disk-brake wheel-ejection question

Quoted message said:

While I was browsing the familiar disk-brake wheel-ejection
thread, a question occurred to me that must have been
answered somewhere, but I can't find it.

Various diagrams show that a trailing-caliper disk-brake
will try to lever the axle down and out of the U-shaped
dropouts.

But does the braking force that tries to eject the axle also
try to retain the disk?

That is, does the geometry require the axle to move downward
and away (at some angle) from the caliper pads, which are
squeezing the disk and trying to keep the disk in place?

Yes to the latter question at least. Although it does vary with disk
size. But it doesn't really have any effect. The disk is already
slipping through the pads, in order for the wheel to be ejected, the
direction of travel of the disk edge only needs to change very slightly.

There's a very simple way to show this. Remove the skewer entirely, and
lock the brake on very hard so the disk cannot move. If you try push the
bike forwards on the ground, it will be very hard to eject the wheel (at
least unless the angle from pad to axle happens to be very close to
perpendicular to the dropout slot). If, however, you release the brake,
and gently squeeze it while the bike is already rolling, the wheel will
pop out very easily.

James
--
If I have seen further than others, it is
by treading on the toes of giants.
http://www.ne.jp/asahi/julesandjames/home/

James Annan said:

Quoted message said:

While I was browsing the familiar disk-brake wheel-ejection
thread, a question occurred to me that must have been
answered somewhere, but I can't find it.

Various diagrams show that a trailing-caliper disk-brake
will try to lever the axle down and out of the U-shaped
dropouts.

But does the braking force that tries to eject the axle also
try to retain the disk?

That is, does the geometry require the axle to move downward
and away (at some angle) from the caliper pads, which are
squeezing the disk and trying to keep the disk in place?

Yes to the latter question at least. Although it does vary with disk
size. But it doesn't really have any effect. The disk is already
slipping through the pads, in order for the wheel to be ejected, the
direction of travel of the disk edge only needs to change very slightly.

There's a very simple way to show this. Remove the skewer entirely, and
lock the brake on very hard so the disk cannot move. If you try push the
bike forwards on the ground, it will be very hard to eject the wheel (at
least unless the angle from pad to axle happens to be very close to
perpendicular to the dropout slot). If, however, you release the brake,
and gently squeeze it while the bike is already rolling, the wheel will
pop out very easily.

James

Dear James,

The two examples seem to be extremes that will support
either outcome, with light braking ejecting wheels and heavy
braking locking them in place.

At one extreme, the disk is locked by the brake and the
wheel won't eject at all.

At the other extreme, the disk is barely touched by the
brake and the wheel ejects almost effortlessly.

There's no skewer, and I'm guessing (but perhaps mistaken)
that you mean testing by rolling the bike along by hand with
no rider's weight on it.

What I'm wondering is what happens in-between when the disk
is moving (which is toward the extreme of the light braking
that pops the wheel out) but is also being given a serious
squeeze by the pads (which is toward the extreme of the
heaviest possible braking that locks the wheel in place).

I'm guessing that equally serious forces are needed to pull
the disk in any direction, whether in the normal rotating
direction or "sideways" by dragging the axle down and out of
the dropouts.

That is, there's considerable force resisting the disk's
movement under normal braking in the normal direction
because the pads squeeze it, so it seems likely that just as
much force will be needed to move it in any direction in the
same plane, including downward.

Of course, this is resistance in terms of movement, so maybe
it works out that there's lots of travel (a full spin, say)
of the disk in normal movement, but only half an inch of
linear downward movement to let the axle duck out.

Or, as you suggest, the necessary movement is very slight in
terms of an angle.

What would the unskewered wheel do under moderate real-world
braking? I think that you're saying that it would pop out
just as easily as if light braking were applied.

If so, am I right that your theory is that an unskewered
wheel will pop out easily at all braking levels--until the
brake pads lock the wheel and stop it not only from
rotating, but also from moving in any other direction?

I'm really not at all sure about the physics here, so I'm
hoping that you (or anyone else) can work through some kind
of example illustrating what (if any) resistance the pads
squeezing the disk provide when a wheel tries to pull the
disk (already braking) at roughly 90 degrees to its normal
direction.

(I'm having trouble just describing it. There's a tendency
to say that the axle is trying to "climb" out of the
dropouts, when it's actually moving downward.)

Carl Fogel

Quoted message said:

James Annan said:

Quoted message said:

While I was browsing the familiar disk-brake wheel-ejection
thread, a question occurred to me that must have been
answered somewhere, but I can't find it.

Various diagrams show that a trailing-caliper disk-brake
will try to lever the axle down and out of the U-shaped
dropouts.

But does the braking force that tries to eject the axle also
try to retain the disk?

That is, does the geometry require the axle to move downward
and away (at some angle) from the caliper pads, which are
squeezing the disk and trying to keep the disk in place?

Yes to the latter question at least. Although it does vary with disk
size. But it doesn't really have any effect. The disk is already
slipping through the pads, in order for the wheel to be ejected, the
direction of travel of the disk edge only needs to change very slightly.

