18 replies to this topic

[murpia]

I was wondering what is the correct approach to modelling the power lost in a tyre when driving?

Tyres are always more complicated than you think, so it seems naive to assume they work like a clutch whereby the efficiency of power transfer is equal to the slip ratio.

Also, if by performing a coastdown one can approximately measure the rolling resistance of an undriven tyre, how should the driven case be handled? Is the tyre 'driving' against it's own rolling resistance and losing power that way? Is that equivalent to the power loss of a driving tyre?

To the best of my memory I have not seen this treated in any textbook I know of, but if it is I would appreciate a reference.

Thanks, Ian

[Charles E Taylor]

Tyre (Tire) Rolling Resistance.

Not Racing Cars, but This is a great site for a better understanding of this issues involved.



Charlie

[munks]

Also, if by performing a coastdown one can approximately measure the rolling resistance of an undriven tyre, how should the driven case be handled? Is the tyre 'driving' against it's own rolling resistance and losing power that way? Is that equivalent to the power loss of a driving tyre?

To the best of my memory I have not seen this treated in any textbook I know of, but if it is I would appreciate a reference.

Thanks, Ian

I've certainly seen graphs of rolling resistance vs. traction/braking, I believe in Wong's "Theory of Ground Vehicles" (a pretty good book, but I don't have it in front of me at the moment). IIRC, both traction and braking increase the rolling resistance, presumably because there's more deformation happening in the carcass and tread. Probably more squirming in the contact patch, too.

Edited by munks, 21 July 2011 - 18:37.

[pugfan]

I was wondering what is the correct approach to modelling the power lost in a tyre when driving?

I don't know about the correct approach but this is the first order approach I would take:

The rim is driven by a torque (Don't you start) at a certain rotational speed. The tyre exerts a force on the road, which should be able to be calculated from the torque, at a given speed.

Calculate the power for both which will be different (use longitudinal tyre slip as the parameter) and subtract the power at the rim from the power at the road to find the power going into the tyre.

[gruntguru]

How would you express longtitudinal slip? The standard measure is tyre circumferential speed/road speed but tyre radius is greater at the unloaded areas away from the contact patch so the standard measure exaggerates the actual amount of slip ocurring in the contact patch. Of course the amount of slip varies with location in the contact patch too.

[murpia]

How would you express longtitudinal slip? The standard measure is tyre circumferential speed/road speed but tyre radius is greater at the unloaded areas away from the contact patch so the standard measure exaggerates the actual amount of slip ocurring in the contact patch. Of course the amount of slip varies with location in the contact patch too.

I think this is where I need guidance...

The conventional longitudinal slip might not be the right value to use in my 'clutch slip' analogy.

An approach could be:

Calculate driveshaft torque * driveshaft speed = driveshaft power.
Apply driveshaft torque at a 'particular' radius to get contact patch force on the road.
Contact patch force * road speed = road power.
Difference between the two = power loss in tyre.

So the question is, how are the 'particular' radius, the circumferential speed and the road speed related? Is any of those equivalent to 'slip ratio'?

Regards, Ian

[desmo]

Maybe one could do it more simply by measuring the tire temperature as it is run on a dyno and calculating from that? Power loss should equal heat above ambient.

[Lee Nicolle]

Wider tyre = more drag but generally increases cornering force

Larger diameter= more drag and more power to turn it, but a longer contact patch makes more forward bite.

Stiff case= generally equals less rolling resistance and far less ride quality.

Low profile= generally better handling,, on smooth bitumen roads. Worse rough road ability yet alone dirt and ride quality is severely compromised, as is tyre and rim [and suspension] life from kerbs, potholes, etc.

So if you want a decent riding car with respectable handling buy the standard model car. If you want poor ride, good smooth road performance buy the performance models.

If you want terrible ride, blowouts every month, constant suspension damage, bent, out of round wheels etc buy the kiddy car with 20" + wheels that are about 12" wide and go slow on a racetrack but look COOL !!

A lot of power and torque is sucked up by the larger, heavier, wider tyres. What works on a racetrack may still be exceeded in lap times by a smaller tyre that lets the car go faster in a straight line at least for a short period. But a widish tyre of a bigger diameter is generally more consistent. Generally puts the power down better and depending on aspect ratio generally steers and brakes better too.

