Bike stability physics

On Jul 24, 3:32*pm, Phil W Lee wrote:
"T°m Sherm@n" " considered
Sun, 24 Jul 2011 05:31:31 -0500 the perfect time to write:



On 7/23/2011 10:20 AM, Joe Riel wrote:
"T°m " *writes:

On 7/22/2011 1:29 PM, thirty-six aka Trevor Jeffrey wrote:
On Jul 22, 4:13 pm, Joe * wrote:
* writes:
On Jul 22, 2:39 pm, Peter * wrote:

Say a cyclist is approaching an intersection at 18 mph and is confronted
by a "left hook" hazard. Should he brake, or attempt an "instant turn"
(to the right)?

Can turn within 7feet. *Braking takes another 4feet (typically).

Minimum turning radius at 8m/s is 6.4m at 1g (v^2/g).
Minimum stopping distance at 1g is 3.2m (v^2/2/g), but for an upright the
typical max deceleration is more like 0.7 g, so the minimum stopping
distance is 4.6m. *Peter's estimates are the ones to use.

--
Joe Riel

I'm not looking at a math's exercise, but 3.2m is 10'8" , I said
11feet was the practical expected limit. *It's obvious then that I can
reproducibly generate 1g deceleration under braking. * Cornering is
tighter, although on reflection is over 7feet because I was
considering the inside of the turn as determined by the end of the
handlebar. *I don't know how to explain it completely but there is a
delay when braking to get the hands over the levers, and this delay is
eliminated. *Perhaps it just that the rubber is optimised to provide
the most grip through interlocking when cornering so can deliver
cornering forces in excess of 1g

Ok, so it's probably around 10feet radius, taken at the wheel, at
18mph. *A little shorter than that acheived by straight line braking.
The technique though does take out of the equation the possibility of
a less than optimum braking system or a snapping cable(which
unfortunately just sometimes cannot be prevented). *And, getting a
bike side on into a slide is no big deal, I was controlling those
slides before the age of ten, a bike on the smaller size is preferable
to ease this technique.

The planet Trevor lives on is an amazing place. *We wish it was
possible to visit.

* *v^2 = g*tan(theta)*r
* * *r = R - L*sin(theta)

with

* *v = velocity of CG (18 mph = 8 m/s)
* *g = gravita accel (9.8 m/s^2)
* *R = radius of turn at the tire patch (10 ft = 3.05 m)
* *r = turning radius at CG
* *L = height of CG above road with bike vertical (~ 1.1 m)
* *theta = lean from vertical

Solving for theta gives 73 degrees (from vertical, not horizontal)
with a lateral acceleration of 3.3 g. *Very impressive, indeed.

Maybe Trevor has one of those huge "barn door [1]" spring car wings on
his bike for generating down-force?

[1] E.g. http://padirtreport.com/images/Susquehanna_4-13-06_1.jpg.

I was thinking more of a pram handle:

http://www.carspotting.com/s/sport-c...mpreza-WRX-200...

Well spoiled.

On Jul 23, 8:02 pm, Frank Krygowski wrote:
On Jul 23, 2:35 pm, Peter Cole wrote:

On 7/23/2011 1:22 PM, Frank Krygowski wrote:

On Jul 23, 12:47 pm, Peter wrote:

Your use of the term "high lateral acceleration" is confusing...

Ah well. Don't feel bad. Mechanical physics isn't everyone's cup of
tea. There are probably other areas in which you do better.

- Frank Krygowski

You really can't resist, can you?

There was a weirdly obnoxious little kid in our grade school.
Occasionally he would nag, harass, taunt and annoy kids much bigger
than him. He'd keep it up, on and on, until the bigger kid would
explode. He wouldn't let up until that happened. Occasionally, the
bigger kid might even smack him one.

It's nearly 60 years later, but I still remember him sitting on the
ground with a bloody lip, smirking as if he'd won something. I never
understood that.


If this is supposed to be analogous to something, you've lost me.

You've got issues, my friend - but you can't seem to see them. I
guess to be that way at this point in your life they must be soundly
ingrained - complete with mechanisms to rationalize them. But, never
say never. I wish you well - even though you are often not very nice
to me.

