The MYTH Of Wider Tire Contact Patch Dispelled!

[CapriRacer]

Originally Posted by RidingOnRailz

Apologies: Both my father and I always had issues with context when we were communicating with others, so it wasn't always clear what we were talking about. We also had zero concept of something called a segue, or, verbal indication that we were about to switch subjects.


When I say "wheel/rim" or "tire", 90% of the time I'm referring to the whole package.

Whenever I say "wider" I am referring to the lateral width of both tire and the rim its mounted to. "Greater or lesser diameter" is self-explanatory.

So in that context(!) let me ask you this:

We have presently seen, over the last twenty years, the following trends:

Tires: given the same overall diameter, sidewall height has decreased, and tire width increased.

Wheels: Both diameter and width increased.


I have an alternative vision of what could have happened during that same time period:

Tires: Sidewall height decreases, but width remains the same.

Wheels: Diameter increases, but width remains the same.

Benefits: Stronger self-aligning/self-centering force preserved. Weight concentration of a narrower tire/wheel package down through rain and snow is maintained.
Aesthetics: Those who prefer to see less sidewall and more wheel have their needs met.

I hope this clears up some of the terms I have been using.

I think you are referring to a thing that started when they took asbestos out of brake pads. The easy fix was to increase the diameter of the rotor - and that meant increasing the diameter of the rim - and since you don't want the load carrying capacity of the tire to go down, AND you don't want the diameter of the tire to change (might interfere with suspension components), the only solution is to go wider with a lower aspect ratio. And if that meant changing the fender a bit, that is a whole lot easier than trying to redesign the suspension.

[blazee]

Originally Posted by Blue Bowtie

One critical qualifier is missing from that statement. It would be important to mention "At the same inflation" because that is the only condition under which it is true.

As experienced people may have come to know, wider tires require less inflation to support a given load. THAT'S where the footprint increases and the traction advantage begins.

This is the most accurate information in the whole thread.


The pressure inside the tire matches the pressure outside the tire. The main factors in determining the size of the contact patch are the tire pressure and the weight of the vehicle. A 3000lb car with 30psi of pressure in each tire is going to have a total of 100 square inches of contact. With equal weight distribution, that's 25 square inches per tire. If you took that same car, and lowered the rear tires to 20psi, they'd have 37.5 square inches of contact patch... a 50% increase.

[RidingOnRailz]

Originally Posted by fredjacksonsan

Almost ANY car with sufficient power can spin the rear (or drive) wheels.

I agree with some of your points, thinner tires generally track better and have more predictable handling. But probably not at the limit. And comparing how two different (make) cars handle and basing your proof on that, well, nope sorry.

But if you were right about thinner tires having a bigger contact patch, hell we'd all be driving on tires as wide as some road bicycle.

Your assertions have

1) not been backed up and
2) you're pretty much arguing an invalid position at this point and
3) I am wondering why you're having such a hard time just telling CapriRacer where you found the book!

“ But if you were right about thinner tires
having a bigger contact patch, hell we'd
all be driving on tires as wide as some road
bicycle. “

What I was postulating in my original post was that, all other things equal, widening the wheel & tire package changed not the area of contact patch, but the orientation of said patch. So no, I do not believe that a narrower combo would have a larger area of contact.

[RidingOnRailz]

Originally Posted by CapriRacer

I think you are referring to a thing that started when
they took asbestos out of brake pads. The
easy fix was to increase the diameter of the rotor -
and that meant increasing the diameter of the rim -
and since you don't want the load carrying capacity
of the tire to go down, AND you don't want the
diameter of the tire to change (might interfere with
suspension components), the only solution is to go
wider with a lower aspect ratio. And if that meant
changing the fender a bit, that is a whole lot easier
than trying to redesign the suspension.

I am familiar with that aspect of the whole wider, larger diameter wheel to accommodate larger brake elements. I did not know the additional reasoning behind it - replacing asbestos brakes with non-asbestos that had less braking ability per square inch. Larger discs to help control and disperse heat buildup from braking.

It is unfortunate, though, that it led to giant rims/wheels becoming ‘a thing’ in the off-track passenger car segment.

