Braking while turning

MY suggestion is to brake like others said before a corner, but if your in the corner, i use a small
combination of both brakes.

Recently I got to learn this, i was going down a dirt road pushing 25mph and went into a dirt
corner, my bike has slicks in the middle of the tire and small knobs on the outsides. Well I pushed
the tires too far and started to lose traction, so I tried to feather both brakes in a no turning
back attemp to slow down. The rear end broke lose which I thought i could handle because me and
friends do it all the time for fun, it was all working out but with the bike sideways like I wanted
it, I hit the raised portion of the road, and all hell broke loose.

No damage to the bike or me!

TJ Poseno said:

MY suggestion is to brake like others said before a corner, but if your in the corner, i use a
small combination of both brakes. far and started to lose traction, so I tried to feather both
brakes in a no turning back attemp to slow down. The rear end broke lose which I thought i could
handle because me and friends do it all the time for fun, it was all working out but with the bike
sideways like I wanted it, I hit the raised portion of the road, and all hell broke loose.

"I had an accident using both brakes, so I suggest you use both brakes in the corners, too."
--
David Damerell <[email hidden]> Distortion Field!

[email hidden] wrote ...

Quoted message said:

One thing I have noticed is how differently bicycles handle when the rear brake is applied
compared to the front brake.

While braking with the front wheel and turning, the rear wheel naturally swings around follows
through the turn. The steering feels the same as if the brake wasn't applied.

When the rear brake is applied while turning it feels like a force is pulling rear tire in a
straight line backwards causing it to not want to swing around and follow the path of the front
tire. The bike plows though the turn. Even when the bike is traveling in a straight line the bike
doesn't feel nearly as controllable as it would with the front brake is applied.

Is there a physics explanation for this? Even the seasoned riders at the bike shop give me a wierd
look when I tell them I hardly ever use the rear brake.

Another thing to consider is changing the bike's track through the turn WITHOUT braking. I use 2
techniques to tighten my turn radius:

1) Countersteer. It's counter-intuitive, but if you slightly steer toward the OUTSIDE of the turn,
the bike leans IN more and tracks INTO the apex. It's a bit scary to try at first, but you get
used to it, and I find it corrects my line throught the turn - allowing me to hit apexes I can't
hit merely by leaning - and it feels like I retain all my speed through the turn.

2) Lean the bike. Just push the handlebars down toward the ground underneath you. This also tightens
my line, but feels like it reduces speed more than countersteering. But it's easier and safer.

Michael

All these numbers are quite unnecessary. Look at it this way: In a race in any vehicle:

- at the apex of the corner, you want to be using up 99% (or a bit more) of your available traction
from turning. If you are not, you could be going faster.
- Before entering a turn, you want to be using up 99% of your available traction due to braking. If
you are not, then you could have held your speed longer.

What you want to do, therefore, is match the two so that they add up to almost 100%. You brake as
hard as you can before a turn, and you back off the brakes as cornering force increases.

Doug

Christopher Brian Colohan <[email hidden]> wrote in message
news:<[email hidden]>...

Quoted message said:

Joe Riel said:

... Do you corner anywhere near that lean angle? Few do.

Ok, so you first establish that on flat pavement with no surface gunk friction is very high. And
you argue that few folks ride anywhere near this limit on corners. So for most people fc is not
that high.

Quoted message said:

Second, because the total force on the contact patch is the vector sum of the cornering force
and the braking force, which are applied perpendicular to each other, it is possible to apply
considerable braking force while barely changing the total force. For example, let the braking
force (fb) be 20% of the cornering force (fc). The total force is then

You then argue that 20% of fc (which we established is not that high) is a "significant braking
force". For some reason, I disagree. :-)

Quoted message said:

ftot = sqrt(fc^2 + fb^2)
= fc*sqrt(1+(fb/fc)^2)
~ fc*(1+(fb/fc^2)/2) for fb << fc
= fc*(1+(2/10)^2/2) = 1.01*fc

Here is a much easier way of solving (and understanding) this, with the bonus of getting a more
accurate answer:

ftot = sqrt(fc^2 + fb^2)

Assume that fc = 1unit. Assume that fb = .2*fc = .2units

ftot = sqrt(1^2 + .2^2) = 1.02

Now, if you assume that the rider is cornering somewhat conservatively, because they fear that
there might be a patch of sand or tar on the corner somewhere, then you can assume that fc is not
too high. What happens if they are also going down a steep hill, and want to brake to maintain
their speed? If fc=fb, then this changes to:

ftot = sqrt(1^2 + 1^2) = 1.41

You have just lost 40% of the margin of safety you planned to have in the corner.

