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#1 castercambertoe

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Posted 11 December 2003 - 21:57

Apologies for the simplistic nature of the question, but I'm wondering how the position of the CG along the car (fore and aft) affects handling. I assume a forward CG will cause understeer-is this correct? and is it due to saturating the tire as its producing Fy? Any help would be greatly appreciated-the more background info the better. Thanks all.

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#2 red300zx99

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Posted 12 December 2003 - 01:15

I will try to give the simplest definition I know. Imagine a long beam with a pivot in the center(CG). Now if you place a force a the pivot(CG) in order to equalise this force so that there is no yaw you would need 2 equal forces acting at the ends of the beam. Now if you move the pivot forward and apply a force to the pivot the force needed to keep the beam from rotating or yawing about the pivot would be greater in the front then in the back as we know that torque equals force * radius. Moving the CG forward would promote US because of this, and the front tires inability to produce a force proportionally larger in some cases. Sorry if I wasnt to clear in my explanation.

#3 RDV

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Posted 12 December 2003 - 16:46

Moving the CG forward would promote US because of this, and the front tires inability to produce a force proportionally larger in some cases. Sorry if I wasnt to clear in my explanation.



...I'll have to tell my tires this, they obviously don't know.... front weight distribuition can help turn in , by causing front tire quick turn-in response, as cornering force is dependent on Fz.

After the tires are saturated a forward CG will tend to promote understeer, all things being equal.

Do a thought experiment, and assume a car with 1% ft /99% rr.... you will have a great deal of difficulty turning in ... likewise , a car with 99%ft will turn on a dime... its only in steady state that this quote would apply, and in racing you are never in steady state , its all about transients.

#4 castercambertoe

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Posted 12 December 2003 - 18:16

Thank you both for the info. Both replies get at the heart of the difficulty I was having with this. I know that lateral force is a function of vertical load (among other things), and that the increase in lateral force is not linear with an increase in Fz, but I admit to being confused as to the effect the position of the CG might have. Thanks again guys.

#5 red300zx99

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Posted 12 December 2003 - 22:18

A racecar shouldn't stay intransients very long. If so you probably have a bad setup or a bad driver. A smooth driver keeps his car in steady-state for as long as possible, this is where the limit is

#6 Greg Locock

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Posted 12 December 2003 - 22:50

" A racecar shouldn't stay intransients very long. If so you probably have a bad setup or a bad driver. A smooth driver keeps his car in steady-state for as long as possible, this is where the limit is"

In the 1950s most F1 people would have agreed with you ie the job of the driver was to transition smoothly between a set of steady state accelerations.

However, Moss and others realised that the optimum is a continuous exploitation of the friction circle. The most efficient way to do this is to keep the dot moving, except when on a pure straight or a long constant radius corner. I think there is a lot about this in RCVD.

#7 red300zx99

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Posted 12 December 2003 - 23:07

Milliken is where i got the knowledge from. if u think about it every course is nothing but straights and constant radius turns. The driver doesn't always treat them as such. i must agree that transients in a straight are about all you are going to get given the fact that you are power limited and not traction limited, but for turns it's get to the steady state as fast as possible and stay there until not needed. Ahhhh,I think I am gettingur point now, may start a discussion over the American method and the European method, that would be a good one. I'm down

#8 Ben

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Posted 13 December 2003 - 11:52

Originally posted by red300zx99
A racecar shouldn't stay intransients very long. If so you probably have a bad setup or a bad driver. A smooth driver keeps his car in steady-state for as long as possible, this is where the limit is


Have you ever looked at a data trace from a race car? Steady-State is a useful approximation for understanding vehicle behaviour, (hence why I'm writing a steady-state lap time sim at the moment). However in reality the car is always transitioning between states.

As to the limit, if you're at the limit you should be transitioning around the limit of the g-g envelope which means you should be constantly trading lateral for longitudinal grip. This is most definitely not steady-state, but is very definitely the quickest way to drive a race car.

Ben

#9 RDV

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Posted 14 December 2003 - 14:35

may start a discussion over the American method and the European method, that would be a good one.



A good point, my emphasis on road courses and ovals where you would actually lift-off and re-accelerate... in the case of ovals such as Fontana and MIS you would be pretty much on a steady state for long periods as power limited on long, fast corners, but they are exceptions.

Oval racing has a dynamic of its own, due to long, big radius corners in relation to track width, where you are at a constant radius and speed , therefore steady state, maybe Buford could elaborate in more detail from his experience...

My experience of ovals limited to those two, Motegi and testing at Indy in Cart plus Charlotte, Rockingham(US not UK) and Bristol with Nascar.. where you definitely have transients for most of the time.. As Greg and Ben point out, if you are running on the edge of the G-G diagram ergo trading lateral for longitudinal accel & vice-versa, the data logging shows it distinctly....

#10 Paolo

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Posted 15 December 2003 - 13:17

Originally posted by RDV

Do a thought experiment, and assume a car with 1% ft /99% rr.... you will have a great deal of difficulty turning in ... likewise , a car with 99%ft will turn on a dime...


If you are talking about a RWD car with no or little downforce, probably this good "turn in" offered by front weight bias just depends on terminal oversteer caused by spinning rear tyres.
Are you sure it is really "turn in" and not "mid turn" ?
Or better, does the "bite" come when you leave the brake and reapply power ?

It's true one is seldom in a steady state in a car, but there is no abrupt steady state - transient state separation : a car in a moderately changing transient state will behave in a very similar way to a car in steady state.
And tyres "saturate" fast.

#11 RDV

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Posted 15 December 2003 - 16:02

I would postulate that there is no "true" steady state in road circuit racing, even in a straight line, as very seldom cars achive power limited top speed, but I'm niggling a bit here.