There's a very simple way to show this. Remove the skewer entirely, and
lock the brake on very hard so the disk cannot move. If you try push the
bike forwards on the ground, it will be very hard to eject the wheel (at
least unless the angle from pad to axle happens to be very close to
perpendicular to the dropout slot). If, however, you release the brake,
and gently squeeze it while the bike is already rolling, the wheel will
pop out very easily.

James

Dear James,

The two examples seem to be extremes that will support
either outcome, with light braking ejecting wheels and heavy
braking locking them in place.

At one extreme, the disk is locked by the brake and the
wheel won't eject at all.

At the other extreme, the disk is barely touched by the
brake and the wheel ejects almost effortlessly.

There's no skewer, and I'm guessing (but perhaps mistaken)
that you mean testing by rolling the bike along by hand with
no rider's weight on it.

What I'm wondering is what happens in-between when the disk
is moving (which is toward the extreme of the light braking
that pops the wheel out) but is also being given a serious
squeeze by the pads (which is toward the extreme of the
heaviest possible braking that locks the wheel in place).

I'm guessing that equally serious forces are needed to pull
the disk in any direction, whether in the normal rotating
direction or "sideways" by dragging the axle down and out of
the dropouts.

That is, there's considerable force resisting the disk's
movement under normal braking in the normal direction
because the pads squeeze it, so it seems likely that just as
much force will be needed to move it in any direction in the
same plane, including downward.

Of course, this is resistance in terms of movement, so maybe
it works out that there's lots of travel (a full spin, say)
of the disk in normal movement, but only half an inch of
linear downward movement to let the axle duck out.

Or, as you suggest, the necessary movement is very slight in
terms of an angle.

What would the unskewered wheel do under moderate real-world
braking? I think that you're saying that it would pop out
just as easily as if light braking were applied.

If so, am I right that your theory is that an unskewered
wheel will pop out easily at all braking levels--until the
brake pads lock the wheel and stop it not only from
rotating, but also from moving in any other direction?

I'm really not at all sure about the physics here, so I'm
hoping that you (or anyone else) can work through some kind
of example illustrating what (if any) resistance the pads
squeezing the disk provide when a wheel tries to pull the
disk (already braking) at roughly 90 degrees to its normal
direction.

(I'm having trouble just describing it. There's a tendency
to say that the axle is trying to "climb" out of the
dropouts, when it's actually moving downward.)

Carl Fogel

Put the bike in a stand, lock the brake with a tie wrap, remove the
skewer, yank down on the wheel..or, if you are feeling scientific,
hang ever increasing weights on it until it pulls out. I'd be happy
to do this, except I don't own a disc equipped bike (nor, by the way
would I ever until they move the caliper to the front.)

Alternatively ,if your feeling brave, take a disc equipped bike (minus
skewer) for a slow ride on a nice grassy surface (slowly) and apply
the brakes.

My bet is the force the caliper exerts clamping the disc is minimal,
but that is just a guess. Anyone up for the test?

Regards,

Bob

Quoted message said:

Of course, this is resistance in terms of movement, so maybe
it works out that there's lots of travel (a full spin, say)
of the disk in normal movement, but only half an inch of
linear downward movement to let the axle duck out.

Or, as you suggest, the necessary movement is very slight in
terms of an angle.

What would the unskewered wheel do under moderate real-world
braking? I think that you're saying that it would pop out
just as easily as if light braking were applied.

Yes, and I think your previous two sentences explain it well enough.

I'll try a different analogy. Consider a brick sliding down a planar
slope due to gravity. Even though there is a considerable frictional
force involved, even the slightest side force on the brick will cause
its direction of slip to diverge from the direct downhill direction -
even blowing firmly on it would have a (small) effect. However, if the
frictional force is so high that the brick does not slip at all and is
sitting still on the slope, then a large side force may be required to
make it move at all.

Under normal conditions, the disk slips through the pads in the
direction perpendicular to the pad-axle line. In order for the wheel to
slip out, there only needs to be a very small side force which causes
the angle of slip to change almost imperceptibly. (This side force is
generated by the reaction of rearward portion of dropout on the axle, if
you want to look at it at this level of detail).

James
--
If I have seen further than others, it is
by treading on the toes of giants.
http://www.ne.jp/asahi/julesandjames/home/

James Annan said:

Quoted message said:

Of course, this is resistance in terms of movement, so maybe
it works out that there's lots of travel (a full spin, say)
of the disk in normal movement, but only half an inch of
linear downward movement to let the axle duck out.

Or, as you suggest, the necessary movement is very slight in
terms of an angle.

What would the unskewered wheel do under moderate real-world
braking? I think that you're saying that it would pop out
just as easily as if light braking were applied.

Yes, and I think your previous two sentences explain it well enough.

I'll try a different analogy. Consider a brick sliding down a planar
slope due to gravity. Even though there is a considerable frictional
force involved, even the slightest side force on the brick will cause
its direction of slip to diverge from the direct downhill direction -
even blowing firmly on it would have a (small) effect. However, if the
frictional force is so high that the brick does not slip at all and is
sitting still on the slope, then a large side force may be required to
make it move at all.

