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July 18th 03, 04:43 PM
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.
--
----

Phil, Squid-in-Training
July 19th 03, 03:16 AM
> 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.

FWIW, if you lose traction on the front in a turn, it is very likely you
will crash. If you lose traction on the rear in a turn, it is usually
recoverable unless your lean angle is extreme.

Braking the rear in a turn may cause the rear to skip causing the effects
you mentioned. But when you brake with the front, weight is pushed forward
onto the front wheel and giving it heavier steering, etc.

--
Phil, Squid-in-Training

David L. Johnson
July 19th 03, 04:17 AM
On Sat, 19 Jul 2003 02:16:20 +0000, Phil, Squid-in-Training wrote:

>> 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.

You are doing it right, by the way.
>
> FWIW, if you lose traction on the front in a turn, it is very likely you
> will crash. If you lose traction on the rear in a turn, it is usually
> recoverable unless your lean angle is extreme.

Yes, if you lose traction on the front in a turn, you will go
down. But you will usually go down if you skid the rear as well, and it
is much, much easier to skid the rear in any circumstances. In good
weather on a good road, it is impossible to skid the front.
>
> Braking the rear in a turn may cause the rear to skip causing the effects
> you mentioned. But when you brake with the front, weight is pushed
> forward onto the front wheel and giving it heavier steering, etc.

Any braking of either wheel will shift the weight distribution
forward. That is why the rear wheel skids so easily when you apply the
rear brake, since the amount of your weight on the rear can go to near 0.

--

David L. Johnson

__o | The trouble with the rat race is that even if you win you're
_`\(,_ | still a rat. --Lilly Tomlin
(_)/ (_) |

Phil, Squid-in-Training
July 19th 03, 05:27 PM
"g.daniels" > wrote in message
om...
> WE ARE LOOKING FOR THE LAB'S COMPUTER PRINTOUTS GENERATED BY THE LAB
> VEHICLE!!
> this inorder! to! pin down the language into numbers. then we can
> conceptualize from the other side.
> anybody who goes reaaaaalllly deeeeeep into the corners care to
> cawment???

LOL - reading this was just too much. Not to be condescending, but is
English your first language?

--
Phil, Squid-in-Training

Doug
July 20th 03, 06:30 AM
On Fri, 18 Jul 2003 23:17:12 -0400, "David L. Johnson"
> wrote:

>>> 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.
>
>You are doing it right, by the way.
>
>Yes, if you lose traction on the front in a turn, you will go
>down. But you will usually go down if you skid the rear as well, and it

It's an important discussion, I don't mean to be flippant. But I find
there is more to worry about than skidding with the front. If, for
example, a stone large enough to lift the wheel off the ground is hit,
the front wheel will stop. When it hits the ground again, an end over
is a given. So on turns on steep grades coming out of hills like in
Malibu where rocks are prevalent, I am very reluctant to use the front
brake.

>is much, much easier to skid the rear in any circumstances. In good
>weather on a good road, it is impossible to skid the front.

Good road, good weather. But even then, that lone stone. I just see
too many. I practice rear wheel skids actually to reduce trouble, but
yes, they can be nasty too, although never has one brought me down
(all handful of them).

Doug

Chalo
July 20th 03, 11:48 AM
"Phil, Squid-in-Training" > wrote:

> FWIW, if you lose traction on the front in a turn, it is very likely you
> will crash.

Yes, but it will always be a lowside. Because of the forward weight
shift, front wheel traction increases as the brake is applied. This
effect permits more braking while steering than most folks think
possible-- provided the tires are adequate to the task.

> If you lose traction on the rear in a turn, it is usually
> recoverable unless your lean angle is extreme.

In my experience, if you go into a turn hot enough, or downhill
enough, to have to brake while turning, a slip at either end will put
you on the ground. In those circumstances, it is much easier to lose
the rear. Furthermore, slipping the rear can cause a highside,
flinging you into the land of broken clavicles.

Chalo Colina

A Muzi
July 20th 03, 04:17 PM
> 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.
>
> 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.


We've discussed that here before and you're right. Note Sheldon's comments
on applying the rear brake of a tandem with nobody in back.

For a more dramatic example, just increase the speed and vehicle weight. Try
tapping the brakes during an aggressive corner in a rear engine car. You'll
only try that once!

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

Phil, Squid-in-Training
July 20th 03, 07:06 PM
> the rear. Furthermore, slipping the rear can cause a highside,
> flinging you into the land of broken clavicles.

Really now? In my experience, highsides have been limited to the domain of
motorcycles, where the speeds of the bikes and massive grips of the tires
have the potential to cause a highside.

