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.Rider and bike parameters
Weight of bike (lb)
Total weight W (lb):
Drag coefficient C_{d}
C_{d} · A (ft^{2}):
Speed of headwind (negative for tailwind) (mph)
Coefficient of rolling resistance C_{rr}
Air density Rho (lb/ft^{3})
If you apply
watts of power, you will ride at groundspeed
velocity
mph.
If you want to ride at groundspeed velocity
mph, you must apply
watts of power.
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/s^{2}).
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 (m^{2}) 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/m^{3}) 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:
\( \qquad F_{\mbox{drag}} = 0.5 \cdot C_{d} \cdot A \cdot \mbox{Rho} \cdot V_{as}^2 \)
\( \qquad F_{\mbox{resist}} = F_{\mbox{gravity}} + F_{\mbox{rolling}} + F_{\mbox{drag}} \)
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:
\( \qquad Q = (3ac - b^2) / (9a^2) \)
\( \qquad R = (9abc - 27a^2d - 2b^3) / (54a^3) \)
\( \qquad S = \sqrt[3]{R + \sqrt{Q^3 + R^2}} \)
\( \qquad T = \sqrt[3]{R - \sqrt{Q^3 + R^2}} \)
\( \qquad V_{\mbox{gs}} = S + T - (b / (3a)) \)