From: jobst.brandt@stanfordalumni.org
Subject: Re: Slipping cranks on hills
Newsgroups: rec.bicycles.misc
Message-ID: <xJMk9.31029$Ik.669476@typhoon.sonic.net>
Date: Thu, 26 Sep 2002 23:27:25 GMT

Ron Hardin writes:

>>> You can see sunlight in the shadow of a chainwheel with a
>>> stretched chain on it between chain and chainwheel, with
>>> increasing obviousness as the stretch increases.

>> Yes, and that is the difference I pointed out in earlier items on
>> this subject.  Old chains will run on new driven sprockets and new
>> chains will run on old driving sprockets, but not the converse.

> You can run old chains on new chainwheel (driving sprockets) but you
> have sun under the chain.  The teeth are still tall however and are
> above the roller midpoints.  Think of it as drilling the
> circumference of the chainwheel, saving weight.

That is not my experience and here is why it does not work.  The worn
chain, with its elongated pitch, cannot engage the new sprocket that
has the correct pitch because succeeding links approach under high
tension.  The chain being too long, progressively rides up the teeth
and jumps over.  In contrast, on driven sprockets, links of an
elongated chain can settle into position on an in-pitch driven
sprocket before exiting under load.

> The curious thing is that the math doesn't work out as to gear ratio
> based on radius and based on tooth count, a paradox I think resolved
> in hammer blows to the freewheel cogs.

I think you need to handle this digitally.  The sprockets have integer
numbers of teeth that will give the gear ratio to great precision.
Looking at the correct parameters is essential to solving these
problems.

Jobst Brandt  <jobst.brandt@stanfordalumni.org>  Palo Alto CA

From: jobst.brandt@stanfordalumni.org
Subject: Re: Slipping cranks on hills
Newsgroups: rec.bicycles.misc
Message-ID: <zkOk9.31066$Ik.670736@typhoon.sonic.net>
Date: Fri, 27 Sep 2002 01:17:19 GMT

Ron Hardin writes:

>> That is not my experience and here is why it does not work.  The
>> worn chain, with its elongated pitch, cannot engage the new
>> sprocket that has the correct pitch because succeeding links
>> approach under high tension.  The chain being too long,
>> progressively rides up the teeth and jumps over.  In contrast, on
>> driven sprockets, links of an elongated chain can settle into
>> position on an in-pitch driven sprocket before exiting under load.

> The pitch is just right once it rides up on the teeth some amount.
> That's why it rides up.  It stays there and works fine until the
> teeth wear down to it or the chain stretches more.

Did you try this with, for instance, 1% elongated chain?  I have and
it skipped.  I believe the reason for this is that subsequent
partially engaged links ride high enough on new sprocket teeth, to be
on the exit ramp, so that a taught chain is ejected.  This may be true
only for sprockets with more than 50t.  In any case, I had the
opportunity to try this and was surprised at the results.

Jobst Brandt  <jobst.brandt@stanfordalumni.org>  Palo Alto CA

From: jobst.brandt@stanfordalumni.org
Subject: Re: Slipping cranks on hills
Newsgroups: rec.bicycles.misc
Message-ID: <Mg2l9.31526$Ik.678399@typhoon.sonic.net>
Date: Fri, 27 Sep 2002 19:25:32 GMT

Ron Hardin writes:

> My chains are hugely elongated when I change them (20k miles usually).

> There's no particular reason for the chain to go higher, on the exit
> ramp or not, because it's at the right radius.  That is, it's not a
> stretched chain on the chainwheel.  The circumference of its circle
> is exactly right, and every tooth is under load.

That isn't possible because the pitch of the chainwheel is unchanged,
meaning its an in-pitch chain will lie in the base circle of its
teeth.  An elongated chain even under the tension of the derailleur
will statically sit in the base circle only in the forward quadrant
and rise up at the engagement and disengagement area.

It is this rising up that wears a ramp part way up sprocket teeth when
transmitting load.  That is the location that chainsuck hooks on teeth
are formed by a worn chain.

> There could be an instability where its circle is not centered at
> the center of the chainwheel, I suppose.

The chainwheel is not in the center of the chain circle that is larger
by pi x pitch elongation.  The chain rides in the chainwheel root
diameter only at one point.

