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You just stated that pressure is force/area.
Therefore if you have a constant pressure acting on a larger area, the force acting on the larger must be greater. i.e. 60psi acting on 10square inches = 600lbs force, whilst 60psi acting on 20 square inches=1200lbs force.
If you increase the volume of a tyre, the surface area of the tyre also increases - e.g. a 2" tyre has approximately twice the surface area as a 1" tyre. Therefore the total force acting on the tyre surface is also approximately double, for a given pressure.
Now, the only thing stopping the tyre expanding sideways at the bead is the rim. This means that the tyre exerts a force at the rim.
This force is effictively caused by the pressure acting on the tyre surface, which as I just explained above will be greater for a wider volume tyre.
Or, another way of thinking about it:
Raising the pressure of a gas takes work/energy.
If you have a 2" tyre and a 1" tyre, the 2" tyre will contain approximately four times the volume of air compared to the 1" tyre.
If you inflate both tyres to 60psi, then for the 2" tyre you had to inflate four times as much air to 60psi as for the 1" tyre! That means that you have to use a lot more energy/work to inflate the 2" tyre, ultimately resulting in the greater forces felt at the rim.
And if those arguments don't convince you, then I am afraid I must admit defeat!
a 2" tyre has approximately twice the surface area as a 1" tyre. Therefore the total force acting on the tyre surface is also approximately double, for a given pressure.
Well, that just contradicts itself; the pressure acting on any square inch of the tyre or rim is always the same; 100psi.
Your tyre inflation idea actually conflicts with your previous explanations...
Ok, sorry to say this but you seem to have a fundamental misunderstaning of the difference between force and pressure, and how the two are related. I give up.
Ok. Anyone else want to give it a go?
Nope.
One last go.
Total force isn't the same as pressure.
As you say, the pressure is the same.
But the total force is pressure times area.
Hence greater for a larger (volume, therefore area) tyre.
But the total force is pressure times area.
????
Ok, I need someone who can explain this clearly, as I keep asking. I hope none of you are teachers, because a) you don't seem convinced yourselves, judging by all the conflicting 'explanations', and b) none of you are explaining things in a manner that can be understood by an idiot, ie me. 😀
a) Everyone has been explaining the same thing in different ways, you are just interpreting them as being conflicting when they are, on the whole, not.
b) You need to lay the foundations before you can build the house.
Ok, I need someone who can explain this clearly, as I keep asking.
why don't you go away and read up on the theory, then you might be able to understand what has been written above
the explanation doesn't have to be perfect to be understood
Nobody's explained this clearly and effectively. People ramble on about 'force' and this and that, but a lot of it is conflicting ideas.
I think that you just don't undertstand the basic concepts and that you're don't understand what actually causes rims to fail due to excess pressure int the tyre.
One last try
Thing that causes a rim to break/blow/whatever is the tyre pulling it apart, not the air pressure directly acting on the rim for which you're right, the size of the tyre has no influence.
Imagine that you're holding two pieces of material that are being pulled apart and you're trying to stop them moving. That's the equivalent of the rim trying to stop the tyre beads being pulled apart due to tyre pressure.
The force that the rim feels is caused by the air pressure inside it but not directly due to it - eg the pressure is causing the beads to want to pull away from eachother which the rim is resisting.
If you double the pressure in the tyre then the force with which the beads are trying to move away from eachother will also double and as such, the force that is applied to the rim is doubled. Make sense?
Assuming so, then if you keep the same pressure but double the surface that the pressure is acting on (eg double the size of the tyre) the you'll get the same result - twice the force since force = pressure x area.
That's as clear as I can think of making it.
OK. My turn to try.
We've already said that force is not the same as pressure. Pressure acts over an area, force just acts in a direction (at one point). I'll try to focus on force.
I think the reason rim manufacturers have limits is to prevent the rim sides being split apart - too much force against the walls from the inside will press them away from each other. The problem is [u]not[/u] that the rim will be squashed to be a smaller circle, but that the rim in cross-section will become less like a U shape (where the sides of the U are the braking surfaces, if you had rim brakes) and more like a big V as the walls get pulled apart. Also that the hooky bit of the rim that holds the tyre on will get pulled away from the rest of the rim.