On Jul 24, 10:57*am, thirty-six wrote:
On Jul 24, 3:32*pm, Phil W Lee wrote:



"T°m Sherm@n" " considered
Sun, 24 Jul 2011 05:31:31 -0500 the perfect time to write:

On 7/23/2011 10:20 AM, Joe Riel wrote:
"T°m " *writes:

On 7/22/2011 1:29 PM, thirty-six aka Trevor Jeffrey wrote:
On Jul 22, 4:13 pm, Joe * wrote:
* writes:
On Jul 22, 2:39 pm, Peter * wrote:

Say a cyclist is approaching an intersection at 18 mph and is confronted
by a "left hook" hazard. Should he brake, or attempt an "instant turn"
(to the right)?

Can turn within 7feet. *Braking takes another 4feet (typically).

Minimum turning radius at 8m/s is 6.4m at 1g (v^2/g).
Minimum stopping distance at 1g is 3.2m (v^2/2/g), but for an upright the
typical max deceleration is more like 0.7 g, so the minimum stopping
distance is 4.6m. *Peter's estimates are the ones to use.

--
Joe Riel

I'm not looking at a math's exercise, but 3.2m is 10'8" , I said
11feet was the practical expected limit. *It's obvious then that I can
reproducibly generate 1g deceleration under braking. * Cornering is
tighter, although on reflection is over 7feet because I was
considering the inside of the turn as determined by the end of the
handlebar. *I don't know how to explain it completely but there is a
delay when braking to get the hands over the levers, and this delay is
eliminated. *Perhaps it just that the rubber is optimised to provide
the most grip through interlocking when cornering so can deliver
cornering forces in excess of 1g

Ok, so it's probably around 10feet radius, taken at the wheel, at
18mph. *A little shorter than that acheived by straight line braking.
The technique though does take out of the equation the possibility of
a less than optimum braking system or a snapping cable(which
unfortunately just sometimes cannot be prevented). *And, getting a
bike side on into a slide is no big deal, I was controlling those
slides before the age of ten, a bike on the smaller size is preferable
to ease this technique.

The planet Trevor lives on is an amazing place. *We wish it was
possible to visit.

* *v^2 = g*tan(theta)*r
* * *r = R - L*sin(theta)

with

* *v = velocity of CG (18 mph = 8 m/s)
* *g = gravita accel (9.8 m/s^2)
* *R = radius of turn at the tire patch (10 ft = 3.05 m)
* *r = turning radius at CG
* *L = height of CG above road with bike vertical (~ 1.1 m)
* *theta = lean from vertical

Solving for theta gives 73 degrees (from vertical, not horizontal)
with a lateral acceleration of 3.3 g. *Very impressive, indeed.

Maybe Trevor has one of those huge "barn door [1]" spring car wings on
his bike for generating down-force?

[1] E.g. http://padirtreport.com/images/Susquehanna_4-13-06_1.jpg.

I was thinking more of a pram handle:

http://www.carspotting.com/s/sport-c...mpreza-WRX-200...

Well spoiled.

Do we need "SPOILER ALERT" in this thread title, too?

- Frank Krygowski

On Jul 23, 8:02 pm, Frank Krygowski wrote:
On Jul 23, 2:35 pm, Peter Cole wrote:

On 7/23/2011 1:22 PM, Frank Krygowski wrote:

On Jul 23, 12:47 pm, Peter wrote:

Your use of the term "high lateral acceleration" is confusing...

Ah well. Don't feel bad. Mechanical physics isn't everyone's cup of
tea. There are probably other areas in which you do better.

- Frank Krygowski

You really can't resist, can you?

There was a weirdly obnoxious little kid in our grade school.
Occasionally he would nag, harass, taunt and annoy kids much bigger
than him. He'd keep it up, on and on, until the bigger kid would
explode. He wouldn't let up until that happened. Occasionally, the
bigger kid might even smack him one.

It's nearly 60 years later, but I still remember him sitting on the
ground with a bloody lip, smirking as if he'd won something. I never
understood that.


If this is supposed to be analogous to something, you've lost me.

You've got issues, my friend - but you can't even see them. To be
that way at this point in your life they must be deeply ingrained -
complete with rationalization mechanisms. But, never say never. I
wish you well - even though you have often been not very nice to me.

