Safe Speed Forums

With all the talk these days complaining about urban 4x4s, it got me thinking, what about mountain bikes?

They are often heavier than a standard road bike and require more energy, in the form of food, to propel them. With so many people starving in the world, how can this be allowed?

These are expensive vehicles designed for off road use, yet 90% of them will never see a non-tarmacced surface (except, of course, for paving slabs). Most of them are owned by poseurs purely for the rugged off-roading image whose practical needs would be better served by a road bike.

I propose that a £300 tax be applied to mountain bikes (raising to £400 in 2008) in order to address this problem. Bugger those who use them in the countryside

And what about pedestrians who wear heavy boots when a pair of trainers would suffice!

I can see a pattern developing.... not having short hair, thereby increasing wind resistance, and simply being too fat, which is unhealthy AND an unecessary darin on the worlds resources.
After all weren't we being encouraged to remove unecessary clutter from our car boots not so long back?

_________________
Time to take responsibility for our actions.. and don't be afraid of speaking out!

I know this is a joke, but PLEASE let's not join the 'divide and conquer' brigade.

What's the latest group to be branded anti-social?

Smokers?
4x4 owners?
Speeders?
Fat people?

It's bloody endless, but rest assured, if you're not in one of those groups, you'll be in the next wave. Start fighting back. Now.

_________________
Paul Smith
Our scrap speed cameras petition got over 28,000 sigs
The Safe Speed campaign demands a return to intelligent road safety

It was an attempt to point out the rediculousness of the anti-4x4 argument rather than to attempt to divide and conquer cyclists. Last time I checked this campaign was not anti-cyclist in any way.

That said, it is funny to watch them squirm when the argument is turned around on it's head and used against something they like

Absolutely. I know. I understood. But casual visitors might not.

_________________
Paul Smith
Our scrap speed cameras petition got over 28,000 sigs
The Safe Speed campaign demands a return to intelligent road safety

Strangely, I did some calculations after a fellow cyclist turned out to be an anti-4x4 bigot. He insisted that my 4x4 (on 265R70s) does more damage to the road than his road bike that he runs on 700Cx25s. He agreed that "road damage" is proportional to the fourth power of loading (i.e. pressure exerted on the road) and directly proportional to contact area (IIRC, it was he who claimed that). At the beginning of the exercise, I only had an inkling of the way the results would turn out but I'll summarise them here:

• The tyres on my 4x4 are run at 30 psi, he runs his tyres at 120 psi.
• On measuring the contact areas, we found that this was
  • 265mm (tyre width) by 150mm on my car, so with four wheels that's a total contact area of 0.159 sq m;
  • on his bike it's 25mm (tyre width) by 50mm and with two wheels that's 0.0025 sq m
• Now the force on the outside of a tyre must be the same as the force on the inside of the tyre (if it were not, the tyre would change shape until it was), so the pressure exerted on the road is equal to the tyre pressure. So
  • for the 4x4, the "pressure factor" is 30^4, or 810,000
  • for the bike it's 120 ^ 4, or 207,360,000
• This gives "damage factors" (pressure factor x contact area) of 128,790 for the 4x4 and 518,400 for the bike!

IOW a road bike causes about four times more damage to the roads than a 4x4

However and thankfully, our roads are actually designed to withstand the even higher loadings imposed by LGVs, and so neither cars nor pedal cycles actually cause significant damage to the road!

Nevertheless the look of horror on his face when the results dropped out of the end of the spreadsheet was a picture

_________________
Will

I think there are a number of factors missing here. As you say quite rightly the withstand factor on the road is much higher than any static figures the spreadsheet you've constructed produces. So I suggest we have to concentrate on dynamics.

1) Braking redistributes weight dramatically. Since that aspect is fourth power, damage under hard braking - and I'm assuming that the 4x4 can pull a lot more G force than the bike - will go up dramatically more than that from the bike.

2) Power put down too fast - wheel-spinning - will generate local heat on the road....

