Physics of traction
Discussion
I have come across two references to a theory of traction which I do agree with. One is here - http://www.ffcobra.com/FAQ/handling102.html - and another is in Speed Secrets by Ross Bentley.
The explanation says that loss of traction occurs because the traction:weight relationship is non-linear. So as you corner, more vertical load goes to the outside wheel, and this increases traction until a point where it declines and then the car slides. The Speed Secrets book says this is why adding a massive weight to the car won't increase cornering ability.
I do not agree with this. Firstly, the graph says 'Vertical Load'. If you put a big weight on the car and go round a corner, the centrifugal acceleration of the weight becomes lateral rather than vertical. For instance, if you put a six pack of beer cans on the roof and go round a corner, the cans do not fall straight to the floor they fly laterally.
So my understanding is this, if you put a big weight on the car, at a certain lateral acceleration speed the traction increasing vertical load becomes a traction decreasing lateral load. So that is why in principle a lighter car has increased cornering speed.
I agree with the graph in that vertical load increases friction, and that this relationship is non-linear. But this would only apply for instance with aerodynamic downforce, where there would be (in this explanation) no extra lateral acceleration just increased vertical load. But if you keep increasing downforce (which is probably impractical), the tyres will lose effectiveness and lose traction at a certain level regardless.
Thanks,
The explanation says that loss of traction occurs because the traction:weight relationship is non-linear. So as you corner, more vertical load goes to the outside wheel, and this increases traction until a point where it declines and then the car slides. The Speed Secrets book says this is why adding a massive weight to the car won't increase cornering ability.
I do not agree with this. Firstly, the graph says 'Vertical Load'. If you put a big weight on the car and go round a corner, the centrifugal acceleration of the weight becomes lateral rather than vertical. For instance, if you put a six pack of beer cans on the roof and go round a corner, the cans do not fall straight to the floor they fly laterally.
So my understanding is this, if you put a big weight on the car, at a certain lateral acceleration speed the traction increasing vertical load becomes a traction decreasing lateral load. So that is why in principle a lighter car has increased cornering speed.
I agree with the graph in that vertical load increases friction, and that this relationship is non-linear. But this would only apply for instance with aerodynamic downforce, where there would be (in this explanation) no extra lateral acceleration just increased vertical load. But if you keep increasing downforce (which is probably impractical), the tyres will lose effectiveness and lose traction at a certain level regardless.
Thanks,
Dac10 said:
If you put a big weight on the car and go round a corner, the centrifugal acceleration of the weight becomes lateral rather than vertical. For instance, if you put a six pack of beer cans on the roof and go round a corner, the cans do not fall straight to the floor they fly laterally.
Centrifugal force does not exist. The beer cans do not fly sideways, the car moves sideways (turns) and the beer continues in a straight line.Mr Will said:
Centrifugal force does not exist. The beer cans do not fly sideways, the car moves sideways (turns) and the beer continues in a straight line.
That is why I did not say 'force' but acceleration, I hope that is still accurate?What you say is what I mean, the beer cans do not suddenly shoot to the floor, they continue in a lateral direction of travel. Just like a sling sends a stone through the air rather than directly to the floor.
The simple schoolboy model of friction is sufficient to explain why increasing weight does not increase traction or lateral acceleration; the amount of force you can apply to the car through the friction of of the tyres is proportional to the mass of the car. The amount of acceleration you get for a given amount of force is inversely proportional to the mass of the car. The mass term cancels out, you don't need anything more complicated to explain it.
Have a look at 'mu-slip curves' and 'slip ratio'. The only real complexity in traction is modelling tyres. There are two sorts of grip, sticking friction (stiction) and interference friction from the tread pattern on a road. If you can work out how these two sorts of friction work for all load cases, temperatures, pressures, road surfaces etc, you will be a rich man!
Sorry if I'm missing your point, but from the beer cans point of view there is no difference between the car turning and the car vanishing from existence.
Starting from first principals an object will only change speed or direction if there is a force acting upon it. It does not need a constant force to keep moving. The beer cans continue to move forwards because there is nothing to slow them down apart from a very small amount of wind resistance.
Starting from first principals an object will only change speed or direction if there is a force acting upon it. It does not need a constant force to keep moving. The beer cans continue to move forwards because there is nothing to slow them down apart from a very small amount of wind resistance.
alephnull said:
Have a look at 'mu-slip curves' and 'slip ratio'. The only real complexity in traction is modelling tyres. There are two sorts of grip, sticking friction (stiction) and interference friction from the tread pattern on a road. If you can work out how these two sorts of friction work for all load cases, temperatures, pressures, road surfaces etc, you will be a rich man!