Uh... wait. I recall highsiding on my MTB commuter once. I landed on my
forearms, so nothing broken. I now have a dolphin-shaped scar on my arm.
Nevermind what I said.

--
Phil, Squid-in-Training

Chris Zacho The Wheelman
July 20th 03, 08:43 PM
Just how hard are you applying the rear brake? If it feels like your
rear wheel is going straight while the rest of the bike is going through
a turn, you're skidding the rear wheel. And yes, there is a "physics"
reason for this. It's called inertia. An object in motion tends to move
in a straight line unless acted upon by another force. (sic.)

If your front wheel slides out, it too will go in a straight line. With
disastrous results. Actually, the best thing is not to brake at all in a
turn, Do all your braking before.

The tires have a limited amount of "grip" on the pavement, any that's
diverted towards slowing the bike down is being taken away from the
amount holding it in a curved path.

If you must apply brakes in a turn, do so very gently, and it's best, in
this case, to favor the rear. If that gives away, you have a much better
chance of riding it out.

May you have the wind at your back.
And a really low gear for the hills!
Chris

Chris'Z Corner
"The Website for the Common Bicyclist":
http://www.geocities.com/czcorner

Chris Zacho The Wheelman
July 20th 03, 08:47 PM
>Good road, good weather. But even
>then, that lone stone. I just see too
>many. I practice rear wheel skids
>actually to reduce trouble, but yes, they
>can be nasty too, although never has one
>brought me down (all handful of them).

>Doug

A stone, sand, water, oil, tar (just as Beloki).

May you have the wind at your back.
And a really low gear for the hills!
Chris

Chris'Z Corner
"The Website for the Common Bicyclist":
http://www.geocities.com/czcorner

Joe Riel
July 21st 03, 04:45 AM
(Chris Zacho "The Wheelman") writes:

> If your front wheel slides out, it too will go in a straight line. With
> disastrous results. Actually, the best thing is not to brake at all in a
> turn, Do all your braking before.

It may be the safest thing to do, but if you want to descend fast,
it's poor technique.

> The tires have a limited amount of "grip" on the pavement, any that's
> diverted towards slowing the bike down is being taken away from the
> amount holding it in a curved path.

Not really. First off, it is unlikely that you will be riding
right at the limits of adhesion. For a simple test, with your
bicycle beside you, lean the bike over, with one hand on the
stem and the other on the seat. Push into the bike. See how
how far over you can lean the bike until the tires slip. You
should be able to get the bike to around 45 degrees. Do you
corner anywhere near that lean angle? Few do.

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

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

With a mere 1% increase in contact patch force,
you can be slowing considerably.

> If you must apply brakes in a turn, do so very gently, and it's best, in
> this case, to favor the rear. If that gives away, you have a much better
> chance of riding it out.

This is poor advice. Better to learn and practice skills so that
you can corner at a comfortable speed with safety. If you only
learn to brake with the rear, you won't be able to stop or slow
down when you really need to.

Joe Riel

Christopher Brian Colohan
July 21st 03, 03:20 PM
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
--
Chris Colohan Email: PGP: finger
Web: www.colohan.com Phone: (412)268-4751

Joe Riel
July 21st 03, 03:31 PM
"Mike S." <mikeshaw2@coxDOTnet> writes:

> "> It may be the safest thing to do, but if you want to descend fast,
> > it's poor technique.
> >
> Actually, its better to brake and accelerate in straight lines. Don't ask
> me to explain, its physics...
>
> I have a nice scar on my left bicep from braking in a corner and running
> wide into a barbed wire fence. Grab brakes in the middle of a turn and the
> bicycle will have a tendency to stand up and widen the arc of the corner
> you're taking. If you don't know its going to happen, you too can end up
> with a nifty scar like mine.

I doubt that you would have been better off without braking. The point
being, unless you are willing to go slow all the time, you are better
off learning how to brake in turns. The time to learn is not when
you need it...rather practice under known conditions.

> If you have to brake, do it just like in cars and motorcycles: brake before
> the corner, corner, then accelerate as you're exiting.

You think cars and motorcycles don't brake in turns? That's nonsensical.
Furthermore, the amount of acceleration available to a cyclist is trivial
compared to the deceleration. Learn to use.

> If you can help it, don't brake. That's the fastest way down the
> mountain...

Not one with real turns.

