prathmann

A link to one FEA was provided in an early page of this thread and I believe one by Jobst Brandt was also mentioned. As I wrote previously (#170) in response to 'waterlaz':
"A paper giving an FEA of bicycle wheels was already presented at:
https://www.ewp.rpi.edu/hartford/~ern...inalReport.pdf
As expected it shows large reductions in tension of a few spokes at the bottom of the wheel and much smaller increases in other spokes. There's a table with the forces on p. 17 and you could calculate the corresponding stored energies. But just glancing at the table makes it clear that the sum of the spoke stored energies went down when the load was applied."

tomato coupe

Okay, agreed.

waterlaz

It's actually really close. I've taken the time to calculate the squares of spokes' tension (the energy is proportional to that) for a 28 spoked wheel from that FEA.
The spokes that became more slack have lost about 41000.
The spokes that became more tense have gained about 30000.

waterlaz

You can safely assume that the tire pressure doesn't change when the wheel is loaded. But I would ignore the tire altogether when considering spoke tension of the wheel.

MikeyBoyAz

Except you can't. A force is applied, which affects spokes within a specific region. Without accounting for these forces, and the work performed, the model is incomplete.

waterlaz

Yes, you can. The tire pressure does not change when the wheel is loaded.

FBinNY 

I agree with you but let's nail it closed for those who like to split hairs.

Tire pressure change is inversely proportional to volume change (at same temp) So when the loaded tire sags, there's a small drop in volume, and comparable increase in pressure. This is tiny at high pressures, but might be a few percentage points with lower pressure tires which sag more. Keep in mind that even if the tire sags all the way to the rim, it's still only a few inches out of a circumference of about 80".

While tire pressure does affect spoke tension somewhat, you still ignore it when building the wheel. That's because whatever guidelines or standards you used to determine optimal tension were based on the wheel before the ir was mounted.

Guidelines based on mounted nd inflated tires are problematic for two reasons. First, it would impose an added burden to have to mount and inflate tires every time you needed to check tension. More important, since multiple width tires at various pressures may be used, any standard based on inflated tires would be useless because it would call for unknown data.

wphamilton

Well if we're splitting hairs, that would depend on the shape of the cross section of the unloaded tire wouldn't it? My 23's looks a little elliptical, so the volume would decrease at some sag (making it more spherical), thus lowering the pressure. But to split it again, I'm looking at the outside of the mounted tire, not the inside where the air is so I can't really tell for sure.

Either way, the first guy is right that the pressure changes when loaded. Which you confirmed. An increase in pressure will compress the rim, decreasing spoke tension, so he's right about that also. My intuition however, is that the change there is very much less than changes due to the lower rim distorting (squashed) under load. Yet I'd want to see credible measurements or calculations to be confident of that.

waterlaz

No! No matter how you design the tire, the volume will decrease when load is applied. Pneumatic tires act like a spring by compressing air inside.

And no, the pressure will not change in any measurable amount, since the change in volume is minimal.

wphamilton

Yes! If it is circular and squishes to elliptical, then volume decreases. If it is elliptical and squishes to circular, then the volume increases. (A circle encloses the largest area of all plane shapes having the same perimeter measure).

Construction of the casing and sidewalls are among the factors which determine the unloaded shape.

And you should have said "And no yes the pressure will not change in any measurable amount" since you're agreeing with me that the resultant change in spoke tension is insignificant.

waterlaz

When there is no load applied the tire is at it's maximum possible volume.

Every system tries to get into a state with minimal energy.
If you can reshape the casing in such a way that the volume increases, the air inside will lose some energy (pressure times the change in volume). When you apply load to the wheel, you give some energy to the air inside the tire, which causes it to compress and reduce in volume just a bit.

Bike Gremlin

Pressure of tabular tyres affects the rim a bit.
Pressure of tubeless and tube tyres does not - since the same force that pushes against the rim is also, at the same time, pushing the tyre off the rim, while tyre sits in the bead and holds on. So the two forces dismiss each other - law of action and reaction.

FBinNY 

The entire tire volume / energy debate is an example of looking too closely and seeing dots instead of the painting.

First of all, don't mix energy and static load considerations. They're entirely separate issues.

As for tire volume/pressure changes and tire shape, that too is unrelated to the question of how a tire supports load.

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The air is pressing on the walls of the tire (through the tube) making it a semi rigid tube the same way a canvas firehose becomes round and rigid when charged with water.

In the unloaded wheel, the air is pressing in all directions (radially in and out, and to both sides) equally, so the entire system is in equilibrium. When that wheel is placed on the ground and loaded, the tire distorts until the contact patch assumes an area such that it's area X the pressure = the load. (discounting any support from the tire's stiffness)

If you were to draw a free body diagram of the loaded wheel, you'd see that the new equilibrium is achieved by the ground supporting the contact patch and the load pressing down. Everything else is unaffected, and the slight change in air pressure from the tires distortion isn't relevant.

BTW- while bicycle tires may appear elliptical, the air chamber (beyond the rim) is very close to round because the pressure forces the tire to assume the maximum volume profile. Of course a non-round profile is possible, but that requires a more rigid shaped body and/or stiffening belts to shape the tread area. Since it's preferable for bicycle tires to be as light and supple as possible, making them stiff enough to resist the tendency to be round would be counter productive.

What appears to be an oval profile is the result of the added thickness of the tread plies outside of the circular profile tire body.

FBinNY 

No, No, NO.

Both tires cause compression forces on the rim, though a different mechanism is involved.

