> Now, freight and commercial trucks are effectively the biggest contributor to this tax plus they have to pay for weight. Consumers are the ones getting freebies (especially electric, hybrid etc.)
I couldn't disagree more! Wear on roads is to the fourth power of pressure. And while trucks might have 10x the contact patch your car does, they might weigh 30x as much. So the pressure might be 3x and thus the wear 3 x 3 x 3 x 3 = 81 times as much.
If you look at road costs as largely maintenance then cars are subsidizing trucks because one truck could do orders of magnitude more damage, but have to buy orders of magnitude more fuel. Why is that? Because past a certain point fuel consumption is dominated by drag, and thus, frontal area. 8-10 MPG for an 18-wheeler is reasonable. That's not 81 times worse fuel consumption, perhaps only 2x or 3x compared to modern cars.
> Wear on roads is to the fourth power of pressure.
The rule of thumb in civil engineering is actually fourth power of axle load, per axle.
> So the pressure might be 3x and thus the wear 3 x 3 x 3 x 3 = 81 times as much.
Considering the rule of thumb is by axle load (not pressure density) you're off by about two orders of magnitude when comparing a large sedan to an 18-wheeler:
* A Model S is 2T over 2 axles, or 2 * 1T/axle
* A fully loaded 18-wheeler can go up to 36T (legal limit) over 5 axles, or 5 * 7.2T/axle[0]
That means the road wear per mile of an 18-wheeler is up to 2.5 (number of axles) * 7.2 (axle load) ^ 4 (4th power), or 6700 times the road wear of a Model S.
Now that's a worst case scenario (unloaded model s versus fully loaded 18-wheeler), but you get the point.
[0] on average, the front axle is usually ~5.5T with the rear axles being in the 7.5~8T range which increases the road wear difference quite a bit, to about 7500
Trucks pay fuel tax on diesel fuel (can be higher rate than gasoline), surcharges on fuel tax in some states, weight-distance tax in some states, and heavy vehicle use tax.
The federal weight limit for a truck--including the weight of the truck and trailer--is 80,000 pounds (without special permits), but they rarely haul that much.
8-10 MPG for a truck under real-world conditions is extremely uncommon. Typical efficiency falls in the 5-8 range, depending on the type of cargo + weight, geographic area, etc.
Comparing MPG between cars and trucks is not appropriate. It should be weight per mile per gallon or something accounting for the massive amount of supplies the truck is carrying.
In this case MPG is being used as a proxy for the amount of tax paid. Since weight was already accounted for during the analysis of the road wear per truck, it isn't meaningful for this comparison.
> Wear on roads is to the fourth power of pressure.
This isn't quite right. The pressure on the outside of the tire is the same as the pressure on the inside, and truck tires aren't inflated to 3x the psi of car tires - they might be 55 psi instead of 35. If pressure were really the issue, bicycles would be much more damaging to the road. Racing bikes are typically inflated to 100-150 psi.
As your link points out, the correct relationship is linear in the fourth power of the weight per axle. Your conclusion is correct of course.
(Doesn't change your assertion materially, of course. Your comment didn't scan correctly to me because I've seen video of arm injury caused when someone punctured a truck tyre with a knife. Meanwhile, I've deliberately punctured motorcycle tyres by hand at 36psi while performing a repair (boring out a hole for a plug).)
> What does the tyre inflation pressure have to do with the pressure applied to the road due to the gross vehicle mass?
>
> Nothing.
Well, the "pressure applied to the road" is approximately the same as the tire "inflation pressure"... It's not exactly the same because you need to take into account that the pressure in the tires increases (a bit) when you load the truck, and you have to be clear that you're talking about "overpressure", i.e. the relative pressure difference between the inside of the tire and the atmosphere.
What I think you meant to say is "What does the tyre inflation pressure have to do with the force applied to the road due to the gross vehicle mass?" Then your conclusion ("Nothing") is correct.
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.
I couldn't disagree more! Wear on roads is to the fourth power of pressure. And while trucks might have 10x the contact patch your car does, they might weigh 30x as much. So the pressure might be 3x and thus the wear 3 x 3 x 3 x 3 = 81 times as much.
If you look at road costs as largely maintenance then cars are subsidizing trucks because one truck could do orders of magnitude more damage, but have to buy orders of magnitude more fuel. Why is that? Because past a certain point fuel consumption is dominated by drag, and thus, frontal area. 8-10 MPG for an 18-wheeler is reasonable. That's not 81 times worse fuel consumption, perhaps only 2x or 3x compared to modern cars.
https://en.wikipedia.org/wiki/AASHO_Road_Test