Joe Riel

Chalo
July 21st 03, 09:39 PM
"Phil, Squid-in-Training" > wrote:

Chalo wrote:
> > Furthermore, slipping the rear can cause a highside,
> > flinging you into the land of broken clavicles.
>
> Really now? In my experience, highsides have been limited to the domain of
> motorcycles, where the speeds of the bikes and massive grips of the tires
> have the potential to cause a highside.
>
> Uh... wait. I recall highsiding on my MTB commuter once. I landed on my
> forearms, so nothing broken. I now have a dolphin-shaped scar on my arm.
> Nevermind what I said.

Though I've not actually seen video of it, the crash that took Beloki
out of the Tour this year (which commands its own thread right now)
has been described as a highside.

I've highsided both pushbikes and motorbikes, and motorbikes are
worse. ;^)

Chalo Colina

Joe Riel
July 22nd 03, 04:28 AM
Christopher Brian Colohan > writes:

> 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:

The reason I expanded the term was so that the math would be easier
(i.e. I wouldn't have to do the square root in my head). Alas,
I apparently can't do squares in my head. My formula gives 1.02
when done properly.

> 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.

That's why it's called margin 8-).

Consider this. The maximum braking force on dry asphalt is limited
to approximately 0.7g, to prevent flipping over. Assume that the
maximum contact force (where sliding initiates) is about 1g. It
is then possible to be simulateously braking at the maximum (0.7g),
while cornering at 0.7g! Cornering at 0.7g corresponds to a lean
angle, from upright, of approximately 35 degrees. I have crudely
measured my lean angles and know that 35 degrees feels quite aggressive;
I consider myself a reasonably fast descender---certainly one of
the faster in my club. The point being, the typical margins
riders deal with are wide indeed. Naturally, stuff in road
can reduce the margins in a hurry, but one of the points of practicing
is to learn to maneuver at speed.

>
> I do agree that the best way to learn this is not through math, but
> through practice.

Agreed, but I like doing the math anyway...

> 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...

I guess that's one of the downsides to living in San Diego; we don't
get much snowfall here 8-). To improve my cornering I have
used stiff wire attached to the rear dropout and set to touch the
ground at a preset lean angle. Start with a lean angle of about 30 degrees,
ride and corner progressively harder until you just hear the wire
scraping on the ground. Then gradually increase the lean angle.

Joe Riel

Chris B.
July 22nd 03, 07:31 AM
On Tue, 22 Jul 2003 03:28:19 GMT, Joe Riel > wrote:

>Christopher Brian Colohan > writes:

[...]

>> 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...
>
>I guess that's one of the downsides to living in San Diego; we don't
>get much snowfall here 8-). To improve my cornering I have
>used stiff wire attached to the rear dropout and set to touch the
>ground at a preset lean angle. Start with a lean angle of about 30 degrees,
>ride and corner progressively harder until you just hear the wire
>scraping on the ground. Then gradually increase the lean angle.
>
>Joe Riel

I have to try this...

--

Chris Bird

TJ Poseno
July 22nd 03, 01:48 PM
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!

David Damerell
July 22nd 03, 04:37 PM
TJ Poseno > wrote:
>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 > Distortion Field!

Michael
July 22nd 03, 04:51 PM
wrote ...
> 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

Douglas Landau
July 23rd 03, 01:55 AM
> > 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
July 23rd 03, 02:02 AM
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

Trentus
July 23rd 03, 12:30 PM
> 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

Peter Cole
July 23rd 03, 01:29 PM
"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."

Christopher Brian Colohan
July 23rd 03, 05:21 PM
"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

Mike S.
July 23rd 03, 05:30 PM
>
> "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

Peter Cole
July 23rd 03, 05:57 PM
"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.

Peter Cole
July 23rd 03, 05:59 PM
"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.

Mike S.
July 23rd 03, 06:27 PM
> > 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

Jay Beattie
July 23rd 03, 07:34 PM
"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.

Douglas Landau
July 24th 03, 12:05 AM
> 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.

Douglas Landau
July 24th 03, 12:09 AM
Joe Riel > wrote in message >...
> (Douglas Landau) writes:
>
> > - 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

Mark McMaster
July 24th 03, 02:12 AM
Douglas Landau wrote:
>>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.

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.

> 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.

> 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.

> 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.


> 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

Joe Riel
July 24th 03, 06:01 AM
"Jay Beattie" > writes:

> ... 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

Rick Onanian
July 24th 03, 01:17 PM
On Thu, 24 Jul 2003 05:01:34 GMT, Joe Riel > wrote:
> 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.

True, but you also couldn't blast into that turn going the full speed that
you judge you can handle for the section you CAN see, assuming that you'll
deal with the section you CAN'T see when it's too late.