3) Cornering forces on a bike, thanks to lean and round profile "chicken strips" on the tyre will not appreciably reduce contact area. In the 4x4, apart from the 4th power weight shift sideways, contact area on the outermost tyre and drag sideways of both innermost tyres can only reduce and increase respectively due to suboptimum camber/castor angle.

But I agree the best part of all this is watching a jaw drop

Not strictly true to my mind.

The force on the outside must be less than the force on the inside.

If the bike tyre was inflated to 600psi (yes, I know it'll probably explode but, bear with me) it wouldn't suddenly be exerting 4 times the force on the road............
In other words, you forgot to factor weight into your equation.

Mike.

See: http://www.sightline.org/daily_score/ar ... /heavy_man which contains:



I believe this to be absolutely correct. (I confess to not following the link I'm posting - I got that bit quoted to me by Google.)

_________________
Paul Smith
Our scrap speed cameras petition got over 28,000 sigs
The Safe Speed campaign demands a return to intelligent road safety

If Mr Brownstuff really wanted to influence the buying decision of people with this unfair tax on vehicles in band G, he would have said any band G vehicle registered after this budget will incur the new tax instead he back dated it. When Landrover goes bust and the caravan industry is gone and people won’t be able to give away a 5 year old band G car what will he (if in power) pick on next Band F?

Not really. The difference is marginal. The weight of the vehicle must be borne by an equal and opposite force from the road. This force will be equal to the pressure multiplied by the area of contact and it is the pressure and the weight of the vehicle that determines the contact area. For a normally inflated tyre, if the product of pressure and contact area is less than the weight it's supporting, the tyre "squishes" to increase the contact area until equilibrium is obtained. Conversely, if the product of pressure and contact area is greater than the weight, the tyre tends to restore it's unloaded shape so reducing the contact area until equilibrium is again restored.

Now attending to your example:
Since pressure (p) = force (w) / area (a), it follows that:
a = w/p

Let the contact area at 120 psi be a1 and at 600 psi be a2:

a1 = w/120 (1)
a2 = w/600 (2)

Therefore, from (1), w = 120 * a1 (3)

Substituting for w from (3) in (2):

a2 = 120 * a1 / 600

So a2 = a1 / 5

So, no, it wouldn't exert five times the force because the contact area would reduce to a fifth of what it was at the lower pressure. IOW, the tyre deforms to produce the contact area that results in the required force on the road.

HTH,

_________________
Will

So, supposing you run your tyres flat, how does your equation work then?

The contact area is constrained by the material and design of the tyre though. If the bike has a 23 stone man on it the tyre can't spread enough to adjust for the incresed weight as it can't expand that far. Conversly you can't pump it up enough to reduce the contact patch sufficiently to make the figure the same with a small child on board.

To my mind, no matter what you do to the tyres a bike with a 23 stone bloke on it will exert more force on the road than with a 6 stone child.

Another example.

I have a van which weighs 2 tons, it has a 1 ton payload capability. If I add the 1 ton into the van I find it hard to believe that the contact patch of the tyre will increase by 50% to compensate. So the added weight means more force is exerted on the road.

Maybe I'm missing something here or I'm looking at it in a too simplistic manner but I can't see it.

I'm happy to concede your figures may probably work within a very narrow band.

Mike.

If the rim is in contact with the inner wall of the tread area, the weight is no longer wholly supported by the cushion of air and so the tyre pressure will no longer equal the pressure on the road. However, provided the tyre isn't "fully flat", it just spreads until the product of contact area and pressure are equal to the weight that the tyre supports.

_________________
Will

If a tyre was a perfectly elastic balloon, then you might be correct. However, tyres are anything but perfectly elastic balloons, and most of the pressure most of the time braces the carcass.

_________________
Paul Smith
Our scrap speed cameras petition got over 28,000 sigs
The Safe Speed campaign demands a return to intelligent road safety

Methinks that's a bit like the argument about whether wheels hang from or stand on their spokes!

