Best RPM for most efficient fuel consumption.

[Paganguy]

Okay, I think the information I was trying to determine has gotten
buried beneath a physics discussion of drag, areodynamics and
friction.

What I am more concerned about however is what RPM my 2.5 litre engine
is happiest at. Each engine is unique. Run it too slow and it labours
using more fuel for less output. Run it too quickly and it burns more
fuel than it needs to for the speed at which you travel.

Think for a moment about riding your ten speed bicycle (more likely 18
speed now). If you pedal in too low a gear you pedal like mad and go
slowly but get tired quickly. Pedal in too high a gear and you don't
go as fast as you are capable of and still tire quickly. Find the
optimum gear to pedal in and you go quickly with moderate effort.

Translate this to my 2003 Outback and under flat dry highway driving
what RPM would my engine put out maximum horsepower while using
minimum fuel? My Toyota Corolla used to purr at 3000 RPM. The specs
for the Outback give maximum HP at 4200 RPM. Don't think I want to be
running at 4200 RPM. Shell, Esso and Texaco might be happy if I did
(for that matter so would Subaru) but I would be broke in no time and
looking to drop a new egine in after about 150,000 km.

Any more thoughts folks?
BTW, thanks for the feedback so far.

Best regards,
Paul

 

[Bruce Hoult]

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.

Indeed you are 100% correct and are saying many of the same things that
I am.

So I'm just wondering whether your response to my message is intented to
agree with me (which you appear to) or disagree.

 

[Bruce Hoult]

indeed they are.

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

You too.

Just thought I'd add one point...


Cars have blunt backs while planes have sharp ones and some people might
be confused and think this means cars are unaerodynamic. But for drag
it's mostly the shape at the front that matters. Smooth at the back is
a *little* bit better than blunt but the problem is that planes and
wings are sharp at the back because that is what causes lift. Planes
want lift, and control the angle of attack very carefully to get the
right amount. Cars don't want lift so they are much more stable with
blunt back ends.

 

[Grolsch]

I bvelieve my Forester manual says it is most efficient to shift to 5th gear
at 77kph in my 04 forester, thats about 2050rpm(+/-). However, if I put my
foot down or start climbing I need to downshift because theres very little
torque left. Now if I'm driving and know I will need a little reserve
(torque) I generally run at about 2500rpm, but the sweet spot is 3500+,
thats where the most power is, but if I don't need it, then I just put it my
back pocket by upshifting to cruise at about 2000rpm.

Slight declines, I leave in 5th down to, say 1400rpm.

The slower the rpm, without lugging the engine is most efficient IN TERMS OF
FUEL CONSUMPTION, NOT HORSE POWER.

I have no idea how all this translates to automatics because I actually
drive my car.

 

[Bruce Hoult]

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?[/QUOTE]

Grr, stupid arithmetic. The correct statement is "So if you need 5 hp
to overcome aerodynamic drag at 40 mph then you'll need 40 hp to
overcome aerodynamic drag at 80 mph.". I.e. twice the speed is 8 times
the horsepower. Not 25 times.

Which will be most efficient when you have an engine capable of
grnerating anything from 165 to 300 hp? And when it probably takes
several hp just to turn the motor over? Note that starter motors draw
around 200 amps at 12V *after* the motor is already turning (more
initially). That's 2400W or more than 3 hp.

And don't forget that you're only running the engine for half as long at
80 mph.

 

[Bruce Hoult]

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!

Look up "induced drag". It's the primary source of drag for sailplanes
at under about 100 km/h, rising to infinity as you drop below about 60
or 70 km/h, and dropping to nothing at high speeeds.

http://en.wikipedia.org/wiki/Induced_drag

 

[JohnO]

Think for a moment about riding your ten speed bicycle (more likely 18

speed now). If you pedal in too low a gear you pedal like mad and go
slowly but get tired quickly. Pedal in too high a gear and you don't
go as fast as you are capable of and still tire quickly. Find the
optimum gear to pedal in and you go quickly with moderate effort.

It all depends. Are you going uphill, downhill, headwind, tailwind, and on
what surface. With a tailwind going downhill, 21st gear (they added a third
front gear) isn't very hard. Then, how strong are your legs?

Translate this to my 2003 Outback and under flat dry highway driving
what RPM would my engine put out maximum horsepower while using
minimum fuel?

It isn't about RPM, it's about speed, and the most significant factor (by
far) is the wind. Wind resistance isn't any big deal until you hit about 35
mph. And as others noted in varying degrees of detail, wind resistance
increases *dramatically* from there. There's a tool on the web somewhere
that lets you determine your optimum speed for max mpg, but I can't find it.
The tool bases its answer on the size of your engine and the shape of the
vehicle, and maybe something else, I forget.

Bottom line is that you're going to get the best mileage near the lowest rpm
in your highest gear. The optimum mpg is at higher speeds for smaller
vehicles with larger engines, but we're talking about 55-60 mph at the high
end of this scale, IIRC.

