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Help Me Decide Between Long Range and Performance

sroy

Closed
Mar 13, 2021
542
227
New Jersey
I hear you, just wondering the relationship of mass and the coefficient of friction of the tire itself; is this linear? Exponential? If its linear and 1:1 proportional then the loss of coefficient of friction due to mass/load on the tire itself would negate the argument that weight doesn’t impact stopping distance.
That's the opposite. unless it's linear, the mass won't cancel out in the equation. It takes a greater force to stop decelerate as much a heavier vehicle and the higher the mass, the higher the friction henve the higher the force. There might be 2nd order mass corrections but the coefficient is negligible (unless you're an F1 or a massive loaded truck).
 

Mahamilto

Member
Feb 20, 2021
397
324
New York, NY
I’m basically arguing that if the loss of coefficient of friction in tires proportional to weight is 1:1, then the friction would be constant after all due to intrinsic properties of the tire itself.
 

sroy

Closed
Mar 13, 2021
542
227
New Jersey
I’m basically arguing that if the loss of coefficient of friction in tires proportional to weight is 1:1, then the friction would be constant after all due to intrinsic properties of the tire itself.
No. The friction scales linearly with mass.

F = ug = ma
a = ug/m

no mass left if the friction (u) is linear in mass.
 

Mahamilto

Member
Feb 20, 2021
397
324
New York, NY
Friction = coefficient x normal Force(based on massxgravity). So If the coefficient drops 1:1 linear on mass (asking, not stating) as in coefficient friction of tire = mass x (compound coefficient if unloaded), then this cancels out the loss of the friction based on having less normal more due to the absence of as much normal force.
 

sroy

Closed
Mar 13, 2021
542
227
New Jersey
Friction = coefficient x normal Force(based on massxgravity). So If the coefficient drops 1:1 linear on mass (asking, not stating) as in coefficient friction of tire = mass x (compound coefficient if unloaded), then this cancels out the loss of the friction based on having less normal more due to the absence of as much normal force.
Ok there's some syntaxical confusion. There is the friction force and the friction coefficient. The fiction force dictates the deceleration (negative acceleration). There is the aerodynamics at play as well but it's the same for a SR vs LR. If the mass is higher, the friction force is higher as the the friction coefficient of each car is the same and the friction force varies linearly with mass. So the higher the mass, the higher the force. but the acceleration is the same. The distance traveled depends only on velocity and acceleration, not mass. As result, the braking distance is the same because the respective - yet different for each car - forces are different proportional to mass differences.
 

Mahamilto

Member
Feb 20, 2021
397
324
New York, NY
Im speaking directly to the initial basis of the argument; the force of friction (only thing stopping the car) is 100% F = coefficient of friction x mg. No argument.

but the element that the coefficient of friction of an air filled tire is independent and inversely proportional to the mass/load on said tire, changes this entire equation. The reason the cars would stop at the same distance, from our original discussion, was because the friction force is different due to the normal force being less on a lighter car. And assuming the coefficient of friction (originally attributed to tire compound alone) was the only constant, this would mean that mass cancels out of the kinetics equation; the two cars stop in the same time and distance.

But if the coefficient of friction loss due to mass/load on a tire negates the mass effect on the normal force, suddenly the Force of friction IS constant, and mass remains in the kinetics equation.

F=ma
F=force friction
F friction = coeff x mg
coeff = compound coeffecient / masss of object loading tire (again... I don’t know if this is 1:1 and if you know it isn’t or it’s scaled this would change)

F friction = (compound tire coeff / mass of object) x mg

Force of friction in this instance then being only compound of tire x gravity

compound tire x gravity = mass x acceleration...

mass still would be a factor in stopping distance
 

Mahamilto

Member
Feb 20, 2021
397
324
New York, NY
Well that’s just it. Friction is NOT constant. Also the force in each case is different. The acceleration is the same which is why weight is not a factor. This cease to be true if you are skidding in which case the weight does play a role (lighter better). But under “ABS” braking, the tire quality is all that matters.
Ok there's some syntaxical confusion. There is the friction force and the friction coefficient. The fiction force dictates the deceleration (negative acceleration). There is the aerodynamics at play as well but it's the same for a SR vs LR. If the mass is higher, the friction force is higher as the the friction coefficient of each car is the same and the friction force varies linearly with mass. So the higher the mass, the higher the force. but the acceleration is the same. The distance traveled depends only on velocity and acceleration, not mass. As result, the braking distance is the same because the respective - yet different for each car - forces are different proportional to mass differences.
I’m saying if the friction coefficient becomes a variable, dependent on mass, then the friction coeffecient is NOT the same on both cars...
 

Mahamilto

Member
Feb 20, 2021
397
324
New York, NY
There’s no syntax error here... I’m simply asking if the decrease of the friction coeffecient due to the mass of the car is proportional to the increase of normal force due to the mass of the car.

If so, then the force of friction becomes constant, and the mass of the car again becomes a major factor in stopping distance.
 
Last edited:

Knightshade

Well-Known Member
Jul 31, 2017
12,841
19,047
NC
AFAIK the formula for stopping distance is:


d= V^2/2(u)(g)

Where
d= stopping distance (m)
v= velocity of the car (m/s)
u- coefficient of friction between tire and road (unitless)
g= acceleration due to gravity (9.8 m/s^2)

Nothing in there about mass of vehicle.
 
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Mahamilto

Member
Feb 20, 2021
397
324
New York, NY
AFAIK the formula for stopping distance is:


d= V^2/2(u)(g)

Where
d= stopping distance (m)
v= velocity of the car (m/s)
u- coefficient of friction between tire and road (unitless)
g= acceleration due to gravity (9.8 m/s^2)

Nothing in there about mass of vehicle.
Except that we have now established that u (coeffecient of friction between tire and road) decreases as mass of the car increases.

By how much, I have no idea, that’s what I’m asking.

“For rubber tyres on cars, the coefficient of friction (μ) decreases as the mass of the car increases. Additionally, μ depends on whether the wheels are locked or rolling during the braking, and a few more parameters such as rubber temperature (increases during the braking) and speed” from the article earlier cited
 

Mahamilto

Member
Feb 20, 2021
397
324
New York, NY
If the coeff of friction is no longer a constant, but a variable dependent on mass, mass renters the equation...

all I’m trying to garner is how much mass of the car influences the coeffecient of friction of a tire; we’ve established that it does, but not established by how much.
 

MikeyMikeD

Member
Jun 7, 2020
161
118
Ohio
I'm still trying to figure out what in the heck all of this has to do with deciding between LR and AWD...

I know weight is an issue, but to go from the physics discussion to geometry, we are now on quite the tangent 🤓
 

sroy

Closed
Mar 13, 2021
542
227
New Jersey
The rest of the conversation has gone right over my head, but this I can get onboard with.

Lighter cars stop more quickly. Yep, sounds correct.

Right, I’ll you guys get back on with your equations. 😁
Except this is false. Stopping distance for cars is only dependent on the tires. Not the mass of the car. The stopping distance is independent from mass.
 

MikeyMikeD

Member
Jun 7, 2020
161
118
Ohio
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