Tell me your fuel economy in MPG or gallons per mile at two speeds, your fuel cost, and I’ll tell you how much you could save per increased unit of time by driving slower. Tell me what your fuel economy is at 60 MPH on a flat stretch of road, and then reset the computer and tell me what it is at a higher speed like 65 or 70. If you don’t have a flat stretch of road, then drive the exact same distance twice on the exact same section of road at both speeds and record the fuel economy for each trip.
For most cars slowing down to less than 60 MPH isn’t going to save you money unless you bring home less than something in the ballpark of $12 per hour with gasoline prices at $4 per gallon. This assumes that time spent driving takes time away from work. If you have a fixed salary then you have to decide if you want that extra tax free money that comes from spending more time on the road. Driving slower also significantly increases safety, so the advantage isn’t just extra money.
Below is an explanation of wind resistance and an example of a car that uses 1/30 of a gallon per mile at 55 MPH with $4 per gallon fuel. The fuel economy at another speed isn’t known so it is estimated.
Look at slowing down to save fuel like this. Wind resistance is the cube of the speed, so if you go twice as fast there is 8 times as much wind drag. But you get there twice as fast, so that brings the cube down to a square. Driving twice as fast uses four times as much fuel, and eight times the engine power, if wind resistance alone is considered.
Some of the friction from driving a vehicle, such as drive train efficiency with a locked torque converter or manual transmission, is mostly linear, so speed doesn’t have much effect on efficiency. Gasoline engines tend to be a bit more efficient at higher load (torque at constant RPM), so driving faster increases engine efficiency. Then there is wind resistance loss which increases with the square of the speed. If your torque converter isn’t locked, then the loss from that will increase too as drag increases, increasing the exponential loss even more.
The goal here is to calculate the decrease in travel time due to increased speed compared to the increased cost. For instance, if you’re pulling a big trailer with lots of wind resistance, driving faster than 55 MPH could cost $20 in fuel per hour of saved travel time. Are you missing work due to travel time that pays $50 per hour after all taxes? Then it is worth it for you to drive faster in this situation.
The fuel price in this example is $4 per gallon.
Figuring out how much loss is due to wind drag is the tricky part. Say I get 30 MPG at 55 MPH ($0.1333 per mile) , and I get 60 MPG at 55 MPH ($0.0667 per mile) with a 55 MPH tail wind. In this case half of my energy is used to overcome wind resistance. I’ll ignore the increased engine efficiency at higher load in this example.
60^2 / 55^2 = 1.19 Increasing from 55 to 60 MPH increases my wind loss by 1.19. If that is half of my energy, then it becomes 1.095, so 1.095 (109.5%) as much fuel used to go 60 MPH. $0.1333 per mile at 55 MPH now costs $0.146 per mile at 60 MPH. That’s a $0.012667 per mile cost increase. 55 MPH is 0.0181818 hours per mile, and 60 MPH is 0.0166667 hours per mile. That’s a 0.0015151 hour decrease to go a mile at 60 MPH, and you save $0.012667. Now do $0.012667 / 0.0015151 hours = $8.36 per hour saved. So in that example if you’re making minimum wage then going 55 instead 60 is probably worth it for you. Any decrease in speed down to 55 will be worth it for you. As you go past 60 MPH the wind resistance becomes a bigger factor and that $8.36 figure grows. Saving as much as $30 per hour by driving 69.9 MPH instead of 70 MPH while pulling a big box trailer wouldn’t surprise me.