Under normal conditions, the disk slips through the pads in the
direction perpendicular to the pad-axle line. In order for the wheel to
slip out, there only needs to be a very small side force which causes
the angle of slip to change almost imperceptibly. (This side force is
generated by the reaction of rearward portion of dropout on the axle, if
you want to look at it at this level of detail).

James

Dear James,

I'm not sure that analogies will give us the actual force,
but consider this one, which seems to be closer to a disk
being squeezed between two pads than a brick sliding down a
slope.

Squeeze a brick hard between two planks, hard enough that it
takes a strong pull to move the brick steadily between the
two planks.

Won't it take just as strong a pull (though shorter) to pull
the brick out from between the two planks?

That is, the force and friction applied by the pads to the
disk is enormous and, as far as I know, has no directional
preference. Shouldn't the disk resist moving in any
direction equally, whether rotating as intended, or at
roughly 90 degrees toward the dropouts from the calipers?

Where I'm dubious is when you say that only a trifling force
is needed to pull the piece of metal pinched between two
pads in one direction, but not the other.

If this were true, wouldn't only a trifling force also be
sufficient to accelerate the disk to a higher rotational
speed? How would the pads distinguish between the direction
of the force, normal or 90-degrees to normal?

There's certainly less surface to slip in the ejection
direction, so maybe that explains things. Pulling out will
drag the pads across a much shorter surface than normal
braking.

But it seems to me that the same force must be applied to
overcome the pad friction and move an inch of disk surface
in any direction, whether rotating normally or being tugged
toward the dropouts.

Carl Fogel

bobqzzi said:

Quoted message said:

James Annan said:

[email hidden] wrote:

> While I was browsing the familiar disk-brake wheel-ejection
> thread, a question occurred to me that must have been
> answered somewhere, but I can't find it.
>
> Various diagrams show that a trailing-caliper disk-brake
> will try to lever the axle down and out of the U-shaped
> dropouts.
>
> But does the braking force that tries to eject the axle also
> try to retain the disk?
>
> That is, does the geometry require the axle to move downward
> and away (at some angle) from the caliper pads, which are
> squeezing the disk and trying to keep the disk in place?

Yes to the latter question at least. Although it does vary with disk
size. But it doesn't really have any effect. The disk is already
slipping through the pads, in order for the wheel to be ejected, the
direction of travel of the disk edge only needs to change very slightly.

There's a very simple way to show this. Remove the skewer entirely, and
lock the brake on very hard so the disk cannot move. If you try push the
bike forwards on the ground, it will be very hard to eject the wheel (at
least unless the angle from pad to axle happens to be very close to
perpendicular to the dropout slot). If, however, you release the brake,
and gently squeeze it while the bike is already rolling, the wheel will
pop out very easily.

James

Dear James,

The two examples seem to be extremes that will support
either outcome, with light braking ejecting wheels and heavy
braking locking them in place.

At one extreme, the disk is locked by the brake and the
wheel won't eject at all.

At the other extreme, the disk is barely touched by the
brake and the wheel ejects almost effortlessly.

There's no skewer, and I'm guessing (but perhaps mistaken)
that you mean testing by rolling the bike along by hand with
no rider's weight on it.

What I'm wondering is what happens in-between when the disk
is moving (which is toward the extreme of the light braking
that pops the wheel out) but is also being given a serious
squeeze by the pads (which is toward the extreme of the
heaviest possible braking that locks the wheel in place).

I'm guessing that equally serious forces are needed to pull
the disk in any direction, whether in the normal rotating
direction or "sideways" by dragging the axle down and out of
the dropouts.

That is, there's considerable force resisting the disk's
movement under normal braking in the normal direction
because the pads squeeze it, so it seems likely that just as
much force will be needed to move it in any direction in the
same plane, including downward.

Of course, this is resistance in terms of movement, so maybe
it works out that there's lots of travel (a full spin, say)
of the disk in normal movement, but only half an inch of
linear downward movement to let the axle duck out.

Or, as you suggest, the necessary movement is very slight in
terms of an angle.

What would the unskewered wheel do under moderate real-world
braking? I think that you're saying that it would pop out
just as easily as if light braking were applied.

If so, am I right that your theory is that an unskewered
wheel will pop out easily at all braking levels--until the
brake pads lock the wheel and stop it not only from
rotating, but also from moving in any other direction?

I'm really not at all sure about the physics here, so I'm
hoping that you (or anyone else) can work through some kind
of example illustrating what (if any) resistance the pads
squeezing the disk provide when a wheel tries to pull the
disk (already braking) at roughly 90 degrees to its normal
direction.

(I'm having trouble just describing it. There's a tendency
to say that the axle is trying to "climb" out of the
dropouts, when it's actually moving downward.)

Carl Fogel

Put the bike in a stand, lock the brake with a tie wrap, remove the
skewer, yank down on the wheel..or, if you are feeling scientific,
hang ever increasing weights on it until it pulls out. I'd be happy
to do this, except I don't own a disc equipped bike (nor, by the way
would I ever until they move the caliper to the front.)

Alternatively ,if your feeling brave, take a disc equipped bike (minus
skewer) for a slow ride on a nice grassy surface (slowly) and apply
the brakes.

My bet is the force the caliper exerts clamping the disc is minimal,
but that is just a guess. Anyone up for the test?

Regards,

Bob

Dear Bob,

If the clamping force that the caliper exerts is minimal,
how does it provide far from minimal braking and ejection
forces?

That is, if the pads were greased, then I could see how the
disk would pull out easily--but then greased brake pads
wouldn't provide much braking or ejection force, would they?

And the clamping/braking/ejection forces might scale up
together, or one might outstrip the other two.

This is like those beastly aerodynamic situations. I want
someone to draw me some extremely clear diagrams with
beautifully simple, well-chosen numbers that will leave me
in no doubt about how much force is needed to yank a disk
out from between two pads when the pads are pressing hard
enough to slow a bike down with their friction.

Everything else in the diagrams about trailing brakes has
arrows and numbers that I can pretend to follow, even though
sometimes my private efforts to work through things end up
with the braking forces flipping the rider over backwards.

Shouldn't the clamping effect of the pads resist movement by
the disk equally in any direction? If so, maybe the effect
still wouldn't matter because the disk has to move only a
little bit in the wrong direction (pulling out), compared to
moving a lot in the normal rotating direction.

But I think that it would be hard to yank a motionless disk
out from between a pair of pads--harder and harder as brake
pressure increases. I think that James is saying that things
somehow become easier when the disk is sliding between the
pads, but I kinda-sorta think that it should be just as hard
to make the disk move the right way (rotating) as it would
be to make it move about 90 degrees the wrong way
(ejecting).

Carl Fogel

Quoted message said:

I'm not sure that analogies will give us the actual force,
but consider this one, which seems to be closer to a disk
being squeezed between two pads than a brick sliding down a
slope.

Squeeze a brick hard between two planks, hard enough that it
takes a strong pull to move the brick steadily between the
two planks.

Won't it take just as strong a pull (though shorter) to pull
the brick out from between the two planks?

No, as long as you keep up the main pull, even a very small side force
will make the plank slide at a small angle, sufficient that it will
eventually escape from between the bricks.

For another look at the same thing, if the angle of pull is changed very
slightly, the plank will follow the line of pull and start to move
somewhat sideways, even though the lateral component of force can be
extremely small.

Quoted message said:

Where I'm dubious is when you say that only a trifling force
is needed to pull the piece of metal pinched between two
pads in one direction, but not the other.

No, what I am saying is that only a trifling side force is needed to
change the direction of travel by a trifling amount. We already know the
total force is greater than the frictional reaction, the only question
is in which direction the (net) force acts, as this is the direction in
which the slip occurs. You can't resolve the force in two arbitrary
directions, state that one is below the frictonal threshold and thus no
slip occurs in that direction. What if the total force available is only
marginally above the frictional reaction, and you resolve at +-45
degrees to the main pull? Using your logic, each component is too small
to cause any motion at all.

James
--
If I have seen further than others, it is
by treading on the toes of giants.
http://www.ne.jp/asahi/julesandjames/home/

Carl Fogel said:

While I was browsing the familiar disk-brake wheel-ejection thread,
a question occurred to me that must have been answered somewhere,
but I can't find it.

Quoted message said:

Various diagrams show that a trailing-caliper disk-brake will try to
lever the axle down and out of the U-shaped dropouts.

Quoted message said:

But does the braking force that tries to eject the axle also try to
retain the disk?

Quoted message said:

That is, does the geometry require the axle to move downward and
away (at some angle) from the caliper pads, which are squeezing the
disk and trying to keep the disk in place?

Quoted message said:

I'm not sure if the typical geometry requires the axle to move
downward very much in relation to the caliper to escape from the
dropouts.

Quoted message said:

Nor am I sure how much force is needed to pull the disk downward
enough to clear the dropouts...

Rather than postulate and hypothesize on what occurs, try it. Loosen
the skewer so that it is out of the problem and, without even sitting
on the bicycle, push it forward and apply the brake. If you don't
have such a bicycle, I'm sure there is a bicycle shop nearby where you
can do the test with the benefit of an observer from the shop.

The wheel will be canted and jam in the fork whether you brake hard or
lightly, no one having a step function hand clasp that can cause the
effect which you propose. This is just another diversion from the
main topic, that the brake caliper is in the wrong place for a bicycle
with a manually removable front wheel (which the ones in question are).

Jobst Brandt
[email hidden]

Quoted message said:

Dear James,

I'm not sure that analogies will give us the actual force,
but consider this one, which seems to be closer to a disk
being squeezed between two pads than a brick sliding down a
slope.

Squeeze a brick hard between two planks, hard enough that it
takes a strong pull to move the brick steadily between the
two planks.

Won't it take just as strong a pull (though shorter) to pull
the brick out from between the two planks?

