Crazy oversteer / wheelspin with Michelin Pilot Super Sport

[s2d4]

Originally Posted by Ubersuber

None of this makes any sense.

Should've left it at this...

Originally Posted by Ubersuber

Oversteer is defined by reference to slip angles. Lower tire pressures result in higher slip angles for the same side force. If the car oversteers to begin with lowering the tire pressure will increase the oversteer.

Hmmmmmmmmmm.

[stonenewt]

Originally Posted by ZDan

What assumptions are you making to get a contact patch area? An ellipse 265mm wide (or some fraction of 265mm) with length dictated by 90% sidewall deformation for each size? I don't think that's valid. Contact patch area should be roughly equal in each case, given same make/model tire with the same internal construction, ply count, sidewall stiffness, etc.

The larger-diameter tires with the same profile have greater internal volume and hence greater load-carrying capacity.

OK, so contact patch area is weight divided by pressure. So for a given inflation pressure and the same weight, the tires you list above should have the *same contact patch area*. The reason the load-carrying capacity is different is due to internal volume. The larger-diameter tires have greater internal volume, and are rated for higher load.

Pay attention in your mathematics classes? What you’re missing here is that the chord length of an arc for a given sagitta varies with diameter - http://en.wikipedia.org/wiki/Sagitta_(geometry)

If we keep the same tyre profile (265/30) this fixes sagitta, but increase the wheel diameter & the outside diameter of the tyre increases the arc radius. This basically defines the tyre deformation formed by tyre load, as the length of tyre contact patch is the chord the length. That contact patch is going to be approximately elliptical. So from tread width & chord length we can get a rough idea of the maximum contact patch. We also know that load carried is directly proportional to the tyre pressure, so I’ve also factored in the tyre pressure to the calculation.

As to your previous statement about the increase in load between 225/50R16 & 225/45R17 tyre. Using the (XL) load indexes from the same line of tyres I have the 16” tyre at 650kg (710kg) & the 17” at 615 (670kg) max load. Based on the geometry above I get a calculated 618kg (674kg) max load for the R17 tyre based on the 16” tyre’s load rating. Based on a direct volumetric calculation I get an estimated load of 560kg (632kg).

For shit’s n’ giggles I calculated the max load for a 265/45R20 tyre from the load index of a 125/82R12 wheel, smallest & largest diameter tyres I could find. This is calculating the load rating from sing my maximum contact patch method and the 272kg load of the little 12” tyre I calculated load of the R20 tyre at 892kg, it’s actual load is 900kg, calculated is nearly 1200kg load calculated by air volume.

[ZDan]

Originally Posted by stonenewt

Pay attention in your mathematics classes? What you’re missing here is that the chord length of an arc for a given sagitta varies with diameter - http://en.wikipedia.org/wiki/Sagitta_(geometry)

I don't know why you'd think I missed that, it's pretty clear in my post that I knew how you were calculating contact patch length based on an assumption, and what that assumption was.

You're assuming that those tires will have the same sidewall squish and at the same pressure. That's not necessarily true. You're also assuming that all of those tires will have the same contact patch width. Also not necessarily true. Contact patch width is not the section width, and it's not the tread width either, and it's not necessarily going to be the same for a given tire section profile (section width/aspect ratio) with different wheel diameters and outside tire diameters.

From your post:

We also know that the weight supported by the tyre is contact patch area multiplied by tyre pressure.

If *THIS* statement is true, then all of the tire sizes you listed will have the SAME contact patch area for a given load and pressure.
area = load/pressure.

So in the same post, you're saying that the taller tires will have a greater contact patch area (assuming that the taller tires will have a longer contact patch length and the same contact patch width as the shorter tires), AND that the tires will all have the SAME contact patch area.

Both of these cannot be true.

You are making assumptions about the contact patch size to come up with a greater area for the larger tire, and then multiplying the areas by the same pressure to get a "load" to incorrectly conclude that the larger tire can carry greater load because the contact patch area is bigger.

But this is inconsistent. Again, if the load is simply pressure times area, then the contact areas MUST be the same for a given load and pressure.

[wparsons]

Originally Posted by Ubersuber

None of this makes any sense.

Did I use too many big words for your simple brain?

Originally Posted by Ubersuber

The link I posted doesn't have a page 17 to begin with.

The pages I referenced are what my PDF viewer was saying for the page, not the page number that is part of the graphic in the document. Either way, by simply scanning the document you can easily find the parts I was quoting, no one else had issues with that.