-John O

 

[JohnO]

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

Grr, stupid arithmetic. The correct statement is "So if you need 5 hp
to overcome aerodynamic drag at 40 mph then you'll need 40 hp to
overcome aerodynamic drag at 80 mph.". I.e. twice the speed is 8 times
the horsepower. Not 25 times.

OK, but that doesn't change the equation. If I can cruise in a car at
40mph/5 hp, and you cruise in the exact same car at 80mph/40 hp, with one
gallon of gas are you going to travel *more than eight times* the distance I
travel?

-John O

 

[David Patnaude]

An interesting study was done here:
http://www.fsec.ucf.edu/pubs/energynotes/en-19.htm

The testing used a VW Jetta which had 45 mph as it most efficent operating
speed.

 

[Cam Penner]

@newssvr19.news.prodigy.com>, "JohnO" <johno@@&%
heathkit##.com> says...

OK, but that doesn't change the equation. If I can cruise in a car at
40mph/5 hp, and you cruise in the exact same car at 80mph/40 hp, with one
gallon of gas are you going to travel *more than eight times* the distance I
travel?

And more relevant to everyday driving is the fact that you
don't usually get to pick your speed on the road. You can
vary about 5mph or so, but much more than that and you are
massively increasing your risk of a traffic accident. It's
not much good "saving gas" at 45mph on the interstate when
traffic is going by you at 65mph. You'll likely spend
those savings and more in repair bills when you get run
over.

So, more relevantly, the question should be - what is the
optimum gear to be in for the speed I am driving. If I'm
doing 30mph, should I be in 5th or 3rd?

 

[Larry Van Wormer]

Look up "induced drag". It's the primary source of drag for sailplanes
at under about 100 km/h, rising to infinity as you drop below about 60
or 70 km/h, and dropping to nothing at high speeeds.

http://en.wikipedia.org/wiki/Induced_drag

An interesting link! However, it contains a more complete description,
and that includes:

"induced drag becomes less of a factor the faster the aircraft flies
because at higher speeds a smaller angle of attack is required for the
same amount of lift. The opposite occurs with parasitic drag (the drag
caused simply by pushing the aircraft through the air), which increases
with speed. The combined overall drag curve therefore shows a minimum at
some airspeed — an aircraft flying at this speed will be at or close to
its optimal efficiency. Pilots use this speed when it is necessary to
maximise endurance (minimum fuel consumption and maximum time aloft), or
maximise gliding range in the event of an engine failure or when engaged
in the sport of gliding. In practice,this airspeed corresponds to that
resulting from the minimum power required to maintain level flight."

So induced drag is related not directly to speed, but to the angle of
attack. As such, it does not directly reduce with increased speed, it's
simply that the angle of attack is reduced.

Not especially relevant to automobiles, I believe. Does underline that
Wikipedia is a good source of info, though.

Larry Van Wormer

 

[Larry Van Wormer]

An interesting study was done here:
http://www.fsec.ucf.edu/pubs/energynotes/en-19.htm

The testing used a VW Jetta which had 45 mph as it most efficent operating
speed.

Apparently the lowest speed at which they could use fifth gear?

 

[Mike Lloyd]

but induced drag is relevant because as the relative wind increases (i.e.
forward velocity = oncoming air) there are vortices created by the car.
Albeit small, these create a drag force and if the angle remains constant on
the producing edge, which it does, then as the airflow changes so does the
relative wind and hence the angle of attack and thus the drag.

all this in a very nonspecific sense, of course.

 

[Larry Van Wormer]

but induced drag is relevant because as the relative wind increases (i.e.
forward velocity = oncoming air) there are vortices created by the car.
Albeit small, these create a drag force and if the angle remains constant on
the producing edge, which it does, then as the airflow changes so does the
relative wind and hence the angle of attack and thus the drag.

all this in a very nonspecific sense, of course.

Okay, fair enough, I accept that can occur. Any data available as to how
much of the total drag this component is likely to make up?

(I've found this interesting. The various references seem to leave it
clear: For a specific vehicle, the lowest speed that can be maintained
in top gear is most likely going to give the best fuel economy, because
engine efficiency is thus maximized (lower pumping losses, mainly) and
oeverall drag is minimized.)

 

[Clifford Heath]

...better fuel economy at speeds around
120 - 130 km/h than it does at 100 - 110.

I don't believe you either.

Required power to overcome air resistance varies with the *fourth power*
of the speed - so the efficiency of using the *delivered* power varies
with the third power. Travelling slower makes a dramatic difference in
reducing fuel usage - 60Km/h is better than 90, but try getting people
to drive that slowly on long-haul trips.

As far as the best RPM for the engine to deliver power efficiently,
all combustion engines are most efficient at their maximum-torque
point. I did careful testing a few years ago with a 4-cyl Mazda and
found that driving at low RPM, coasting downhill, accelerating *really*
slowly, etc, actually used up to 20% *more* fuel than when I drove by
hammering to the *same top speed* at much higher RPMs. Both tests were
conducted over months (many tanks) of driving to work on the same roads.

Clifford Heath.