Jobst Brandt  <jobst.brandt@stanfordalumni.org>  Palo Alto CA

From: jobst.brandt@stanfordalumni.org
Subject: Re: Slipping cranks on hills
Newsgroups: rec.bicycles.misc
Message-ID: <WG7l9.31627$Ik.679480@typhoon.sonic.net>
Date: Sat, 28 Sep 2002 01:34:46 GMT

Ron Hardin writes:

>>> There's no particular reason for the chain to go higher, on the
>>> exit ramp or not, because it's at the right radius.  That is, it's
>>> not a stretched chain on the chainwheel.  The circumference of its
>>> circle is exactly right, and every tooth is under load.

>> That isn't possible because the pitch of the chainwheel is
>> unchanged, meaning its an in-pitch chain will lie in the base
>> circle of its teeth.  An elongated chain even under the tension of
>> the derailleur will statically sit in the base circle only in the
>> forward quadrant and rise up at the engagement and disengagement
>> area.

> That doesn't happen though.  It's away from the chainwheel base even
> in front.

Try it.  rotate the chainwheel manually by the pedal and notice when
it comes to rest that the chain rests on the base circle in the front.
This place shifts to the center of tension as more tension arises on
the upper run of the chain under pedaling torque.

> The forward force is exerted tangentially (indeed that's why it's
> work to pedal.)  A force diagram would be complicated to do as a
> thought experiment, I think.  Is there rearward tangential force at
> the bottom?  How far does the load on the chain distribute, in other
> words.  There's certainly a net forward force but only the
> derailleur force can come _radially_ forward from the base of the
> chainwheel.  I think.

With an elongated pitch the chain bears heavily on only two or so
teeth, which accelerates sprocket wear the greater the chain pitch.
The sprocket is always in pitch, albeit with sloppier tooth profile.
Therefore, an out of pitch chain will never fit well in more than a
couple of teeth.

> An elongated chain finds its pitch in the increased tooth spacing
> away from the base of the chainwheel, ie. the teeth get further
> apart the farther from the base of the chainwheel you go.

Tooth spacing on the sprocket is always the same regardless of wear.
When the chain has tension, it cannot mesh because its rollers are
against one tooth and too far for the next.  Sprocket pitch being
constant and chain pitch not.

> On the internet, nobody knows you're a jerk.

I don't believe that either.

Jobst Brandt  <jobst.brandt@stanfordalumni.org>  Palo Alto CA

From: jobst.brandt@stanfordalumni.org
Subject: Re: Slipping cranks on hills
Newsgroups: rec.bicycles.misc
Message-ID: <%Ttl9.32018$Ik.691208@typhoon.sonic.net>
Date: Sun, 29 Sep 2002 02:50:35 GMT

Ron Hardin writes:

>>>  An elongated chain finds its pitch in the increased tooth spacing
>>>  away from the base of the chainwheel, ie. the teeth get further
>>>  apart the farther from the base of the chainwheel you go.

>> Tooth spacing on the sprocket is always the same regardless of
>> wear.  When the chain has tension, it cannot mesh because its
>> rollers are against one tooth and too far for the next.  Sprocket
>> pitch being constant and chain pitch not.

> The heart of the disagreement then.

Possibly, but I contend that running an elongated chain on a new
sprocket will skip.  As I said, I've only experienced this on a road
bicycle with a 50t chainring.  I don't know whether this effect occurs
on chainwheels from about 32t or fewer.

> Tooth spacing varies depending on what height the teeth engaged at,
> a consequence of geometry (not tooth shape, so it's nothing
> complicated).  All spacings scale up like the radius.  That's why
> the chain is out there when it's elongated.  It's finding its own
> pitch in the teeth.

I agree, but this exceeds the tooth height if the sprocket is new and
the chain is, say, 1% elongated.  Try it some time and I think you
will discover a different effect at work.

> Imagine foot-long teeth and the tooth spacing at their tips.

More than what you can get for a bicycle.

