Angular acceleration and rolling resistance

I'm trying to understand the resulting torques that act back on the
axle. For a free rolling wheel travelling purely in the longitudinal
direction and with a slip ratio > 0, it will output a force of say
100N. As well as pushing the vehicle forward it also creates a torque
on the axle which is then used to calculate the angular acceleration.
The forward velocity also creates a rolling resistance moment as a
function of vertical load and forward rolling speed. There is also a
moment produced by the internal bearing friction trying to slow it
down.

I understand this, however if the vehicle were to rotate
instantaneously such that the the longitudinal velocity = 0, the
longitudinal slip ratio = 0, the angular velocity > 0, and the lateral
velocity was 10mts/s and hence the lateral slip ratio > 0, the tyre is
not outputting any force in the longitudinal direction thus meaning
the only forces acting against the angular velocity of the wheel are
the internal bearing friction and rolling resistance moment. However
the rolling resistance moment is a function of forward rolling speed
which if equal to 0 there is no moment produced. Thus in the
simulation if travelling purely in a lateral direction in the tyre's
frame of reference, the only force acting against the angular velocity
of the wheel is internal bearing friction, however this is wrong as it
is not the only force acting on it. The wheel would only slow down as
a result of the bearing friction which is not correct There must be
another additional force derived from lateral slip that adds to the
rolling resistance moment.

In real life if a vehicle were to instantaneously rotate and travel in
a purely lateral direction, the wheels would stop rolling from the
lateral slip. I'm trying to work out what that force is. I hope this
post is clear. Any help is much appreciated!

"Zach" > wrote in message
...
> I'm trying to understand the resulting torques that act back on the
> axle. For a free rolling wheel travelling purely in the longitudinal
> direction and with a slip ratio > 0, it will output a force of say
> 100N. As well as pushing the vehicle forward it also creates a torque
> on the axle which is then used to calculate the angular acceleration.
> The forward velocity also creates a rolling resistance moment as a
> function of vertical load and forward rolling speed. There is also a
> moment produced by the internal bearing friction trying to slow it
> down.
>
> I understand this, however if the vehicle were to rotate
> instantaneously such that the the longitudinal velocity = 0, the
> longitudinal slip ratio = 0, the angular velocity > 0, and the lateral
> velocity was 10mts/s and hence the lateral slip ratio > 0, the tyre is
> not outputting any force in the longitudinal direction thus meaning
> the only forces acting against the angular velocity of the wheel are
> the internal bearing friction and rolling resistance moment. However
> the rolling resistance moment is a function of forward rolling speed
> which if equal to 0 there is no moment produced. Thus in the
> simulation if travelling purely in a lateral direction in the tyre's
> frame of reference, the only force acting against the angular velocity
> of the wheel is internal bearing friction, however this is wrong as it
> is not the only force acting on it. The wheel would only slow down as
> a result of the bearing friction which is not correct There must be
> another additional force derived from lateral slip that adds to the
> rolling resistance moment.
>
> In real life if a vehicle were to instantaneously rotate and travel in
> a purely lateral direction, the wheels would stop rolling from the
> lateral slip. I'm trying to work out what that force is. I hope this
> post is clear. Any help is much appreciated!

Tire friction with the road surface? Higher vehicle weight = more road
friction and bearing friction. Drivetrain friction? Even air resistance to a
minimal degree. A good example of decreasing rolling resistance is when
Nascar crews pry the brake pads 1/2" away from the rotors before
qualification at Daytona and Talledega and the driver is always reminded not
to touch the brake pedal. Higher tech F1 braking systems keep the pads a
good distance from the rotors when not braking and even take into
consideration slight warping of the rotors over the course of a race. Taking
into consideration your slip ratios, your main consideration is forward
rolling resistance and there are many factors to consider. Even the
viscosity of lubrication fluids is a factor.

