Most of the Friction In A Bicycle Chain

[Bretcahill]

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

 

[Dan Musicant]

On 09 Sep 2003 14:20:25 GMT, [email protected] (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

 

[Tim McNamara]

In article <[email protected]>, [email protected] (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.

 

[Jobst Brandt]

Tim McNamara writes:

> 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.

I doubt that, because as tension increases, lubrication films decrease and friction increases.
However, that me be the result of there being loss even with a slack chain and that overhead becomes
insignificant at higher loads.

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).

Sprocket engagement friction is insignificant for an in-pitch chain, because its roller are engaged
before they bear load. This is not true for a worn chain that has an elongated pitch.

Jobst Brandt [email protected]

 

[Bill Davidson]

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

 

[Joe Riel]

[email protected] 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

 

[Jobst Brandt]

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 [email protected]

 

[Joe Riel]

[email protected] writes:

> Thanks. I think you might have written that before some time and I forgot it. It rings familiar.

You're welcome. I may have, but cannot recall. Hopefully the result is the same .

Joe Riel

 

[Joe Riel]

[email protected] 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.

 

[Peter Cole]

"Joe Riel" <[email protected]> wrote in message news:[email protected]...
> [email protected] 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/home99/aug99/bike.html

 

[Joe Riel]

"Peter Cole" <[email protected]> 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

 

[Joe Riel]

Joe Riel <[email protected]> writes:

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

Here is a slightly more detailed estimation of the chain loss.

Terms
-----
Nr = number of teeth of rear sprocket Nf = number of teeth of front chainwheel mu = coefficient of
friction in bushing p = chain pitch pi = 3.14159... Pl = power loss in chain Ptot = total power
transfer Rb = inner radius of chain bushing Rr = radius of rear sprocket T = chain tension wr =
angular velocity of rear sprocket wf = angular velocity of front sprocket

Assume that the primary power dissipation in the chain occurs due to rotational friction in the
links as they disengage the rear sprocket and engage the front chainwheel. This power loss is

(2) Pl = mu*Rb*T*(wr + wf)
(3) = mu*Rb*T*wr*(1+Nr/Nf)

The power transferred through the chain is

(4) Ptot = T*wr*Rr

The radius of the rear sprocket is

(5) Rr = Nr*p/2/pi

Combining (3)--(5) gives

(6) Pl/Ptot = 2*pi*mu*Rb/p*(1/Nr + 1/Nf)

which essentially agrees with Kyle's formula (1). To minimize chain loss one wants to maximize the
diameter of both the front and rear sprockets.

Joe Riel

 

[John Dacey]

On Tue, 09 Sep 2003 23:49:21 GMT, [email protected] wrote:

>Tim McNamara writes:

>> ISTR that an MIT research project, published in the last few years, showed that chain efficiency
>> increased with tension rather than decreasing, rather counterintuitively.
>
>I doubt that, because as tension increases, lubrication films decrease and friction increases.
>However, that me be the result of there being loss even with a slack chain and that overhead
>becomes insignificant at higher loads.

Is the type of chain construction much of a factor? In what ways will a chain with bushings for its
rollers be better/worse than its bushingless counterpart?
-------------------------------
http://www.businesscycles.com John Dacey Business Cycles, Miami, Florida 305-273-4440 Now in our
twenty-first year. Our catalog of track equipment: eighth year online
-------------------------------

 

[Tim McNamara]

In article <[email protected]>,
[email protected] wrote:

> Tim McNamara <[email protected]> wrote:
>
> : 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.
>
> If we shield the chain from road dirt, will it give us the optimum efficiency?

Well, there are a number of issues besides cleanliness, as has been discussed in other posts. The
roller chain was intended by Mr. Renolds (sp?) to run in an oil bath, if I understand correctly. So
a chain in a chain case with an oil bath, over largish cogs and a straight chainline ought to give
you the best efficiency. Not to mention lasting much, much longer than exposed bicycle chains.

 

[Jobst Brandt]

John Dacey writes:

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

They didn't say on what basis this relied. Was this constant bicycle speed or was it percentage
loss of input.

>> I doubt that, because as tension increases, lubrication films decrease and friction increases.
>> However, that me be the result of there being loss even with a slack chain and that overhead
>> becomes insignificant at higher loads.

Three things are varying in these comparisons, angular articulation of the chain, chain tension,
and chain speed. As I see it, a chain running between two 20t sprockets in comparison to one
running on two 40t sprockets, transmitting the same rotational torque and speed to the rear wheel
have these effects:

chain tension 2:1 chain bend 2:1 chain speed 1:2

Since it is chain articulation that causes friction in the links, the smaller sprocket pair has
larger bend angle under higher tension although at half the speed. It still worse than the larger
sprocket pair. Besides, lubrication failure is greater at higher tension.

