A sliding tyre...

Ski's develop lateral force by action of their edges. They are, in fact, carving a curve instead of simply sliding, and really don't belong in this discussion. A slide is distinct from slip. Slip is OK and normal; slide is not good. A clear definition might help. The tire is sliding when no part of its contact patch is in static contact with the road surface. In a slip, lateral forces deform the tire carcass to the side. Each point on the contact patch in static contact is slightly offset from the next point to come down as the tire rolls. Thus, the wheel track is offset from the plane of rotation by an amount known as the slip angle. So long as some part of the contact patch is holding the road, some part of the tire is acting as a tire and not as a *** eraser. A tire locked under braking is sliding just as surely as one overwhelmed by lateral force, or spinning from power. What little directionality remains is miniscule, and due more to the shape of the contact patch and weathervaning than virtue of its being donut shaped and mounted on a wheel. The lateral force available is very close to that in fore-aft direction, and is equal by definition to the sliding coefficient of friction integrated over the sliding contact area. Michael.

: Ski's develop lateral force by action of their edges. They are, in fact, : carving a curve instead of simply sliding, and really don't belong in this : discussion. That's true of a ski on a person but can it be true of a ski on a skidoo/ I'm not sure. : A tire locked under braking is sliding just as surely as one overwhelmed by : lateral force, or spinning from power. What little directionality remains is : miniscule, and due more to the shape of the contact patch and weathervaning : than virtue of its being donut shaped and mounted on a wheel. The lateral : force available is very close to that in fore-aft direction, and is equal by : definition to the sliding coefficient of friction integrated over the : sliding contact area. Sure - I agree that the directionality of a sliding tyre has nothing to do with its "tyre nature" and you would get a similar directionality from (say) a huge pencil eraser mounted on our front axle... But I'm just interested in how miniscule that is - is it significant and do sims model it? Richard G. Clegg Only the mind is waving Networks and Non-Linear Dynamics Group Dept. of Mathematics, Uni. of York www: http://manor.york.ac.uk/top.html

Richard G Clegg <[email hidden]> wrote in message news:88p220$iq3$[email hidden]... > Hmm.., I was just wondering really how accurate the GPL "lock the > wheels and off you go" model was... From casual observation, it's pretty close. :-) The transition from almost sliding to fully sliding is gradual; this is reflected in the gradation of skid marks. Without more rigorous verification, I'm pretty satisfied that they have a good understanding of the forces acting at the contact patch. > : Aside from the obvious academic interest, why do you ask? > > I have a sick and perverse interest in the physics of racing cars. > :-) (True, you can't expect me to have anything other than an > academic interest - I'm a professional academic). Milliken & Milliken's "Race Car Vehicle Dynamics", SAE, ISBN 1-56091-526-9. GPL is a wonderful workshop for these explorations. Michael.

Richard G Clegg wrote: > > Hmm.., I was just wondering really how accurate the GPL "lock the > wheels and off you go" model was... Sounds pretty accurate. Last time I checked, locking up the wheels can send you off the track in real life to :) I assume that what you mean is that you have trouble recovering from locking the wheels. Once you are able to read the cues from GPL, that's not a problem -- Pat Dotson IMPACT Motorsports http://www.impactmotorsports.com/pd.html

Michael Young <[email hidden]> wrote: : Richard G Clegg <[email hidden]> wrote in message : news:88mmh1$puf$[email hidden]... :> :> Sure - I agree that the directionality of a sliding tyre has nothing :> to do with its "tyre nature" and you would get a similar directionality :> from (say) a huge pencil eraser mounted on our front axle... But I'm :> just interested in how miniscule that is - is it significant and do :> sims model it? : The contact patch is somewhat of a malformed ellipse, blunter in front than : the rear; the major axis typically running across the width of the tire, That's kinda what my pencil eraser looks like... well it was the nearest everyday object I could think of. : depending mostly on the tire radius to width ratio, and also influenced by : inflation pressure. Pressure distribution varies across the contact patch. I : am not aware of any quantitative studies of how the scrub loads vary on a : sliding contact patch. My guess, based on nothing more than conjecture, is : less than 10% variance from strongest to weakest angle of presentation. I : further expect that the small difference wouldn't be apparent at all from : casual observation (as when reviewing a game replay). I would also guess : that, in a proper sim, the sliding condition is not treated much differently : from a more "normal" condition. Thus, it's entirely likely that a reasonably : sophisticated physics model would use different values for lateral and : longitudinal forces acting on a sliding contact patch. For the definitive : answer, however, you'll have to ask the game authors if they do indeed : account for this. Hmm.., I was just wondering really how accurate the GPL "lock the wheels and off you go" model was... : Aside from the obvious academic interest, why do you ask? I have a sick and perverse interest in the physics of racing cars. :-) (True, you can't expect me to have anything other than an academic interest - I'm a professional academic). -- Richard G. Clegg Only the mind is waving Networks and Non-Linear Dynamics Group Dept. of Mathematics, Uni. of York www: http://manor.york.ac.uk/top.html

Richard G Clegg <[email hidden]> wrote: : That's one way of looking at it - but the other way of looking at it : is that it's a fixed "slab" of rubber at an angle being driven across : the ground. I'd have thought this would have some boat rudder like : turning effect - experiments with pushing my rubberised mouse-mat : across a desk (my favourite "locked tyre" contact patch experiment) seem : to show that it does try to turn around a drawing pin pushed through : its centre. I believe it turns because your pushpin is not perfectly centered on the pad. If it is in the center, then integrating the friction force on either side of the pushpin shows that the forces are balanced and there is no net turning moment trying to twist the pad. If you move the pushipn off center, then there is a greater friction force to one side of the pin that the other (area x contact pressure x mu) and there is a turning moment which realigns the pad until the area on either side of the pin ("sides" here defined by the direction of travel) is equal, and then it stops turning. Of course, you could probably set up some good oscillations depending on how fast you can "drive" your eccentric mousepad. As for a tire, that's an interesting question. I'm still puzzling over it. If an individual wheel is loaded through its hub, which is centered w.r.t. the tire contact patch, then the orientation of the tire w.r.t. the direction of travel shouldn't matter, as there will be no net turning moment trying to twist the tire. But not all wheels are centered. Now my brain is puzzling about weight distribution in a car, and polar moments of inertia. I'll get back to you. It's a nice challenge for myself to see if I can dust out the cobwebs and get back to some old textbook stuff. Stephen

