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Comment by wolverine876 | Hacker News Reader

Thank you, but maybe I'm being dense:

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.

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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.

The bump is basically static.

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