Source: http://www.lefigaro.fr/conjoncture/2016/06/08/20002-20160608...
It's fine for public transportation to not make back the investment if the advantages to the public are high enough (though with the Franc that's easier to deal with than the Euro)
Sounds like the broken window fallacy argument to me.
In Germany the public railway company Deutsche Bahn makes more than 1 billion in profits. All the profits come from freight, transporting people is done at a loss. The problem is that not enough money gets invested into upgrading lines highly frequented by freight.
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.)
https://cdn.theatlantic.com/assets/media/img/posts/2015/05/r...
And of course that doesn't take the environmental externalities into account.
That makes sense, since the road wear increases superlinearly with the vehicle weight - so commercial trucks are also the one generating most road maintenance costs.
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.
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.
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.
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.
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]
R = [pressure on road] * [contact surface area]
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.
Ehuu... So why didn't you/they just upvote dmurray who said that 10 hours ago...?
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.
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.
http://www.honestjohn.co.uk/forum/post/index.htm?t=79414
http://www.trucknetuk.com/phpBB/viewtopic.php?f=4&t=89338
(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).)
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.
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
They contribute more to the tax because they are bigger, more powerful, potentially less efficient.
They have to pay for weight because the increased weight has a direct correlation in actively shortening the life of the road, above and beyond "vehicle-miles covered".
Not really the case any more, I just renewed my tabs on our EV and there's now a $150 EV tax per year in WA state.
Except the number of miles driven (or where you drive them) has very little relation to how much damage you do to the road, and thus how much of the maintenance costs you generate.
The eventual beneficiary of cheap transportation is the consumer. I suppose this should considered a regressive tax.
Amazingly, this is not categorically true:
https://en.m.wikipedia.org/wiki/Great_Northern_Railway_(U.S....
On the one hand, rail is much cheaper per-mile than traditional roads, for both construction and maintenance.
Rail is also more efficient in terms of goods hauled per unit of fuel spent, and controlling carbon emissions from rail is massively easier than doing so from individual trucks.
However, rail itself is fundamentally... unegalitarian, maybe? Rail lines can only be used by those with the capital to run rolling stock between well-specified endpoints (typically large urban centers), and on specified timetables.
The system as it exists in the US right now is less efficient, but opens up a lot of possibilities. For example, I live in Japan and don't own a car right now, which means that I basically can't go backpacking. But if I rent a car, then I can go out to more remote places that are less frequented by the non-hiking masses, and actually enjoy some of the sheer beauty of the Japanese countryside.
Self-driving cars and trucks really do seem to bridge the gap between "cheap rail transport" and "private vehicles for everything", but I worry that they will continue just to be a convenience for the well-off (which is what Uber and Lyft currently are, frankly), rather than a means by which to make transportation available to everybody.