Car Physics Question from the Pub??
28 August 2003 | 11:37 AM
#1
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In the pub the other night a great debate erupted over the physics of a car sliding around a bend.
Group 1 Once sliding a car can only go round a bend at x MPH this speed is determined only by the dynamic friction of the tyres and the bend radius.
Group2 Argued that Mass is part of the equation to as a heavier car would not be able to go round as quickly.
Group 1 are insistent that mass does not come into it when you are in the realms of dynamic friction on the tyres.
Group 2 insist that a heavier car will require more friction from the tyres to go round at the same MPH to balance the forces.
What does scoobynet think?
Group 1 Once sliding a car can only go round a bend at x MPH this speed is determined only by the dynamic friction of the tyres and the bend radius.
Group2 Argued that Mass is part of the equation to as a heavier car would not be able to go round as quickly.
Group 1 are insistent that mass does not come into it when you are in the realms of dynamic friction on the tyres.
Group 2 insist that a heavier car will require more friction from the tyres to go round at the same MPH to balance the forces.
What does scoobynet think?
28 August 2003 | 12:02 PM
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Group 2 is correct.
In order to change the direction of an object a force must be applied to it
Newtons second law applies: F=ma
m is the mass of the object
Therefore mass is a factor
It is irrelevant whether the car is sliding or not
In order to change the direction of an object a force must be applied to it
Newtons second law applies: F=ma
m is the mass of the object
Therefore mass is a factor
It is irrelevant whether the car is sliding or not
28 August 2003 | 12:23 PM
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of course mass comes into it...
This is a simplification, but remember from school science or physics that acceleration is defined as a change in velocity, not just speed, and that velocity is speed with direction.
Cornering involves a change in velocity, so acceleration is involved, so in a Newtonian world, force=mass x acceleration... Hence the force that the tyres need to exert is directly proportional to mass of the car.
Frictional force is also dependent upon mass... The coefficient of friction for a surface is defined by the resistive force of friction divided by the force applied to the surface's normal. heavier car = bigger force applied to normal.
You could try and argue that frictional force being proportional to mass and force required being proportional to mass means you can cancel mass from both sides, but that doesn't work because the force available is distinctly non-linear... Furthermore, increasing the load on a tyre decreases the coefficient of friction.
This is a damn complex subject of which I know nothing. I can see why a little knowledge (F=MA and u=Fr/Fn, with Fn being directly related to mass) could cause the argument, but the problem with simplified models is in the very name of the model... It's simplified and doesn't work in all circumstances.
If Simon de Banke gets time he could write essays on how complex this whole subject is. Before you blink, he'll be mentioning slip angles and contact patches, then the age old argument of wider tyres give more grip will raise its ugly head and we'll be debating this for pages and pages and pages and pages and...
Cheers,
Nick.
This is a simplification, but remember from school science or physics that acceleration is defined as a change in velocity, not just speed, and that velocity is speed with direction.
Cornering involves a change in velocity, so acceleration is involved, so in a Newtonian world, force=mass x acceleration... Hence the force that the tyres need to exert is directly proportional to mass of the car.
Frictional force is also dependent upon mass... The coefficient of friction for a surface is defined by the resistive force of friction divided by the force applied to the surface's normal. heavier car = bigger force applied to normal.
You could try and argue that frictional force being proportional to mass and force required being proportional to mass means you can cancel mass from both sides, but that doesn't work because the force available is distinctly non-linear... Furthermore, increasing the load on a tyre decreases the coefficient of friction.
This is a damn complex subject of which I know nothing. I can see why a little knowledge (F=MA and u=Fr/Fn, with Fn being directly related to mass) could cause the argument, but the problem with simplified models is in the very name of the model... It's simplified and doesn't work in all circumstances.
If Simon de Banke gets time he could write essays on how complex this whole subject is. Before you blink, he'll be mentioning slip angles and contact patches, then the age old argument of wider tyres give more grip will raise its ugly head and we'll be debating this for pages and pages and pages and pages and...
Cheers,
Nick.
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28 August 2003 | 12:25 PM
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It is irrelevant whether the car is sliding or not
28 August 2003 | 01:18 PM
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agree with what Chiark has said. But need to add the below:
2 forces have to equal each other on the limit of slip -
The horizontal force which the tyres and weight of the car can produce - F = (mu) * N. mu = coeffient of friction = how sticky the tyres are, and N is the force pushing down (vertically) on the tyres (= ma = mass * acceleration (gravity!))
This force must be sufficient to overcoem the Centrifugal force which 'pull' you away from the centre of the bend. THis force =
mv2/r. V is the velocity, and R is the radius of the bend.
So mass cancels out. Cornering speed is only dependent on the radius of the curve, and the speed.
Remember this is on the limit of slipping when the forces are equal.
I'm remebering physics from 10yrs ago, so feel free to correct, but the theory is there.
edited to say, Group 1 were right
[Edited by Woz - 8/28/2003 1:19:34 PM]
2 forces have to equal each other on the limit of slip -
The horizontal force which the tyres and weight of the car can produce - F = (mu) * N. mu = coeffient of friction = how sticky the tyres are, and N is the force pushing down (vertically) on the tyres (= ma = mass * acceleration (gravity!))
