Correct Braking Proceedure

Alexander Rice said:

In my personal instance of riding a recumbent bike a lot of the time
it's virtually impossible to lift the back wheel under braking and the
disc brakes are capable of locking either wheel at will.

In my attempts at a brief reply I missed out the caveats of recumbents,
tandems and bicycles with ten tons of bricks on the rear carrier. ;-)

--
Tony

"I did make a mistake once - I thought I'd made a mistake but I hadn't"
Anon

Tosspot said:

So, the question is, is it impossible to lift the rear wheel before
losing traction for some bikes?

What about wet road surface?

For my tuppence worth, for me, it's front brake only. The rear is only
used to impress 16yo girlies at bus stops.

My experience is the rear brake is good for balancing the bike during
braking, esp on loose off-road corners, and for testing traction on
slippery surfaces but if you are braking in slippery conditions and you
use the back brake to brake, it will lock very easily at which point the
back will skid round on you. So IME the back brake is something to be
very wary of in slippery conditions and of little use in dry conditions.
--
Tony

"I did make a mistake once - I thought I'd made a mistake but I hadn't"
Anon

Bertie Wiggins said:

On Tue, 25 Oct 2005 19:05:14 +0100, Tony Raven <[email hidden]>

Quoted message said:

But that will not slow the bike as fast as the front brake alone purely
because the geometry means that the maximum deceleration you can achieve
will have transferred all the weight to the front wheel only.

Then shift your weight back and use the back brake too.

Has very little effect. Getting your centre of gravity down lower would
help more in that situation but putting your weight back does very
little to change the point of weight transferring completely to the
front wheel. Braking downhill off-road where you are trying to
increase rear wheel traction, not achieve maximimum braking is another
story.

--
Tony

"I did make a mistake once - I thought I'd made a mistake but I hadn't"
Anon

Bertie Wiggins said:

It seems to me that the most effective theoretical braking is when the
cyclist adjusts their mass so there is equal weight on both wheels so
as much tyre is in contact with the ground as possible, the brakes are
applied with a force at a point before a skid, so the turning of the
two wheels are being slowed at maximum rate without locking.

That would be when the cyclists is hanging upside down from the down
tube so their centre of gravity is as close vertically to the contact
patch as they can make it. Moving backwards doesn't actually change
much the turning moment of the CoG about the contact patch. How far
back, in percentage terms can you move your weight relative to the
contact patch to saddle position?

--
Tony

"I did make a mistake once - I thought I'd made a mistake but I hadn't"
Anon

elyob said:

On road ... you should be covering your brakes at all times.

Riding on the tops (road bars) or bar ends (flat bars) is out then?

--
Tony

"I did make a mistake once - I thought I'd made a mistake but I hadn't"
Anon

Jon Senior said:

Think how far back from the front contact point you have to be in order
to not go over the bars. You need to be most of the distance behind the
rear contact point in order to keep the weight on the rear wheel. How
long are your arms?

I've noticed that as I've got older my arms have got shorter. They are
now so short I can barely hold things far enough away to read them ;-)

--
Tony

"I did make a mistake once - I thought I'd made a mistake but I hadn't"
Anon

Tony Raven said:

Has very little effect. Getting your centre of gravity down lower would
help more in that situation but putting your weight back does very
little to change the point of weight transferring completely to the
front wheel. Braking downhill off-road where you are trying to
increase rear wheel traction, not achieve maximimum braking is another
story.

I guess the way to think about it is:

Think how far back from the front contact point you have to be in order
to not go over the bars. You need to be most of the distance behind the
rear contact point in order to keep the weight on the rear wheel. How
long are your arms?

Jon

Tony Raven said:

Bertie Wiggins said:

On Tue, 25 Oct 2005 19:05:14 +0100, Tony Raven <[email hidden]>

Quoted message said:

But that will not slow the bike as fast as the front brake alone purely
because the geometry means that the maximum deceleration you can achieve
will have transferred all the weight to the front wheel only.

Then shift your weight back and use the back brake too.

