[Closed] Anyone running Conti Grand Prix's on their MTB?

The rim manufacturer is setting an upper limit. Stick to it.

I didn't know GPs came in a 1.5"?

I usually run tyres really hard to avoid punctures so they would not be the rim for me

mavic rims just come apart at high pressures...so beware

The GP's are 1 1/8". Its DT that are recommending not to go under 1.5".

I ran my 1.2" GPs at 90 PSI on Sunn rims which I'm sure would have had a similar rating with no issue.

i ran conti ultra gators on mavic 717s at 120psi for a while with no problems

was hellish uncomfotable though, so dropped it to about 90

I used to use the 1" GP's at over 100psi, if you're using rim brakes you'll soon find out if the braking surface is too worn, wear earplugs and safety goggles when pumping them up 😀 my ears are still ringing now. 😳

Contacted DT and the official response was not to use them. They aren't designed for it and rims would not be covered by warranty if they do give way.

mavic rims just come apart at high pressures...so beware

Do they? What, all of them?

Shit! I've been running Mavic rims on my road bike, with pressures of well over 100psi, for years! To think, just how much I've cheated death all this time... 😯

I shall change them immediately. What should I replace them with?

My audax bike uses 26" wheels with not very special CR18 rims - been running 25m GPs at 115psi without problems for a couple of years

Haven't had any problems with Mavic xm 317 & Conti GP's at 105psi yet 😕

High pressures and large volume tyres = lots of force pushing rim walls apart
High pressures and low volume tyres = not quite so much force pushing rim walls apart

IIRC my Mavic rims give two recommended pressures, one for narrow tyres and one for wide!

High pressures and large volume tyres = lots of force pushing rim walls apart
High pressures and low volume tyres = not quite so much force pushing rim walls apart

Eh? I don't get that. Pressure is pressure, no? Regardless of the space it's in?

IE, 100psi in 10cc is the same as 100 psi in 1000cc?

No? Am I missing something?

I thought larger volume = less pressure needed, at the expense of greater weight and rolling resistance.

I'm confuddled now... 😕

Pressure is force divided by area, therefore if you have the same amount of pressure (e.g. 60psi) acting on a larger surface area (e.g. the surface of a 2.1" tyre as opposed to that of a 1" tyre), then you will have a greater total force acting on the tyre bead....

At least I think that's the way it works.

No, I still don't think that's right. I don't think the volume has anything to do with it.

A skinny road tyre is designed to hold much higher pressure (up to 140psi), because a skinny hard tyre is mechanically more efficient; smaller contact patch, much less rolling resistance. If you factor in the additional pressure caused by the riders weight, and sudden spikes caused by bumps etc, then it would be safe to assume the tyre and rim are designed to withstand even greater pressure, maybe 200psi+.

A fatter, larger volume tyre is designed for comfort and shock absorption, rather than optimum mechanical efficiency, so lower pressures are sufficient. This would surely allow for the greater bump forces suddenly increasing the pressure within. If you land from a jump, then the pressure inside the tyre could spike from 40 to 80 psi, or perhaps even more. So, the tyre and rim would surely be designed to withstand higher pressures than the indicated guide. So, I suppose in theory you could have a rim of the same dimensions/materials/proportions as a road tyre, and a fat tyre which could withstand the same pressures as a road tyre. It just wouldn't be very comfortable off road.

I'm at a loss as to how increasing volume somehow would put more strain on tyre and rim, if the pressure remains the same.

Here's my reasoning...

If you have a tyre that measures, say, 50mm from bead to bead and is 26" in diameter, it will have a surface area of (approximately) 0.033m2.

If you have a tyre that measures 25mm from bead-to-bead and is 26" in diameter it will have a surface area approximately half this amount.

If you inflate both tyres to the same pressure, the larger tyre will have double the total force acting on the surface, since force=pressure x area. The area is double for the wider tyre. That is fairly fundamental physics, I think.

Since the pressure will be acting normal to the tyre surface, some of it will be trying to expand the tyre radially, and some will be acting lateral to the wheel (i.e. push the rim walls apart).

This basically means that for two tyres of differing volume, under static conditions and [i]inflated to the same pressure[/i], the larger tyre will exert more force on the rim.

This would explain why my Mavic 719 rims has max. safe pressures of 48psi for 2.3" tyres and 113psi for 1" tyres. (see here: http://www.mavic.com/mtb/products/xm-719.323318.2.aspx)

If that theory were true, then the force on the rim would be approx. proportional to the tyre width for a given pressure. This would mean that the max force should be constant, so 2.3*48 should be about the same as 113*1, which it almost exactly is. Hurrah.

If you inflate both tyres to the same pressure, the larger tyre will have double the total force acting on the surface, since force=pressure x area.

Surelyt that's a contradiction? The pressure remains the same, regardless of the surface. The only thing that changes is the volume.

Imagine you have a tin can, of 500ml. You fill it with 100psi. Then, get another can, this time 5000ml, and fill it with the same pressure. The same forces are acting within both the cans. So, they could both be constructed of the same materials with the same properties and dimensions.

Are there any phyisicisticianists out there can clear this one up, or am I going to have to start a whole new thread?

Force=Newtons
Pressure=Newtons/Metre^2

To get the force you integrate pressure over the area of the vessel, be it a tyre or tin can.

Since the pressure is the same for both cases, but the surface area changes, the total force (in Newtons) acting on the surface will also change.

I think i've said all I can on the matter now!

I still think you're wrong. I think the pressure guidelines are more to do with the design of different types of tyre than they are to do with the design of the rim. MTB tyres aren't designed to withstand the higher pressures of road tyres. If you could construct a fat tyre of the same properties of a road tyre, then surely it would be able to withstand the pressure?

