does lowering weight affect cornering speed?

[KnifeEdge_2K1]

lemme run this by you

ive just been working the numbers in my head and it seems that weight doesnt affect the cornering speed (assuming uniform cornering speed)

Fnet=MA where Fnet points towards center of the turn M being mass and A being acceleration

assuming constant speed acceleration is just direction so A = Ac which = (V^2)/R

so FR/M = V^2
now since F is the force of friction which = Miu(mg) the equation now reads

Miu(MG)R/M = V^2

now let my original weight be M, the above equation remains the same
if i lower my weight by 50% my weight is now .5M
miu(.5Mx9.8)xRadius/.5M = V^2

as you can see the masses cancel out so cornering speed remains the same

as long as 0<m<infinity

this relationship remains the same

amazing huh ?

[d-quik]

why...do you do this to yourself in your spare time :(

[6'bore]

I suppose a huge reduction in weight would affect the cornering speed, But small changes wouldn't, as the same forces are still being put onto the car and causing it to roll

[Coventrysucks]

Mass, and therefore weight, affects momentum, so while in theory cornering a fixed radius at fixed speeds would not be affected any attempt to change direction will depend on mass.

[ScionDriver]

Wow, you lost me.

[Matra et Alpine]

The force required to slide a tire is called the adhesive limit of the tyre. This law, in mathematical form, is F <= mu * W where F is the force with which the tyre resists sliding; mu is the coefficient of static friction or coefficient of adhesion; and W is the weight or vertical load on the tyre contact patch.

This equation states that the sideways force a tire can withstand before sliding is less than or equal to mu times W . Thus, mu*W is the maximum sideways force the tire can withstand and is equal to the stiction.

THUS W (= weight) DOES affect cornering by affecting the tyre grip

[CdocZ]

What about momentum? Momentum is affected by weight. The more weight, the harder it is to move your momentum's direction. So if a car weighs 10 tons, it will not handle as well as something that is the same shape but weighs 5 tons.

[Egg Nog]

The force required to slide a tire is called the adhesive limit of the tyre. This law, in mathematical form, is F <= mu * W where F is the force with which the tyre resists sliding; mu is the coefficient of static friction or coefficient of adhesion; and W is the weight or vertical load on the tyre contact patch.

This equation states that the sideways force a tire can withstand before sliding is less than or equal to mu times W . Thus, mu*W is the maximum sideways force the tire can withstand and is equal to the stiction.

THUS W (= weight) DOES affect cornering by affecting the tyre grip

That post was a brilliant work of friction.

[Matra et Alpine]

What about momentum? Momentum is affected by weight. The more weight, the harder it is to move your momentum's direction. So if a car weighs 10 tons, it will not handle as well as something that is the same shape but weighs 5 tons.
KE was keeping it simple and only using constant angular velocity.
It gets MUCH more complicated when you try to do all the equations for a car turning IN to a corner - or accelerating or decellerating. There were many excellent articles written on this in the 90s when lots of people were trying to write accurate computer sims.
NOW we're finally getting enough computer power at our desks where we can do it the RIGHT way. Which is to treat the cara as a collection of parts and for each part to obey the laws governing IT and the sim to determine what the car does by the summation of the interaction of it's parts :) Richard Brusn rally is the first to coem from completely that approach.


That post was a brilliant work of friction.
excellent ROFL :)

[KnifeEdge_2K1]

i also forgot to calculate weight transition

all i was taking into account was uniform speed (therefore no force acting on the tyre forward and back, like during acceleration) so all of the tyre's "grip" will be in the sideways direction

and static friction (when there is no tyre slip) is techically always higher then kenetic friction (when the tyre DOES slip) which is given by the equation Miu(mg) = fsmax

the acceleration of any object under going centripetal acceleration (no change in speed just direction of velocity) is given by Ac = (V^2)/R
and since its uniform motion Ac = A = Fnet/Mass and if we sub in Fnet using the grip equation Miu(MG) the masses cancel out

this works perfectly well in theory because as mass increases the grip from the tyres increase by the exact same ammount which means they cancel out

[KnifeEdge_2K1]

The force required to slide a tire is called the adhesive limit of the tyre. This law, in mathematical form, is F <= mu * W where F is the force with which the tyre resists sliding; mu is the coefficient of static friction or coefficient of adhesion; and W is the weight or vertical load on the tyre contact patch.

This equation states that the sideways force a tire can withstand before sliding is less than or equal to mu times W . Thus, mu*W is the maximum sideways force the tire can withstand and is equal to the stiction.

