if you wanted maximum braking, where would you sit?
wle
maximum braking power, where would it be?
obviously over the front wheel is no good, you would flip.
there is a point, leaning either ahead of the front wheel, or
behind the back wheel, that the opposite wheel is off the ground.
clearly those are 2 limits, the answer must lie between them.
if there were very little friction, it would hardly matter.
assume a level road, brakes that can cause a skid no matter what.
ok, so where do you sit?
state assumptions, like coefficient of friction between tire and road,
weight of bike and rider.
show your work. :)
now for extra credit, make it a function of road slope.
wle.
Rick Onanian
>if you could position your center of gravity anywhere, to ensure
>maximum braking power, where would it be?
On a flat piece of rubber in direct contact with the pavement.
>wle.
--
Rick Onanian
Ken
news:60402c15.04010...@posting.google.com:
> there is a point, leaning either ahead of the front wheel, or
> behind the back wheel, that the opposite wheel is off the ground.
Unless you're climbing a steep hill, you're going to have a real hard time
lifting the front wheel. If the hill is that steep, stopping shouldn't be a
problem.
Tom Sherman
> if you could position your center of gravity anywhere, to ensure
> maximum braking power, where would it be?
>
> obviously over the front wheel is no good, you would flip.
>
> there is a point, leaning either ahead of the front wheel, or
> behind the back wheel, that the opposite wheel is off the ground.
>
> clearly those are 2 limits, the answer must lie between them.
>
> if there were very little friction, it would hardly matter.
>
> assume a level road, brakes that can cause a skid no matter what.
>
> ok, so where do you sit?
>
> state assumptions, like coefficient of friction between tire and road,
> weight of bike and rider.
I would sit near or below the level of the front axle to reduce or
elminate the tendency for the rear wheel to lift off of the ground, and
I would also be near the rear wheel for more favorable weight
distribution for braking. This describes a recumbent lowracer. Braking
can be improved by adding a second wheel in front (tadpole trike) as
weight transfers forward to two contact patches during braking, and
locking one or both front wheels under braking will not lead to a loss
of balance.
Tom Sherman - Quad Cities
R15757
<< if you could position your center of gravity anywhere, to ensure
maximum braking power, where would it be? >>
<snip>
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.
Very hard braking is an athletic maneuver that requires a quick, exaggerated
body movement. There is no sitting whatsoever.
Robert
Steve Juniper
wheels. Because the rear tends to lose braking power as weight shifts
forward, moving back just far enough to balance the effect ought to produce
the max. I suppose you could tell where that is by trial and error,
determining how far back you need to move so that both wheels skid at the
same time.
No, wait a minute - on reflection it seems moving back would reduce front
braking power. Maybe it doesn't make any difference at all, so long as both
wheels stay on the ground or, theoretically, maybe not even then, until
braking itself causes the rear wheel to lift.
--
Steve Juniper
"Donald Rumsfeld took that job as Secretary of Defense because
he couldn't get a job with a death squad."
-- Barbara
Bush --
"wle" <w...@mailinator.com> wrote in message
news:60402c15.04010...@posting.google.com...
Tim McNamara
> << w...@mailinator.com (wle) >>
>
> << if you could position your center of gravity anywhere, to ensure
> maximum braking power, where would it be? >>
>
> <snip>
>
> 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.
Mike Beauchamp
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..
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.
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.
"wle" <w...@mailinator.com> wrote in message
news:60402c15.04010...@posting.google.com...
Ted Bennett
No. The important dimension is where the center of mass is in relation
to the front wheel's contact patch. An endo is possible with the weight
below the axle, although not likely because of insufficient traction.
--
Ted Bennett
Portland OR
Sheldon Brown
>>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.
If this angle is steeper than a critical value, locking up the front
wheel will cause an endo. This is the case for typical upright bikes.
If the angle is shallower than the critical angle, locking up the front
wheel will cause it to skid. This is the case for tandems and many
recumbents.
The value of the critical angle depends on the coefficient of friction
between the tire and the road surface.
For vehicles where the angle is steep, the rear brake is useless at
maximal braking.
For vehicles where the angle is shallow, both brakes are needed to
achieve maximal braking.
Note that the _length_ of the line doesn't matter, only the angle.
See also http://sheldonbrown.com/brakturn.html
Sheldon "Geometry And Physics" Brown
+------------------------------------------+
| So we'll go no more a roving |
| So late into the night, |
| Though the heart be still as loving, |
| And the moon be still as bright. |
| |
| For the sword outwears its sheath, |
| And the soul wears out the breast, |
| And the heart must pause to breathe, |
| And Love itself have rest. |
| |
| Though the night was made for loving, |
| And the day returns too soon, |
| Yet we'll go no more a roving |
| By the light of the moon. |
| --Lord Byron |
+------------------------------------------+
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
Ted Bennett
> 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
R15757
extremely important factors in hard braking, but these never seem to be
accounted for in anybody's theorems. The rider is always assumed to be still as
a statue during the act. I think there is another order of complexity that is
being missed.
Robert
stu
> > 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.
>
> No. The important dimension is where the center of mass is in relation
> to the front wheel's contact patch. An endo is possible with the weight
> below the axle, although not likely because of insufficient traction.
a locked front wheel does not undo you, you skid and fall of sideways.
ZeeExSixAre
> gravity below the front axle. You could lock the wheel up tight and
> still not do an endo.
Almost anybody can lock up the front wheel tight by leaning back far enough
(meaning butt just over the rear wheel). You don't need to build any kind
of special bike... I do it on my MTB.
--
Phil, Squid-in-Training
Pete Biggs
> if you could position your center of gravity anywhere, to ensure
> maximum braking power, where would it be?
As far back as possible, and brace hard against the bars.
~PB
dianne_1234
Underground. ;-)
David Damerell
>if you could position your center of gravity anywhere, to ensure
>maximum braking power, where would it be?
Evidently maximum braking power is a function of the angle described
between the ground and a line drawn through the contact patch and the CoG.
Hence for a given angle the CoG may move freely along that line without
influencing the braking power. Therefore to maximise braking power I would
position the CoG at the bottom bracket, assuming we are constrained to
having it within the bike-rider structure; otherwise, at ground level
below the BB or thereabouts (to have an even load on each wheel when not
braking or accellerating).
>assume a level road, brakes that can cause a skid no matter what.
It is extremely unlikely that front brake will cause a skid no matter
what. If tyre traction is that poor then that is evidently the limiting
factor on braking, and the CoG may roam freely.
--
David Damerell <dame...@chiark.greenend.org.uk> Distortion Field!
Pete Biggs
> Almost anybody can lock up the front wheel tight by leaning back far
> enough (meaning butt just over the rear wheel). You don't need to
> build any kind of special bike... I do it on my MTB.
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).
~PB
H. M. Leary
Sheldon Brown <Capt...@sheldonbrown.com> wrote:
snip
>
> See also http://sheldonbrown.com/brakturn.html
>
> Sheldon "Geometry And Physics" Brown
> +------------------------------------------+
> | So we'll go no more a roving |
> | So late into the night, |
> | Though the heart be still as loving, |
> | And the moon be still as bright. |
> | |
> | For the sword outwears its sheath, |
> | And the soul wears out the breast, |
> | And the heart must pause to breathe, |
> | And Love itself have rest. |
> | |
> | Though the night was made for loving, |
> | And the day returns too soon, |
> | Yet we'll go no more a roving |
> | By the light of the moon. |
> | --Lord Byron |
> +------------------------------------------+
> 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
>
Thanks for the Lord Byron - made my day.
