Pedersen self energizing brakes.

On Mon, 06 Jun 2005 20:52:53 GMT, [email hidden]

Quoted message said:

... When a RR
wheel skids, it loses traction as it glides on molten metal.

ITYM molten rubber.

Once the skid starts, the traction's lost in any event.
--
Typoes are a feature, not a bug.
Some gardening required to reply via email.
Words processed in a facility that contains nuts.

Quoted message said:

Dan Burkhart said:

Every day is an adventure for a new bike mechanic. There seems to be
no end to the exotic componentry out there. I just had a unit come
in with the Pedersen cantilever. It was easy enough to figure out
how they work, the only real mystery is, well, why? I am certainly
no engineer,so maybe someone who is can explain the advantage if
there is one. How does moving the pads forward as the brake arms
rotate toward rim contact energize the brake? Was this just another
answer to a question no one asked?

I think brake subjects on this NG easily become lost. We had a long
discussion about this a while back showing that a non-linear brake
(one that has a variable brake force to hand force ratio) is dangerous
and reduces control. It was for this reason that the Campagnolo Delta
brake was scrapped. It had a greatly variable mechanical advantage as
pads wore, even varying within brake pad compression. This caused
crashes when riders over-braked unintentionally.

The reason we don't use drum brakes on passenger cars any more is that
they are also "self-servo" (self activating) brakes. Those who recall
driving with them may recall that on foggy wet days, car brakes
occasionally locked solid with a light touch of the pedal. It is for
this non-linearity that we use disc brakes today even though disc
brakes have less heat capacity than drum brakes and why highway trucks
still use drums... often leaving long dual skid marks on highways.

A self-servo brake as you describe is useless because the user cannot
anticipate how much braking will occur from a given hand force. The
brake can lock up since it is its own application force multiplier.
The multiplier depends on the coefficient of friction of brake pads,
which most riders are aware, changes with temperature and humidity.
Because self servo brakes are unpredictable, you will not see a drum
brake on railways where a skid permanently damages wheels. When a RR
wheel skids, it loses traction as it glides on molten metal.

DON'T USE THIS BRAKE!

IIRC Scott-Pedersen sold these brakes for mountain bikes with a disclaimer not
to use them on the front. Of course no one needs the extra power on the rear,
which is usually skidding along already anyway.

Matt O.

In article <[email hidden]>,

Matt O'Toole said:

IIRC Scott-Pedersen sold these brakes for mountain bikes with a disclaimer not
to use them on the front. Of course no one needs the extra power on the rear,
which is usually skidding along already anyway.

Incorrect- Scott-Pedersen made them for the front as well (I have one).
Suntour only sold them for the rear.

--
Mike DeMicco <[email hidden]>

Matt O'Toole said:

IIRC Scott-Pedersen sold these brakes for mountain bikes with a disclaimer not
to use them on the front. Of course no one needs the extra power on the rear,
which is usually skidding along already anyway.

Of course, assuming they were manufactured for the rear they would not
work as intended on the front anyhow (unless riding backwards).

DR

Dirtroadie said:

Matt O'Toole said:

IIRC Scott-Pedersen sold these brakes for mountain bikes with a
disclaimer not to use them on the front. Of course no one needs the
extra power on the rear, which is usually skidding along already
anyway.

Of course, assuming they were manufactured for the rear they would not
work as intended on the front anyhow (unless riding backwards).

This is true, but the front ones were available too, and were not recommended.
They were pretty much pulled from the market for being too dangerous.

Matt O.

Quoted message said:

Quoted message said:

I think I understand your example, but we are talking about two
mechanisms here not one. If you were to take your thumb and push hard
on the trailing end of the brake pad without loading the lever/cable at
all, you would see (if your thumb was strong enough) the body of the
canti twist and move in towards the rim. similarly if you grab the
lever as hard as you can when the wheel is still the self energizing
mechanism doesn't come into play.. the only force you can exert on the
rim comes from the cable. The two mechanisms are related but
essentially independent in their operation.

OK when a brake is actuated it "pushes" on the rim. According to
Newton, the rim must also be "pushing" back i.e. trying to move the pad
away from the rim.
The quantity of these balanced forces is reflected in the tension on
the brake cable.

Not necessarily, and in particular, not with these brakes (or, as Jobst
points out, with drum brakes).

Imagine brake pads in the form of wedges between the fork and rim, with
the narrow end of the wedge pointing forward. If the wedges are pushed
forward so that they jam between the fork and rim, motion of the wheel will
pull them forward even farther without any additional outside force, and
the wheel will lock up.

