Steve Gribble   ·   gribble [at] gmail [dot] com

Cycling power and speed

Explore the relationship between your cycling power (wattage) and speed. Move your cursor over the graph, or tap on it, to explore specific points. Modify the fields below and the graph will update. Solve for specific power or speed

The physics affecting cycling at constant speed

This web page uses physical models of forces on a cyclist to help you estimate the relationship between power P (watts) and groundspeed velocity V_gs (kph or mph) of a cyclist. To do this, you need to estimate several parameters; reasonable defaults are given.

There are three primary forces that you, as a cyclist, must overcome in order to move forward:

• Gravity: If you're cycling uphill, you're fighting against gravity, but if you're cycling downhill, gravity works for you. This page measures the steepness of a hill in terms of percentage grade G: rise divided by run, multiplied by 100. The heavier you and your bike are, the more energy you must spend to overcome gravity. The combined weight of you (the cyclist) and your bike is W (kg). The gravitational force constant g is 9.8067 (m/s²).

  
  The formula for gravitational force acting on a cyclist, in metric units, is:
  
  \( \qquad F_{\mbox{gravity}} = 9.8067 \cdot \sin(\arctan(\frac{G}{100})) \cdot W \)
  

• Rolling resistance: Friction between your tires and the road surface slows you down. The bumpier the road, the more friction you'll experience; the higher quality your tires and tube, the less friction you'll experience. As well, the heavier you and your bike are, the more friction you'll experience. There is a dimensionless parameter, called the coefficient of rolling resistance, or C_rr, that captures the bumpiness of the road and the quality of your tires.

  
  The formula for the rolling resistance acting on a cyclist, in metric units, is:
  
  \( \qquad F_{\mbox{rolling}} = 9.8067 \cdot \cos(\arctan(\frac{G}{100})) \cdot W \cdot C_{rr} \)
  

• Aerodynamic drag: As you cycle through the air, your bike and body need to push the air around you, similar to how a snowplow pushes snow out of the way. Because of this, the air exerts a force against you as you ride. There are a few things that dictate how much force the air exerts against you. The faster your groundspeed (V_gs (m/s)), the more force the air pushes against you. The stronger the headwind (V_hw (m/s)), the more force the air pushes against you. As well, you and your bike present a certain frontal area A (m²) to the air. The larger this frontal area, the more air you have to displace, and the larger the force the air pushes against you. This is why cyclists and bike manufacturers try hard to minimize frontal area in an aerodynamic position. The air density Rho (kg/m³) is also important; the more dense the air, the more force it exerts on you.
  
  

  Finally, there are other effects, like the slipperyness of your clothing and the degree to which air flows laminarly rather than turbulently around you and your bike. Optimizing your aerodynamic positions also help with this. These other effects are captured in another dimensionless parameter called the drag coefficient, or C_d. Sometimes you will see people talking about "C_d · A", or CdA. This is just the drag coefficient C_d multiplied by the frontal area A. Unless you have access to a wind tunnel, it is hard to measure C_d and A separately; instead, people often just measure or infer C_d · A as a combined number.

  Your airspeed V_as (m/s) is the speed that the wind strikes your face, and it is the sum of your groundspeed V_gs (m/s) and the headwind speed V_hw (m/s):
  
  \( \qquad V_{\mbox{as}} = V_{\mbox{gs}} + V_{\mbox{hw}} \)
  
  The formula for the aerodynamic drag acting on a cyclist, in metric units, is:
  
  \( \qquad F_{\mbox{drag}} = 0.5 \cdot C_{d} \cdot A \cdot \mbox{Rho} \cdot V_{as}^2 \)
  

The total force resisting you, the cyclist, is the sum of these three forces:

\( \qquad F_{\mbox{resist}} = F_{\mbox{gravity}} + F_{\mbox{rolling}} + F_{\mbox{drag}} \)

For each meter that you cycle forward, you spend energy overcoming this resistive force. The total amount of energy you must expend to move a distance D (m) against this force is called the Work (Joules) that you do:

\( \qquad \mbox{Work} = F_{\mbox{resist}} \cdot D \)

