if you wanted maximum braking, where would you sit?

ZeeExSixAre wrote:
You won't be able to get butt far back enough and still reach the bars on
a road bike to skid the front wheel (on clean dry road).
True, but braking on a road bike isn't usually as nearly effective as on an
MTB because you can't really pull big leverage while on the hoods.

Bunk. I can lift the rear wheel from the hoods. How much more braking can
I have?
--
David Damerell Kill the tomato!

On Thu, 08 Jan 2004 19:34:05 -0600, Tom Sherman
wrote:
My trike [2] has two front brakes (Avid mechanical disc on each wheel -
one lever for each brake) and no rear brake. Stopping quickly takes as

Wow! Cool...fail-safe steering.

Tom Sherman - Quad Cities
--
Rick Onanian

On Fri, 09 Jan 2004 10:52:59 -0500, Rick Onanian
wrote:
wrote:
My trike [2] has two front brakes (Avid mechanical disc on each wheel -
one lever for each brake) and no rear brake. Stopping quickly takes as

Wow! Cool...fail-safe steering.

Nevermind; I just looked at the picture, and it was the ONLY
steering.

Okay...Wow! Cool...integrated steering.

Okay, on a second look, I think I see spindles and control arms. Do
you steer with the brakes, with the front wheels, or with the rear
wheel? What control do you operate to steer? Do you steer with the
handlebars, which then aim the front wheels through control arms?
--
Rick Onanian

Thanks for correcting every instance of me typing "Breaking" with
"Braking".. my apologies.

Mike
http://mikebeauchamp.com

"Ted Bennett" wrote in message
...
I find the idea position changes depending on speed and deaccelleration
while breaking.
So, that means it's constantly changing...
It's hard to put into words actually, so I'm trying to think about it..

Yes, a force vector diagram is the clearest way to see the problem.

First of all, I wouldn't "sit" anywhere. I"d definately get out of the
saddle, but I'd stay very low. I'd apply the breaks lightly and fairly
evenly with more front than rear. Then, I'd increase pressure and the
faster I de accellerate the farther I'd slide my ass (which is still off
the
seat) rearwards. Under the heaviest breaking, Ideally my weight would
probably want to be above the rear axle althought I doubt I could get it
that far back on even ground.

You meant, "apply the brakes", right?
No, you can't get your center of mass anywhere near the rear axle as
long as you have your hands on the bars.

While breaking, I think it's impossible to lean far enough back where
you
would not put enough weight on the front wheel (which should do the
majority
of the breaking). So, when you're leaning as far back as you can under
the
hardest breaking you're putting your physical weight onto the rear wheel
so
you can break with that better than normally, plus because of the
deaccelleration a LOT of weight is still on the front wheel so you can
get
as much out of it as well.

You meant "braking", didn't you? (Yes, I know that breaking can be
related to braking, as in, failure of.)
Assuming you have the traction, at the point of greatest deceleration
all of the weight is on the front wheel. The smaller the angle between
the line from the center of mass to the front contact patch and the
road, the greater the possible deceleration.

Mike
http://mikebeauchamp.com

--
Ted Bennett
Portland OR

"David Damerell" wrote in message
...
ZeeExSixAre wrote:
You won't be able to get butt far back enough and still reach the bars
on
a road bike to skid the front wheel (on clean dry road).
True, but braking on a road bike isn't usually as nearly effective as on
an
MTB because you can't really pull big leverage while on the hoods.

Bunk. I can lift the rear wheel from the hoods. How much more braking can
I have?

I can too, but I have to lean way forward. I was thinking more of the
insta-endo that most people can do with flat bars. My point is that it's
easier to pull a stoppie on a MTB than a road bike.

--
Phil, Squid-in-Training

Rick Onanian wrote:

On Fri, 09 Jan 2004 10:52:59 -0500, Rick Onanian
wrote:

wrote:

My trike [2] has two front brakes (Avid mechanical disc on each wheel -
one lever for each brake) and no rear brake. Stopping quickly takes as

Wow! Cool...fail-safe steering.


