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Braking while turning



 
 
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  #21  
Old July 23rd 03, 02:02 AM
Douglas Landau
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Default Braking while turning

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 wrote in message ...
Joe Riel writes:
... 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.

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. :-)

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

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  #22  
Old July 23rd 03, 12:30 PM
Trentus
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Default Braking while turning

wrote in message
...
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


  #23  
Old July 23rd 03, 01:29 PM
Peter Cole
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Default Braking while turning

"Douglas Landau" wrote in message
om...
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
0.1 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."



  #24  
Old July 23rd 03, 05:21 PM
Christopher Brian Colohan
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Default Braking while turning

"Peter Cole" writes:
"Take for example a rider cornering on good traction, leaning at 45
degrees. With this 1 G centrifugal acceleration, he can still apply
0.1 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: PGP: finger
Web:
www.colohan.com Phone: (412)268-4751
  #25  
Old July 23rd 03, 05:30 PM
Mike S.
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Default Braking while turning


"Take for example a rider cornering on good traction, leaning at 45
degrees. With this 1 G centrifugal acceleration, he can still apply
0.1 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


  #26  
Old July 23rd 03, 05:57 PM
Peter Cole
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Default Braking while turning

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

"Take for example a rider cornering on good traction, leaning at 45
degrees. With this 1 G centrifugal acceleration, he can still apply
0.1 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.


  #27  
Old July 23rd 03, 05:59 PM
Peter Cole
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Default Braking while turning

"Christopher Brian Colohan" wrote in message
.. .
"Peter Cole" writes:
"Take for example a rider cornering on good traction, leaning at 45
degrees. With this 1 G centrifugal acceleration, he can still apply
0.1 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.


  #28  
Old July 23rd 03, 06:27 PM
Mike S.
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Posts: n/a
Default Braking while turning


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.


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




  #29  
Old July 23rd 03, 07:34 PM
Jay Beattie
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Default Braking while turning


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

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.


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...


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.


  #30  
Old July 24th 03, 12:05 AM
Douglas Landau
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Posts: n/a
Default Braking while turning

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
0.1 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.
 




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