Quoted message said:

Carl Fogel

Good analogy. Now lets think of the brick sliding between the planks
at a steady rate over a fair distance (like the rotor going through
the pads) Now introduce a small force pushing the brick out of the
planks. I don't believe it would take long for the brick to move out
of the planks. Remember, that X "plank pressure" produces Y clamping
force which provides Z resistance to movement in any direction.

If a new force moving the brick outward is introduced it is added to
the force sliding it upward, the planks really have no more force to
resist the outward movement.

Are you familiar with the Friction Circle?

http://www.nyracer.com/friction.htm

Quoted message said:

Carl Fogel said:

While I was browsing the familiar disk-brake wheel-ejection thread,
a question occurred to me that must have been answered somewhere,
but I can't find it.

Quoted message said:

Various diagrams show that a trailing-caliper disk-brake will try to
lever the axle down and out of the U-shaped dropouts.

Quoted message said:

But does the braking force that tries to eject the axle also try to
retain the disk?

Quoted message said:

That is, does the geometry require the axle to move downward and
away (at some angle) from the caliper pads, which are squeezing the
disk and trying to keep the disk in place?

Quoted message said:

I'm not sure if the typical geometry requires the axle to move
downward very much in relation to the caliper to escape from the
dropouts.

Quoted message said:

Nor am I sure how much force is needed to pull the disk downward
enough to clear the dropouts...

Rather than postulate and hypothesize on what occurs, try it. Loosen
the skewer so that it is out of the problem and, without even sitting
on the bicycle, push it forward and apply the brake. If you don't
have such a bicycle, I'm sure there is a bicycle shop nearby where you
can do the test with the benefit of an observer from the shop.

The wheel will be canted and jam in the fork whether you brake hard or
lightly, no one having a step function hand clasp that can cause the
effect which you propose. This is just another diversion from the
main topic, that the brake caliper is in the wrong place for a bicycle
with a manually removable front wheel (which the ones in question are).

Jobst Brandt
[email hidden]

Dear Jobst,

These things may be tricky, so bear with me for a moment.
You may have missed my point.

Elsewhere, James has indicated that he expects the wheel
will remain in place and immobile if the pads stop the
disk--no disk motion through the pads, no ejection (given,
of course, the right geometry).

Is motion necessary to your explanation? That is, when you
say "whether you brake hard or lightly," is it understood
that it's not hard enough to stop the motion?

In any case, my question is whether there is a significant
resistance to ejection caused by the clamping of the pads.
(Maybe "clamping" will turn out to be a misleading term.)

Whether this pad clamping, when coupled with the weight of
the rider and the clamping of the quick release, would
provide enough resistance to explain why so many wheels fail
to eject is another matter.

That is, I'm not asking if the pad clamping by itself is
enough to retain the wheel. I'm asking how much resistance
(if any) it provides. I'm asking if it's an overlooked
force, just as Jim Beam suggested that the resistance of the
serrated washers biting into fork metal is an overlooked
force.

If pad clamping provides a significant retaining force, it
could still not be enough to retain a wheel with no quick
release clamped--the wheel would still eject, but we would
not be able to tell how much overlooked resistance the
clamping provided. The greater the ejection force,
presumably the greater the braking force--and presumably
also the greater the clamping force.

I see figures and calculations for other forces, but not for
whatever resistance pad-clamping might provide. If you (or
anyone) has some figures, equations, or diagrams, I'm hoping
to see them soon. But I don't think that rolling a bike with
an open quick-release across the floor and braking will tell
tell me much about how much resistance (if any) the pads
clamping the disk are providing to wheel ejection, other
than that it's not enough by itself to hold the wheel in.

Is the resistance 0.1% of the ejection force? One per cent?
Ten per cent? Twenty? Since I have no idea how to measure or
calculate this, I'm asking other people to tackle it.

Is it overcome instantly by the ejection force? Or does the
wheel sort of spiral out over several revolutions?
Elsewhere, James is suggesting that a very small force will
be enough to move the disk out from between the pads if it's
moving, but I'm hoping for figures or an example showing how
small or large a force over how great a rotating distance.

Does any clamping resistance scale up with braking and
ejection force? That is, how are the three forces related?
If I clamp twice as hard, does it resist twice as hard, slow
down twice as hard, and eject twice as hard? (In contrast,
the rider's weight resisting ejection remains the same.)

Carl Fogel

bobqzzi said:

Quoted message said:

Dear James,

I'm not sure that analogies will give us the actual force,
but consider this one, which seems to be closer to a disk
being squeezed between two pads than a brick sliding down a
slope.

Squeeze a brick hard between two planks, hard enough that it
takes a strong pull to move the brick steadily between the
two planks.

Won't it take just as strong a pull (though shorter) to pull
the brick out from between the two planks?

Quoted message said:

Carl Fogel

Good analogy. Now lets think of the brick sliding between the planks
at a steady rate over a fair distance (like the rotor going through
the pads) Now introduce a small force pushing the brick out of the
planks. I don't believe it would take long for the brick to move out
of the planks. Remember, that X "plank pressure" produces Y clamping
force which provides Z resistance to movement in any direction.

