Energy Use in Cars 1: Stop-and-Go Driving

Printer-friendly versionPrinter-friendly version Share this

When does cycling cost more in fuel than driving?

Big Ideas: 
  • A comparison of different transportation options - when a cyclist makes the bus stop at a red light, the fuel the commuter saves by cycling might be burnt to get the bus up to speed again.

When does cycling cost more fuel than driving? People recommend commuting by bicycle rather than by car to save energy. However sometimes when a cyclist pushes the button to make a traffic light change, it means that a bus full of people has to stop. When a cyclist makes the bus stop like this, is the extra fuel it has to burn to get back up to speed MORE than the fuel saved by choosing to commute by bicycle rather than car?

To tell whether stopping the bus uses more fuel than one person driving to work we need to figure out how much fuel it costs to get the bus back up to speed AND how much fuel is burnt in a typical daily commute. Let's look at the bus first.

We know that driving uses fuel, because we keep having to buy it at the gas station! We burn the fuel to get heat energy, and then the bus's engine converts heat energy into motion of the bus and some other things. The energy from the fuel goes into four main places:

  1. Accelerating the bus up to its cruising speed. A moving car has kinetic energy, and it needs to get this energy from the engine. Once we are at a constant speed, we don't need to spend any more energy accelerating but we still need our foot on the gas because of the next two issues.
  2. Air resistance. Driving a bus makes the air around it swirl around, and this takes energy. Driving faster makes the air swirl much more.
  3. Rolling resistance. This accounts for all of the small bits of friction within the bus, as well as resistance due to the tires on the road.
  4. Heat. Burning fuel doesn't make the bus move directly; it creates a lot of heat, and then the engine has to convert that heat into motion. However there is still a lot of heat in the exhaust gases that gets pumped out the back of the bus, so not ALL of it gets converted into motion.

Question: Where do you think most of the energy from the fuel ends up? (choose 1-4 above)

Answer: 4. Heat. For a typical gasoline engine, only around 25% of the heat energy from the fuel gets converted into mechanical energy which gets used for the first three items on this list. One way of thinking about an engine is that it transforms the chemical energy from the fuel into mechanical energy and heat, like this:

We would say that this engine has an efficiency of 25%.

$ \textnormal{Engine Efficiency} = \dfrac{ \textnormal{Mechanical Energy Out}}{\textnormal{Fuel Energy In}} $

To answer our question about the bus we are mostly concerned with how much energy it takes to get back up to speed so let's just look at the energy cost for that. In city driving, this is a very important part of your fuel consumption, especially if you are moving at typical traffic speeds (below 50 km/h). Then, once we know how much mechanical energy is required, we can use the efficiency of the engine to determine how much fuel needs to be burned.

Imagine the bus is initially moving at 50 km/h. When the cyclist makes the light change it decelerates using the brakes. This converts ALL of the kinetic energy of the bus into heat energy in the brakes. Then after it stops it must re-accelerate.

The mass of an empty 99 bus is 19,820 kg1

The passengers weigh an average of 70 kg each, and at rush hour the bus is likely full to capacity with 120 passengers1.

So the total mass of the bus is

$ \textnormal{mass} = 19,820 \textnormal{ kg} + (120 \textnormal{ passengers})(\dfrac{70 \textnormal{ kg}} {\textnormal{passenger}}) = 28,220 \textnormal{ kg} $

Now we need to know how much energy it takes to get this mass moving again at 50 km/h (14 m/s). The kinetic energy of a moving object is given by KE = ½mv2, so the total kinetic energy required is

<br />
\begin{eqnarray}<br />
   KE &=& \dfrac{1}{2} m v^2 \nonumber \\<br />
   &=& \dfrac{1}{2} (28,220 \textnormal{ kg})(50 \textnormal{ km/h})^2 \nonumber \\<br />
   &=& \dfrac{1}{2} (28,220 \textnormal{ kg})(14 \textnormal{ m/s})^2 \nonumber \\<br />
   &=& 2.77 \times 10^6 \textnormal{ kg}\cdot \textnormal{m}^2 \textnormal{/s}^2 \nonumber \\<br />
  &=& 2.77 \times 10^6 \textnormal{ J} \nonumber<br />
   \end{eqnarray}<br />

In order to create this much mechanical energy, we need to burn additional fuel. We can see from the efficiency diagram that

$ \textnormal{Mechanical Energy} = \textnormal{(Engine Efficiency)(Fuel Energy)} $

The energy content of diesel is 38.7 x 106 J / litre2 and since Diesel engines are significantly more efficient than gasoline engines, we use an efficiency estimate of 37% for the bus3. Now we can calculate how much Fuel Energy is required.

