Aero Basics
The science around cycling and the fundamental equations that govern how fast we can go are usually obscured by terms such as Watts and CdA that are not all that well explained. Our goal here is to break that down a bit, or really, add more detail to the math that tells us how fast we can go.
I - Terminology
| Symbol | Term | Units |
|---|---|---|
| $a$ | Acceleration | $\text{m/s}^2$ |
| $A$ | Area | $\text{m}^2$ |
| $C_d$ | Coefficient of Aerodynamic Drag | unit-less |
| $C_{rr}$ | Coefficient of Rolling Resistance | unit-less |
| $E$ | Energy | J |
| $F$ | Force | N |
| $g$ | Acceleration due to Gravity | $9.8\ \text{m/s}^2$ |
| $G$ | Grade or Slope | unit-less |
| $h$ | Height | m |
| $x$ | Distance | m |
| $m$ | Mass | kg |
| $M$ | Molar Mass | $\text{kg/mol}$ |
| $n$ | Amount of Substance | mol |
| $p$ | Momentum | $\text{kg}\cdot\text{m/s}$ |
| $P$ | Pressure | Pa, $\text{J/m}^3$ |
| $R$ | Ideal Gas Constant | $8.314\ \text{J/(mol}\cdot\text{K)}$ |
| $Re$ | Reynolds Number | unit-less |
| $t$ | Time | sec |
| $T$ | Temperature | K |
| $v$ | Velocity | $\text{m/s}$ |
| $V$ | Volume | $\text{m}^3$ |
| $W$ | Power | W, $\text{J/s}$ |
| $\theta$ | Angle | radians |
| $\epsilon$ | Vertical Velocity Fraction | unit-less |
| $\rho$ | Density | $\text{kg/m}^3$ |
| $\mu$ | Dynamic Viscosity | $\text{Pa}\cdot\text{s}$ |
1.0 Fundamentals of Energy
In order to understand the equations that govern cycling, it is essential to start from the basics and define what energy and power fundamentally are. In physics, there are Fundamental Quantities, which include mass, distance, temperature, and time, which are used to define the subsequent Derived Quantities. The Derived Quantities include velocity, acceleration, force, energy, and power. These are defined in the following paragraphs.
Table 1 - Fundamental Quantities
| Quantity | Quantity Symbol | Unit | Unit Symbol |
|---|---|---|---|
| Mass | m | Kilograms | kg |
| Distance | x | Meters | m |
| Temperature | T | Kelvin | K |
| Time | t | Seconds | s |
The Derived Quantities are defined below. Velocity ($v$) is the change in distance with respect to time:
$$(1.00) \quad v = \frac{dx}{dt}$$Acceleration ($a$) is then the change in velocity with respect to time:
$$(1.01) \quad a = \frac{dv}{dt}$$Momentum ($p$) is defined as the product of mass and velocity:
$$(1.02) \quad p = mv$$Force ($F$), from Newton's Second Law of Motion, is the change in momentum with respect to time.
$$(1.03) \quad F = \frac{dp}{dt}$$With constant mass this becomes:
$$(1.04) \quad F = m\frac{dv}{dt}$$Which is:
$$(1.05) \quad F = ma$$Change in Energy ($E$) is force applied over a distance, or simply the product of force and distance. Using the force equation this becomes:
$$(1.06) \quad dE = F \, dx$$Using equation (1.04) this becomes:
$$(1.07) \quad dE = m\frac{dv}{dt}dx = m\frac{dx}{dt}dv = mvdv$$Integrating both sides of the equation gives the definition of Kinetic Energy:
$$(1.08) \quad E_k = \frac{1}{2}mv^2$$Potential energy is found in a similar way, where equation (1.06) becomes:
$$(1.09) \quad dE = mg\,dh$$Where $g$ is acceleration due to gravity and $h$ is elevation. Integrating gives the definition of potential energy:
$$(1.10) \quad E_p = mgh$$Next Power ($W$) is the rate of energy use, or the change in energy with respect to time:
$$(1.11) \quad W = \frac{dE}{dt}$$2.0 Power Balance
The motion of bicycles is governed by the conservation of energy, or to put it more simply, the energy in is equal to the energy out.
$$(2.00) \quad E_{in} = E_{out}$$The $E_{in}$ is the power provided by the cyclist.
$$(2.01) \quad E_{in} = E_{\text{cyclist}}$$Giving:
$$(2.02) \quad E_{\text{cyclist}} = E_{out}$$The energy balance must consider all forms of energy lost to the surroundings as well as both kinetic and potential energy. That lost energy includes rolling resistance ($E_{rr}$), aerodynamic drag ($E_{C_d A}$), and drivetrain losses. Here drivetrain losses are neglected as they are both small and difficult to measure in this setting.
$$(2.03) \quad E_{\text{cyclist}} = E_k + E_p + E_{rr} + E_{C_d A}$$Because bicycles use power-meters, it is more convenient to look at the rate of energy consumption, that is power, as opposed to total energy consumption. This is found by taking the derivative of energy with respect to time. Taking the derivative of each term on the right side of the equation gives the following equations for the different forms of power. First taking the derivative of Potential Energy:
$$(2.04) \quad \frac{dE_p}{dt} = \frac{d(mgh)}{dt} = mg\frac{dh}{dt}$$This gives:
$$(2.05) \quad W_p = mg v_{\text{vertical}}$$The calculation of $v_{\text{vertical}}$ in the equation above is shown in Appendix A. Kinetic Energy follows the same pattern as potential energy:
$$(2.06) \quad \frac{dE_k}{dt} = \frac{d(\frac{1}{2}mv^2)}{dt} = \frac{1}{2}m\frac{dv^2}{dt} = \frac{1}{2}m 2v \frac{dv}{dt}$$This produces:
$$(2.07) \quad W_k = mva$$Next rolling resistance is:
$$(2.08) \quad W_{rr} = C_{rr} mg v_{\text{abs}}$$As noted previously, $C_{rr}$ is the coefficient of rolling resistance. Aerodynamic losses are found by starting with Bernoulli's equation. It is worth noting that while pressure is typically viewed as force per unit area ($\text{N/m}^2$) it can be helpful to view it as energy per unit volume ($\text{J/m}^3$), especially when dealing with fluids.
