Best RPM for most efficient fuel consumption.

[Paganguy]

I own a 2003 5 speed manual outback. I usually run my tacho at 3200 to
3500 RPM. Is this in fact the most efficient rpm for the engine? I
know that typically once you go over 110 kph fuel efficiency drops but
it depends on what rpm the engine in running at the point where most
horsepower is put out without consuming too much fuel.

Any thoughts on this?

Paul

 

[BobN]

My 98 OBW seems to get the best mileage around 2500 rpm, but that's only
around 60 MPH which is dangerously slow on US highways. When trying for
good mileage, I try to keep the RPMs just under 3000 (around 73 MPH) and
drive proactively to avoid using the brakes. Even in moderate traffic, I
can go over 100 miles on the highway without touching the brake even once.
This gets me 26-27 mpg when the car is running right, and the most I've had
is 28 with new tires.

 

[Bruce Hoult]

I own a 2003 5 speed manual outback. I usually run my tacho at 3200 to
3500 RPM. Is this in fact the most efficient rpm for the engine? I
know that typically once you go over 110 kph fuel efficiency drops but
it depends on what rpm the engine in running at the point where most
horsepower is put out without consuming too much fuel.

Any thoughts on this?

I have a 1995 2.5l Legacy wagon with 4EAT. It's doing 3000 at 120 km/h
(75 mph). Unfortunately the NZ speed limit is 100 km/h, but I've had
indications that on the few trips where I've been able to do significant
miles on back roads the car gets better fuel economy at speeds around
120 - 130 km/h than it does at 100 - 110. It's hard to say because
those roads have more hills and corners that the main highway, but if
I'm getting the *same* fuel economy going fast on twisty back roads at
120+ as I am going slowly on the highway then there must be *something*
interesting going on.

 

[JohnO]

but if

I'm getting the *same* fuel economy going fast on twisty back roads at
120+ as I am going slowly on the highway then there must be *something*
interesting going on.

You are breaking the laws of physics, that's all.

I'd suggest measuring your fuel consumption more carefully. <g>

-John O

 

[Edward Hayes]

My 2.5 L Forester AT converter locks up at 38 mph (1700 rpm) so I
assume that will be my best gas mileage. Do not confuse best gas
mileage with best efficiency which is max HP per pound of fuel
consumed which relates to Maximum bmep ( brake mean effective
pressure) in the cylinder.

 

[Bruce Hoult]

but if
I'm getting the *same* fuel economy going fast on twisty back roads at
120+ as I am going slowly on the highway then there must be *something*
interesting going on.

You are breaking the laws of physics, that's all.[/QUOTE]

How so?

We're talking about finding the speed for maximum fuel economy. That
implies that going either faster or slower than that gives you worse
fuel economy.

There is nothing magic about 50 or 55 mph that makes that the most
efficient speed. Aerodynamics and engines are both much better now than
they were in the 1970s and with so many countries around the world
having speed limits around 75 - 80 mph it would not surprise me at all
if Subaru had designed their car for best fuel economy at that speed.

 

[Grolsch]

I call bullshit!

Best efficiency for an engine does not translate into best fuel economy. The
enngine may be most efficient making 84 hp or whatever, but it would only
take a fraction of that to maintain a given highway speed. Wind, friction,
rolling resistance etc increase exponentially with speed, therefore go
faster = more fuel. It's simple physics.
--

You are breaking the laws of physics, that's all.

How so?

We're talking about finding the speed for maximum fuel economy. That
implies that going either faster or slower than that gives you worse
fuel economy.

There is nothing magic about 50 or 55 mph that makes that the most
efficient speed. Aerodynamics and engines are both much better now than
they were in the 1970s and with so many countries around the world
having speed limits around 75 - 80 mph it would not surprise me at all
if Subaru had designed their car for best fuel economy at that speed.
[/QUOTE]

 

[Bruce Hoult]

I call bullshit!

Best efficiency for an engine does not translate into best fuel economy. The
enngine may be most efficient making 84 hp or whatever, but it would only
take a fraction of that to maintain a given highway speed. Wind, friction,
rolling resistance etc increase exponentially with speed, therefore go
faster = more fuel. It's simple physics.

I'm not sure whether you didn't read what I said or didn't understand it
so I'll attempt to put it into even more basic terms for you.

You suggest that the slower you go the better the fuel economy.

But it's obvious that there is a limit to this. In particular, sitting
still gives you the worst possible fuel economy. Therefore the best
fuel economy is at some finite speed that is greater than zero. By the
very definition of "best" and of "maximum" this means that travelling
either faster than this best speed OR SLOWER than this best speed will
give worse fuel economy.

