# Optimum speed for mileage per gallon

I imagine that this magnitude is dependent on a lot of factors, but does anybody know a good source to get the relationship betwen speed and mpg for at leas families of cars?

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Go to CarTalk.com and look at December, 1995.

russ_watters
Mentor
The optimum speed (with no air conditioner running) is generally the lowest speed that you can easily maintain in your highest gear.

rbj
The optimum speed (with no air conditioner running) is generally the lowest speed that you can easily maintain in your highest gear.
what exactly do you mean by "maintain", Russ? i don't think "lugging an engine" is the most efficient use of it. even on perfectly flat and smooth pavement.

when the gearbox doesnt make weird sounds, i guess russ meant this

rcgldr
Homework Helper
The optimum speed is generally the lowest speed that you can easily maintain in your highest gear.
I remember this same belief from the 1970's, but it wasn't true. For cars back then the optimum speed was around 45mph. This is because gasoline engines are very inefficient at producing small amounts of power, and running at higher speeds, with more of a power load, results in better milage, because the increase in rate of fuel consumed is less than the increase in the rate of speed.

Modern cars have fairly tall gearing in the form of overdrive, and better aerodynamics than cars of the 1970's, so they should be more efficient at higher speeds. Considering that EPA measures highway milage at 55mph (this has just changed), the car designers may have designed their cars to get the best milage at 55mph.

So other than a 1300cc VW bug from the 1970's, or a hybrid, the optimum speed for most cars will be between 45mph and 55mph.

Integral
Staff Emeritus
Gold Member
..snip...

So other than a 1300cc VW bug from the 1970's, or a hybrid, the optimum speed for most cars will be between 45mph and 55mph.

Which is about the slowest you can comfortably go in the highest gear.

russ_watters
Mentor
what exactly do you mean by "maintain", Russ? i don't think "lugging an engine" is the most efficient use of it. even on perfectly flat and smooth pavement.
That's where the "easily" part comes in.

russ_watters
Mentor
Modern cars have fairly tall gearing in the form of overdrive, and better aerodynamics than cars of the 1970's, so they should be more efficient at higher speeds. Considering that EPA measures highway milage at 55mph (this has just changed), the car designers may have designed their cars to get the best milage at 55mph.

So other than a 1300cc VW bug from the 1970's, or a hybrid, the optimum speed for most cars will be between 45mph and 55mph.
I drive stick, but don't most automatics only have 4 gears? Also, gear ratios vary a lot by engine and type of car. Ie, an SUV will be geared a lot lower (or at least be able to handle a lower rpm). So I think that's too narrow a range - maybe 35-55.

Last edited:
stewartcs
I drive stick, but don't most automatics only have 4 gears?
My automatic has 6 gears.

So, there's an easy way to do this. Some cars have a built-in mileage indicator. Rent one and drive it in all different ways and watch the indication. When I did this, I found I got the best mileage at about 1500 RPM in the highest gear. This was on level ground (such as we have in western PA) and corresponded to about 40-55 mph. Presumably this would vary a bit in different cars, but that answer seems to "ball-park" it.

RonL
Gold Member
For my first post i'll give a firsthand experience.
Last year i purchased a used F350 truck, with a 351 V-8, the transmission is a 5 speed ( the final gear is overdrive ).

Two trips out of town, with speeds of about 60 MPH, produced a fuel average of 11.25 MPG, my big supprise came when i started a part time job, about 13 miles from home, and traveling thru town in stop and go traffic ( top speed between 35-40 MPH ) average speed while moving is about 30 MPH, produced a fuel average increase to about 12.5 MPG.
My thoughts are less wind resistance, and lower average speeds, actually produced an increase in MPG, where i least expected it.

In line with what Russ said.

If someone will point me to the proper place, i have a few thoughts and questions, about how to add a power system ( electric ) to this truck, that will provide 13 miles each way. The current power system will remain intact for heavy or long haul use.

Thanks
RonL

This is because gasoline engines are very inefficient at producing small amounts of power, and running at higher speeds, with more of a power load, results in better milage, because the increase in rate of fuel consumed is less than the increase in the rate of speed.
Then, best mileage will be around the rpm's which give the highest torque with the highest gear. Does it make sense?

no...

No? Just no?

