> in theory, wider tires are faster due to their shorter contact patch, which deforms less as they roll
What is meant by "contact patch"? If it means 'portion of tire in contact with the ground': Assuming the tires are the same circumference, I'd expect the front-to-back dimension of the contact area to be the same, and being a wider tire, the side-to-side contact dimension to be larger, creating a larger contact area.
And given equal force on the tires, wouldn't the same force deform a smaller contact area more, not less? What am I misunderstanding?
> Laboratory tests on steel drums eliminate the rider and thus the suspension losses. If you look at hysteretic losses alone, narrow tires run at higher pressures and thus flex less, meaning they absorb less energy.
> We tested on real roads, with a rider on the bike, and found that the increased vibrations of the narrower tires caused energy losses that canceled out the gains from the reduced flex. These suspension losses are mostly absorbed in the rider’s body.
How do hysteretic losses apply here?
And, why does the increased vibration cause energy losses (which I take to mean reduced efficiency of energy used for movement)? If the energy is absorbed by the tire or is transferred via other bicycle components to the rider, what's the difference in energy loss/efficiency?
The contact patch is the portion of tire that is touching the road.
A wider tire should have a smaller contact patch... Maybe. A wider tire means the force can be spread out more and should deform the tire less. So the front to back dimension will differ.
Tire hits bump, causing energy B to impact tire. The whole bike and rider go 'up' or just the tire goes 'up'. How is the former causing more energy loss? The amount of energy is B either way.
I can see how one is more comfortable, because the tire takes the hit and not me, but I don't grasp the difference in energy.
When the pressure is lower, less energy is deflected upwards. The tyre flexes over the hit.
_ \________/
\___/x\___/ versus x
Less energy is transferred to the 'up' meaning the energy is still going forward. You still have upward motion, but that's less energy taken away from forward motion.
The human body is not perfectly elastic, as it is made out of meat. When the bike goes up and down, it makes the meat jiggle or flap which dissipates energy in the flesh as heat. So taking the hit at the tire which is much more elastic instead of in the heavily damped rider is more efficient.
Take a look a slow motion video of riders on a rough road, say some Paris-Roubaix footage and you can see this.
Ah, thank you; that makes sense to me. I didn't anticipate a big difference between a tire and a person in that regard. FWIW, from the article:
> Studies by the U.S. Army found that the more discomfort vibrations cause, the more energy is being absorbed. And the amount of energy that a vibrating human body can absorb is significant – the U.S. Army’s study measured up to 2000 Watt!
Let's not imagine how that study was conducted ...
There is no fixed energy B. If the bump only deforms a small part of the tire, that's little energy. If the bump makes the whole rider jump, thats lots of energy.
What is meant by "contact patch"? If it means 'portion of tire in contact with the ground': Assuming the tires are the same circumference, I'd expect the front-to-back dimension of the contact area to be the same, and being a wider tire, the side-to-side contact dimension to be larger, creating a larger contact area.
And given equal force on the tires, wouldn't the same force deform a smaller contact area more, not less? What am I misunderstanding?
> Laboratory tests on steel drums eliminate the rider and thus the suspension losses. If you look at hysteretic losses alone, narrow tires run at higher pressures and thus flex less, meaning they absorb less energy.
> We tested on real roads, with a rider on the bike, and found that the increased vibrations of the narrower tires caused energy losses that canceled out the gains from the reduced flex. These suspension losses are mostly absorbed in the rider’s body.
How do hysteretic losses apply here?
And, why does the increased vibration cause energy losses (which I take to mean reduced efficiency of energy used for movement)? If the energy is absorbed by the tire or is transferred via other bicycle components to the rider, what's the difference in energy loss/efficiency?