Yes I know that drag increases with speed. And, no, it is *not*
exponential. It is a polynomial function, not an exponential one. your
saying such a thing immediately indicates that you are not qualified to
talk about a mathematical subject involving calculus, such as
maximization or minimization of a function.

Best efficiency for an engine does not translate into best fuel economy.

No one said that it did.

The enngine may be most efficient making 84 hp or whatever, but it
would only take a fraction of that to maintain a given highway speed.

Indeed, but using the engine at the 84 hp level during acceleration or
climbing steep hills will be both possible and most efficient. The back
roads I mentioned have a *lot* of hills, and the road is such that even
though it consists nearly 100% of corners you can take them all at 120+
km/h without discomfort or danger.

I don't know whether mid-90's Subarus cut the fuel off entirely with a
closed throttle -- both my 1986 and 1995 model BMW motorcycles do -- but
if so then using the most efficient hp level to climb a hill at 120 km/h
in top gear and then descending the other side with closed throttle and
zero fuel flow may well be a very efficient way to travel.

 

[JohnO]

Indeed, but using the engine at the 84 hp level during acceleration or
climbing steep hills will be both possible and most efficient.

But it still uses far more fuel than traveling on a straight and flat road.
There's simply no way you get identical fuel use while doing more work.

The back
roads I mentioned have a *lot* of hills, and the road is such that even
though it consists nearly 100% of corners you can take them all at 120+
km/h without discomfort or danger.

You lose a lot of momentum going around corners unless they are highly
banked. That requires fuel to overcome, fuel which is not required on a
relatively flat highway. I forget the name of the principle involved, but
objects in motion don't want to change direction.

Unless you have a torpedo-shaped car with a very slippery surface, such as
those aerodynamic bicycles, your best fuel efficiency will be around 50 mph,
plus-or-minus 5. Wind resistance is THE major factor.

-John O

 

[CompUser]

I'm not sure whether you didn't read what I said or didn't understand it
so I'll attempt to put it into even more basic terms for you.

You suggest that the slower you go the better the fuel economy.

But it's obvious that there is a limit to this. In particular, sitting
still gives you the worst possible fuel economy. Therefore the best
fuel economy is at some finite speed that is greater than zero. By the
very definition of "best" and of "maximum" this means that travelling
either faster than this best speed OR SLOWER than this best speed will
give worse fuel economy.

Yes I know that drag increases with speed. And, no, it is *not*
exponential. It is a polynomial function, not an exponential one. your

Oh, criminy!

Where's that guy who was talking about the laws
of physics?

 

[Edward Hayes]

Generally the power needed to double the speed is a squared function.
If I drive 40 mph and it requires 5 horsepower and if I drive 80 mph
I will need 5*5 or 25 HP. I think this is correct if I remember my
first year physics. Of course this is subject to the changes in
aerodynamics as the car goes faster. i.e. a drag coefficient of 0.3 at
50 mph may change to 0.35 at 100 mph.

 

[Bruce Hoult]

Indeed, but using the engine at the 84 hp level during acceleration or
climbing steep hills will be both possible and most efficient.

But it still uses far more fuel than traveling on a straight and flat road.
There's simply no way you get identical fuel use while doing more work.[/QUOTE]

No, climbing a hill and then descending to the original level is an
identical amount of work to taking a flat road of the same length,
providing only that you travel at the same constant speed and use the
same gear both up and down the hill as you would have on the flat road
and don't use the brakes.

The physics clearly says it's the same amount of work, and requires the
same amount of energy, so it may well use the same or even less fuel if
the engine is wroking at a power level that is more efficient on the
climb.

Notice that this is exactly how almost all airliners work on short hauls
(up to 500 km or so). They climb continually until about half way to
the destination and then throttle back to idle and glide the other half.
It's also the same with sailplanes (I fly them) that have engines. In
some of them you can cruise long distances with the engine on at around
160 - 200 km/h but it's *far* more efficient to run the engine at full
power to climb to 10,000+ ft (which takes 15 min or so, covering 30 - 40
km) and then turn the engine off and glide for the next 200 km.

You lose a lot of momentum going around corners unless they are highly
banked. That requires fuel to overcome, fuel which is not required on a
relatively flat highway. I forget the name of the principle involved, but
objects in motion don't want to change direction.

Given an "elastic collision" between the car and the corner, cornering
requires force but does not require a change in either energy or
momentum. There is no energy change in changing direction and the
momentum change is balanced out by the momentum change of the earth. If
you corner too fast then you'll wear your tyres, but normal cornering
wears tyres pretty much the same as driving straight.

Unless you have a torpedo-shaped car with a very slippery surface, such as
those aerodynamic bicycles, your best fuel efficiency will be around 50 mph,
plus-or-minus 5. Wind resistance is THE major factor.

