More on fuel efficiency improvements and reductions in gas consumption.
What is the amount of fuel saved per mile driven when the fuel efficiency of a car is improved from 20 mpg to 25 mpg?
How much does one need to improve the fuel efficiency of a 30-mpg vehicle in order to achieve the same reduction in fuel consumption?
Answer: We know that gas consumption is equal to miles driven divided by mpg.
Gallons = Miles / (Miles/Gallon)
So the gas saved per mile driven from an improvement in fuel efficiency from 20 to 25 miles per gallon is
Reduction in Gas Consumption = (1/20) – (1/25) or 0.05 -0.04 or 0.01 gallon.
Now assume the initial fuel consumption for the car is 30 mpg and the reduction in gas consumption from a one-mile trip is 0.01 gallons
Reduced Gas Consumption = 0.01 = (1/30) – (1/NFE)
Where NFE is the new fuel efficiency.
1/NFE = (1/30) – 0.01 = 0.023333
Take the reciprocal of both sides and get
NFE = 42.857.
We can check our answer by observing that
(1/30) – (1/42.852) is equal to 0.01.
0.033333 – 0.023333 is equal to 0.01.
A Note: This post shows that a relatively small improvement in fuel efficiency from a very fuel inefficient car can result in large reductions in gasoline. However, returns from improvement in fuel efficiency will eventually diminish. Consider a 1000-mpg car. The amount of gas needed for this car to travel 1 mile is 1/1000 or 0.001 gallon. An increase in the fuel efficiency of this car to 2000 mpg would only result in a 0.0005 reduction in gas consumed for each mile traveled.
Note also we are a long way from 1000 mpg.
This issue was previously looked at more informally in the post below.