I imagine that regime for highest torque is likely to be the most efficient (power vs fuel), and using the highest gear will produce the lowest number of explosions for a given mileage.
Am I wrong?

rcgldr
Homework Helper
I imagine that regime for highest torque is likely to be the most efficient (power vs fuel), and using the highest gear will produce the lowest number of explosions for a given mileage.
The amount of power generated for the amount of fuel consumed is best, but because of aerodynamic drag, it's consuming more fuel per mile traveled.

from what i have read, wind resistance is not linear. meaning that wind resistance at 60 mph is not 2x the resistance at 30mph, it is lower. although the engine is most efficent at it highest torque rpm, the increase in air resistance at this speed and in top gear more than compensates to make lower mph driving more efficent.

DaveC426913
Gold Member
This chart was done a few years ago on my 1993 Chrysler Intrepid:

http://www.davesbrain.ca/miscpix/mileage.gif

What I find odd about it is:
- the lack of curvature. I would have thought it would level out to the left.
- the minimal effect of the AC. It has a smaller effect than changing highway speed by 20kph.

DaveC426913
Gold Member
wind resistance at 60 mph is not 2x the resistance at 30mph
The formula is such that 2x velocity will result in 4x wind resistance.

good to know daveC, i didnt know what the actuall formula was. thanks

mgb_phys
Homework Helper
The drag force is velcotiy squared but power used to overcome drag is velocity cubed. So the engine power is proportional to v^3 * density of air * front area * drag coefficent.
Which is why planes like to fly high, SUVs don't get good gas milage and removing roof racks is a good idea.

As a rule of thumb I think that minimal speed in top gear is normally the most efficent for most cars. The real world best technique is to try not to brake and accelarate, so leave a larger gap to the car in front so you can keep a constant speed.

Speed per RPM

Since RPMs have a pretty close correlation to fuel consumed, don't we want the maximum speed per RPM? So if 2000 RPMs get me 60mph that is .030mph per RPM. At 70mph, my tach reads 2500 so that's .028mph per RPM, showing 60mph is more efficient than 70mph. So with this, you can record speed and tach at various speeds and see which one gives the lowest mph to rpm ratio to determine the best cruising speed for optimum mileage. I've also noticed that with an automatic transmission I get spoiled and lazy and don't realize that cruising at 35 mph has the engine revving up close to where it's ready to shift but not doing so. By increasing to 36, it shifts and becomes more efficient. With a stick I wouldn't think of driving at that revved up speed, with a small car auto trans I barely even notice it. Just some thoughts I've had. Do they make sense to anyone else or am I missing something?

mgb_phys
Homework Helper
Since RPMs have a pretty close correlation to fuel consumed, don't we want the maximum speed per RPM? So if 2000 RPMs get me 60mph that is .030mph per RPM. At 70mph, my tach reads 2500 so that's .028mph per RPM
Almost correct - except you are interested in DISTANCE/gallon not TIME/gallon.

So at 60 you use a certain amount of fuel per min and at 70 you use 2500/2000 = 125% as much fuel but only go 70/60 = 116% as far per min so you are loosing.

But if you went at 2000RPM at 60 and 2300RPM at 70 you would be doing better at 70.
Since you would only be using 2300/2000 = 115% as much fuel to go 116% as far.

RPM are not directly correlated to fuel consumption. You can be going at 50mph at 2200 in 5th gear both uphill and downhill. The RPM is the same, but the amount you have to open the throttle is different.

Also, assuming no change in gear, you should get the same mph/rpm for a whole gear, thats the whole point of a gear ratio. Its locked. (non cvt). I wonder if you could just get a measure of Throttle Angle vs. RPM (or mph, same thing). I think theres a 1:1 correspondence between throttle angle and fuel injector flow.

Almost correct - except you are interested in DISTANCE/gallon not TIME/gallon.

So at 60 you use a certain amount of fuel per min and at 70 you use 2500/2000 = 125% as much fuel but only go 70/60 = 116% as far per min so you are loosing.

But if you went at 2000RPM at 60 and 2300RPM at 70 you would be doing better at 70.
Since you would only be using 2300/2000 = 115% as much fuel to go 116% as far.
If you went 2000rpm at 60 then you MUST be doing 2333 at 70. Which must give the same ratio. The gears are locked. For every rotation of the engine it does a certain rotation of the wheel.