Anybody who isn't sure when to brake should listen to their stomach and
their fear, and do what's required to make them comfortable. If you're
tense and scared, you'll definately wipe out.

There are just too damn many variables to make hard-and-fast rules and
mathematical formulas...but I guess this thread isn't about reality, but
rather, theory.

> Joe
--
Rick Onanian

Peter Cole
July 24th 03, 01:36 PM
"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...

I don't know what you mean by "corner harder". The example was given to
explain the technique for getting through a corner *fastest*. If you don't
care about speed, keep doing it your way, it doesn't harm anything, it's just
unnecessarily conservative.

Joe Riel
July 24th 03, 03:19 PM
Rick Onanian > writes:

> True, but you also couldn't blast into that turn going the full speed
> that you judge you can handle for the section you CAN see, assuming
> that you'll deal with the section you CAN'T see when it's too late.

Of course, but no one has suggested that braking should only be done
while turning. The converse, that braking should never be done
while turning, is frequently suggested. The purpose of the theory
is to show that significant braking can be accomplished while turning.

Joe

Mike S.
July 24th 03, 05:39 PM
"Joe Riel" > wrote in message
...
> Rick Onanian > writes:
>
> > True, but you also couldn't blast into that turn going the full speed
> > that you judge you can handle for the section you CAN see, assuming
> > that you'll deal with the section you CAN'T see when it's too late.
>
> Of course, but no one has suggested that braking should only be done
> while turning. The converse, that braking should never be done
> while turning, is frequently suggested. The purpose of the theory
> is to show that significant braking can be accomplished while turning.
>
> Joe

There's no question that you CAN brake and turn at the same time. Everyone
that's gone down a hill with turns in it has done it at some point.

I'd suggest a "coast off" next time you're going down your favorite hill.
Try the braking and turning style, then the brake, turn, accelerate style
and see which one's faster.

Mike

Benjamin Lewis
July 24th 03, 08:03 PM
Mike S. wrote:

>> I don't know what you mean by "corner harder". The example was given to
>> explain the technique for getting through a corner *fastest*. If you
>> don't care about speed, keep doing it your way, it doesn't harm
>> anything, it's just unnecessarily conservative.
>>
> That's something considering that you have no clue how fast most of the
> guys here can get down a mountain.
>
> Keep up the blanket statements!

Okay. It's true for everyone, now matter how fast they can get down a
mountain.

Some blanket statements are true.

--
Benjamin Lewis

On a paper submitted by a physicist colleague:
"This isn't right. This isn't even wrong." -- Wolfgang Pauli

Peter Cole
July 25th 03, 03:11 PM
"Mike S." <mikeshaw2@coxDOTnet> wrote in message
news:50UTa.15258$Bp2.7531@fed1read07...
> > I don't know what you mean by "corner harder". The example was given to
> > explain the technique for getting through a corner *fastest*. If you don't
> > care about speed, keep doing it your way, it doesn't harm anything, it's
> just
> > unnecessarily conservative.
> >
> >
> That's something considering that you have no clue how fast most of the guys
> here can get down a mountain.

I don't have to. I know that if you corner the way you say you do you're not
as fast as you could be.

> Keep up the blanket statements!

Yeah, vector math = "blanket statement". Sure.

August 1st 03, 08:36 AM
Subject: 9.15 Descending II
From: Jobst Brandt >
Date: Fri, 11 May 2001 16:35:42 PDT

Descending and Fast Cornering

Descending on mountain roads, bicycles can reach speeds that are more
common on motorcycles. Speeds that are otherwise not attainable, or
at least not continuously. Criterium racing also presents this
challenge, but not as intensely. Unlike a motorcycle, the bicycle is
lighter than the rider and power cannot be applied when banked over
when cornering hard. Because narrow bicycle tires inflated hard have
little traction margin, a slip on pavement is usually unrecoverable.

Drifting a Road Bicycle on Pavement

Riders have claimed they can slide a bicycle on dry pavement in curves
to achieve greater cornering speed, as in drifting through a turn. A
drift, in contrast to a slide, means that both wheels slip, which is
even more difficult. This notion may come from observing motorcycles,
that can cause a rear wheel slide by applying power when banked over.
Besides, when questioned about how this is done, the proponent says
that the ability was observed, done by others.

A bicycle can be pedaled only at lean angles far less than the maximum
without grounding a pedal, so hard cornering is always done coasting,
hence, there is no power in hard cornering. Although bicycles with
high ground clearance have been built, they showed only that pedaling
imbalance has such a disturbing influence on traction, that pedaling
at a greater lean angle than that of a standard road racing bicycles
has no benefit. That is why road bicycles are built the way they are,
no higher than is useful.