Jobst Brandt  <jobst.brandt@stanfordalumni.org>  Palo Alto CA

From: jobst.brandt@stanfordalumni.org
Subject: Re: Slipping cranks on hills
Newsgroups: rec.bicycles.misc
Message-ID: <UGJl9.32311$Ik.695435@typhoon.sonic.net>
Date: Sun, 29 Sep 2002 20:48:52 GMT

Ron Hardin writes:

>> I agree, but this exceeds the tooth height if the sprocket is new
>> and the chain is, say, 1% elongated.  Try it some time and I think
>> you will discover a different effect at work.

> A 1% elongation on a (huge) 6.25" radius would give a .0625 = 1/16
> inch gap between chain and chainwheel.

I don't know where you get these numbers but I have a well worn 50t
chainring on which a 1% wear chain leaves about 1/8" gap at both
ends of the engagement while it rests in the base circle at the
midpoint.  Would you be good enough to take a break from theoretical
considerations and look at real hardware.

I'm sure there are others, closer to hardware who might speak up on
this matter.  I can't imagine that all bicycle mechanics are blind to
these issues.

> The problem is to account for what happens.  "Chain suck" is another
> problem to account for on small chainwheels.

Chain suck occurs when a new chain is used with a worn sprocket.  What
we have been discussing here is the converse.

Jobst Brandt  <jobst.brandt@stanfordalumni.org>  Palo Alto CA

From: jobst.brandt@stanfordalumni.org
Subject: Re: Slipping cranks on hills
Newsgroups: rec.bicycles.misc
Message-ID: <i4ul9.32022$Ik.691199@typhoon.sonic.net>
Date: Sun, 29 Sep 2002 03:03:42 GMT

Ron Hardin writes:

>>> That doesn't happen though.  It's away from the chainwheel base
>>> even in front.

>> Try it.  rotate the chainwheel manually by the pedal and notice
>> when it comes to rest that the chain rests on the base circle in
>> the front.  This place shifts to the center of tension as more
>> tension arises on the upper run of the chain under pedaling torque.

> That's from derailleur force, which is in fact supported by the
> front of the chainwheel (that's the only force that's supported
> there, otherwise it would not be work to pedal).

I suggested this test to make clear what the chain does when idling.
when pedaling force is added the situation doesn't change but the
engagement force is taken by the first couple of teeth on the exit
ramp (of a new chainwheel) high enough to never engage deep enough to
hold the chain tension, so it skips.

Again, I ask, have you tried it?

> Under load, the chain is not free to slip down to the front of the
> chainwheel and that small derailleur force is supported on the slant
> in the teeth at the height the chain is fixed by friction against
> the pedaling load.

It will sink to the root diameter of the chainwheel, there being
nothing to prevent it from doing so, the chain being slack in the
forward quadrant of the chainwheel.  A chain with an elongated pitch
must be synchronous with the chainwheel pitch somewhere.  This is true
for a matched worn pairing as well as a new chainwheel and an old
chain, the point where this discussion began.

> The derailleur force is equal on top and bottom chain, so no work is
> done against it by tangential forces.  (And, there being no radial
> movement, no work is done by radial forces either.)

So?  This tension is enough to take the slack out of the chain and
demonstrate the tooth climbing effect at the engagement and
disengagement point on the chainwheel.

> Pedaling force is only on the top chain, and only tangential forces
> can do work with it.

I don't understand what this has to do with a non engaging chain.

Jobst Brandt  <jobst.brandt@stanfordalumni.org>  Palo Alto CA

From: jobst.brandt@stanfordalumni.org
Subject: Re: Slipping cranks on hills
Newsgroups: rec.bicycles.misc
Message-ID: <9eqk9.30479$Ik.658588@typhoon.sonic.net>
Date: Wed, 25 Sep 2002 21:52:05 GMT

Ron Hardin writes:

>> Chainring skips are cause by new sprockets with old chains.

> Faster with old chainrings and old chains.  The teeth on the driving
> wheel wear down.  When their circumference competes with the
> stretched chain's circumference, the chain pops off under load.

Not from my experience.  I ride chainwheels until the teeth are nearly
gone.  As the teeth wear thin from an off-axis chainline, tooth faces
become so thin that they deform on hard pedaling.  That is the time I
change chainwheels.

> I suppose a really really stretched chain could do that with a new
> chainwheel.

No.  That supposition is incorrect.

Jobst Brandt  <jobst.brandt@stanfordalumni.org>  Palo Alto CA