> Is the type of chain construction much of a factor? In what ways will a chain with bushings for
> its rollers be better/worse than its bushingless counterpart?

Bushings give roughly four times the bearing area to the pins than side plates with upset
collars give. This alone causes higher wear, but in addition, water intrusion and lubricant loss
is accelerated by a fairly direct path into the bearing area of the pin through the gap under
the roller.

Jobst Brandt [email protected]

 

[Risto Varanka]

Tim McNamara <[email protected]> wrote:

: 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.

If we shield the chain from road dirt, will it give us the optimum efficiency?

--
Risto Varanka | http://www.helsinki.fi/~rvaranka/hpv/hpv.html varis at no spam please iki fi

 

[N2vx Jim]

On Wed, 10 Sep 2003 20:14:14 GMT, [email protected] wrote:

>John Dacey writes:
>
>Bushings give roughly four times the bearing area to the pins than side plates with upset
>collars give. This alone causes higher wear, but in addition, water intrusion and lubricant loss
>is accelerated by a fairly direct path into the bearing area of the pin through the gap under
>the roller.
>
>Jobst Brandt [email protected]

Are there any chains with bushings for 7/8 or 9 speed drivetrains in current production? Are
they any good?

 

[Joe Riel]

Joe Riel <[email protected]> writes:

> Terms
> -----
> Nr = number of teeth of rear sprocket Nf = number of teeth of front chainwheel mu = coefficient
> of friction in bushing p = chain pitch pi = 3.14159... Pl = power loss in chain Ptot = total
> power transfer Rb = inner radius of chain bushing Rr = radius of rear sprocket T = chain tension
> wr = angular velocity of rear sprocket wf = angular velocity of front sprocket
>
>
> (6) Pl/Ptot = 2*pi*mu*Rb/p*(1/Nr + 1/Nf)

While this formula is rather simple, it does give results that are consistent with my crude
knowledge of chain efficiency.

Consider that

Rb ~ 0.11 inch p = 0.50 inch mu ~ 0.16 (lubricated steel on steel)

Then

Pl/Ptot = (0.22)(1/Nr + 1/Nr)

The extremes for a typical road bike are

Pl/Ptot|min = (0.22)(1/13 + 1/39) = 2.3% Pl/Ptot|max = (0.22)(1/23 + 1/53) = 1.4%

which is about what I would expect, probably lower than achievable.

Joe Riel

 

[Jobst Brandt]

Jim Cronin writes:

>> Bushings give roughly four times the bearing area to the pins than side plates with upset collars
>> give. This alone causes higher wear, but in addition, water intrusion and lubricant loss is
>> accelerated by a fairly direct path into the bearing area of the pin through the gap under the
>> roller.

> Are there any chains with bushings for 7, 8 or 9 speed drivetrains in current production? Are they
> any good?

No. As I mentioned, with 20t chainwheels and heavy riders, a 5-element derailleur chain one
with full sleeves, is probably not safely possible, the sleeve cutting a large hole in the
inner side plates.

The ones I have are excellent but I can only put less that half the tension in these chains with the
gears I ride. I don't have a 20t chainwheel.

Jobst Brandt [email protected]

 

[Jobst Brandt]

Joe Riel writes:

>> Terms
>> -----
>> Nr = number of teeth of rear sprocket Nf = number of teeth of front chainwheel mu = coefficient
>> of friction in bushing p = chain pitch pi = 3.14159... Pl = power loss in chain Ptot = total
>> power transfer Rb = inner radius of chain bushing Rr = radius of rear sprocket T = chain
>> tension wr = angular velocity of rear sprocket wf = angular velocity of front sprocket

>> (6) Pl/Ptot = 2*pi*mu*Rb/p*(1/Nr + 1/Nf)

> While this formula is rather simple, it does give results that are consistent with my crude
> knowledge of chain efficiency.

> Consider that

> Rb ~ 0.11 inch p = 0.50 inch mu ~ 0.16 (lubricated steel on steel)
>
> Then

> Pl/Ptot = (0.22)(1/Nr + 1/Nr)

> The extremes for a typical road bike are

> Pl/Ptot|min = (0.22)(1/13 + 1/39) = 2.3% Pl/Ptot|max = (0.22)(1/23 + 1/53) = 1.4%

> which is about what I would expect, probably lower than achievable.

There are so many constants floating around in these equations that it muddles the picture. Items Nr
Nf mu p pi Rb Rr are fixed for the obvious test two cases I proposed, that of a 20t to 20t pair and
a 40t to 40t drive train. This gets rid of most of the fog and gets right down to the issue of power
loss due to sprocket size... for the same ratio.

This would be like a 52t-13t and a 44t-11t for instance but far simpler to analyze in the 20 and 40
arrangement. So what's your take on that?

Jobst Brandt [email protected]