The last time I locked my front wheel, I ended up hugging a pine tree.
HAND
Stay warm
--
"Freedom Is a Light for Which Many Have Died in Darkness"
- Tomb of the unknown - American Revolution
S o r n i
Bill "landing, that is" S.
Zog The Undeniable
As you still have to hold on to the bars, throwing yourself back off the
saddle is the best you can do. The back wheel still skids but the bike
shouldn't cartwheel if you go back and low enough. Failing that, slide
the bike sideways at 90 degrees to the direction of travel. This may hurt.
Bill Putnam
> recumbents.
Have you ever skidded a front tire on clean dry pavement with a
tandem? Or has anyone else?
I have never been able to skid the front tire on my tandem on clean
dry pavement despite practiing emergency stops to the point where I
worried about the fork failing, though I haven't bent a fork or
observed any obvious damage from this maneuver. I know Chalo's bent
plenty of forks but I'd like to see him do that with my Santana.
Bill Putnam, who on second thought wouldn't like to see Chalo bend the
fork on his tandem even if he could.
Zog The Undeniable
> It is extremely unlikely that front brake will cause a skid no matter
> what. If tyre traction is that poor then that is evidently the limiting
> factor on braking, and the CoG may roam freely.
I've done long 2-wheel skids on dry tarmac. A panic snatch of the brake
will get it sliding, after which the friction is greatly reduced.
Steering is *difficult*.
Rick Onanian
<tshe...@qconline.com> wrote:
>I would sit near or below the level of the front axle to reduce or
>elminate the tendency for the rear wheel to lift off of the ground, and
>I would also be near the rear wheel for more favorable weight
>distribution for braking. This describes a recumbent lowracer. Braking
>can be improved by adding a second wheel in front (tadpole trike) as
>weight transfers forward to two contact patches during braking, and
>locking one or both front wheels under braking will not lead to a loss
>of balance.
I imagine that all of those conditions would result in fishtailing
under very hard braking. Is that the case?
>Tom Sherman - Quad Cities
--
Rick "Sounds like fun!" Onanian
Pete Biggs
>> It is extremely unlikely that front brake will cause a skid no matter
>> what. If tyre traction is that poor then that is evidently the
>> limiting factor on braking, and the CoG may roam freely.
>
> I've done long 2-wheel skids on dry tarmac.
Dry tarmac coated with dust or gravel? Otherwise I find it hard to
believe. A panic snatch on my bikes will result in me going over the
bars.
~PB
(Pete Cresswell)
>ok, so where do you sit?
I can't imagine sitting when braking as hard as I can. I'd be afraid of
endo-ing.
It's more like crouching with my butt somewhere between the back of the saddle
and the rear axle.
--
PeteCresswell
David Damerell
>David Damerell wrote:
>>It is extremely unlikely that front brake will cause a skid no matter
>>what. If tyre traction is that poor then that is evidently the limiting
>>factor on braking, and the CoG may roam freely.
>I've done long 2-wheel skids on dry tarmac.
With slick tyres? That seems highly unlikely.
[Note the words "no matter what" copied directly from the original post,
so I think I can specify tyre design.]
If I panic snatch the front brake I get a free flying lesson.
--
David Damerell <dame...@chiark.greenend.org.uk> Kill the tomato!
Tom Sherman
Fishtailing only happens on the recumbent lowracer [1] when the rear
wheel is locked up, and this takes considerably more braking force than
it does to lock up the rear wheel on an upright.
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
much skill as stopping an automobile without ABS and the brakes biased
towards the front.
Subjectively, the stopping power does not feel as great as it is, since
the rider is naturally braced by his/her legs being on the pedals,
unlike an upright bike where bracing is done with the (much weaker) arms.
[1] This does not apply to all recumbent bikes. Some are poorly designed
with a high seat and too much weight on the front wheel, and may have
braking performance that is worse than an upright bicycle.
[2] < http://home.mindspring.com/~kb7mxu/images/DragonFlyer.jpg > This
is the same model as my trike.
Tom Sherman
As an owner of a bicycle with a 12-inch (30.5 cm) seat height, I can
verify this (at least on a dry, broom finish Portland cement concrete
surface, which is certainly a high traction situation). Something beyond
friction between the tire and the pavement surface (such as wedging the
wheel in a storm drain grate) would have to occur to get the rear wheel
off the ground.
Tom Sherman
A good example of this would be an ordinary bicycle, which has a long
line from the contact, patch to the combined rider/bicycle center of
mass that is at a steep angle (from horizontal). It is the general
consensus that these bikes are relatively easy to do "endos" on.
Tom Sherman
> Sheldon Brown <Capt...@sheldonbrown.com> wrote in message > If the angle is shallower than the critical angle, locking up the front
>
>>wheel will cause it to skid. This is the case for tandems and many
>>recumbents.
>
> Have you ever skidded a front tire on clean dry pavement with a
> tandem? Or has anyone else?...
I have locked up both front wheels on my recumbent tadpole trike. It
should be noted that I have Avid mechanical disc brakes that are
designed for use on wheels of about 27" (~68 cm) in diameter on wheels
that are about 20" (~51 cm) in diameter.
ZeeExSixAre
> 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.
--
Phil, Squid-in-Training
Phil Holman
"S o r n i" <so...@bite-me.san.rr.com> wrote in message
news:2hgLb.45507$C87....@twister.socal.rr.com...
> Cockpit of a 747?
>
> Bill "landing, that is" S.
>
takeoff and have been known to put a wrinkle in the forward fuselage
crown skin panels. Braking forces on most commercial airplanes are
limited by system design but generally this is the load case
(compression buckling) that determines the skin gage of the upper
panels.
Phil Holman
Ted Bennett
> Rick "Sounds like fun!" Onanian
No, it isn't the case on any of my trikes. Yes, it is fun.
Charles Beristain
tires).. lock up the front wheel and continue to pedal.. the rear
wheel has more than enough traction on ice to move the bike forward
with the studs scratching the ice while being locked up.. But it takes
practice keeping the front wheel straight... a little movement in the
handlebars and down you go. Wheelies on ice is another way to put some
excitement in your life.
charlie
Douglas Landau
> 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
meb
> if you could position your center of gravity anywhere, to ensure maximum
> braking power, where would it be?
> obviously over the front wheel is no good, you would flip.
> there is a point, leaning either ahead of the front wheel, or behind the
> back wheel, that the opposite wheel is off the ground.
> clearly those are 2 limits, the answer must lie between them.
> if there were very little friction, it would hardly matter.
> ok, so where do you sit?
> weight of bike and rider.
> show your work. :)
> now for extra credit, make it a function of road slope.
The bikes two tires will have their maximum adhesion sum if weight under
braking is evenly distributed on front and rear tires. The more weight
on the front, the more front braking, the less rear braking and vice
versa. The endo issue occurs when the moment caused by the decel of the
bike+rider about the front contact patch exceeds the product of distance
the combination cg trails the front patch times the mass- a point at
which the weight is entirely on the front tire. If selecting position
for maximum tire adhesion, you’ve already removed endo as an issue.
The front/rear weight distribution of the bike under constant speed is
not the same as the weight distribution under deceleration. Under
decell, a moment is created which is the product of the height of the
combined (bike+rider) cg above the road times the mass of the
rider+bike. That moment will shift weight rear tire to front tire by
the moment/(cg distance behind the front contact patch). The lower you
are, the less forward shift in weight under braking. The further back
you are, the more weight on the rear to start with and you can
distribute the weight more evenly under decel to a point. Beyond that
point, you have too much weight on the rear and the front tire is
underutilized in braking.
