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> bicycle and pedestrians could be exempted from a road tax based on fairly simple assumption: the wear and tear is minimal compared to a car. It wouldn't even surprise me if the wear and tear is so small that it can't even be measured.

That is the case, which came as a surprise to me.

For things like the road bed, where the force from the surface is going to be spread out more, I'd expect the car to do more damage because total weight would be all that mattered.

But for damage at or near the road surface I'd have expected bikes to do more.

My reasoning was simple: although cars weigh a lot more they are spreading that weight over a larger contact area.

My car for example has four tires each at a pressure of 32 PSI. My bike, before I replaced the tires with wider tires, had two tires each at 110 PSI. Any given small patch of road I drive or ride over would only get about 29% as much force on it from the car as it would from the bike at any given time a tire is on that patch of road.

It would get that force for more time from the car than from the bike if the car is going less than ~3.4 times the speed of the bike and for less time if the car is going faster than that, and more different small patches of road would get force from the car than would get force from the bike.

This is similar to why if I were to put a flat metal plate on my chest and set a bowling ball on top it would not hurt, but if I were to set hold a dagger against my chest and set a bowling ball on top it would likely hurt a lot.

But something is wrong with my reasoning, and it turns out that road damage goes as something like the fourth power of the vehicles axel weight. Whether the tires are narrow high pressure tires or wide low pressure tires, from what I've read, is not relevant.

I'm not sure if that is because I was just wrong about top layer damage, or if it is just that lower level damage is just much more important and expensive, so any difference in top layer damage due to tire size is lost in the noise.




> Any given small patch of road I drive or ride over would only get about 29% as much force on it from the car as it would from the bike at any given time a tire is on that patch of road.

...

> This is similar to why if I were to put a flat metal plate on my chest and set a bowling ball on top it would not hurt, but if I were to set hold a dagger against my chest and set a bowling ball on top it would likely hurt a lot.

This is because your chest is weak and soft in comparison to the incompressible metal plate or dagger blade, which is not analogous to the situation on the road. A better analogy would be poking yourself with the tip of an al dente noodle compared to the load transmitted by a large, spongy slice of bread. Neither is going to hurt you in any way.

Whether your tire is transmitting 110 psi or 32 psi doesn't matter much because the compressive strength of Portland cement concrete is ~4000 PSI and the compressive strength of bituminous pavement is ~500 PSI.

However, soil bearing loads are on the order of 15-50 psi under constant loading, and can act as a spring or can yield at much lower transitory loads. The dirt beneath the pavement (or rather, the dirt beneath the gravel beneath the pavement) gives easily, so instead of asking the concrete to survive the compression forces caused by the contact patch, you're asking it to form a beam/plate that spreads the loads over a broad area, and asking that beam to resist deflection rather than crushing. The real damage, and the reason that damage scales with the fourth power of axle loading, is caused when a large truck rolls over the road, the ground shakes, and the roadbed flexes down into the soil under the axle and back up after it passes.

Try to imagine riding your road bike over a piece of cardboard laid on the sand. That wrinkling and creasing is much closer to actual road wear patterns than a dagger cutting your chest.


This analysis based on tire pressure seems faulty. If you place your hand on a bike tire, you do not experience thousands of pounds of force.

To estimate the pressure exerted on the road, you want to compare vehicle system weight/tire contact area. Velocity also plays a role.


Sure, the contact patch is a good bit smaller on a bike than a car, but people don't usually weigh 5,000lbs.

I'd much rather get run over by average person on a bicycle than get run over by a car.


That's because you aren't a smooth surface :) Because of your shape (and I admit I'm making an assumption here) you will experience pressure spikes due to the fact that different bits of you stick out. E.g. knees due to their shape will experience much higher pressure. A road won't do that.


>My car for example has four tires each at a pressure of 32 PSI.

The air pressure of your car/bike tires doesn't matter - all tires could be flat. Try comparing (vehicle weight + your weight) / (vehicle contact surface area) where vehicle is either your car or your bike.


For tires that support the vehicle via air pressure in the tires, contact area is roughly weight supported by tire / air pressure in tire [1].

[1] http://ffden-2.phys.uaf.edu/211_fall2013.web.dir/connor_matt...


When the tire is flat it is being supported by the sidewalls and by the wheel.

And if you notice the contact patch of a flat tire is much larger.

The ground pressure of any vehicle is almost exactly equal to the psi. With the exception of force carried by the sidewalls.




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