An overlooked point in fuel-efficient car driving speed

[Nantes]

I started commuting 65 km (40 miles) each day to go to work. This got me intrigued with the question of which speed I should drive into get the best mileage. The theory is straightforward: at low speeds, the driving is inneficient because the engine must use a set minimum amount of gas to function, so driving at 10 mph is not significantly more efficient than at say, 15 mph. At high speeds, various resistance forces increase exponentially. So you just need to find a sweet spot between these considerations, which for most cars seem to be about 90 km/h (55 mph). All of this is explained in more detail in this great HowStuffWorks article: http://auto.howstuffworks.com/fuel-efficiency/fuel-economy/question477.htm

But this and other websites I consulted fail to consider an important aspect: driving faster than 90 km/h does decrease fuel efficiency while it runs, but also gets you to the destination faster, meaning the car spends less minutes running than it otherwise would. To illustrate, consider a commute of 90 km:

at 90 km/h = commute takes 60 minutes, with fuel consumption X

at 120 km/h (75 mph) = commute takes 45 minutes, with fuel consumption Y

But true fuel consumption at 120 km/h = Y - 1/4 X (which is 1/4 of an hour you don't spend driving at 90 km/h)

Why is this aspect seemingly not taken into account when calculating fuel efficiency? It's apparently assumed that you're driving on an infinite road, forever, rather than eventually getting somewhere.

Finally, since there are more possible speeds than just 120 km/h and 90 km/h, can someone devise a formula that is able to take this into account for all speeds, calculating thus the true most efficient speed to drive in?

 

[billy_joule]

You are talking about fuel ECONOMY. Which is generally measured in volume fuel/distance traveled (eg l/km) which obviously does take speed into account.

eg using your own figures

X/(90km/hr * 1hr) = X litres/90 km
Y/(120km/hr * .75hr) = Y litres/90 km

If you put a dollar value on your time you can calculate which speed is better for you.
There is no formula to account for all speeds. You can change economy without changing speeds, just change gear.

 

[russ_watters]

Youvmight try thinking of it this way: one you are in your top gear, the cylinders move the same number of times for a given distance, regardless of speed. So you want to minimize the force applied at the cylinders by the fuel.

 

[Nantes]

You are talking about fuel ECONOMY. Which is generally measured in volume fuel/distance traveled (eg l/km) which obviously does take speed into account.

eg using your own figures

X/(90km/hr * 1hr) = X litres/90 km
Y/(120km/hr * .75hr) = Y litres/90 km

If you put a dollar value on your time you can calculate which speed is better for you.



There is no formula to account for all speeds. You can change economy without changing speeds, just change gear.

Yeah I wrote fuel economy instead of consumption initially, but then it wouldn't make sense to subtract 1/4 of X, I'd have to add it, and it would be even more confusing.

I'm not trying to take time savings into account, i. e. I'm not considering that these 15 minutes are worth something to me because I can use them to do whatever I want. From a purely fuel savings standpoint, the extra fuel cost of driving a higher speed is, in practice, compensated by not driving at the lower speed for some minutes, which you otherwise would if you drove at said lower speed. Do you know of any website that takes this into consideration for calculating fuel cost, or a formula? Thanks!

P.S: I can't change gear because my car is CVT automatic. It always has the best possible gear at all speeds, because it has infinite gears.

 

[cjl]

But that's already accounted for, since fuel usage is pretty much always specified in terms of distance per unit fuel, not fuel per unit time.

 

[AlephZero]

Most of the world except the US measures fuel per unit distance, not distance per unit fuel. That change might make the OP less confused.

You can easily convert fuel per unit distance to fuel per unit time - just multiply by the speed. (Fuel / distance) x (distance / time) = (Fuel / time).

But the OP's idea of "Y - 1/4 X" doesn't make any sense to me.

 

[Darryl]

but is there not an efficiency curve for the motor, as well as a torque curve, that is a RPM value that is most efficient for the motor, so it would make sense to me that driving at the speed that relates to this efficiency peak would make for more efficient travel, your gearbox and diff ratio should make this speed close to the legal cruising speed.

https://www.physicsforums.com/showthread.php?t=218847