The optimal point will vary on a given bike on level ground vs.
downhill (although your posted question specified level ground)l, and
since the traction varies from surface to surface, the optimal point
will vary as well. In fact, the optimal position is not a constant on a
given bike as you have to factor in the aero drag of the rider at say
30mph vs. 10 mph.
If you neglect the air drag, since it will be a minority component in
the calculation, the optimal position for a max G decel will be a
continuous series of points along a path going back and upwards,
although lower and further back than typical seat positions.
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?
--
Chalo
> Sheldon Brown <Capt...@sheldonbrown.com> wrote:
> > If the angle is shallower than the critical angle, locking up the front
> > wheel will cause it to skid. This is the case for tandems and many
> > recumbents.
>
> Have you ever skidded a front tire on clean dry pavement with a
> tandem? Or has anyone else?
>
> I have never been able to skid the front tire on my tandem on clean
> dry pavement despite practiing emergency stops to the point where I
> worried about the fork failing, though I haven't bent a fork or
> observed any obvious damage from this maneuver. I know Chalo's bent
> plenty of forks but I'd like to see him do that with my Santana.
The implication is that you *can* use even more braking power than
that before absolute limits apply. Tandems can often stop much harder
than singles, though, and shorter stops (than you can get already)
might not be very useful.
I face this issue sometimes, where the real limitation on my braking
is how much I can stomach the fork bending back down there. More
often, polluted braking surfaces or less-than-optimal brakes impose a
limit first.
I took my rain bike on my recent holiday vacation, where I discovered
that stopping on dry downhills steeper than about 10% causes me to max
out the drum brakes and call for divine intervention. I would not
mind if _those_ brakes taxed the fork's strength a little more!
Chalo Colina
JeffP
postion...
Right foot clipped in
Left foot out front lower than the front axel
No appreciable rear brake, the rear wheel about 1ft off the ground
Front brake locked
Rear of truck in front of me going from 45mph to zero, blue smoke as it starts
to jack-knife
Me encountering the rear liftgate at less than 1mph, no damage to me or the bike
Clearly my CofG had most of my weight pressing straight down on the front wheel,
my left foot lower than the axel provided additional downward force to
counteract a total flipover.
My skinny slick sew-ups provided the maximum friction and slip enabling me to
skid, stop and not flip over.
JeffP....
"wle" <w...@mailinator.com> wrote in message
news:60402c15.04010...@posting.google.com...
(Pete Cresswell)
>We skid our front wheels all the time on our icebikes ( studded
>tires).. lock up the front wheel and continue to pedal.. the rear
>wheel has more than enough traction on ice to move the bike forward
>with the studs scratching the ice while being locked up.. But it takes
>practice keeping the front wheel straight.
That sounds like good practice for surviving front wheel washout in regular
riding.
--
PeteCresswell
David Damerell
Bunk. I can lift the rear wheel from the hoods. How much more braking can
I have?
Rick Onanian
<tshe...@qconline.com> 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
Rick Onanian
wrote:
><tshe...@qconline.com> 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
Mike Beauchamp
"Braking".. my apologies.
"Ted Bennett" <tedbe...@earthlink.net> wrote in message
news:tedbennett-8DAEC...@news03.west.earthlink.net...
> > 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
ZeeExSixAre
"David Damerell" <dame...@chiark.greenend.org.uk> wrote in message
news:EKf*B7...@news.chiark.greenend.org.uk...
> ZeeExSixAre <philMuN...@hahtmail.com> 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
Tom Sherman
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/PracticalInnovations/index.html >
meb
in fishtailing under very hard braking. Is that the case?”
Tom Sherman responded: ”Fishtailing only happens on the recumbent
lowracer [1] when the rear wheel is locked up, and this takes
considerably more braking force than it does to lock up the rear wheel
on an upright.”
The last generation of street automobile pre-abs brakes were forward
biased so as to have the fronts locked first, brought the rear tires
into the lower sliding coefficient of friction relative their higher
static coefficient of friction before the fronts locked, and thereby
brake in a straight line. 1970’s And earlier brakes were more balanced
resulting in the rears locking first as weight was transferred forward
in braking. With those rears locked first, rear tires reached sliding
coefficient of friction before the fronts, so the vehicle fishtailed or
had the rear come around. In a most race cars, your taught if you get
the out of control sliding/spinning car sliding in a safe direction, you
lock the wheels hard and it vectors in a straight direction-all tires
are in sliding coefficient of friction.
Tom Sherman wrote: ”Stopping quickly takes as much skill as stopping an
automobile without ABS and the brakes biased towards the front.”
At least you didn’t identify as being as difficult as those pre-forward
biased non-ABS brakes.
--
meb
The op didn't mistate his problem-he dictated the problem scope and some
parameters, leaving some assumptions open.
Some subsequent posters mistated the issue and a solution to the
mistated issue with some misattribution in the quotes concerning who
made earlier quotes.
Sheldon pointed out the err of those tangential postings.
With a sufficiently high coefficient of friction a longitudinally
elongated object (whether it has locked wheels or no wheels) can still
endo even if its cg is very low, even a height proximate the contact
surface. If the moment about the lead contact point by the forward force
on the cg times the height above the contact point is more than the
moment about that same point of the weight of the object times the
longitdunal length of the distance the cg is behind the lead contact
point, the object rolls forward.
--
Sheldon Brown
>>
>>>>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
Nemo
"meb" <usenet...@cyclingforums.com> wrote in message
news:toqLb.59151$0Q2....@fe03.usenetserver.com...
>
> 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.
meb
Hence, the equal load on the front and rear tires, since optimal
traction occurs when both tires are equally loaded.
--
meb
deceleration rate to the tune of about .6g's.”
That’s in accord with the .57g figure the article I later cited
indicated as a strong braking level.
”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.”
I initially though the coefficient of friction might be that high, but
when I calculated rollovers at those levels, I though it might be off
since I remember skidding fronts on road bikes on occasion without doing
endos. So the presented tabulations were for lower decel rates assuming
the adhesion limits were close to the .6G. Apparently, we need plug in
higher decel rates into the optimal seat placement formula I derived for
more realistic seat placement on the braking specialized bike.
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.” Yep, that’s why I calculated seat placement spreading
load over both tires equally under decel.
“(I believe this has something to do with the thermal properties of
rubber.)” Nope, it has to do with the flexibility and distortion of the
rubber letting the tire embed itself into small irregular contours in
the pavement and then increasing tractions due the tires horizontal
abutting of those irregularities.
“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. “
I made bike and rider cg independent variables.
Since the OP was interested in the maximum braking position, I
calculated optimal seat position premised upon the even weight
distribution under decel, independent of whether deviations from that
position might be small or large. The non-linear coefficient of friction
not trivial with automobile, although I’m not sure whether the round top
profile of the bike tire reduces or enhances this effect relative the
flat top profile of a car tire or if the lighter weight of the bike tire
would place the function of coefficient of friction as a function of
weight in a flatter curve.
Obviously, positioning a seat for maximum braking means ignoring ideal
handling positioning, so this is a braking purpose only solution
equation, and except for the low cg recumbents, is so poorly weight
distributed outside of decel as to be a very poor handling bike. A road
bike optimized for a coefficient of friction of 1, will be doomed to
wheelies when seated and not braking.