So I'm going to try to explain this while thinking about a cross-section of the rim with the tyre and tube fitted - don't worry about the whole rim as a circle, that's not the problem, just think about the cross-section of the rim like this:
[img] 
[/img]
When you fit a tyre to a rim, as you inflate the tyre the bead (the bright red dots) hooks inside the rim wall to hold the tyre on. As you increase the pressure in the tube (the pink circle), the tube expands and applies more force to hold the bead into the groove inside the rim wall. However, as the tube expands, it also applies more force to the rest of the tyre (the bit at the bottom of the picture above), which tries to pull the bead out of the groove, or pull the hooked bit of rim away from the rest of the rim.
The way the air pressure acts to try and expand the tyre (and pull it away from the rim) is important here. [b]The bigger the tyre, the more tyre material there is for the air pressure to act on.[/b] So a small tyre will only have a small cross-section of material for the air pressure to pull on, whereas if you try to imagine the picture above with a tyre double the size, you can imagine that there will be lots more tyre material that the air pressure is pressing against.
That's the important bit - if you fit a small tyre and you inflate it to, say 30psi, the air pressure will only be acting on a small area of tyre so it will not provide a lot of force against the side walls of the rim or try to pull the hooked part away. If you fitted a big tyre and inflated it to 30psi, that pressure will be acting on a lot more tyre so there will be more force trying to pull the hook away from the rest of the rim.
The "problem" force is not the force between the rim walls, it's the force that the rest of the tyre (not inside the rim) is exerting on the rim. So if you keep the air pressure the same, more tyre means more force trying to pull the rim apart.
I guess if you want to bring the difference between pressure and force into it, think of one side of a tug-of-war with one man per metre pulling on a fixed length of rope. The force on the rim is the overall pull on the rope. You can increase the force on the rope by increasing the number of men - say two men per metre - and this is similar to increasing the pressure in the tyre. Or you can increase the force on the rope by adding on more rope and keeping one man per metre, effectively adding more men. This is like fitting a bigger tyre at the same pressure.
Hope this makes sense.
: P
I think what you're explaining is this:
[img] 
[/img]
The bit where I'm getting stuck, however, is that the rim is designed to withstand a certain maximum force over it's area, but if the area remains constant, then the pressure/force exerted on each square inch of it always remains the same, no?
By your explanation, the force generated by the pressure inside the tyre is spread out all over the entire surface area of the tyre and rim, not concentrated on any one part of the rim. So, the force being applied overall, within the entire volume, is greater in the fatter tyre, but it's pushing out in all directions, not just against the rim sidewalls.
that the rim is designed to withstand a certain maximum force over it's area,
No! The rim is designed to resist a force splitting it apart. Area has nothing to do with it.
The tyre is trying to split it apart. The rim doesn't care what pressure is in the tyre, only what force the tyre is applying to it, pulling it apart. Imagine that you were the rim and you were holding the two beads together. You'll just have to take the word of people that twice the pressure in teh tyre leads to twice the force pulling the beads apart.
Or you can increase the force on the rope by adding on more rope and keeping one man per metre, effectively adding more men. This is like fitting a bigger tyre at the same pressure.
That makes more sense, to my feeble little mind, but that translates also as increased pressure on one square inch of the rim, to me.
See, the larger tyre is expanding outwards away from the rim, surely absorbing all this extra force?
[i]The bit where I'm getting stuck, however, is that the rim is designed to withstand a certain maximum force over it's area, but if the area remains constant, then the pressure/force exerted on each square inch of it always remains the same, no?[/i]
This is true - but the problem is not the air pressure's [u]direct force on the rim[/u], it's the air pressure pushing on the bit of the tyre that's outside the rim - the direction of the forces that result from this means that the tyre tries to pull the hooky bit away from the rim and make it more of a V-shape.
: P
Sorry but you clearly just don't understand how force/pressure and so on work...
PRESSURE ON THE RIM IS IRRELEVANT!!!!
THE FORCE GENERATED BY THE PRESSURE IN THE TYRE ACTING ON THE RIM IS WHAT'S IMPORTANT!!!!
and breathe.