I do agree that the best way to learn this is not through math, but through practice. I found that
a great way to learn how to deal with low traction conditions in corners is to go out and ride
just after a fresh snowfall (before the plows come by). Very low traction, loads of fun, and if
you happen to fall then the snow offers some padding...

Chris

Quoted message said:

Quoted message said:

Actually, its better to brake and accelerate in straight lines. Don't ask me to explain, its
physics...

The physics have been explained. It doesn't support your claim.

Actually, the physics do, if 'better' is defined as 'safer'. One has less chance of losing traction
if one stays off the brakes.

As others have said, though, it's not the fastest.

Doug

(Douglas Landau) said:

All these numbers are quite unnecessary. Look at it this way: In a race in any vehicle:

- at the apex of the corner, you want to be using up 99% (or a bit more) of your available
traction from turning. If you are not, you could be going faster.

Agree, more or less.

Quoted message said:

- Before entering a turn, you want to be using up 99% of your available traction due to braking.
If you are not, then you could have held your speed longer.

Disagree. It depends on the corner. For some corners you don't have to brake at all. For others you
don't brake until you are well into the corner. From a theoretical standpoint, you want to maximize
the magnitude of the acceleration vector. That means, whether braking or turning or a combination,
keeping the contact patch right at the limit. From a practical standpoint that isn't necessarily the
best way to go.

Joe Riel

<[email hidden]> wrote in message "]news:[email hidden]...

Quoted message said:

One thing I have noticed is how differently bicycles handle when the rear brake is applied
compared to the front brake.

While braking with the front wheel and turning, the rear wheel naturally swings around follows
through the turn. The steering feels the same as if the brake wasn't applied.

If I brake while turning, the front suspension compresses and the entire steering geometry changes
drastically. The first ride after getting the RS Pilot XC's put on, I damned near came unstuck on
the first corner that I applied even light braking on the front. I now try to avoid the front brake
whilst in turns.

Trentus

"Douglas Landau" <[email hidden]> wrote in message
"]news:[email hidden]...

Quoted message said:

All these numbers are quite unnecessary. Look at it this way: In a race in any vehicle:

- at the apex of the corner, you want to be using up 99% (or a bit more) of your available
traction from turning. If you are not, you could be going faster.
- Before entering a turn, you want to be using up 99% of your available traction due to braking.
If you are not, then you could have held your speed longer.

What you want to do, therefore, is match the two so that they add up to almost 100%. You brake as
hard as you can before a turn, and you back off the brakes as cornering force increases.

From the FAQ:

"Take for example a rider cornering on good traction, leaning at 45 degrees. With this 1 G
centrifugal acceleration, he can still apply .1 G braking and hardly increase the load on the tires,
which is given by the square root(1^2+0.1^2)=1.005 or 1/2%. In other words, you can brake
substantially near maximum cornering. The centrifugal acceleration changes as the square of the
speed, so braking rapidly reduces the required lean angle and allows increased braking. Being aware
of this relationship should leave no doubt why racers are nearly always applying brakes at the apex
of max speed turns."

Peter Cole said:

"Take for example a rider cornering on good traction, leaning at 45 degrees. With this 1 G
centrifugal acceleration, he can still apply .1 G braking and hardly increase the load on the
tires, which is given by the square root(1^2+0.1^2)=1.005 or 1/2%.

A bit of a tangent: does this mean there are bike tires with a coefficient of friction larger than
1? If so, how do they do it---glue? super sticky rubber?

Just curious...

Chris
--
Chris Colohan Email: [email hidden] PGP: finger [email hidden] Web: www.colohan.com Phone:
(412)268-4751

Quoted message said:

"Take for example a rider cornering on good traction, leaning at 45 degrees. With this 1 G
centrifugal acceleration, he can still apply .1 G braking and hardly increase the load on the
tires, which is given by the square root(1^2+0.1^2)=1.005 or 1/2%. In other words, you can brake
substantially near maximum cornering. The centrifugal acceleration changes as the square of the
speed, so braking rapidly reduces the required lean angle and allows increased braking. Being
aware of this relationship should leave no doubt why racers are nearly always applying brakes at
the apex of max speed turns."