Changing the decel rate to 1G, gives the following seat positions for
the previously posted bikes:
Road bike 77.11 inches behind patch Second lowracer (9” seat height)
35.38 behind patch (that puts us in the rear tire). LWB recumbent 53.28
“behind patch (that also puts us in the rear tire). MWB recumbent 56.14”
behind patch (above the rear, 2” back of axle, so no wheelies.) Carbon
Prone 61.11” behind patch.
--
Tom Sherman
> ...
> Changing the decel rate to 1G, gives the following seat positions for
> the previously posted bikes:
>
> Road bike 77.11 inches behind patch Second lowracer (9?seat height)
> 35.38 behind patch (that puts us in the rear tire)....
On my recumbent lowracer, the (estimated) combined CG is about 24"
behind the front tire contact patch.
Tom Schneider
wrote:
clipped
inch would have a CG at 16 to 18 inches high or 16-18 inches behind
the contact patch for 1 G braking. Practically, the front wheel can
do 100% of the braking without changing the CF much, using automobiles
(front or rear engine) as an example.
Tom Schneider
meb
The equation I derived had seperate bike and rider cg premised upon the
rider location being adjustable.
If the 24" coincides with the bike cg position, the optimal braking seat
position would shift rearwards.
Obviously, the designer of your lowracer was not focussing on best
braking as the ultimate design objective to sacrifice all other
attributes in hopes of achieving the optimal braking bike. :D
--
David Damerell
>"David Damerell" <dame...@chiark.greenend.org.uk> wrote in message
>>>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.
Er, if I can lift the rear wheel from the hoods in a controlled fashion,
but getting a free flying lesson is easier from flat bars, I think I would
describe the road bicycle's braking as more effective.
--
David Damerell <dame...@chiark.greenend.org.uk> flcl?
Benjamin Lewis
That's just because you've defined effective as "easily able to stop as
quickly as possible" rather than "easily able to brake so hard I can kill
myself." What on earth are you thinking?
--
Benjamin Lewis
Everything that can be invented has been invented.
-- Charles Duell, Director of U.S. Patent Office, 1899
Simon Brooke
> if you could position your center of gravity anywhere, to ensure
> maximum braking power, where would it be?
>
> obviously over the front wheel is no good, you would flip.
>
> there is a point, leaning either ahead of the front wheel, or
> behind the back wheel, that the opposite wheel is off the ground.
>
> clearly those are 2 limits, the answer must lie between them.
>
> if there were very little friction, it would hardly matter.
>
> assume a level road, brakes that can cause a skid no matter what.
>
> ok, so where do you sit?
When you brake, you decelerate; this causes the net vector of force on
each of the masses which make up the bike and rider to swing forward,
so it is equivalent to going down a hill. Essentially you need your
body weight as far aft and as low down as possible consistent with
(a) maintaining control of the bike - so your hands have to stay on
the grips and your feet on the pedals;
(b) not doing your posterior a severe mischief - so your arse has to
stay off the back tyre.
Generally this means sliding off the back of the saddle and lowering
yourself behind it which in my case at the limit equates to resting my
sternum on the sadde.
Note that, on a normal 26", 27" or 700c wheeled diamond frame bicycle
it is impossible on level ground for a normal rider to get their
weight so far aft as to tip the bicycle over backwards while
maintaining grip on the bars and feet on the pedals, still less when
braking or descending; so that's a non-issue.
> state assumptions, like coefficient of friction between tire and road,
> weight of bike and rider.
It really doesn't matter. On wet roots the coefficient is nearly
nil. On dry, course-grained sandstone it's effectively infinite. In
order to keep the wheels rolling (which you need to do to get most
efficient stopping power) you obviously need much less braking force
on the former than on the latter, but it still does no harm to get
your weight back in an 'oh-shit-I-need-to-stop-NOW' situation.
Of course in an 'oh-shit-this-one's-going-to-hurt' situation it often
makes sense to get off the bike... *before* it hits whatever's going
to stop it dead. Less damage to you and usually less damage to the
bike. Note that if you are in mid air at the time this may not be
relevent...
--
si...@jasmine.org.uk (Simon Brooke) http://www.jasmine.org.uk/~simon/
;; Want to know what SCO stands for?
;; http://ars.userfriendly.org/cartoons/?id=20030605
Douglas Landau
> > 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.
I don't think the wheel has to stop - so every nose wheelie is
preceeded by a wheel locking? A relative slowing would suffice,
presuming the necessary rotation got it's start somehow...
> ...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?
From rider movement. The same way that riders do nose wheelies, or
"stoppies" -- by throwing ones weight forward into the bars, or by
picking up the rear wheel by lifting one's feet -- except unintentionally
-- an accident of rapid deceleration.
Surface irregularities in the road could do it as well.
> 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.
Agreed. There has to be something else there to get it started.
Doug
Rick Onanian
<tshe...@qconline.com> wrote:
>The following website will be instructive to mechanically inclined
>people interested in trike construction details.
That is so cool, it makes me want to go and build one right now.
Unfortunately, there are any number of limiting factors, such as my
lack of equipment, supplies, knowledge, ability, money, and time.
I've got the enthusiasm, although with my short attention span I
don't know if it would last long enough to finish the project.
In
http://www.ihpva.org/com/PracticalInnovations/weld.html
under the section titled "Trike steering geometry", sub-title
"Ackerman steering compensation", steering is treated as if there
was a live axle with no differential for the front wheels. I think
such trikes are nearly all rear wheel drive, leaving no need for a
live axle; why would there be front wheel scrubbing?
Does your trike use Ackerman steering compensation?
>Tom Sherman - Quad Cities
--
Rick Onanian
Rick Onanian
Okay, but optimal _braking_ traction would have to take into account
how the load distribution changes under hard braking -- the front
tire ends up with [nearly] the complete load on an average upright
bike with acceptably good brakes, for example.
--
Rick Onanian
Rick Onanian
<si...@jasmine.org.uk> wrote:
>bike. Note that if you are in mid air at the time this may not be
>relevent...
Since when was relevance a relevant concern here?
--
Rick Onanian
Tom Sherman
> ... In
> http://www.ihpva.org/com/PracticalInnovations/weld.html
> under the section titled "Trike steering geometry", sub-title
> "Ackerman steering compensation", steering is treated as if there
> was a live axle with no differential for the front wheels. I think
> such trikes are nearly all rear wheel drive, leaving no need for a
> live axle; why would there be front wheel scrubbing?
>
> Does your trike use Ackerman steering compensation?
I am aware of only one tadpole [1] trike with front wheel drive (a low
production model made in Russia), so I would estimate that almost all
are rear wheel drive.
If a tadpole trike did not have Ackerman steering, there would be tire
scrub even though the wheels are free to rotate at different speeds.
Since the wheels are traversing arcs of different radii in a turn, the
inner wheel must be turned more sharply than the outer wheel for both
tires to have a zero slip angle [2].
The Ackerman compensation on my trike is visually obvious at large
steering angles.
[1] Two front wheels, one rear wheel
[2] Slip angle is the difference between the direction in which the tire
points and in which it travels.
Tom Sherman
> Obviously, the designer of your lowracer was not focussing on best
> braking as the ultimate design objective to sacrifice all other
> attributes in hopes of achieving the optimal braking bike. :D
Agreed. Moving the seat back farther would result in crank/front wheel
overlap (NOT a good thing) unless the crank bearing was about 18"
(46-cm) above seat level, which would be very bad ergonomically. (I had
a bike with the crank bearing about 13" (33-cm) above seat level, and I
would not want a difference greater than that.)
ZeeExSixAre
> quickly as possible" rather than "easily able to brake so hard I can kill
> myself." What on earth are you thinking?