Originally Posted by Ubersuber

Oversteer is defined by reference to slip angles. Lower tire pressures result in higher slip angles for the same side force. If the car oversteers to begin with lowering the tire pressure will increase the oversteer.

I don't think you understand what slip angles actually are, if you did you wouldn't have posted that.

Slip angle is the difference between the path the tire would take if rolling compared to the trajectory it's actually moving on. At slow speeds, the slip angle of the tires is essentially zero. At the limit they'll start to slide sideways a bit, that's when you get slip angle. If you surpass the limit you get much bigger slip angles.

Why does lower pressure automatically mean more slip angle? Slip angle comes from the tire sliding, not from the sidewall rolling over. Lower pressures will allow the sidewall to roll more, but that doesn't automatically mean it's going to slide more/easier. To get any slip angle, the tire needs to be sliding sideways slightly. Too much pressure will actually make it slide more/easier since you've reduced the contact patch.

Yes, if you go way too low it'll lose grip from deforming too much, but this is all situational. You keep posting blanket statements that might be right 5% of the time, but state them as if they are the only way that can ever work.

As for the typo, how do you know which one has the typo? You're assuming it's the pdf because that supports your argument, but what if tirerack is the one that's wrong?

Further, you're assuming that 35psi at all corners is optimal, but did you see what the tire tirerack page suggests?

Front Engine/Rear DriveFront
Rear35-45 psi
30-40 psi

Notice that the rear pressure is lower than the front?

The problem is that you're arguing that running too low of pressure (below optimal pressure) will reduce grip, while everyone else is telling you that sometimes optimal pressure means having lower pressure in the rear, and sometimes it means having more pressure in the rear.

You REALLY need to start realizing that nothing about ANY of what you argue on here is black and white. It's all very contextual based on the car, the driver, the driving conditions, etc. Your blanket statements just don't work in reality.

Originally Posted by Ubersuber

The reference in the article to higher cold pressures being needed for autocross as compared to track work results from the lower temperature rise experienced in autocross than on the track. Even lower temperature rises are experienced in normal driving require even higher cold temperatures than that.

But that's not what you said, originally. So by changing your story you were somehow never wrong in the first place? Target pressures for max grip are about the same in any driving condition, the only thing that varies is the cold pressure to get you to that target pressure when up to operating temperature.

Also, autox typically builds less heat than street driving does due to the low speeds and very short runs.

Originally Posted by Ubersuber

Quoting out of context can get you into all sorts of logical problems.

Reading what is actually written is a good start.

You should try your own advice sometime, might make you look like less of a blathering idiot.

[stonenewt]

Originally Posted by ZDan

You're assuming that those tires will have the same sidewall squish and at the same pressure.

I’ve made the tyre pressure & the side wall deformation, as a % of the side wall height, constants in the equation. Those are not assumptions, they are facts because those are input constants.

You're also assuming that all of those tires will have the same contact patch width. Also not necessarily true. Contact patch width is not the section width, and it's not the tread width either, and it's not necessarily going to be the same for a given tire section profile (section width/aspect ratio).

I’m using the Bridgestone 2004 Technical Tyre Data Book for the Max. Operational Contact Width which for all the 265/30R## tyres is listed at the same 267mm wide & 84mm tall. I could have chosen 205/50 tyres but those vary in width (200-209mm) & 97-108mm tall which make the numbers less intuitive to deal with.

If *THIS* statement is true, then all of the tire sizes you listed will have the SAME contact patch area for a given load and pressure.
area = load/pressure.

Yeah so as there’s an increased surface area available with larger tyres means that the maximum load is higher.

So in the same post, you're saying that the taller tires will have a greater contact patch area (assuming that the taller tires will have a longer contact patch length and the same contact patch width as the shorter tires), AND that the tires will all have the SAME contact patch area.

Both of these cannot be true.

NO! I’m saying that BECAUSE for the same deformation larger tyres have a higher surface area they can support more weight.

You are making assumptions about the contact patch size to come up with a greater area for the larger tire, and then multiplying the areas by the same pressure to get a "load" to incorrectly conclude that the larger tire can carry greater load because the contact patch area is bigger.

Well this is constant with proven laws of physics & geometry when all see & done.

But this is inconsistent.

HOW?

Again, if the load is simply pressure times area, then the contact areas MUST be the same for a given load and pressure.

Yeah exactly, So when one increases the area in contact with the ground you end up with a higher load capacity. You’re making the case against your self very well here!

