How a Pneumatic Tire Works

A car tire supports the weight of the car by the difference in tension in the top and bottom tire walls.

Figure 1: An idealized model of a tire, from my 2018 notes.

Background

About December 2, 1981¹ I was asked in a undergraduate University interview how a pneumatic tire holds up a car. At the time, the best I could come up with was that it wasn’t because of the air pressure on the axle, because that operated equally in all directions. I thought about this problem on and off for many years afterwards but never got anywhere. I generally got confused about \(PV/T\) being a constant for a gas and wondered about how best to measure the quantity of air in the tire. When I finally semi-retired, I decided the time had come to figure out the answer. Strangely, Google failed me.

I decided to start from first principles and read Feynman’s Lecture Notes on Physics, Chapter 39 The Kinetic Theory of Gasses. The second paragraph is not very promising

It is obvious that this is a difficult subject, and we emphasize at the beginning that it is in fact an extremely difficult subject, and that we have to deal with it differently than we have dealt with the other subjects so far. In the case of mechanics and in the case o flight, we were able to begin with a precise statement of some laws, like Newton’s law, or the formula for the field produced by an accelerating charge, from which a whole host of phenomena could be essentially understood, and which would produce a basis for our understanding of mechanics and of light from that time on. That is, we may learn more later, but we do not learn different physics, we only learn better methods of mathematical analysis to deal with the situation.

We cannot use this approach effectively in studying the properties of matter. We can discuss matter only in a most elementary way; it is much too complicated a subject to analyze directly from its specific basic laws, which are none other than the laws of mechanics and electricity. But these are a bit too far away from the properties we wish to study; it takes too many steps to get from Newton’s laws to the properties of matter, and these steps are, in themselves, fairly complicated. We will now start to take some of these steps, but while many of our analyses will be quite accurate, they will eventually get less and less accurate. We will have only a rough understanding of the properties of matter.

He explains this is because we need an understanding of probability, and that the actual behavior of atoms is governed by quantum mechanics. However, Feynman then goes on to explain a model of an ideal gas that was just what I needed for this problem. The key idea is to think about gas one molecule at a time. The molecule is the appropriate measure of the “amount” of a gas. From there the other variables largely take care of themselves. For example, the constant \(PV/T\) equals \(Nk\) (Eq. 39.22) where \(N\) is the number of molecules and \(k\) is a constant, better written \[ PV = NkT. \] Remarkably, this means that “at the same temperature and pressure and volume, the number of atoms is determined; it too is a universal constant!” These ideas are explained more below in a section Chat GPT helped write.

From here, two observations got me to a good model of “how a tire works”.

Two Observations

Tires are not like balloons

The first observation is that tires are not like balloons: although they are made of rubber they aren’t really elastic. In fact, a modern radial tire contains quite a lot of steel, and it can be regarded as largely inelastic.

Watching a bicycle tire inflate

Secondly, I observed happens when inflating a flat bicycle tire—something I’d done recently. There are four distinct phases of inflation:

0. The tire is fully deflated, and the rim rests on the ground.
1. The tire partially inflates, gradually filling the “top” of the tire.
2. The tire continues to inflate and tension builds in the upper wall, but the rim stays on the ground.
3. A critical pressure is reached and the rim lifts off the ground.
4. Further inflation increases the volume and introduces tension into the lower tire wall.

Eventually, the tire reaches its natural volume and is essentially round with a very small flat contact surface with the ground. Further inflation increases pressure but the tire’s volume does not change materially.

A Simple Tire Model

Figure 2: A schematic of a tire. Most forces occur in cancelling pairs except above the contact footprint.

Most of the tire doesn’t matter to solve this problem, Figure 2. You can symmetrically remove most of the tire without changing the essential features we want to understand. The model I initially worked with is shown in Figure 3.

Figure 3: A tire stripped to the essential components needed to understand how it supports a car.

In the simple model, there are two reservoirs with rigid tops and bottoms, and flexible walls that are “scrunchable” but not elastic. Each reservoir has a natural volume. The two are connected by a tube to equalize pressure. This provides a model of the tire above the patch where it touches the ground, removing the parts to the sides. These are symmetric and contain exactly opposite forces, so they can be ignored. This “tire” will support a load hung from the middle of the tube connecting the top and bottom reservoirs but obviously won’t go round nor function as a wheel. Figure 4 illustrates the forces acting in this model.

