Since writing about hoop stress in tires in March, I have had some back-and-forth email exchanges with some readers about it. It’s evident that how I’ve explained it thus far has not been clear enough — some obviously intelligent readers remain unconvinced that increasing the rim width increases the stress on the casing of the same tire run at the same pressure.
First, here is an interesting graphic on this subject sent in by a reader named Ross, followed by distilled versions of email exchanges with two other readers. I’m hoping that Ross’s graphic as well as some of the alternative ways of explaining it that I came up with will clarify it for those who still don’t see how increasing the rim width increases the stress on the casing of the same tire run at the same pressure.
A few years ago, when wide rims and tires were getting a lot of press and hype with no apparent scientific evidence/backing, I went looking but came up short. I was most interested in quantifying the width and volume increase in going from a 20mm to 30mm inner-width MTB rim. Or road rims and 25mm tires. Long story short, I did a little calculation where I assumed that a tire would take a circular (this is a “first principle” type calculation) profile and have a constant distance from rim hook to rim hook for any rim (constant casing length). Then I grabbed a 25mm tire and measured the distance from bead to bead (basically just the casing “circumference”) and plotted the portion of the circular arc that would be formed for different rim widths. I’ve attached this plot for you.
In your column, you say that removing a 23C tire from a narrow rim and placing it in a wider one makes the tire wider, and then you must reduce the pressure slightly to have the same outward force on the rim.
I disagree, and here’s why: Yes, the tire gets wider. But I think it surely also must get lower, meaning, smaller in overall diameter. When you pull the sidewalls farther apart, the outside area of the tire shouldn’t change. Instead, the center of the tread is pulled downward, in the direction of the hub, if that makes sense.
If I’m right about that, the tire changes shape, but not volume. It just becomes more squat in section. So, the pressure should remain the same, and the force on the bead doesn’t change.
Just thinking of a tire here, unmounted; the outside surface of it has x area. No matter how you wrinkle or fold or flatten it, if you don’t use tension that is constant. So why would it be different when you mount it on a rim and inflate? If the substance from which it’s made has, say, 3 percent elasticity in all directions at 100 psi it stretches to a set surface area that is y.
Whether the rim has an internal width of 17mm or 20mm, the distance and construction of tire doesn’t change just because you mount and inflate it.
Here is the simple image that troubles me: Take a rectangular piece of cloth (as a former member of a composites engineering team I can’t help but think about, say, biaxial dry carbon, but any cloth will do); set it flat on a table. Then push the long sides in toward each other until in section it is an upside-down “U”, similar to a tire. Now, slowly pull the edges back out — similar to a tire on ever-wider rims. The “U” loses height slowly at first, but eventually it is TOTALLY flat. It has zero height. Meaning, at the absolute maximum rim width possible the tire adds nothing to the height of the rim, except the thickness of the carcass.
That’s one thing. Now, as to interior volume, that is reduced until it no longer has ANY when the cloth is flat on the table again. Zip.
The cloth is of course an oversimplification; the channel in the rim BETWEEN the rim walls gives a wheel/tire permanent volume no matter the shape of the tire, and in theory if it is deep enough (say, if the tiny center channel on my Pacentis was eight inches deep) could add interior volume to the tire/rim combination as the rim was widened, even if the rim became so wide the shape of the tire was perfectly flat. But of course that volume would matter little since the tire wouldn’t be able to function as a tire.
These are great questions and mental images to ponder. We are in agreement that the total area of the tire itself does not change depending on what rim it is mounted on. But the total area of the rim plus tire does change with differences in rim width, and that makes all of the difference.
First, in response to your contention that the tire height necessarily becomes lower with the wider rim, I thought you’d be interested to see the above analysis sent to me by another reader. As you can see, Ross’s calculation shows that the height of the tire actually changes very little as the rim width increases, within a normal range of rim widths. You can also see that the volume contained inside the rim/tire combination also increases, but that is only relevant here in terms of what formula you use to compute the stresses on the tire wall; if the tire were really to flatten way out, a formula for a cylinder would no longer be representative. You are correct in your point with your image of the flat piece of cloth that as the rim becomes extremely wide, the tire’s height has to come down. In physics and mathematics, one way to test a theory is to “evaluate it at the limits” (i.e., see what happens when it gets very big or very small), and your example of making the rim progressively wider until the tire profile is flat is a good one.
After I discuss the forces on your flat piece of fabric, I’m going to show that the arguments I have made about tire stress still apply, even if the height of the tire had become lower with the wider rim. I also want to point out that much of your argument relates to air volume, but for the discussion of stress on the tire casing, the surface area inside of the rim and tire is the salient feature, not the volume.
But first, given that you are an engineer, I will take on your flat piece of cloth in a way that I’m hoping will make the point very clear. Let’s consider the rim that you describe that is almost as wide as the flattened-out tire so that the tire lies almost completely flat, and let’s ignore the point that is probably obvious to all of us as bike riders that the tire bead would not stay hooked in the rim in that circumstance.