If you are moving forward at groundspeed V_gs (m/s), then you must supply energy at a rate that is sufficient to do the work to move V_gs meters each second. This rate of energy expenditure is called power, and it is measured in watts. The power P_wheel (watts) that must be provided to your bicyle's wheels to overcome the total resistive force F_resist (Newtons) while moving forward with groundspeed V_gs (m/s) is:

\( \qquad P_{\mbox{wheel}} = F_{\mbox{resist}} \cdot V_{\mbox{gs}} \)

You, the cyclist, are the engine providing this power. The power that must be provided to your bicycle's wheels comes from your legs, but not all of the power that your legs deliver make it to the wheels. Friction in the drive train (chains, gears, bearings, etc.) causes a small amount of loss, usually around 2%, assuming you have a clean and nicely lubricated drivetrain. Let's call the percentage of drivetain loss Loss_dt (percent).

So, if the power that your legs provide is P_legs (watts), then the power that makes it to the wheel is:

\( \qquad P_{\mbox{wheel}} = \left(1 - \frac{\mbox{Loss}_{\mbox{dt}}}{100}\right) \cdot P_{\mbox{legs}} \)

Putting it all together, the equation that relates the power produced by your legs to the steady-state speed you travel is:

\( \qquad P_{\mbox{legs}} = \left(1 - \frac{\mbox{Loss}_{\mbox{dt}}}{100}\right)^{-1} \cdot \left[F_{\mbox{gravity}} + F_{\mbox{rolling}} + F_{\mbox{drag}}\right] \cdot V_{\mbox{gs}} \)

More fully, the following formula calculates the power that your legs must produce to maintain a given groundspeed velocity V_gs (m/s):

\( \qquad P_{\mbox{legs}} = \left(1 - \frac{\mbox{Loss}_{\mbox{dt}}}{100}\right)^{-1} \cdot \left[ \left(9.8067 \cdot W \cdot \left[\sin(\arctan(\frac{G}{100})) + C_{\mbox{rr}} \cdot \cos(\arctan(\frac{G}{100}))\right] \right) + \left(0.5 \cdot C_{d} \cdot A \cdot \mbox{Rho} \cdot \left(V_{\mbox{gs}} + V_{\mbox{hw}}\right)^2\right) \right] \cdot V_{\mbox{gs}} \)

One of the scary implications of this equation is that at high speed, the power you have to produce is proportional to the cube of your velocity. So, to increase your speed by 25%, you need to nearly double your wattage!

We may also want to calculate the groundspeed velocity that we will achieve if our legs produce a given power P. To do this, we rewrite the formula in cubic form:

\( \qquad aV_{\mbox{gs}}^3 + bV_{\mbox{gs}}^2 + cV_{\mbox{gs}} + d = 0 \)

    where:

\( \qquad a = 0.5 \cdot C_{d} \cdot A \cdot \mbox{Rho} \)
\( \qquad b = V_{\mbox{hw}} \cdot C_{d} \cdot A \cdot \mbox{Rho} \)
\( \qquad c = \left(9.8067 \cdot W \cdot \left[\sin(\arctan(\frac{G}{100})) + C_{\mbox{rr}} \cdot \cos(\arctan(\frac{G}{100}))\right]\right) + \left(0.5 \cdot C_{d} \cdot A \cdot \mbox{Rho} \cdot V_{\mbox{hw}}^2 \right) \)
\( \qquad d = -\left(1 - \frac{\mbox{Loss}_{\mbox{dt}}}{100}\right) \cdot P_{\mbox{legs}} \)

We can now use Cardano's formula to calculate the first real root of the equation, and hence the value of V that solves the cubic equation. (Thanks go to Csaba Toth for the hint to use Cardano!) Concretely:

\( \qquad Q = (3ac - b^2) / (9a^2) \)
\( \qquad R = (9abc - 27a^2d - 2b^3) / (54a^3) \)
\( \qquad S = \sqrt[3]{R + \sqrt{Q^3 + R^2}} \)
\( \qquad T = \sqrt[3]{R - \sqrt{Q^3 + R^2}} \)

\( \qquad V_{\mbox{gs}} = S + T - (b / (3a)) \)