Does anyone know how to calculate a Friction/Traction Circle

[blank]

Coefficient of friction will be around 1.1-1.2 for a decent dry surface and normal road tyres.

You need the total vertical force on the tyre. Which is mostly just the corner weight. If it's something which generates a lot of down force you need to add that on.

To be honest the one you linked to is about right for a normal road car.

[blank]

It depends how deep you want to go. As the longitudanal friction for accelration and braking will be different to the lateral friction aswell.

It can get very complicated!

[MintSprint]

• You wouldn't calculate it, you'd measure it.
• In the olden days, when all this were green fields and PistonHeads was mainly populated by people with unreliable plastic penis extensions, the way to work out your maximum lateral grip was to mark out a big circle (preferably as a multiple of 32 feet, to make the sums easier) on some tarmac, then drive round it at ever increasing speed until you fell off. With a bit of calculation, you could then work out the cornering force in g. These days, the easy way to do it (especially since Tesco seems to object to people marking out cornering circles on their car parks) is to download an appropriate app onto your iPhone, then find an empty roundabout and try not to kill yourself or attract police attention.
• In reality, it's not a circle, it's a sort of funny asymmetric egg shape.

A perfect circle would imply that the maximum lateral force that the tyre can generate under steady state cornering is exactly the same as the maximum longitudinal force that it can generate under braking or acceleration, and that any combination of cornering/braking/acceleration forces would also add up to exactly the same figure. In reality, tyres just don't work like that.

The traction circle is a useful approximate analogy to help understand tyre behaviour, but it only bears a passing resemblance to the reality.

HTH

Edited by MintSprint on Saturday 6th December 15:16

[MintSprint]

maverickdrives said:

I can try and calculate the g's due to acceleration and braking but I still don't understand how to calculate the tyre limit/friction circle's radius?

It depends on speed. You can go round a tight radius, slowly, or you can go round a wide radius, fast.

The fact that you've (presumably?) managed to pass a driving test without grasping this is actually quite scary...

[MintSprint]

maverickdrives said:

I still don't understand how to calculate the tyre limit/friction circle's radius?

See above. There isn't a fixed radius.

Assuming you have data on the maximum cornering force that a McLaren is capable of generating (on a given set of tyres, on a road surface with a given coeffcient of friction, at a given temperature) then the best you would achieve would be to plot a graph of speed vs. radius of turn... because like I said the tighter the radius, the lower the speed will be (this is why you have to slow down more for tight corners ).

But even that would be a simplification, because in reality there isn't even a linear relationship between speed and cornering force.




edited to add: try asking this question in the Lounge, but tell them there's a conveyor belt involved... it'll keep them going for months.

Edited by MintSprint on Saturday 6th December 15:55

[Ozzie Osmond]

blank said:

You need the total vertical force on the tyre. Which is mostly just the corner weight. If it's something which generates a lot of down force you need to add that on.

Which is why cars turn corners better if the brakes are applied - weight is transferred to front tyres, and especially the one at the outside of the turn.

In other words, braking through a corner is fine. Just don't suddenly release the brakes half way round or you're likely to fly off the road.

[littleredrooster]

Ozzie Osmond said:

In other words, braking through a corner is fine.

Not entirely true if you're wanting to find the maximum lateral g. Braking (or accelerating) will generate a compound slip-angle.