YACT: If a car has an acceleration of 1G+, it accelerates faster then free-fall, right?

[Eli]

Originally posted by: LordMorpheus

Originally posted by: Hector13

Originally posted by: Imdmn04
1G is pretty fast accleration, a car that goes from 0 to 60 mph in 6 secs is only about .45G, if i calculated correctly.

not to mention that it is pretty much impossible for a car to reach 1g. It's been a while since I took physics or anything, but I am pretty sure that a car's max acceleration is limited by the friction between the car's wheel and the road.

Assuming a rubber tire has a coeffecient of friction around .9 or so and that half of the car's weight is over its rear tires (if it is rwd), than the max possible acceleration should be only .45g.

true, but if the coefficient of friction between the tires and the road is equal to or greater than 1, then an AWD vehicle could accelerate at one g without spinning tires (spinning tires decreases acceleration)

for a two wheel drive vehicle, assuming the weight is 50% front wheels, 50% back wheels, the coefficient of friction would have to be at least 2.

A cf of 1 is easy, 2 . . . maybe racing slicks on special pavement, but I dunno.


to see if it has a coefficent of on, put it on a 45 degree slope. If it doesn't slide down, cf>= 1. easy as that.

AWD vehicles can potentially accelerate twice as fast as anything else on the road . . . I wonder why big street ricers don't get an AWD subaru and drop some huge engine into it. It woudln't slip its tires until it was accelerating twice as fast as the little Honda that is now behind it . . . .

but you would need to mod the engine enough so that it was signifigantly more powerful. . . . Most cars rarely burn out, until you get to the drag racers Mustang Cobra types, although I still thing a track car would be better off with AWD. Mor expensive . . . but much . . . . better.

edit: Mass of a car has absolutely nothing to do with max acceleration without spinning the tires. friction varies directly with mass, but acceleration varies inversly. The two masses cancel. Work it out, if you don't believe me.

300HP and 300ftlbs of torque isn't enough for you? (WRX STi)

That's just stock though, I bet if you had the $$ you could double it.

 

[Hector13]

Originally posted by: Imdmn04
coeffiecienct of friction of 2? is that even possible? doesnt that mean the friction force would be twice the weight of the car? provided that the car is on a leveled track. what kinda accleration do u need to oppose this kinda of friction force?

friction=2mg
forward force has to be greater than 2mg (to make the netforce more than 0) for the car to accelerate, so ma>2mg, cancell out the m's,which yields to accleration of at least 2g in order for the thing to acclerate because there is such a big friction force, is that even possible for linear acceleration?, thats like having a car going from 0-60 faster than 1.4 secs.
and we arent even counting the drag force, which will be significant when the car reaches high speeds.

the nice thing about tires is that they roll. So while the c.f. may be 2, you don't need 2g of force to move forward as your tires will roll with you (ie, "rolling friction").

 

[WinkOsmosis]

Originally posted by: Howard

Originally posted by: Hector13

Originally posted by: Imdmn04
1G is pretty fast accleration, a car that goes from 0 to 60 mph in 6 secs is only about .45G, if i calculated correctly.

not to mention that it is pretty much impossible for a car to reach 1g. It's been a while since I took physics or anything, but I am pretty sure that a car's max acceleration is limited by the friction between the car's wheel and the road.

Assuming a rubber tire has a coeffecient of friction around .9 or so and that half of the car's weight is over its rear tires (if it is rwd), than the max possible acceleration should be only .45g.

Some Vipers and Corvettes have made it over 1G.

Maybe that's because their bodies, wings, etc, create downforce

 

[DrPizza]

I'll hit the accessories/calculator button and do the math...
60 m/h *5280 ft/m = 316800 feet per hour
316800 feet per hour divided by 3600 seconds per hour = 88 feet per second.

So, at 60mph, you are travelling 88 feet per second.
At constant acceleration, over 4.9 seconds (Bullhonkie's time above), 88ft/sec divided by 4.9 seconds = 18.0 ft/sec/sec

acceleration from gravity is around (just a hair under, and it depends on geographical location) 32.2 ft/sec/sec


I just want to ask, "who thinks that 18.0 is more than 32.2????"
Granted, the acceleration isn't constant, but that means that some of the time it would be less than 18, and some of the time it's have to be more than 1 1/2 times the average acceleration... I doubt that happens

To accelerate at 32.2 from 0 to 60mph, 88/32.2 = 2.73 seconds. So, you would have to go 0 to 60 in 2.73 seconds.

 

[silverpig]

Originally posted by: Hector13

Originally posted by: Imdmn04
coeffiecienct of friction of 2? is that even possible? doesnt that mean the friction force would be twice the weight of the car? provided that the car is on a leveled track. what kinda accleration do u need to oppose this kinda of friction force?

friction=2mg
forward force has to be greater than 2mg (to make the netforce more than 0) for the car to accelerate, so ma>2mg, cancell out the m's,which yields to accleration of at least 2g in order for the thing to acclerate because there is such a big friction force, is that even possible for linear acceleration?, thats like having a car going from 0-60 faster than 1.4 secs.
and we arent even counting the drag force, which will be significant when the car reaches high speeds.

the nice thing about tires is that they roll. So while the c.f. may be 2, you don't need 2g of force to move forward as your tires will roll with you (ie, "rolling friction").

That's the thing... because they roll, they are always using static friction against the road (the tires don't slip). This means the surface of the tire doesn't move w.r.t. the surface of the road.