[ZDan]

Originally Posted by stonenewt

I’ve made the tyre pressure & the side wall deformation, as a % of the side wall height, constants in the equation. Those are not assumptions, they are facts because those are input constants.

FACT is actual measurement of sidewall deformation under a given load and pressure. ASSUMING 10% squish and setting that as a *constant* in your equations does not dictate reality!

If you run the numbers on tire OD and revs per mile, you'll find that squish varies quite a bit. You're in the ballpark, but from the handful of tires I've run, I'm seeing 10%-12% squish even for the same section width and aspect ratio, same make/model tire, differing only in wheel diameter.

I’m using the Bridgestone 2004 Technical Tyre Data Book for the Max. Operational Contact Width which for all the 265/30R## tyres is listed at the same 267mm wide & 84mm tall. I could have chosen 205/50 tyres but those vary in width (200-209mm) & 97-108mm tall which make the numbers less intuitive to deal with.

"Max Operational Contact Width" (link or scan of that for context?) is not going to be the same as *actual* contact patch width under a given load.

Yeah so as there’s an increased surface area available with larger tyres means that the maximum load is higher.

Not really. Easy enough to get more contact patch area with the smaller tire, you just reduce the pressure. But that's not what you do if you want to increase load. You INCREASE pressure to increase load-carrying capacity, which REDUCES contact patch area.

HOW?

You say that load is pressure times area. OK, so area is load divided by pressure, right?
So for a given load and a given pressure, area is a constant, right?
So, what's the area for a 215/45-17 tire supporting, 700 lb. at an inflation pressure of 35psi?
And what's the area for a 255/40-17 tire supporting 700 lb. at an inflation pressure of 35psi?

Yeah exactly, So when one increases the area in contact with the ground you end up with a higher load capacity. You’re making the case against your self very well here!

No, you are saying the one hand that a bigger tire has a larger contact area, and on the other hand you're saying that area is purely a function of load and pressure. These two can't both be true.

Obviously, for a given inflation pressure, if you add load you add contact patch area.
That doesn't mean that you WANT greater contact patch area, it's just a natural result of it.
In fact, you will want to INCREASE tire pressure with increased load to prevent the contact patch from getting bigger and to reduce deformation and heat into the tire.

You don't increase contact patch area to increase load-carrying capacity.

[stonenewt]

Originally Posted by wparsons

Slip angle is the difference between the path the tire would take if rolling compared to the trajectory it's actually moving on. At slow speeds, the slip angle of the tires is essentially zero. At the limit they'll start to slide sideways a bit, that's when you get slip angle. If you surpass the limit you get much bigger slip angles.

Why does lower pressure automatically mean more slip angle? Slip angle comes from the tire sliding, not from the sidewall rolling over. Lower pressures will allow the sidewall to roll more, but that doesn't automatically mean it's going to slide more/easier. To get any slip angle, the tire needs to be sliding sideways slightly. Too much pressure will actually make it slide more/easier since you've reduced the contact patch.

Yes, if you go way too low it'll lose grip from deforming too much, but this is all situational. You keep posting blanket statements that might be right 5% of the time, but state them as if they are the only way that can ever work.

As for the typo, how do you know which one has the typo? You're assuming it's the pdf because that supports your argument, but what if tirerack is the one that's wrong?

I think this shows what's going on

Sam_68 on PH explains on the 5th post

[7thgear]

Originally Posted by wparsons

Why does lower pressure automatically mean more slip angle? Slip angle comes from the tire sliding, not from the sidewall rolling over.

well, not quite true.. a "no slip" angle would be the rim.. a softer sidewall allows the contact patch to deviate with greater freedom from it's intended path.. hence the higher net slip angle..

[Trettiosjuan]

Indeed, slip angle has nothing to do with sliding the tyre per se...

[stonenewt]

Originally Posted by ZDan

FACT is actual measurement of sidewall deformation under a given load and pressure. ASSUMING 10% squish and setting that as a *constant* in your equations does not dictate reality!

If you run the numbers on tire OD and revs per mile, you'll find that squish varies quite a bit. You're in the ballpark, but from the handful of tires I've run, I'm seeing 10%-12% squish even for the same section width and aspect ratio, same make/model tire, differing only in wheel diameter.

"Max Operational Contact Width" (link or scan of that for context?) is not going to be the same as *actual* contact patch width under a given load.

Not really. Easy enough to get more contact patch area with the smaller tire, you just reduce the pressure. But that's not what you do if you want to increase load. You INCREASE pressure to increase load-carrying capacity, which REDUCES contact patch area.


