You are reading a single comment by @gbj_tester and its replies.
Click here to read the full conversation.
London Fixed Gear and Single-Speed is a community of predominantly fixed gear and single-speed cyclists in and around London, UK.
This site is supported almost exclusively by donations. Please consider donating a small amount regularly.
moocher
@gbj_tester
Since there are always 100 ways to skin any particular cat, you can look at the problem from an energy perspective. The energy stored in a pressure vessel (idealised isothermal system) is
In this case:
pA = 0.1MPa (1 bar ambient pressure)
pB = 0.3MPa (3 bar absolute pressure in tyre)
VB = 0.004m³ (4 litres tyre volume)
If we evaluate
0.3 MPa × 0.004 m³ × ln(0.1 MPa/0.3 MPa) + (0.3 MPa − 0.1 MPa) ×0.004m³ = 518J
The pump stroke energy input will have to be a bit more than that to account for losses, but the integral of force over distance must be at least 518J. For the practical maximum stroke for a manual floor pump of about 0.5m, the force will ramp up from zero to just over 2kN. Obviously, if you weigh 2kN you will need more pressure in your 29"×2"tyres that 2 bar gauge, so you end up chasing yourself in a circle 🙂
This leads to a practical guide to #whatpressureyourunning , since the energy stored in your compressed air will always be quite close to the product of your weight and height, e.g. 100kg and 1.85m gives 1800J, compared with 1700J for 2.5 bar gauge in a pair of 55-559 tyres or 1500J in a pair of 25-622 tyres at 6 bar gauge