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# Technical FAQ: The science of tire pressure, rim width, and heat buildup

## This week, Lennard Zinn explains the science — complete with equations — regarding tire pressure and the width of your rims.

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## Do wider rims ultimately lower heat buildup?

Dear Lennard,
I am considering carbon clinchers again but have concerns about the heat buildup. I loved the way carbon wheels feel and am considering a set that is fairly wide. Is it safe to assume that with a wider rim you can run lower tire pressure, which then would result in less outward force on the braking-surface walls? Would this be safer on a long descent, or is that 10 psi difference inconsequential? I am 175 pounds and currently run 25mm tires at 90f/100r psi.
— Steve

Dear Steve,
Ooh, I love this question!! It makes a flawed assumption that I think a lot of people make, namely that decreasing the pressure with a wider rim and the same tire would necessarily decrease the stress on the rim and tire walls. I further think that people largely assume that the recommendation for running bigger tires at lower pressures than smaller tires is a choice based entirely on considerations like the likelihood of pinch flats, comfort, perhaps rolling resistance (all of which are valid) and don’t consider it a necessity in terms of safety (i.e., not over-stressing the tire).

So here’s the thing: When you use a wider rim with the same tire, you increase the internal diameter of the tire/rim combination — you effectively have a bigger tire. I believe you understand that, or you wouldn’t have asked if it allows you to lower the pressure.

“So what?” you might say. Well, here’s the part that may not be obvious: if you do not lower the pressure (i.e., the tire and the pressure inside it are the same with either rim), the stress on the rim walls (and on the tire casing) is HIGHER with the wider rim than with the narrower rim. You MUST lower the pressure simply to get to the same level of stress on the rim walls as you had before with the narrower rim. Here’s why:

We can’t have this discussion without talking about stress (σ), which has the same units as pressure and is defined as “the force per unit area on a body that tends to cause it to change shape.” I will show why you must lower the pressure as the rim gets wider in order not to increase the stress on the rim and tire walls.

With a cylindrical thing like a tire/rim combination, we are talking about hoop stress — check out the illustration of the cylinder section. Hoop stress is what is trying to tear your tire casing apart, fold your rim walls outward like a limp taco shell, and yank your tire beads out of the rim, so pay attention.

The hoop stress on the wall of the cylinder is:

σ = force per unit area = F/A

The area (A) of the material being stressed in this case is the length (L) of the cylinder times its thickness (T), or A = TL.

The force (F) being applied by the air inside the tire and rim is equal to the air pressure (P) multiplied by the cross-sectional radius (R) of the tire/rim cylinder multiplied by the length (L) of the cylinder. But the radius (R) is equal to half of the diameter (D), so F = PDL/2, and:

σ = F/A = PDL/2TL = PD/2T

The tire is the same in both cases, so its thickness is the same in both scenarios, and we will assume that the rim wall thickness is the same with both rims as well. Thus, the thicknesses cancel, and, with subscript “n” referring to the narrow rim, and subscript “w” referring to the wide rim, we have:

σw/σn = PwDw/PnDn

If we want to keep the stress on the tire casing and on the rim walls the same with either rim, then σw = σn , and σw/σn = 1, so:

PwDw = PnDn

Now, let’s assume that your 700 X 25C tire has a cross-sectional diameter (Dn) of 25mm when it is on a typical narrow rim — a 19.5mm-wide rim whose inner rim width is 14mm. Now, if you put it on a wider but equally thick 24mm-wide rim, its inner rim width is 18.5mm, and you have increased the cross-sectional circumference (C) by 18.5 – 14 = 4.5mm. Since C = πD, you have Cn = π Dn = π * 25mm = 78.5mm.

We know that Cw = Cn + 4.5mm, so:

Dw = Cw/π = (Cn + 4.5mm)/π = (78.5 + 4.5)/π = 83/π = 26.4mm

So, by going from a 19.5mm rim to a 24mm rim, you have essentially made your 25C tire go from being 25mm in diameter to 26.4mm in diameter.

Let’s look at your rear tire, in which you were running 100psi in your 25C tire with the narrower rim. Since maintaining the same stress (σ) on the rim and tire walls requires that:

PwDw = PnDn,

then Pw = PnDn/Dw = (100 * 25)/26.4 = 94.7 psi

Thus, just to maintain the same stress on the walls of the wider rim (and on the tire casing), you must decrease your tire pressure by 5.3psi. Using the same analysis, the pressure in your 90psi front tire must drop to 85.2psi with the wider rim. For this reason, a 33mm cyclocross tire at 70psi or a 5-inch (127mm) fat-bike tire at 18psi feels just as hard as a 23mm tire at 100psi. They feel just as hard because they are just as hard — the tension on the casing is the same in each case (assuming the tire casing is the same thickness, that is). And furthermore, if they are clinchers, the outward force on their rim walls is the same as well.

So, in effect, when you lower your pressure by the 10psi you suggested, you are having the same reduction in stress on the rim walls as if you had the same rim width and had only reduced the pressure by 5psi.

And then, when we are talking about rim heating due to braking, we are talking about dramatic increases in pressure. Many of you have had tires explode in a hot car that you know would have been fine to ride on or leave inflated in your house without incident, and rim heating due to braking can drive temperatures even higher yet. So I would tend to think that the 5psi you would effectively be lowering your tire pressure by would make a relatively minor difference (relatively “inconsequential,” as you described it) in whether your rim was going to fail and your tire was going to blow off due to braking.

In the end, I think this means that what will be a lot more important than that rim width and air pressure difference will be the rim construction, and, particularly, the resin used. As I discussed here, different wheel brands use different resins with different Glass Transition Temperature (Tg), the temperature region in which a non-crystalline material (like the hardened resin in a carbon matrix) transitions from a hard, glassy material to a soft, rubbery material. If you pick a rim with a high enough Tg, you will not have to worry about your rim failing on a long descent. And perhaps even more important yet is your braking technique on a long descent: squeeze and release the brakes, squeeze and release, squeeze and release, etc., and avoid “riding” your brakes (holding them on constantly). With your 175-pound body weight, you will also heat a rim faster than a 150-pound rider will and less rapidly than a 200-pounder will, so ride steep descents with heavier riders and pull over immediately if their rim or tire fails. And, of course, you can get disc brakes and never worry about rim heating during braking again.
― Lennard