I believe it's because he might not exhibit the kind of control that a
trials rider such as myself might have. Anyone can stop more quickly on
V-brakes with flat bars than a road bike with drop bars, with 10 minutes of
practice and instruction on body positioning.
--
Phil, Squid-in-Training
ZeeExSixAre
> believe. A panic snatch on my bikes will result in me going over the
> bars.
Try it sometime on a MTB with good V-brakes or disc brakes. Pull the brake
at the same time that you throw yourself back, extending your arms fully and
putting your butt just over the rear wheel. It's doable.
--
Phil, Squid-in-Training
meb
Exactly. The posting you quoted was my followup response to my
immediately preceding posting. The immediate preceding posting showed
that derivation of the optimal seat position (seat behind front contact
patch as a dependent variable, seat hieght as an independent variable)
with the weight transfer effect included for differing decel rates. I
had omited the relevance to the equal weight distribution from the same
post since I had earlier posted the fact tjhat the optmized seat
position will distribute the load equally over both tires under decel
not static speed conditions. I followed up the posting a few minutes
later since it was an earlier post a few days earlier, I thought it
better to reiiterate for clarity reason for equalizing the weight
distribution on each tire under decel since the earlier posting was so
disjoined in time and lineage to the posting of several days earlier.
--
meb
Ton Schneider wrote:
“CG is approximately at the navel I've read, so the lowracer with 9 inch
would have a CG at 16 to 18 inches high”
My independent variable RhcgvsS reflects that vertical difference in the
CG. My mention of the subsequently added offset: “Rider center of
gravity for back/forward leaning added later since it is a simple
negative offset insertable with the solution above.” Reflecting on some
bikes the rider is leaning forward or backwards.
I plugged into the equation a rider cg height of 5” vertical above the
seat and 4” horizontal behind the seat on both the 12” high and 9”
high examples.
That is a seat position 39 degrees back from vertical.
Tom further wrote: “or 16-18 inches behind the contact patch for 1 G
braking. Practically, the front wheel can do 100% of the braking without
changing the CF much, using automobiles”
The object of the OP was to find the max (or peak) braking performance
seat positon rather than a braking performance plateau edge.
Automobiles are a bad example for a pro plateau argument.
Rear engined cars outbrake midengined cars which outbreak front engined
cars. Ever notice how well the rear engined Porsches and Volkswagons
braked relative to similar sized and equipped cars of the same era. Or
compare the low tech ‘87/88 midengined Fiero with no ABS to a high tech
front engined ‘87/88 Corvette with ABS.
Handling optimization, is also related to that same nonlinearity on
coefficient of friction as a function of weight on tires as braking,
just left right instead. A front engined car generally handles poorer
than a rear car which in turn is poorer handling than a mid engine car
because the front/rear weight distribution is poorer on the front
engined car.
Many automobile oval racing bodies limit maximum static weight
distribution on the left side tires.
There is even an interesting story of one driver’s attempt to circumvent
this limitation. At a local NASCAR event in Bakersfield's Mesa Merin
Speedway, one ingenious driver developed a system with left-right frame
rails slightly higher on the left than the right. The tubes were filled
with mercury. When on the high 36 degree banks on a slow pace lap, the
driver opened the valve, the mercury flowed to the left, the valves were
closed locking in the weight on the left side of the car. After the
race, the valve was opened, and on level ground the mercury flowed back
to the right to comply with the weight distribution load requirement.
One race they spotted water on the track. The safety crew got out to
sweep away the water and found for some strange reason the water
wouldn’t sweep away. On further inspection, it wasn’t water but mercury.
Now you know why NASCAR banned closed framerails after the multimillion
dollar cleanup.:D
--
Benjamin Lewis
As has been continually pointed out, those of us on road bikes with drop
bars can easily brake hard enough to raise the rear wheel. You can't stop
more quickly than that -- at least not by braking harder.
--
Benjamin Lewis
Now is the time for all good men to come to.
-- Walt Kelly
(Pete Cresswell)
>As has been continually pointed out, those of us on road bikes with drop
>bars can easily brake hard enough to raise the rear wheel. You can't stop
>more quickly than that -- at least not by braking harder.
Even with somebody's butt hanging back over the rear axle?
--
PeteCresswell
David Damerell
>>That's just because you've defined effective as "easily able to stop as
>>quickly as possible" rather than "easily able to brake so hard I can kill
>>myself." What on earth are you thinking?
>I believe it's because he might not exhibit the kind of control that a
>trials rider such as myself might have.
As we know, road cycling in traffic rarely demands that one stop with any
urgency... wait, no...
>Anyone can stop more quickly on
>V-brakes with flat bars than a road bike with drop bars, with 10 minutes of
>practice and instruction on body positioning.
How can one stop faster than lifting the rear wheel?
--
David Damerell <dame...@chiark.greenend.org.uk> Distortion Field!
David Damerell
>>Dry tarmac coated with dust or gravel? Otherwise I find it hard to
>>believe. A panic snatch on my bikes will result in me going over the
>>bars.
>Try it sometime on a MTB with good V-brakes or disc brakes.
I think the issue here is tyres with poor traction, not brake design at
all. With proper road tyres you will have greater braking available.
David Damerell
Partly I suspect this serves to compensate for the larger CoG/contact
point/ground angle caused by a shorter wheelbase and more upright
position.
If we are to believe some of the posters in this thread the limiting
factor on MTBs is actually tyre traction, so they may not get as much
braking as a naive examination of the geometry would suggest.
Rick Onanian
<bcl...@cs.sfu.ca> wrote:
>philMuN...@hahtmail.com wrote:
>> trials rider such as myself might have. Anyone can stop more quickly on
>> V-brakes with flat bars than a road bike with drop bars, with 10 minutes
>> of practice and instruction on body positioning.
>
>As has been continually pointed out, those of us on road bikes with drop
>bars can easily brake hard enough to raise the rear wheel. You can't stop
>more quickly than that -- at least not by braking harder.
Right, but a mtb has different geometry than a road bike, allowing
you to transfer your CG further back (and lower), so the amount of
braking force available before raising the rear wheel is more. Given
the same pavement and rider (and speed, etc), a MTB with slicks can
probably be stopped in a shorter distance.
That all said, I can stop as short as I've ever wanted to on my road
bike.
--
Rick Onanian
Rick Onanian
<dame...@chiark.greenend.org.uk> wrote:
>>Anyone can stop more quickly on
>>V-brakes with flat bars than a road bike with drop bars, with 10 minutes of
>>practice and instruction on body positioning.
>
>How can one stop faster than lifting the rear wheel?
By lifting the rear wheel with different geometry and set-up
(handlebars, wheels, saddle height) which allows the rider to
achieve a lower, more rearward CG.
My MTB is longer than my road bike, I sit farther back, I can get
behind the saddle easier, and I can get my butt lower due to the
smaller wheels. The CG probably gets lower as a result of nose dive
in suspended MTBs, although FS models might have zero net effect if
the rear suspension lifts by as much as the front lowers.
--
Rick Onanian
Benjamin Lewis
Even if this is the case, it's not because the MTB has better brakes. You
can still do the same thing on a MTB equipped with cantilevers.
--
Benjamin Lewis
Accordion, n.:
A bagpipe with pleats.
Benjamin Lewis
You may be able to stop faster with different bike/rider geometry, but even
taking that into account, road bike brakes are powerful enough to lift the
rear wheel. Things may be different with tandems or recumbents; I have no
experience with these.
