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