--
Benjamin Lewis

Seeing is deceiving. It's eating that's believing.
-- James Thurber

Dirtroadie said:

Quoted message said:

I think I understand your example, but we are talking about two
mechanisms here not one. If you were to take your thumb and push hard
on the trailing end of the brake pad without loading the lever/cable at
all, you would see (if your thumb was strong enough) the body of the
canti twist and move in towards the rim. similarly if you grab the
lever as hard as you can when the wheel is still the self energizing
mechanism doesn't come into play.. the only force you can exert on the
rim comes from the cable. The two mechanisms are related but
essentially independent in their operation.

OK when a brake is actuated it "pushes" on the rim. According to
Newton, the rim must also be "pushing" back i.e. trying to move the pad
away from the rim.
The quantity of these balanced forces is reflected in the tension on
the brake cable. Now when we add the additional force of the "self
energizing" effect, where does that additonal force go if not into
cable tension? Of course that is in a system where we are assuming
non-real things such as a friction-free system.

As we have been bantering with this I think I am seeing anther
explanation for why these brakes are generally perceived as having a
benefit in actual use. The additional force which is presumably applied
to the brake pad does not have to be applied through movement of the
cable, although the cable *system* still must counteract or balance any
such additional force. Since the cable does not need to move, some of
that force can be balanced by static friction within the system. It's a
bit like throwing a rope over a tree branch and being able to
statically support a weight (at the weighted end) using a force less
than the weight being supported (at the supporting end) by relying on
the friction of the rope on the branch to make up the difference. Under
that scenario it would also be possible to have situations where
additional weight could be added without any perceived change in the
force required at the supporting end. Sailors do the same thing when
handheld lines are wrapped around winches.

This may also explain the Pedersen brakes have been generally praised
for tandem use where longer cables runs may create greater cable
friction as well (but not necessarily so where cables are not housed).

DR

The total force on the pad is not just from cable tension, it's also
from torque resulting from pushing a female helical thread against a
male helical thread. If the thread is steep enough not to be locked
by its own friction, torque results from the axial force. This is no
different in principle than other self-servoing brakes, such as
leading-shoe drum brakes. Hydraulic pressure does not rise due to
the self-servoing; the extra force comes from the component of the
reaction force to the pivot pin which is perpendicular to the brake
shoe.

This brake should have the same problems as other self-servoing
brakes; if designed to be safe it isn't very effective, and if
designed to be effective it isn't very safe. If you can predict the
highest coefficient of friction between the pads and rim that could
be generated in the worst case, you can design the brake to never
lock from self-servoing, but it won't give a lot of boost, and the
amount of boost changes as the coefficient of friction changes.

As for the effects of static friction, in the helical pivots or
elsewhere, it should make the brakes less linear and predictable,
with small changes in lever pressure making no difference, until the
static friction is broken and motion occurs, and the brake force
jumps up or down.

Dave Lehnen

hatrack & millinery said:

Quoted message said:

When a RR wheel skids, it loses traction as it glides on molten
metal.

Quoted message said:

ITYM molten rubber.

Not so. Rubber may melt but it is left behind while the skidding rail
wheel was originally round and now had a flat spot. Liquid metal is
noted for its lubrication and in the days of steam engines the sound
of spinning drivers on a steam engine, long after the throttle was
shut, were common.

Quoted message said:

Once the skid starts [on steel rails], traction is lost in any event.

[email hidden]

Benjamin Lewis said:

Imagine brake pads in the form of wedges between the fork and rim, with
the narrow end of the wedge pointing forward. If the wedges are pushed
forward so that they jam between the fork and rim, motion of the wheel will
pull them forward even farther without any additional outside force, and
the wheel will lock up.

No question. But you have added the assumption that you are now bracing
the pads against an essentially immovable surface - the fork. It is
also clear that in this scenario the pads exert a force against the
fork in order to be able to "wedge" the wheel to a stop. It is exactly
that force (that *additional* force) that must also be resisted by the
arms of the cantilevers which are NOT immovable but are braced against
movement by the brake cable, which is, in turn, braced by the grip of
the rider.

Picture gripping a rim as tightly as possible with your thumb and
forefinger as a brake caliper. Can you then grip the rim *tighter* if
you create a similar wedging effect between your finger tips and the
rim? If so, where do you get the extra strength?

DR

Dave Lehnen said:

The total force on the pad is not just from cable tension, it's also
from torque resulting from pushing a female helical thread against a
male helical thread.

Just how is this helicallly created force applied to the pad/rim if
there is no corresponding tension in the cable?

DR

Quoted message said:

A self-servo brake as you describe is useless because the user cannot
anticipate how much braking will occur from a given hand force.

