Carlton Reid on QR safety

Jay Beattie wrote:
. The more
important point was that from a wheel ejection/disc brake
standpoint, if your are sliding either wheel, then you are not
getting the ejection forces theorized in this thread which,
apparently, assume infinite traction.

Forces at the front contact patch can be greater than what's usually
assumed, especially off road.

Usually, people think in terms of a deceleration limit imposed by
pitchover. "You can't apply that much force, because you'll go over
the bars."

But you can have transient forces that are much larger than the force
necessary for pitchover. On the road, it's much more unlikely, but on
rough off-road, you can envision near-tangential impacts of rocks, etc.
with the tire's perimeter. The force goes away as soon as the tire
clears the rock, so the rotational inertia of the rider&bike prevents
instantaneous pitchover. But the force the QR has to deal with (if
it's a typical disk brake) can be momentarily very large.

- Frank Krygowski

Tony Raven wrote:

Tim McNamara wrote:

Those calculations were based on real
world numbers, published measurements of pull-out resistance, and
minor things like the laws of physics.


Your calculations are all fine except that you keep considering pull out
forces in the absence of lawyers lips. Now most disk brake forks have
lawyers lips. With them the pull out forces are probably at least an
order of magnitude higher because you need to physically push them out
the way or stretch the skewer so it passes over them.


Actually, it's been explained in some detail how a bolt will likely
loosen under transverse slipping, and there have been several
descriptions of riders who have regularly found their QRs loosened in
exactly this way - ie with the lever still closed, but when installed on
the LHS, rotated in an anticlockwise direction:


"Both my G/f and I have had problems with Hopes.

1) They were done up f@cking tight.

2) Every time they've come loose, the lever has been shut, but instead
of being next to the fork leg, its pointing straight down to the floor,
implying they've "wound" loose. Spin the lever back through 180° to its
usual position and its "normal tight" again."


James
--
James Annan
see web pages for email
http://www.ne.jp/asahi/julesandjames/home/
http://julesandjames.blogspot.com/

Jay Beattie wrote:
"James Annan" wrote in message
...

dvt wrote:


Tim McNamara wrote:


One thing I wondered about was instantaneous loading versus

static

loading, if those are the correct terms, which I have no

idea how to

calculate. I would think- but don't know- that a quick jam

on the

brake at 25 mph would result in a high sharp force compared

to my more

static force based on a .6 g deceleration. Would the

magnitude of the

force be raised with higher speeds, or just the time

interval over

which the load develops? My understanding of physics

suggests the

latter.


It is possible to generate more than 0.6g deceleration

momentarily.

Others have written in this NG about the peak forces possible

on rough

terrain. But I don't think many people ride that hard.


The cannondale "tests" measured a peak 235 ft-pounds of braking

torque

on the front wheel, fromw hich you can work out about 950N

deceleration

and 3800N ejection force, far in excess of the ballpark

estimates I and

others have produced based on a steady 0.6g braking.



http://www.ne.jp/asahi/julesandjames...annondale.html

Do these calculations assume infinite traction and occupant
restraint? There is a point at which braking force will eject
the ride, and the ejection of the wheel will be irrelevant since
the rider already is airborne. I also wonder whether you can
generate those super high braking forces when your front wheel is
skidding down the road or sliding in soft dirt. In fact, most of
the hard braking on an MTB is on the rear wheel. Somebody should
rig a real bike with strain gauges or accelerometers (or whatever
the instrumentation should be) and find out what the real world
forces are. I had a broken frame case where we did that. Very
enlightening. -- Jay Beattie.


The bit you apparently missed in my post:

The cannondale "tests" MEASURED a peak 235 ft-pounds of braking torque

If you'd bothered clicking on the link, you'd have seen what they did.

James
--
James Annan
see web pages for email
http://www.ne.jp/asahi/julesandjames/home/
http://julesandjames.blogspot.com/

Jay Beattie writes:

In fact, most of the hard braking on an MTB is on the rear wheel.

Dude! Most of you braking should be applied to the front wheel. If
you're jamming on the rear brake, you're not slowing you're
sliding.

Oh and the rest of your post was wrong too.

Dude! You are not going to lock up your front wheel on loose dirt
because it is really hard, dude, to control a front wheel skid.
More back brake than front on loose terrain, dude. That's what I'm
getting at. In fact you lock up your rear even on purpose sometimes
to carve that big, dirt churning turn, dude. Try that on the front
wheel for a little fun sometime, dude.

That may be your perception but in fact most heavy braking is done
with the front wheel. If traction is miserable, then no heavy braking
is done with either wheel. Just the same 1g braking occurs often on
descending hard and dry trails with embedded river bottom (rounded)
rocks. On such descents the bicycle is barely on the ground between
bumps and on bumps maximum braking occurs as the suspension fork
absorbs the rock. These are the jarring force reversals that cause
axles to move in the dropout and only on the hub brake side, the other
end having no ejection forces.

Jobst Brandt

Tony Raven writes:

Tim McNamara wrote:
Those calculations were based on real world numbers, published
measurements of pull-out resistance, and minor things like the laws
of physics.