 

[WinkOsmosis]

Originally posted by: silverpig

Originally posted by: Hector13

Originally posted by: Imdmn04
coeffiecienct of friction of 2? is that even possible? doesnt that mean the friction force would be twice the weight of the car? provided that the car is on a leveled track. what kinda accleration do u need to oppose this kinda of friction force?

friction=2mg
forward force has to be greater than 2mg (to make the netforce more than 0) for the car to accelerate, so ma>2mg, cancell out the m's,which yields to accleration of at least 2g in order for the thing to acclerate because there is such a big friction force, is that even possible for linear acceleration?, thats like having a car going from 0-60 faster than 1.4 secs.
and we arent even counting the drag force, which will be significant when the car reaches high speeds.

the nice thing about tires is that they roll. So while the c.f. may be 2, you don't need 2g of force to move forward as your tires will roll with you (ie, "rolling friction").

That's the thing... because they roll, they are always using static friction against the road (the tires don't slip). This means the surface of the tire doesn't move w.r.t. the surface of the road.

Static coefficient of friction can be higher than 1?

 

[DrPizza]

Heyyyyyyyyy, I know!! When they tested those cars to get 1.1G acceleration, they were going down a really steep hill. Hmmmmm, and since the weight of the car pushing against the surface of the road would be less than its weight on a level road.... (by the cosine of the angle of decent), they loaded up the car with a lot of rocks to make sure they had enough grip on the road (enough static friction between the tire and road surface). Yeahhhhh, that's how they did it.
Hey.... one more post to 1000

 

[silverpig]

Originally posted by: WinkOsmosis

Originally posted by: silverpig

Originally posted by: Hector13

Originally posted by: Imdmn04
coeffiecienct of friction of 2? is that even possible? doesnt that mean the friction force would be twice the weight of the car? provided that the car is on a leveled track. what kinda accleration do u need to oppose this kinda of friction force?

friction=2mg
forward force has to be greater than 2mg (to make the netforce more than 0) for the car to accelerate, so ma>2mg, cancell out the m's,which yields to accleration of at least 2g in order for the thing to acclerate because there is such a big friction force, is that even possible for linear acceleration?, thats like having a car going from 0-60 faster than 1.4 secs.
and we arent even counting the drag force, which will be significant when the car reaches high speeds.

the nice thing about tires is that they roll. So while the c.f. may be 2, you don't need 2g of force to move forward as your tires will roll with you (ie, "rolling friction").

That's the thing... because they roll, they are always using static friction against the road (the tires don't slip). This means the surface of the tire doesn't move w.r.t. the surface of the road.

Static coefficient of friction can be higher than 1?

Using certain materials/adhesives yes. When they test these cars they usually lay down a strip of rubber first, spin the tires to get them hot and sticky, and then run them.

 

[KenGr]

Originally posted by: DrPizza
Heyyyyyyyyy, I know!! When they tested those cars to get 1.1G acceleration, they were going down a really steep hill. Hmmmmm, and since the weight of the car pushing against the surface of the road would be less than its weight on a level road.... (by the cosine of the angle of decent), they loaded up the car with a lot of rocks to make sure they had enough grip on the road (enough static friction between the tire and road surface). Yeahhhhh, that's how they did it.
Hey.... one more post to 1000

I don't think this one should count.

 

[FracturedSoul]

As an FYI most prototype sportscars(Daytona Prototypes don't count) will produce on average 2-3G's under heavy breaking, acceleration or laterally from cornering. Of course there is always the combo of acceleration coupled with cornering that leads to sustained G's and different angles.

 

[LordMorpheus]

Originally posted by: WinkOsmosis
I'm still trying to figure out if a car's tires can have more friction force than 1G...

I told you before.

Car tires do not have a coefficient of friction.

The cf between the tires and the road is greater than 1 if you can put the tire on a 45 degree incline made outta 'road stuff' and not have it slide downwards.

here is why.

tire mass = m
acc. due to gravity = g
cf = u

normal force between tire and pavement = mg*sin(45)=mg/sqr(2)
this is also the force parallel to the surface

therefore, for the force parallel to the surface not cause it to accelerate, friction must be equal to that.

mg/sqr(2)=normal force * cf = mg/sqr(2) * u

not too hard to say: hmm, u must equal 1.

I know it is possible: I've tested it and seen my calculator do 60 degrees on top of my physyics book one day in class.

 

[HokieESM]

Originally posted by: WinkOsmosis

Static coefficient of friction can be higher than 1?

Well, the "coefficient of friction" is a VERY limited concept... its not exactly the physics behind what's happening. Useful in nice little "slide" tests.... but limited in situations with varying force or area or applied stress. The "slip/stick" effect also occurs at the microscopic level.

Can the coefficient of friction be higher than one? It depends on what you determine "friction". You stick things to your wall with stickers, right? Well, that's still "friction"... although, with an adhesive.

In a dynamic situation, such as a rolling tire, you have some really odd things going on--loading and unloading of multiple materials in the tire... incompressibility constraints on the rubber... tread blocks deforming.... so the "gross" coefficient of friction is higher than 1. Locally, actually, you can see SOME tread blocks sliding (meaning that they've broke static friction) and some tread blocks static... enough so that the wheel doesn't "break loose" overall. Note also, that its hard to define "sliding" and "staying still"--the tire is rotating also. So yes, cars can easily have overall coefficients of friction over 1--I've personally driven a car pulling 1.2 g's on the skidpad.

Also note that the F1 cars claiming 4 or 5 g's in lateral acceleration is very true (when you're looking at RACE conditions).... but you have to take into account the downforce as well. While they DO have good tires (with good "coefficients of friction"), you must also realize that there is frequently as much as twice the weight of the car in aerodynamic downforce when these measurements are recorded. Going around a 200' skidpad, I seriously doubt they could get that number--because the speed isn't high enough to generate those levels of downforce.

Also--F1 cars can break 1.0 g in linear acceleration for brief stints (at the power band in a low gear). And braking--even some street cars (like the Porsche 911 turbo) EASILY decelerate at over 1 g.