Why Such Sudden Oversteer....?

[David]

Solid physics. I like it. I always try to explain polar moment to people, but its hard without math. Good examples.

There is another practical reason some cars are more prone to trailing throttle oversteer than others - gearing. Sports cars tend to have relatively high gearing (3.5+ rear ends) and that greatly increases the effect of lifitng the throttle while the suspension is already working hard.

[This message has been edited by David (edited 30 March 2002).]

 

[nsx4fun]

Time for a little more physics. As someone mentioned, yes, if it takes less energy to spin a vehicle with a low Moment of Inertia, then it also takes less energy to stop the spin. Now a little on the traction side. There are 2 coefficients of Friction, static and dynamic, and the difference is always a step function. In other words, one coefficient is higher than the other, and the transition is abrupt. And Static is always higher. So as you push the envelope, once you cross the border from static friction (aka no sliding) to dynamic friction (sliding), you just lost some friction and so you slip even easier. That is also why, the shortest stop you can make is the one where the brakes are on the verge of locking up, as soon as they lock you lose friction between the tire and the pavement and you skid farther.

 

[8000RPM]

I am impressed by the level of analytical discussion related to the physics behind the dynamics of motion presented in this thread.

Keep it up guys!

 

[Simon]

This physics talk is all very interesting espceially the points about the polar moment of inertia and the distribution of mass within the car. The word 'centroid' was also mentioned.

Unfortunately when a car turns a corner, the point about which the car rotates is not the centre of the car. i.e. halfway between the axles and centre of the car.

Has everyone forgotten that a car actually has four wheels! The point about which the car rotates is not a point in the centre of the car. There are actually two points - the front wheels and the rear wheels. Hence distributing mass over these points (i.e. engine over the front and gearbox over the rear axle) may actually improve the dynamics. The car is not even rotating at all ? The car only experiences a centripetal force.

In a spin, the point about which the car rotates would be closer to the centre of the car and in this situation would accept your points about this polar moment of inertia.

 

[sjs]

Originally posted by Simon:
...Has everyone forgotten that a car actually has four wheels!...

Well, not everyone! If you read my posts you would see that I mentioned it more than once. In my first post:

...That too is over simplified because a car has tires at the corners rather than a literal fulcrum at the center...

And in my most recent post:

The fact that a car has it's wheels at the extreme four corners (in terms of mass) seems like an important factor that hasn't been thoroughly incorporated into the analogies here. Can you help to analyze their impact on the overall physics?

So I agree that it seems like an important issue.


[This message has been edited by sjs (edited 30 March 2002).]

 

[Simon]

Apologies sjs.

sjs, I think we may have pointed out an important point that contradicts the physicists on this topic.

Waiting for a reply??

 

[Number9]

Originally posted by Simon:
This physics talk is all very interesting espceially the points about the polar moment of inertia and the distribution of mass within the car. The word 'centroid' was also mentioned.

Unfortunately when a car turns a corner, the point about which the car rotates is not the centre of the car. i.e. halfway between the axles and centre of the car.

Has everyone forgotten that a car actually has four wheels! The point about which the car rotates is not a point in the centre of the car. There are actually two points - the front wheels and the rear wheels. Hence distributing mass over these points (i.e. engine over the front and gearbox over the rear axle) may actually improve the dynamics. The car is not even rotating at all ? The car only experiences a centripetal force.

In a spin, the point about which the car rotates would be closer to the centre of the car and in this situation would accept your points about this polar moment of inertia.

Uhm, you haven't stated a coherent theory to argue against. If in a spin, assuming loss of wheel traction, it's incorrect to say that you would rotate closer to the center of the car (but rather the centroid which is not at the center of the car in an NSX). Not rotating at all? Then no centripetal acceleration either. Please articulate a better question and we math/physics types will be happy to oblige you.

--twc

99 NSX-Z SC (360 RWHP)
95 Dinan M3 SC (290 RWHP)
95 BMW M3 (Bone Stock)

[This message has been edited by Number9 (edited 01 April 2002).]

 

[sjs]

Originally posted by Number9:
Uhm, you haven't stated a coherent theory to argue against.

Uhm, I don't understand your confusion. The person asking the question was not attempting to state any theory, but rather wanted someone to examine those statements made by others and modify or replace them as necessary to take into account the presence of tires at the four corners. Real world problems are not always as clearly defined as those in a text book.

