Tyre inflation pressure is irrelevant. We could assume the tyres were solid rubber for the purposes of this conversation and the wear on the road would be the same due to the gross vehicle mass devided by the number of tyres.
As another commenter pointed out, I've ridden my bike with 150psi tyre inflation pressure but the weight per wheel is my weight + bike weight devided by contact area of each tyre. Sure, higher pressure tyres typically have less contact area, but the gross weight of the bike is tiny compared to a truck the road wear from cycling is pretty much zero.
> I couldn't disagree more! Wear on roads is to the fourth power of pressure.
No, it's probably to the fourth power of force, not pressure. These are distinct and separate concepts in physics.
> We could assume the tyres were solid rubber for the purposes of this conversation and the wear on the road would be the same due to the gross vehicle mass devided by the number of tyres.
Ok, but in that case we must be talking about force, not pressure. Because the pressure is force divided by contact area...
> As another commenter pointed out, I've ridden my bike with 150psi tyre inflation pressure but the weight per wheel is my weight + bike weight devided by contact area of each tyre. Sure, higher pressure tyres typically have less contact area, but the gross weight of the bike is tiny compared to a truck the road wear from cycling is pretty much zero.
Now you're mixing pressure and force in a way that doesn't make any sense to me...
So dmurray is saying that "correct relationship is linear in the fourth power of the weight per axle" which IMHO is correct (but to fit with my argument above I would use the words "force exerted by tire on road" rather than "weight per axle", it's the same).
You (TheSpiceIsLife) are talking about something else... I don't know what... I can't say if you're right or wrong, but I feel quite certain that what you say is beside the point.
My (only) point was that the force that a wheel can exert on the road (what Newton called "action") is
A = [current "overpressure" inside tire] * [contact surface area]
and the force that the road exerts on the tire (what Newton called "reaction") is
R = [pressure on road] * [contact surface area]
Since [contact surface area] is one and the same in both expressions and Newton says R = A it must follow that [pressure on road] = [current "overpressure" in tire].
EDIT: Extended to better explain context of my argument.
They're right. Civil engineering estimates of road wear for wheeled vehicles[0] is the fourth power of axle load (the weight borne by the axle) per axle, tire pressure does not enter the equation.
[0] following extensive testing in the 60s, repeated a few decades later, the exact exponent is variable but 4 has proven pretty good for a rule of thumb.
> They're right. Civil engineering estimates of road wear for wheeled vehicles[0] is the fourth power of axle load (the weight borne by the axle) per axle
Ehuu... So why didn't you/they just upvote dmurray who said that 10 hours ago...?
Inflation pressure isn't the same as running pressure!
Once a vehicle is loaded the air pressure of the tires can go up substantially relative to what they're inflated to. Like you said, action and reaction.
That's why under-inflated tires will have a very large contact patch; increasing the contact patch increases the area in the area * pressure equation. But it also reduces the volume of air inside the tire/tube and that raises the pressure, also.
True. I was talking about [current "overpressure" inside tire] (see my other answer) and not "inflation pressure", so I have taken that effect into account. :P
Pressure on two sides of a surface don't have to be equal. The overall forces do, but there's an additional component being applied by the surface itself. An alternate way of looking at it : consider how the space station internal pressure differs from space.
> Pressure on two sides of a surface don't have to be equal.
Absolutely, and Newton surely didn't say so. :P
(A more close to home example is that the pressure inside a tire surely doesn't have to be the same as the pressure outside the tire. In fact, if it is then we call that a "flat tire". :P)
But the force that the tire exerts on the road (=the "action" as Newton called it) has to be equal to the force that the road exerts on the tire (=the "reaction" as Newton called it). Since the contact surface between tire and road is the same that means that the "overpressure" in the tire (=the pressure the tire exerts on the road) has to be the same as the pressure that the road exerts on the tire.