50 mph? Where is that figure from? It may have been correct in the
1950's but it's not a law of nature. Car aerodynamics have improved
immensely since then. The average modern car *is* "torpedo shaped"
compared to just about anything from back then.

 

[Bruce Hoult]

Generally the power needed to double the speed is a squared function.
If I drive 40 mph and it requires 5 horsepower and if I drive 80 mph
I will need 5*5 or 25 HP.

No, the drag is proportional to the square of the speed, so because work
is force times distance the energy required to overcome drag while
travelling a given distance is also proportional to the square of the
speed. But because you are covering that distance in less time
(inversely proportional to the speed) the power needed is proportional
to the cube of the speed.

So if you need 5 hp to overcome aerodynamic drag at 40 mph then you'll
need 125 hp to overcome aerodynamic drag at 80 mph.

 

[Dave - Dave.net.nz]

50 mph? Where is that figure from? It may have been correct in the
1950's but it's not a law of nature. Car aerodynamics have improved
immensely since then. The average modern car *is* "torpedo shaped"
compared to just about anything from back then.

indeed they are.

sorry to interject...
Bruce... you're a long way from nz.comp... nice to see you here.

 

[Edward Hayes]

Thanks Bruce: I guess I forgot allot of physics as I've been out of
class for 40+ years. Time to hit the books again before I goof again
and mislead someone. ed

 

[Mike Lloyd]

There are multiple types of drag: induced, parasite, friction. Yes, one
type is exponentially proportional to speed, others are not... Some are
even inversely proportional to speed. The *total* drag is what matters, so
if the car is very aerodynamic it's shape will effect its mileage. If you
neglect these principles than you'd be correct, but otherwise look at
airplane and car aerodynamics and realize that you will use much more gas to
push a barn door than a "torpedo" to the same speed. So you can find an
"optimum" drag to thrust ration -- just like in all modern airplanes where
they have an optimal speed for mileage range and time aloft (which are not
the same speed, mind you).

Think of it this way. Your car engine will undoubtedly run many many
hours on end at idle in a parking lot (more than 6 -- ask how I know), but
you'll get now where. Idle around town and burn gas faster but make
distance. Now go 75 MPH on the freeway and you run out of gas in 4 hours,
but cover 300 miles... See? There are *many* factors that play into maximum
fuel economy. Essentially it is where the car gets the most distance per
fuel used. Go too fast and you use fuel faster per mile. Go too slow and
you use more fuel per mile. Find your middle ground and *that* is your best
fuel mileage.


Mike

http://www.centennialofflight.gov/essay/Theories_of_Flight/drag/TH4.htm
http://navygouge.com/api/basics/aero1-5.html

 

[Dave Morrison]

Aero bikes are designed to go as quickly as possible. Efficiency has
nothing to do with it.

Shell fuelathon vehicles, which in the early '80s started setting
records of 1600 mpg travelled at 15 mph to demonstrate efficiency
and look very much like the faired cycles used to set speed records.

50 mph? Where is that figure from? It may have been correct in the
1950's but it's not a law of nature. Car aerodynamics have improved
immensely since then. The average modern car *is* "torpedo shaped"
compared to just about anything from back then.

When a car is more aerodynamic it just means that the drag
is less than it would be if it were less aerodynamic.
But the drag still increases with the square of the speed.

A car is most efficient from an aerodynamic point of view
when it is moving as slowly as possible.

A car is most efficient from the motor point of view
when the motor is running most efficiently. I would
expect (as I have seen in tests) that any car's most
efficient speed is in top gear just over idle or at
the lowest point where speed can be maintained and
all losses overpowered, often at full throttle.
Since in most cars (pick your era),
gear ratios are too low for that speed to be practical for
travelling on the highway, nobody goes at that speed
other than in the city where the results are ruined
by having to slow down or stop.

 

[Larry Van Wormer]

There are multiple types of drag: induced, parasite, friction. Yes, one
type is exponentially proportional to speed, others are not... Some are
even inversely proportional to speed.

I'd not heard of any form of drag that's inversely proportional to
speed, and indeed have a hard time believing in it. Please enlighten me
with more details!

Larry Van Wormer

 

[Larry Van Wormer]

There are multiple types of drag: induced, parasite, friction. Yes, one
type is exponentially proportional to speed, others are not... Some are
even inversely proportional to speed.

I'd not heard of any form of drag that's inversely proportional to
speed, and indeed have a hard time believing in it. Please enlighten us
with more details!

Larry Van Wormer

 

[JohnO]

So if you need 5 hp to overcome aerodynamic drag at 40 mph then you'll

need 125 hp to overcome aerodynamic drag at 80 mph.

OK, so you'll get more miles out of a gallon when generating 5 hp or 125 hp?

-John O