On Sun, 20 Feb 2000, Michael Young wrote: > Richard G Clegg <[email hidden]> wrote in message > news:88p220$iq3$[email hidden]...... > > I have a sick and perverse interest in the physics of racing cars. > > :-) (True, you can't expect me to have anything other than an > > academic interest - I'm a professional academic). > > Milliken & Milliken's "Race Car Vehicle Dynamics", SAE, ISBN 1-56091-526-9. > GPL is a wonderful workshop for these explorations. Thanks for the plug! SAE gives a good price on shipping, if you pay with a credit card. <http://www.sae.org/products/books/R-146.htm> -- Doug Douglas Milliken <[email hidden]> Milliken Research Associates Inc.

Richard G Clegg wrote:

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> > ... just wondering about sliding tyres... The standard line is "a > sliding tyre has no directional integrity". I was thinking about this > the other day and it strikes me as extremely unlikely. A tyre at an > angle would surely produce a small amount of steering force in it's > correct direction - otherwise a skidoo couldn't steer (if a sliding tyre > has no directional integrity why should a sliding ski?) > > I was also wondering how various sims modelled a sliding tyre. > No direction? Some direction but much reduced? > > -- > Richard G. Clegg Only the mind is waving > Networks and Non-Linear Dynamics Group > Dept. of Mathematics, Uni. of York > www: http://manor.york.ac.uk/top.html

As you are more of a science buff than an engineer, let a simple yet similarly oriented mind try to give you a simple and hopefully explanation. First of all, the truth is that a LOCKED tyre has no directional integrity. A locked tyre is just a piece of rubber in a fixed position sliding along the ground, and it doesn't matter much in which direction it slides, as the friciton force is always pointing in (approximately) the oposite direction to the velocity of the tyre relative to the ground. Consider the example of a tyre that is freely rolling, but is pointing (sliding) at an angle (say 'alpha'-the slip angle) relative to its velocity with respect the ground. There can be no force in the direction of the orientation of the tyre, as the tyre is freely rolling, there is just a (friction, F) force perpendicular to it (in the direction of the tyre axis). The component of the force in the direction of the velocity is then -|F|*Sin(alpha), and the perpendicular component is |F|*Cos(alpha). If alpha is not too big but still large enough so thatthe tyre is sliding (really a hard to define term, though), MOST of the force is still in the lateral direction, so a sliding tyre (but not a locked one) still has lots of directional integrity. Gregor Veble

Gregor Veble <[email hidden]> wrote: : As you are more of a science buff than an engineer, let a simple yet : similarly oriented mind try to give you a simple and hopefully : explanation. Thanks - much appreciated. : First of all, the truth is that a LOCKED tyre has no directional : integrity. A locked tyre is just a piece of rubber in a fixed position : sliding along the ground, and it doesn't matter much in which direction : it slides, as the friciton force is always pointing in (approximately) : the oposite direction to the velocity of the tyre relative to the : ground. That's one way of looking at it - but the other way of looking at it is that it's a fixed "slab" of rubber at an angle being driven across the ground. I'd have thought this would have some boat rudder like turning effect - experiments with pushing my rubberised mouse-mat across a desk (my favourite "locked tyre" contact patch experiment) seem to show that it does try to turn around a drawing pin pushed through its centre. Anyway, this is all getting waaaay off topic. I had just come to wonder about how accurate the statement "a locked tyre has no directional integrity" was.... I guess the conclusion from this group is "very accurate indeed" -- Richard G. Clegg Only the mind is waving Networks and Non-Linear Dynamics Group Dept. of Mathematics, Uni. of York www: http://manor.york.ac.uk/top.html

Gregor, what you say of a free rolling tire would be true for a tire slipping, not sliding. We don't normally encounter sliding tires in our day to day experience, so the misconception is understandable. The difficulty is that the condition is transient; changing the direction of applied force from purely lateral -- perpendicular to the plane of rotation -- to something else, allows the tire in common experiments to recover from the slide, and gives the false impression that there is some directional integrity. This indicates only that the sliding tire predisposes itself to recovering to its original free rolling state. My assertion is that a sliding tire has very little directional integrity. Any variation is due only to the assymetry of the contact patch. I realize this is counter-intuitive, given the discussion above. Try the following thought experiment. The stated condition for a slide is that no part of the contact patch is in static contact with the road surface. Driving on normal dry pavement, the sliding coefficient of friction is such that a true slide is difficult to achieve, let alone maintain. If we reduce the coefficient of friction, however, we begin to see that a slide truly doesn't match our expectations. It's easier to imagine how a polished steel drum, instead of sticky ***, will slide freely once we apply a large enough side force to interrupt its free rolling state. The same is true when driving on an oil slick, or on ice. We've all witnessed the effects of oil on the race track, and probably experienced the terror of driving on ice. Let me know if these examples don't hold up to your scrutiny. Michael.

