I drive a standard, although I assume the situation would be similar for an automatic. Basically: when I drive my standard, is my gas mileage always better the lower I keep the RPMs? Will the mpg always be best if the car is driven in the highest gear you can use without lugging the engine?
I was never quite clear on this, but it seems that less turns of the crankshaft would equal less gas being burned; anyone care to weigh in?
-occ
The turns of the crankshaft don’t determine how much gas goes into the cylinders. The throttle does. For a given setting of the throttle, if the crankshaft is spinning twice as fast, then half as much gas goes into the cyliders on every cycle. However, the total amount of gas burned per second is going to be the same. You can have less power per cycle and more cycles, or more power per cycle and less cycles. Either way you get about the same power overall per unit of time.
The engine itself though does not linearly convert the amount of gas exploded into movement. The typical gasoline engine is most efficient at somewhere around 2000 rpm, IIRC (varies quite a bit between engine designs), which would definately put you in 5th gear on the highway.
I believe that your engine will be most efficient when manifold vacuum is greatest, and this is almost never at low RPMs.
Engine efficiency (aka: brake specific fuel consumption) changes with engine RPM, mainly because the intake and exhaust are tuned for a specific RPM range. Engines are designed use the intake and exhaust mixtures’ inertia to improve efficiency, and the driver has to open the throttle further to allow more fuel in at a nonoptimum RPM. The open throttle will cause the vacuum in the intake tract to be reduced.
In modern engines with electronic fuel-injection, I have been told that extra fuel is injected if the engine is labouring, i.e. if you are in too high a gear.
NO!
In modern cars, the mixture is more-or-less stoichometric at the whole RPM range. So, what ultimately determines fuel consumption is the crankshaft RPM.
Always is a tough word, but just about always would be accurate.
That’s not right at all. Your fuel consumption is going to depend mostly on your power output, not RPM.
If you sit in neutral spinning the engine at 3000 rpm, do you think it will use exactly as much fuel as if you were in gear climbing a hill at 3000 rpm? No way.
Miles per gallon?
Break it down.
Miles->velocity->top gear
Gallon->RPMs->bottom throttle
Thus, the best mileage comes from the highest gear that can be maintained at the lowest throttle.
This generally is around 45 miles per hour…YMMV 
Una can certainly explain it better.
I agree that engine load is important, but I am not sure about the mostly part. I think mostly should go to the RPM instead.
You go up a hill at 3000 RPM. Are you making a lot of power or a little? Are you burning a lot of gas?
You get to the top (flat section) of the hill and maintain 3000 RPM. Are you making more or less power now? Are you burning more or less gas than when you were climbing the hill?
You roll down the other side of the hill in gear, maintaining 3000 RPM. How much power are you making now? How much gas are you burning?
So why are you saying that RPM ultimately determines fuel consumption? The RPM stays the same, but the amount of fuel burned varies with how much power you are producing.
Your engine’s efficiency may rated in terms of brake specific fuel consumption. BSFC is a measure of how much fuel your engine burns for power produced. The more power you make, the more fuel you burn.
HOWEVER, it’s not unusual at all to have an engine that has a better BSFC at lower RPM. But does not mean that the RPM ultimately determines fuel consumption.
one model of our stickshift cars had a shift indicator system to help the driver improve gas mileage.
A light would flash when it was time to shift to a higher gear.
The documantion said that best fuel mileage was attained with low RPM and wide throttle openings.
One of the side benefits of driving a car with 22 different computers is that there is an engine load signal sent to the instrument cluster for fuel mileage calcuations both instant and average. When I go out tomorrow, I will try driving the same speed on the same road in a couple of different gears. I will report the results.
Well power is equivelant to rpm anyway. I’d have to say fuel consumption is closest to amount of power being produced than anything. The most important thing that affects power is the amount of air flowing through the engine, and the amount of air is equivelant to the amount of fuel. The throttle position mostly affects torque, which is the other think that’s equivelant to power.
power = torque * rpm / 5252
No, it is not. Let’s take another example: a turbine generator at a power station. The rotatinal speed of these typically is held fairly constant at 3600 RPM (there are counterexamples, but let’s not quibble), which provides 60 Hz AC power. The amount of power one produces, on the other hand, depends entirely on demand + losses.
RPM is the speed of the engine. If I’m idling at a stop sign at 2,000 RPM my engine requires far less power to maintain that speed of rotation than if I were climbing a hill in third gear. In the first scenario the only power necessary is what’s required to overcome the frictional drag of the engine. In the second, the engine is doing additional external work driving the car up the hill. If in the second scenario I were to step on the clutch and suddenly free the engine from the work of climbing the hill, the RPMs would shoot into the red.
Parasitic losses from internal friction and driving all the various accessory loads such as alternator and oil/water/air pumps increase proportional to RPM (or perhaps even more than that). Combustion efficiency increases with percentage of maximum load(at that particular RPM). Because your typical gas car has a power peak at relatively high RPM, you will be using a greater portion of your available power when driving at low RPM (because the car doesn’t make that much power at low RPM). So, low RPM driving = operating closer to maximum load = greater efficiency - lower parasitic loss = even greater efficiency.
If you’re idling at 2,000 RPM at a stop sign, you need to adjust your idle. (Unless you drive a formula car around.) That aside, when you’re idling at 2,000 RPM, you’re using very little torque to overcome friction. Like I said, torque is the other thing power is equivelant to. If you got on the gas a little and brought it to 4,000 RPM, you would require twice the torque to overcome friction (we’ll pretend it’s that simple anyway) and it would be twice the rpm, thus four times the power being used.
But you’re getting much better miles per gallon in the second case. 
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It’s not unusual to have fun, with any one (dah-do-do-daaaaah)!
This may be true in a general sense, but lugging your engine will not necessarily or always improve mileage (graphic from this article.)