The tyre doesn't absorb _all_ the extra force from the air pressure. Tyre casings have lots of threads that are designed _not_ to expand too much. What happens is that extra air pressure is converted into extra force trying to pull the tyre away from the rim.
: P
Hang on a minute, I'll try and draw a picture.
: P
Sorry but you clearly just don't understand how force/pressure and so on work...
Do you? Because if you do, you're not explaining it very well...
Hang on a minute, I'll try and draw a picture.
Goody! I like pictures! 😀
Maybe not but it's proving difficult 😉
OK, here we go...
[img] 
[/img]
Those are the rims in black biro, and the tyres in blue biro. I've assumed that the inner tube will inflate to fill the entire space (as it does) so the little arrows show force.
The "problem" forces are the ones I've drawn in pencil. They're the ones that will try and pull the rim apart. They're caused by the tyre pulling on the rim in that direction.
As I hope you can see, the arrows inside the tyre are pretty evenly and regularly spaced. Imagine both tyres are at the same pressure. There are the same number of green arrows in both diagrams, because the tyres are at the same pressure so there are the same forces acting on the [b]inside[/b] bit of each rim.
However, the red arrows should also be pretty evenly spaced. You don't need to count them to see that the bigger tyre has more red arrows because the air pressure is acting over more of an area of tyre material. As the tyre doesn't expand (very much) under this pressure, this is effectively the same as adding more rope, with more men at one man per metre, to the tug-of-war analogy. The pressure itself (one man per metre) hasn't increased, but the amount of stuff that the pressure is acting on (the rope, or the tyre) has increased.
And that's the bit that means there is more force being exerted, in the direction of the pencil arrows, on the rim. Not from the amount of air inside the rim (which stays the same), but through the tyre pulling on the rim edges. A bigger tyre at the same pressure means more pulling.*
I hope this makes sense!
: P
* or, what I haven't shown, more air pressure means more pulling in both cases. It would increase the number of all arrows.
That's a lovely picture, Pierre, and I really do appreciate your efforts to help me understand this conundrum, but I'm afraid I remain unconvinced. You're explanations have definitely been by far the best though. I suspect some others have blundered along, then reached a dead end, as they don't fully understand it themselves...
To me, it just seems as though the force is being dissipated over the larger area of the bigger tyre, rather than exerting more leverage on the beads of the rim.
The force isn't dissapated in the larger tyre - or only a small amount - that's what the threads in the tyre are are for - stopping the rubber stretching.
Try it yourself - pump up your tyre to say 30 psi, measure the circumference (roll it along the ground) then inflate to 60psi and remeasure. While it may expand a small amount, the size of the tyre certainly hasn't doubled.
It's not that the force is being dissipated, it's that there is more force, because there is more air at that pressure.
Each of those arrows is meant to represent the same amount of force. Pressure is force per amount of area, so if the pressure stays constant then more area means more overall force. If you count one arrow as one unit of force, then inside the rim there are 11 arrows - 11 units of force - in both cases. This makes sense because there is the same area, the same amount of air, at the same pressure, inside the rim itself in both cases.
But in the tyre, outside the rim, the small tyre has 16 arrows - 16 units of force - acting on it. The large tyre has 26 arrows - 26 units of force - acting on it. Those forces don't just get dissipated or go away. Because the tyre can't expand, those forces pull on the tyre so that they are transferred to the rim (the tyre is the rope in the tug-of-war analogy).
Now, granted, the rim is pretty stiff. In most cases, it doesn't move. It would be like the tug-of-war rope being attached to a brick wall. Because there is an equal and opposite force being exerted, the brick wall exerts a force on the rope so that it doesn't go anywhere. Likewise, the rim exerts a force right back against the tyre to stop it going anywhere.
But just like the wall is only so strong, a rim can only provide a finite amount of force to resist the air pressure trying to pull the tyre and trying to force the rim apart. If you add enough men, or enough rope with one man per metre, you can pull the wall down. Likewise, if you put in enough pressure, or make the tyre _really_ big and keep the pressure low, that resultant force will pull the rim apart.