We need a proper scientist on here. Where's that Barnes, he's quite clever.

I am happy to be proven wrong though, and I apologise if I'm wrong at all.

It's bugging me now, and I need to know!

Bigger tyres do exert a greater force on the rim/bead for a given pressure. This should make it clear.

http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/pressure_vessel.cfm

Fair enough! I think pressure guidelines for rims are telling you the safe operating conditions for the rims, whilst pressure guidelines for tyres are telling you the safe operating conditions for tyres though.

To answer the question without jabbering on about high school physics, yes i've put those same tyres on my mountain bike to use as a commuter. I run them at 80psi and it's very fast on the road if a little harsh. My rims are the original Mavics on a very old skool XC bike, they were probably a bit narrower back in the day so more suitable. Of current 26" Mavic rims I think I only spotted the XC717 that would accept down to a 1" tyre. Hope I don't ever puncture, they were harder to get on the rim than a normal mtb tyre.

I don't understand that stuff, ac282. 🙁

I still don't understand how a simple increase in volume somehow puts a tyre or rim under more strain.

Poppa is right. The surface area inside the tyre is greater on a fat tyre and force = Pressure x area. The force is trying to separate the 2 beads. The only thing resisting this force is the rim.
It is the force not the pressure that is trying to split your rim apart.
If you have 2 pneumatic rams, 1 small 1 large, at a given pressure the large one will exert more force. Your tyre is effectively a annular pneumatic ram.
HTH

I wouldn't worry too much anyway, just stay within the safety margins that the rim and tyre manufacturers recommend and you should be fine.

If you have 2 pneumatic rams, 1 small 1 large, at a given pressure the large one will exert more force.

How?

100psi is 100psi, regardless of the size of the ram, surely?

i ran some 1" slicks on both mavic 317, 717 and some cheapo aluminium unbranded rims at over 100 psi with no problems for 2 years.
Rims were only changed after crashing into a pedestrian who ran across the road!

100psi is 100 pounds per square inch. If your ram has a piston of 1 sq in in area the force is 100 x 1 = 100lbs force
If your ram is 2 sq inch in area the force is 100 x 2 = 200 lbs force.

That still doesn't make sense, in relation to tyres. The 2sq. in ram is still only pushing 100psi.

If you get a set of scales, and put a 100lb weight on it, that is 10cmx10cm, it still weighs the same as a 100lb weight measuring 100x100cm.

Or; if your weights are both 1x1", but one is only 10" long, and the other 100", they still apply a pressure of 100psi.

Can anyone explain this in Idiot please, because I'm struggling here... 🙁

psi doesn't matter to the rim. Force will break the rim (assuming that it's going to break from bending rather than bursting), not pressure - force is the result of pressure x area.

A bigger area (bigger tyre) will result in a greater force applied to the rim.

In your analogy you would have to say that the scale's area is 10x10cm and that if you applied a fixed pressure and then wanted to double it, you'd have to double the weight - eg twice the force.

I'm still not convinced at all.

I want a simple, clear explanation in language I can understand, and preferably graphs and diagrams.

[img] [/img]

Leg pulling now, shurely?

Well, the question I want answering, is 'If the tyre volume is increased, but the pressure remains the same, will this put the rim under more strain?'.

This has not been answered sufficiently. Just a load of confusing and conflicting babbling. 🙁

Nice graph though! 😀

Talkemada,
Forget about the scales in your recent post.
Back to those rams. The 2 sq in one does, as you say, see 100psi. But that's 100 pounds force on every inch, 2x100=200 pounds force.
(likewise, a 1sq inch ram will also see 100psi, but have a force of 1x100 = 100 pounds force.
Pressure isn't force (unless area is 1)

2x100=200 pounds force.

On what? The same square inch? No; its the same psi, isn't it?

And what's this 'force' business? P = F/A, so P = 100 psi.

So, by my reckoning, the pressure in a 1" tyre would be applying the same force on any given part of a rim, as the same pressure in a 2" tyre. So if a rim can withstand 100psi, the volume of the tyre is therefore irrelevant, surely?

I'm sticking with my theory, until someone can prove otherwise.

On what? The ram, or the rim.
PSI isn't pounds per sq inch. It's pounds force per sq inch. Hence 100 pounds force per square inch is a total of 200 pounds force if the area is 2 sq inchs.

Lord Above... 🙄

Yes, but 100psi will only exert 100 pounds force per SQUARE INCH of the rim. Not 200, not 3, not banana. 100.

Soooooooo........

Will a larger volume tyre at the SAME PRESSURE be theoretically possible on the same rim?

I refer you to the many answers above.

The air pressure creates a force on the tyre that pulls on the rim. It's not just direct force of the air pressure on the rim you have to worry about.

Shirley the point is that the tyre carcas at the rim exerts a much greater force when the tyre is wider?

wot he said ^

I think poppa has maybe explained better what I was trying to say.

It's not the pressure of the inner tube/air on the rim that's important - it's the force of the tyre pulling the rim sideways that counts and that force is a product of the pressure and the surface area of the tyre (ok, for the pedants, I know it's not quite that simple but it is near enough proportional) so a tyre with twice the surface area will be pulling with twice the force on the rim.

Nope. Don't believe you.

Nobody's explained this clearly and effectively. People ramble on about 'force' and this and that, but a lot of it is conflicting ideas.

Force is any influence which causes a free body to undergo an acceleration (according to Wiki). Mass/velocity, no? Pressure is Force/area. So, how can increasing the volume in which the pressure exists, as long as that pressure remains the same, make any difference?

I suspect there is another factor here, which no-one has yet explained...

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...