email replies. I'm a 17 year veteran of usenet -- you'd think I'd be over it by now |
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Most of the Friction In A Bicycle Chain
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Most of the Friction In A Bicycle Chain
Joe Riel writes:
What is usually overlooked is that the sprocket size has more to do with losses than is suspected, since the articulation angle of the chain decreases as the number of teeth increases. That is, the angle through which the chain bends under load is 360/t (t = number of sprocket teeth). Therefore a chain running on a 13t sprocket has twice the frictional loss of running over one with 26t (for the same tension). Note that tension is not power but that power is tension times the speed of chain (in whatever units you prefer). The power transfer through a chain is P = T*v = T*R*w where P = power T = chain tension R = sprocket radius v = chain velocity w = sprocket angular velocity For a given operating point, P and w are fixed, so T*R = P/w = constant As R goes up, T goes down proportionally. So, to first order, the frictional loss in the chain is independent of the sprocket size. Thanks. I think you might have written that before some time and I forgot it. It rings familiar. In any case, it makes me feel better about using my 13t sprocket while cruising down the road. Not that it makes any performance difference, but it isn't wasting power. Jobst Brandt |
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Most of the Friction In A Bicycle Chain
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Most of the Friction In A Bicycle Chain
"Peter Cole" writes:
I don't see the connection to frictional losses. Yes, there was some hand waving there. Incorrect, as usual. See my follow-up post. Joe Riel |
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