Tyre pressure?

[ScienceIsCool]

True only for an isolated system and it also doesn't include the entire energy picture such as the tires and tubes. You're getting losses is the tire casing as well as outside the wheel system you defined. E1 does not equal E0.

BTW, I think I'm going to remove myself from this thread for a while. Seeya.

John Swanson
www.bikephysics.com

Yeah, you're wrong, and apology accepted.

See, as I am sure you know and appreciate, the beauty of energy conservation laws is that I don't need to worry about the kinds of details you mention. Just as you can present the most wondrously complicated contraption to me, claiming it's a perpetuum mobile, and I can tell you straight to your face that you are wrong, without wasting my time having to examine the details of what all those cute little levers, wheels and pulleys in your machine may do.

So, in the case of our bicycle, if we have the following scenario: At time t0 a bicycle with rider, total mass m, speed of their center of gravity v, the effective inertial moment of the rotating wheels I, altitude of the center of gravity h0. This system will have the total energy E0=1/2 m v^2 + 1/2 I \omega^2 + m g h0, with g the gravitational acceleration, and \omega the rotational speed of the wheels (\omega=v/r, r the radius of the wheel, if both wheels are identical).

Now we imagine the bicycle has traveled some distance on a level road, so that at time t1, we still have the same altitude h1=h0. The conclusion is that, unless the system of bicycle plus rider has some internal damping mechanism that dissipates energy, at time t1 the system will still have the energy E1=E0. Note that this statement is exact, and does not depend on any details of the motion of the bicycle between t0 and t1. In the above, all I have left out are additional degrees of freedom of the bike motion, such as vibration. Thus, the implicit assumption is that those contributions do not change between t0 and t1, which is again exact under the assumption of a steady-state system (so, the bicycle does not somehow vibrate more at t1 than it did at t0; this seems to correspond to reality, at least for my bike ).

 

[Dietmar]

True only for an isolated system and it also doesn't include the entire energy picture such as the tires and tubes. You're getting losses is the tire casing as well as outside the wheel system you defined. E1 does not equal E0.

Yes, I have neglected aerodynamic drag, since the discussion was focussing on rolling resistance. So, yes, E1=E0 without an atmosphere, and no internal damping. The tire losses are part of the damping that I mentioned.

 

[mikesbytes]

At time t0 a bicycle with rider, total mass m, speed of their center of gravity v, the effective inertial moment of the rotating wheels I, altitude of the center of gravity h0. This system will have the total energy E0=1/2 m v^2 + 1/2 I \omega^2 + m g h0, with g the gravitational acceleration, and \omega the rotational speed of the wheels (\omega=v/r, r the radius of the wheel, if both wheels are identical).

This is the first time I have ever seen a formula and I have no idea what it means.

The reason that we keep seeing posts on tyre pressure is that there is no decent guide on how to set them, the closest I've seen this the mitchlen guide, which in itself is unclear as you can't tell if the weight includes the bike or excludes the bike. The other solution I've seen to measure the size of the contact patch, something thats rather difficult to do when you are on top of the bike.

What would be great is 2 formulas, say a simple one containing bike weight, person weight, tyre diameter and a complex one containing all of the other factors - tyre construction, road surface, bike ridigity, whatever. Put them in a spreadsheet for dummies like me.

One silly question, when you take the pump off the tube, how much pressure do you loose?

 

[Dietmar]

This is the first time I have ever seen a formula and I have no idea what it means.

It's not really important for any practical purposes such as the ones you mention below. So I'd say, don't worry about it. Not to be misunderstood, if your comment above was meant as a request to explain these formulas in more detail, I'd be happy to oblige. Just let me know.

The reason that we keep seeing posts on tyre pressure is that there is no decent guide on how to set them, the closest I've seen this the mitchlen guide, which in itself is unclear as you can't tell if the weight includes the bike or excludes the bike. The other solution I've seen to measure the size of the contact patch, something thats rather difficult to do when you are on top of the bike.

The size of the contact patch, to a good approximation, is simply the weight on that particular wheel divided by the tire pressure. If we call the size of the contact patch A, the weight on the wheel W, and the pressure p, we have A=W/p. In terms of the weight of bicycle and rider, the contact patches for the front and rear wheel then depend on the rider's position, but as a rough estimate you could use half the sum of rider plus bike weight for W (which would ignore that for a normal sitting position, more of the weight will be on the rear wheels.

I just bought a new set of Michelin's, and they had a pressure guide on the package of the tire, with a simple curve for recommended tire pressure as a function of the weight of the rider. Is that what you are looking for?

But, unless someone proves me wrong, my feeling is that you will be very close to optimal if you simply pump up to the highest recommended tire pressure, at least if you ride on reasonably smooth asphalt roads. My gut feeling is that with an optimal tire pressure, if you can find it, you will only save rolling resistance in the single digit percentages, meaning the total resistance of your bike (rolling resistance plus aerodynamic drag) will change at most by a fraction of a percent. That should be pretty much irrelevant for just about anybody. I wouldn't even worry about that if I'd race for a living, and I don't (thank god, I'd probably be starving then...)

What would be great is 2 formulas, say a simple one containing bike weight, person weight, tyre diameter and a complex one containing all of the other factors - tyre construction, road surface, bike ridigity, whatever. Put them in a spreadsheet for dummies like me.

Ah, but the second one is really hard to find, and depends on a large number of parameters, including the rigidity of your butt, which is hard to measure...

One silly question, when you take the pump off the tube, how much pressure do you loose?

Not silly at all. I would say it depends a bit on the construction of your pump, and the way it measures the pressure. My guess is, though, that if the pump measures the pressure in the tube connecting to the valve, and if you wait a second or so after the last pump stroke before disconnecting the pipe (for my pump at least, the pressure it shows does not drop significantly during a couple of seconds after the last stroke), then the tire valve will be closed before you disconnect the tube, and you should not loose any pressure at all compared to the one you just saw before you disconnected he tube.
But this might be a great little experiment for our firend ScienceIsCool to do.

 

[mikesbytes]

No need to explain the forumla, I was just thinking that it would be great to have it in a spreadsheet, where you put in your figures, perhaps using drop down list box's.

The Mitchlen guide is what I base my pressure on, however I have applied some assumptions;
1. Bicycle = 6.9kg. As my bike is around 10kg depending on how much water I am carrying, I add 3kg to my body weight.
2. Pro race 2 is a quality tyre. I add 5psi to compensate for my existing tyres. When my current tyres wear out, I'll be switching to pro race 2's which are sitting in my cupboard.
3. I am lazy. Optimum tyre pressure should be at the mid point between pumping it up. If I accidently put slightly more in, I don't correct.
4. Loss when pump is removed. as per 3.

Also down under its being a bit wet on the winter rides over the last 2 months, so I've often opted not to pump them up and go out on softer tyres.

I don't know how accurate my method is and to be honest it is very subjective. I'm not as senstitive to tyre pressures as I was say 2 years ago, perhaps because I'm a little stronger.

Should also mention that I'm not sure whether to run the front at 5psi or 10psi softer. If you get off the seat and push it when riding up a hill, I load up the front more, but most of the time you are seated.

 

[Dietmar]

Here's what I do: I pump my Pro 2 Races up to 115 psi, and be done with it. Works fine.

 

[mikesbytes]

Here's what I do: I pump my Pro 2 Races up to 115 psi, and be done with it. Works fine.

My calc for pro race 2's came out at 115psi rear and 105psi front with my current body weight. I might be a little lighter over summer and in that case I'll probably go for 110psi in the rear. Still it would be nice to get the figure from a formula to back up or dispell the subjective assumptions I have used.