Has very little effect. Getting your centre of gravity down lower would
help more in that situation but putting your weight back does very
little to change the point of weight transferring completely to the
front wheel. Braking downhill off-road where you are trying to
increase rear wheel traction, not achieve maximimum braking is another
story.

shifting your weight backwards is actually more about bracing your arms
against the handlebars to prevent your body shifting forwards relative
to the bike as the bike decelerates. It is more likely to be this
forward shift of the body that either makes you lose balance and fall
off or (less likely) shifts your CoG far enough fowards that what would
have been safe braking force causes an endo

best wishes
james

Bertie Wiggins said:

Sheldon argues that maximum braking force is at the point when the rear
wheel is about to lift from the ground, and the rear wheel has no
traction, therefore applying the rear brake has no effect, thus for
maximum braking the front brake is all that is required in most
situations.

Here's what I say...

Braking with both brakes is more effective because the braking force is by
slowing both wheels by pushing the pads onto the rims, and neither brake
need be fully applied to slow the bike as fast as applying the front brake
alone.

Sheldon is taking into account something that you might not be aware of.

Although your _weight_ may be evenly distributed between the wheels, when
braking the _load_ where the tyre touches the road is no longer even.
This movement of the _load_ is a product of the braking, and although you
can alter its value a little by moving your weight you can't stop the
basic process happening.

For most calculations involving movement and forces any "rigid" body can
be considered to have all its mass at a notional point called the "Centre
of Gravity" (sometimes Centre of Mass would be more accurate, but C0G is
the convention). OK, you don't have to, but working out the relationship
of the road/brakes/wind etc to each
arm/leg/finger/fingernail/tooth/cell/atom would be exceeding tedious and
can be proven to be unnecessary. For practical purposes you and the bike
are a rigid body, and the combined CoG is going to be around your
belt-buckle if you're in a touring/commuting position, and lower and
further forward if you're in a Time Trial position.

When you are braking the relationship between you and the Earth is
changing. You were going at some speed around it, and now you are
changing that speed. Changing speed ("acceleration"😉 requires a force
between you and the Earth, and in this case it is generated where the
tyres touch the ground and applied to the C0G of each body. (NB I am
ignoring air resistance in this explanation.)

So we have a force at tyre/ground level which affects the relationship
between the C0G of the Earth 6370km below, and your C0G, let's say 1.4m
above. The force is sufficiently smaller than the Earth's mass that the
Earth won't notice it, but it's close enough to yours that you will.

Now, the force is generated at the tyre/road junction, but it's _applied_
to your CoG. These two points are 1.4m apart, so as well as "pushing" you
back, the combination of force and distance will try to rotate your CoG
relative to the tyre/road. Tyre/road force pushes back, your mass at the
CoG pushes forward. But you do not actually rotate because that
rotational force is balanced by the vertical load on the front tyre
getting bigger. It can only do this at the expense of the vertical load
on the back wheel getting smaller. If the load at the back did not
diminish you and the bike would benefit from a vertical push from the road
greater than your weight, and you'd leap into the air. (Thereby inventing
anti-gravity and becoming very, very rich!).

The magnitude of the braking effort governs the amount of load transfer
and the load transfer governs how much "weight" is still on the back wheel.
A great enough braking effort leaves no "weight" on the back wheel, and
there's nothing you can do about it, other than let off the brakes.
Even a moderate braking effort could shift half the back wheels load to
the front, which is where the 75:25 split comes from.

And remember that this load transfer will happen even if only the back
brake is applied. It comes from the vertical distance between the road
and the CoG and applies no matter which wheel is braked. Load _will_ be
transfered from back to front, up to the point at which the back wheel
slides.

So, in principle you're right to say that dividing the effort as
evenly as possible between the wheels is best, but for an upright bike
you can't achieve it. Only by moving the CoG back or down, or increasing
the wheelbase can you do anything to get closer to this.

Mike

"Bertie Wiggins" <cycling_remove_bertie@yahoo_dot_co_dot_uk> wrote in
message news:[email hidden]...

Quoted message said:

Clive George said:

Quoted message said:

I've no idea where he gets the 75:25 figure from. It suggests to me
that the rider's mass should be positioned so that under maximum
braking 75% of the weight is on the front wheel and 25% on the rear.

What if it's not possible to position the rider's weight there? Under
properly heavy braking on an upright bike, the rider's weight on the rear
wheel is about zero. And that's even if you're trying.

It seems dangerous to teach people to brake so heavily that the rear
wheel is on the point of getting airborne.

