# Turn off daytime running lights, or reduce speed? Which saves more?

Which wastes more fuel? (and thus produces more carbon dioxide). This is a difficult to question to answer for a variety of reasons. The main reason is that a speed change from 71 mph to 70 mph is different than a reduction from 56 to 55 mph.

First, let me be clear that the question of how much fuel is wasted using daytime running lights (or DRL as they are called) has already been addressed. The first source I found was howstuffworks.com

**Assumptions**

• The daytime running lights on a car run at about 100 watts (for the pair)
• The energy density of gasoline is 1.21 x 108 Joules/gallon.
• A car is 20% efficient at converting this energy to mechanical energy.
• The alternator is 70% efficient at converting mechanical energy into electrical.
• At highway speeds, air resistance is the dominating factor in fuel efficiency (this might be wrong)
• The air resistance can be modeled as Fair = (1/2)?CAv2
• I will assume an “average” car that has combined CdA of 9 ft2 or 0.84 m2 (where Cd is the coefficient of drag and A is the cross sectional area. Also ? is the density of air, about 1.2 kg/m2)
• An average trip of 50 miles (I completely made this up).
• My mythical “average” car gets 25 mpg when going 70 mph

**Now to Calculate**

First, for this 50 mile trip going at 70 mph, how much gas is “spent” on daytime running lights? Well, a 50 mile trip at 70 mph would take:

![Page 0 Blog Entry 59 1](http://blog.dotphys.net/wp-content/uploads/2008/10/page-0-blog-entry-59-1.jpg)

During this time, the lights are running at 100 watts, this is the same as:

![Page 0 Blog Entry 59 2](http://blog.dotphys.net/wp-content/uploads/2008/10/page-0-blog-entry-59-2.jpg)

How much gas does this require? Using the information from the above, I can write an expression for how much energy a certain number of gallons of gas would produce using the alternator. Note in this expression, the 0.2 and 0.7 are the efficiencies of the engine and alternator.

![Page 0 Blog Entry 59 3](http://blog.dotphys.net/wp-content/uploads/2008/10/page-0-blog-entry-59-3.jpg)

Solving this for “gallons of gas”

![Page 0 Blog Entry 59 4](http://blog.dotphys.net/wp-content/uploads/2008/10/page-0-blog-entry-59-4.jpg)

Note: this wouldn’t change much if driving 71 mph, the time would be about the same.

Now for the slowing down part.

How much energy would be saved by driving 70 mph instead of 71 mph? What is the energy lost due to air resistance? To calculate this, I can calculate the work done by the air resistance (which would be equivalent to the energy lost). Remember that work is

![Page 0 Blog Entry 59 5](http://blog.dotphys.net/wp-content/uploads/2008/10/page-0-blog-entry-59-5.jpg)

In the case of air resistance, the force of air and the direction of motion are in the opposite directions. This means that the work done by friction is:

![Page 0 Blog Entry 59 6](http://blog.dotphys.net/wp-content/uploads/2008/10/page-0-blog-entry-59-6.jpg)

where these are the magnitudes of the vectors (d is the distance traveled).
Since I am not really concerned with the work done by air resistance, but rather the CHANGE in work done in going 71 mph vs. 70 mph, I can just calculate:

![Page 0 Blog Entry 59 7](http://blog.dotphys.net/wp-content/uploads/2008/10/page-0-blog-entry-59-7.jpg)

Now to convert this into gallons of gas. Note: I only need to include the 20% efficiency of the engine and not the 70% efficiency of the alternator.

![Page 0 Blog Entry 59 8](http://blog.dotphys.net/wp-content/uploads/2008/10/page-0-blog-entry-59-8.jpg)