If you are talking about a RWD car with no or little downforce, probably this good "turn in" offered by front weight bias just depends on terminal oversteer caused by spinning rear tyres.



As its quarter to one AM local time , this would be too long and complex to go into detail Paolo, but from your previous posts I think you know enough to think it out.

Check out Pacejka plots of a typical tire to see raise in CF coupled to slip angles and Fz. To initiate turn in the sum total of loads ft and rr axle (Fx,Fy,Fz) mandate front tires to provide turning couple; after this has initiated then presumably we can assume spinning wheels will carry on to oversteer... but initially you must get front of car turning.... your example is more in the lines of cars in west-coast or Japanese "drifting' where rr wheels are spinning all the time, but to turn you still need aplication of lock to generate ft tire slip and so ft cornering force.

There seems to be an ideal balance for each track... in a fairly defined window, and really is more a function of tires than anything else. Optimal ft weight % has shifted on cars depending on tire technology dictated by development, rules or aero influence.

Or better, does the "bite" come when you leave the brake and reapply power ?



Rather the reverse in my experience; either on cars with a lot or little downforce, and even applicable to FWD. Use of brakes ( again assuming tires not saturated or locking) to transfer load to front wheels to help initiate turn in as applying lock quite usual, in fact understeer when 'off' brakes shows not enough Fzft for Fyft values to be >MftV^2 , thus understeer. (assuming not an aero effect due to rising of ft RH thus diminishing ft wing/difuser ground effect in case of non droop-limited cars, or limited slip locking giving understeer couple from rear axle).

We can go on at lenght but really time to sleep.. catch you later, tell me how you get on with thought experiment.

#12 Dmitriy_Guller

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Posted 16 December 2003 - 08:05

I'm not a professional, but here's how I view the weight distribution question. If you add weight with mass to one end of the car, the corresponding end would not work as well. If you add or transfer just weight on that end, then it would work better. For example, if you have 2000 lb car with 50-50 weight distribution, the front tires would have 1000 lbs on them, and the rear tires would have 1000 lbs on them. If you move CG back to 25-75 distribution, then there would be 1500 lbs on the rear tires. There is now 50% more mass supported by the rear tires, but that weight generates less than 50% more grip, if we assume less than linear weight to grip relationship. So, the rear tires have to generate 50% more force to keep the new car on the same arc, but since they don't, the rear of the car has to take a wider arc, and hence you got oversteer. You can reverse the numbers to get the understeer behavior.

However, if you take the 50-50 car and drive it into a corner, and touch the brakes a little, so now 1200 lbs are on the fronts and 800 lbs on the rears, you give your car extra front bite. That's because the front tires still support just 1000 lbs of mass, but now have 1200 lbs of weight on them, so they have extra grip that they do not spend to keep extra mass going on the arc. Therefore, you can use that extra grip to reduce the radius of the arc. You will also get extra front bite if you add a spoiler to the front of your car, to give your front tires more weight without adding more mass.

#13 RDV

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Posted 16 December 2003 - 08:17

Precisely... as long as we dont saturate tire, that is lateral force begins to tail off either due to overloading carcass, thus generating less side force, overheating compound, ditto, or sliding too much.

Note definition of over or under steer is difference between front slip angle and rear , and relative steering angle to maintain trajectory.

#14 Yelnats

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Posted 16 December 2003 - 12:08

Exactly Dmitriy_Guller.

Front weight transfer doesn't effect the center of gravity (or mass) and this is why the transient condition of weight transfer during deceleration for corner entry often assists with turn in. In effect you get a free ride for a few seconds as the weight on the front tires increases without ANY increase in inertial mass as no mass transfer has occured (except for a bit of fuel slosh).

#15 Cappy

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Posted 17 December 2003 - 23:06

Ratio of center of mass height to track:

This ratio of height divided by width determines the lateral (side to side) weight shift at a given acceleration. Coefficient of friction decreases with increasing contact pressure. The Inside tires' coefficient of friction increases during weight shift, but this has little effect since they support a smaller percentage of the car's weight. The outside tires' coefficient decrease during weight shift, and this decreases the car's lateral acceleration to a noticable degree since most of the weight is on the outside tires. If the cg height to track ratio is .5, there will be 100% weight on the outside tires during a 1 g lateral acceleration turn. 100% weight shift not only means reduced traction, but it also means a roll over is imminent.

Ratio of center of mass height to wheelbase:

This effects the weight shift during longitudinal acceleration. The center of mass may not be equidistant from the front and rear axles, so front-to-rear and rear-to-front weight shift may be different at equal amounts of acceleration.

Center of mass longitudinal location:

The front to back location of the center of mass effects the average pressure on the contact patches. If the contact patches are of equal area, a 50F/50R distribution will produce even pressure at the contact patches, for the greatest traction. A 50/50 static weight distribution is optimum when front and rear contact patches are of equal area and there is no longitudinal acceleration.

Yaw moment of inertia and wheelbase:

The average distance of mass from the center of yaw (turning) rotation affects how easily it will turn. If mass is distributed mostly at the nose and tail of the car, it will be more difficult to change rotational speed. Conversely, mass distributed near the center of the car (mid-engine) will hav a lower moment of inertia. Higher moment of inertia increases stability, negatively affecting turn in and transitioning ability. Longer wheelbase provides a more torque to rotate the car. but it sometimes means a longer more massive car (with higher moment of inertia).

#16 Greg Locock

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Posted 18 December 2003 - 04:51

It might be worth downloading the Bosch lap time simulator and having a play with CG position. The URL is given in a thread below with the obvious name. Switch expert mode on to play with some of the details of the setup, but I strongly advise you to make one change at a time..