We're talking about the most effective braking here, right? That which
brings you to a stop in the shortest possible distance (or time)? You did
say "maximum braking" - this definitely implies you're talking about the
same thing.

If so, there's no choice but to brake heavily. In fact, you're going to be
wanting to brake as heavily as is possible.

If you're teaching people, it's probably a good idea for them to become
accustomed to the idea that the rear wheel is almost airborne under maximum
braking. It'll come under teaching emergency stops. If you don't teach them
that, then the time they do need to do a full-on emergency stop, they'll
either not put the brakes on hard enough and end up under the bus, or
they'll put them on too hard and end up in the road. (or they'll learn how
to do it properly themselves, but that'll be unlearning your lessons).
Of course this sort of thing isn't suitable for the first lesson. But I do
think it's an important skill.

Quoted message said:

It also seems inefficient
to dissipate energy through 2 brake blocks instead of 4.

It seems inefficient to put the forces which propel you through one tyre
instead of two, but that's what we've got. Even in cars, which don't tip
over forwards, the rear brakes are typically a lot less powerful than the
fronts. Geometry limits you - for a hard stop, you have no choice but to use
those two brake blocks. Fortunately 2 brake blocks can provide sufficient
force - and if they can't then the entire discussion is irrelevant.

Now if you're talking riding down steep hills with heavily laden bikes,
where heat dissipation can become important, then you will want to consider
the usefulness of putting half the energy through each wheel. But in these
situations you're not braking to stop, you're braking to slow down. And yes,
I do ride bikes where this is a concern (and have got it wrong more than
once).

Quoted message said:

Quoted message said:

Are you arguing that the design of the bike should be changed, or are you
trying to work within the limitations provided by conventional bicycle
technology?

I'm trying to argue the case for the most effective braking.

What's your definition of 'the most effective braking'? Is it that which
brings you to a stop in the shortest distance/time? That's the one I think
everybody else is working with.

If it isn't, please tell us what you think it means. And why "maximum
braking" isn't.

cheers,
clive

Rich said:

If you haven't hit the brick wall yet, you're doing it right. Stop
worrying.

I have injured myself on a couple of occasions, one meant four stitches in
my upper lip, which wasn't pleasant!

Both incidents were nothing to do with braking, they were to do with not
looking (lack of concentration).

I might have the best brakes in the world, but I'm afraid not the best
brain!

Bertie Wiggins said:

James Annan said:

I repeat: it is a matter of elementary geometry and mechanics. If you
can't do it, stop pretending that you know better than those of us who
can!

I am not so arrogant as to insist that I am correct; I am not so
conceited as to claim superior mathematical, geometric or mechanical
knowledge. However, I am not going to accept, on your word, that
applying the front brake alone is more effective than applying front
and rear brake together. As the geometry and mechanics is so
"elementary" you may provide a mathematical model to support your
claim, if I cannot follow the "elementary" mathematics I am sure
others can.

Okay, here goes:

Imagine a cyclist with mass 100kg whose centre of mass (it's about where
your belly button is) is 1m in the air and precisely in the centre of a
bike with a 2m wheelbase [I'm making this up for mathmatical convenience].

Our cyclist is traveling at constant velocity of 10m/s (okay, he's
really fit) on a level road. The only acceleration acting on him is
gravity of 10ms^-2 (metres per second per second) meaning that his
weight is 1000N (Newtons) and it is distributed equally between his
front and rear wheel - 500N per wheel.

Imagine now that the traffic lights far ahead turn red and our cyclist
wants to slow down because he knows RLJs are devilspawn condemned to
being tortured by disgruntled London cabiies for all eternity. He
applies his brakes to cause decelleration of 1ms^-2 so it will take him
10s to come to a standstill.

This means he is acted on by gravity (10ms^-2 downwards and his
decelleration of 1m2^-2 backwards). This can be resolved as an
acceleration of 10.05ms^-2 at 5.71 degrees to vertical. A quick bit if
trig shows that a line drawn through his center of mass in the direction
of the net acceleration now touches the ground 10cm further forward than
it did meaning his weight is now 55% on the front wheel (550N) and 45%
on the back wheel (450N). If he's using both brakes equally then there's
a retarding force of 50N at each contact point.