If a new force moving the brick outward is introduced it is added to
the force sliding it upward, the planks really have no more force to
resist the outward movement.

Are you familiar with the Friction Circle?

http://www.nyracer.com/friction.htm

Dear Bob,

I think that I see your point. You're saying that given
steady friction and pressure, the pads limit or match the
force of the disk as it rotates, so any new force will meet
no additional resistance.

I'm just not sure that this is the case (I'm groping in
unfamiliar territory, not arguing against you).

If a disk whose force would normally spin it at 20 mph is
squeezed steadily by brake pads and slows to a steady 10
mph, then the forces are matching at that velocity.

But will the disk spin back up as if there is no increased
friction with increased speed? That is, will adding the
amount of force that would normally accelerate it from 10
mph back to 20 mph produce 20 mph against the 10 mph of
braking? Or does the braking friction increase with speed?

If so, then there's resistance to sideways movement, too,
and it will be hard to drag the disk sideways from between
the pads.

If not, then the pads are functioning like a puck on an
air-hockey table, and the disk will slide out effortlessly.

Has the disk broken free of the pads like a hockey puck on
an ice rink, with no significant frictional resistance in
any direction at any speed?

Or does the braking friction of pads on disks work
differently?

I don't know, so please imagine a badly furrowed brow here.

I kinda-sorta lean toward your theory, but then I keep
thinking that brake pads keep producing friction at all
speeds.

Carl Fogel

Quoted message said:

bobqzzi said:

Quoted message said:

Dear James,

I'm not sure that analogies will give us the actual force,
but consider this one, which seems to be closer to a disk
being squeezed between two pads than a brick sliding down a
slope.

Squeeze a brick hard between two planks, hard enough that it
takes a strong pull to move the brick steadily between the
two planks.

Won't it take just as strong a pull (though shorter) to pull
the brick out from between the two planks?

Quoted message said:

Carl Fogel

Good analogy. Now lets think of the brick sliding between the planks
at a steady rate over a fair distance (like the rotor going through
the pads) Now introduce a small force pushing the brick out of the
planks. I don't believe it would take long for the brick to move out
of the planks. Remember, that X "plank pressure" produces Y clamping
force which provides Z resistance to movement in any direction.

If a new force moving the brick outward is introduced it is added to
the force sliding it upward, the planks really have no more force to
resist the outward movement.

Are you familiar with the Friction Circle?

http://www.nyracer.com/friction.htm

Dear Bob,

I think that I see your point. You're saying that given
steady friction and pressure, the pads limit or match the
force of the disk as it rotates, so any new force will meet
no additional resistance.

I'm just not sure that this is the case (I'm groping in
unfamiliar territory, not arguing against you).

If a disk whose force would normally spin it at 20 mph is
squeezed steadily by brake pads and slows to a steady 10
mph, then the forces are matching at that velocity.

But will the disk spin back up as if there is no increased
friction with increased speed? That is, will adding the
amount of force that would normally accelerate it from 10
mph back to 20 mph produce 20 mph against the 10 mph of
braking? Or does the braking friction increase with speed?

If so, then there's resistance to sideways movement, too,
and it will be hard to drag the disk sideways from between
the pads.

If not, then the pads are functioning like a puck on an
air-hockey table, and the disk will slide out effortlessly.

Has the disk broken free of the pads like a hockey puck on
an ice rink, with no significant frictional resistance in
any direction at any speed?

Or does the braking friction of pads on disks work
differently?

I don't know, so please imagine a badly furrowed brow here.

I kinda-sorta lean toward your theory, but then I keep
thinking that brake pads keep producing friction at all
speeds.

Help! Is there a tribologist in the house?

Jim Smith said:

Quoted message said:

bobqzzi said:

On Sat, 02 Oct 2004 19:48:03 -0600, [email hidden] wrote:

>
>Dear James,
>
>I'm not sure that analogies will give us the actual force,
>but consider this one, which seems to be closer to a disk
>being squeezed between two pads than a brick sliding down a
>slope.
>
>Squeeze a brick hard between two planks, hard enough that it
>takes a strong pull to move the brick steadily between the
>two planks.
>
>Won't it take just as strong a pull (though shorter) to pull
>the brick out from between the two planks?

>Carl Fogel

Good analogy. Now lets think of the brick sliding between the planks
at a steady rate over a fair distance (like the rotor going through
the pads) Now introduce a small force pushing the brick out of the
planks. I don't believe it would take long for the brick to move out
of the planks. Remember, that X "plank pressure" produces Y clamping
force which provides Z resistance to movement in any direction.

If a new force moving the brick outward is introduced it is added to
the force sliding it upward, the planks really have no more force to
resist the outward movement.

Are you familiar with the Friction Circle?

http://www.nyracer.com/friction.htm

Dear Bob,

I think that I see your point. You're saying that given
steady friction and pressure, the pads limit or match the
force of the disk as it rotates, so any new force will meet
no additional resistance.

I'm just not sure that this is the case (I'm groping in
unfamiliar territory, not arguing against you).

