For a math guy like me, this one will be strictly fun. In "Use of time" I discussed various matters related to the time "lost" by traveling at 55 m.p.h. on the freeway. One of the comparisons I developed was that, at the fuel prices in effect at that time (April of 2006), the time I lost in a day was valued by my company at $12.39 whereas I saved fuel valued at $4.57.
This would seem to indicate that, from a purely "dollars and cents" point of view, the faster I go, the better. Let's take an analytical look at that. I'll ignore limits on engine performance, speed limits and law enforcement, the physics of negotiating curves, and all other real-world matters.
The aerodynamic force resisting my vehicle's forward motion is proportional to the square of velocity (I insist) so additional speed increases fuel use dramatically. The time gained increases as my speed increases. Can I go so fast that I burn extra fuel worth more than the dollar value of my time savings?
Since I do in fact have to go to work and must use fuel, I can't merely calculate when fuel burned per minute equals my salary per minute. I have to use a baseline. I'll choose to use 55 m.p.h. So the problem is to determine how fast I must go to burn so much more fuel than I would burn at 55 m.p.h. that it is worth more than the value of my salary for the time that I'm traveling at the high speed.
The reader should note that I typically contemplate the problems about which I write "on the fly," hence I don't know what the answer will be until I have set it up in the blog. This is no exception, but I anticipate that the the speed will be a ludicrously high one.
So let's get started. I can easily calculate that the value of the time saved by driving faster than 55 m.p.h., as determined by my salary (and hence by my company's valuation of my time) to be $62.95-(3462/x) where x is my speed in miles per hour and the result is the savings for a single day's driving to and from work. So, for example, at 75 m.p.h. I save time worth $16.79.
The excess gasoline costs are not nearly so easy. To get a handle on this, I started with the model on the "How Stuff Works" web site. There, Marshall Brain models the power required as p = a*v + b*v^2 + c*v^3. Now, using some physics definitions and the chain rule from elementary calculus, we get to c = d + e*v + f*v^2, where c is "consumption" in appropriate units (say, gallons per mile) and d, e, and f are constants for a given vehicle. To determine the three constants, I need to know the consumption in gallons per mile (the inverse of miles per gallon) at three different speeds.
Well, I have 31 m.p.g. or 1/31 gal./mile at 55 m.p.h. so there's one. I can't use my idling fuel consumption since it is at zero m.p.h. and consequently would lead to an undefined consumption in gallons per mile, since we would be dividing by zero. So I acquired two more points, one at 40 m.p.h. and one at 70 m.p.h. Utilizing those numbers in the same way and solving the three simultaneous linear equations for the constants d, e, and f, and plugging them into the equation enables me to equate the dollars saved on my salary to the dollars spent in extra fuel, assuming gasoline at $3.40/gallon. At that speed, any faster and I would start losing money since losses on the fuel side would exceed gains on the salary side.
I hope I've built the suspense enough. The break even point occurs at 170 m.p.h. Now, if someone were to think that a Jeep Grand Cherokee Limited isn't capable of those speeds, that person would be correct. So what does it mean? It further emphasizes the need to be productive on the road, because gas isn't expensive enough, by far, to make up for the time lost in driving slowly.
Of course, I'm relatively well paid and gas could go up, so I guess the next step would be to develop a table for the break even speed at different salaries and fuel costs. But even then, a change in the miles driven at freeway speeds (such as 170 m.p.h.) would change the table. But it is clear that, from a purely economic point of view, I'm not doing myself any favors.