Thanks for your reply, it did make me think, especially about definitions. I like your definiton of a sliding tire, namely that all (or at least most) of the contact patch is sliding, so only the coefficient of friction (and not stiction applies). As you stated this can be better observed not with *** but a metal drum. My analysis is, however, still valid for this case, in fact, it is easier to show its point here. When a metal drum acquires a nonzero slip angle, the stiction immediately ceases and the force on it becomes purely frictional, thus the wheel develops a proper slide in accord with the definition above. The frictional force always shows in the opposite direction to the relative velocity between the rubbing surfaces. The main point is that, with a free rolling drum in a sliding equlibrium, the relative velocity of the contact point on the bottom of the drum with respect to the ground is always in the direction of the wheel axis, i.e. perpendicular to the plane of rotation. This happens because, unless there is rotational acceleration involved which must be 0 in equlibrium anyway, there can be no torque along the wheel axis and consequently no friction force component in the wheel plane of rotation. The friction force in equlibrium thus always points along the wheel axis and so does the relative slip direction. The magnitude of this force is just the coefficient of friction times the normal force, and is the magnitude of the force |F| in my previous post (see below). With a metal wheel, the slip angle alpha can be set to almost 0 and still only friction is occuring. The only loss of lateral force in this case is the difference between the coefficient of friction and the coefficient of stiction. But the force is still purely lateral at small angles, so directional integrity is indeed preserved. What is lost is something more subtle. While the lateral forces are still preserved, although a bit smaller than when the wheel is still sticking, the steering looses any practical effect. While the car may still be executing a turn due to the preservation of the lateral forces, when the slide of the steered tyres occurs the input of the steering wheel does not have any (linear) effect. So what is lost is the control, but not (a great part of) the turning ability. I know from first hand experience, I did a lot of wonderful long turning drifts with my little Renault Clio in the snow this winter. Only when you aplly the brakes to lock the wheels or the accelerator to spin them does the directional integrity gradually cease relative to the amount of the longitudinal slip of the corresponding tires. I really enjoy these kinds of discussions, so please, anyone, be critical about what is written here! Greetings, Gregor Veble

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> Gregor, what you say of a free rolling tire would be true for a tire > slipping, not sliding. We don't normally encounter sliding tires in our day > to day experience, so the misconception is understandable. The difficulty is > that the condition is transient; changing the direction of applied force > from purely lateral -- perpendicular to the plane of rotation -- to > something else, allows the tire in common experiments to recover from the > slide, and gives the false impression that there is some directional > integrity. This indicates only that the sliding tire predisposes itself to > recovering to its original free rolling state. > My assertion is that a sliding tire has very little directional integrity. > Any variation is due only to the assymetry of the contact patch. I realize > this is counter-intuitive, given the discussion above. Try the following > thought experiment. > The stated condition for a slide is that no part of the contact patch is in > static contact with the road surface. Driving on normal dry pavement, the > sliding coefficient of friction is such that a true slide is difficult to > achieve, let alone maintain. If we reduce the coefficient of friction, > however, we begin to see that a slide truly doesn't match our expectations. > It's easier to imagine how a polished steel drum, instead of sticky ***, > will slide freely once we apply a large enough side force to interrupt its > free rolling state. The same is true when driving on an oil slick, or on > ice. We've all witnessed the effects of oil on the race track, and probably > experienced the terror of driving on ice. > Let me know if these examples don't hold up to your scrutiny. > Michael. > > Consider the example of a tyre that is freely rolling, but is pointing > > (sliding) at an angle (say 'alpha'-the slip angle) relative to its > > velocity with respect the ground. There can be no force in the direction > > of the orientation of the tyre, as the tyre is freely rolling, there is > > just a (friction, F) force perpendicular to it (in the direction of the > > tyre axis). The component of the force in the direction of the velocity > > is then -|F|*Sin(alpha), and the perpendicular component is > > |F|*Cos(alpha). If alpha is not too big but still large enough so that > > the tyre is sliding (really a hard to define term, though), MOST of the > > force is still in the lateral direction, so a sliding tyre (but not a > > locked one) still has lots of directional integrity.

: : That's one way of looking at it - but the other way of looking at it : : is that it's a fixed "slab" of *** at an angle being driven across : : the ground. I'd have thought this would have some boat rudder like : : turning effect - experiments with pushing my ***ised mouse-mat : : across a desk (my favourite "locked tyre" contact patch experiment) seem : : to show that it does try to turn around a drawing pin pushed through : : its centre. : I believe it turns because your pushpin is not perfectly centered on the : pad. If it is in the center, then integrating the friction force on : either side of the pushpin shows that the forces are balanced and there is : no net turning moment trying to twist the pad. If you move the pushipn : off center, then there is a greater friction force to one side of the pin : that the other (area x contact pressure x mu) and there is a turning : moment which realigns the pad until the area on either side of the pin : ("sides" here defined by the direction of travel) is equal, and then it : stops turning. Being a thorough experimenter I tried to compensate for that by (without moving the pin) making my mousemat execute a left then a right turn. However, retrying this I think my results were not very reproducible. (I think there was a significant amount of seeing "what I expected" in the experiment). Hmm... point taken about integrated frictional forces though. I must rethink. Instinctively I had thought of a locked tyre across a road being "like" a rudder on a boat going through water. : Of course, you could probably set up some good : oscillations depending on how fast you can "drive" your eccentric : mousepad. It's much better now : As for a tire, that's an interesting question. I'm still puzzling over : it. If an individual wheel is loaded through its hub, which is centered : w.r.t. the tire contact patch, then the orientation of the tire w.r.t. the : direction of travel shouldn't matter, as there will be no net turning : moment trying to twist the tire. But not all wheels are centered. I think in my earlier experiment with the mousemat, my drawing pin was slightly to the rear of the "tyre". This would lead to a "tyre" which when turned and locked would have greater friction on the side it turned to. (Actually, if you designed a race car wheel like that, you might get a wheel which would turn the car while locked up - a positive boon for ***poor drivers who like to go into a corner with the front wheels locked). Are real wheels not pivotted more or less about the centre of the contact patch then? I'd always assumed that they were - wouldn't turning be a bit peculiar if they weren't. (grin) and, of course, if we have a force trying to twist the tyre and we hold the tyre steady (using the wheel) then we will have a force trying to twist the car (which is what I thought might happen). : Now my brain is puzzling about weight distribution in a car, and polar : moments of inertia. I'll get back to you. It's a nice challenge for : myself to see if I can dust out the cobwebs and get back to some old : textbook stuff. Heh... well, I'm glad to see that at least I've interested someone else in the problem.... let me know your conclusions. Now, of course, having thought about the problem deeply, I've no idea why a rolling tyre produces a turning effect (or rather don't understand it in the depth I'd like to). -- Richard G. Clegg Only the mind is waving Networks and Non-Linear Dynamics Group Dept. of Mathematics, Uni. of York www: http://manor.york.ac.uk/top.html