Figure 4: Forces acting on each component in the simple tire model.

Figure 5 illustrates the four phases of inflation using the two-reservoir model.

Figure 5: The four phases of inflation using a simplified tire model.

A Simpler Model

On further thought, I realized that the model in Figure 3 can be simplified further. A simpler model is shown in Figure 6. Here, we have a piston, connected to a middle plate with holes, through which the air can freely flow, which in turn is connected to the bottom of the cylinder. The middle plate (the axis) supports the mass. The connectors (sides of the tire) are light, in-extensible strings.

Figure 6: A simpler tire model.

The four phases of inflation in the simpler model are shown in Figure 7.

Figure 7: The four phases of inflation in the simpler model.

In phase 1, air pressure forces the piston up, but its travel is constrained by outside air pressure and there is no tension in the upper string. Tire air pressure equals atmospheric pressure (because the piston doesn’t move). At some point, phase 2, the unconstrained volume implies a height greater than the connecting string and tension builds up in the string, restricting further increase in volume. Now pressure increases but volume stays constant, with the piston constrained by tension in the upper string. The mass on the axle acts opposite to the tension and the lower platform stays at the bottom of the cylinder. In phase 3, the tension equals the force exerted by gravity on the supported mass and the lower platform starts to rise. Now, pressure stays the same and volume increases. Finally, in phase 4, further increase in volume of the tire is constrained by tension in the lower string.

Four Phase Inflation Model

Table 1 defines the variables used in the simpler model. Pressure is absolute, measured relative to a perfect vacuum, not “guage” pressure relative to ambient atmospheric pressure. Remarks in the table give values and relationships for the simpler model.

Table 1: Variables, base values, and SI units.

Variable	Meaning	Units	Value
\(a\)	contact area of tire; area of piston	\(m^2\)	0.0225
\(M\)	mass supported on axle	\(kg\)	500
\(w\)	wall height of tire	\(m\)	0.075
\(g\)	gravity acceleration	\(m / s^2\)	9.8
\(k\)	Boltzmann’s constant	\(J/K\)	\(1.38 \times 10^{-23}\)
\(A\)	Avogadro’s Number		\(6.022\times 10^{23}\)
\(R\)	Universal gas constant \(=kA\)	\(J/K\)	8.31446
\(N\)	number of molecules of gas inflating tire		variable
\(\alpha\)	atmospheric pressure in Pa	\(Pa=kg /ms^2=N/m^2\)	101,325
\(T\)	temperature in Kelvin, taken as fixed	\(K\)	293
\(M\)	mass supported by tire	\(kg\)	340
\(g\)	gravitational acceleration	\(m/s^2\)	9.8
\(P\)	absolute pressure in tire	\(Pa\)	variable
\(h_1\)	wall height of upper wall, \(h_1\le w\)	\(m\)
\(F_1\)	tension in upper wall	\(N=kg m/s^2\)
\(h_2\)	wall height of lower wall, \(h_2\le w\)	\(m\)
\(F_2\)	tension in lower wall	\(N=kg m/s^2\)
\(V\)	volume of tire; \(a(h_1+h_2)\le 2aw\)	\(m^3\)
\(N'\)	\(N' := kNT\)

Table 2 shows the state of the system during each phase of inflation, and gives the critical values of \(N'\) when the system changes from one phase to the next. The tire “leaves” phase four when the tire wall fails because of pressure and the tire explodes. The state of the system is determined by the value of \(N':=kNT\), proportional to the number of molecules of gas in the tire \(N\) (or number of moles). In each phase, one variable is increasing linearly with \(N'\): in phase 1 it is \(h_1\), the height of the upper chamber, in phase 2 the tension \(F_1\) in the upper connector (tire wall), in phase 3 the height \(h_2\) of the lower chamber, and in phase 4 the tension \(T_2\) in the lower connector.

• Phase 0: Tire completely empty.
• Phase 1: Upper reservoir inflates.
• Phase 2: Tension increases in upper side walls, no change in volume.
• Phase 3: Lower reservoir inflates, no tension in lower side walls.
• Phase 4: Tire fully inflated (has reached its maximum volume), pressure increases.

Table 2: State of the system during each phase of inflation. Pressure is absolute, not above atmospheric and \(N':=kN\).