I’d summarize, but I’m out of time tonight.

 

[nsx4fun]

Mr. Simon et al,
Yes, when going around a corner, the center of the radius of the turn is somewhere off in the distance on the inside of the curve you are taking. But as soon as you lose control (aka go into a spin or sudden oversteer as this topic began), for the most part, your car is now sliding in a straight line and spinning like a top about the center of mass of the vehicle (more or less). To some extent this is oversimplified, but if you watch cars that lose traction in curves on the racetrack, you will see they pretty much stop turning around the curve and slide in a straight line eventually impacting the wall. Yes, the wheels are not equidistance from the center of mass, but I don't know that is a big factor in why the NSX seems to spin so quick and easy once traction is lost, whereas the Moment of Inertia is directly related to the ease of spin. If you want to try a little experiment, take 2 empty spools of thread(new lightweight styrofoam will be better thanold solid wood ones). Drill them both out with a 1/2 in drill. Now get some solder and wrap solder around the outside of one just below the lip of the spool so it still rolls on the spool edge and not the solder. Take an equal amount of solder and cram it into the center of the other spool. Now both spools have the same mass, but the distribution is different so the Moment of Inertia is different. Now side by side roll the 2 spools down a ramp. The spool with the mass in the center will roll faster than the spool with the mass on it's circumference.

Fritz

 

[Simon]

Fantastic, at least nsx4fun and sjs can understand what I was trying to explain. Being one of the engineering and analytical types it is sometimes a little difficult to put into words what the mind is thinking.

So the whole theory about polar moment of inertia and the benefits of having the mass near the centre of the car do not seem to be explainable by the above theories. It is a great theory if we are discussing which car may be best to do some donuts with or handbrake turns or when you spin off the track whose car will spin faster. Yes, agree.

Unfortunately where it matters is on the limits of adhesion - the old point of static friction as someone else pointed out. In that situation the car is not spinning at all. Merely experiencing a centripetal force as the car turns on an arc.

This can proabably be explained using a simple engineering 'statics' type analysis. Ahhh the old "If body A exerts a force on body B, then body B exerts an equal and opposite force on body A" theory. Lets assume the car is not braking or accelerating hence we will not consider the forces in that plane. Lets only look at the lateral forces that the car would experience.

There are two forces resisting the centripetal force in the opposite direction - an equal and opposite force. This force would be the front wheels and the rear wheels - two forces. The sum of these forces, ie the resultant force, is approximately in the centre of these two points and twice the value (roughly .. although with differnet size tyres, weight distribution etc etc it would vary between the force exerted by the front wheels and rear wheels.) Lets assume that the resultant force acts through the centre of the vehicle.

The other equal and opposite force is the centripetal force of the mass of the vehicle as it changes direction. Quite straightforward ... the force acts through the centre of mass of the vehicle. If the vehicle is say like a 911 this may be slightly towards the rear, whereas if it is front engined the mass may be slightly more inclined to the front - depending on the specifics of the vehicle etc etc.

If the car has an equal weight distribution between the front wheels and rear wheels, regardless of where this mass is distributed, then the centripetal force would act at the midpoint of the front and rear axle. Approximately the same point as the equal and opposite point force exerted by the tires.

The difference between where these two resultant points act in relation to the vehicle may result in a torque applied to the vehicle, ie - force x perpendicular distance to the fulcrum.

Thats about where my theory ends ... but some further points ... for further investigation ...

I wonder if this torque was acting in the right direction, then maybe it could balance out other factors :
i.e. if the car had insufficient front-end grip, then maybe this torque could balance the car back into a more neutral handling vehicle, or even lead to oversteer. Maybe its not due to the excessively hard swaybars or springs, but the distribution of mass within the vehicle. Definately a combintaion of many factors.

Guess at the end of the day it is a combination of the whole package - and two wrongs can make actually make a right. Correction, there probably are no wrongs - it is the whole package?

 

[Number9]

Originally posted by Simon:
If the car has an equal weight distribution between the front wheels and rear wheels, regardless of where this mass is distributed, then the centripetal force would act at the midpoint of the front and rear axle. Approximately the same point as the equal and opposite point force exerted by the tires.