THUS W (= weight) DOES affect cornering by affecting the tyre grip

that part is true, but since Ac = V^2/R and Ac is given by Fnet/M and your equation F<= MuW the weights cancel out

and as you said when you're deccelerating or accelerating out of a corner my simple equation gets blown to hell

[KnifeEdge_2K1]

in simple words basically is that if you have a car with mass M, the grip it has is directly related to its M

but the acceleration is inversely proportional to M which means the two cancels out

[KnifeEdge_2K1]

now you may say that "ok KE lets assume you're right, why then do racing teams try to lighten their car as much as possible if cornering speeds remain constant"

well the answer is obvious, if you lower the weight, and power/torque remains constant you accelerate much faster, and since you have less weight it takes less energy to slow the car down improving strait line acceleration/speed, corner entry and exit speeds, and lowers the ammount of time needed to decellerate into the corner giving you a couple milliseconds every corner

and the reason why formula 1 and GT cars can corner at 4 and 3 gs respectively (at their best, normal cornering speeds are closer to 3 and 2.3 respectively) when theoretically the tyre's girp is limited to 1 g is the use of downforce which creates .. well more downforce then it does drag (which acts as a force limiting acceleration)

[Rockefella]

your example works in a theoretical environment, but not in real life. the coefficient of friction will go up regardless of the mass due to downforce, but the radial acceleration towards the center of the circle will remain dependent on the mass of the object, so yes, a lighter car will corner better
double posts ^, lock/ban. Here comes a warning for spam on a Saturday afternoon.
EDIT: a mod must have cleaned up or the guy deleted the second post. :(

[KnifeEdge_2K1]

your example works in a theoretical environment, but not in real life. the coefficient of friction will go up regardless of the mass due to downforce, but the radial acceleration towards the center of the circle will remain dependent on the mass of the object, so yes, a lighter car will corner better

im assuming no aerodynamic aid or any other forces other then the ones ive mentioned

im assuming your "downforce" is the same as my weight mass x gravity
and in the equation the mass variable is canceled out

[KnifeEdge_2K1]

note: coefficient of kenetic and static friction are constant in an object, it doesnt change ever (in the case of a tyre it depends on the tempreature and road surface but they really have nothing to do with our model so assume they remain constant)

[Matra et Alpine]

double posts ^, lock/ban. Here comes a warning for spam on a Saturday afternoon.
EDIT: a mod must have cleaned up or the guy deleted the second post. :(
He cleared it up himself AND as it is SOOO easy for UCP to double post then it should NEVER be a reason fro banning.

Mistakes happen sometimes - sometimes they even happen twice :cough: bush :cough: :) :) :)

[KnifeEdge_2K1]

watch out for that cough buddy ;)

[Egg Nog]

Mistakes happen sometimes - sometimes they even happen twice :cough: bush :cough: :) :) :)

If this thread goes completely haywire/off topic, you're totally the one to blame :D ;)

[KnifeEdge_2K1]

i cant believe there's been this many posts in liek 3 hours

[Matra et Alpine]

There's another thing on the grip in real -life and I can't remember how it's quantifies - except it's different for every car ( I usually do this be tweakign in the corner weights durign testing )

It's the weight trasnfer from the inner to the outer wheel.
It is NOT a linear process adn IIRC the difference between the upper and lower segments of the graph are different for different car weights. So a heavier car is penalised worse.

That is that as the weight trasnfers to the outer wheel, in a perfect world the increase in contact patch on the outer tyre that caises is matched by the decreas in the inner and the overal contact area for the car syats the same. Sadly it doesnt' work like that. at the bottom end of the line of weight trasnfer it is pretty much 1 for 1 , but as the load increases it's rate reduces.
http://www.ffcobra.com/FAQ/friction.jpg
This says, when weight transfer to the outside wheel, the grip on the outside wheel is increased, but not increase as much as the grip loss on the inside wheel.

As most car sim develoeprs find out, the world is a much more complex place than some of those early models portrayed :)

[Matra et Alpine]

i cant believe there's been this many posts in liek 3 hours
good mental stimulation :)

I'm annoyed that I can't remember the equation for the variation of the friction in a compliant item. There some differential that captures the way an object deforms to "bite" a surface - which a tyre does with more weight. And again in the real world that is missing from your equation. The inital equations work for non-deforming surfaces. Rubber IS a deformabel object but as I said, I cant' remember the theory on that.. have to make do with that graph :)

[KnifeEdge_2K1]

yup you're right again matra, this is why i said i ignored weight transfer

i think its something to do with camber change as the weight transfers, cuz when you camber to one side there's more weight on the one side of the contact patch on the other and the overall center of the contact patch is shifted to one side, and we all know the contact patch is not uniform and grip (coefficient of friction) changes accordingly

the thing is if we lowered the center of gravity this effect would be lowered also since weight transfer depends on acceleration and the location of cog above ground level

like matra has said, in a perfect world the change in frictional force would remain 1:1 with weight but alas this is not true

[Matra et Alpine]

yup you're right again matra, this is why i said i ignored weight transfer
or our heads explode :)

i think its something to do with camber change as the weight transfers,
No because as long as the tyres are 'normal' then the weight transfer increases the contact area, that part is linear or it's not obeying the laws of physics. Again however, in the real world the walls of the tyre come to play and their construction adn ..... I prefer to do this by trial and error on the track :)

It niggled me enough to fidn the core maths on the physics of friction.