Imagine instead that a blind iPod wearing pedestrian numpty steps out
right in front of him and he wants to stop (okay, he'd quite like to run
the old bat over but hey...) as quickly as is possible. There are three
scenarios that can limit his braking

a) the friction between brake pads and rims, if this is the case his
bike needs servicing

b) the friction between his tyres and the road... either his tyres are
made of cheese or he's riding on ice/greasy road

c) the point at which he does a face plant in the tarmac

Let us imagine that we are in situation c

He applies a deceleration of 10ms^-2 meaning it will take him 1s to
stop. This means the net acceleration is 14.1ms^-2 at an angle of 45
degrees to the vertical. A line drawn in the direction of this
acceleration through his center of mass now touches the ground 1m
further forward at exactly the contact point of the front tyre.

All the cyclists weight is now on the front wheel (1000N) and none on
his rear wheel (0N). Since there's no contact force between the rear
tyre and the road the braking done by the rear wheel must be zero. The
front wheel alone is applying a retarding force of 1000N. Any further
force applied by the front wheel will mean that the line contacts the
ground ahead of the front wheel and the cyclist will go over the bars.

Alex

Clive George said:

What's your definition of 'the most effective braking'? Is it that which
brings you to a stop in the shortest distance/time? That's the one I think
everybody else is working with.

Yes. The theoretical fastest way to bring a bicycle to a standstill.

"Bertie Wiggins" <cycling_remove_bertie@yahoo_dot_co_dot_uk> wrote in
message news:[email hidden]...

Quoted message said:

Clive George said:

What's your definition of 'the most effective braking'? Is it that which
brings you to a stop in the shortest distance/time? That's the one I think
everybody else is working with.

Yes. The theoretical fastest way to bring a bicycle to a standstill.

Right. Then forget trying to say "the ideal weight distribution is this",
and accept that for maximum braking on an upright bike, the weight
distribution is 100% on the front wheel. If you've got any weight on the
back wheel, you're not braking as hard as possible.

Others have posted the sums demonstrating this - I think it's time you
either accept them, or sat down and worked it out properly for yourself.

cheers,
clive

James Annan said:

Ok, here is a simple version.

Let's go with it then.

Quoted message said:

If you slow down too fast, you'll go over the handlebars. For a normal
riding position, the limiting deceleration is about 0.6g (and you can't
do a great deal about shifting your CofG). That corresponds to a CofG
that is at an angle of about 30 degrees behind vertical, above and
behind the front wheel's contact point with the ground.

So at constant speed the CoG and two contact points with the ground
form a 30,60,90 triangle? Does that mean that for a 100Kg cyclist and
bike there will be a 634N reaction at the rear contact point and a
366N reaction at the front contact point?

Quoted message said:

So, once you are braking at this theoretical limit of 0.6g, there is no
weight on the back wheel. Say you want to be a bit more conservative,
and only brake at 0.5g. In this case, you'll have about 9/10 of your
weight on the back wheel and 1/10 on the front

(I think you mean 1/10 back and 9/10 front.)

Quoted message said:

(can't be bothered doing
the sums properly). Since the coefficient of friction is unlikely to be
significantly above 1, this limits the braking on the rear wheel to
0.1g at best, ie 20% of your total. Energy dissipation simply doesn't
come into it.

Without an accurate figure for the weight on each wheel the rest is
pretty much useless.

Quoted message said:

Forrester's 75%/25% is not too unreasonable in practice, especially
given that a rear wheel skid is rarely fatal and generally easy to
recover from. Attempting a ratio of 50%/50% will substantially limit
the deceleration you can achieve, because the rear wheel will start to
skid as your effective weight distribution changes under moderate
braking.

OK, I'm beginning to see why 50:50 braking is not the fastest way to
bring a cycle to a complete standstill. I'd still like to see why
100:0, 75:25, 80:20 or whatever is fastest.

Let's go with the earlier suggestion that at constant speed a 100Kg
weight exerts a force of 634N on the rear wheel and 366N on the front
wheel, and that the centre of gravity and two contact points form a
30,60,90 triangle.

Let's also assume a coefficient of friction of 1 between tyre and
tarmac, i.e. the tyre won't skid.

What is the maximum deceleration possible so that the force on the
rear wheel is 0?

Is there sufficient information to solve?