The trouble is with force is that it can't be dissipated. It can be opposed: if I push against a wall, the wall opposes with the same force. But it can't just go away. The force isn't dissipated, it's still there, as the pressure senses in my hands will tell me.
: P
To me the point is that it's the tension in the tyre carcass that is trying to pull the bead off the rim.
1. The tension is greater for a wider tyre at the same pressure (I think) and
2. The direction of that tension (i.e. in line with the carcass) on the narrower tyre makes it more difficult for the bead to be pulled off?
Talkemada,
The dead end you mention wasn't the limit of my understanding but when I realised that however good the explaination you were never going to understand some pretty basic science.
I suspect some others have blundered along, then reached a dead end, as they don't fully understand it themselves...
are you trolling? people are trying to help you out here
it is possible to fully comprehend something without being able to articulate your understanding of it
I think Talkemada is up to his elbows in leg-pulling.
are you trolling? people are trying to help you out here
No, I do appreciate it, especially Pierre, who has done a lovely picture, and made more sense than anyone else really.
however good the explaination you were never going to understand some pretty basic science.
Well, it doesn't seem that basic to me. The explanations have been contradictory and confusing.
I stil don't get it. 🙁
I'm gonna have another try.
Do you accept that a larger tyre has more surface area and force is pressure x area?
If so you will understand that there is more force acting on the tyre.
The tyre is very flexible so the only thing that is stopping it opening up is the rim.
Try pumping up a inner tube inside a tyre off a rim. The beads get further apart as the tube inflates
The force is NOT dissipated. Force does not dissipate.
So therefore is nothing to stop the tyre beads spreading apart other than the rim. What is making the tyre open up is the FORCE created by the pressure x area so the rim feels more force with a large tyre.
Ok, so the only way I can rationalise it is like this:
Imagine a bridge that is suspended over a gorge, held up by steel cables. The bridge is 10m wide, 100m long, and weighs 100 tonnes. So, that's 0.1 tonnes per sq metre, pulling down on the cables, yes? Or have I got my maths wrong?
So imagine if you doubled the width of the bridge, but kept the same cables in place. Still 0.1 tonnes/sq. m, yes? But twice as much weight, or 'force', pulling down on the cables? If you kept widening the bridge, without reinforcing the cables, then eventually their limit would be reached and overcome, no?
Is this the kind of thing you're all trying to explain?
No, that has nothing to do with what we were talking about.
Pft, so explain it then!
All I've had so far, is a load of guff saying that the 'force' is increased on the rim, without any actual understandable (in Idiot) explanation as to why this is so.
I'm sorry, but I'm not going to accept something is 'so', without understanding it.
Carry on...
No thanks...
Oh no!
Talkemada won't accept some basic physics!
You have kind of got it there. The weight of the bridge is the equiv of the force from the tyre (each unit area of the bridge exerts a force due to mass same as each unit area of the tyre exerts as force due to pressure). i.e. double the width and the brdge weighs twice as much. The cables are the equiv of the rim. This is fixed in size/strength. The only thing resisting the force from the weight of the bridge is the cables so now they are twice has heavily loaded, same as the rim is twice as heavily loaded as the tyre exerts twice the force.
Actually the bridge analogy is not bad. Part of what I tried to explain is that more tyre with the same pressure acting on it (per unit area of tyre) will mean more force on the rim. Which is quite a lot like more bridge of the same density means more weight on the cables.
So to carry on that analogy, the rim strength is the strength of the cables holding up the bridge. A small tyre is a narrow bridge; a big tyre is a wide bridge. Increasing the pressure is like adding more material uniformly to the bridge: you can make the narrow bridge much thicker before the cables snap, whereas because the wide bridge has much more area, you can only make it a little thicker (all over) before the cables snap.
Kind of convoluted, but hope that helps.
: P
Talkemada won't accept some basic physics!
Why accept something you don't fully understand?
See, in just a few posts, one person has said 'no that's not what we're talking at all', to two others telling me that I'm on the right track! Can you see how confuddling this is for me? 😀
Seriously, this is not a troll. I simply don't understand it, that's all. I'm sure there's probably other stuff that I have in-depth knowledge of, that others might not 'get'. Make sense?