DON'T USE THIS BRAKE!

I used Scott Pedersen Self-Energizing brakes for several years and tens
of thousands of miles before they became near-unobtainable. They
worked well and proved to be safer for me (because they gave me the
option of quick stops) than any other brake I used during the same span
of time.

You may be able to lift your rear wheel with ancient sidepulls, but I
couldn't do that even when I weighed 150 lbs less than I do now. With
SE cantilevers, I was able to slide back, put my chest on the saddle,
and decelerate hard (hard enough to bend unicrown forks, which I did
many times before getting Bontrager forks).

For such powerful brakes, Scott SE brakes were comparatively easy to
set up, since they required no toe-in.

The best linear-pull brakes have come to equal the stopping power of SE
cantilevers, and are even simpler to set up and more consistent in wet
weather. But SE brakes were vastly superior in their stopping power to
all other brakes available at the time, and would still be an excellent
choice for use with drop bars if they were available today.

Those who need abundant stopping power must be discriminating about
their brakes. For such riders, SE brakes are appropriate if they can
be had. Those who don't need much stopping power can make do with
whatever pleases them.

Chalo Colina

Dirtroadie said:

Dave Lehnen said:

The total force on the pad is not just from cable tension, it's also
from torque resulting from pushing a female helical thread against a
male helical thread.

Just how is this helicallly created force applied to the pad/rim if
there is no corresponding tension in the cable?

The reaction force is exerted by the bolt heads on the ends of the
brake studs, since the helix's axis is perpendicular to the tension on
the wire.

They work. Find some and try them if you don't believe it. I bent
many forks under braking force alone, and it did not take unusual lever
input to do so.

Chalo Colina

Chalo said:

The reaction force is exerted by the bolt heads on the ends of the
brake studs, since the helix's axis is perpendicular to the tension on
the wire.

That doesn't account for any force component which would create an
increased force against the rim.

Quoted message said:

They work.

I have acknowledged that the general consensus is that they work. What
I am not entirely willing to accept is the commonly accepted REASON why
they work. It makes little sense despite being oft repeated.

Quoted message said:

Find some and try them if you don't believe it.

I have some but I have never used them.

Quoted message said:

I bent many forks under braking force alone, and it did not take unusual
lever input to do so.

I'm not sure what that establishes. I am reasonably sure that I would
have difficulty bending a fork, yet I have locked up front wheels
quickly enough to sommersault gracefully over the bars. Might it be
that you are harder on equipment than I am?

DR

Dirtroadie said:

Chalo said:

The reaction force is exerted by the bolt heads on the ends of the
brake studs, since the helix's axis is perpendicular to the tension on
the wire.

That doesn't account for any force component which would create an
increased force against the rim.

Quoted message said:

They work.

I have acknowledged that the general consensus is that they work. What
I am not entirely willing to accept is the commonly accepted REASON why
they work. It makes little sense despite being oft repeated.

One consideration is that brake systems are not entirely efficient since
some of the force applied at the lever is lost to friction in the brake
handle and in the cable. When you apply force (F) to a normal brake
then only some percentage of that force is effectively applied to the
rim brakes. But in the case of self-energizing brakes, the friction in
the system actually makes the brakes work 'better' (or at least apply
more stopping force - controllability is impaired) since you're still
applying force F to the lever, but now that force plus the friction in
the system act together to balance the self braking force of the brake.

I don't have any numbers on how much of a factor such frictional losses
are, but let's assume 30%. In that case the normal brake only sees 0.7
x F applied while the self-energizing brake would be applying 1.43 x F
to the rim brake which is then reduced by a factor of 0.7 to F at the
lever. So if friction is this high then the self-energizing brake could
apply just over twice as much stopping power assuming other factors (pad
material, leverage, etc.) are equal.

Dan Burkhart said:

Every day is an adventure for a new bike mechanic. There seems to be no
end to the exotic componentry out there.
I just had a unit come in with the Pedersen cantis. It was easy enough
to figure out how they work, the only real mystery is, well, why? I am
certainly no engineer,so maybe someone who is can explain the advantage
if there is one. How does moving the pads forward as the brake arms
rotate toward rim contact energize the brake?
Was this just another answer to a question no one asked?

There have been several versions of that idea over the
years, from your SE brakes to a CLB ramped shoe/holder. All
are dangerous IMHO but some riders insist on them (??!?!?)

--
Andrew Muzi
www.yellowjersey.org
Open every day since 1 April, 1971

Quoting Matt O'Toole <[email hidden]>:

Quoted message said:

IIRC Scott-Pedersen sold these brakes for mountain bikes with a disclaimer not
to use them on the front. Of course no one needs the extra power on the rear,
which is usually skidding along already anyway.

