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" <[email hidden]> wrote in message news:[email hidden]...

Quoted text (31 lines)

> 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" <[email hidden]> wrote:

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> "Zach" <[email hidden]> wrote in message > > news:[email hidden]... > > > > > 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" <[email hidden]> wrote in message news:[email hidden]... On Jan 19, 12:45 am, "Ed Medlin" <[email hidden]> wrote:

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> "Zach" <[email hidden]> wrote in message > > news:[email hidden]... > > > > > 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" <[email hidden]> wrote in message news:vXqel.43$[email hidden]...

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> >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

On Jan 18, 7:45 am, "Ed Medlin" <[email hidden]> wrote:

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> "Zach" <[email hidden]> wrote in message > > news:[email hidden]... > > > > > 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! >

Hi Zach :-) Guess you guys won't be using my tire model after all then, eh? ;-) Just to make sure I understand: You're talking about simply moving the tire sideways (90 degree slip angle)? The tire should not spin indeed of course. The force that will cause this is just the longitudinal force from slip ratio. Tire rotates forward, it gets a backwards torque. Tire rotates backward, it gets a forwards torque. That should keep the angular velocity at or near enough to zero in most tire models, although without a relaxation approach of some kind it may still jitter a little bit. There isn't any "missing force" here. What's happening wrong in the simulation? Do the wheels continue to spin even at 90 degrees slip angle? Concerning the other posts: There is indeed a change in rolling resistance with slip angle. Also with slip ratio. The example of an IndyCar losing speed doesn't have much to do with that, though. That's induced drag caused by the component of the lateral force pointing rearwards because there's a slip angle, which slows the car down. Two different things, and I suspect that's not really what Zach was asking about. Correct me if I'm mistaken..

> Concerning the other posts: There is indeed a change in rolling > resistance with slip angle. Also with slip ratio. The example of an > IndyCar losing speed doesn't have much to do with that, though. > That's induced drag caused by the component of the lateral force > pointing rearwards because there's a slip angle, which slows the car > down. I assume aerodyanmic drag is also increased by the overall slip angle. Perhaps this is why the tires used on Indy and the former Champ and Cart race cars are so stiff compared to the tires used on other racing cars. The "working" (max g turn) slip angle is about 2 degrees (I think "working" slip angle means the total yaw angle of the car versus the direction the car is going, essentially the slip angle of the rear tires).

On Jan 29, 5:40 pm, "jeffareid" <[email hidden]> wrote:

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> > Concerning the other posts:  There is indeed a change in rolling > > resistance with slip angle.  Also with slip ratio.  The example of an > > IndyCar losing speed doesn't have much to do with that, though. > > That's induced drag caused by the component of the lateral force > > pointing rearwards because there's a slip angle, which slows the car > > down. > > I assume aerodyanmic drag is also increased by the overall slip angle. > Perhaps this is why the tires used on Indy and the former Champ and > Cart race cars are so stiff compared to the tires used on other racing > cars. The "working" (max g turn) slip angle is about 2 degrees (I > think "working" slip angle means the total yaw angle of the car versus > the direction the car is going, essentially the slip angle of the > rear tires).

Yes, the drag will change with vehicle sideslip angle, but I'm not sure in which direction it goes on an Indycar over that first few degrees. I saw one aero CFD test that showed a very slight increase in front downforce on a formula style car (I was late to the lecture so don't know what kind of car it was exactly). The vertical fins in the front wing were designed to produce that increased downforce with slip angle, interestingly enough, by getting rid of a little low air pressure bubble that (simulation-wise at least) developed behind and to the outside of the front wing. What happens to drag I don't know. It's possible it increases a bit on some cars and decreases on others. On a GT car with a spoiler you can wind up with quite a bit less downforce, but it's probably not safe to say that the drag would also decrease along with it. Different cars, different effects...