That bicycle tires have no margin for recovering a slip at maximum
lean angle, has been tested in lean-slip tests on roads and testing
machines. For smooth tires on pavement, slipout occurs at slightly
less than 45 degrees from the road surface and is both precipitous and
unrecoverable. Although knobby tires have a less sudden slipout and
can be drifted around curves, they begin to side-slip at a more
upright angle as their tread fingers walk rather than slip. For this
reason, knobby tires cannot achieve lean angles of smooth tires and
offer no cornering advantage on pavement.

How to Corner

Cornering requires estimating the required lean angle before reaching
the apex of the turn where the angle with the road surface is the
critical parameter rather the angle with the vertical, as is evident
from banked curves. Lean angle is limited by the available traction
that must be assessed from velocity and appearance of the surface.
For good pavement, this angle is about 45 degrees, in the absence of
oil, water, or smooth and slick spots. Therefore, a curve banked
inward 10 degrees, allows a lean of up to at least 55 degrees from the
vertical, while a crowned road with no banking, where the surface
falls off about 10 degrees, would allow only up to 35 degrees.

Banked curves have a greater effect than just adding to the maximum
lean angle, because with a steeper banking, more of the centripetal
cornering force goes into increasing traction directly into the
banking up to the point of a vertical wall where only the maximum
G-forces limit what speed a bicyclists can attain. In contrast, an
off banked curve makes cornering progressively more difficult until
the bicycle will slip even at zero speed. This effect is more
naturally apparent to riders who exceeded these limits early in life
and have added the experience to expected natural phenomena.

The skill of visualizing effects of speed, traction, braking, and
curvature are complex, but is something humans and other creatures do
regularly in self propulsion. The difficulty arises in adapting this
to higher speeds. When running, we anticipate how fast and sharply to
turn on a sidewalk, dirt track, or lawn, to avoid sliding. The method
is the same on a bicycle although the consequences of error are more
severe.

Cornering requires reflexes to dynamics that are easily developed in
youth, while people who have not exercised this in a long time find
they can no longer summon these skills. A single fall strongly
reinforces doubt, so cautious practice is advisable if returning to
bicycling after a long time.

Countersteer

Countersteer is a popular subject for people who belatedly discover or
rediscover how to balance. What is not apparent, is that two wheeled
vehicles can be controlled ONLY by countersteer, there is no other
way. Unlike a car, a bicycle cannot be diverted from a straight path
by steering the wheel to one side. The bicycle must first be leaned
in that direction by steering it ever so slightly the other way. This
is the means by which a broomstick is balanced on the palm of the hand
or a bicycle on the road. The point of support is moved beneath the
mass, in line with the combined forces of gravity and cornering, and
it requires steering, counter and otherwise. It is so obvious that
runners never mention it, although football, basketball, and ice
hockey players conspicuously do it.

Braking

Once the basics of getting around a corner are developed, doing it
fast involves careful use of the brakes. Besides knowing how steeply
to lean in curves, understanding braking makes the difference between
the average and the fast rider. When approaching a curve with good
traction, the front brake can be used almost exclusively, because it
is capable of slowing the bicycle so rapidly that nearly all weight
transfers to the front wheel, at which point the rear brake is nearly
useless. Once in the curve, more and more traction is used to resist
lateral slip as the lean angle increases, but that does not mean the
brakes cannot be used. When banked over, braking should be done with
both brakes, because now neither wheel has much traction to spare and
with lighter braking, weight transfers diminishes. A feel for how
hard the front brake must be applied for rear wheel lift-off, can be
developed at low speed.

Braking in Corners

Why brake in the turn? If all braking is done before the turn, speed
will be slower than necessary before the apex. Anticipating maximum
speed for the apex is difficult, and because the path is not a
circular arc, speed must be trimmed all the way to that point. Fear
of braking in curves usually comes from an incident of injudicious
braking at a point where braking should have been done with a gentle
touch to match the conditions.

Substantial weight transfer from the rear to the front wheel will
occur with strong use of the front brake on good traction just before
entering the curve. When traction is poor or the lean angle is great,
deceleration cannot be large and therefore, weight transfer will be
small, so light braking with both wheels is appropriate. If traction
is miserable, only the rear brake should be used, because although a
rear skid is recoverable, a front skid is generally not. An exception
to this is in deep snow, where the front wheel can slide and function
as a sled runner while being steered.