This problem is actually a more complex than it appears. As you correctly noted, the tires are important here, and since this centripetal force is acting on the tires, the mass distribution, the relative loading and grip of the tires matter, so you can't make the simplifying assumption that the centripetal force acts at the midpoint. In the NSX case, we have rear bias mass distribution plus lateral weight transfer, etc.

So SJS, are you telling me I should drop out of school and live in the real world? No thanks!

[This message has been edited by Number9 (edited 03 April 2002).]

 

[Simon]

Number9 - where these two points acts was the kind of the important point of analysis. I think I said the centripetal force acts through the centre of the mass of the car - that would vary say in a 911, NSX, and Ford Falcon.

Was trying to dispute the fact that the reduced moment of inertia improves the handling of the car. If the mass was distributed evenly between the front and rear it would make little difference - until the car goes into a spin.

Although - after all of this, I note that the guys original posting was asking about the "sudden spin" - which would be related to the NSX's more centred mass - as opposed to say a 944 or 968 with the weight in the front and rear.

 

[Number9]

Originally posted by Simon:
Was trying to dispute the fact that the reduced moment of inertia improves the handling of the car. If the mass was distributed evenly between the front and rear it would make little difference - until the car goes into a spin.

The perceived handling improvement from a low polar moment car arises from the fact that for a given turn-in force (so to speak) at the front wheels, it will rotate quicker. Or to rotate at the same rate, you need less force. When you turn the car, even if you haven't lost traction, you are in fact rotating the car - for example, in a skidpad, the car would rotate once per loop.

 

[Simon]

Number9

Yes, in your example I certainly agree that the car would rotate one full turn relative to an fixed external point.

Unfortunately relative to the car and the road, the car does not rotate at all.

I think some of the other people above have also missed the theories of relativity in their explanation.

 

[Edo]

So basically you are saying that all cornering is based on centripetal motion and very little on Polar Inertia until you lose traction..ok. I think I understand. I am not disagreeing since I am no physicist by any means but can you answer this for me...

How can cornering be based in Centripetal motion if you are trying to change the arc of your centripetal motion via your tires? We're not talking about mainting static cornering as mentioned earlier but increasing or decreasing a corner mid corner.

The amount of force necessary to rotate a car with the tires, while cornering (Centripetal Motion) would differ greatly based on where the bulk of the mass is located correct?

A car with a heavy object in the front when cornering steadily (Like a ball on then end of a string) is not going to respond well steering input since the bulk of the mass sits up front. yes it will Rotate around the center regardless of where the mass lies, but when changing the arc of the corner mass does matter.Having a stick with a weight on either end, being swumg in an arc by the center of its length would rotate (In relation to its center of itself) much slower correct?

If the heavy object is in the back, it will rotate at the front quicker (Less work on the tires) and the back will move outward correct?

Which comes back to the whole concept I was originally explaining..I think. Now I'm confusing even myself.
Although a Car does not rotate around the center of its mass, since its pivot points are at the tires cornering would be near equal parts centripetal force AND polar intertia...especially since cornering is not too related to static "Skidpad" type forces.

From a Pure physics point I can understand how Polar intertia may have little to do with cornering, but at various points in a corner you are not just maintaining speed and cornering at a given rate. Its quite a bit more dynamic...

When you need to rotate the car inward, less mass near the edges is still better right?

Also someone mentioned that since the car has 4 tires, that the center of rotation is actually halfway between the axles and the tires...How can something have 2 rotation points? I don't understand...

Do I make any sense? Am I totally off base? Am I arguing a point that does not require arguing since you already agree??

Someone educate me!!!! I feel dumb.
(Frequent occurence when I am around other NSX owners)



[This message has been edited by Edo (edited 03 April 2002).]

 

[Number9]

Originally posted by Simon:
Unfortunately relative to the car and the road, the car does not rotate at all.

This is why you're having problems understanding the argument. For instance under your theory, in an S-shaped curve (negotiated properly), you would argue that the car hasn't rotated. I'd say that's a peculiar frame of reference.