$$(2.09) \quad P_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2$$Here there is no change in elevation so this simplifies to:
$$(2.10) \quad P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$$In this case $v_2$ is the stagnation velocity of air as it hits the rider, which is zero.
$$(2.11) \quad P_1 + \frac{1}{2}\rho v_1^2 = P_2$$Solving for change in pressure gives:
$$(2.12) \quad \Delta P = P_2 - P_1 = \frac{1}{2}\rho v_1^2$$Multiplying each side by frontal area gives the total force:
$$(2.13) \quad F = \frac{1}{2}\rho v_1^2 A$$From equations $dE = F\,dx$ and $W = \frac{dE}{dt}$ this becomes:
$$(2.14) \quad W_{C_d A} = C_d A \frac{1}{2}\rho v_{\text{relative}}^2 v_{\text{abs}}$$In which $v_{\text{abs}}$ is the absolute speed of the bike and $v_{\text{relative}}$ is the speed of the air relative to the bike or the air speed of the bike. $C_d A$ is the product of $C_d$, the coefficient of drag, and $A$, the frontal area of the bicycle and cyclist. Because it is difficult to isolate $A$, it is simpler to look at $C_d A$ as a single term.
In many methods $C_d A$ is assumed to be a true constant, but in reality it is a function of the Reynolds Number ($Re$), which is a dimensionless quantity defining the ratio of inertial forces to viscous forces of the system. Low $Re$, typically below 2300, characterizes Laminar Flow and high $Re$, above 10000, characterizes Turbulent Flow with a transitional region existing between Laminar and Turbulent flows (2300 to 10000).
$$(2.15) \quad Re = \frac{\rho v_{\text{relative}} L}{\mu}$$Where $\rho$ is the density of air, $L$ is the characteristic length, and $\mu$ is the dynamic viscosity of air. Both $Re$ and $W_{C_d A}$ require the density of air which is found from the Ideal Gas Law, this is shown in Appendix B. Putting all of this together gives the power balance equation for cycling.
$$(2.16) \quad W_{\text{cyclist}} = m v_{\text{abs}} a + m g v_{\text{vertical}} + C_{rr} m g v_{\text{abs}} + C_d A \frac{1}{2} \rho v_{\text{relative}}^2 v_{\text{abs}}$$Appendix A: Vertical Velocity
For potential energy:
$$(A.1) \quad W_p = mg v_{\text{vertical}}$$$v_{\text{vertical}}$, the rate of climbing in $\text{m/s}$, can be found multiplying the Vertical Velocity Fraction ($\epsilon$) by $v_{\text{abs}}$.
$$(A.2) \quad W_p = mg \epsilon v_{\text{abs}}$$Here the Vertical Velocity Fraction is the ratio of the vertical velocity to the absolute velocity. This can be viewed simply as the ratio of the height of a right triangle to its hypotenuse. $\epsilon$ is found by determining the angle ($\theta$) of the grade or slope ($G$) in radians and then taking the sine of that angle.
$$(A.3) \quad \epsilon = \sin(\theta)$$ $$(A.4) \quad \theta = \tan^{-1}(G/100)$$ $$(A.5) \quad G = \frac{\text{rise}}{\text{run}} \cdot 100$$Or putting this all together:
$$(A.6) \quad W_p = mg v_{\text{abs}} \sin\left(\tan^{-1}(G/100)\right)$$Appendix B: Air Density
To get air density start with the Ideal Gas Law:
$$(B.1) \quad PV = nRT$$Where $P$ is pressure, $V$ is volume, $n$ is the moles of gas, $R$ is ideal gas constant, and $T$ is the absolute temperature. Rearranging this provides:
$$(B.2) \quad \frac{n}{V} = \frac{P}{RT}$$Multiplying both sides by Molar Mass ($M$) gives:
$$(B.3) \quad \frac{Mn}{V} = \frac{PM}{RT}$$Since molar mass is:
$$(B.4) \quad M = \frac{m}{n}$$The equation becomes:
$$(B.5) \quad \frac{m}{V} = \frac{\Sigma PM}{RT} = \rho$$Here $\Sigma PM$ is the sum of the partial pressure of air and the partial pressure of water vapor:
$$(B.6) \quad \rho = \frac{P_{\text{air}} M_{\text{air}} + P_{\text{vapor}} M_{\text{vapor}}}{RT}$$The partial pressure of water vapor is found by taking the product of saturation pressure of air and the relative humidity ($\phi$):
$$(B.7) \quad P_{\text{vapor}} = P_{\text{saturation}} \phi$$Where $P_{\text{saturation}}$ can be found from the August-Roche-Magnus equation:
$$(B.8) \quad P_{\text{saturation}} = 610.94 e^{\frac{17.625T}{T + 243.04}}$$The partial pressure of air is then found by subtracting $P_{\text{vapor}}$ from atmospheric pressure, $P_{\text{atm}}$:
$$(B.9) \quad P_{\text{air}} = P_{\text{atm}} - P_{\text{vapor}}$$