"When either or both of the surfaces are polymers, Equation 3.1 no longer applies. Polymers deform viscoelastically: the deformation depends not only on the normal load N but also on the geometry and time of loading. With fixed geometry and duration of loading, the area of true contact is proportional to N^(2/3) for rubber [Lincoln, 1952]. "

So now we can plot that graph with scales :)

hands up if you DON'T know by now that I'm actually a geek :) :) ????

[KnifeEdge_2K1]

bigger geek then i obviously

but then again u prolly are liek twice my age :P so u are allowed to be twice as geeky

[KnifeEdge_2K1]

as long as i get this correct im fine

if you lower the weight of the car it should corner a bit faster because the tyres deform less right?

[Spastik_Roach]

I think hes about 3 times ur age KnifeEdge... no offence Peter :)

[Matra et Alpine]

If this thread goes completely haywire/off topic, you're totally the one to blame :D ;)
I'm applying for Australian citizenship :)

I'm UCPs CCOS


Clan Chieftan of Spam :)

[Matra et Alpine]

as long as i get this correct I'm fine

if you lower the weight of the car it should corner a bit faster because the tyres deform less right?
No the lower the weight the more ( I think ) you are playing in the bottom end of the graph and hence the more likely that the weight transfer has a lesser effect and hence you don't LOSE grip the way a heavier car would.
With a heavier car as the weight transfers it moves the car faster and further up onto the flat part of the graphs and the huge differential then comes into play.

My brains hurting - it comes with the senility :)

[KnifeEdge_2K1]

No, my downforce is different from your m*g.
let me explain.
the forces on an objectr that just sits on a table is the force due to gravity (mg), and the opposing force of the floor on the object, which keeps it from falling downward, called the normal force. when a car creates downforce, the force pushing it down can be stronger than m*g, thats why some cars, like the saleen s7 or mclaren or whatever can theoretically roll on a ceiling at high speed, because the force due to downforce is greater than the force due to m*g.

So, when a car is going around a radius on a track, the force radially on it is V^2/r * m, while the force pushing it down is downforce + M*g.

So, a lighter car will have less force pulling it radially around the corner, and will also have an easier time overcoming that force with as much downforce as necessary.

dude i know what a normal force is
but normal force is equal in magnitude to mg
and centripetal acceleration is V^2/r

so cornering force is equal to V^2(M)/r not r*m

and cornering force is equal to miu mg and the masses cancel out but as matra explained already this only happens in a perfect enviroment and when tyres arnt made of rubber

[RacingManiac]

Without aero download a car CAN exceed more than 1 G of lateral acceleration, as modern high performance(and racing) tire have Mu > 1. The construction of the pneumatic tires, their slip angle characteristic, and camber thrust will result in great force generated than the load applied.

[85RX7]

it really does depend if the weight is off blaance or not. If 80% of the weight is on one side of the car the effects would be terrable.

[Alastor]

In regards to rubber tires the relationship between normal load and lateral force is non-linear. Therefore, a constant coefficient of friction does not apply. Not only does the coefficient change with applied load, but the change is also non-linear. Generally, the coefficient of friction (Cf) is highest at one particular load but the exact shape of the Cf curve varies with tire construction.

Due to the non-linearity as the normal load is increased the resulting lateral load also increases but at a reduced rate. Below is an old post from another forum in which I did the same calculation demonstrating the effect of tire load sensitivity.

http://racingknowledge.org/topic.asp?TOPIC_ID=772

==========================

If the Cf is defined as:

Cf = Ff / Fn

where the fictional force (Ff) is the maximum force generated by the tire and the normal force (Fn) is the load on the tire. Rearranging the Cf equation to:

Ff = Cf * Fn

it can be seen that although the Cf is not constant and in fact decreasing, Ff can still continue to increase with an increase in Fn but at a reduced rate. In other words even though the coefficient of fiction has been reduced the maximum fictional force generated by the tire can still increase with an increase in load on the tire.