Braking at maximum lean

For braking in a curve, take the example of a rider cornering with
good traction, leaning at 45 degrees, the equivalent of 1G centrifugal
acceleration. Braking with 1/10g increases the traction demand by one
half percent. The sum of cornering and braking vectors is the square
root of the sum of their squares, SQRT(1^2+0.1^2)=1.005 or an increase
of 0.005. In other words, there is room to brake substantially during
maximum cornering. Because the lean angle changes as the square of
the speed, braking can rapidly reduce the angle and allow even more
braking. For this reason skilled racers nearly always apply both
brakes into the apex of turns.

Suspension

Beyond leaning and braking, suspension helps substantially in
descending. For bicycles without built-in suspension, this is
furnished by the legs. Standing up is not necessary on roads with
fine ripples, just taking the weight off the pelvic bones is adequate.
For rougher roads, enough clearance must be used so the saddle carries
no weight. The reason for this is twofold. Vision will become
blurred if the saddle is not unloaded, and traction will be
compromised if the tires are not bearing with uniform force on the
road while rolling over bumps. Ideally the tires should bear on the
road at constant load. Besides, if the road has whoop-de-doos, the
seated rider will get launched from the saddle and possibly crash.

Lean the Bicycle, the Rider, or Both

Some riders believe that sticking the knee out or leaning the body
away from the bicycle, improves cornering. Sticking out a knee is the
same thing that riders without cleats do when they stick out a foot in
dirt track motorcycle fashion. On paved roads this is a useless but
reassuring gesture that, on uneven roads, even degrades control. Any
body weight that is not centered over the bicycle (leaning the bike or
sticking out a knee) puts a side load on the bicycle, and side loads
cause steering motions over uneven road. Getting weight off the
saddle is also made more difficult by such maneuvers.

To verify this, coast down a straight but rough road, weight on one
pedal with the bike slanted, and note how the bike follows an erratic
line. In contrast, if you ride centered on the bike you can ride
no-hands perfectly straight over the same road. While leaning off the
bike, trail of the front wheel causes steering on rough roads.

Outside Pedal Down

It is often said that putting the outside pedal down in a curve
improves cornering. Although most experienced riders do this, it is
not because it has anything to do with traction. The reason is that
it enables the rider to unload the saddle while standing with little
effort on a locked knee, cushioning his weight on his ankle. This can
only be done on the outside pedal because the inside pedal would hit
the road. However, standing on one extended leg does not work on
rougher roads, because the ankle cannot absorb large road bumps nor
raise the rider high enough from the saddle to avoid getting bounced.
Rough roads require rising high enough from the saddle to avoid hard
contact while the legs supply shock absorbing knee action, with pedals
and cranks horizontal.

Body Contortions

Most of the "body English" riders display is gratuitous gesturing,
much like the motorcyclists who stick their butt out in curves while
their bikes never get down to 45 degrees (the angle below which hiking
out becomes necessary to keep hardware from dragging on the road). In
fact, in a series of tight ess bends, there's no time to do any of
this. It's done by supporting weight on the (horizontally positioned)
pedals, and unless the road is rough, with a light load on the saddle.
On rough roads, the cheeks of the saddle, (the ones that went away
with the Flite like saddles) are used to hold the bicycle stably
between the legs while not sitting.

The path through a curve is not symmetrical for a bicycle, because it
can slow down much faster than it can regain speed. Thus the
trajectory is naturally asymmetric. Brakes are generally used to the
apex (that is usually not the middle) of the curve, where pedaling at
that lean angle is not possible, nor does pedaling accelerate as fast
as braking decelerates.

Hairpin Turns

Although the railroad term switchback arises from early mountain
railroading where at the end of a traverse, a switch is turned to back
up the next traverse, after which another switch is turned to head up
the next, on roads these are hairpin turns. In such turns trajectory
asymmetry is most conspicuous, because braking can be hard enough to
raise the rear wheel when entering but one cannot exit with such
acceleration. For this reason, riders often find themselves with
extra road on the exit of such turns, having slowed down too much.

Vision

Where to direct vision is critical for fast cornering. Central vision
should be focused on the pavement where the tire will track, while
allowing peripheral vision, with its low resolution and good
sensitivity to motion, to detect obstacles and possible oncoming
traffic. Peripheral vision monitors surroundings anyway, so the
presence of a car in that "backdrop" does not require additional
consideration other than its path.

If central vision is directed at the place where an oncoming vehicle
might appear, its appearance presents a new problem of confrontation,
stopping image processing of the road surface for substantial time.
Because the color or model of car is irrelevant, this job can be left
to peripheral vision in high speed primitive processing, while
concentrating on pavement surface and composition.