Your calculations are all fine except that you keep considering pull
out forces in the absence of lawyers lips. Now most disk brake
forks have lawyers lips. With them the pull out forces are probably
at least an order of magnitude higher because you need to physically
push them out the way or stretch the skewer so it passes over them.

I considered the lawyer lips in their correct role: they are there in
case of *failure* of the QR to retain the axle in the dropout and are
not part of the retention system proper. If they lawyer lips are
what's holding your axle in the dropouts, that is because of a
*failure* of the QR to retain the axle. Relying on the lawyer lips to
hold in the wheel is simply horrendous incompetent design.

I didn't even have to appeal to the issue of transverse cyclic
forces to show that the problem exists.

Only because you and everyone else keep ignoring the presence and
influence of the lawyers lips. If you include them you have to
resort to transverse cyclical forces and QR unscrewing or some other
mechanism to allow wheel ejection. That is the critical bit that
everyone avoids dealing with and has yet to show clear demonstration
of a mechanism preferring instead to rely on the post hoc fallacy
that because a wheel was lost it must have been ejected by the
brakes. When the mechanism can be demonstrated I will be persuaded
but at present it's the elephant in the room everyone is pretending
is not there.

Given how often lawyer lips have come up in this and the previous
threads on the topic, there is no chance that anyone has forgotten the
lawyer lips. The intent of laywer lips is not to provide axle
retention for disk brakes, it is to prevent people who don't know how
to use QRs from losing their front wheels.

The elephant is actually the size of a cat, and we all know it's
there. Those of us advocating change are just prudent enough not to
rely on the cat to save us.

Going back to the first principles of the discussion: there is a
reactive force that results from using disk brakes. That force has a
specific direction dictated by the location of the brak caliper. With
many if not most current designs which mount the front brake on the
trailing side of the fork, the direction of that reactive force is
towards the opening of the dropouts. This force attempts to push the
axle out of the dropout, hence it is called an "ejection force." The
magnitude of the force is large- per Cannondale's numbers up to 854
pounds. Even using my more conservative numbers, as corrected by dvt,
the net ejection force can easily be over 400 lbs acting on one
dropout. That's more than the pull-out resistance measured by Howat
et al for quite a few skewers.

"James Annan" wrote in message
...
Jay Beattie wrote:
"James Annan" wrote in
message
...

dvt wrote:


Tim McNamara wrote:


One thing I wondered about was instantaneous loading versus

static

loading, if those are the correct terms, which I have no

idea how to

calculate. I would think- but don't know- that a quick jam

on the

brake at 25 mph would result in a high sharp force compared

to my more

static force based on a .6 g deceleration. Would the

magnitude of the

force be raised with higher speeds, or just the time

interval over

which the load develops? My understanding of physics

suggests the

latter.


It is possible to generate more than 0.6g deceleration

momentarily.

Others have written in this NG about the peak forces
possible

on rough

terrain. But I don't think many people ride that hard.


The cannondale "tests" measured a peak 235 ft-pounds of
braking

torque

on the front wheel, fromw hich you can work out about 950N

deceleration

and 3800N ejection force, far in excess of the ballpark

estimates I and

others have produced based on a steady 0.6g braking.




http://www.ne.jp/asahi/julesandjames...annondale.html

Do these calculations assume infinite traction and occupant
restraint? There is a point at which braking force will
eject
the ride, and the ejection of the wheel will be irrelevant
since
the rider already is airborne. I also wonder whether you can
generate those super high braking forces when your front
wheel is
skidding down the road or sliding in soft dirt. In fact,
most of
the hard braking on an MTB is on the rear wheel. Somebody
should
rig a real bike with strain gauges or accelerometers (or
whatever
the instrumentation should be) and find out what the real
world
forces are. I had a broken frame case where we did that.
Very
enlightening. -- Jay Beattie.


The bit you apparently missed in my post:

The cannondale "tests" MEASURED a peak 235 ft-pounds of braking
torque

If you'd bothered clicking on the link, you'd have seen what
they did.

I don't want to dig myself in deeper here, but I DID read that
link, and that is why I asked the question -- but my attention
was drawn to the fixture with the rotating drum and trying to
figure out what that was about. Also, the prior discussion of
the force generated by a "quick jam" on the brakes resonated with
me because I rode in to work yesterday on ice. A quick jam on
the brake (I ride a cross-bike with a disc) would have generated
little force at the drop outs and would have landed me on my ass
or somewhere else. That is why I thought traction was missing
from the equation, but apparently it is not, and now I see why
(actually a few posts ago). -- Jay Beattie.

In article
.com,
"David Martin" wrote:

Mike Causer wrote:
On Thu, 16 Feb 2006 12:36:31 -0800, Jay Beattie wrote:

The more important
point was that from a wheel ejection/disc brake standpoint, if your are
sliding either wheel, then you are not getting the ejection forces
theorized in this thread which, apparently, assume infinite traction.