Don't forget that I was using claims the other cyclist made. It was he who asserted that pressure inside is the same as pressure outside, that road damage is proportional to the fourth power of pressure on the road and directly proportional to contact area. Also, those measurements were very approximate and I've made some further calcs since using a wide range of what seems reasonable - and it doesn't look good for bikes. At best (assuming that tyres have a 100% tread pattern and using the weights and estimated contact areas to calculate pressure on the road) the bike causes about three times the damage of the car, and at the worst about twenty times the damage. In every approximation I've tried, the bike does more damage than the car!

FWIW, the claim about fourth power of axle weight only starts to make sense when you realise that heavier vehicles run at much higher tyre pressures. Many LGVs, for example, run with 100 psi or more in their tyres (according to a tyre fitter I cheekily rang and asked!) and axle weight is roughly proportional to tyre pressure over the range of vehicles where this holds true. If you want evidence that fourth power of axle weight doesn't universally work, consider two Icelandic vehicles identical except that one has normal tyres and the other has those special balloon tyres they use on snow. The one with normal tyres cuts straight through the snow and gets bogged down while the other exerts a much lower pressure and so damages the road surface far less and rides over the snow - yet they have the same axle weights and if the fourth power rule was universal they'd damage the road equally.

_________________
Will

Nah. Hold a wheel in mid air. Pump the tyre up to 60psi.

What's taking the force now?

Fit it to a vehicle.

How much does the tyre pressure change when the vehicle's weight is put on the tyre? (I believe it must rise, but I bet it rises less than 1psi.)

_________________
Paul Smith
Our scrap speed cameras petition got over 28,000 sigs
The Safe Speed campaign demands a return to intelligent road safety

It's only an approximation - but a surprisingly good one. The pressure stresses on an unloaded tyre are distributed throughout the carcass. Now apply a point load (e.g. push the tyre against a surface, like a road) and initially there is an inequality of pressure inside to out and the tyre deforms to restore pressure equilibrium with the pressure stress at the point of contact now being taken by the supporting surface (the road) instead of the carcass.

Revisiting the problem, I calculated pressure on the road by again approximating the contact area and assuming that the vehicle weight is evenly distributed among the road wheels (which isn't actually true for either the bike or the car). For the car, I'd used the manufacturer's declared kerbweight and for the bike I'd assumed the standard 86kg "eurobod" and a 14kg bike. For the car the recalculated pressure on the road came to 18 psi and for the bike it came to 57 psi - it seemed that the naysayers were right...

... However, the tread on the car is such that only about two thirds of the contact footprint is actually in contact with the road, so I had to recalculate the pressure after decreasing the contact area by a third to allow for the tread pattern. The 27 psi that this gave is close enough to the 30 psi shown by my pressure gauge to suggest that the approximation is valid and measurement errors with the assumption that the entire tread width is in contact for the entire footprint is more than enough to account for the discrepancy.

When I considered the bike it seemed that tread pattern wasn't going to explain the difference because road tyres are normally almost slick. However, I got my bike out (on 40mm tyres at 60 psi), and checked the contact width by riding across a sheet of A4 - approximately 20mm, or about half the bead width. Now a 25mm road tyre has a more rounded profile than my hybrid tyres, so a contact width no greater than half the bead width seems reasonable. Plug that into the approximation, and you end up with a pressure on the road of 114 psi - which isn't a bad approximation for the 120 psi claimed!

_________________
Will

You're still regarding tyres as 'perfectly elastic' - which they aren't.

Repeat the experiment with a stainless steel pressure vessel. (approximates to perfectly inelastic). Whatever the pressure the contact patch is unchanged.

Tyres are a long long way from perfectly elastic.

_________________
Paul Smith
Our scrap speed cameras petition got over 28,000 sigs
The Safe Speed campaign demands a return to intelligent road safety

Heh, pressure = force/area is not valid to describe tyre pressures and how they affect contact areas, its about whether the elephant or the woman in stilettos does more damage to the wooden floor!

Force = mass * acceleration, so the force exerted on the ground by the car/bike is equal to its mass multiplied by 9.8.

Area is the contact area of the tyres. This is the only way in which tyre pressure affects the outcome, and its not plugged into any equations.