In conventional wired on tires the air presses directly against the rim (through the rim band) which is part of the air chamber.

Tubulars are a bit different because if they were rigid they wouldn't press on the rim. However tubulars are built like those "Chinese finger traps" we played with as children. The tire is built of bias plies and pressure within makes them fatter and shorter, the net effect of which is a constricting force around the rim. It's that constriction in the curved belly of the rim that holds them fast, with the glue only acting to stabilize them from sliding or rolling to either side.

Either way, air inside the tire adds compression force to the rim, and will cause some slight loss of spoke tension.

Bike Gremlin

A wire bead tyre is pushed off the rim by the pressure. Isn't that correct? If it isn't seated well in the bead, it blows off.

What holds it on the rim? Bead. It pushes on the bead, resisting the force to be blown off. So the same pressure that compresses the rim, also pushes the tyre out, but the tyre holds onto the bead and stretches the rim out. These two forces cancel each other out from my understanding of physics. Am I missing something?

wphamilton

With all due respect, I think you've missed what the discussion is about. People are questioning where, how much and why the tension of spokes may increase or decrease. The only relevance of a change in pressure in tires, in this context, is how much it might affect spoke tension.


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I doubt that the oval shape is due solely to the added tread thickness, although you're kind of repeating me, when I said that I'm not looking inside where the air is. It is possible that part of the elliptical shape is due to the construction of the carcass. Especially in light of a couple of obvious advantages in doing it that way, against the likely downside of higher rolling resistance.

FBinNY 

There's a reason tires have a wire or Kevlar bead. It's that non-stretchable bead which keeps the tire from stretching over the rim flange and blowing off. The force on the flange is outward to the side and not away from the axle.

The fact that the tire cannot pull the rim flange upward is apparent with straight side rims where there's no mechanism for the tire to do so. On hook edge rims, the rim hood does provide some help to the bead, so there will be some tension, but this is small in comparison to the radial force exerted by the air acting on the belly.

FBinNY 

No offense taken, but I was aware of the earlier context vis a vis spoke tension, and had posted to that earlier. This particular post was a direct response to the more recent back and forth about tire shape, volume changes, etc. and the air inside acting like a spring.

I don't know if we're agreeing or disagreeing, so I'll simply repeat that the tire wants to assume a profile such that the air chamber, or more precisely, the two bias plies have a circular cross section. This is the most relaxed condition, and resisting that to form an oval in either direction would require bracing of some kind. In any case, whatever benefit would arise from an oval profile would be highest if the oval were the flat way (like car tires) to minimize the radial distortion required to achieve the necessary contact patch area.

waterlaz

I guess this is subjective but I think the volume/energy point of view is looking at the big picture. I don't have to think of the details.

Static loads and energy are closely related. This is a typical method of analyzing statical loads. Just pretend everything to move just a bit and see how energy distributes. Once you do that the forces are usually easy to work out.

PS I'm not sure if "Virtual work method" is a correct name in English but it looks like it.

Bike Gremlin

If there is no force pushing the tyre outward, then it wouldn't hold the weight of the bicycle. Simple example of action-reaction law IMO.

So while the grip and traction on the bead does come from side way pressure, there is also force pushing the tyre outward, off the rim - equal to the force compressing the rim.

FBinNY 

Interesting that you quote my passage describing the role of the bead then totally ignore it.

Yes, air pressure pushes the tire outward, but the tire isn't pilling on the rim, it's pulling on the non-stretchable bead. Without that bead, the tire would blow off. So, the air pushes the tire out, and likewise pushes the belly of the rim inward.

Another way to look at it is that the tire and rim are two parts of a single pressure vessel with the air inside pushing the walls outward in all directions.


BTW- since you opened this post with the reference to Newton's 2nd Law, and correctly stated that if the air didn't push the tire out it couldn't support the weight of the bicycle. By extension of that logic, if the air didn't push the rim inward it couldn't support the weight of the bicycle either.

wphamilton

Sure. If you pump air into a steel box it wants to become a sphere. But it also resists changing its shape. A child's balloon wants to retain its shape. It may be that all bicycle tires assume a perfect circular cross section in the interior, but I'm not ready to assume that just because we pump air into it. Are you?

The benefit I was thinking of, perhaps not so obvious after all, had to do with absorption of minor impulses from the road surface.

FBinNY 

Unlike a metal pressure vessel, bicycle tire bodies are made of 2 loosely attached bias plies. These are designed to be very flexible and supple and offer very little rigidity compared the stresses of containing air at 100psi. Balloons retain shape somewhat because the pressures involved are so low compared to the tension of the elastic membrane.

So, I don't assume tires will be perfectly round, I only assume that they'll be close enough for it not to matter. OTOH- since the normal state would be a circular profile, I'd expect that unless there were specific design factors to lead me to think otherwise. I don't see any in bike tires, so I figure nature will take it's course.

As to the benefit of shock absorption, as I pointed out earlier, that would be best if the oval were oriented so the flatter surface is to the road, simulating the effect of a wider tire.

But all this doesn't matter in the scheme of things until/unless some tire maker intentionally produces oval profile tires and claims an advantage. Then it would be worth revisiting to try to separate hype from fact.

wphamilton

I sense that you're trying to withdraw (and I'm off-topic here) so I won't go into it, but I believe the bolded part to be incorrect.

FBinNY 

I have no problem agreeing to disagree on that point. As I said earlier, it doesn't matter until/unless someone decides to market and make claims about oval profile tires.