No. A disclaimer would be of no use, because a brake of this type for the
front is differently arranged to one for use on the rear; a front one is
only ever intended for front use.

Pedersen made front and rear; Suntour only rear.
--
David Damerell <[email hidden]> Kill the tomato!
Today is First Leicesterday, June.

Quoting Werehatrack <[email hidden]>:

Quoted message said:

On Mon, 06 Jun 2005 20:52:53 GMT, [email hidden]

Quoted message said:

... When a RR
wheel skids, it loses traction as it glides on molten metal.

ITYM molten rubber.

In my experience railway trains rarely have any rubber to melt.
--
David Damerell <[email hidden]> Kill the tomato!
Today is First Leicesterday, June.

Dirtroadie said:

Chalo said:

The reaction force is exerted by the bolt heads on the ends of the
brake studs, since the helix's axis is perpendicular to the tension on
the wire.

That doesn't account for any force component which would create an
increased force against the rim.

Quoted message said:

They work.

I have acknowledged that the general consensus is that they work. What
I am not entirely willing to accept is the commonly accepted REASON why
they work. It makes little sense despite being oft repeated.

Be cool. There is a force vector that combines the rotational
component of the thrust on the helix and the rotational component of
the pull on the wire. That's how self-servo systems work. The drum
brakes in an old car don't kick your foot back off the pedal when they
engage, right?.

In a normal cantilever brake, the force generated by friction against
the rim is wholly transferred to the heads of the pivot bolts. In an
SE brake, some amount of this rim drag force is transferred to the pad
against the face of the rim. The amount transferred depends on the
pitch of the helix and its change in friction as it is loaded.

If you acknowledge that SE brakes exhibit power boosting compared to
normal cantilevers, then you must accept that the added braking force
is attributable to deflection of axial force at the pivot. There is no
other mechanism that distinguishes SE brakes from other cantis.

The trait of these brakes I have wondered about is why they never ever
fail to release when the lever is released. It doesn't matter how hard
you brake, they always stop biting when you let go of the lever.
That's a good thing, I just wonder why. I guess it would be possible
to make SE brakes that self-locked by steepening the pitch of the helix
past a certain angle.

Quoted message said:

Quoted message said:

I bent many forks under braking force alone, and it did not take unusual
lever input to do so.

I'm not sure what that establishes. I am reasonably sure that I would
have difficulty bending a fork, yet I have locked up front wheels
quickly enough to sommersault gracefully over the bars. Might it be
that you are harder on equipment than I am?

The point is that SE brakes will supply enough braking torque to bend
the fork, unlike almost all other species of bicycle brake. The fact
that you, specifically, go over the bars first just means that you
would never be able to use that amount of braking torque, because you
reach another kind of limit. My statement turns out to be irrelevant
to your question; I mentioned it because I thought you were suggesting
that self-energizing did not take place at all.

If you are not able to "max out" a brake, then you can only evaluate
its response curve and not its gross power. I think this partially
accounts for the popular rejection of SE brakes. Their response curve
tends to be somewhat abrupt and inconsistent-- it is their gross power
that is unusually good. If you can't use such a brake's primary
benefit, then you are just coping with its tradeoffs.

Chalo Colina

Quoted message said:

Benjamin Lewis said:

Imagine brake pads in the form of wedges between the fork and rim, with
the narrow end of the wedge pointing forward. If the wedges are pushed
forward so that they jam between the fork and rim, motion of the wheel
will pull them forward even farther without any additional outside
force, and the wheel will lock up.

No question. But you have added the assumption that you are now bracing
the pads against an essentially immovable surface - the fork. It is
also clear that in this scenario the pads exert a force against the
fork in order to be able to "wedge" the wheel to a stop. It is exactly
that force (that *additional* force) that must also be resisted by the
arms of the cantilevers which are NOT immovable but are braced against
movement by the brake cable, which is, in turn, braced by the grip of
the rider.

The brake cable is not the only force bracing them against movement. If
you apply an inward force on the brakes, the mechanism transforms some of
this force into force in the forward direction; conversely, if you (or in
this case, the rim), applies a forward force to the brakes, the mechanism
transforms some of that force to an inward force.

From a given fore-aft position, the pads are *not* free to move inwards and
outwards.

--
Benjamin Lewis

Seeing is deceiving. It's eating that's believing.
-- James Thurber

add a smallish gas shock each side?
equalize or reduct the tendency to lock up and send the rider into the
trees?
old stuff adapts to new stuff and revives!