Ed

On Jan 19, 12:45*am, "Ed Medlin" > wrote:
> "Zach" > wrote in message
>
> ...
>
>
>
> > I'm trying to understand the resulting torques that act back on the
> > axle. For a free rolling wheel travelling purely in the longitudinal
> > direction and with a slip ratio > 0, it will output a force of say
> > 100N. As well as pushing the vehicle forward it also creates a torque
> > on the axle which is then used to calculate the angular acceleration.
> > The forward velocity also creates a rolling resistance moment as a
> > function of vertical load and forward rolling speed. There is also a
> > moment produced by the internal bearing friction trying to slow it
> > down.
>
> > I understand this, however if the vehicle were to rotate
> > instantaneously such that the the longitudinal velocity = 0, the
> > longitudinal slip ratio = 0, the angular velocity > 0, and the lateral
> > velocity was 10mts/s and hence the lateral slip ratio > 0, the tyre is
> > not outputting any force in the longitudinal direction thus meaning
> > the only forces acting against the angular velocity of the wheel are
> > the internal bearing friction and rolling resistance moment. However
> > the rolling resistance moment is a function of forward rolling speed
> > which if equal to 0 there is no moment produced. Thus in the
> > simulation if travelling purely in a lateral direction in the tyre's
> > frame of reference, the only force acting against the angular velocity
> > of the wheel is internal bearing friction, however this is wrong as it
> > is not the only force acting on it. The wheel would only slow down as
> > a result of the bearing friction which is not correct There must be
> > another additional force derived from lateral slip that adds to the
> > rolling resistance moment.
>
> > In real life if a vehicle were to instantaneously rotate and travel in
> > a purely lateral direction, the wheels would stop rolling from the
> > lateral slip. I'm trying to work out what that force is. I hope this
> > post is clear. Any help is much appreciated!
>
> Tire friction with the road surface? Higher vehicle weight = more road
> friction and bearing friction. Drivetrain friction? Even air resistance to a
> minimal degree. A good example of decreasing rolling resistance is when
> Nascar crews pry the brake pads 1/2" away from the rotors before
> qualification at Daytona and Talledega and the driver is always reminded not
> to touch the brake pedal. Higher tech F1 braking systems keep the pads a
> good distance from the rotors when not braking and even take into
> consideration slight warping of the rotors over the course of a race. Taking
> into consideration your slip ratios, your main consideration is forward
> rolling resistance and there are many factors to consider. Even the
> viscosity of lubrication fluids is a factor.
>
> Ed

Right. The bearing friction I'm modelling as a function of radial load
and coefficients for viscosity/friction etc.

I would have thought that the rolling resistance would be increased
dramatically if there was any degree of lateral slip or is it totally
independent of it? The equation for rolling resistance taken from
Pacejka's Tyre & Vehicle Dynamics is:

My = -Vertical Load * Unloaded Tyre Radius * (scalar value * arctan
(Forward Rolling Speed / Reference Velocity) + scalar value * Fx /
Adapted Vertical Load) * scalar value

which would suggest that it is independent of lateral slip

Thanks for you help Ed

"Zach" > wrote in message
...
On Jan 19, 12:45 am, "Ed Medlin" > wrote:
> "Zach" > wrote in message
>
> ...
>
>
>
> > I'm trying to understand the resulting torques that act back on the
> > axle. For a free rolling wheel travelling purely in the longitudinal
> > direction and with a slip ratio > 0, it will output a force of say
> > 100N. As well as pushing the vehicle forward it also creates a torque
> > on the axle which is then used to calculate the angular acceleration.
> > The forward velocity also creates a rolling resistance moment as a
> > function of vertical load and forward rolling speed. There is also a
> > moment produced by the internal bearing friction trying to slow it
> > down.
>
> > I understand this, however if the vehicle were to rotate
> > instantaneously such that the the longitudinal velocity = 0, the
> > longitudinal slip ratio = 0, the angular velocity > 0, and the lateral
> > velocity was 10mts/s and hence the lateral slip ratio > 0, the tyre is
> > not outputting any force in the longitudinal direction thus meaning
> > the only forces acting against the angular velocity of the wheel are
> > the internal bearing friction and rolling resistance moment. However
> > the rolling resistance moment is a function of forward rolling speed
> > which if equal to 0 there is no moment produced. Thus in the
> > simulation if travelling purely in a lateral direction in the tyre's
> > frame of reference, the only force acting against the angular velocity
> > of the wheel is internal bearing friction, however this is wrong as it
> > is not the only force acting on it. The wheel would only slow down as
> > a result of the bearing friction which is not correct There must be
> > another additional force derived from lateral slip that adds to the
> > rolling resistance moment.
>
> > In real life if a vehicle were to instantaneously rotate and travel in
> > a purely lateral direction, the wheels would stop rolling from the
> > lateral slip. I'm trying to work out what that force is. I hope this
> > post is clear. Any help is much appreciated!
>
> Tire friction with the road surface? Higher vehicle weight = more road
> friction and bearing friction. Drivetrain friction? Even air resistance to
> a
> minimal degree. A good example of decreasing rolling resistance is when
> Nascar crews pry the brake pads 1/2" away from the rotors before
> qualification at Daytona and Talledega and the driver is always reminded
> not
> to touch the brake pedal. Higher tech F1 braking systems keep the pads a
> good distance from the rotors when not braking and even take into
> consideration slight warping of the rotors over the course of a race.
> Taking
> into consideration your slip ratios, your main consideration is forward
> rolling resistance and there are many factors to consider. Even the
> viscosity of lubrication fluids is a factor.
>
> Ed