You say that load is pressure times area. OK, so area is load divided by pressure, right?
So for a given load and a given pressure, area is a constant, right?
So, what's the area for a 215/45-17 tire supporting, 700 lb. at an inflation pressure of 35psi?
And what's the area for a 255/40-17 tire supporting 700 lb. at an inflation pressure of 35psi?



No, you are saying the one hand that a bigger tire has a larger contact area, and on the other hand you're saying that area is purely a function of load and pressure. These two can't both be true.

Obviously, for a given inflation pressure, if you add load you add contact patch area.
That doesn't mean that you WANT greater contact patch area, it's just a natural result of it.
In fact, you will want to INCREASE tire pressure with increased load to prevent the contact patch from getting bigger and to reduce deformation and heat into the tire.

You don't increase contact patch area to increase load-carrying capacity.

Okay one last go at this.

The force triangle -

   F
———————
 P | A

F = Tyre load
P = Tyre pressure
A = Contact patch

If you know two of them you can calculate the other.

If I have
P as the maximum tyre pressure
A as the maximum contact patch
F becomes apparent through simple multiplication.

If I know
F as the tyre load
P as the tyre pressure
A is the division of F by P

When I know
the tyre load (F)
contact patch (A)
I can then calculate the tyre pressure (P)


With the help of a data book I can find enough information to calculate A, at maximum tyre deformation, with bit of simple geometry gives. From the same data book I can find out P which is maximum pressure of a tyre. That gives me the two bits of information I need to calculate F, the load needed on the tyre to cause A & P to be those values.

If you don't get that you're either trolling or simply not that bright. I'm going to give you the doubt and assume the former.

Bye!

[tYtEn86]

Hey OP did u get your answer?

[wparsons]

Originally Posted by 7thgear

well, not quite true.. a "no slip" angle would be the rim.. a softer sidewall allows the contact patch to deviate with greater freedom from it's intended path.. hence the higher net slip angle..

Originally Posted by Trettiosjuan

Indeed, slip angle has nothing to do with sliding the tyre per se...

Slip angle is the difference between the direction the tread is pointing and the trajectory of the wheel. It can be from tread deformation or from the tire sliding slightly (or most likely, a bit of both).

That's why running too much or too little pressure at one end of the car can both reduce grip.

[ZDan]

Originally Posted by stonenewt

Okay one last go at this.

You're not teaching anyone anything about load, pressure and contact patch.

I don't see why you'd think I don't grok the relationship between pressure, force, area. Elementary and obvious (and also an approximate *estimate* that doesn't take into account stiffness of the carcass).

So, tell me *approximately* what the contact patch area should be for a 700 lb. load with:
A) 215/45-17 at 35psi
B) 265/30-22 at 35psi

They will have roughly the same contact patch area.

With the help of a data book I can find enough information to calculate A, at maximum tyre deformation, with bit of simple geometry gives. From the same data book I can find out P which is maximum pressure of a tyre. That gives me the two bits of information I need to calculate F, the load needed on the tyre to cause A & P to be those values.

You're saying that the data book your looking at gives max tire deformation and/or contact patch length, maximum contact patch width, but doesn't give a maximum load rating for the tire? That's weird...

But again, the inflation pressure=force/area is not accurate, it's an approximation. It doesn't account for the stiffness of the tire.

If you don't get that you're either trolling or simply not that bright. I'm going to give you the doubt and assume the former.

Neither.

Bigger tires with greater internal volume can support greater loads at a given pressure. Obviously, at the same pressure and a greater load, the contact patch area is greater. That does *NOT* mean that the greater load capacity is due to the greater contact patch area. Greater contact patch area is a result of greater load at a given pressure.

I got into this when this question:
why do heavy cars use larger tires but roughly the same tire pressure?
was incorrectly answered thusly:
Because they have tires with significantly larger contact patches.
(I now see that it wasn't even you who gave that answer)
Anyway, that's incorrect. If larger contact patches increased load carrying capacity, you would lower pressures for higher loads. But you INCREASE pressure for higher loads, which reduces contact patch area.

[Ferrari]

UPDATE: Lowered the rear pressure to 34 psi. I get less oversteer. I'm happy.

Although, I also changed the rear diff fluid from Motul Gear 300LS to Motul Gear 300. It's a less viscous fluid with no friction modifiers. The diff lockup seems to be less aggressive, so that also helped with the oversteering tendency.