This force must be sufficient to overcoem the Centrifugal force which 'pull' you away from the centre of the bend. THis force =
mv2/r. V is the velocity, and R is the radius of the bend.
So mass cancels out. Cornering speed is only dependent on the radius of the curve, and the speed.
Remember this is on the limit of slipping when the forces are equal.
I'm remebering physics from 10yrs ago, so feel free to correct, but the theory is there.
edited to say, Group 1 were right
[Edited by Woz - 8/28/2003 1:19:34 PM]
28 August 2003 | 01:28 PM
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Sorry to be nerdy but there is no such thing as 'Centrifugal force' the law is that of objects (mass) want to travel in straight lines, 'Centrifugal force' is just a convienient name given to it (not a real physics term) unfortunatly im not nerdy enough to actually answer the actual question 
Simon
Simon
28 August 2003 | 01:34 PM
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So mass cancels out
The frictional force available from a tyre is non-linear with respect to mass... The basic formula doesn't work in this instance as there's lots more variables necessary to model it with a degree of accuracy (mu * N just doesn't work for this)... hence as mass increases the force needed to accelerate (turn) the car increases linearly but the frictional force available is less than linear WRT mass. Hence slipping. Hence all sorts of things happening that further affect the friction...
Hence, all things being equal, heavier cars cannot corner as fast as light cars.
You've never seen a stripped out Civic Type-R (the old shape) at a track day embarassing seriously well sorted scoobs have you?
[Edited by chiark - 8/28/2003 1:39:39 PM]
28 August 2003 | 02:25 PM
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From: In the corner with a bottle of wine
You just can't leave it in the pub can you Nick?
Anyway i wasn't listeing to that drivel i was listening to Dickie and Kate on about........ /thinks better of it and shuts up/
Anyway i wasn't listeing to that drivel i was listening to Dickie and Kate on about........ /thinks better of it and shuts up/
28 August 2003 | 02:29 PM
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cheer Simon, I'll replace Centrifugal with Centripedal force (always mixed those up)
Nick, I agree the frictional force is not linear, but this is because the co-efficient of friction varies. The vertical force is mg, that is not disputed. The horizontal friction force is some non-linear co-efficient times the vertical force. Mass never changes, and gravity never changes (well it does, but not for this argument!). So only mu can vary.
The coefficent of friction is very complex - agreed, and is a function of compound, temperature, deformability among other factors. But i'm not an expert on this.
I still believe that mass can be cancelled in each formula, as mass is not variable in this case.
Back in the real world (everyone sighs in relief!) big cars don't handle as good as lighter cars - agreed. But I believe this is due to how the weight is tranferred in the vehicle, when cornering. Weight transfer deforms tyres, which alters their coefficient of friction i.e. use low profile tyres. Rigidity limits weight transfer e.g strut braces etc. but bigger forces need hugely stronger framework to give the same rigidity, and less weight transfer.
all the above comparison only works for cars that are coasting round corners, to remove the traction 4wd/fwd/rwd factor - totally
ignored in all of the above. One issue at a time eh!
edited to say, time to go back to the pub!
[Edited by Woz - 8/28/2003 2:32:08 PM]
Nick, I agree the frictional force is not linear, but this is because the co-efficient of friction varies. The vertical force is mg, that is not disputed. The horizontal friction force is some non-linear co-efficient times the vertical force. Mass never changes, and gravity never changes (well it does, but not for this argument!). So only mu can vary.
The coefficent of friction is very complex - agreed, and is a function of compound, temperature, deformability among other factors. But i'm not an expert on this.
I still believe that mass can be cancelled in each formula, as mass is not variable in this case.
Back in the real world (everyone sighs in relief!) big cars don't handle as good as lighter cars - agreed. But I believe this is due to how the weight is tranferred in the vehicle, when cornering. Weight transfer deforms tyres, which alters their coefficient of friction i.e. use low profile tyres. Rigidity limits weight transfer e.g strut braces etc. but bigger forces need hugely stronger framework to give the same rigidity, and less weight transfer.
all the above comparison only works for cars that are coasting round corners, to remove the traction 4wd/fwd/rwd factor - totally
ignored in all of the above. One issue at a time eh!
edited to say, time to go back to the pub!
[Edited by Woz - 8/28/2003 2:32:08 PM]
28 August 2003 | 02:55 PM
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One further correction to Woz's argument:
According to a recent article I read in New Scientist, I think...
Gravity is an illusion and does not exist.
The argument goes like this:
Nothing can escape a black hole, not even light which has no mass. Therefore gravity can not escape either. Therefore gravity can not exist.
However, humans use gravity as a physical model to correct for our inability to perceive the curvature of space time caused by massive objects.
Some scientists think that this may explain man's failure to factor gravity in to the unified field theory.
I'll get my coat............ see you in the pub. I just hope the curvature of space time keeps my beer from drifting away from the glass........
According to a recent article I read in New Scientist, I think...
Gravity is an illusion and does not exist.
The argument goes like this:
Nothing can escape a black hole, not even light which has no mass. Therefore gravity can not escape either. Therefore gravity can not exist.