On Wed, 26 Oct 2005 13:10:50 +0100, Alexander Rice <[email hidden]>

Quoted message said:

Bertie Wiggins said:

James Annan said:

I repeat: it is a matter of elementary geometry and mechanics. If you
can't do it, stop pretending that you know better than those of us who
can!

I am not so arrogant as to insist that I am correct; I am not so
conceited as to claim superior mathematical, geometric or mechanical
knowledge. However, I am not going to accept, on your word, that
applying the front brake alone is more effective than applying front
and rear brake together. As the geometry and mechanics is so
"elementary" you may provide a mathematical model to support your
claim, if I cannot follow the "elementary" mathematics I am sure
others can.

Okay, here goes:

Imagine a cyclist with mass 100kg whose centre of mass (it's about where
your belly button is) is 1m in the air and precisely in the centre of a
bike with a 2m wheelbase [I'm making this up for mathmatical convenience].

OK so far.

Quoted message said:

Our cyclist is traveling at constant velocity of 10m/s (okay, he's
really fit) on a level road. The only acceleration acting on him is
gravity of 10ms^-2 (metres per second per second) meaning that his
weight is 1000N (Newtons) and it is distributed equally between his
front and rear wheel - 500N per wheel.

OK so far.

Quoted message said:

Imagine now that the traffic lights far ahead turn red and our cyclist
wants to slow down because he knows RLJs are devilspawn condemned to
being tortured by disgruntled London cabiies for all eternity. He
applies his brakes to cause decelleration of 1ms^-2 so it will take him
10s to come to a standstill.

OK so far.

Quoted message said:

This means he is acted on by gravity (10ms^-2 downwards and his
decelleration of 1m2^-2 backwards). This can be resolved as an
acceleration of 10.05ms^-2 at 5.71 degrees to vertical. A quick bit if
trig shows that a line drawn through his center of mass in the direction
of the net acceleration now touches the ground 10cm further forward than
it did meaning his weight is now 55% on the front wheel (550N) and 45%
on the back wheel (450N). If he's using both brakes equally then there's
a retarding force of 50N at each contact point.

OK so far.

Quoted message said:

Imagine instead that a blind iPod wearing pedestrian numpty steps out
right in front of him and he wants to stop (okay, he'd quite like to run
the old bat over but hey...) as quickly as is possible. There are three
scenarios that can limit his braking

a) the friction between brake pads and rims, if this is the case his
bike needs servicing

b) the friction between his tyres and the road... either his tyres are
made of cheese or he's riding on ice/greasy road

c) the point at which he does a face plant in the tarmac

Let us imagine that we are in situation c

He applies a deceleration of 10ms^-2 meaning it will take him 1s to
stop. This means the net acceleration is 14.1ms^-2 at an angle of 45
degrees to the vertical. A line drawn in the direction of this
acceleration through his center of mass now touches the ground 1m
further forward at exactly the contact point of the front tyre.

OK so far.

Quoted message said:

All the cyclists weight is now on the front wheel (1000N) and none on
his rear wheel (0N). Since there's no contact force between the rear
tyre and the road the braking done by the rear wheel must be zero. The
front wheel alone is applying a retarding force of 1000N. Any further
force applied by the front wheel will mean that the line contacts the
ground ahead of the front wheel and the cyclist will go over the bars.

Beautifully put.

Bertie Wiggins said:

James Annan said:

Ok, here is a simple version.

Let's go with it then.

Quoted message said:

If you slow down too fast, you'll go over the handlebars. For a normal
riding position, the limiting deceleration is about 0.6g (and you can't
do a great deal about shifting your CofG). That corresponds to a CofG
that is at an angle of about 30 degrees behind vertical, above and
behind the front wheel's contact point with the ground.

So at constant speed the CoG and two contact points with the ground
form a 30,60,90 triangle? Does that mean that for a 100Kg cyclist and
bike there will be a 634N reaction at the rear contact point and a
366N reaction at the front contact point?

not quite. I didn't give enough information to determine the load on the
rear wheel, but I think about 55/45 rear/front is generaly considered
typical (maybe 60/40 at most), which implies the CoG is slightly
rearward of the mid-point between the wheels.