On Sun, 24 Jul 2011 10:42:47 -0700 (PDT), thirty-six
wrote:

On Jul 24, 6:25*pm, "T°m Sherm@n" ""twshermanREMOVE\"@THI
$southslope.net" wrote:
On 7/24/2011 8:41 AM, thirty-six aka Trevor Jeffrey wrote:



On Jul 24, 11:31 am, "T°m Sherm@n"""twshermanREMOVE\"@THI
$southslope.net" *wrote:
On 7/23/2011 10:20 AM, Joe Riel wrote:

"T°m " * *writes:

On 7/22/2011 1:29 PM, thirty-six aka Trevor Jeffrey wrote:
On Jul 22, 4:13 pm, Joe * * wrote:
* * writes:
On Jul 22, 2:39 pm, Peter * * wrote:

Say a cyclist is approaching an intersection at 18 mph and is confronted
by a "left hook" hazard. Should he brake, or attempt an "instant turn"
(to the right)?

Can turn within 7feet. *Braking takes another 4feet (typically).

Minimum turning radius at 8m/s is 6.4m at 1g (v^2/g).
Minimum stopping distance at 1g is 3.2m (v^2/2/g), but for an upright the
typical max deceleration is more like 0.7 g, so the minimum stopping
distance is 4.6m. *Peter's estimates are the ones to use.

--
Joe Riel

I'm not looking at a math's exercise, but 3.2m is 10'8" , I said
11feet was the practical expected limit. *It's obvious then that I can
reproducibly generate 1g deceleration under braking. * Cornering is
tighter, although on reflection is over 7feet because I was
considering the inside of the turn as determined by the end of the
handlebar. *I don't know how to explain it completely but there is a
delay when braking to get the hands over the levers, and this delay is
eliminated. *Perhaps it just that the rubber is optimised to provide
the most grip through interlocking when cornering so can deliver
cornering forces in excess of 1g

Ok, so it's probably around 10feet radius, taken at the wheel, at
18mph. *A little shorter than that acheived by straight line braking.
The technique though does take out of the equation the possibility of
a less than optimum braking system or a snapping cable(which
unfortunately just sometimes cannot be prevented). *And, getting a
bike side on into a slide is no big deal, I was controlling those
slides before the age of ten, a bike on the smaller size is preferable
to ease this technique.

The planet Trevor lives on is an amazing place. *We wish it was
possible to visit.

* * v^2 = g*tan(theta)*r
* * * r = R - L*sin(theta)

with

* * v = velocity of CG (18 mph = 8 m/s)
* * g = gravita accel (9.8 m/s^2)
* * R = radius of turn at the tire patch (10 ft = 3.05 m)
* * r = turning radius at CG
* * L = height of CG above road with bike vertical (~ 1.1 m)
* * theta = lean from vertical

Solving for theta gives 73 degrees (from vertical, not horizontal)
with a lateral acceleration of 3.3 g. *Very impressive, indeed.

Maybe Trevor has one of those huge "barn door [1]" spring car wings on
his bike for generating down-force?

[1] E.g.http://padirtreport.com/images/Susquehanna_4-13-06_1.jpg.

--
Tºm Shermªn - 42.435731°N, 83.985007°W
I am a vehicular cyclist.

No, but I do throw my inner knee out, get off the saddle weighting the
outer foot (getting the saddle to the outside of my backside), Lift
the bike further, letting go of the outer end of the h'bar and twist
in to the corner as the bike dives. *Each contributes to harder
cornering. *The body twist thing is reserved for emergency turns and
for tightening up quickly a miscalculated bend. *The twist is most
helpful on adverse camber after crowning the road a bit early.

All that will not change the fact that the maximum force that can be
generated at the contact patch is a function of normal force, contact
patch area, and coefficient of friction between the tire and pavement
(USian usage of pavement).

In our Universe, at least.

--
Tºm Shermªn - 42.435731°N, 83.985007°W
I am a vehicular cyclist.