Phase	0	1	2	3	4
description	fully deflated	upper inflation	upper tension	lower inflation	lower tension
\(P\)	0	\(\alpha\)	\(N'/aw\)	\(\alpha + Mg/a\)	\(\alpha + F_1/a\)
\(V\)	0	\(ah_1\)	\(aw\)	\(a(w+h_2)\)	\(2aw\)
\(h_1\)	0	\(N'/\alpha a\le w\)	\(w\)	\(w\)	\(w\)
\(F_1\)	0	0	\(N'/w-\alpha a\le Mg\)	\(Mg\)	\(Mg+F_2>Mg\)
\(h_2\)	0	0	0	\(N'/(Mg+\alpha a) - w\le w\)	\(w\)
\(F_2\)	0	0	0	0	\(N'/2w - Mg - \alpha a> 0\)
Critical \(N'\)	\(>0\)	\(\alpha aw\)	\((Mg+\alpha a)w\)	\(2(Mg+\alpha a)w\)	explosion

In phase 1: \(P=\alpha +F_1 /a = \alpha + (N'/w - \alpha a)/ a = N'/aw\). Other phases are similar: start with known values and apply \(PV=N'\).

A normal car tire contains around 115g of air or about 3.95 moles or about \(2.38\times 10^{24}\) molecules, see estimates below.

The compressed air in a tire has a explosive force if released suddenly, as several (gruesome) on-line videos show. This is why large truck tires are often inflated in cages.

Numerical Example

Figure 8 shows how wall tension, wall height, and tire pressure vary with the amount of air in the tire. In each plot the dashed vertical lines delineate the phases. The wall tension plot shows that initially both walls are slack. As the tire inflates the upper side will extends to its natural height (0.075m) at which point tension starts to build in the upper wall (\(F_1\), blue line). When it exceeds the force \(500g=4900N\) on the axis the lower “rim” lifts off the ground and the lower wall height increases while the lower wall remains slack. When it is fully extended pressure increases but volume stays constant. The tire pressure is shown in gauge psi on the left and absolute Newtons on the right. The horizontal line shows 33psi standard operating pressure is just enough to create a small amount of tension in the lower wall. The simpler model tire has a volume of \(2aw=0.0035\), or about \(1/8.6\) as much as a realistic tire, so it reaches operating pressure with about 0.45 moles of air and \(0.45 \times 8.6 = 3.9\), consistent with the estimate given above.

import matplotlib.pyplot as plt
import numpy as np
import pandas as pd

A = 6.022e23

def tire(N, a=0.0225, w=0.075, M=500, T=293):
    '''
    Contact patch 6" x 6" = approx. 0.0225 sq meters
    Wall height = 3" = 7.5 cm = 0.075 meters
    Mass approx 4000 lbs over 4 wheels = 500 kg per wheel
    20C = 293K

    Given the number of atoms or molecules of gas N output

    * moles of gas
    * P pressure
    * h1 upper wall height
    * h2 lower wall height
    * F1 = upper wall tension
    * F2 = lower wall tension
    * step = 0, 1, 2, 3, 4: empty; inflating upper; increasing pressure; inflating lower; increasing pressure
    * Pressure in PSI adjusted for atmospheric pressure
    * N1 molecules to fully inflate upper reservoir
    * N2 molecules to start inflating lower reservoir
    * N3 molecultes to fully inflat lower reservoir
    '''
    # physical constants
    global A
    k = 1.38e-23
    g = 9.8
    alpha = 101325
    pa_psi = 6895

    # figure critical values of N
    N1 = alpha * a * w / (k * T)                # upper tire fully inflated
    N2 = (alpha * a + M * g) * w / (k * T)      # lower tire starts to life
    N3 = (alpha * a + M * g) * w * 2 / (k * T)  # lower tire fully inflated

    if N <= 0:
        P = 0
        h1 = 0
        h2 = 0
        F1 = 0
        F2 = 0
        step = 0
    elif N <= N1:
        # phase 1
        P = alpha
        h1 = N * k * T / (alpha * a)
        h2 = 0
        F1 = 0
        F2 = 0
        step = 1
    elif N <= N2:
        # phase 2
        F1 = N * k * T / w - alpha * a
        P = F1 / a + alpha
        h1 = w
        h2 = 0
        F2 = 0
        step = 2
    elif N <= N3:
        # phase 3
        h1 = w
        F1 = M * g
        h2 = N * k * T / (alpha * a + M * g)  - w
        P = F1 / a + alpha
        F2 = 0
        step = 3
    else:
        # phase 4
        h1 = w
        h2 = w
        F2 = N * k * T / (2 * w) - M * g - alpha * a
        F1 = F2 + M * g
        P = F1 / a + alpha
        step = 4

