# Bicycling and Calories

Printer-friendly version

How long do you have to ride your bicycle to burn off a doughnut?

Big Ideas:
• Food (like gasoline) is chemical energy that the body transforms into heat and mechanical energy.
• The mechanical work done on the bicycle is linked to the metabolic energy of the rider.

How long do I have to ride my bicycle to burn off a doughnut? This example looks at the relationship between physical exercise and calorie consumption.

In magazine articles and on the web, you often find numbers relating exercise and calorie consumption. In this example, we will explore where the numbers come from.

For example, you find that a 68 kg person cycling at 15 km/h for one hour burns approximately 400 Calories1,2 (corresponding to approximately 1.5 - 2.0 donuts3). Let's check.

Our calculation is based on the principle that the energy in food is like other forms of energy, so it can be transformed into mechanical energy. However, like other real-life engines, our body cannot transfer 100% of the chemical energy in food into mechanical energy (motion). We need two additional bits of information: the amount of energy that the body consumes during sedentary periods (typically 100 W; this is mostly needed to keep the internal organs functioning) and the efficiency to convert chemical energy into mechanical energy (approximately 25%2). The two numbers given above vary somewhat with the person and the activity but can conveniently be used to get a good estimate.

So why does bicycling consume energy, even on a flat road and a day without wind? You have to overcome rolling friction and air drag, since the relative wind speed is 15 km/h.

Assumptions:

• level road, no wind (relative wind speed = 15 km/h)
• 68 kg person plus 10 kg bike
• rolling friction: rubber on concrete: μ r = 0.02 (table 5.12).

Calculations:

• Force due to air drag: FD = ¼ ρ A v 2, with density of air ρ = 1.28 kg/m 3
• (The simplified equation above follows from the more general formula for air drag FD = ½ ρ CD A v 2 by using CD = 0.5, which is typical for everyday moving objects, see p. 156 of ref.2.)

We need to estimate the frontal area A of person (+ bike)
A = (1.5 m) · (0.6 m) = 0.9 m 2.
v = 4.2 m/s, so the distance traveled is d = 4.2 m each second.
So F = 5.1 N and the work done is W = F d = 21 J, so P = 21 W

• Rolling friction: f r = μ r m g = (0.02) · (68 kg+10 kg) · (9.8 m/s2) = 15.3 N,
• so the work done by rolling friction is W = (15.3) · (4.2 m) = 64 J and the corresponding power is P = 64 W.

• So we need a total of 85 W of mechanical power to maintain our speed.
• Now we want to relate this to the total metabolic power. We need to multiply our result by 4 and add 100 W to account for our efficiency and the energy consumed by our internal organs. Result = 440 W.
• Finally, we have to perform a unit conversion to relate our result to food calories: 1 Cal = 1kcal = 4200 J.
• 440 W = 440 J/s correspond to (440 J/s) · (3600 s/h) = 1.58 MJ/h or 377 Calories burned in one hour.
• You can replenish your energy by eating almost two doughnuts (62 g each) or six timbits (17 g each)3.

Interpretation:
Our result is close to the published value showing that the calorie consumption in bicycling is mainly due to overcoming air drag and rolling friction.

Finally, how long will someone have to bike to burn off eating a donut?  A chocolate glazed donut from Tim Horton's has 260 Calories and is 70 g3.  If you eat one it will take you (260 Cal)(1 h/377 Cal)(60 min/h) = 41 min of biking  at 15 km/h to burn it off.

Resources
• Browse by Theme:
• Keywords:

## Post new comment

Please note that these comments are moderated and reviewed before publishing.

By submitting this form, you accept the Mollom privacy policy.

C21: Physics Teaching for the 21st Century
UBC Department of Physics & Astronomy
6224 Agricultural Road
Vancouver, BC V6T 1Z1
Tel 604.822.3675
Fax 604.822.5324
Email: