**How
to calculate the speed of a commuting bicyclist**

**Equations**

Cyclist
reach a steady speed when the motive power that they produce balances the
rolling and air friction; in other words when the friction drags balance
the motive power.

The
air drag is given by the well known equation:

Here
A is the frontal cross section area, Cd is the drag coefficient, ris
the density of air, The air drag is proportional to the velocity v squared.
The dependence on the cross sectional area should make intuitive sense;
its harder to shove a big object through the air than a thin object. The
density comes into the air friction because the moving object has to shove
the air out of the way.

The
rolling drag is given by the equation:

Here
M is the mass of the rider and bike, g is the acceleration of gravity,
and Crr is the coefficient of rolling friction. This is simply the weight
(mg) of the rider and bike times the coefficient of friction.

The
total drag is the sum of these two drags:

The
power required to overcome these drags is the velocity times the drag:

Thus
the power required to overcome the drag scales as the velocity cubed.

To
calculate some actual values, we need to know the values of constants.
The density of air and the acceleration of gravity are well known.

The
rolling coefficient is more obscure, but various sources give it as approximately

Certainly
one can due better than this with a tuned racing bicycle, but the average
commuter bike isn't in great shape. Fortunately the exact value doesn't
matter much because the air drag is much more important than the rolling
drag at high velocities.

The
drag coefficient is commonly take to be 0.9

The
cross sectional area is more problematic. Most of the measurements have
been taken for racing cyclists, not commuters. Typical values quoted are
0.4 to 0.6m^2. Since a commuting cyclist rarely crouches, and is probably
more upright than a racer using the top of her handlebars, I will use 0.67m^2.

Finally,
we need the mass of the cyclist and bicycle. Take a 150lb cyclist, with
a heavyish bike of 28lbs.

**Velocity
vs. Power curves**

Using
these equations, we can easily calculate the speed of a cyclist as a function
of the power the cyclist is putting out. For example, if the cyclist puts
out 100w

Here
are two graphs of the power that a cyclist has to put out as a function
of the cyclist's velocity.

Thus
much over 20mph (10m/s) is very hard.

There
are many calculators on the web that will tell you the power for any given
velocity. Be aware that they do not always state the constants that they
are employing. Two such calculators are at:

A
very good shareware program to calculate the power, (and many other things)
is at

**Stop
signs and bicyclists**

Frequent
stop signs dramatically decrease the average speed of bicyclists. For instance,
it turns out that the to maintain as speed of 12.7mph, the cyclist would
have to increase her power output from 100 to 500watts on a street that
has a stop sign every 300ft. Calculating this exactly is messy, so heres
a simplified version:

Defining
the distance between stop signs

The
time between stop signs is

So
every 16.1 seconds she would have generate, and then lose, all her kinetic
energy. Now since she slows down near every stop sign, her peak speed must
be higher than her average speed. Lets say she can accelerate and deaccelerate
at a maximum rate of 0.15g. For comparison, a car going from zero to sixty
in 13 seconds has an acceleration of 0.2g

Acceleration

The
approximate peak velocity can then be found by solving the equation:

The
solution is

Her
kinetic energy at this speed would be

She
has to recreate this energy every 16 seconds, so her average power for
this alone must be about

At
this speed, the power necessary to compensate for the drag would be

Thus
the total power she would have to supply would be

Note
that this is very approximate. In particular, she is not at her peak speed
all the time, do the drag power would be lower. Nonetheless, approximately
the same answer is obtained when the calculation is done more exactly.