No, it depends on how it is applied, particularly on greasy roads.
Ride like a strap on anvil and apply sudden steering and you are
creating unecessary stress levels. Buckling the bike to rider
interface allows not only for a greater shock capability but also
greater cornering ability as the rider compresses down during the
sharpest part of the turn.


But as Sherman said, the ultimate turn is still governed by the size
of the tire patch, force applied and coefficient of friction for the
particular tire/pavement combination.

--
John B. Slocomb
(johnbslocombatgmaildotcom)

On Jul 25, 1:34*pm, John B. Slocomb wrote:
On Sun, 24 Jul 2011 10:42:47 -0700 (PDT), thirty-six



wrote:
On Jul 24, 6:25*pm, "T°m Sherm@n" ""twshermanREMOVE\"@THI
$southslope.net" wrote:
On 7/24/2011 8:41 AM, thirty-six aka Trevor Jeffrey wrote:

On Jul 24, 11:31 am, "T°m Sherm@n"""twshermanREMOVE\"@THI
$southslope.net" *wrote:
On 7/23/2011 10:20 AM, Joe Riel wrote:

"T°m " * *writes:

On 7/22/2011 1:29 PM, thirty-six aka Trevor Jeffrey wrote:
On Jul 22, 4:13 pm, Joe * * wrote:
* * writes:
On Jul 22, 2:39 pm, Peter * * wrote:

Say a cyclist is approaching an intersection at 18 mph and is confronted
by a "left hook" hazard. Should he brake, or attempt an "instant turn"
(to the right)?

Can turn within 7feet. *Braking takes another 4feet (typically).

Minimum turning radius at 8m/s is 6.4m at 1g (v^2/g).
Minimum stopping distance at 1g is 3.2m (v^2/2/g), but for an upright the
typical max deceleration is more like 0.7 g, so the minimum stopping
distance is 4.6m. *Peter's estimates are the ones to use.

--
Joe Riel

I'm not looking at a math's exercise, but 3.2m is 10'8" , I said
11feet was the practical expected limit. *It's obvious then that I can
reproducibly generate 1g deceleration under braking. * Cornering is
tighter, although on reflection is over 7feet because I was
considering the inside of the turn as determined by the end of the
handlebar. *I don't know how to explain it completely but there is a
delay when braking to get the hands over the levers, and this delay is
eliminated. *Perhaps it just that the rubber is optimised to provide
the most grip through interlocking when cornering so can deliver
cornering forces in excess of 1g

Ok, so it's probably around 10feet radius, taken at the wheel, at
18mph. *A little shorter than that acheived by straight line braking.
The technique though does take out of the equation the possibility of
a less than optimum braking system or a snapping cable(which
unfortunately just sometimes cannot be prevented). *And, getting a
bike side on into a slide is no big deal, I was controlling those
slides before the age of ten, a bike on the smaller size is preferable
to ease this technique.

The planet Trevor lives on is an amazing place. *We wish it was
possible to visit.

* * v^2 = g*tan(theta)*r
* * * r = R - L*sin(theta)

with

* * v = velocity of CG (18 mph = 8 m/s)
* * g = gravita accel (9.8 m/s^2)
* * R = radius of turn at the tire patch (10 ft = 3.05 m)
* * r = turning radius at CG
* * L = height of CG above road with bike vertical (~ 1.1 m)
* * theta = lean from vertical

Solving for theta gives 73 degrees (from vertical, not horizontal)
with a lateral acceleration of 3.3 g. *Very impressive, indeed.

Maybe Trevor has one of those huge "barn door [1]" spring car wings on
his bike for generating down-force?

[1] E.g.http://padirtreport.com/images/Susquehanna_4-13-06_1.jpg.

--
Tºm Shermªn - 42.435731°N, 83.985007°W
I am a vehicular cyclist.

No, but I do throw my inner knee out, get off the saddle weighting the
outer foot (getting the saddle to the outside of my backside), Lift
the bike further, letting go of the outer end of the h'bar and twist
in to the corner as the bike dives. *Each contributes to harder
cornering. *The body twist thing is reserved for emergency turns and
for tightening up quickly a miscalculated bend. *The twist is most
helpful on adverse camber after crowning the road a bit early.

All that will not change the fact that the maximum force that can be
generated at the contact patch is a function of normal force, contact
patch area, and coefficient of friction between the tire and pavement
(USian usage of pavement).