So, I'm trying to understand how it works. And maybe others would find it interesting, I don't know. But to say 'oh you're stupid you don't understand just accept it' isn't really very helpful. Why not go away and have a think about your abilities to explain stuff?
I'm sort of starting to understand it, but I'm still having difficulty in getting how the bridge analogy applies to a rim. In my head, the bridge analogy is sort of opposite to the tyre/rim scenario. Because I'm thinking that the increased load on each cable is like an increase in pressure. Can you see where I'm coming from? The 'bridge' thing is my way of trying to understand the force/area/pressure thing that youse have explained to me. The next step is trying to work out ho that applies to a structure like a rim and tyre combo. I'm having difficulty in understanding why the tyre beads would be pulling on the rim walls more, if the volume is increased. Surely that force would be the same regardless of the tyre volume, given the same pressure?
I think that's the bit we're all trying to explain and why people keep repeating that pressure and force are very different things.
I weigh 75kg. That means the FORCE I exert on the earth due to gravity is 75kg (assuming g=10ms^-2 and not using Newtons as a unit for the moment)
If I stand on the floor in my Vans, the surface area of the soles is, let's say (25x10cm^2 x2 = ) 500cm^2. So the PRESSURE I exert on the floor is 75kg per 500cm^2, or 0.15kg per square centimetre.
However, if I put on pair of stiletto heels and stand on just one heel, I haven't changed my weight, so the FORCE I exert is exactly the same. But this time all that force is concentrated on an area, let's say, 1 sqaure centimetre in size. The PRESSURE I'm exerting is now 75kg in just one square centimetre.
Does this make sense so far?
: P
It's seems CRC would like me to run a Grand Prix on my rear wheel, having sent me a 700c example instead of the Speed King 2.3 I ordered. Taking the lower tread / fasting running rear tyre concept to the logical conclusion?
However, if I put on pair of stiletto heels
Kinky! 😮
I understand that, I just don't understand how a fatter tyre at the same pressure can exert a greater force on the rim, at the same pressure as as skinny one.
My way of looking at it is like this: The rim always has the same surface area, so the only change in surface area, is in the tyre. The increased volume of air may well be exerting more force because of the increase in surface area of the tyre, but that extra force is exerted outwards in all directions, not just against the rim. So, the force created by the pressure, is still the same regardless of tyre size, no? How does the bigger tyre then become a stronger lever to push/pull against the rim's edge, if the pressure is the same? Surely all that extra force is being used to push against the extra surface of the tyre?
You've got to think of that force becoming tension in the tyre carcass which is then pulling the bead off the rim.
Oh so now 'force' becomes 'tension'?
And you wnder why I'm confuddled?? 😯
You're right that the forces from the air inside the rim do not change - after all, as you say, the rim has the same surface area, the air pressure is the same, so the force it exerts on the rim is the same.
I think you may not be seeing how the forces on the tyre are turned into forces pulling on the rim.
The bigger tyre, as you say, has more surface area so has more force acting on it. However, because the tyre does not expand (or, at least, doesn't expand very much - as someone else said, double your air pressure and your tyre will not double in size), those increased forces don't just act to expand the tyre.
OK, an imagination experiment:
Imagine you have a small cereal box, one of those little variety pack ones. You've taken all the cereal out and left the flaps open. You turn it upside down and sellotape the flaps to a table so it's "open" side down and the flaps are sticking out, taped to the table. Now you drill a hole through the bottom of the table under the box and put a balloon through it. Then you try to inflate the balloon. It will need quite a bit of puff, but sooner or later as you blow up the balloon it will force the box off the table. The force from the air pressure inside the balloon will not be able to expand the box, so it will be released in the only remaining direction, and lift the box away from the table.
Does that make sense?
Where the tyre analogy comes in is that if you instead use a full size cereal box, or maybe a shoe box, instead (and a bigger balloon), you won't need as much air pressure. As before, the air pressure will push against the sides of the box and they'll push back, meaning that the force ends up pushing against the table and lifting the box up.
Granted, this is not an exact analogy, the forces inside the tyre and rim are in slightly different directions, but the general principle is the same...
: P