 

[CompUser]

Not especially relevant to automobiles, I believe. Does underline that
Wikipedia is a good source of info, though.

Haven't looked lately, but wikipedia used to
say deionized and distilled water were the same
thing, and deionized was good for batteries. :-/

 

[Mickey]

...
As far as the best RPM for the engine to deliver power efficiently,
all combustion engines are most efficient at their maximum-torque
point. ....

Clifford Heath.

Bingo! I think some one has got it.

When talking about efficiency of an internal combustion engine we have
to deal with several things. there is a parasitic drag from friction,
and external accessories and they are somewhat a consistent but there
is also the volumetric efficiency of an engine and that is not a
linear function, If one were to look at a curve that is based on
energy in (fuel consumption) Vs power out, one would see it change
with speed. The curve will decline to a point and then start
increasing again. That lowest point corresponds with peak torque.

To know the exact speed with the lowest fuel consumption while
driving, one has to sum up a number of components in the overall
equation. CoF of drag, drivetrain losses, wind resistance, both from
speed and and wind one is having to deal with, rolling resistance,
incline, etc. An awful lot of things to consider while sitting behind
the wheel.

The one thing that should stick in your mind is the power needed to
move you down the road varies with the cube of the speed as has
already been pointed out. If memory serve me right the last time I
did look at a fuel consumption Vs power curve, fuel consumption is
more linear below the peak torque RPM.

I have a large spreadsheet for calc power to move down the road.
Takes into consideration, weight, frontal area, head wind, rolling
resistance & drivetrain efficiency.

Here are the power numbers for a 3500 lb vehicle, 66" wide, 54" wide,
no head wind.

10mph = .6hp
20mph = 1.6
40mph = 5.9
60mph = 15.8
70mph = 23.7
80mph = 33.9

These number are only taking in air resistance and rolling resistance.
Add in drivetrain losses and engine losses you need to add another
..5 to 4 HP depending upon speed.

Mickey

 

[Gilles Gour]

Bingo! I think some one has got it.

When talking about efficiency of an internal combustion engine we have
to deal with several things. there is a parasitic drag from friction,
and external accessories and they are somewhat a consistent but there is
also the volumetric efficiency of an engine and that is not a linear
function, If one were to look at a curve that is based on energy in
(fuel consumption) Vs power out, one would see it change with speed.
The curve will decline to a point and then start increasing again. That
lowest point corresponds with peak torque.

To know the exact speed with the lowest fuel consumption while driving,
one has to sum up a number of components in the overall equation. CoF
of drag, drivetrain losses, wind resistance, both from speed and and
wind one is having to deal with, rolling resistance, incline, etc. An
awful lot of things to consider while sitting behind the wheel.

The one thing that should stick in your mind is the power needed to move
you down the road varies with the cube of the speed as has already been
pointed out. If memory serve me right the last time I did look at a
fuel consumption Vs power curve, fuel consumption is more linear below
the peak torque RPM.

I have a large spreadsheet for calc power to move down the road. Takes
into consideration, weight, frontal area, head wind, rolling resistance
& drivetrain efficiency.

Here are the power numbers for a 3500 lb vehicle, 66" wide, 54" wide, no
head wind.

10mph = .6hp
20mph = 1.6
40mph = 5.9
60mph = 15.8
70mph = 23.7
80mph = 33.9

These number are only taking in air resistance and rolling resistance.
Add in drivetrain losses and engine losses you need to add another .5
to 4 HP depending upon speed.

Mickey

Just an honest question here : if it takes 34 hp to go 80 mph, what is
my engine doing with the other 100 hp it has in reserve? I only tap into
it when accelereating? TIA.

 

[Dave Morrison]

so to take your table,
here is an illustration of how high the engine, drivetrain
and other losses are-
If you take the HP and convert it to gallons per second
consumption and then use the MPH to get miles per second -
divide miles per second by gallons per second and you get the
MPG if all other losses did not exist. It shows why for
slower vehicles, the other losses are so significant.

MPH HP MPG
10 0.6 273
20 1.6 204
40 5.9 111
60 15.8 62
70 23.7 48
80 33.9 38

Which is why vehicles designed to run maximum MPG for
demonstration purposes are slow speed,
with all the other losses attacked to
try and reduce them to minimum.

Just an honest question here : if it takes 34 hp to go 80 mph, what is
my engine doing with the other 100 hp it has in reserve? I only tap into
it when accelereating? TIA.

The throttle of course controls the horsepower output. The engine is only
being asked to provide 34 HP by your right foot. If you ask for more
power, you accelerate.

 

[Dave Morrison]

MPH HP MPG
10 0.6 273
20 1.6 204
40 5.9 111
60 15.8 62
70 23.7 48
80 33.9 38

Forgot to add a constant 5 HP *load* to show what
that does-

MPH HP MPG MPG(HP+5)
10 0.6 273 29
20 1.6 204 49
40 5.9 111 60
60 15.8 62 47
70 23.7 48 40
80 33.9 38 33

More data points from your table would be interesting to
see to show where the actual peak MPG is for various parasitic
losses.