Originally Posted by rumrunn6

oh ok. don't know why a bigger tire CAN run at a lower pressure. gimme 10 yrs & I'll get back to ya

but I don't think my new 45mm riddlers (whose max is 55 lbs) could be run at 120 psi

Originally Posted by 69chevy

I am an engineer. You are overthinking it.


Lower pressure can be run with wider clincher rims, because the wider rim minimizes the "pinch" on the tube. Just widening the tire doesn't help here.


Pressure is a measurement of weight acting on an area.


If you increase the area (wider tire) then less "weight" (force of air pushing tire out) is necessary to support the weight of the rider pushing down. As a bonus, when encountering a bump that compresses the tire (and air inside), the lower pressure air will allow more deformation of the tire.


So in a nutshell, the wider tire can support the same weight at a lower pressure, yet can act as a better "suspension" over bumps. But... if the rim bed isn't wider, it will be even more prone to pinching than a narrower tire on the same wheel.

Originally Posted by rpenmanparker

Nor would you want them to. I'm just saying that larger tires are best inflated to lower pressure for ride characteristics. Asking why that is.

Originally Posted by Doge

Because they can, not because they have to.
A smaller tire would be flexing the casing more and not giving back. A larger tire inflated to the same pressure as a smaller one has less casing flex.

Originally Posted by Doge

Geometry.
Pretend that the contact area needed to be the same for a 20mm and a 25mm tire.
The 25mm would have a rounder contact patch as the downward pressure increased, there was a wee bit of casing on the side - and forward and rear of the contact point.
The 20mm would contact and there would be less casing on the sides to create the same contact area. If contact area is to be the same it would have to deform more case in front and behind the contact point. That would mean more total drop, more case flex/deformation.

Under the same pressure thinner drops more and deforms the case more.

To me, depending on road, that does not mean a whole bunch. If you don't like drop, pump up the PSI.

Originally Posted by rpenmanparker

I like 15% drop just fine.

Originally Posted by Doge

At the same PSI the wider tire drops less.

So lets say you like 15% drop.

Tire A is a 23 mm pumped to 130PSI and Tire B is a 25.5 mm pumped to 105PSI. Both have 15% drop. Both have (imaginary) same contact patch area, not shape.

What is the difference?

Originally Posted by Doge

So lets say you like 15% drop.

Tire A is a 23 mm pumped to 130PSI and Tire B is a 25.5 mm pumped to 105PSI. Both have 15% drop. Both have (imaginary) same contact patch area, not shape.

What is the difference?

Originally Posted by rpenmanparker

Nor would you want them to. I'm just saying that larger tires are best inflated to lower pressure for ride characteristics. Asking why that is.

Originally Posted by rumrunn6

must be due to the weight of the vehicle & rider

Originally Posted by Dean V

Why is a wider rim less prone to pinching?
The "bite" on the tube would be closer together than with a narrow rim but can't see how it is less likely to puncture.

Originally Posted by rpenmanparker

Would an engineer care to explain what higher air volume in wider tire has to do with being able to run lower pressure? I will tell you that while thinking about this recently, I came to the conclusion it has nothing to do with contained air volume. Assuming you want a uniform 15% tire depression ("drop") when the bike is loaded, two things happen when the tire is depressed. First the pressure increases inside the tire just s tiny bit. But the contact patch area increases significantly as the tire is squashed. It seems to me the reason you can lower pressure on a wider tire is that the contact patch with the ground is bigger on wider tires. The force at the contact patch has to support the bike. So a lower pressure times a larger area will give the necessary force (F=P X A).

Think about this as well: a lower profile tire would contain less air but have a larger contact patch. You should be able inflate it to lower pressure than a narrower, taller tire with a higher air volume.

Comments? I really want to know if I am making an obvious mistake.

Originally Posted by 69chevy

I re-read your post now that I have a little more time.


I think I am understanding what you are asking, and with all variables aside, I agree with your "assessment".


At the same inflation pressure, a wide and a narrow tire have the same contact area.


A wide tire is flattens over its width. A narrow tire has a skinnier but longer contact area, which means as you lower pressure proportionally, the narrow tire will move the rim closer to the ground than the wider tire.


This is assuming the same width rim and same tire casing.


So... wider tires can be run at a lower pressure without risk of bottoming out a rim, because of the shape of the contact patch. If that answers what I think your question is.

Originally Posted by 69chevy

A wide tire is flattens over its width. A narrow tire has a skinnier but longer contact area, which means as you lower pressure proportionally, the narrow tire will move the rim closer to the ground than the wider tire.

This is assuming the same width rim and same tire casing.

So... wider tires can be run at a lower pressure without risk of bottoming out a rim, because of the shape of the contact patch.

Originally Posted by wphamilton

The answer is, a wide contact point as opposed to a narrow one has less vertical depression for a given force. In other words, the rim is further from the ground when we hit the pothole, so it takes a bigger hit to get the pinch flat. "Narrow" being a misnomer IMO because it's really the length in the radial plane that's relevant. It just happens to also be narrower for bike tires. Just draw a horizontal line through the tire between the ground and rim. When that line is longer, the rim is lower.

Originally Posted by 69chevy

Not a perfect illustration, but the best I can find. The wider the mounting base for the tire, the less the tube has to contort. This means on bumps, the tube isn't as likely to be bound up and become pinched.

Originally Posted by rpenmanparker

I agree, but I usually focus on tire drop rather contact patch shape. The wider tire would most likely have a lower % drop, probably a lower absolute drop.

Originally Posted by rpenmanparker

A number of folks have provided support for the best answer I have seen, that the larger contact patch needs less pressure to provide the force against the road required to support the system weight without accidental stresses causing the tire to bottom out.

Originally Posted by 69chevy

And I still agree with my own assessment that you are overthinking it.

Originally Posted by Kopsis

This is fundamentally correct. You can even look at "accidental stresses" as just a greater (temporary) % drop. But because volume increases if you hold diameter constant and increase the width, people often mistake the direct correlation between volume and minimum pressure for a cause. It really is as simple as the formula for pressure (p = f/a). The physics of pneumatic tires, like the physics of a tensioned spoke, is something people seem to want to make much more complicated than it really is

Originally Posted by Gyro_T

I have done some research on vibration damping using tire volume and inflation pressure in wheelchairs. I have used 26" bike tires in the research. Most "ultralight" manual wheelchairs have a 26" wheel available so it opens up a wealth of tire choices. My engineer and physics aquaintences have problems with my idea of "pressure equivalencies". This is the premise: “Let's say Brand X has a popular narrow (28mm diameter) tire they determined has an optimal inflation pressure of 50-75 psi. They have decided to make the same model in a 44mm diameter with the same construction features, only the diameter and therefore volume has changed. They ask the engineer to calculate the analogous operating pressure range for the 44mm model so they can emboss it into the sidewall”. How would that be properly calculated?

I am inclined to think they use a proportionality using inflation pressure and volume that ignore the number of molecules re: the gas laws. I think Robert and 69chevy may be looking at other factors such as drop and contact patch.

I am hoping some of the great minds three years ago on this thread are still available to help.

Mike "Gearloose" B.