Nevermind; I just looked at the picture, and it was the ONLY
steering.

Okay...Wow! Cool...integrated steering.

Okay, on a second look, I think I see spindles and control arms. Do
you steer with the brakes, with the front wheels, or with the rear
wheel? What control do you operate to steer? Do you steer with the
handlebars, which then aim the front wheels through control arms?
--
Rick Onanian

The "U" handlebars are connected to a pivot (headset) with a short stem.
There is "Y" shaped steering arm attached to the stem that is connected
with tie rods to the steering knuckles.

The following website will be instructive to mechanically inclined
people interested in trike construction details.

http://www.ihpva.org/com/PracticalIn...ons/index.html

Tom Sherman - Quad Cities

Douglas Landau wrote:
Sheldon Brown wrote in message
...
Tim McNamara wrote:

Under "maximum braking" the front wheel stops dead, the bicycle
flips, and the rider is ejected. If you are talking about keeping
both wheels on the ground that is far less braking force than
maximum.

That can be circumvented. Build a bike that positions your center of
gravity below the front axle. You could lock the wheel up tight and
still not do an endo.

It actually has nothing to do with the axle, since a locked-up wheel
doesn't rotate, so it effectively stops being a wheel.

The critical thing is the angle of a line drawn from the tire contact
patch to the center of mass of the bike-and-rider.
I think you are correct because the OP misstated the problem. In reality
however Tim M is correct. In practice, when we go over the bars it is
not [usually] because the front wheel stops and the bike and rider pivot
around the contact patch. More often we go over the bars because the
frame and fork and rider and rear wheel pivot around the front axle.
If you build a bike like Tim says, you are right that it could still
endo around the contact patch. However, such a bike would not endo
around the front axle. I have seen riders endo after dropping the front
wheel in a pothole. Tim's bike would not do that. The rider would come
off the seat in an even more painful way.
Doug


The op didn't mistate his problem-he dictated the problem scope and som
parameters, leaving some assumptions open

Some subsequent posters mistated the issue and a solution to th
mistated issue with some misattribution in the quotes concerning wh
made earlier quotes

Sheldon pointed out the err of those tangential postings

With a sufficiently high coefficient of friction a longitudinall
elongated object (whether it has locked wheels or no wheels) can stil
endo even if its cg is very low, even a height proximate the contac
surface. If the moment about the lead contact point by the forward forc
on the cg times the height above the contact point is more than th
moment about that same point of the weight of the object times th
longitdunal length of the distance the cg is behind the lead contac
point, the object rolls forward


-

Rick Onanian wrote: “I imagine that all of those conditions would resul
in fishtailing under very hard braking. Is that the case?

Tom Sherman responded: ”Fishtailing only happens on the recumben
lowracer [1] when the rear wheel is locked up, and this take
considerably more braking force than it does to lock up the rear whee
on an upright.

The last generation of street automobile pre-abs brakes were forwar
biased so as to have the fronts locked first, brought the rear tire
into the lower sliding coefficient of friction relative their highe
static coefficient of friction before the fronts locked, and thereb
brake in a straight line. 1970’s And earlier brakes were more balance
resulting in the rears locking first as weight was transferred forwar
in braking. With those rears locked first, rear tires reached slidin
coefficient of friction before the fronts, so the vehicle fishtailed o
had the rear come around. In a most race cars, your taught if you ge
the out of control sliding/spinning car sliding in a safe direction, yo
lock the wheels hard and it vectors in a straight direction-all tire
are in sliding coefficient of friction

Tom Sherman wrote: ”Stopping quickly takes as much skill as stopping a
automobile without ABS and the brakes biased towards the front.

At least you didn’t identify as being as difficult as those pre-forwar
biased non-ABS brakes


-

Tim McNamara wrote:

Under "maximum braking" the front wheel stops dead, the bicycle
flips, and the rider is ejected. If you are talking about keeping
both wheels on the ground that is far less braking force than
maximum.