Rick Onanian
<bcl...@cs.sfu.ca> wrote:
>Rick Onanian wrote:
>>> philMuN...@hahtmail.com wrote:
>>>> trials rider such as myself might have. Anyone can stop more quickly
>>>> on V-brakes with flat bars than a road bike with drop bars, with 10
>>>> minutes of practice and instruction on body positioning.
>>
>
>can still do the same thing on a MTB equipped with cantilevers.
Was Phil implying anything about power available from any given type
of brakes? I thought he was just referring to the issue with the CG
on which I elaborated.
--
Rick Onanian
Benjamin Lewis
Maybe he wasn't, but then why mention V-brakes?
Jose Rizal
Once again we are privvy to Onanian's loopy version of reality.
Geometry of bike type plays a lot less importance than geometry of bike,
period. A cyclist can achieve the same shift in CG on a road bike as on
a mountain bike, because it depends on the measurements of the bike only
as related to the rider's body measurements. To make that as simple as
possible, the ability to shift one's behind further back depends on the
rider's reach to the handlebars.
> That all said, I can stop as short as I've ever wanted to on my road
> bike.
Thereby contradicting what he previously stated. Picture a dog chasing
its own tail...
Jose Rizal
> On 14 Jan 2004 15:14:08 +0000 (GMT), David Damerell
> <dame...@chiark.greenend.org.uk> wrote:
> >How can one stop faster than lifting the rear wheel?
>
> By lifting the rear wheel with different geometry and set-up
> (handlebars, wheels, saddle height) which allows the rider to
> achieve a lower, more rearward CG.
Gobbledygook nonsense.
> My MTB is longer than my road bike, I sit farther back, I can get
> behind the saddle easier,
Effects which are characteristic of the particular bike, not bike type.
> and I can get my butt lower due to the
> smaller wheels.
By a full half inch. Wow.
> The CG probably gets lower as a result of nose dive
> in suspended MTBs, although FS models might have zero net effect if
> the rear suspension lifts by as much as the front lowers.
More gobbledygook nonsense.
Nemo
> Try it sometime on a MTB with good V-brakes or disc brakes. Pull the
brake
> at the same time that you throw yourself back, extending your arms fully
and
> putting your butt just over the rear wheel. It's doable.
>
sudden stop is the fact that most MTB's have front suspension. If you jam on
the front brake before the forward weight shift has made it's way through
the springs and into the contact patch, available traction will be reduced.
Bikes with strong damping will be less effected by this.
Nemo
> compare the low tech '87/88 midengined Fiero with no ABS to a high tech
> front engined '87/88 Corvette with ABS.
All other things being equal: a skilled driver can stop a car faster without
ABS than with its "assistance".
Also at issue in the 'vette vs. Fiero braking match: I don't know the
numbers, but I'd bet the poor little Pontiac is quite a bit lighter than the
'vette.
A final random tid-bit: 50/50 front-to-rear weight balance can be achieved
in a front engine car by putting the gearbox/transaxle in the rear.
Examples: C5 and C6 Corvette and Porsche 924, 944, 928, and 968.
meb
> I'm getting a bit off topic.
Not much, it is relevant to the braking issues that this thread is
about.
“All other things being equal: a skilled driver can stop a car faster
without ABS than with its "assistance". No!
No driver can make the 50+ pedal corrections per second (and that was
using the actuator technology in use 14 years ago when I wrote abs
software as an engineer at General Motors). In 1993, Roger Penske made a
public rebuttal against competitor accusations his Indy cars were using
abs to pick up an unfair advantage.
Plus the abs system has the ability to correct each wheel slip
independently through varying coefficients of friction and even split
coefficient of friction surfaces (tires on differing surfaces) found in
contact with the tires.
Only on surfaces in which a locked wheel forward packs the surface
(snow, sand, loose gravel) in front of the tire can a non abs vehicle be
brought to a stop as quickly or quicker as an abs equipped vehicle. On
water or ice the abs stopping advantage gets overwhelming. In all cases,
the steering control issue favors abs.
The major motors sports organization have long recognized the
technological superiority of abs even with the most skilled of drivers,
and most have barred this competitive tool either on expense grounds or
due to taking a driver skill element out of competition.
”Also at issue in the 'vette vs. Fiero braking match: I don't know the
numbers, but I'd bet the poor little Pontiac is quite a bit lighter than
the 'vette.”
The Fierro was about half the weight, but also much narrower tires.
”A final random tid-bit: 50/50 front-to-rear weight balance can be
achieved in a front engine car by putting the gearbox/transaxle in the
rear. Examples: C5 and C6 Corvette and Porsche 924, 944, 928, and 968. “
Surprises me that those cars have a 50/50 balance since the engine is
still much bigger than the gearbox & transaxle, but I’ll accept your
report. That makes a nice handling collection of vehicles with
sophisticated suspension and low cg become a nicer handling vehicles.
The rearward weight movement helps under braking with further braking
optimization at the expense of handling achieveable if more rearward
weight bias were also achieved!
--
Mark Hickey
>My MTB is longer than my road bike, I sit farther back, I can get
>behind the saddle easier, and I can get my butt lower due to the
>smaller wheels.
I did some graphing, and estimate that relative to the front tire's
contact patch, the saddle is about 3" / 7cm closer on the road bike,
and the bars are about 2" / 5cm further forward.
OTOH, if I'm riding in the drops, my hands are probably starting out
about 4" / 10cm lower on the road bike. That's not enough to offset
the more rearward position on the MTB, but doesn't hurt.
>The CG probably gets lower as a result of nose dive
>in suspended MTBs, although FS models might have zero net effect if
>the rear suspension lifts by as much as the front lowers.
The front suspension can actually interfere with stopping power, since
it delays full weight transfer to the front tire... the more damping,
the longer this process will take (though the less the effect then
too). From that point on, I suspect the compliance gained with the
shock would help, especially on the kind of terrain you NEED a shock
for in the first place!
Mark Hickey
Habanero Cycles
http://www.habcycles.com
Home of the $695 ti frame
Nemo
>> "A final random tid-bit: 50/50 front-to-rear weight balance can be
>> achieved in a front engine car by putting the gearbox/transaxle in the
>> rear. Examples: C5 and C6 Corvette and Porsche 924, 944, 928, and 968. "
>
> Surprises me that those cars have a 50/50 balance since the engine is
> still much bigger than the gearbox & transaxle
Of the cars mentioned, the only ones that I know for a fact to have 50/50
wieght balanve are the Porsche 924, 944, and 968. All of these have a 2.0L
to 3.0L 4cly up front. Beyond the weight of the transaxle, the balance is
also acheived by pushing the passenger compartment back over the rear axle.
Another example of this is the freont-drive Miata: it has a 52/48 front-rear
balance.
David Damerell
I was thinking about this, and I think these gymnastics are not pertinent.
I can raise the back wheel with laden rear panniers; I suspect with a
heavy load there (whose own CoG is much lower than that of a hanging-over
arse) the overall CoG is comparable to that of an unladen MTB rider
performing this stunt.
Nemo
> achieved in a front engine car by putting the gearbox/transaxle in the
> rear. Examples: C5 and C6 Corvette and Porsche 924, 944, 928, and 968. "
>
> Surprises me that those cars have a 50/50 balance since the engine is
> still much bigger than the gearbox & transaxle, but I'll accept your
> report. That makes a nice handling collection of vehicles with
> sophisticated suspension and low cg become a nicer handling vehicles.
> The rearward weight movement helps under braking with further braking
> optimization at the expense of handling achieveable if more rearward
> weight bias were also achieved!