[snip] My interpretation is that what make the car steer is really not changing the direction of the forces generated by the front wheels, but changing the magnitude of the force component lateral in car coordinates (i.e. the amount of sideways force). This means that when you are not sliding and increase the steering input the total increase in force will be large enough to increase the lateral force component even though turning the wheels is rotating the total force vector backwards. But when you are near the maximum force, increasing steering input will only rotate the force vector backwards leaving the lateral component constant or even smaller. _ Mats Lofkvist

[snip] On the contrary, turning would be quite a bit peculiar (i.e. unstable) if caster angle didn't put the pivot point ahead of the contact patch :-) _ Mats Lofkvist

> Are real wheels not pivotted more or less about the centre of the > contact patch then? I'd always assumed that they were - wouldn't > turning be a bit peculiar if they weren't. Never. The pivot axis around which the front wheels are turned to steer a car is determined by the caster angle (longitudinal plane) and the king pin angle (lateral plane). This is necessary to create forces that make the car stable and provide feed back through the steering wheel...the car wants to go straight and the force required to turn it is proportional to the steering angle. Marty

Yes, non-zero slip angles occur in *** tires due to their deformation. Steel drums have a much smaller range of deformation. :-) This is where intuition departs for me. To demonstrate your point, say I drag a steel drum behind me as I cruise down the road. The harness allows me to "steer" the drum. The edges of the drum are rounded to prevent edge effect, which would negate the conditions under discussion. Thus, only sliding friction of the contact patch and the pinned connection at the axle acts on the wheel. The frictional force acts to directly oppose the direction of travel. This force can be decomposed to components perpendicular and parallel to the plane of wheel rotation. Intuitively, I expect the wheel will spin up to road speed (unless this force is reduced to zero by steering it perpendicular to the direction of travel). The component of force in the plane of rotation is entirely dissipated by accelerating the wheel's rotation about its axis. The perpendicular component is the resultant lateral force. No other result is possible. The difficulty with this explanation is what happens when the wheel is already rotating at its steady state speed? The frictional force no longer accelerates the wheel's rotation, and must be dissipated somewhere else. So, refining the mental picture of what's happening: As the wheel accelerates (spins faster), the apparent slide angle differs from the actual steer angle to account for the motion of the face of the wheel across the road. At steady state, each infinitesimal element on the face of the drum has a component of velocity equal to the road speed. Thus, incredibly, the drum is actually turning faster than it would if the steering angle was held at zero. My head hurts from trying to reconcile this, so I'll just offer a guess on the nature of the new resultant force. The rotation speed is such that the perpendicular component just cancels the lateral force arising from the steering angle. Thus, the net resultant frictional force is still directly behind the direction of travel, regardless of steering angle, meaning there is no directional integrity. I expect I'm more surprised than you at this result. :-) I drive a late model Cobra all year round; its fat tires have almost zero traction in snow. Forgive me for presuming; I believe I am better acquainted with true slides versus simply extreme slip (which I also play with from time to time). So long as you still have directional control, you are not yet sliding. :-) Michael.

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> > Are real wheels not pivotted more or less about the centre of the > > contact patch then? I'd always assumed that they were - wouldn't > > turning be a bit peculiar if they weren't. > Never. The pivot axis around which the front wheels are turned to steer a car is > determined by the caster angle (longitudinal plane) and the king pin angle > (lateral plane).

...err, I think I meant to say longitudinal AXIS & lateral AXIS.

Richard G Clegg wrote:

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> > Gregor Veble <[email hidden]> wrote: > : As you are more of a science buff than an engineer, let a simple yet > : similarly oriented mind try to give you a simple and hopefully > : explanation. > > Thanks - much appreciated. > > : First of all, the truth is that a LOCKED tyre has no directional > : integrity. A locked tyre is just a piece of rubber in a fixed position > : sliding along the ground, and it doesn't matter much in which direction > : it slides, as the friciton force is always pointing in (approximately) > : the oposite direction to the velocity of the tyre relative to the > : ground. > > That's one way of looking at it - but the other way of looking at it > is that it's a fixed "slab" of rubber at an angle being driven across > the ground. I'd have thought this would have some boat rudder like > turning effect - experiments with pushing my rubberised mouse-mat > across a desk (my favourite "locked tyre" contact patch experiment) seem > to show that it does try to turn around a drawing pin pushed through > its centre. >