When following another bicycle or a car downhill, the same technique
is even more important, because by focusing on the leading vehicle,
pavement and road alignment information is being obscured giving a
tendency to mentally become a passenger of that vehicle. Always look
ahead of the vehicle, observing it only peripherally.

Riders often prefer to keep their head upright in curves, although
leaning the head with the bicycle and body is more natural to the
motion. Pilots who roll their aircraft do not attempt to keep their
head level during the maneuver, or in curves, for that matter.

The Line

Picking the broadest curve through a corner may be obvious by the time
the preceding skills are mastered, but that may not be the best line,
either for safety or because the road surface is poor. Sometimes
hitting a bump or a "Bott's dot" is better than altering the line,
especially at high speed. Tires should be large enough to absorb the
entire height of a lane marker without pinching the tube. This means
that a minimum of a 25mm actual cross section tire is advisable. At
times, the crown of the road is sufficient to make broadening the
curve, by taking the curve wide, counterproductive because the crown
on the far side gives a restricted lean angle.

Mental Speed

Mental speed is demanded by all of these. However, being quick does
not guarantee success, because judgment is even more important. To
not be daring but rather to ride with a margin that leaves a feeling
of comfort rather than high risk, is more important. Just the same,
do not be blinded by the age old presumption that everyone who rides
faster than I is crazy. "He descends like a madman!" is one of the
most common descriptions of fast descenders. The comment generally
means that the speaker is slower.

Braking Heat on Steep Descents

Although tandems with their higher weight to wind drag ratio have this
problem more often, steep mountain roads, especially ones with poor or
no pavement require so much braking that single bicycles blow off
tires from overheating. For tubulars the problem is not so much over
pressure than rim glue melting as all pressure sensitive glues do with
heating. As glue softens, tires slip on the hot rim and pile up on
the valve stem. This is the usual indicator that tubular tire wheels
are too hot. The next is that the tire arches off the rim in the area
just before the stem.

This is a serious problem both for tubulars and clinchers because most
clincher tires, given enough time on a hot rim will blow off if
inflated to recommended pressure. Pressure that gives good rolling
performance (hard) while tubulars roll off from lack of adhesion to
the rim. The faster the travel, the more descending power goes into
wind drag and the better the rims are cooled. Going slowly does not
help, unless speed is reduced below walking pace.

On steep descents, where rims stay too hot to touch for more than a
minute, reducing tire inflation pressure is a sure remedy. However,
tires should be re-inflated once the rims cool to normal. The
blow-off pressure is the same for small and large tires on the same
rim, it being dependent only on the opening of the rim width. Also,
tires with a smaller air volume become hot faster than larger ones.

There is no way of descending continuously and steeply without
reducing inflation pressure, unless there is an insulator between the
tube and rim of a clincher. Insulating rim strips are no longer
offered because they were an artifact of dirt roads that often
required riders to descend so slowly that all potential energy went
into the brakes and almost none into wind drag. These rim strips were
cloth tubes filled with kapok, their insulating purpose being unknown
to most people when they were last offered.

Jobst Brandt

Palo Alto CA

August 1st 03, 10:01 AM
Chalo Colina writes:

> Though I've not actually seen video of it, the crash that took
> Beloki out of the Tour this year (which commands its own thread
> right now) has been described as a highside.

> I've highsided both pushbikes and motorbikes, and motorbikes are
> worse.

That's why he hit so hard and it was from losing traction on a tar
stripe, that with the hot weather was gooey. It was a classic
slippery surface type crash as we see more often in Paris-Roubaix on
wet cobbles. Motos just throw the rider farther and with greater
speed.

Jobst Brandt

Palo Alto CA

g.daniels
August 1st 03, 03:59 PM
ah yes:
that's one way to solve the question. drive a bald tired whatever but
several extensively framewheel tire combos on a geased surface or put
snow tires with studs on all two or four's or 18 if available.
but yagotta work up, gently whatever your relative ground speed(as
opposed to ur mitty speed) to the experimental level in the first
paragraph. experience it. develop awareness of what your doing. hey,
the dead are buried just outside the wall, dude.
and some can't get there. they're not fast enough to reach a
conclusion. they think they're fast enough and that's ok 'cause it
saves the ins. burial expenses we all share when budman meets live
axle and wooden wheels.

g.daniels
August 1st 03, 04:03 PM
one question? front grip minus turning friction minus brake friction
gives us X grip. but the front does go in the "right direction" more
or less whereas the rear has no differential-the rear gets dragged
along. What does "drag along" mean in terms of grip and brake balance?