>
roughly the same weight distrobution and steady-state handling potential as
a similarly sized mid-engined car, it's high polar moment of inertia makes
it slower to react to yaw and pitch accelerations. This does compromise
absoulute performance potential, though it makes the vehicle ride smoother
and more stable when turning in quickly: If you overwork the steering wheel,
you are more likely to slide the front tires before the car has gained
enough yaw momentum. If, however you do manage to get the car to over-yaw
quickly, good luck catching it as that high PMI is now working against you.
To tie this back in to the thread: I have a 35lb MTB and a 20lb road bike. I
can stop both while holding the rear wheel about 1" off the ground. I find
that it is much easier to do this consistantly with my MTB. Most of the
reasons for this have already been discussed, so I'll not further belabor
them.
The factor pertinate to what I ahve already mentioned is PMI in pitch. My
roadbike has almost all of it's mass located in one spot, me. The mass is
more spread out with my MTB simply because the bike is heavier so it
off-sets my mass more. (I weigh about 130lbs, so the 15 difference is
significant.) Also, my mass and the mass of the bike is distrubuted further
from the front contact patch. This high PMI means that the rear wheel lifts
(and drops) less quickly, resulting in a bike that feels much less nervous.
Very closely related to this is that the further back CG is, the less the
angle between the CG,Contact patch, and the ground, changes as the rear
wheel starts to lift. This means that you only have to let up on the brakes
a little to get the rear wheel to start moving back towards the ground.
(sidenote: If at the end of a stop, you brake hard enough to stand the bike
completely on its nose and then set it back down after you stop, you can
shorten your stopping distance by about a foot. If you mess up this manuver,
you can shorten your nose by about an inch.)
Rick Onanian
<dame...@chiark.greenend.org.uk> wrote:
>I was thinking about this, and I think these gymnastics are not pertinent.
>I can raise the back wheel with laden rear panniers; I suspect with a
>heavy load there (whose own CoG is much lower than that of a hanging-over
>arse) the overall CoG is comparable to that of an unladen MTB rider
>performing this stunt.
The unladen MTB has less inertia for the brakes to stop, so given
that CoG, can probably stop in a shorter distance.
--
Rick Onanian
Benjamin Lewis
Even if this is so, this is clear evidence that this shorter distance is
not due to "better brakes", since *more* brake power is required, and
clearly available, in David's example.
--
Benjamin Lewis
A small, but vocal, contingent even argues that tin is superior, but they
are held by most to be the lunatic fringe of Foil Deflector Beanie science.
Rick Onanian
<bcl...@cs.sfu.ca> wrote:
>Rick Onanian wrote:
>> The unladen MTB has less inertia for the brakes to stop, so given
>> that CoG, can probably stop in a shorter distance.
>
>Even if this is so, this is clear evidence that this shorter distance is
>not due to "better brakes", since *more* brake power is required, and
>clearly available, in David's example.
No disagreements here. Sheesh, one guy accidentally gives more
information than was required, and everybody jumps to conclusions
about it's implications...
--
Rick Onanian
Jose Rizal
> I did some graphing, and estimate that relative to the front tire's
> contact patch, the saddle is about 3" / 7cm closer on the road bike,
> and the bars are about 2" / 5cm further forward.
But how did you find equivalency between your MTB and road bike
dimensions, especially with the dropbar and flatbar effect on reach?
> OTOH, if I'm riding in the drops, my hands are probably starting out
> about 4" / 10cm lower on the road bike. That's not enough to offset
> the more rearward position on the MTB, but doesn't hurt.
Your BB on the road bike is lower, however, and provided your seat-to-BB
length is the same as in the MTB, your CG will be a little lower on the
road bike.
> The front suspension can actually interfere with stopping power, since
> it delays full weight transfer to the front tire...
How is that interference? Full weight transfer to the front wheel isn't
necessary for the front wheel to make its contribution to braking.
Tom Sherman
> ...
> ?lso at issue in the 'vette vs. Fiero braking match: I don't know the
> numbers, but I'd bet the poor little Pontiac is quite a bit lighter than
> the 'vette.?
>
> The Fierro was about half the weight, but also much narrower tires....
The Corvette has additional mass from all the gold chains worn by the
driver.
Tom Sherman - Quad Cities
S o r n i
> The Corvette has additional mass from all the gold chains worn by the
> driver.
Ah, but isn't that more than offset by his* much-smaller-than-average
penis?!?
Bill "typing in stereo" S.
*assumption alert!
Ted Bennett
> The Corvette has additional mass from all the gold chains worn by the
> driver.
>
> Tom Sherman - Quad Cities
Q: What's the difference between a cactus and a Corvette?
A: A cactus has its pricks on the outside.
(Works for Camaros too.)
--
Ted Bennett
Portland OR
Mark Hickey
>Mark Hickey:
>
>> I did some graphing, and estimate that relative to the front tire's
>> contact patch, the saddle is about 3" / 7cm closer on the road bike,
>> and the bars are about 2" / 5cm further forward.
>
>But how did you find equivalency between your MTB and road bike
>dimensions, especially with the dropbar and flatbar effect on reach?
It's not difficult - just start at the front tire contact patch and
plot backward from there. I took into account a 9cm reach (typical)
for my bars.
>> OTOH, if I'm riding in the drops, my hands are probably starting out
>> about 4" / 10cm lower on the road bike. That's not enough to offset
>> the more rearward position on the MTB, but doesn't hurt.
>
>Your BB on the road bike is lower, however, and provided your seat-to-BB
>length is the same as in the MTB, your CG will be a little lower on the
>road bike.
It's a give and take - if I'm riding off-road, my saddle is a little
lower (relative to the BB), and in the end, with the shock preloaded,
my MTB BB is only about 1/2" lower. Under braking I suspect it's
about the same as the road bike ('til it endos, that is...). ;-)
>> The front suspension can actually interfere with stopping power, since
>> it delays full weight transfer to the front tire...
>
>How is that interference? Full weight transfer to the front wheel isn't
>necessary for the front wheel to make its contribution to braking.
More weight on the tire allows it to provide more braking force.
Think about it - the rear wheel doesn't provide much braking in a stop
for the same reason - not enough weight on it. While the front shock
is compressing, the rear wheel will have a bit more weight on it which
would help make up some of the difference, but still the end result is
that there will be a very brief period where a front shock will delay
the onset of "full braking".
meb
> Omitted assumtion: the tires are at or near the limits of adhesion.
> Hence, the equal load on the front and rear tires, since optimal
> traction occurs when both tires are equally loaded.