Only now am I able to see what your original idea was. I am sorry, I should have read your post more carfully. Indeed I can only agree with you that as long as the tyre or any other rubber contact patch is not fully symmetric there can exist forces and torques perpendicular to the direction of the travel due to the deformations and elastic properties of the material (rubber, that is). Now, if the total contact patch would be sliding uniformly all the time, no lateral force is possible as the direction of the force on any little part of the patch is exactly in the opposite direction to the velocity vector. But, since rubber doesn't exactly behave this way as it tends to stick and then elastically slip with a high frequency which is percieved as the tyre squeal, the time average of the lateral force when sticking and then slipping can be nonzero and an overall lateral force can occur. A nonzero torque (the turning of your mousepad) around a certain point can occur even if no stick-slip scenario occurs, it really depends on where you put that point and on the manner of how you apply the vertical forces on the patch. In the limit when the surface of the contact patch goes to zero, however, all lateral forces and twisting torques cease so no directional integrity exists anymore. > > Anyway, this is all getting waaaay off topic. I had just come to > wonder about how accurate the statement "a locked tyre has no > directional integrity" was.... I guess the conclusion from this group > is "very accurate indeed" Of course, only a scientist can ask such a question, any engineer or driver will tell you that when you lock, you plow ;). The fact remains that this can only be a minor effect which can probably not be felt when driving. > > -- > Richard G. Clegg Only the mind is waving > Networks and Non-Linear Dynamics Group > Dept. of Mathematics, Uni. of York > www: http://manor.york.ac.uk/top.html

Michael Young wrote:

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> Gregor Veble <[email hidden]> wrote in message > news:[email hidden]... > > When a metal drum acquires a nonzero slip angle, the stiction > > immediately ceases and the force on it becomes purely frictional, thus > > the wheel develops a proper slide in accord with the definition above. > Yes, non-zero slip angles occur in rubber tires due to their deformation. > Steel drums have a much smaller range of deformation. :-) > > The frictional force always shows in the opposite direction to the > > relative velocity between the rubbing surfaces. > > The main point is that, with a free rolling drum in a sliding > > equlibrium, the relative velocity of the contact point on the bottom of > > the drum with respect to the ground is always in the direction of the > > wheel axis, i.e. perpendicular to the plane of rotation. > This is where intuition departs for me. To demonstrate your point, say I > drag a steel drum behind me as I cruise down the road. The harness allows me > to "steer" the drum. The edges of the drum are rounded to prevent edge > effect, which would negate the conditions under discussion. Thus, only > sliding friction of the contact patch and the pinned connection at the axle > acts on the wheel.

Let us define the experiment a bit further. All measurements are done at a constant velocity of the tyre so no translational acceleration accurs (i.e. accelerating the center of gravity of the tyre), but accelerations around the axis of the wheel are possible. All velocities considered here will be measured with respect to the ground (this is important).

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> The frictional force acts to directly oppose the direction of travel. This > force can be decomposed to components perpendicular and parallel to the > plane of wheel rotation. Intuitively, I expect the wheel will spin up to > road speed (unless this force is reduced to zero by steering it > perpendicular to the direction of travel). The component of force in the > plane of rotation is entirely dissipated by accelerating the wheel's > rotation about its axis. The perpendicular component is the resultant > lateral force. No other result is possible.

To be more exact, the frictional force acts to directly oppose the direction of the slipping, and not the direction of travel of the center of the wheel. If the tyre would not be rotating, the direction of the travel would be the same as the direction of the slipping and the frictional force would then be opposing the direcion of the travel of the wheel (i.e. its velocity vector). As soon as the wheel rotation is present, the direction of slipping does not match the direction of travel anymore and things get a bit more involved. For example, when the rotation of the wheel is faster than the corresponding road travel velocity but the wheel still points in the direction of travel, the friction force does not oppose the direction of travel but actually points in the same direction (this happens in real cars when you have, for example, wheel spin on acceleration - bit let's forget about real cars for now). However, the forces do not get dissipated in the way, for example, energy does. The component of the frictional force that is used to accelerate the rotation of the wheel through the torque it produces on the rotational axis when the wheel is pointing at a slip angle, is also in effect when considering the overall force on the wheel. The fact that it is used for rotational acceleration does not negate ('dissipate') its effect when translational effects are in question. So if you press a stationary yet freely rotating (i.e. no brakes on the axis) wheel onto a moving road surface, the frictional force - at that exact instant - points exactly in the opposite direction to the road travel as the slipping is done in that direction, and there is NO translational component at this moment. Still, the component of the frictional force in the plane of rotation produces a torque which starts accelerating the wheel, but this does not have any effect on the total force felt by the wheel, i.e. the wheel stil feels that component of force. > The difficulty with this explanation is what happens when the wheel is > already rotating at its steady state speed? The frictional force no longer > accelerates the wheel's rotation, and must be dissipated somewhere else. As soon as the wheel is rotating, the velocity vector of the point on the wheel at its bottom (at the contact point) is not the same as the one at the center of the wheel, but gets a component that is proportional to the angular velocity of the wheel and is pointing in the plane of rotation of the wheel. To put it in mathematical terms, the components of the total velocity of the point at the bottom in the plane of the wheel are (after a little bit of geometry) V_in_plane = V_travel * Cos(alpha)-Omega * R (1) V_perpendicular = V_travel * Sin(alpha) (2) where alpha is the slip angle, Omega the angular velocity and R the radius of the wheel. As this is the velocity as measured with respect to the ground, this are exactly the components of the slipping velocity, because it is the point that is in contact with the ground. The friction force points exactly opposite to this velocity but its magnitude is always the same, namely the coefficient of friction times the normal force. Now, as long as V_in_plane is not 0, there is a component of the friction force opposing it which in turn produces the torque accelerates the rotation of the wheel (changes the Omega) until V_in_plane=0. In this case, Omega = (V_travel/R) * Cos(alpha). (3) This means that the slipping velocity in equilibrium always points perpendicular to the plane of rotation (as its component in the plane of rotation is zero), and from this it follows that the friction force is also perpendicular to the plane of rotation as it points exactly the opposite way.

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> So, refining the mental picture of what's happening: As the wheel > accelerates (spins faster), the apparent slide angle differs from the actual > steer angle to account for the motion of the face of the wheel across the > road. At steady state, each infinitesimal element on the face of the drum > has a component of velocity equal to the road speed. Thus, incredibly, the > drum is actually turning faster than it would if the steering angle was held > at zero.