[RidingOnRailz]

Originally Posted by blazee

This is the most accurate information in the whole thread.


The pressure inside the tire matches the pressure outside the tire. The main factors in determining the size of the contact patch are the tire pressure and the weight of the vehicle. A 3000lb car with 30psi of pressure in each tire is going to have a total of 100 square inches of contact. With equal weight distribution, that's 25 square inches per tire. If you took that same car, and lowered the rear tires to 20psi, they'd have 37.5 square inches of contact patch... a 50% increase.

Makes sense, assuming the cold tire pressures do not change.

In my case, a 2010 Honda Accord EX with a wider 50R17 package, the specified cold tire pressure is 2psi higher(32) than for the 2010 base Accord LX: 30psi in 60R16.

So it is quite likely that the contact patch area under both above Accords is equal, or possibly 5-10% more area in the LX, with its lower cold pressure spec of 30psi.

Another interesting fact: The 2008-2010 Accord LX OEM tire load index was 94, vs 93 for the HIGHER trim EX.

I sought out the EX because it has niceties like a moonroof. But I suspect that the LX trim, with its narrower, higher profile tires and a higher load index, was the more practical version of the Accord for those model years.

Question: as far as the whole “wider lower profile tires require lower tire pressure” line of thinking, then why does Honda specify, on several models, not just Accord above, higher cold pressures on trims with bigger wheels and wider, lower profile rubber?

[CapriRacer]

Originally Posted by blazee

.....The pressure inside the tire matches the pressure outside the tire. The main factors in determining the size of the contact patch are the tire pressure and the weight of the vehicle. A 3000lb car with 30psi of pressure in each tire is going to have a total of 100 square inches of contact. With equal weight distribution, that's 25 square inches per tire. If you took that same car, and lowered the rear tires to 20psi, they'd have 37.5 square inches of contact patch... a 50% increase.

It is such a common misconception that you can calculate the size of the footprint knowing only the inflation pressure and the load on the tire, I wrote a webpage debunking it.

http://barrystiretech.com/Footprint.html

[shorod]

Originally Posted by RidingOnRailz

In my case, a 2010 Honda Accord EX with a wider 50R17 package, the specified cold tire pressure is 2psi higher(32) than for the 2010 base Accord LX: 30psi in 60R16.

So it is quite likely that the contact patch area under both above Accords is equal, or possibly 5-10% more area in the LX, with its lower cold pressure spec of 30psi.

I sought out the EX because it has niceties like a moonroof. But I suspect that the LX trim, with its narrower, higher profile tires and a higher load index, was the more practical version of the Accord for those model years.

Question: as far as the whole “wider lower profile tires require lower tire pressure” line of thinking, then why does Honda specify, on several models, not just Accord above, higher cold pressures on trims with bigger wheels and wider, lower profile rubber?

Power sunroofs, leather seating, added sound deadening material, GPS navigation systems, etc. that also often come with the more upscale trims add weight. So maybe some of the reasoning behind the higher inflation pressures is due to increased curb weight + optional equipment? I'm just speculating here, I don't have any professional basis in this area.

-Rod

[RidingOnRailz]

Originally Posted by shorod

Power sunroofs, leather seating, added sound deadening
material, GPS navigation systems, etc. that also often come with the
more upscale trims add weight. So maybe some of the reasoning
behind the higher inflation pressures is due to increased curb weight
+ optional equipment? I'm just speculating here, I don't have any
professional basis in this area.

-Rod

Good points.

Although the EX trim level across Honda's lineup has historically been more of a 'mid' trim level. In my Accord, the only extra weight might be the moonroof unit. 30-40lbs? I don't know how much heavier a 6-CD changer is compared to a single disc AM/FM stereo unit. And the difference in the weight of a power driver's seat vs a manual one is probably negligible.

Cloth interior on the EX, unlike EX-L(eather) or EX-V6.

[Stealthee]

Power seat to manual can be as much as 40lbs.

[RidingOnRailz]

Originally Posted by Stealthee

Power seat to manual can be as much as 40lbs.

Shayzusss..! It's a power automobile seat, not a space shuttle seat.