From the RCVD book the following data is presented in graph form for a P215/60 R15 Goodyear Eagle GT-S (shaved for racing) tire at a pressure 31 psi. The data plots are for lateral force (and normalized lateral force) vs. slip angle for a load of 900 lbs, 1350 lbs, and 1800 lbs. I have not included the corresponding slip angles.

Load (Fn) : Lateral Force (Ff) : Effective Cf

900 lbs : 1000 lbs : 1.11
1350 lbs : 1475 lbs : 1.09
1800 lbs : 1750 lbs : 0.97


So lets use this data to look at the performance of 4 different vehicles on a 300 ft diameter skidpad. For simplicity use the same approach that was taken in load transfer write-up, and idealize the vehicles. That is let them be perfect with 25% of load distrusted to each wheel, assume no load is transferred longitudinally, assume that all 4 four wheels reach their peak lateral force at the same time, and etc.

First, look at a car with a gross weight of 3600 lbs which corresponds to corner weights of 900 lbs. In this case we will assume that the car does not transfer load laterally, because of an insufficient amount of data but the point will be the same nonetheless. The table shows that for a load of 900 lbs a lateral force of 1000 lbs can be generated. From our assumptions we can say that the total vehicle lateral force is 4000 lbs. Using ‘F = m * a’ , we can divide 3600 lbs by 32.2 ft/s^2 and we get the vehicles mass to equal 111.8 slugs:

F = m * a , m = F / a

3600 lbs = m * 32.2 ft/s^2 , m = 3600 lbs / 32.2 ft/s^2 , m = 111.8 slugs

Now using ‘F = m * a’ again we can calculate the vehicles maximum lateral acceleration:

4000 lbs = 111.8 slugs * a2 , a2 = 35.8 ft/s^2

Next we can use the centripetal acceleration equation ‘a = V^2 / r’, to find the vehicles velocity:

a2 = V ^2 / r , V = sqrt ( a2 * r )

V = sqrt ( 35.8 ft/s^2 * 150 ft ) , V = 73.3 ft/s (49.9 MPH)

Secondly, lets perform the same calculations on vehicle with a gross weight of 5400 lbs. With a corner weight of 1350 lbs each tire can generate 1475 lbs of lateral force. Again assuming no longitudinal load transfer we find that:

m = 167.7 slugs
a2 = 35.2 ft/s^2
V = 72.6 ft/s (49.5 MPH)

Lastly, assume the gross weight to be 7200 lbs producing 1800 lbs of load on each tire. The result will be:

m = 223.6 slugs
a2 = 31.3 ft/s^2
V = 68.5 ft/s (46.7 MPH)

Summarizing the results we see:

Vehicle Weight : Cornering Velocity

3600 lbs : 49.9 MPH
5400 lbs : 49.5 MPH
7200 lbs : 46.7 MPH

So in terms of the skidpad test the heavy car indeed does well, although the lighter cars do better for this particular tire profile.

Finally, we can look at a limited load transfer example to see how the corning velocity would be effected. In the previous three example we assumed no lateral load transfer but for this case we will assume 450 lbs is transferred from the inside tires to the outside tires. If our vehicle gross weight is 5400 lbs this corresponds to a load of 900 lbs on the inside tires and 1800 lbs on the outside tires. The result is:

m = 167.7 slugs
a2 = 32.8 ft/s^2
V = 70.1 ft/s (47.8 MPH)

In other words from our data, calculations, and assumptions we find that lateral load transfer reduced our maximum cornering velocity by 1.7 MPH.
======================

[Alastor]

A good book on tire dynamics is the The Racing and High Performance Tire by Paul Haney. It not only covers the physics behind the pneumatic tire as they are currently understood but also basic vehicle dynamics.

An excerpt form the book on tire friction can be found here:

http://www.insideracingtechnology.com/tirebkexerpt1.htm

Overview:

The two (although there are believed to be more) primary components of tire friction are adhesion and deformation. Adhesion is the natural ability of rubber to stick or adhere to another surface. Deformation occurs as rubber is forced in between microscopic road irregularities.

http://img217.exs.cx/img217/9199/rubber0kt.jpg (http://www.imageshack.us)

The diagram above shows road irregularities deforming the rubber body. The dashed lines indicate the amount of deformation as the load is increased. It can be seen that as the load is increased deformation increases as does the contact area between the road and rubber. This increase in contact area allows for more adhesion. In dry conditions both adhesion and deformation contribute to grip. However, in wet conditions adhesion is significantly reduced leaving deformation as the predominate friction mechanism.

As the load on the tire is increased eventually the rubber will completely fill the spaces created by the road irregularities. This reduction in available irregularity volume as load increases results in the reduction of effective grip. Eventually, as the load becomes sufficient less deformation occurs and instead the rubber itself is compressed.