The recommended safe distance to drive behind another vehicle traveling 55 mph is 150 feet. The risk of something bad happening--like dieing--increases the closer you are to the vehicle in front. On a 100 mile trip at 55 mph in a car that regularly gets 25 mpg, you will save $2.07 if you drive the entire distance 100 feet (1.25 seconds) behind a semi. You will save $3.86 if you drive 10 feet behind the semi (.124 seconds). Oh, and you will be labeled an a$$h@!& driver.

In the comments, John threw down the gauntlet and challenged me to come up with the implied value of a statistical life from these numbers. Not wanting to back down from a challenge I calculate that choosing to drive 10 feet behind a semi versus 100 feet implies a value of statistical life of $120,000. Gory details are below the jump...and believe me, they're gory.

I'm fairly certain there are numerous problems with the following--feel free to attack.

To calculate the value of a statistical life we need two things: 1) a change in risk of death, and 2) a willingness to trade off money for that change in risk. In the tailgating example, we have the dollar figures. Unfortunately, the risk figures are a little harder to come up with.

To calculate the change in risk we first need to know the risk of accident per mile driven. A quick Google search on 'accidents per mile driven' turns up this site--gotta love Google--which claims the risk of accident per mile driven to be around .55 per 100,000 miles or .0000055 per mile.

So let's use that as the baseline and suppose that you are an otherwise safe driver. Your risk of accident is now .0000055 per mile driven. I am going to further assume that your risk of accident is completely independent of the truck you are following as long as you are at least 150 ft behind the truck. This isn't exactly right since the .0000055 figure includes all of the other stupid drivers out there, but if that's really where you're going to find problems with this analysis then...%$#@ you.

Anyway, given this independence assumption, the probability the truck will crash is also .0000055.

But, if you are trailing the truck closer than 150 feet, your risk of accident is now correlated with whether the truck crashes--because it's harder to get out of the way. As the distance you are trailing the truck decreases, the likelihood of crashing if the truck crashes increases until you are certain to crash if you are on the trucks bumper.

Since specific numbers don't exist for the probability of a crash as function of distance to the truck you are following, I'll do what any good economist would do, assume the relationship. Suppose the conditional probability of the trailing car crashing if the truck crashes is:

P(C)=1/(1+exp((f-75)/15))

where f is the number of feet behind the truck you are traveling. The conditional probability is shown in the graph at the right--looks reasonable, doesn't it?

Alright, armed with my assumed probabilities, I can now calculate the probability of my car crashing in a given mile. My assumed independent probability of a crash is .0000055, but I also have to account for the likelihood of crashing if the truck crashes which will be equal to P(C)*(probability of a truck crash). So my probability of a crash is now: .0000055+P(C)*.0000055 or .0000055*(1+P(C)).

Substituting in for P(C) and partially differentiating--this may very well be the first time I've used a derivative on the blog--with respect to f, I find that the probability of an accident increases by:

.0000055*(exp((f-75)/15)/((1+exp((f-75)/15))^2)

per foot.

Now I have everything I need to figure out the change in risk of an accident from tailgating. Let's suppose we go from a fairly safe distance of 100 feet (16% chance of an accident if the truck crashes) to a crazy distance of 10 feet (99% chance of crashing if the truck crashes). The overall probability of an accident increases from 0.0000064 to 0.0000109 or an increase of .0000045 accidents per mile driven.

Multiplying the increase by the probability of death in an accident (about 34 per 1000 accidents), the increased risk of death due to driving at 10 feet versus 100 feet is .00000015 per mile driven. Based on the numbers from yesterday's post, hypermilers will save about $.018 per mile driven. Dividing the $.018 per mile driven by the .00000015 deaths per mile increase yields a value of statistical life of $120,000.

I think I have too much--clap, clap--time on my hands.

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