As an engineer, you know that the force vector of the air pressure against the top of the tire is directed straight up. Now, physics tells us that if the tire is at equilibrium (i.e., the tire is staying put on the rim), then the net force on it is zero. That means that the vector sum of all of the forces working on any given patch of the tire add up to zero. Thus, the inward-directed components of the force vectors of the tension in the tire casing have to add up to the same magnitude as the one created by the air pressure directed outward. I realize that not everyone is familiar with vectors, but it should be obvious to you as an engineer that the flatter the tire profile, the smaller will be the inward-directed component of the casing-tension vector. In other words, as the tire profile becomes closer to flat, the casing tension will approach infinity, because when it is completely flat, there will be no casing-tension vector component in the inward direction, and thus no amount of tension in the casing can balance the outward force. If you “evaluate it at the limits” of flatness as you are asking us to, you simply cannot build a tire strong enough to hold any significant air pressure if its profile is completely flat.
Okay, enough about force vectors; let’s think about something that all of us can visualize, namely a water bottle being crushed by the increase in air pressure bringing it down from high altitude to low altitude. I’m going to use the example of a cylindrical vessel in which the pressure on the outside exceeds the pressure on the inside, because it is easier for people to visualize how much force it will take to crush such a vessel than it is for people to visualize how much force it will take to burst a cylindrical vessel in which the pressure on the inside exceeds the pressure on the outside (like a tire/rim combination).
I wrote a VeloNews article decades ago about Prof. Igor Gamow’s high-altitude-simulating hypobaric (i.e., vacuum) chamber; I slept in said chamber at a simulated altitude of 14,000 feet (by pumping air out of constantly) over a period of time and measured my blood hematocrit before and after that time period. (The Gamow hypobaric chamber is not to be confused with the Gamow Bag, a soft, hyperbaric chamber of similar dimensions used to rescue climbers on Mt. Everest suffering from pulmonary edema by, using a foot pump, pumping the bag up with the stricken climber inside to simulate sea-level air pressure.) The idea of the Gamow hypobaric chamber is to train high-altitude adaptation in athletes; it’s the more literal approach than an altitude-simulation tent, which reduces oxygen inside by flooding it with nitrogen, rather than by decreasing the air pressure inside. The Gamow hypobaric chamber was the approximate size and shape of a coffin (not lost on me at the time), and it was made of extremely thick fiberglass. I thought about this hoop stress issue while trying to fall asleep in the thing.
Bear with me; I’m getting to the point. The highest paved road in the continental USA is up Mt. Evans, and, being close to here, I have ridden and driven up it many times, and I have seen what happens to water bottles on it. Mt. Evans’s summit is at 14,265 feet above sea level, whereas Denver is at 5,280 (one-mile) elevation, so the difference is 9,000 feet. Air pressure at sea level is 14.7psi (1 bar), whereas it is 10.5psi (0.7 bar) at 9,000 feet. Most commercial airplane cabins are pressurized to 8,000-foot elevation, so this altitude difference is probably familiar to most of you reading this.
I think all of you who have done something similar know that if you were to consume the contents of a 1-pint water bottle or pop bottle at the top of Mt. Evans or while at cruising altitude in an airliner and then close it up tightly, when you drove down to Denver or landed at a sea-level airport, the water bottle or pop bottle would have been crushed. On the other hand, if you were to instead eat the contents of a 1-pint plastic peanut-butter jar at the top of Mt. Evans or while at cruising altitude in an airliner and then close it up tightly, when you drove down to Denver or landed at a sea-level airport, the plastic peanut-butter jar might be a bit squeezed in about its waist, but it would not be crushed.
Now let’s again “evaluate it at the limits.” I think all of you can likely imagine that if you instead had a bigger plastic jar made out of the same PET plastic and with the same wall thickness, and if this cylindrical plastic vessel were instead the size of a coffin, that if you were to close it up tightly at the top of Mt. Evans or while at cruising altitude in an airliner, when you drove down to Denver or landed at a sea-level airport, the coffin-sized PET jar would have been crushed. Yes? You agree?
Okay, now let’s say you were going to train a human cyclist and a mouse for high altitude in hypobaric chambers. You put the human in an opaque Gamow hypobaric chamber, and it is thick, because Dr. Gamow did the calculations and knows that the chamber has to be very thick to not collapse upon the athlete inside when air is pumped out to simulate high altitude. On the other hand, we can use the plastic peanut butter jar for the mouse’s hypobaric chamber, because we know that we can drive down from Mt. Evans with it sealed and it won’t collapse. Now, say the cyclist wants a window in his/her hypobaric chamber as big as the one the mouse has. To fulfill this request, we could simply take another peanut butter jar the same as mouse’s jar and cut it vertically down one side, flatten it out, cut a hole in the Gamow hypobaric chamber of that exact size and shape, and rigidly attach it into that hole.
