tIre pressure

Tom and Ken are both right, but Tom's response is more
on the mark.

A higher pressure shifts the tire's natural frequency way
up, so that it is a huge mismatch with the rider's forcing
frequency. The engineering term is "impedance mismatch". The
tire is rebounding even as the rider is trying to put more
energy into it.

By lowering the air pressure, the tire's natural frequency
is much lower, and the rider's forcing activity can fit
"inside" the tire's activity.

It's like getting a car to rock by repeatedly pushing on it.
The frequency you use in pushing it is crucial to getting
the desired outcome of maximum rock.

Other things that affect the tire's natural frequency are
the tire construction and volume, and the rim-tire boundary,
and even the temperature of the air in the tire.

--
U-Turn - Member of Generation XO

Weep in the dojo... laugh on the battlefield.

'29er Tire Study' (u-turn.unicyclist.com29erTireStudyOpen ↗)

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(unicyclist.comalbup39Open ↗)

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U-Turn said:

Tom and Ken are both right, but Tom's response is more on
the mark.

Thanks for being diplomatic, but we were both wrong! Still,
I was closer to providing an explanation.

I was wrong because I looked only at the energy going
into the tire, rather than total hop energy, and didn't
explain that energy stored in the tire comes from the pre-
hop bounce.

Tom was wrong in stating that the tire energy is the same
(it isn't) and in giving bad counterexamples (bouncing ball
and unicycle).

Quoted message said:

A higher pressure shifts the tire's natural frequency way
up, so that it is a huge mismatch with the rider's forcing
frequency. The engineering term is "impedance mismatch".
The tire is rebounding even as the rider is trying to put
more energy into it.

Impedance matching? Sounds nice, but I'd like to know how
you define the impedances (and expecially the boundaries) of
the system. The problem is that the rider gets to yank the
unicycle up by its seat/handle, thereby transfering energy
(at high frequency) at the end of the push phase.

Here's how I see it: You have, essentially, constant
maximum pushing force and work which don't depend on tire
characteristics. In other words, you can only push so hard,
and the energy you put into the system is the integral of
force over extension (distance), and you get to push all
that work regardless of tire pressure. The reason you hop
higher on a soft tire is that the tire compresses at the
end of the pre-hop, before you start pushing upwards. The
energy going into the hop is that prehop energy plus the
work you do pushing. As I pointed out orginally, a softer
tire stores more energy (up to the point friction causes
losses or the tire bottoms out), and it is that energy that
propels you higher.

Quoted message said:

Other things that affect the tire's natural frequency are
the tire construction and volume, and the rim-tire
boundary, and even the temperature of the air in the tire.

The tire's natural frequency - huh? Certainly the tire has
modes of oscillation, but they are high (audio range). The
frequency I believe you are interested in interest is that
of the unicycle, which has period 2pi*sqrt(m/k) (the
undamped, simplified case), where k is the spring constant
of the tire. I am highly skeptical that k depends on air
temperature. Except to the extent that air temperature
affects pressure or heats tire and rim.

Sorry, but I was right all along. I hope I said that nicely.
I am a retired Physics Police Inspector and, as a layperson,
try not to offend.

Ken Cline

In article <[email hidden]>, Ken Cline
<[email hidden]> wrote: ) )In other words, the energy
imparted to a bouncing ball is exactly the )kinetic energy
of the ball. Dampening scrubs energy and dulls the )bounds,
more and more as the air pressure gets lowered. ) )The
physics of the unicycle wheel is different: As the rider
pushes )off, he imparts a force, f, to the wheel which
compresses and storse )energy. Assume for the moment that
the wheel acts as a spring with )constant k. Then the amount
of energy stored is f^2/2k, which )increases as k decreases
(i.e. as pressure is lowered). The wheel is )not a perfect
spring, and dampening plays a role, at some point
)overwhelming the benefit of lower pressure.

The rider doesn't get something for nothing; he imparts a
force in the same way that a bouncing ball imparts a
force. The ground doesn't know whether there's a rider on
the unicycle.

You're leaving out the self-cancelling nature of spring
equations, in your model of the unicycle tire as a spring.
In particular, "k" (the spring constant) is not the same for
a low-pressure and a high-pressure tire, and that difference
cancels out your supposedly greater force returned by the
low-pressure tire.

We can calculate the spring constant with F = -k * x, where
x is the displacement from equilibrium for a given force.
This is *exactly the same equation* we use to calculate the
force returned from a compressed spring. Or in other words,
if you put 100N into a spring, you get 100N back out, minus
inefficiency. Two different tires have spring constants
inversely related to the distance they displace under a
given force, so these equations cancel each other out.

I don't see where the force in the higher-pressure tire (or
tighter spring) could possibly go, other than to be
returned. Heat, sound, and inelastic tire deformation are
all going to be greater on the low-pressure tire. If the higher-
pressure tire returns a lesser force than the low-pressure
tire, where did the difference come from?

If you know of a way to get more than 100N out of spring
that you've put 100N into, I suggest you report it to
<http://www.lhup.edu/~dsimanek/museum/unwork.htm>. -Tom

wow, I feel like I'm in a physics class!

Based on what I've pulled out of all that text, the timing
factor seems to be where you get a difference between low
and high pressure tires.

I am heavier than most Trials riders. Therefore I can get
lots of bounce from a pretty hard tire. If I hop on Zack
Baldwin's unicycle, for instance, I'll mostly just beat the
rim into the ground. I weigh more than he does.

What this means is that exchanging numbers on tire pressure
is useless unless you also include rider weight, tire volume
and rim width.

In the "old days" we did all our unicycling on 1.75" tires.
This meant you basically ran real high pressure all the
time. Otherwise your rim was on the ground. But I used to
ride up stairways and bleachers, and could get a decent
bounce (not like today's Trials experts).

Now that our tires can be a lot fatter, we have much more of
a choice on tire pressure. But the reason there is no
"correct" pressure for a given tire is that it can't be used
without factoring in the rider, tire size, tire shape, and
type of riding to be done.

--
johnfoss - Walkin' on the edge

John Foss, the Uni-Cyclone
"jfoss" at "unicycling.com"
www.unicycling.com

"Hey, could I have some of that spinach? I need to get this pork rind
taste out of my mouth." -- Ryan Atkins to Kris Holm, on the way back
from Moab after sampling some of my pork rinds. They grossed out the
whole van!
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