Have you guys actually DONE this? or are you just repeating things?

Mike

Quoted message said:

From the FAQ:

"Take for example a rider cornering on good traction, leaning at 45 degrees. With this 1 G
centrifugal acceleration, he can still apply .1 G braking and hardly increase the load on the
tires, which is given by the square root(1^2+0.1^2)=1.005 or 1/2%. In other words, you can brake
substantially near maximum cornering. The centrifugal acceleration changes as the square of the
speed, so braking rapidly reduces the required lean angle and allows increased braking. Being
aware of this relationship should leave no doubt why racers are nearly always applying brakes at
the apex of max speed turns."

From the FAQ or not, this paragraph falls apart in multiple ways halfway through.

The first part is correct - up through the statement "you can brake substantially near maximum
cornering". However, second half of the next sentence is incorrect.
1. Reduced speed, not braking itself, reduces required lean angle.
2. Even this is not a result of the fact that centrifugal acceleration changes as the square of the
speed, so the word "so" in the middle of that sentence is incorrect.
3. Reduced lean angle does not permit increased braking, excepting for subtleties arising from, for
example, less-than-perfectly-rigid wheels.
4. Increased braking is not what you want at the apex of a turn.

There is a proper place for braking up until very close to the apex.

Joe Riel <[email hidden]> wrote in message news:<[email hidden]>...

Quoted message said:

(Douglas Landau) said:

- Before entering a turn, you want to be using up 99% of your available traction due to braking.
If you are not, then you could have held your speed longer.

Disagree. It depends on the corner. For some corners you don't have to brake at all. ...

You are correct. What I wrote refers only to one class of turns. If you were to categorize all turns
as slow, medium, or fast, I am discussing slow turns.

Doug

"Mike S." <mikeshaw2@coxDOTnet> wrote in message news:7KyTa.14678$Bp2.738@fed1read07...

Quoted message said:

Quoted message said:

"Take for example a rider cornering on good traction, leaning at 45 degrees. With this 1 G
centrifugal acceleration, he can still apply .1 G braking and hardly increase the load on the
tires, which is given by the square root(1^2+0.1^2)=1.005 or 1/2%. In other words, you can brake
substantially near maximum cornering. The centrifugal acceleration changes as the square of the
speed, so braking rapidly reduces the required lean angle and allows increased braking. Being
aware of this relationship should leave no doubt why racers are nearly always applying brakes at
the apex of max speed turns."

Have you guys actually DONE this? or are you just repeating things?

Since it was (stated as) a quote from the FAQ, I certainly was "repeating things". But it is only
simple vector math, something bicycles (and everything else in the universe) must comply with.

"Christopher Brian Colohan" <[email hidden]> wrote in message
"]news:[email hidden]...

Quoted message said:

Peter Cole said:

"Take for example a rider cornering on good traction, leaning at 45 degrees. With this 1 G
centrifugal acceleration, he can still apply .1 G braking and hardly increase the load on the
tires, which is given by the square root(1^2+0.1^2)=1.005 or 1/2%.

A bit of a tangent: does this mean there are bike tires with a coefficient of friction larger than
1? If so, how do they do it---glue? super sticky rubber?

It's an example. Plug in whatever numbers you want. Maximum centrifugal acceleration may be limited
by other factors than the coefficient of friction of the rubber tread. Read the FAQ for a more
complete description.

Quoted message said:

Quoted message said:

Have you guys actually DONE this? or are you just repeating things?

Since it was (stated as) a quote from the FAQ, I certainly was "repeating things". But it is only
simple vector math, something bicycles (and

everything

Quoted message said:

else in the universe) must comply with.

The math's great, but have you actually DONE what you're preaching?

I've done it both ways. Braking going straight, then turning seems to work better for me. Seems that
I can corner harder when I'm not braking in the middle of the turn. Then again YMMV...

Mike

"Mike S." <mikeshaw2@coxDOTnet> wrote in message news:EzzTa.14689$Bp2.7758@fed1read07...

Quoted message said:

Quoted message said:

Quoted message said:

Have you guys actually DONE this? or are you just repeating

things?

Quoted message said:

Quoted message said:

Since it was (stated as) a quote from the FAQ, I certainly was

"repeating

Quoted message said:

Quoted message said:

things". But it is only simple vector math, something bicycles (and

everything

Quoted message said:

else in the universe) must comply with.