Quoted message said:

Quoted message said:

So, once you are braking at this theoretical limit of 0.6g, there is no
weight on the back wheel. Say you want to be a bit more conservative,
and only brake at 0.5g. In this case, you'll have about 9/10 of your
weight on the back wheel and 1/10 on the front

(I think you mean 1/10 back and 9/10 front.)

Yes, sorry.

Quoted message said:

Quoted message said:

(can't be bothered doing
the sums properly). Since the coefficient of friction is unlikely to be
significantly above 1, this limits the braking on the rear wheel to
0.1g at best, ie 20% of your total. Energy dissipation simply doesn't
come into it.

Without an accurate figure for the weight on each wheel the rest is
pretty much useless.

No it isn't. Any reasonable estimates will show the effect clearly enough.

Quoted message said:

Quoted message said:

Forrester's 75%/25% is not too unreasonable in practice, especially
given that a rear wheel skid is rarely fatal and generally easy to
recover from. Attempting a ratio of 50%/50% will substantially limit
the deceleration you can achieve, because the rear wheel will start to
skid as your effective weight distribution changes under moderate
braking.

OK, I'm beginning to see why 50:50 braking is not the fastest way to
bring a cycle to a complete standstill. I'd still like to see why
100:0, 75:25, 80:20 or whatever is fastest.

Let's go with the earlier suggestion that at constant speed a 100Kg
weight exerts a force of 634N on the rear wheel and 366N on the front
wheel, and that the centre of gravity and two contact points form a
30,60,90 triangle.

Let's also assume a coefficient of friction of 1 between tyre and
tarmac, i.e. the tyre won't skid.

What is the maximum deceleration possible so that the force on the
rear wheel is 0?

Is there sufficient information to solve?

The maximum deceleration is determined purely by the angle of the CoG
behind the front wheel contact point, which is where I started from.
Using the rear brake cannot help to reach or exceed this value, since
the rear wheel wil skid at a lower deceleration.

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

Bertie Wiggins said:

James Annan said:

Ok, here is a simple version.

Let's go with it then.

Or you could try an empirical experiment.

Borrow a bike with front suspension.

Ride at a constant pace. Assuming no bobbing, the front fork
compression is X, determined by the weight applied to the front wheel.

Now brake at a rate Y with the back brake. The forks will compress by a
factor dX due to the momentum of the rider acting as a lever around the
contact point for the back wheel. This serves to show that there is
weight transfer from rear to front under braking.

Now repeat, braking at a rate Y but with the front brake. Will there be
the same degree of compression of the front forks? I'm thinking about
this and without trying to stretch my brain too much, suspect that it
will compress more with braking on the front brake. due to the
difference in position of the pivot (rear wheel vs front wheel).

And then you reach the limit of the performance of the brake. With the
back brake it is when the weight reduces to insufficient to prevent the
tyre sliding. With the front brake it is when the effective CoG becomes
unstably located wrt to the wheels.

...d

Tony Raven said:

Alexander Rice said:

In my personal instance of riding a recumbent bike a lot of the time
it's virtually impossible to lift the back wheel under braking and
the disc brakes are capable of locking either wheel at will.

In my attempts at a brief reply I missed out the caveats of
recumbents, tandems and bicycles with ten tons of bricks on the rear
carrier. ;-)

My bike's broken now I tried to load ten tons of bricks on the rear carrier.
Who can I sue?

--
Ambrose

Tony Raven said:

My experience is the rear brake is good for balancing the bike during
braking, esp on loose off-road corners, and for testing traction on
slippery surfaces but if you are braking in slippery conditions and
you use the back brake to brake, it will lock very easily at which
point the back will skid round on you. So IME the back brake is
something to be very wary of in slippery conditions and of little use
in dry conditions.

But a front tyre skid is harder to recover from, especially on a bend*.
The rear brake is useful when that is a serious risk, be it together with
the front or on its own, depending on the situation.

Personally, I have rear brakes that are difficult to lock up on all but
the very most slippery conditions. In one case deliberately: swapped
dual-pivot rear caliper for a single (has other advantages as well).

* Even if you normally avoid braking at all on a bend, it is sometimes
necessary when misjudging the tightness or when forced to by a
vehicle/person/animal/object. Also it's all too easy to lock up the front
on ice and fall, even in a straight line. Risk of rear skid, even if
greater, is then the lesser of the two evils.

~PB