I think you're probably conflating the overall path of the car and the (distinct) rotational motion of the car along that path. For example, you can have geostationary satellite that orbits the earth and points to a fixed spot on the ground at a fixed distance away or you can have a spinning satellite in orbit (hint: at least one of those is rotating)

Or to try another example, let's suppose I'm on the skidpad and doing perfect arcs with the rear trailing the front wheels perfectly. Based on your argument, no rotation. Let's say I'm on the same skidpad but now using various combinations of steering, brake and throttle inputs to introduce yaw by understeering and/or oversteering (i.e., I can point the nose to the left and right of the velocity vector of the car), but still negotiate the circle. Now even in your frame of reference, I must have rotated the car somewhere along that path.

I'm afraid I'm at a loss to make this any simpler, so I would invite Bob or Fritz to explain this more clearly.

 

[nsx4fun]

Simon, Edo, & Number 9
I'm confused by your posts and am unclear what question to answer, but I think I can at least point out some errors in thinking. Let's take a simple turn. You start the turn pointing North and travelling north. You finish the turn going east and facing east. 2 things have happened in space. The vehicle has changed it's direction of travel and it has rotated so it is pointing the same way it is travelling. The road has exerted a centripetal force (via static friction) through the tires to accomplish this change in direction. Now whether the front tires or the rear tires have more force applied is truly dependent on the location of the Center of mass with respect to the tires. If it is 1/2 way between the axles, then the fronts will have to exert more force to rotate the car, but as the center of mass moves backwards, the torque applied by the front tires increases (length of moment arm has increased) and vice versa for the rear. So to account for the change in moment arm, the force distribution for rotating the car will change, and I would need exact dimensions to calculate the ratio. Does this make sense, or am I confusing you as much as your descriptions have confused me?

Fritz

 

[Simon]

This really takes some thinking doesnt it? Each of your messages takes about 20 minutes to read & understand, let alone the time to put together a response!

If only I was Physics lecturer and could pass to my students as a tutorial

My first year university physics may not be up to this.

Give me a few days.

 

[Edo]

nsx4fun..
I understand what you are saying. It makes sense.

So basically what it comes back down to is that less mass in the front=less work for the front tires to turn the car.

 

[Hrant]

Folks, for those of us who simpletone Newtonians, we were doing fine following this thread and we really got excited but midway, you lost us with physics ......

.... and I almost flunked physics in high school ........

.... remember the kiss principle

 

[David]

Forget Newton. I still think the fundamental reason is that sports cars have numberically high rear ends. If you are loaded up in a corner, lifting the throttle instantly overwhelms the rear tires (exceeds the limits of static friction) and you break them loose. Instant oversteer. How the car behaves at that point is a physics lesson, but the primary force invloved in initiating it is mechanical.

Although I am enjoying a discussion where people actually understand tems like 'polar moment' and 'instaneous center.' ; )

 

[steveny]

I went to the BMW dealer last weekend and drove the new 745i. Maybe one of you physics majors can explain this. When I drove the car in a parking lot in very tight circles at 60-70 MPH there was no body roll at all. To top that off there was not even the smallest amount of force placed on my body either. I was sitting up straight in the seat without any leaning as if I were going down a highway at 55MPH. It seemed to negate the laws of physics. BTW the 745i has to be the most technolicaly advanced car I have ever seen or driven. It is way ahead of its time.

 

[nsxtasy]

Originally posted by steveny:
I went to the BMW dealer last weekend and drove the new 745i. Maybe one of you physics majors can explain this. When I drove the car in a parking lot in very tight circles at 60-70 MPH there was no body roll at all. To top that off there was not even the smallest amount of force placed on my body either. I was sitting up straight in the seat without any leaning as if I were going down a highway at 55MPH. It seemed to negate the laws of physics.

That doesn't make sense. I understand it has an active suspension, which has the capability of eliminating body roll entirely but they designed it to still leave a little bit of body roll in. However, as long as you're going fast enough in a circle that there should be centrifugal force acting laterally on you, the active suspension can't negate that, it only keeps the car body level. Hmmm...

 

[Michigan NSX]

Originally posted by nsxtasy:
However, as long as you're going fast enough in a circle that there should be centrifugal force acting laterally on you, the active suspension can't negate that, it only keeps the car body level. Hmmm...

Not if the inertial dampers have been engaged!

( a little Star Trek lingo there)

------------------
'91 Black/Black

 

[steveny]

I know it makes no sense at all. All I could think was maybe the seats are active as well to offset the g-force. I also did a couple of high speed jerks of the wheel on the highway. Still there was no force on my body what-so-ever.