If a disk whose force would normally spin it at 20 mph is
squeezed steadily by brake pads and slows to a steady 10
mph, then the forces are matching at that velocity.

But will the disk spin back up as if there is no increased
friction with increased speed? That is, will adding the
amount of force that would normally accelerate it from 10
mph back to 20 mph produce 20 mph against the 10 mph of
braking? Or does the braking friction increase with speed?

If so, then there's resistance to sideways movement, too,
and it will be hard to drag the disk sideways from between
the pads.

If not, then the pads are functioning like a puck on an
air-hockey table, and the disk will slide out effortlessly.

Has the disk broken free of the pads like a hockey puck on
an ice rink, with no significant frictional resistance in
any direction at any speed?

Or does the braking friction of pads on disks work
differently?

I don't know, so please imagine a badly furrowed brow here.

I kinda-sorta lean toward your theory, but then I keep
thinking that brake pads keep producing friction at all
speeds.

Help! Is there a tribologist in the house?

Dear Jim,

In case of friction, brake the glass?

Carl Fogel

Carl Fogel said:

Quoted message said:

Rather than postulate and hypothesize on what occurs, try it.
Loosen the skewer so that it is out of the problem and, without
even sitting on the bicycle, push it forward and apply the brake.
If you don't have such a bicycle, I'm sure there is a bicycle shop
nearby where you can do the test with the benefit of an observer
from the shop.

Quoted message said:

Quoted message said:

The wheel will be canted and jam in the fork whether you brake hard
or lightly, no one having a step function hand clasp that can cause
the effect which you propose. This is just another diversion from
the main topic, that the brake caliper is in the wrong place for a
bicycle with a manually removable front wheel (which the ones in
question are).

Quoted message said:

These things may be tricky, so bear with me for a moment.
You may have missed my point.

Quoted message said:

Elsewhere, James has indicated that he expects the wheel will remain
in place and immobile if the pads stop the disk--no disk motion
through the pads, no ejection (given, of course, the right
geometry).

Quoted message said:

Is motion necessary to your explanation? That is, when you say
"whether you brake hard or lightly," is it understood that it's not
hard enough to stop the motion?

Braking is a dynamic action and not, as you propose, static. I
suggest that you do as I described above and resolve this without
hypothesizing about what might be possible if and when etc. I have
done the demonstration on several bicycles and the wheel jams in the
fork solidly.

Jobst Brandt
[email hidden]

Quoted message said:

Carl Fogel said:

Quoted message said:

Rather than postulate and hypothesize on what occurs, try it.
Loosen the skewer so that it is out of the problem and, without
even sitting on the bicycle, push it forward and apply the brake.
If you don't have such a bicycle, I'm sure there is a bicycle shop
nearby where you can do the test with the benefit of an observer
from the shop.

Quoted message said:

Quoted message said:

The wheel will be canted and jam in the fork whether you brake hard
or lightly, no one having a step function hand clasp that can cause
the effect which you propose. This is just another diversion from
the main topic, that the brake caliper is in the wrong place for a
bicycle with a manually removable front wheel (which the ones in
question are).

Quoted message said:

These things may be tricky, so bear with me for a moment.
You may have missed my point.

Quoted message said:

Elsewhere, James has indicated that he expects the wheel will remain
in place and immobile if the pads stop the disk--no disk motion
through the pads, no ejection (given, of course, the right
geometry).

Quoted message said:

Is motion necessary to your explanation? That is, when you say
"whether you brake hard or lightly," is it understood that it's not
hard enough to stop the motion?

Braking is a dynamic action and not, as you propose, static. I
suggest that you do as I described above and resolve this without
hypothesizing about what might be possible if and when etc. I have
done the demonstration on several bicycles and the wheel jams in the
fork solidly.

Jobst Brandt
[email hidden]

Dear Jobst,

So I take it that you have no answer for what the resistance
(if any) is for the disk dragging out from between the pads?

As I wrote (and you snipped), all that rolling across the
floor with a loose skewer and popping the wheel out tells me
is that the ejection force is greater than the retaining
force. It doesn't tell me how much resistance from the
dragging (if any) was overcome.

If, for example, the ejection force was 10 newtons after
taking the weight of the bike into account, watching the
wheel pop out tells me only that the retaining force of any
drag on the disk was less than 10 newtons--somewhere between
0 and 9.9 newtons.

Your rolling test, in short, does not resolve my question.

Carl Fogel

Carl Fogel said:

Your rolling test, in short, does not resolve my question.

Well, go to the bicycle shop of your choosing and do the test. You'll
find that the results depend on the disc width and diameter and the
type of caliper, and even then, it has no bearing on wheel ejection.
Why are you so adamant about this arbitrarily defined static force?

I'm sure there are many other questions that have not been resolved,
but they have little to do with the problem at hand.

Jobst Brandt
[email hidden]

On Sat, 02 Oct 2004 13:28:01 -0600, [email hidden]

Quoted message said:

While I was browsing the familiar disk-brake wheel-ejection
thread, a question occurred to me that must have been
answered somewhere, but I can't find it.

