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?