# Pumping up bicycle tires with helium instead of air [duplicate]

If I pumped up my bicycle tires with helium instead of plain air, what would happen if the applied force on my pedals was constant?

Would I go faster because of the reduced ground friction? Would I go slower because I would have less contact with the surface I'm riding on? Or would it make no difference?

• The only difference is that your tyres will go flat a little sooner, as helium will leak more easily through the valve and other minute apertures. It will make no difference to your performance, as the difference in mass between helium and air will have no measurable effect on friction or weight. – hdhondt Jul 16 '17 at 10:32
• @hdhondt That seems more like an answer than a comment. – rob Jul 16 '17 at 10:35
• If you pump the tire to the same pressure, do you think there will be a difference in performance? – Chet Miller Jul 16 '17 at 11:24
• FYI, the sentence "Would I go faster because of the reduced ground friction?" is a very common misconception. The "ground friction" happening at rolling wheels is not counteracting the cars motion. – Steeven Jul 16 '17 at 13:41
• @hdhondt Yes, helium would leak - but in a much worse way than you described: As far as I know, helium diffuses (leaks) through absolutely any material whatsoever. As in glass, steel, gold, diamond... anything. The reason is that the molecules are really small, and fit into any kind of crystal lattice gaps, etc. It is the gas with the smallest molecules: it is single-atomic. Hydrogen, with smaller atoms, has diatomic molecules. – Volker Siegel Jul 16 '17 at 16:05

First, estimate the volume of a bicycle tire. Most you can encircle with your thumb and forefinger, which gives a cross-sectional area of order $\pi\rm\,cm^2$. A "700c" wheel is named for its diameter, about $700\rm\,mm$, so the circumference is a couple of meters. That gives a bike tire volume of about $600\rm\,cm^3$. There seem to exist mountain bike tires with diameter five inches, so the high end of tire volumes is thirty or forty times this large, perhaps $24\,000\rm\,\rm cm^3 = 24\,liter$. Let's use this volume.
The buoyant force exerted by a fluid is equal to the weight of the displaced fluid. The mass of $24\rm\,liter$ of air is about $24\rm\,g$. This isn't very much lift compared the the mass of a typical cyclist. In fact, switching from typical rims/tires (which was saw would provide roughly $0.6\rm\,g$ of lift, if you'll forgive me providing a force with mass units) would probably increase the mass of the bicycle by more than $24\rm\,g$, so you'd come out heavier rather than lighter.