The math's great, but have you actually DONE what you're preaching?

I've done it both ways. Braking going straight, then turning seems to

work

Quoted message said:

better for me. Seems that I can corner harder when I'm not braking in

the

Quoted message said:

middle of the turn. Then again YMMV...

Most of us do not get off the bike before a turn, measure the friction coefficient of the road
surface, the camber and the angle of the turn and then perform a simple mathematical equation to
determine how hard we can squeeze the brakes -- although I think I saw Beloki with his sliderule out
just before the big crash. Most of us have an experienced-based sense of when an how to brake. Those
lacking that sense -- or those riding on a descent with obscured sight lines -- are best served with
a general rule that braking should be done before the turn. This avoids panic braking during the
turn, which, depending on the angle and pitch of the turn, can be the kiss of death. I don't think
anyone could reasonably argue that you should wait until you are in the middle of a steep, blind or
off-camber corner to brake. -- Jay Beattie.

Quoted message said:

I've done it both ways. Braking going straight, then turning seems to work better for me. Seems
that I can corner harder when I'm not braking in the middle of the turn. Then again YMMV...

Mike

You are correct about the first part. You can corner harder when not braking. However, when you
first enter a turn, you're not cornering as hard as you can anyway, so there is room left in the
equation to get some braking in.

However, about the YMMV - I don't think that applies here... it works the same for all of us whether
we agree or not.

Doug

Douglas Landau said:

Quoted message said:

From the FAQ:

"Take for example a rider cornering on good traction, leaning at 45 degrees. With this 1 G
centrifugal acceleration, he can still apply .1 G braking and hardly increase the load on the
tires, which is given by the square root(1^2+0.1^2)=1.005 or 1/2%. In other words, you can brake
substantially near maximum cornering. The centrifugal acceleration changes as the square of the
speed, so braking rapidly reduces the required lean angle and allows increased braking. Being
aware of this relationship should leave no doubt why racers are nearly always applying brakes at
the apex of max speed turns."

From the FAQ or not, this paragraph falls apart in multiple ways halfway through.

The first part is correct - up through the statement "you can brake substantially near maximum
cornering". However, second half of the next sentence is incorrect.
1. Reduced speed, not braking itself, reduces required lean angle.

This may be true grammatically, but in context, since reduced speed is the both the goal and direct
result of braking, the gist of the sentence remains true.

Quoted message said:

2. Even this is not a result of the fact that centrifugal acceleration changes as the square of
the speed, so the word "so" in the middle of that sentence is incorrect.

Again, since braking is the direct cause of the speed decrease, it is still true.

Quoted message said:

3. Reduced lean angle does not permit increased braking, excepting for subtleties arising from,
for example, less-than-perfectly-rigid wheels.

That's not quite what the sentence says. It does not say that reduced lean angle permits more
braking, it says that there are two results of the reduction in centrifugal acceleration - reduction
of lean angle, and permitting increased braking.

Quoted message said:

4. Increased braking is not what you want at the apex of a turn.

That depends on the turn. In a turn with decreasing radius, or with increasing downslope, you may
indeed want increased braking.

Quoted message said:

There is a proper place for braking up until very close to the apex.

In a perfectly flat, constant radius turn on perfect pavement, that is probably true. However, not
all turns are so perfect.

Mark McMaster [email hidden]

Jay Beattie said:

... I don't think anyone could reasonably argue that you should wait until you are in the middle
of a steep, blind or off-camber corner to brake. -- Jay Beattie.

Of course not. On the other hand, we hear frequent suggestions that one shouldn't be braking while
turning. That, too, is not a particularly effective method. Consider a blind turn with a tightening
radius---how would you negotiate it if you didn't turn while braking? You'd never know if you'd
slowed enough to start turning.

The theory is good for showing that there is a lot of potential for simultaneously braking and
turning. Theory, however, won't make you fast, you have to practice technique.

Joe

Quoted message said:

A bit of a tangent: does this mean there are bike tires with a coefficient of friction larger than
1? If so, how do they do it---glue? super sticky rubber?

Just curious...

Chris

Most tires have a friction coefficient greater than 1. The interface is not as simple as two flat
surfaces in contact. On any surface smoother than glass or polished steel, the rubber deforms into
irregularities, allowing a coefficient > 1.

--
Ted Bennett Portland OR