In our Universe, at least.

--
Tºm Shermªn - 42.435731°N, 83.985007°W
I am a vehicular cyclist.

No, it depends on how it is applied, particularly on greasy roads.
Ride like a strap on anvil and apply sudden steering and you are
creating unecessary stress levels. *Buckling the bike to rider
interface allows not only for a greater shock capability but also
greater cornering ability as the rider compresses down during the
sharpest part of the turn.

But as Sherman said, the ultimate turn is still governed by the size
of the tire patch, force applied and coefficient of friction for the
particular tire/pavement combination.

--
John B. Slocomb
(johnbslocombatgmaildotcom)

I suppose there is rotational inertia to assist the turn. I don't
know why it all works together but it does. When I do everything I
know to assist the turn, I suprise myself how tight the turn is,
despite repating the actions perhaps 200 times. That final twist
seems to increase the available grip. It can be used at any time
there is a hint of slipping in the corners and it gets everything back
on track.

On 7/25/2011 8:34 AM, John B. Slocomb wrote:
On Sun, 24 Jul 2011 10:42:47 -0700 (PDT), thirty-six
wrote:

On Jul 24, 6:25 pm, "T°m Sherm@n"""twshermanREMOVE\"@THI
$southslope.net" wrote:
On 7/24/2011 8:41 AM, thirty-six aka Trevor Jeffrey wrote:

No, but I do throw my inner knee out, get off the saddle weighting the
outer foot (getting the saddle to the outside of my backside), Lift
the bike further, letting go of the outer end of the h'bar and twist
in to the corner as the bike dives. Each contributes to harder
cornering. The body twist thing is reserved for emergency turns and
for tightening up quickly a miscalculated bend. The twist is most
helpful on adverse camber after crowning the road a bit early.

All that will not change the fact that the maximum force that can be
generated at the contact patch is a function of normal force, contact
patch area, and coefficient of friction between the tire and pavement
(USian usage of pavement).

No, it depends on how it is applied, particularly on greasy roads.
Ride like a strap on anvil and apply sudden steering and you are
creating unecessary stress levels. Buckling the bike to rider
interface allows not only for a greater shock capability but also
greater cornering ability as the rider compresses down during the
sharpest part of the turn.


But as Sherman said, the ultimate turn is still governed by the size
of the tire patch, force applied and coefficient of friction for the
particular tire/pavement combination.

All that is true, but general enough to not be helpful.

For a given speed, the maximum turn rate is determined by the lean
angle. As the radius of turn decreases, the magnitude of the combined
force at the contact patch increases while the direction moves away from
vertical towards horizontal, eventually slip is inevitable. The force
vector direction and magnitude are revealed by the lean angle.

As far as technique goes, the rider can shift weight in such a way that
the bike is more upright and the rider less. That may affect tire grip,
but there are trade-offs, especially on rough surfaces. Since the effect
of bumps is to modulate the vertical force at the contact patch,
absorbing bumps with the legs is critical to avoiding slip out.

For constant speed, the most critical thing is to keep a uniform radius
of curvature, there is a best line through a curve. In the real world,
there's often a combination of braking and turning, so both picking the
line and timing the braking are critical. Judging the approaching camber
and surface conditions is also very important in variable conditions. If
you're "cutting corners" through turns (using the entire road width) and
the road is crowned, you're going through camber reversals. Gentle
undulations do modulate force in the vertical direction, so I suppose
there might be a way to exploit that via the line you pick, but that's a
bit beyond my skill, at least on a conscious level, who knows what's
going on subconsciously.

I have seen riders "pump" their bikes vertically, timed with the pedal
stroke, to climb very steep loose surfaces in off-road situations. I
guess the physics works, the technique seemed to.

While you've got to keep gravity and centripetal forces balanced to keep
the bike balanced (not fall over), there are inertias involved and
various couplings and geometries/moments that can be controlled to some
degree by the rider, the effects often being subtle and not intuitive.
The tight rope walker example being given by Dave Lehnen earlier being
an example. The models in physics cited usually make simplifying
assumptions, the less you use those, the more complicated, and accurate,
the model gets.