However, humans use gravity as a physical model to correct for our inability to perceive the curvature of space time caused by massive objects.
Some scientists think that this may explain man's failure to factor gravity in to the unified field theory.
I'll get my coat............ see you in the pub. I just hope the curvature of space time keeps my beer from drifting away from the glass........
28 August 2003 | 02:59 PM
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Definitely a stunning one for the pub, isn't it? I did think it was an open and shut case.
Agreed the mass never changes, but friction is dependent upon weight (force) acting on the surface normal... Does the force acting perpendicularly on the tyre stay constant, because I don't *think* that it does. That might be wrong tho?
Again, I'm way out of my depth here but it's a good 'un to debate. The problem in my case is that a little knowledge is a dangerous thing, but also makes for an interesting diversion on a slow Thursday.
Cheers,
Nick.
[Edited by chiark - 8/28/2003 3:19:58 PM]
I did think it was an open and shut case.
Mass never changes
Again, I'm way out of my depth here but it's a good 'un to debate. The problem in my case is that a little knowledge is a dangerous thing, but also makes for an interesting diversion on a slow Thursday.
Cheers,
Nick.
[Edited by chiark - 8/28/2003 3:19:58 PM]
28 August 2003 | 03:28 PM
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Dr Nick, the words nail and head come to mind. Which nail, I have no idea, but u def hit one ;-)
Nick, I get what u are saying, and I agree with what u are saying.
But the total vert force is always going to be m*g. As Dr N explains g is a bit washy now, but lets go with Newton for a sec.
So this was the weight transfer thing I was not explaining well. So yes the vert force will change on each individual wheel, in the corner, but the 4 will always add up to m*g (both constant).
As we have to consider all forces on the body(car), vertically this will be mg.
conclusion....hoos round is it?
Nick, I get what u are saying, and I agree with what u are saying.
But the total vert force is always going to be m*g. As Dr N explains g is a bit washy now, but lets go with Newton for a sec.
So this was the weight transfer thing I was not explaining well. So yes the vert force will change on each individual wheel, in the corner, but the 4 will always add up to m*g (both constant).
As we have to consider all forces on the body(car), vertically this will be mg.
conclusion....hoos round is it?
28 August 2003 | 04:31 PM
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Hmmm, total force down is obviously going to total mg, but the frictional side of the equation isn't going to be neat enough to allow a straight cancellation of mass I reckon.
Basic friction eqn is f=um but this doesn't model into complex situations well. The case of considering a rapidly rotating band of a fancy blend of rubbers (which may or may not actually adhere to the road surface) which is constantly altering its attitude to the road surface, satisfies the condition of being complex methinks.
The main factor at play I think is (possibly) that we have four contact patches, which due to different loads placed on them by the car, slip at different rates to each other, which in turn alters the loads on the other contact patches as each patch does the grip/slip thing. So the mass loading is shifted constantly around the car.
So the less mass there is the less grief each of the tyres gets at the frictional interface in terms of load altering, which allows the tyre to remain at a higher grip level, more of the time, given the same tyres.
Of course some pillock will probably come along in a minute and esculate this thread into a 15 page monster by claiming that a heavier car pushes the tyres down harder so they grip more
Basic friction eqn is f=um but this doesn't model into complex situations well. The case of considering a rapidly rotating band of a fancy blend of rubbers (which may or may not actually adhere to the road surface) which is constantly altering its attitude to the road surface, satisfies the condition of being complex methinks.
The main factor at play I think is (possibly
) that we have four contact patches, which due to different loads placed on them by the car, slip at different rates to each other, which in turn alters the loads on the other contact patches as each patch does the grip/slip thing. So the mass loading is shifted constantly around the car. So the less mass there is the less grief each of the tyres gets at the frictional interface in terms of load altering, which allows the tyre to remain at a higher grip level, more of the time, given the same tyres.
Of course some pillock will probably come along in a minute and esculate this thread into a 15 page monster by claiming that a heavier car pushes the tyres down harder so they grip more
28 August 2003 | 04:40 PM
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From: "Comfortably Numb" since Aug 2003
Blimey!! what pubs do u guys drink in??? us mortals have problems deciding on crisp flavours to buy in our local let alone directional physics...
mind u we did win £50 on a Mozart opera question last nite, swings & roundabouts i guess...
mind u we did win £50 on a Mozart opera question last nite, swings & roundabouts i guess...
28 August 2003 | 05:00 PM
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Something else that was brought up was....
If a car locks its brakes and SKIDS to a halt its braking distance will be the same no matter how much weight is in the car...
All this because an Accident Investigator supposedly said that the maximum speed a car can round a corner whilst sliding. Is dependent only on the radius and properties of the tyre. And that the number of occupants and weight does not come into it.
Cobblers I say....
If a car locks its brakes and SKIDS to a halt its braking distance will be the same no matter how much weight is in the car...
All this because an Accident Investigator supposedly said that the maximum speed a car can round a corner whilst sliding. Is dependent only on the radius and properties of the tyre. And that the number of occupants and weight does not come into it.
Cobblers I say....
What a Thread!!