    return N/A, P, h1, h2, F1, F2, step, (P - alpha) / pa_psi, N1, N2, N3

# set the output data types else h2 can return as an int
vtire = np.vectorize(tire, otypes=[float, float, float, float, float, float,
                                   np.int64, float, float, float, float])

# figure reasonable ranges, start with critical N values
N1, N2, N3 = tire(0)[-3:]
Ns = np.concatenate((np.linspace(N1 / 100, N1, 11),
                np.linspace(N1 , N2, 11),
               np.linspace(N2, N3, 11),
               np.linspace(N3, N3*1.5,11)))

E = pd.DataFrame({ i: j for i, j in zip(['N', 'P', 'h1', 'h2', 'F1', 'F2', 'Step', 'psi'], vtire(Ns))})
E = E.set_index('N')

fig, axs = plt.subplots(2, 2, figsize=(2*3.5, 2*2.45), constrained_layout=True)
ax0, ax1, ax2, ax3 = axs.flat

ax3.remove()

for ax in [ax0, ax1, ax2]:
    for n in [N1 / A, N2 / A, N3 / A]:
        ax.axvline(n, lw=.5, ls='--')

E[['F1', 'F2']].plot(ax=ax0)
ax0.set(title='Wall tension', xlabel='Moles of air', ylabel='Wall tension in Newtons')
ax0.legend()

E[['h1', 'h2']].plot(ax=ax1)
ax1.set(title='Wall height', xlabel='Moles of air', ylabel='Wall height in meters', ylim=[-0.005, 0.085])
ax1.legend()

E[['psi']].plot(ax=ax2)
ax2.set(title='Tire Pressure', xlabel='Moles of air', ylabel='Pressure in psi');

ax2t = ax2.twinx()
E[['P']].plot(ax=ax2t)
ax2t.set_ylabel('Absolute pressure in Pascals')
ax2.axhline(33, lw=.5, ls=':')
ax2.legend().set(visible=False);
ax2t.legend().set(visible=False);

Figure 8: Plot of wall tension, wall height, and tire pressure against against number of moles of air in the tire.

Properties of an Ideal Gas

GPT Notes

An ideal gas is a theoretical concept in physics and chemistry that simplifies the behavior of gases to facilitate understanding and calculations. It is based on several assumptions:

1. Point Particles: The molecules of an ideal gas are considered point particles with no volume. This means that the volume of the gas particles themselves is negligible compared to the volume the gas occupies.

2. Elastic Collisions: Collisions between gas particles and between particles and the walls of the container are perfectly elastic. This means there is no loss of kinetic energy during collisions.

3. No Intermolecular Forces: Ideal gas particles do not exert any forces on each other, except during collisions. There is no attraction or repulsion between particles.

4. Random Motion: The gas particles are in constant, random motion, and their collisions lead to a uniform distribution of velocities (Maxwell-Boltzmann distribution).

5. Kinetic Energy and Temperature: The average kinetic energy of gas particles is directly proportional to the absolute temperature of the gas.

The ideal gas law, \(PV = nRT\), where \(n\) is the number of moles and \(R\) is the universal gas constant, describes the behavior of an ideal gas. This law is a combination of Boyle’s law, Charles’s law, and Avogadro’s law.

While no real gas perfectly fits the description of an ideal gas, many gases behave approximately ideally under a wide range of conditions, especially at low pressures and high temperatures where the volume of the gas particles is much less significant compared to the volume of the container, and intermolecular forces are weaker.

The relationship \(PV = kNT\) connects pressure (\(P\)), volume (\(V\)), the number of molecules (\(N\)), the Boltzmann constant (\(k\)), and temperature (\(T\)) in a gas. This equation is a form of the ideal gas law, reflecting the kinetic theory of gases. Here’s the breakdown:

1. Pressure and Molecular Impacts: Pressure in a gas arises from the force exerted by molecules colliding with the walls of the container. This force per unit area (pressure) is related to the change in momentum of the molecules upon impact. The more frequent and more forceful these collisions, the higher the pressure.