Various diagrams show that a trailing-caliper disk-brake
will try to lever the axle down and out of the U-shaped
dropouts.

But does the braking force that tries to eject the axle also
try to retain the disk?

That is, does the geometry require the axle to move downward
and away (at some angle) from the caliper pads, which are
squeezing the disk and trying to keep the disk in place?

I'm not sure if the typical geometry requires the axle to
move downward very much in relation to the caliper to escape
from the dropouts.

Nor am I sure how much force is needed to pull the disk
downward enough to clear the dropouts.

Here's some crude ASCII art. To eject, the axle A must move
away from the brake pad P enough to clear the trailing
dropout.

So the disk D must also move toward the original axle
position A to a new position, d.

Does the clamping force of the pad at point P resist the
movement of the disk from D to d?

And does a typical brake/dropout geometry require this kind
of movement?

/ / D<--disk
/ / D
/ / P<--pad
/ fork / d
/ / \
/ / \
/ ___ / disk moved enough
/ / \ / toward A to eject
/ | A | /
/____| |_/
a--ejected

P = disk-brake pad position

A = axle in dropout before ejection
a = axle moved enough to eject

D = disk before ejection
d = disk moved enough to eject

Carl Fogel

Dear James and Bob,

I appreciate the time and trouble that you both took to try
to deal with my ignorance. It's nice to find people
interested in addressing the question.

If I follow your posts, the idea is that a force small
enough to be safely ignored is sufficient to pull the
rotating disk out from between the pressure of the disk
brake pads.

One reason is that even a very small angle away from the
normal path will eventually lead to the disk no longer
tracking between the pads.

This is related to the other reason, which is that to some
extent (how much is still unclear to me) the disk has broken
free and is sliding in one direction with a lower
coefficient of friction, so a smaller force can push it
sideways more easily.

I'm still not sure about how much force is needed, but
you've both been awfully nice. If you do find anything else
about resistance to a sideways pull on the disk, I'll be
glad to hear it.

Thanks,

Carl Fogel

I cant give you a number, except to say that it isn't much.
Think of a tyre(I think the analogue holds). A locked tyre has no
directional stability.(a locked tyre being the same sort of friction as a
disk brake working normally).
Try this, roll your bike along, while trying to push the back wheel sideways
by pushing against the axle(if you prefer just try it with the bike not
moving) . This shows the force required to remove the disk from the calliper
with the brake locked(two surfaces pressing against each other). Now roll
the bike forwards with the back wheel locked while trying to push the back
wheel sideways. This shows the force required to remove the disk from the
calliper with at brake working as normal(to surfaces sliding against each
other)

did I help? or just make things worse?

stu

ducking

stu said:

I cant give you a number, except to say that it isn't much.
Think of a tyre(I think the analogue holds). A locked tyre has no
directional stability.(a locked tyre being the same sort of friction as a
disk brake working normally).
Try this, roll your bike along, while trying to push the back wheel sideways
by pushing against the axle(if you prefer just try it with the bike not
moving) . This shows the force required to remove the disk from the calliper
with the brake locked(two surfaces pressing against each other). Now roll
the bike forwards with the back wheel locked while trying to push the back
wheel sideways. This shows the force required to remove the disk from the
calliper with at brake working as normal(to surfaces sliding against each
other)

did I help? or just make things worse?

stu

ducking

Dear Stu,

I think that your analogy is roughly what others were
suggesting, so it's helpful in that it makes me think about
it some more.

The idea (I think) is that once the disk breaks loose from
the pads, it breaks loose in all directions in the sense
that it goes from a rolling coefficient of friction to a
sliding coefficient of friction.

But since the friction of the sliding disk/pad interface is
still generating enormous resistance (enough to slow the
bike down), I'm wondering if there's also some noticeable
resistance to pulling the disk about 90 degrees sideways (or
out or down or toward the dropout--the terms are a little
confusing).

That a wheel with a loose skewer will pop out fairly easily
when the trailing disk brake is applied does not tell us how
much resistance is overcome--it only tells us that the
ejection force is greater than the retention forces. It
could be that the retention force from any sideways drag is
zero, or it could be some significant amount.

Other than arguments that imply that it's zero or
negligible, no numbers have been offered, possibly because
the question is trickier than it looks to calculate and
equally hard to measure directly. I tried to think of a way
to try to measure the pull necessary remove a disk moving
between two squeezing pads and soon gave up--it's awfully
hard to spin the tire, apply the brake, eliminate gravity,
acount for the ejection, and see what--if anything--is left.

A possible difference between the tire/traction situation
and the disk/pad situation is that the tire may lose almost
all traction when sliding, while the pads are obviously
producing lots of friction even while sliding.

It may be that the coefficients for rolling and sliding work
out differently for rubber and pavement (tires) than for
pads and rotors (disk brakes).

After all, we don't normally skid our tires in order to
stop--we pull on the brake lever and skid our brake pads
against rims or rotors.

I do appreciate your analogy, but I'm still uneasy about how
well it applies. It may just be that I need to learn more
about how brakes work.

Thanks,

Carl Fogel