Most of the Friction In A Bicycle Chain

Discussion in 'Cycling Equipment' started by Bretcahill, Sep 9, 2003.

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  1. Bretcahill

    Bretcahill Guest

    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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  2. Dan Musicant

    Dan Musicant Guest

    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
     
  3. Tim McNamara

    Tim McNamara Guest

    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.
     
  4. Jobst Brandt

    Jobst Brandt Guest

    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]
     
  5. 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
     
  6. Joe Riel

    Joe Riel Guest

    [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
     
  7. Jobst Brandt

    Jobst Brandt Guest

    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]
     
  8. Joe Riel

    Joe Riel Guest

    [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 :cool:.

    Joe Riel
     
  9. Joe Riel

    Joe Riel Guest

    [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.
     
  10. Peter Cole

    Peter Cole Guest

    "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
     
  11. Joe Riel

    Joe Riel Guest

    "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
     
  12. Joe Riel

    Joe Riel Guest

    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
     
  13. John Dacey

    John Dacey Guest

    On Tue, 09 Sep 2003 23:49:21 GMT, job[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
    -------------------------------
     
  14. Tim McNamara

    Tim McNamara Guest

    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.
     
  15. Jobst Brandt

    Jobst Brandt Guest

    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]
     
  16. 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
     
  17. N2vx Jim

    N2vx Jim Guest

    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?
     
  18. Joe Riel

    Joe Riel Guest

    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
     
  19. Jobst Brandt

    Jobst Brandt Guest

    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]
     
  20. Jobst Brandt

    Jobst Brandt Guest

    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]
     
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