2. Temperature and Average Kinetic Energy: The temperature of a gas is a measure of the average kinetic energy (\(\frac{1}{2}mv^2\)) of the molecules. As temperature increases, the average speed of the molecules (\(v\)) increases, leading to more frequent and more forceful collisions, which in turn increases the pressure.

3. Connection between Temperature, Speed, and Pressure: Higher temperatures mean higher average speeds for the gas molecules due to their increased kinetic energy. This results in more frequent and more energetic collisions with the container walls, thus increasing the pressure. Mathematically, this is expressed as \(\frac{1}{2}mv^2 = \frac{3}{2}kT\) for a monatomic gas, linking the average kinetic energy directly to temperature.

4. Ideal Gas Law (PV = NkT): This equation encapsulates the relationships mentioned above. It shows that the product of pressure and volume is proportional to the total kinetic energy of all the molecules in the system, which is a function of the number of molecules and the temperature.

• \(P\) (Pressure) increases with more frequent/energetic collisions.
• \(V\) (Volume) provides the space in which the molecules are contained. More volume means fewer collisions per unit area, thus lower pressure.
• \(N\) (Number of molecules) affects how many collisions occur.
• \(T\) (Temperature) influences the speed and kinetic energy of the molecules, affecting their collisions’ frequency and force.
• \(k\) (Boltzmann constant) is a proportionality factor that relates energy at the molecular level to temperature.

By understanding these connections, you can see how changes in temperature influence molecular speeds and how this affects the pressure in a system, given a constant volume and number of molecules.

Gas molecules have internal energy states, including rotational, vibrational, and electronic energy levels. When a gas is heated, energy can be distributed among these internal degrees of freedom in addition to the translational kinetic energy. The distribution of energy depends on the specific heat capacities of the gas and the type of molecules involved.

The theory that accounts for these additional degrees of freedom is the kinetic theory of gases, extended to include quantum mechanical considerations. This more comprehensive approach is often discussed in the context of the equipartition theorem, which states that energy is distributed equally among all available degrees of freedom that are accessible at a given temperature.

The specific heat capacity of a gas (at constant volume or constant pressure) provides insight into how energy is distributed among the different degrees of freedom. For monatomic gases like helium or neon, the energy is primarily in translational motion since these atoms have no rotational or vibrational modes. Diatomic and polyatomic gases, however, have additional rotational and potentially vibrational modes.

For a diatomic gas, for example, at room temperature:

• Translational: There are 3 translational degrees of freedom (movement in x, y, z directions).
• Rotational: There are 2 rotational degrees of freedom (rotation about two axes; the third axis has negligible moment of inertia for a linear molecule).
• Vibrational: Diatomic molecules have 1 vibrational mode (vibrating along the bond axis), but this mode is usually not excited at room temperature.

At low temperatures, the energy is mostly in the translational modes. As the temperature increases, the rotational modes become more populated. Vibrational modes require higher temperatures to become significantly excited due to their higher energy thresholds.

The equipartition theorem would suggest that each degree of freedom gets an equal share of\(\frac{1}{2}kT\) of energy (where\(k\) is Boltzmann’s constant and\(T\) is temperature). So, for a diatomic gas at a temperature high enough to excite rotational but not vibrational modes, the energy would be split into 5 equal parts: 3 for translation and 2 for rotation.

\(C_v\) and\(C_p\) are specific heat capacities at constant volume and constant pressure, respectively. They represent the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin) under constant volume or pressure.

• \(C_v\) (Specific Heat at Constant Volume): This is the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree while keeping the volume constant. For an ideal gas,\(C_v\) is related to the internal energy and does not consider work done by the system as the volume is constant.

• \(C_p\) (Specific Heat at Constant Pressure): This is the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree while keeping the pressure constant. For an ideal gas,\(C_p\) is related to the enthalpy and takes into account the work done by the system during expansion.

The relationship between \(C_p\) and \(C_v\) for an ideal gas is given by the Mayer’s relation: \(C_p - C_v = R\), where \(R\) is the ideal gas constant.

These specific heat capacities are crucial in thermodynamics for understanding and calculating the energy exchanges in thermal processes.