That can be circumvented. Build a bike that positions your center of
gravity below the front axle. You could lock the wheel up tight and
still not do an endo.

I replied in part:

It actually has nothing to do with the axle, since a locked-up wheel
doesn't rotate, so it effectively stops being a wheel.

The critical thing is the angle of a line drawn from the tire contact
patch to the center of mass of the bike-and-rider.

Douglas Landau wrote:

I think you are correct because the OP misstated the problem. In
reality however Tim M is correct. In practice, when we go over
the bars it is not [usually] because the front wheel stops and
the bike and rider pivot around the contact patch. More often
we go over the bars because the frame and fork and rider and rear
wheel pivot around the front axle.

'Fraid not. For that to happen, the front wheel would have to stop
moving forward, while the frame and fork did the endo. Such a scenario
would actually involve the front hub bearing reversing direction, so the
wheel would be rolling backward with respect to the frame. Where is
there a force that would cause the front wheel to rotate backwards?

If you build a bike like Tim says, you are right that it could
still endo around the contact patch. However, such a bike would
not endo around the front axle. I have seen riders endo after
dropping the front wheel in a pothole. Tim's bike would not do
that. The rider would come off the seat in an even more painful
way.

In a case where the front wheel is forcibly stopped by falling into a
deep pothole or hitting a high curb, it _is_ possible for the bike to
pivot around the front axle, but there's no way this can happen under
the influence of the brake alone.

Sheldon "Not A Velikovskian" Brown
+------------------------------------------------+
| Man will occasionally stumble over the truth, |
| but most of the time he will pick himself up |
| and continue on. -- Sir Winston Churchill |
+------------------------------------------------+
Harris Cyclery, West Newton, Massachusetts
Phone 617-244-9772 FAX 617-244-1041
http://harriscyclery.com
Hard-to-find parts shipped Worldwide
http://captainbike.com http://sheldonbrown.com

"meb" wrote in message
.. .

I started to do some calculations assuming a bike decels perhaps twice
as fast as a street car but calculated endos at much lower g's than
that. So before proceeding further, does anyone know typical best decel
rates on bikes in g's? or times or distance from 20-0mph?



I believe that a standard road bike can has a maximum deceleration rate to
the tune of about .6g's. Because quality tires now have coefficients of
friction around .9 to 1.5, in almost all situations involving a traditional
bike on clean, dry pavement, the deceleration rate is limited by the bike's
tendency to roll over the front tire. As previously stated in the thread,
this roll-over tendency is a function of the angle of the CG relative to the
front contact patch and the ground.



Assuming that it is possible to locate the CG anywhere we like, it is
possible to use the maximum available traction between the tires and the
road. If the coefficient of friction of tires is assumed to be constant
regardless of load, weight distribution is a null factor. Unfortunately,
real tires have a coefficient of friction that drops as load increases. (I
believe this has something to do with the thermal properties of rubber.) It
thus becomes evident that ideal weight distribution minimizes weight on both
tires, implying that they each carry half of the constant load. The easiest
way to do this on paper is to simply center the CG between the tires and put
it at ground level. We all know that the CG will always be some distance
above ground level. In that case, it must be located further back to
compensate for the forward weight shift while braking. In fact, the CG
location can be determined by the following formula:



Df= h*CF+WB/2



Where, Df is the horizontal distance of the CG from the front contact patch,
h is the CG height above the ground, CF is the coefficient of friction
between the tires and the road, and WB is the distance between the front and
rear contact patches.



Having said all that, it may be true that the drop of the CF for a bicycle
with unequally loaded tires is so small that it is not worth considering. In
that case, the weight must only be shifted back to a distance of:



Df=h*CF



to achieve maximum braking. This would leave no weight on the rear tire so
the bike would be a bit unstable. Note: A bicycle can be controlled somewhat
effectively with the rear tire continuously held off the ground.