Sprintcar tyres are an example of width and diameter. And soft flexible cases that wind up making an ever greater contact patch. But consume more than a bit of the 800hp they are transmitting. Same tyre on a 400hp car are a total waste of power, similar construction but smaller diameter and width.

On the other end of the extreme is a fuel economy car with very light rims and tyres, very hard cases and air pressure, very narrow tyres that roll very easily.These tend to be larger diameter than could be expected though as they roll over bumps etc far better offsetting the circumfrence issue. Compare a castor with a bicycle tyre, one rolls over most reasonable surface, one drags on every tiny bump.

[Lee Nicolle]

Wider tyre = more drag but generally increases cornering force

Larger diameter= more drag and more power to turn it, but a longer contact patch makes more forward bite.

Stiff case= generally equals less rolling resistance and far less ride quality.

Low profile= generally better handling,, on smooth bitumen roads. Worse rough road ability yet alone dirt and ride quality is severely compromised, as is tyre and rim [and suspension] life from kerbs, potholes, etc.

So if you want a decent riding car with respectable handling buy the standard model car. If you want poor ride, good smooth road performance buy the performance models.

If you want terrible ride, blowouts every month, constant suspension damage, bent, out of round wheels etc buy the kiddy car with 20" + wheels that are about 12" wide and go slow on a racetrack but look COOL !!

A lot of power and torque is sucked up by the larger, heavier, wider tyres. What works on a racetrack may still be exceeded in lap times by a smaller tyre that lets the car go faster in a straight line at least for a short period. But a widish tyre of a bigger diameter is generally more consistent. Generally puts the power down better and depending on aspect ratio generally steers and brakes better too.

Sprintcar tyres are an example of width and diameter. And soft flexible cases that wind up making an ever greater contact patch. But consume more than a bit of the 800hp they are transmitting. Same tyre on a 400hp car are a total waste of power, similar construction but smaller diameter and width.

On the other end of the extreme is a fuel economy car with very light rims and tyres, very hard cases and air pressure, very narrow tyres that roll very easily.These tend to be larger diameter than could be expected though as they roll over bumps etc far better offsetting the circumfrence issue. Compare a castor with a bicycle tyre, one rolls over most reasonable surface, one drags on every tiny bump.

[Greg Locock]

The contact patch power is the dot product of the horizontal contact patch forces and the horizontal contact patch slip velocity.

[munks]

The contact patch power is the dot product of the horizontal contact patch forces and the horizontal contact patch slip velocity.

Assuming you know the effective rolling radius (which you would need in order to calculate the slip velocity), is this pretty accurate? Or does it get messed up significantly by the distribution of forces and slip velocities within the contact patch?

[gruntguru]

I will pre-empt Greg on the assumption he may be unavailable for a while.
His formula is exact.
Effective rolloing radius is pretty close to true axle height for small camber angles.

[Greg Locock]

It is exact. It may well be wrong.

Couple of thought starters.

a) the rate of energy transfer to the vehicle is related to the contact patches velocity, not the SLIP velocity.

b) why can we ignore the vertical component?

So I think the dot product equation tells you about the local heating effect at the CP, but may mislead you if you are attempting to work out the total drag from that tire/suspension.

[gruntguru]

Can't see why. As soon as you posted it I thought "Ah Ha".

A) Power propelling the vehicle = tractive force x vehicle velocity
B) Total power transferred through wheel assy = hub torque x hub angular velocity
= (tractive force x rolling radius) x (wheel velocity at RR / RR)
= tractive force x wheel velocity at RR

Loss = B - A = tractive force x (Wheel velocity at RR - vehicle velocity)
= tractive force x slip velocity

[murpia]

Just re-joining the thread after a holiday...

Thanks for the reponses, I'm pretty happy with the concept of how this can be done exactly provided the effective rolling radius can be calculated or measured.

However, I have also realised I may be asking the wrong questions...

The problem I am looking at is to model all sources of vehicle losses. The correlation is to be done against coastdown data which should allow an estimate of tyre rolling resistance losses to be separated from the other road load loss sources. However, in a coastdown the driving wheels are also coasting down and contributing to rolling resistance losses. But, when these wheels are driving they won't contribute to the road load losses but instead will lose energy according to the contact patch slip as discussed above.