[RidingOnRailz]

Semi-related to this thread:

What is the actual change in air pressure per 10deg F temperature delta -

1.0psi, or 0.8? I've read read estimates of tire pressure loss/gain from half-PSI up to two-PSI on this objective internet we use!

Which one is iiiiiit??

[CapriRacer]

Originally Posted by RidingOnRailz

Semi-related to this thread:

What is the actual change in air pressure per 10deg F temperature delta -

1.0psi, or 0.8? I've read estimates of tire pressure loss/gain from half-PSI up to two-PSI on this objective internet we use!

Which one is iiiiiit??

First, this deals with the Ideal Gas Law - which references absolute zero temperature and absolute pressure. So the answer depends on how hot you start and what elevation you are at. You'll get a different answer if you start at 30°F than 100°F as well as a difference at sea level or at 10,000 feet.

It also depends on the starting inflation pressure. You'll get a different answer if you are starting with 30 psi than if you start with 100 psi.

Having said that, if you have a 30 psi tire at 70°F at sea level, and the temperature rises to 80°F, the increase in pressure is 0.84 psi.

On the other hand if you start out with 65 psi (like on some LT tires), the pressure increases 1.50 psi.

I've worked with other tire engineers and have worked out a "Rule of Thumb" - 2% for every 10°F. It's close enough and easy to remember!

[RidingOnRailz]

Originally Posted by CapriRacer

First, this deals with the Ideal Gas Law - which references absolute zero temperature and absolute pressure. So the answer depends on how hot you start and what elevation you are at. You'll get a different answer if you start at 30°F than 100°F as well as a difference at sea level or at 10,000 feet.

It also depends on the starting inflation pressure. You'll get a different answer if you are starting with 30 psi than if you start with 100 psi.

Having said that, if you have a 30 psi tire at 70°F at sea level, and the temperature rises to 80°F, the increase in pressure is 0.84 psi.

On the other hand if you start out with 65 psi (like on some LT tires), the pressure increases 1.50 psi.

I've worked with other tire engineers and have worked out a "Rule of Thumb" - 2% for every 10°F. It's close enough and easy to remember!

Thank you for confirming, Barry. My bigger brain(my gut!) always suspected that 1psi per 10°F was only a generalization. Yes, I think with my belly, because prenatal alcohol and smoking destroyed half of what’s upstairs by the time I was born.

My takeaways, and confirmations, of what you just explained:

1. Delta in air pressure per given temperature change varies depending on existing air pressure. The lower the pressure, the less the change. The higher, the more the change.

I can attest to this - bleeding down a 35psi tire for two seconds, with my particular gauge, reduces pressure only around 1psi. But: bleeding down a 60psi spare ‘donut’ for two seconds, using that same gauge, reduces pressure 4-5psi! So I do understand now that an equivalent temperature change will affect the higher pressured tire more drastically than will affect the lower pressured tire.

2. At altitude, I know the cold pressure I set near sea level will increase a bit with every 1,000 feet up I go into the mountains. Because there is less atmospheric pressure imposed on the tire up there.

So at sea level, a 10°F increase in temperature would increase that 35psi tire to about 35.9psi? At, say, 4,000 feet up, that same temperature increase might increase that 35psi to…. 36-37psi?

Interesting stuff to know, and it helps me to estimate how high a pressure I should set it for based upon the altitude or temperature conditions where the tire is being driven to.

[RidingOnRailz]

One more brain cell I'd like to pick of Barry's:

Given:

Two days with the same temperature, same humidity, pre-dawn conditions with no sun hitting the car, cold tire pressure set, IE: 32psi.

Change one thing: Lower or raise the barometric pressure approx. one inch on the second of the two days.

What effect will that one change have on the cold tire pressure?

[CapriRacer]

Originally Posted by RidingOnRailz

One more brain cell I'd like to pick of Barry's:

Given:

Two days with the same temperature, same humidity, pre-dawn conditions with no sun hitting the car, cold tire pressure set, IE: 32psi.

Change one thing: Lower or raise the barometric pressure approx. one inch on the second of the two days.

What effect will that one change have on the cold tire pressure?

So I assume you mean 1" of mercury - which is a lot!

That's 0.49 psi.