<br />
\begin{eqnarray}<br />
   \textnormal{Fuel Energy} &=& \dfrac{\textnormal{Mechanical Energy}}{37 \%} \nonumber \\<br />
   &=& \dfrac{2.77 \times 10^6 \textnormal{ Joules}}{0.37} \nonumber \\<br />
   &=&  7.49 \times 10^6 \textnormal{ Joules}\nonumber<br />
   \end{eqnarray}<br />

To determine how many litres of fuel this is, we calculate

<br />
\begin{eqnarray}<br />
   \textnormal{Energy Content of diesel} &=& \dfrac{\textnormal{\# of Joules}}{\textnormal{\# of litres}} \nonumber \\<br />
   \textnormal{\# of litres} &=& \dfrac{\textnormal{ \# of Joules}}{\textnormal{Energy Content of diesel}} \nonumber \\<br />
   &=&  \dfrac{7.49 \times 10^6 \textnormal{ J}}{38.7 \times 10^6 \textnormal{ J/litre}} \nonumber \\<br />
&=& 0.19 \textnormal{ litres of diesel} \nonumber<br />
   \end{eqnarray}<br />

Now we want to compare this to the amount of fuel that the cyclist is SAVING. To do that we'll just use a simple estimate of typical car efficiency. The average gasoline car consumes 0.076 litres of fuel per kilometer traveled4. So if your commute is 10 km long, you would be saving 0.76 litres of gasoline by riding your bike. If you made the bus stop once on your 10 km trip, you would be costing 0.19 litres in order to save 0.76 litres, so it is still a good idea to ride your bike.

Question: Given that stopping a bus costs 0.19 litres of fuel, and cars consume 0.076 litres/ km, how long would your commute need to be to save an amount of gasoline equivalent to stopping the bus once?

  1. 0.014 km
  2. 0.26 km
  3. 0.42 km
  4. 1.0 km
  5. 2.5 km

Answer:  E. 2.5 km = 0.19 L ÷ 0.076 L /km

So how can we interpret this result?

The first thing we might think is that if your commute is shorter than 2.5 km, then the amount of fuel that you save by riding your bike is smaller than 0.19 litres, so if you cause the bus to stop once during your bike commute then you will have caused a larger consumption of fuel than you save. However there are a couple of other factors to consider.

Firstly, when you hit the light for the crosswalk often you stop MORE than just a bus. For example, there might be a few cars behind the bus that also have to stop. This would make us think that the amount of gas wasted is more than 0.19 litres.

However, sometimes when you push the light and you stop the bus the timing of lights is such that it would have stopped in a block or two anyways. So in this case, pushing the light doesn't really make any difference to the fuel consumption of the bus.

These two effects are very difficult to model, and they act in opposite directions, so we might be tempted to conclude that we can't know what the effect of pushing the crosswalk button is. However, we can see from this calculation that some of the time it will cost a significant amount of fuel, and therefore it is a good thing to avoid if you can avoid it.

It also suggests that in order to design efficient systems for transit and bicycles, it is a good idea to design crossing lights so the cyclists don't cause the bus to stop. One way of doing this is to have transponders in the buses that make sure that lights up ahead stay green for them. Another way is to have completely separate cycling routes that don't interfere with the transit, similar to systems used in many European countries.



Post new comment

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

The content of this field is kept private and will not be shown publicly.
By submitting this form, you accept the Mollom privacy policy.

a place of mind, The University of British Columbia

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

Emergency Procedures | Accessibility | Contact UBC | © Copyright The University of British Columbia