inactiveuser7616

Hey guys...

I thought one of the big factors to move to 19"s was the cost of the tires are supposedly less than 20"s.

I'm on Tirerack and don't see a big difference in pricing.

Am I missing something?

Macca

Phil. The point of moving to 19s was the available of serious semi slick and slick rubber such as Hoosier. Its not a cost choice, its a choice to increase the range of brands due to wider fitment choices. Infact the contact patch of the tyres is smaller on 19 than 20 so most choose a wider 19 than the the stock 20" which compensates for the tyre patch and improves traction. There are lots of fans of R6/7 out there...

inactiveuser7616

Got it. Thank you sir.

I knew the Hoosiers didn't come in 20" sizes at this time. I for some reason also thought there was a reduction in cost of consumables (tires).

Keith Verges - Dallas

I see these comments about larger contact patch bandied about, and in my view, it is myth and it is wrong.

As far as I know, the contact patch would be dictated by the overall diameter and width of the tire, and to some extent, I suspect the construction of the tire, such as side wall stiffness, would affect contact patch.

I don't know anyone running 19 inch wheels that is advocating making a significant change to the overall rolling diameter of the tire, so there should be no contact patch difference appreciable between a 20 inch or 19 inch wheel

Am I missing something?

mdkim

$400 less for a set of Trofeo Rs or $700 less for R7s is substantial IMO (based on Trofeo Rs in 20").

Mech33

You're correct.

Macca

Keith. Contact patch is a function of width AND diameter. A 21' when with the same width tyre as a 19 will have a 10% larger contact patch all other things (such as camber, downforce etc) being equal. Along with increased with this is what helps increase traction on cars like 918 & 991 RS. It's not fiction it's physics. Most race series including F1 have rules around maximum wheel diameter to keep this a level playing field....

Mike in CA

Mark, you're correct.

Bob Loblaw

Try basic geometry before invoking physics.

Drifting

$700 less per set of tires is significant.

From what I've read on here, it seems that most guys who put 19s on their GT3s, do it to run Hoosiers (R7s), correct?

Sounds like you'd make up the cost of the extra wheels with a years worth of track days, and then save a lot more after that. Plus as a bonus, you don't have to stress about whether or not you'll be able to find MPSC2s for your non-track wheels, as MPSS and other tires are easier to find.

inactiveuser7616

Compared to what? Where are you getting your pricing?

According to Tirerack...in 19"...here's what I see:
325/30/19
Cup2 - $479
Trofeo - $582
H R7 - $468
H A7 - $468

Macca

Agree. There's a separate thread of Phils that discusses this...,

mdkim

A set of OE sized 20" Trofeo Rs cost roughly $2400. A few dollars less if you shop.

Prices are roughly $2k for the very same Trofeos in 19" and $1700 for R7s. Let me know if my math is incorrect or if you've found better pricing.

Mech33

Not quite...

Let's agree that the force supported by the contact patch is the same regardless of the tire. For a 60% rear weight biased car, that's roughly 30% of the car's weight W on each of the rear tires. So 30% of the car's weight (0.3*W) is the total force applied from the tire to the ground through the "contact patch".

That total 0.3*W force on the patch creates a pressure distribution over the area of that patch. For simplicity, if we assume the pressure over the patch is roughly constant (a good assumption if your tires wear uniformly), then the force, area, and pressure are simply related by:

Force = Pressure x Area

If you look at a piece of tire in the middle of the contact patch, it has the contact patch ground pressure on one side, internal tire air pressure on the other side, and some shear forces in the rubber around the edges of that section. But if the tire section is sitting flat and properly inflated, these internal tire rubber forces are relatively small the the tire contact patch pressure is essentially the internal tire air pressure.

So if you have 35 psi (pounds per square inch) in your tire, then a 3100 lb car with 30% of the weight over each rear wheel as roughly this size contact patch:

Area = Force / Pressure = (0.3 * 3100 lb) / (35 lb / in^2) = 26.6 square inches

For a 12" wide rear tire and an approximation of a simple rectangular contact patch, that's a 2.2" long x 12" wide patch. Sounds reasonable, no?

So far, tire diameter hasn't factored into any of these calculations...

There is a second order effect on contact patch area for narrow profile tires. For a fixed tire outer diameter, but larger wheel diameter, the profile gets shorter, and the sidewall stiffness goes up. The stiffer the sidewall, the higher the actual pressure at the edges of the contact patch where the sidewall touches the road. In the above calculation, we assumed pressure was uniform, but in actuality the edges are a bit higher. For the same tire air pressure, this causes a little more of the car's weight to be supported at the edges of the patch and less in the center, reducing the "contact patch" area slightly.

Also, the wider the tire, the wider the aspect area of the contact patch area, so the less the impact of the sidewall pressure on the overall contact patch area, so the area goes up slightly for the same tire pressure...

Another second order effect is the overall tire diameter. At the long edges of the contact patch where the curvature of the tire lifts up and off the road, the stiffness there is higher just like at the sidewall, so the local contact patch pressure there is higher. The larger the diameter, the less this effect.

Anyway, in conclusion, contact patch area is virtually completely a function of the tire air pressure. So does a 991 GT3 RS rear 21" tire have a larger contact patch than a 991 GT3 20" tire? Depends on the tire pressure. But does it have a more uniform contact patch pressure? Yes, slightly.

Why is more uniform pressure important? Tire material coefficients of friction are non-ideal and decrease with load. That is, if you take a piece of rubber loaded with force F, vs. two identical pieces of rubber each loaded with half the force F/2, the two half-loaded pieces with have more total friction force available than the single higher loaded piece. This is the fundamental reason why everything in motorsports is about being "smooth" (to reduce peak load transients on the tire that result in far less friction holding ability than the steady-state turning load), why sway bar tuning affects balance (distributing more or less load on the inside or outside tire changes total available grip on that axle), etc...

End rant...

Ryan Gates

Thank you for breaking that down, Mech33.