Derivation: Ls= horizontal location of seat behind front axle Hs=height
of seat above ground RhcgvsS= Rider Center of Gravity relative seat
G=deceleration in G’s Wrd= Weight of rider + seat Wbk = Weight of bike
-seat Wsrear= static weight on rear tire Wsf=static weight on front tire
RotTorq= rotational torque about the front contact patch Hbcg= height of
bicycle center of gravity above road Lbcg= length of bicycle center of
gravity behind front contact patch Wdf=dynamic weight on front tire
RotTorq = G ((Hbcg X Wbk) + ((Hs +RhcgvsS )* Wrd))
Static Wsrear= ((Wbk X Lbcg) + (Wrd* Ls)))/ ((Wheelbase))
Static Wsf = (Wrd+Wbk)- Wsrear
Wdf=Wsf+ RotTorq/wheelbase= (Wrd +Wbk)-Wsrear +RotTorq/wheelbase =
(Wrd +Wbk)- (((Wbk * Lbcg) + (Wrd* Ls))/ (Wheelbase)) + G ((Hbcg X Wbk)
+ ((Hs +RhcgvsS )* Wrd))/wheelbase)
RotTorq + Wdf = Wsrear-RotTorq Wdf=Wsrear-2RotTorq= (Wrd+wbk)/2
Wdf=(Wrd+wb)/2
(Wrd+wbk)/2 = (Wrd +Wbk)- (((Wbk X Lbcg) + (Wrd* Ls))/ ((Wheelbase))) +
G ((Hbcg X Wbk) + ((Hs +RhcgvsS )* Wrd))/wheelbase
(Wrd+wbk) = 2(Wrd +Wbk)-2 (((Wbk X Lbcg) + (Wrd* Ls))/ ((Wheelbase) )) +
2G ((Hbcg X Wbk) + 2((Hs +RhcgvsS )* Wrd))/wheelbase
= (Wrd +Wbk)-2 (((Wbk X Lbcg) + (Wrd* Ls))/ ((Wheelbase))) + 2G ((Hbcg X
Wbk) + 2((Hs +RhcgvsS )* Wrd))/wheelbase
= (Wrd +Wbk)-2 (((Wbk X Lbcg) + (Wrd* Ls))/ ((Wheelbase) )) + 2G (Hbcg X
Wbk) /wheelbase
+ 2G((Hs +RhcgvsS )* Wrd)/wheelbase 2*(Wrd* Ls))/ ((Wheelbase) = (Wrd
+Wbk)-2 (((Wbk X Lbcg))/ ((Wheelbase))) + 2G (Hbcg X Wbk) /wheelbase +
2G((Hs +RhcgvsS )* Wrd)/wheelbase
(Wrd* Ls) = .5* (Wrd +Wbk)wheelbase- ((Wbk X Lbcg)) + G (Hbcg X Wbk) +
G((Hs +RhcgvsS )* Wrd)
Ls= .5*( (Wrd +Wbk)wheelbase/(Wrd))- ((Wbk X Lbcg)/Wrd) + G (Hbcg X
Wbk)/Wrd + G((Hs +RhcgvsS ))
--
meb
> Omitted assumtion: the tires are at or near the limits of adhesion.
> Hence, the equal load on the front and rear tires, since optimal
> traction occurs when both tires are equally loaded.
Calculated examples:
All numbers lbs or inches or G’s
26"cruiser wheelbase 46 rider wt 180 bike wt 30 bike cg ht 15 bike cg
trl 28 decel 0.6 seat ht 42 rider cg over seat rider cg over seat 8 cg
after seat 0 optimal seat positon behind lead patch: 53.66667 wheelbase
46 rider wt 180 bike wt 30 bike cg ht 15 bike cg trl 28 decel 0.6 seat
ht 40 rider cg over seat rider cg over seat 8 cg after seat 0 optimal
seat positon behind lead patch: 52.46667
CARBON prone wheelbase 38 rider wt 180 bike wt 20 bike cg ht 16 bike cg
trl 25 decel 0.6 seat ht 30 rider cg over seat rider cg over seat 3 cg
after seat -8 optimal seat positon behind lead patch: 47.2
roadbike wheelbase 48 rider wt 180 bike wt 20 bike cg ht 16 bike cg trl
30 decel 0.6 seat ht 42 rider cg over seat rider cg over seat 5 cg after
seat -5 optimal seat positon behind lead patch: 57.6
recumbent mwb/clwb wheelbase 54 rider wt 230 bike wt 35 bike cg ht 24
bike cg trl 40 decel 0.6 seat ht 24 rider cg over seat rider cg over
seat 6 cg after seat 3 optimal seat positon behind lead patch: 42.21304
wheelbase 54 rider wt 180 bike wt 35 bike cg ht 24 bike cg trl 40 decel
0.6 seat ht 24 rider cg over seat rider cg over seat 6 cg after seat 3
optimal seat positon behind lead patch: 42.27222
skycycle wheelbase 38 rider wt 180 bike wt 50 bike cg ht 25 bike cg trl
21 decel 0.4 seat ht 80 rider cg over seat rider cg over seat 8 cg after
seat 0 optimal seat positon behind lead patch: 56.42222 wheelbase 38
rider wt 180 bike wt 50 bike cg ht 25 bike cg trl 21 decel 0.6 seat ht
80 rider cg over seat rider cg over seat 8 cg after seat 0 optimal seat
positon behind lead patch: 75.41111
lowracer wheelbase 48 rider wt 180 bike wt 25 bike cg ht 18 bike cg trl
32 decel 0.6 seat ht 12 rider cg over seat rider cg over seat 5 cg after
seat 4 optimal seat positon behind lead patch: 30.58889 wheelbase 48
rider wt 180 bike wt 25 bike cg ht 18 bike cg trl 32 decel 0.7 seat ht 9
rider cg over seat rider cg over seat 5 cg after seat 4 optimal seat
positon behind lead patch: 30.43889
Unicycle wheelbase 0 rider wt 180 bike wt 30 bike cg ht 20 bike cg trl 0
decel 0.5 seat ht 34 rider cg over seat rider cg over seat 8 cg after
seat 0 optimal seat positon behind lead patch: 22.66667 wheelbase 0
rider wt 180 bike wt 30 bike cg ht 20 bike cg trl 0 decel 0.4 seat ht 34
rider cg over seat rider cg over seat 8 cg after seat 0 optimal seat
positon behind lead patch: 18.13333
LWB recumbent wheelbase 63 rider wt 180 bike wt 40 bike cg ht 19 bike cg
trl 47 decel 0.6 seat ht 18 rider cg over seat rider cg over seat 6 cg
after seat 3 optimal seat positon behind lead patch: 41.98889
You might be able to put drag motorcycle anti-wheelie bars behind the
skycycle to go with the obvious rider weight well back of the wheelbase
on this .6 G decell skycycle to keep it from going over before decel!
This is a repost of a 1/10/04 post I made to this thread that appears to
have been lost on some servers, possibly a result of a variable name
triggering a spam filter. I have substituted another name for that
variable so if that was the issue, we are now unfilterred.
Rick O., this is the immediate preceding derivation I referred to
responding to one of your posts that may have been invisible to you in
which I reitterated an earlier tire load balancing desire.
--
David Damerell
><dame...@chiark.greenend.org.uk> wrote:
>>I can raise the back wheel with laden rear panniers; I suspect with a
>>heavy load there (whose own CoG is much lower than that of a hanging-over
>>arse) the overall CoG is comparable to that of an unladen MTB rider
>>performing this stunt.
>The unladen MTB has less inertia for the brakes to stop, so given
>that CoG, can probably stop in a shorter distance.
No, that's not correct. The maximum braking deceleration is a function of
the CoG position, not the maximum braking force or amount of work done.
--
David Damerell <dame...@chiark.greenend.org.uk> Kill the tomato!
Rick Onanian
<dame...@chiark.greenend.org.uk> wrote:
>Rick Onanian <spam...@cox.net> wrote:
>>The unladen MTB has less inertia for the brakes to stop, so given
>>that CoG, can probably stop in a shorter distance.
>
>No, that's not correct. The maximum braking deceleration is a function of
>the CoG position, not the maximum braking force or amount of work done.
So, no matter how much you load a bike, if it's got the same CoG
then it can stop in the same distance? How far can you stretch that
effect?
I'll have to try some experiments if the temperature ever makes it
above freezing. In the meantime, I'll work on that extra front wheel
for my MTB, which will be mounted with a semi-slick.
--
Rick Onanian