Now, similar to the example of using the same tire on rims of different widths, the area of that transparent plastic piece from the peanut butter jar has not changed in either the mouse’s hypobaric chamber or the rider’s hypobaric chamber, nor has the pressure difference inside and out. But I think you can see from our above example (of an entire jar the size of a coffin but the same plastic and the same thickness as the mouse’s jar) that the pressure difference in the coffin-sized hypobaric chamber would cave that window in, whereas the mouse can be snoozing comfortably with that same pressure difference inside and out of an identical piece of plastic that is in the form of a cylinder.
Point is, the total area that matters in the tire/rim example is that of the tire and the rim together, not of just the tire itself; the two of them together comprise the cylinder, just as the walls of the Gamow hypobaric chamber and its plastic window together comprise that cylinder, and the area of just the window alone is not important in determining the hoop stress on the entire cylinder. The hoop stress depends on the surface area, not on its volume.
Finally, play around with this pipe burst-pressure calculator a bit. If you input a 0.5-inch-diameter pipe that is only 0.030 inches (30 thousandths of an inch) thick, it will constrain a pressure of up to 3,600psi (leave the allowable stress at 30,000 and the safety factor at 1.5). Now change the pipe diameter and leave everything else the same. Notice, for instance, that a 6-inch-diameter pipe with the same 0.030-inch wall thickness cannot hold a pressure greater than 300psi!
I hope these explanations and Ross’s great graphic are sufficient for you to see that the casing of a given tire has more stress on it at the same pressure when mounted on a wider rim than when it is mounted on a narrower rim.
In your article about the science of tire pressure and rim width, you made an assertion that the pressure on the tire carcass goes up due to the volume of the rim/tire combo going up. This may be an inaccurate statement.
If you used the same 25mm tire carcass on both the smaller volume rim and a larger volume rim the tire carcass did not change its square foot area. The area of the cylinder went up due to the volume of the rim. Pressure in PSI (pounds per square inch) is a per unit volume measurement. Since the tire carcass did not change its area, the pressure per square inch did not change on the fabric. It is all absorbed by the larger volume of the rim.
Your assertion that the pressure on the rim walls being higher would be correct. But I would believe that the person that designed the rim would account for the higher volume when placing the psi rating for the rim in the product guide.
You may want to recalculate and update the assertions in a new article to set the record straight for your readership. It may keep you from lost credibility.
The tire carcass has the same square inch measurement regardless of what rim you use. Pressure per square inch does not change. 100 Psi is 100 Psi. Since the volume of the cylinder goes up the potential energy within goes up. The TOTAL pressure on the container walls has gone up due to the greater container area. The pressure per square inch on the container walls though is the same, it is still 100 Pounds Per Square Inch.
I will give you that the potential energy released from the larger container has more kinetic energy, but 100 Pounds Per Square Inch is still just that, no more.
Thus if the tire carcass used is still the same, it has not grown since it was on the smaller ID rim, then it bears no more total pressure on the tire itself than it did than before. The rim volume has changed not the tire area. The larger volume rim will bear more total pressure than the smaller volume rim.
As a side note, the tire radius has changed. It is now greater than before changing the contact patch. Maybe you could do an article on that. The contact patch should be shorter than when it was on the smaller ID rim. This will affect rolling resistance…
In an attempt to clarify my position here are some cuts from your article:
“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.”
The tire diameter goes up, yes, but the height of the tire from the rim goes down. The tire does not stretch, it does not grow, it merely changes shape.
“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:”
True, but you are not allowing that the area of the tire is the same, not higher due to the increased diameter. Your formula accounts for area as a whole. Although the area of the cylinder has gone up the area of the cylinder that the tire encompasses has not. Therefore the percentage of the formula that affects the tire is not consistent. You failed to adjust how much of the area the tire covers in the larger cylinder. If the pressure on each square inch of the wall went up by say 10% but the area of the cylinder that the tire covers went down by 10% then the tire itself bears no more stress than before…
“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”
My point here is that with the wider rim you have recalculated the tire diameter. You did not adjust for the tire have a smaller surface area in the larger diameter cylinder. THE TIRE DID NOT GET MORE AREA, IT IS THE SAME TIRE AS BEFORE. IT NOW HAS LESS PERCENTAGE OF THE OVERALL AREA.
“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”
This formula is used for the total pressure applied to the total surface area of the cylinder. Total area! The tire now covers a lower percentage of the TOTAL area.
You’re correct; the pressure on the tire carcass did not go up as the rim width went up. However, the stress on the rim walls and tire carcass did go up at the same pressure as the rim got wider.
As for the tire diameter necessarily decreasing, see the accompanying graphic by Ross. I think you can see that the tire/rim volume does go up, not that that is the relevant quantity here.
And while you are correct that the tire’s area makes up a smaller proportion of the total area when it is on a wider rim, it is irrelevant how much of the total surface area of the cylinder is the tire and how much is the rim. The hoop stress depends on the cylinder’s total surface area. See my above answer to Jay.
In response to your Tech FAQ of March 28, 2017 and follow-ups on April 4 regarding eeBrakes: I just swapped out a set of the original eeBrakes (pre-Cane Creek) on my Paketa tandem precisely because even with the eccentric cam positioned to push the brake up, it was too close to my tire. I picked up a small piece of fresh asphalt in the — John