I am aware of slip angle and ratio, but that does not explain the theory of why a weight on the car does not increase lateral acceleration. Dac10 said:
otolith said:
The amount of acceleration you get for a given amount of force is inversely proportional to the mass of the car.
So the more weight you add, the less friction available to meet the acceleration of the force?Mr Will said:
Dac10 said:
otolith said:
The amount of acceleration you get for a given amount of force is inversely proportional to the mass of the car.
So the more weight you add, the less friction available to meet the acceleration of the force?Dac10 said:
Mr Will said:
Dac10 said:
otolith said:
The amount of acceleration you get for a given amount of force is inversely proportional to the mass of the car.
So the more weight you add, the less friction available to meet the acceleration of the force?The downward force on the car is it's mass*gravity. Gravity doesn't suddenly start pulling sideways so the downward force will not change when the car is turning, and hence neither does the amount of friction "available".
Dac10 said:
Mr Will said:
Centrifugal force does not exist. The beer cans do not fly sideways, the car moves sideways (turns) and the beer continues in a straight line.
That is why I did not say 'force' but acceleration, I hope that is still accurate?What you say is what I mean, the beer cans do not suddenly shoot to the floor, they continue in a lateral direction of travel. Just like a sling sends a stone through the air rather than directly to the floor.
The reason the beer can flies off is that the force applied to it by the laterally accelerating vehicle exceeds the amount of friction available in the interface between the can and the roof.
As said, basically tyre grip increases with verticle force (be it static weight or transferred weight), but does so in a non linear fashion, tailing off as load increases until it becomes practically saturated, at which point it doesn't matter how much more weight you put on it, the friction available wont change significantly.
Dac10 said:
Mr Will said:
Dac10 said:
otolith said:
The amount of acceleration you get for a given amount of force is inversely proportional to the mass of the car.
So the more weight you add, the less friction available to meet the acceleration of the force?The holy grail equation is Force = Mass x Acceleration
Say you have a car that weighs 2 tonnes, and a car that weighs 1 tonne, both with identicle tyre characteristics.
In a given corner the heavy car can apply a maximum of 20,000 Newtons of force through it's tyres.
The light car can apply a maximum of 15,000 Newtons of force, less because the static load on the tyre is lower.
To produce a 10m/s/s acceleration (near enough 1G) around the corner, the heavy car requires 20,000N to be transmitted through the tyres, i.e. the limit of grip. The light car however only requires 10,000N, so there's still 5,000N left in the bag.
That make sense?
Edited by The Wookie on Tuesday 13th April 11:07
Mr Will said:
Dac10 said:
If you put a big weight on the car and go round a corner, the centrifugal acceleration of the weight becomes lateral rather than vertical. For instance, if you put a six pack of beer cans on the roof and go round a corner, the cans do not fall straight to the floor they fly laterally.
Centrifugal force does not exist. The beer cans do not fly sideways, the car moves sideways (turns) and the beer continues in a straight line.Don't entirely understand the question, but here's a few things to help:
1) The lateral force a tyre can generate does indeed increase with vertical load, at least up until a point (tyre dynamics are hugely complicated, when you consider 50+ variables in the mix...)
2) You can think of the centripetal acceleration acting on the car as acting purely on a point mass at the car's centre of gravity. This generates a moment which causes the outside tyres to load up. For a given roll stiffness and C of G/roll centre height more lateral acceleration will equal a greater weight transfer to the outside wheel
3) Adding additional mass to the car would increase the lateral force the tyre could generate. However it would also produce an increase in the centripetal force opposing the car (I seem toi remember the nett improvement is zero?). You'd also have to make the extra mass speed up, slow down and change direction, plus deal with it working the tyres harder and causing them to degrade.
ETA: If you want more info there's a great article by Pat Symonds about the physics of tyres in this month's Race Tech. I think it's fair to say he knows a lot more about the subject than I do.
1) The lateral force a tyre can generate does indeed increase with vertical load, at least up until a point (tyre dynamics are hugely complicated, when you consider 50+ variables in the mix...)
2) You can think of the centripetal acceleration acting on the car as acting purely on a point mass at the car's centre of gravity. This generates a moment which causes the outside tyres to load up. For a given roll stiffness and C of G/roll centre height more lateral acceleration will equal a greater weight transfer to the outside wheel
3) Adding additional mass to the car would increase the lateral force the tyre could generate. However it would also produce an increase in the centripetal force opposing the car (I seem toi remember the nett improvement is zero?). You'd also have to make the extra mass speed up, slow down and change direction, plus deal with it working the tyres harder and causing them to degrade.
ETA: If you want more info there's a great article by Pat Symonds about the physics of tyres in this month's Race Tech. I think it's fair to say he knows a lot more about the subject than I do.
Edited by Chris71 on Tuesday 13th April 12:20
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