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

On 09 Sep 2003 14:20:25 GMT, (BretCahill) wrote:

: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

A lot of that friction is not between the pins and links but between the
links and the cogs. A lot would also depend on the cleanness of the
componentry and the state and type of lubrication.

I don't see why the % friction would remain constant although it may be
not far from constant. 5% sounds great, if it is indeed in that
neighborhood. It would be empirically measurable but it would require
some sophisticated equipment and measuring devices to determine percents
at different gear settings and applications of power.

Dan

On 09 Sep 2003 14:20:25 GMT, (BretCahill) may have
said:

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 . . .

The "over" in that statement is also significant. There's a study
that came up with over 98% (I tend to distrust claims that exceed 99%)
in a significant portion of the instances. All of these were with a
system that was working properly. If something's not right, it can
change the results downwards.

--
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In article ,
(BretCahill) wrote:

Someone on sci.engr.mech said a chain sprocket system was over 95%
efficient at transmitting power -- better than a belt drive.

Much better than a belt drive both in terms of measurement and
real-life use. However, chain drives have been measured with
efficiency as high as 98%. These are under ideal conditions with
clean and properly lubricated components, I'd expect, and real-life
efficiency is probably rather less.

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.

There are several sources of friction, inside the chain as well as
between the chain and the other drive train components: chainrings,
freewheel cogs and jockey wheels. Jockey wheels also have internal
frictional losses independent of the chain. Chain line also has an
effect on efficiency.

ISTR that an MIT research project, published in the last few years,
showed that chain efficiency increased with tension rather than
decreasing, rather counterintuitively.

5% might not sound like much but . . .

In a low power situation like a bicycle, that 5% is worth attending to.

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
--
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I'm a 17 year veteran of usenet -- you'd think I'd be over it by now

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.


Joe Riel

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

writes:

Joe Riel writes:

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.

I may have found the source. Chester Kyle followed a similar line of
reasoning [1]. His derivation, frankly speaking, is hard to follow;
among other atrocities he fails to define terms and mixes units, both
360 degrees and 2*Pi enter and leave in intermediate steps. His
result is that

Cf = k*P*(1/Nr + 1/Nf)

where

Cf = chain friction [I assume he means chain friction loss]
P = rider output power
Nr = number of teeth in rear sprocket
Nf = number of teeth in front chainwheel

This does not agree with mine in that it implies proportionally larger
gears have lower losses.


Joe Riel


[1] Chester R. Kyle, "Chain Friction, Windy Hills, and other Quick
Calculations," Cycling Science, September 1990, pp. 23--26.

"Joe Riel" wrote in message
...
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.

I don't see the connection to frictional losses.

I would think the loss due to articulation friction would be:

a*T*v

where

a = articulation angle
T = chain tension
v = chain velocity (= articulation rate)


a, T, and v are all linear with sprocket size, a and T decreasing, with larger
sprockets, v increasing.

This model predicts linear increase in articulation loss with decreasing
sprocket size, which seems to be borne out experimentally:

http://www.jhu.edu/news_info/news/ho...ug99/bike.html

"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