Yes and no. Infinite traction does not give infinite braking capability.
Once above a coefficient of friction better than 0.6 - 0.7 any extra
does no good because the rider will either be over the bars or using
superior skill to pull a "stoppie". In either case the actual "G" is
going to be 0.6 - 0.7 So assume infinite traction if you wish, it makes
no difference to the forces we are discussing.

What does make a difference is if the force is acting in rotating the
wheel the other way.
Ie landing on a rock with the back part of the front wheel. Given that
any rotation is prevented by the rear wheel, a much larger force can be
applied than 0.6g. An applied brake caliper will act as a pivot and
provide an ejection force. (simple visualisation: Pick up a pen, hold
it a short distance from one end with the long end towards the table at
an angle. Push down. The pen rotates around your grip, pushing the free
end down harder. There is an ejection force that is proportional to the
downward force, ie the greater the effective weight of the rider, the
greater the ejection force. The weight of the rider acts on the brake
caliper, rather than the axle.

As my mechanics is a little rusty, I can only provice a qualitative
analysis right now. A quantitative analysis can be performed by someone
more qualified.

We have already had good analyses. The excess of brake
reaction at the axle increases with rider weight. The
brake reaction force at the axle is a multiple of the
rider weight; around 1.4-1.6. A 140 lb rider might see an
ejection force that the quick release can hold against
where a 190 lb rider braking in the same circumstances as
the lighter rider will see an ejection force against which
the QR cannot hold. For both riders the bumps and braking
act to loosen the QR nut, though the acting does not
always succeed in actually loosening the nut.

--
Michael Press

A Muzi a écrit :

-Rolled thread on brake bolts?-

jim beam wrote:

eh? like this?
http://www.flickr.com/photos/38636024@N00/99524699/

The Shimano bolt referenced is indeed rolled. That is
not at all common to our industry.

As I wrote yesterday, the first three examples of caliper brakes that
came to hand appear to have rolled threads on a stepped shaft. I've
just found a fourth - a modern Shimano BR-A550 57mm dual pivot. The
bolt is of exactly the same form as the Shimano 600 bolt pictured, with
one rolled thread on the narrower, swaged portion of the shaft, and a
second rolled thread on the broader portion.

I pulled one of those Shimano bolts and a loupe. It is also
the only brake bolt here without a center pip for turning.

All three of the Shimano brakes that I have on had (1050, 6208, A550)
have an irregular, concave end, like the end of an ordinary rolled M6
screw. The Superbe Pro BRS shows signs of milling along the shaft and
on both ends. The step in the shaft of the Suntour looks cut, while the
Shimanos look swaged. But all the threads look rolled.

James Thomson

Tim McNamara wrote:

I considered the lawyer lips in their correct role: they are there in
case of *failure* of the QR to retain the axle in the dropout and are
not part of the retention system proper. If they lawyer lips are
what's holding your axle in the dropouts, that is because of a
*failure* of the QR to retain the axle. Relying on the lawyer lips to
hold in the wheel is simply horrendous incompetent design.



Its called fail safe design. If the primary retention device fails, the
secondary takes over preferably in a way that ensures it is clear to the
user that there has been a primary system failure.

For an unsafe condition to occur both the primary and secondary systems
have to have failed. All the calculations you are doing only show how a
primary failure might occur by the QR slipping. What they have failed
to address other than by speculation is how the secondary system also
fails which is a necessary condition for the wheel to be ejected.



Given how often lawyer lips have come up in this and the previous
threads on the topic, there is no chance that anyone has forgotten the
lawyer lips. The intent of laywer lips is not to provide axle
retention for disk brakes, it is to prevent people who don't know how
to use QRs from losing their front wheels.


Correct but by default it should prevent a QR from exiting the fork for
forces many times the highest calculated here for an ejection force.
Unless you can demonstrate rather than speculate the mechanism by which
this also fails you have failed to show the two necessary conditions for
wheel ejection to occur - a slipping QR and a verified mechanism for the
QR to pass the lawyers lips.


--
Tony

"The best way I know of to win an argument is to start by being in the
right."
- Lord Hailsham

In article , Tony Raven
wrote:

Tim McNamara wrote:

I considered the lawyer lips in their correct role: they are there in
case of *failure* of the QR to retain the axle in the dropout and are
not part of the retention system proper. If they lawyer lips are
what's holding your axle in the dropouts, that is because of a
*failure* of the QR to retain the axle. Relying on the lawyer lips to
hold in the wheel is simply horrendous incompetent design.



Its called fail safe design. If the primary retention device fails, the
secondary takes over preferably in a way that ensures it is clear to the
user that there has been a primary system failure.

The primary device, unlike the secondary, does not simply retain, it
secures. The difference between these two functions is the difference
between having and losing control of the bike.


For an unsafe condition to occur both the primary and secondary systems
have to have failed.

snip

That's an amusing contradiction. The secondary retention device,
'preferably' takes over in such a manner as to clearly inform that the
QRs have failed to secure the wheel (what if the Lawyers' Lips cite
client privilege, keeping mum?), which in of itself constitutes an
unsafe condition, but is not considered so until the secondary
retention system fails also.

Luke