Not true, see equation (3). The rotation is maximal when the wheel is pointing in the direction of the travel, and gradually goes to zero when alpha goes towards 90 degrees.

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> My head hurts from trying to reconcile this, so I'll just offer a guess on > the nature of the new resultant force. The rotation speed is such that the > perpendicular component just cancels the lateral force arising from the > steering angle. Thus, the net resultant frictional force is still directly > behind the direction of travel, regardless of steering angle, meaning there > is no directional integrity. > I expect I'm more surprised than you at this result. :-)

Due to the points presented above I cannot agree. The friction force in equilibrium is, written in the components perpendicular and in the wheel plane, F_in_plane = 0 F_perpendicular = F_total_friction, which means that in another coordinate system, F_in_travel = -F_total_friction * Sin(alpha) F_perpendicular_to_travel = F_total_friction * Cos(alpha) So when alpha goes to 0, only the component perpendicular to the direction of travel exists, and the one opposing the travel direction in fact tends to 0. This may sound very surprising, but is a subtle consequence of the fact that the friction force is ideally independent of the slipping velocity. So I stand by my original point, that is that the free rolling and purely frictioanally slipping wheel preserves most of the directional integrity, only degraded by the ratio of the coefficient of friction and the coefficient of stiction and the cosine of the slip angle (which is close to 1 for small angles). I apologize if the sign conventions are not obvious, but I would need a sketch to present them better. But I am sure the general idea can be extracted from the above equations.

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> > I know from first hand experience, I did a lot of wonderful long turning > > drifts with my little Renault Clio in the snow this winter. Only when > > you aplly the brakes to lock the wheels or the accelerator to spin them > > does the directional integrity gradually cease relative to the amount of > > the longitudinal slip of the corresponding tires. > I drive a late model Cobra all year round; its fat tires have almost zero > traction in snow. Forgive me for presuming; I believe I am better acquainted > with true slides versus simply extreme slip (which I also play with from > time to time). So long as you still have directional control, you are not > yet sliding. :-)

I forgive you for presuming ;), if you alow me to presume that, although the practical advantage is no doubt yours, the theoretical advantage is on my side :-). I do have a rather good understaning of a slide versus an extreme slip, although the boundary is fuzzy (and sometimes hairy, as any enthusiastic driver should know ;). Never did I say I have directional control in the snow slides, turning the wheel doesn't do diddly-squat then. Control can be had by pulling on the handbrake which locks the rear wheels and changes the direction of the force on the wheel from perpendicular to the wheel towards the direction opposite to travel, thus diminishing the lateral force and sending the tail further out. Or, in a front wheel drive car like mine, flooring it puts the direction of the friction force directly in the forward direction of the orientation of the tyres, which diminishes the lateral force on the fronts. Note that when in a slide, only the direction of the friction force can be changed, and not much can be done about its ... read more »

Gregor, I am a structural engineer by training and early vocation, but now write (non-engineering) software for a living. This background predisposes me to initially think in terms of non-moving systems, so please bear with me. It's clear to me now that I had presumed the proof when I thought this through. The free-body ignored a lateral force that might be present at the point where the harness attaches to the tow vehicle. This being the whole discussion, we would naturally come to quite different conclusions. :-) Let's try this again. [Clarifying: each element on the drum has a component of velocity equal to the road speed ( plus a perpendicular component). ] Ignoring everything else for the time being, an infinitesimal tire element in contact with the road is acted on by an impulse caused by the sliding friction. At this point, we only know the velocity of the road surface, and the wheel's steer angle. For a wheel not already rotating, the condition I'm most comfortable with, the relative velocity is precisely the same as the road speed. The component of the frictional force in the rotation plane adds to the tire's rotation. The component parallel to the wheel axis is the disputed lateral force. This defines one boundary condition. It appears that we're done at this point, as there clearly is a non-zero lateral force, and your point has already been made. The other boundary condition, steady state rotation, is not so clear in my mind, and I have a need to bring it back to first principles. Simplistically speaking, the force resulting from sliding friction does not depend on speed. That is, we expect it to be the same at 10 ft/sec as at 100 ft/sec. However, frictional force is zero when the velocity is zero. Also, superposition applies. At steady state, wheel rotation is neither increasing nor decreasing, implying that the force in this direction is zero. Thus, steady state rotation occurs when the velocity component on the face of the wheel, in the plane of rotation, matches the road speed. There is also a velocity component parallel to the wheel axis. There are two points to be made about lateral velocity component. First, the magnitude of the resultant of these vectors is clearly larger than that of the road speed alone. In fact, it should become infinite when the wheel is steered perpendicular to the direction of travel, the direct opposite of what you state. Second, I'm having trouble this morning mapping across reference frames. I suddenly can't distinguish if this resultant lateral force directly opposes that in the first case above, or if it is the same. Gregor, since we're so far apart on this one point, I feel it pointless to continue without coming to terms with the discrepancy. Help me, if you can, find the flaw in the thinking. If you ask me, that sounds suspiciously like directional control. :-) All kidding aside on this one; the Cobra just goes in whatever direction it was already going, no matter what I do with my hands or feet, even at very slow speeds. It has so far always recovered in time, but I try to avoid those situations now. The situations being having to change speed or direction of travel on snow covered roads. The only saving grace is that it takes a lot of cussing and begging just to get it going in those conditions. Michael.