So, the question should really be, how best to process coastdown data to account for the fact the driving wheels are also coasting down, so as to subtract that element from the driving road load?

Put another way, I would expect the road load loss model to be different for the driving / coasting cases, so how to generate this?

Thanks, Ian

[cheapracer]

Just re-joining the thread after a holiday...

Hmmm, Ian comes back and Greg goes on holiday and I have never seen the 2 together in the same room ....

[gruntguru]

So, the question should really be, how best to process coastdown data to account for the fact the driving wheels are also coasting down, so as to subtract that element from the driving road load?

Put another way, I would expect the road load loss model to be different for the driving / coasting cases, so how to generate this?

If you were lucky enough for the vehicle to have 50:50 weight distribution and equal tyre pressures it would be reasonable to assume each tyre contributes equally to coast down losses. No doubt there are formulae for estimating the effect of varying load and tyre pressure on coasting tyre losses. If your coast down data gives an accurate total tyre loss, the inaccuracy present in such formulae would be less detrimental than otherwise.

[murpia]

If you were lucky enough for the vehicle to have 50:50 weight distribution and equal tyre pressures it would be reasonable to assume each tyre contributes equally to coast down losses. No doubt there are formulae for estimating the effect of varying load and tyre pressure on coasting tyre losses. If your coast down data gives an accurate total tyre loss, the inaccuracy present in such formulae would be less detrimental than otherwise.

From the (brief) search I have just done, tyre rolling resistance seems usually to be expressed as a coefficient of tyre normal force (i.e. it varies linearly with tyre normal force). Therefore even if the vehicle doesn't have 50:50 weight distribution, the relative driven and driving axle tyre rolling resistances could be worked out, if the same tyre construction & inflation pressure is used on both axles.

This formulation seems to appear in certain versions of the Pacejka formulae and is presumably measured on a tyre test rig?

Also, there is a linear region of the tyre longitudinal force vs. longitudinal slip curve, which is where I can assume my modelled vehicle is operating as I'm not looking at a traction limit case but a general 'driving around' case. However, I cannot (yet) track down any coefficient for tyre normal force affecting the tyre longitudinal force vs. longitudinal slip curve. Does this exist?


It may be that data exists that might assist with understanding the relative magnitudes of these 2 effects? Maybe an assumption could be made that the tyre power loss when driving in the linear region is a certain proportion of the rolling resistance of the same tyre when coasting? And if that proportion was close to 1:1 then in fact the 2 effects are self-compensating?

Wishful thinking I know but data might support it...

Thanks, Ian

[RDV]

Formula used on F1 tyres from manufacturer, presumably empiricaly derived, radial, medium compound=
Rolling resistance total(N)=(2*front vehicle mass(kgf)*((1.5E-3+1.5E-4)/(front tyre pressure in bar+1E-5*V²(kph)/front tyre pressure)))+(2*rear vehicle mass(kgf)*((1.5E-3+1.5E-4)/(rear tyre pressure in bar+1E-5*V²(kph)/rear tyre pressure)))

...this disregarded driven or front axle as differences too small to bother, same formula used for LMP tyres, with correcting factor .87, as 18" rims made for stiffer carcass compared to 13" F1 rims, also compounds harder. On most simulation gives adequate values, and as surmised above, is a function of Fz.

Note that there is a definite camber element in real life, but as most cars run in a definite region, comes out ok in calc from manufacturers data. Another coefficient to bear in mind is the compound one (not in formula above, we used to correct separately in calcs...softs use more power (1.02 to 1.05, qualifiers in gumball days could be 1.13, hence wing down when qualies on )

edit- some more info on SAE=

The basic model equation for SAE J2452 is:

Rolling Resistance ( N / Lbs) = Pα x Zβ x (a + bxV + cxVxV)

where:

P is the tire inflation pressure ( kPa / psi)
Z is the applied load for vehicle weight ( N /Lbs)
V is the vehicle speed ( km/h / mph)
alpha, beta, a, b, c are the coefficients for the model.

Edited by RDV, 16 September 2011 - 20:45.