Right. The bearing friction I'm modelling as a function of radial load
and coefficients for viscosity/friction etc.

I would have thought that the rolling resistance would be increased
dramatically if there was any degree of lateral slip or is it totally
independent of it? The equation for rolling resistance taken from
Pacejka's Tyre & Vehicle Dynamics is:

My = -Vertical Load * Unloaded Tyre Radius * (scalar value * arctan
(Forward Rolling Speed / Reference Velocity) + scalar value * Fx /
Adapted Vertical Load) * scalar value

which would suggest that it is independent of lateral slip

Thanks for you help Ed

I think they would be totally independent of each other. Both would affect
momentum, but in entirely different directions.....I guess that is the
correct word.....:-). I hope you get my drift....


Ed

>rolling resistance
>which would suggest that it is independent of lateral slip

In real life it isn't. Most of rolling resistance is due to
deformation at the contact patch, combined with the fact that
hysteresis is involved when rubber is compressed or stretched
and then returns to it's former state. The force during the
deformation is greater than the force during the recovery,
which is why rubber is good for reducing vibration.

Lateral slip increases the rubber deformation, and it's enough
to slow down a race car pulling a high g turn. In the case
of a Formula 1 car, top speed at full throttle might be around
190+ mph on a straight, but this gets reduced to about 160mph
in a 4 g turn.

"jeffareid" > wrote in message
...
> >rolling resistance
>>which would suggest that it is independent of lateral slip
>
> In real life it isn't. Most of rolling resistance is due to
> deformation at the contact patch, combined with the fact that
> hysteresis is involved when rubber is compressed or stretched
> and then returns to it's former state. The force during the
> deformation is greater than the force during the recovery,
> which is why rubber is good for reducing vibration.
>
> Lateral slip increases the rubber deformation, and it's enough
> to slow down a race car pulling a high g turn. In the case
> of a Formula 1 car, top speed at full throttle might be around
> 190+ mph on a straight, but this gets reduced to about 160mph
> in a 4 g turn.
>

Yea, I was kind of "toungue in cheek" on that one.....:-). Where you can
really see these effects is at Indy where the IRL cars are flat out all the
time and the lateral force slows them in the four corners. They do a lot
with camber where on the straights, there is a much smaller contact patch
from the tires and thus causing less friction there. The by product of this
is more wear and heat on the inside of the right tires and outsides of the
lefts. This causes a much larger contact patch in the corners due to the
distortions from the lateral Gs which will place the entire tire into
contact with the track surface. There is also the effect of the extra Gs
causing compression and therefore more friction. How to determine and
calculate the extent of both the Gs and added tire friction have with
forward rolling resistance is a tough one. I am sure the F1 and IRL (and
even Nascar today) engineers have some sort of formula for this.



Ed

> Lateral slip increases the rubber deformation, and it's enough
> to slow down a race car pulling a high g turn. In the case
> of a Formula 1 car, top speed at full throttle might be around
> 190+ mph on a straight, but this gets reduced to about 160mph
> in a 4 g turn.

Correction, the speed loss is there, but it's not 30mph, more
like in the range 10 mph to 20 mph (perhaps some CART cars
back in the 1990's that went over 255 mph on the straights).

The 160mph 4 g turn was a comment in this video, but the driver
slowed before entering "pouen":

David Coulthard in F1 McLaren, 2002 (remember automatic shifters?):

http://jeffareid.net/real/spaf1.wmv

Onboard lap from 1998 CART car, you can hear the engine rpms drop,
but it isn't a lot.

http://www.youtube.com/watch?v=T_kt2T6HM3A&fmt=18