Gregor Veble <[email hidden]> wrote in message news:[email hidden]... > I thought I had an engineer on the other side of the net :). It brings > back all those engineer/scientist jokes, in which the engineer is > invaribly always better off. One wonders why. :-) It isn't so much what we know or don't know; it's our proximity to the guy paying for services. I presume you meant "paid more" when you say better off. I can honestly say an academic's stress level appears to be far lower than an engineer's. What price sanity? > > Thus, > > steady state rotation occurs when the velocity component on the face ofthe

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> > wheel, in the plane of rotation, matches the road speed. > > This is incorrect. This is only true when the wheel is oriented in the > direction of travel. Once the wheel is turned at an angle to the > direction of travel, this then in fact contradicts the previous > statement of the zero component of force in the rotation plane. > > Let me try to demonstrate this with the equations I posted earlier, > > V_in_plane = V_travel * Cos(alpha)-Omega * R , (1) > > V_perpendicular = V_travel * Sin(alpha) . (2) >

... > I hope you agree that the above equations are always valid for the I believe V_in_plane is where we differ. The cos(alpha) term is not right; it should be V_travel/cos(alpha). Steady state rolling behind the tow vehicle with 0 alpha, V_in_plane is precisely V_travel. At alpha approaching pi/2, V_in_plane approaches infinity to maintain a component of velocity equal to V_travel. Force F_in_plane diminishes with larger alpha, while V_in_plane increases.

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> The main point is, in order to get the overall zero velocity component > relative to the road in the plane of the wheel (for a nonslip condition > in that direction, and therefore no corresponding friction force > component), the wheel doesn't need to rotate so as to fully match the > road speed, but only the component of the road velocity in the plane of > the wheel (i.e. V_travel * Cos(alpha)). This is where our discrepancy > comes from, I believe.

Yes it is; I'm fairly confident that V_in_plane has to increase with alpha. > I will not admit it, but I am a bit envious about your Cobra. The fact > that you cannot do much about the slide is that it is a rear wheel drive > thingy. I was talking about something far more dull. In 2" of slushy snow at walking speed, every change in speed or direction is a challenge. Brakes lock instantly, the rears spin at the slightest provocation, and the car just plows straight ahead when I turn the wheel. I've had trouble in flat parking lots where the surface was just packed snow, not even really ice. The tires, Firestone SZ50's, simply don't work in the snow. It might be a temperature thing; the tread is plenty aggressive, and work ok when the road is simply wet. > Have you ever seen any footage from any touring car championship > with front wheel drive cars? They can get into up to 90 degrees slides > (especially after bumps with other cars), and when in any RWD racing > series a spin is inevitable, these guys and gals just floor the gas and > beautifully recover. Looks almost impossible, but is quite obvious from > just simple physics. They are spectacles to behold. :-) I've been saying for three winters now that I need a second car for winter. I truly miss thrashing the parking lots in the early morning after a good snow. Michael.

Michael Young wrote: > > Gregor, I am a structural engineer by training and early vocation, but now > write (non-engineering) software for a living. This background predisposes > me to initially think in terms of non-moving systems, so please bear with > me. I thought I had an engineer on the other side of the net :). It brings back all those engineer/scientist jokes, in which the engineer is invaribly always better off. One wonders why.

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> > It's clear to me now that I had presumed the proof when I thought this > through. The free-body ignored a lateral force that might be present at the > point where the harness attaches to the tow vehicle. This being the whole > discussion, we would naturally come to quite different conclusions. :-) > > Let's try this again. > > Gregor Veble <[email hidden]> wrote in message > news:[email hidden]... > > > > road. At steady state, each infinitesimal element on the face of the > drum > > > has a component of velocity equal to the road speed . Thus, incredibly, > the > > > drum is actually turning faster than it would if the steering angle was > held > > > at zero. > > [Clarifying: each element on the drum has a component of velocity equal to > the road speed ( plus a perpendicular component). ] > > > > > Not true, see equation (3). The rotation is maximal when the wheel is > > pointing in the direction of the travel, and gradually goes to zero when > > alpha goes towards 90 degrees. > > Ignoring everything else for the time being, an infinitesimal tire element > in contact with the road is acted on by an impulse caused by the sliding > friction. At this point, we only know the velocity of the road surface, and > the wheel's steer angle. For a wheel not already rotating, the condition I'm > most comfortable with, the relative velocity is precisely the same as the > road speed. The component of the frictional force in the rotation plane adds > to the tire's rotation. The component parallel to the wheel axis is the > disputed lateral force. This defines one boundary condition. It appears that > we're done at this point, as there clearly is a non-zero lateral force, and > your point has already been made.

In the case of the nonrolling wheel, my point was actually that no lateral force exists. If you reread my previous post on this topic, the fact that the force component of the rotation plane produces a torque on the wheel axis thus starting to rotate the wheel does not mean this force is not felt by the wheel when considering its translational (as opposed to rotational) effects (i.e. the force felt by the harness). To use your choice of words, the force is not 'dissipated' for rotation. As the element at the bottom of the wheel that is in the contact with the road slides in the direction of the travel, the friction force is in parallel with the travel and no lateral force exists. It does not matter whether the tyre is forced to a nonspinning condition (i.e. locked) or is freely rolling but at the moment of contact its rotation is zero, in both cases the (initial) friction force has no lateral component.

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> > The other boundary condition, steady state rotation, is not so clear in my > mind, and I have a need to bring it back to first principles. Simplistically > speaking, the force resulting from sliding friction does not depend on > speed. That is, we expect it to be the same at 10 ft/sec as at 100 ft/sec. > However, frictional force is zero when the velocity is zero. Also, > superposition applies. At steady state, wheel rotation is neither increasing > nor decreasing, implying that the force in this direction is zero.