Werehatrack
August 1st 03, 05:16 PM
On Sun, 20 Jul 2003 19:44:15 -0000, Ray Heindl > may
have said:

>Doug > wrote:
>
>> It's an important discussion, I don't mean to be flippant. But I
>> find there is more to worry about than skidding with the front.
>> If, for example, a stone large enough to lift the wheel off the
>> ground is hit, the front wheel will stop. When it hits the ground
>> again, an end over is a given. So on turns on steep grades coming
>> out of hills like in Malibu where rocks are prevalent, I am very
>> reluctant to use the front brake.
>
>Why does the front wheel stopping cause an endo? I would think that
>once it hits the ground again, it will either skid or start to rotate
>again, depending on the amount of traction available. Is it because
>the coefficient of friction between the brake and the wheel is higher
>once the wheel stops rotating?

Yes. It requires more rotational force to break the grip of pads that
are in contact with a motionless wheel; this is a well-known principle
in physics. The gradient can be remarkable.

--
My email address is antispammed;
pull WEEDS if replying via e-mail.
Yes, I have a killfile. If I don't respond to something,
it's also possible that I'm busy.

August 1st 03, 06:01 PM
anonymous snipes surreptitiously:

>>> If, for example, a stone large enough to lift the wheel off the
>>> ground is hit, the front wheel will stop. When it hits the ground
>>> again, an end over is a given. So on turns on steep grades coming
>>> out of hills like in Malibu where rocks are prevalent, I am very
>>> reluctant to use the front brake.

>>Why does the front wheel stopping cause an endo? I would think that
>>once it hits the ground again, it will either skid or start to
>>rotate again, depending on the amount of traction available. Is it
>>because the coefficient of friction between the brake and the wheel
>>is higher once the wheel stops rotating?

> Yes. It requires more rotational force to break the grip of pads
> that are in contact with a motionless wheel; this is a well-known
> principle in physics. The gradient can be remarkable.

I don't know where you get these mechanical models on which these
theoretical results are based but they are not true. One can easily
stop the front wheel on s spot of gravel and without letting up on the
brake continue a descent with the wheel again rotation when solid
pavement is reached. This assumes the length of the skid is not long
and not at a strong lean angle. This is not difficult for a ride who
knows how to descend.

The reference to static friction being higher is commonly called
stiction but it does not have significant effect on bicycle braking.
Riding over a wet tar stripe or a bit of ice is a more common form of
a wheel stopping but it isn't even slightly as treacherous as it is
made out to be by the worry warts.

I don't see why writers to these forums take pleasure in creating
horror scenarios supported by pseudo-science. They seem to themselves
be inept enough that they crash and and find pleasure in transferring
their pain and fear to others so they won't dare have it any better
than they.

Jobst Brandt

Palo Alto CA

August 1st 03, 06:56 PM
anonymous snipes surreptitiously:

>>> If, for example, a stone large enough to lift the wheel off the
>>> ground is hit, the front wheel will stop. When it hits the ground
>>> again, an end over is a given. So on turns on steep grades coming
>>> out of hills like in Malibu where rocks are prevalent, I am very
>>> reluctant to use the front brake.

>>Why does the front wheel stopping cause an endo? I would think that
>>once it hits the ground again, it will either skid or start to
>>rotate again, depending on the amount of traction available. Is it
>>because the coefficient of friction between the brake and the wheel
>>is higher once the wheel stops rotating?

> Yes. It requires more rotational force to break the grip of pads
> that are in contact with a motionless wheel; this is a well-known
> principle in physics. The gradient can be remarkable.

I don't know where you get these mechanical models on which these
theoretical results are based but they are not true. One can easily
stop the front wheel on a spot of gravel, and without letting up on
the brake, continue a descent with the wheel again rotating when solid
pavement is reached. This assumes the length of the skid is not long
and not at a strong lean angle. This is not difficult for a rider who
knows how to descend.

The reference to static friction being higher is commonly called
stiction but it does not have significant effect on bicycle braking.
Riding over a wet tar stripe or a bit of ice is a more common form of
a wheel stopping but it isn't even slightly as treacherous as it is
made out to be by the worry warts.

I don't see why writers to these forums take pleasure in creating
horror scenarios supported by pseudo-science. They seem to themselves
be inept enough that they crash and and find pleasure in transferring
their pain and fear to others so they won't dare have it any better
than they.

Jobst Brandt

Palo Alto CA

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