The specific heat capacities (\(C_v\) and\(C_p\)) can give you an approximate idea of the split:

• For monatomic gases, \(C_v = \frac{3}{2}R\) and \(C_p = \frac{5}{2}R\).
• For diatomic gases (at temperatures where rotations but not vibrations are excited), \(C_v = \frac{5}{2}R\) and \(C_p = \frac{7}{2}R\).

These values reflect the degrees of freedom active in the energy distribution of the molecules.

The more comprehensive theory that includes quantum mechanical effects for energy distribution in gases is called quantum statistical mechanics, which provides a detailed explanation of how energy is distributed among all possible degrees of freedom according to quantum mechanics.

In gases, some of the energy absorbed by the molecules can be diverted into changing electron orbits, which means exciting electrons to higher energy levels. However, this aspect is more significant in the context of quantum mechanics and is typically observed at higher energy levels than those usually involved in thermal motion.

For most gases at ordinary temperatures and pressures, the energy associated with thermal motion primarily affects translational, rotational, and vibrational modes of the molecules. Electronic excitation typically requires much higher energies and becomes significant at high temperatures or under the influence of electromagnetic radiation with sufficient energy, such as ultraviolet light.

The theory that extends the ideal gas model to include internal energy changes, like electronic excitation, is called the quantum theory of gases. It considers the discrete energy levels associated with electronic, rotational, and vibrational states of molecules.

In summary, at typical conditions for gases, the majority of the thermal energy goes into translational, rotational, and, to some extent, vibrational motion, rather than electronic excitation. The partitioning of energy into these different modes is a key aspect of the quantum theory of gases and is crucial for understanding the specific heat capacities and other thermodynamic properties of real gases.

Air content of an inflated car tire

GPT Notes

To calculate the mass of air in a car tire, we first need to establish the volume of the tire and use the ideal gas law. A typical car tire pressure is around 2.2 bar (which is 2.2 times atmospheric pressure or about 220 kPa or 33psi) when measured relative to the outside air. In absolute terms, this is 220 kPa + 101.3 kPa (atmospheric pressure) = 321.3 kPa.

Let’s assume a typical car tire’s volume is around 30 liters (0.03 m³)². Use the ideal gas law: \(PV = nRT\), where\(P\) is pressure, \(V\) is volume, \(n\) is the amount of substance (in moles), \(R\) is the ideal gas constant, and \(T\) is temperature in Kelvin.

1. Calculate the amount of substance (n):

   \[n = \frac{PV}{RT}\]

   Convert the temperature to Kelvin: \(T = 20°C + 273.15 = 293.15 K\)

   Use \(R = 8.314 \, \text{J/(mol·K)}\) (universal gas constant).

   \[n = \frac{(321.3 \times 10^3 \, \text{Pa}) \times 0.03 \, \text{m}^3}{8.314 \, \text{J/(mol·K)} \times 293.15 \, \text{K}} = 3.95\]

2. Calculate the mass of the air:

   The molar mass of air (considering it as 78% nitrogen, 21% oxygen, and 1% argon) is approximately 28.97 g/mol.

   Mass (m) is given by \(m = n \times \text{molar mass} = 114.6g\).

3. Calculate the number of molecules:

   Use Avogadro’s number \((6.022 \times 10^{23} \text{ molecules/mol})\) to find the number of molecules:

   Number of molecules \(= n \times 6.022 \times 10^{23} = 2.38\times 10^{24}\).

The mass of air in a typical car tire at typical operating pressure and 20°C is approximately 0.115 kg. This amount of air corresponds to about \(2.38 \times 10^{24}\) molecules.

Figure 1: An idealized model of a tire, from my 2018 notes. Figure 2: A schematic of a tire. Most forces occur in cancelling pairs except above the contact footprint. Figure 3: A tire stripped to the essential components needed to understand how it supports a car. Figure 4: Forces acting on each component in the simple tire model. Figure 5: The four phases of inflation using a simplified tire model. Figure 6: A simpler tire model. Figure 7: The four phases of inflation in the simpler model. Figure 8: Plot of wall tension, wall height, and tire pressure against against number of moles of air in the tire.

Footnotes

1. I know the date because I remember buying a book there, and I wrote the date in it.

2. The tires on my car are about 0.6m in diameter with a 0.45m hub, and they are 0.2m wide, giving a volume of 0.025m³.