Correct, the component of the friction force in the plane of rotation is zero in the steady state. > Thus, > steady state rotation occurs when the velocity component on the face of the > wheel, in the plane of rotation, matches the road speed. This is incorrect. This is only true when the wheel is oriented in the direction of travel. Once the wheel is turned at an angle to the direction of travel, this then in fact contradicts the previous statement of the zero component of force in the rotation plane. Let me try to demonstrate this with the equations I posted earlier, V_in_plane = V_travel * Cos(alpha)-Omega * R , (1) V_perpendicular = V_travel * Sin(alpha) . (2) These equations show the velocity of the point at the bottom of the wheel with respect to the ground (i.e. the slip velocity) in the components of the wheel coordinate system (in and perpendicular to the plane of the wheel). Both parameters V_travel and alpha are controlled from the vehicle onto which the wheel is connected, the only free variable is Omega, the angular velocity of the wheel with radius R. I hope you agree that the above equations are always valid for the bottom point of the wheel under the conditions we put it to, no matter what the friction or any other force on the wheel, as they purely state the wheel dynamics without any kinematic effects. When you mention the velocity component on the face of the wheel, I guess you refer to the velocity of the point at the bottom of the wheel due to the rotation of the wheel, expressed in the system of the wheel (i.e. Omega * R in the above equations). The main point is, in order to get the overall zero velocity component relative to the road in the plane of the wheel (for a nonslip condition in that direction, and therefore no corresponding friction force component), the wheel doesn't need to rotate so as to fully match the road speed, but only the component of the road velocity in the plane of the wheel (i.e. V_travel * Cos(alpha)). This is where our discrepancy comes from, I believe. These equations also show what you stated earlier, namely that each part on the wheel has the component of velocity that is equal to the road speed (the part proportional to the V_travel), plus a NONperpendicular component proportional to the angular velocity of the wheel. The fact that these contributions are not perpendicular (in fact, when the wheel is parallel to the ground, they are also parallel for the point at the bottom of the wheel) means that the absolute value of the total velocity can actually be smaller than the road speed (is actualy 0 for alpha=0, as is expected for a fully nonslipping tyre).

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> There is also a > velocity component parallel to the wheel axis. There are two points to be > made about lateral velocity component. First, the magnitude of the resultant > of these vectors is clearly larger than that of the road speed alone. In > fact, it should become infinite when the wheel is steered perpendicular to > the direction of travel, the direct opposite of what you state. Second, I'm > having trouble this morning mapping across reference frames. I suddenly > can't distinguish if this resultant lateral force directly opposes that in > the first case above, or if it is the same. > > Gregor, since we're so far apart on this one point, I feel it pointless to > continue without coming to terms with the discrepancy. Help me, if you can, > find the flaw in the thinking. > > > Never did I say I have directional control in the snow slides, turning > > the wheel doesn't do diddly-squat then. Control can be had by pulling on > > the handbrake which locks the rear wheels and changes the direction of > > the force on the wheel from perpendicular to the wheel towards the > > direction opposite to travel, thus diminishing the lateral force and > > sending the tail further out. Or, in a front wheel drive car like mine, > > flooring it puts the direction of the friction force directly in the > > forward direction of the orientation of the tyres, which diminishes the > > lateral force on the fronts. Note that when in a slide, only the > > direction of the friction force can be changed, and not much can be done > > about its magnitude unless severe weight shifting is present. > > If you ask me, that sounds suspiciously like directional control. :-) All > kidding aside on this one; the Cobra just goes in whatever direction it was > already going, no matter what I do with my hands or feet, even at very slow > speeds. It has so far always recovered in time, but I try to avoid those > situations now. The situations being having to change speed or direction of > travel on snow covered roads. The only saving grace is that it takes a lot > of cussing and begging just to get it going in those conditions.

I will not admit it, but I am a bit envious about your Cobra. The fact that you cannot do much about the slide is that it is a rear wheel drive thingy. Have you ever seen any footage from any touring car championship with front wheel drive cars? They can get into up to 90 degrees slides (especially after bumps with other cars), and when in any RWD racing series a spin is inevitable, these guys and gals just floor the gas and beautifully recover. Looks almost impossible, but is quite obvious from just simple physics. Greetings, Gregor

At the point where only one tire is on the ground, the car is a unicycle. At this and all other times, the loads are transferred through the spindle arm. The spindle arm is in turn connected to the suspension, which moves in accordance with its geometry and the applied loading. Roll center might be a useful concept in calculating roll coupling and distribution, but is not a fundamental concept in itself. Again, Milliken's and Gillespie's books are good sources of information. Michael. Carol Y. Bekker <[email hidden]> wrote in message news:CBiu4.10675$55.61608@news2-hme0... > As a side question, the forces generated through the tire, how are they > transmtted to the vehicle. Is it through the roll center of thesuspension? > And if so what if only one tire is on the ground, is the roll center > calculated as the intersection of the CG center of the car, or do theforces > go through the intersection point of the aarms (if it has a arms). Ifanyone > has any insight on this I would love to know.

As a side question, the forces generated through the tire, how are they transmtted to the vehicle. Is it through the roll center of the suspension? And if so what if only one tire is on the ground, is the roll center calculated as the intersection of the CG center of the car, or do the forces go through the intersection point of the aarms (if it has a arms). If anyone has any insight on this I would love to know. Regards Carol

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> At the point where only one tire is on the ground, the car is a unicycle. At > this and all other times, the loads are transferred through the spindle arm. > The spindle arm is in turn connected to the suspension, which moves in > accordance with its geometry and the applied loading. Roll center might be a > useful concept in calculating roll coupling and distribution, but is not a > fundamental concept in itself. Again, Milliken's and Gillespie's books are > good sources of information. > Michael.

The "Zero Gee News" is running a 12 part series on "The Physics of Racing" by Brian Beckman. This month is the section on Tires, Check it out.. http://bovineracing.com/zgnews.htm in the Editorial section. Section one was on Weight Transfer and relates directly to the forces upon each tire, so.. The ZGN is a comic and serious paper for everything GPL and I understand that Real Racing may be added when the real racing season starts, but don't quote me. And I'd like to see more articles wrote up so if you have something worthy to be posted I'm sure the editors would be more then happy to post it for you. Mike Barlow