MOT exempt

Joe_Horner

jack_pott wrote: »

Erm, the damage to the road is determined by the fourth power of the weight, not the pressure.

Erm, no. It's the weight per unit area (otherwise known as the pressure). Big wright spread over big area doesn't matter.

Remember that the weight of air is about 1kg on every square cm of ground. A Yaris sized patch of road, 3.95 x 1.7 metres, has an area of 67150 square cm, and is supporting 67 tonnes of air. Big weight but no damage!

[Deleted User]

Joe_Horner wrote: »

Erm, no. It's the weight per unit area (otherwise known as the pressure). Big wright spread over big area doesn't matter.

Remember that the weight of air is about 1kg on every square cm of ground. A Yaris sized patch of road, 3.95 x 1.7 metres, has an area of 67150 square cm, and is supporting 67 tonnes of air. Big weight but no damage!

From the American Association of State Highway and Transportation Officials:

the damage caused by a particular load is roughly related to the load by a power of four (see table 1)

Road damage is calculated in ESALs (equivalent single axle loads), not from the tyre pressure.

Atmospheric pressure is irrelevant. It's static, and doesn't cause the road to flex as passing vehicles do.

Joe_Horner

All good, except if you notice I wasn't talking about tyre pressure, I was talking about the pressure exerted by the tyre against the road. That is the same as the load per unit area. Static or dynamic, that's what causes damage - which is why big baloon tyres do less damge off road than normal onez. They spread the load over a larger area.

if you also notice my reply to Car54 (before you "corrected" me) I was also being slightly tongue in cheek!

Gloomendoom

jack_pott wrote: »

Erm, the damage to the road is determined by the fourth power of the weight, not the pressure.

Just a hunch, but I think you need to update your Photobucket account to enable 3rd party hosting.

Herzlos

ilikewatch wrote: »

Most won't, but I can personally think of 3 mid-seventies vehicles (Capri, Cortina, VW camper) which have been parked up for years since failing their tests due to corrosion. All of them start and drive and I would be amazed if all 3 weren't back on the road as soon as they are MOT exempt with nothing more than a fresh tank of petrol, a wash and a jump start to "recommission them".

It'll still be illegal to drive a car that isn't roadworthy, though. There just won't be an annual MOT for them.

There aren't that many pre-77 cars on the road currently, and those that do are unlikely to cover anything more than negligible mileage.

Herzlos

50Twuncle wrote: »

why should an old banger that puts out unlimited amounts of CO and pollutants, get a waiver from tax -

Because (a) most of the emissions are produced on production and (b) emissions from driving are negligible when you only do 100 miles a year. It's completely dwarfed by even the most efficient modern car doing 10,000 miles a year.

[Deleted User]

Gloomendoom wrote: »

Just a hunch, but I think you need to update your Photobucket account to enable 3rd party hosting.

I've no idea what that means, when I followed the link it looked as if I was expected to open a subscription account. (The link was working fine for several hours after it was posted.)

[Deleted User]

Joe_Horner wrote: »

All good, except if you notice I wasn't talking about tyre pressure, I was talking about the pressure exerted by the tyre against the road. That is the same as the load per unit area. Static or dynamic, that's what causes damage - which is why big baloon tyres do less damge off road than normal onez. They spread the load over a larger area.

if you also notice my reply to Car54 (before you "corrected" me) I was also being slightly tongue in cheek!

You have a misconception of how vehicles damage roads. It's not like a stiletto heel on a soft wooden floor, it's by fatigue fractures caused by the repeated flexure of the road. A stiletto damages a floor because the compression strength of the heel is greater than the shear strength of the wood, so the wood shears at the periphery of the heel and you're left with a depression.

On a road, the compression strength of a tyre is much lower than the shear strength of the road, so that shear force distributes the pressure from the wheel over a much wider area than just the area of contact.

Think about a diver standing on the end of a springboard, the board doesn't flex any further if the diver increases the pressure by standing on tiptoe, but it will flex further if a heavier diver uses it. The board spreads the load sideways via the shear strength of the steel.

Balloon tyres off road are another example like the stilettos, as the mud and snow are soft surfaces, unlike tarmac.

The pressure exerted by the tyre on the road is the same as the pressure in the tyre, it can't be anything other. Old Ike, he say: "Every action has an equal and opposite reaction", so if there's 28psi of air pushing the rubber onto the road, then there has to be 28 psi from the road pushing the rubber against the air.

bigadaj

jack_pott wrote: »

You have a misconception of how vehicles damage roads. It's not like a stiletto heel on a soft wooden floor, it's by fatigue fractures caused by the repeated flexure of the road. A stiletto damages a floor because the compression strength of the heel is greater than the shear strength of the wood, so the wood shears at the periphery of the heel and you're left with a depression.

On a road, the compression strength of a tyre is much lower than the shear strength of the road, so that shear force distributes the pressure from the wheel over a much wider area than just the area of contact.

Think about a diver standing on the end of a springboard, the board doesn't flex any further if the diver increases the pressure by standing on tiptoe, but it will flex further if a heavier diver uses it. The board spreads the load sideways via the shear strength of the steel.

Balloon tyres off road are another example like the stilettos, as the mud and snow are soft surfaces, unlike tarmac.

The pressure exerted by the tyre on the road is the same as the pressure in the tyre, it can't be anything other. Old Ike, he say: "Every action has an equal and opposite reaction", so if there's 28psi of air pushing the rubber onto the road, then there has to be 28 psi from the road pushing the rubber against the air.

Yes but the damage to the road often isn't to the tarmac, or more correctly asphalt nowadays, Tarmac is not only a trademark but most material produced as such is now classed as a contaminated material.

It has some relationship to the base course but in many instances is a function of the sub base and formation materials on which the wearing course is laid.

Any design will derive from full or equivalent cbr testing of the in situ materials and or the road building materials as they are emplaced, together with a best guess of the equivalent axle loads that will be incumbent on that surface for its design life. There is very little useful theory behind the road design process, if you look at the equivalent axle load derivation in the highways brown book for example it's a rough approximation of the impact of different lorry weights with different axles and load characteristics, for design purposes car loading let alone cyclists are largely irrelevant.

Also teh concept if shear strength isn't really apt in this instance, we don't have a sudden failure in mist cases but suffer from serviceability limit issues. These are derived from slow rutting of materials, Tarmac or asphalt is an odd material in that it acts like a slightly fluid solid and so may be subject to creep over time. Continued loading of a specific area means this creep is exacerbated and concentrated. So rather than teh shear strength of the material were more interested in the relevant modulus, potentially shear or secant, as well as the poissons ratio to a lesser extent.

Manxman_in_exile

jack_pott wrote: »

Erm, the damage to the road is determined by the fourth power of the weight, not the pressure.

http://pedalfortcollins.com/greatest-demand-on-tax-dollars/

I'm familiar with second power equating to area (eg pressure) and third power equating to volume (eg weight, mass x g) but what is the fourth power? Are we talking time?


I have looked at your link but it's not clear to me(?) where the fourth power comes from.