Best RPM for most efficient fuel consumption.

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

Reply to
Paganguy
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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.

Reply to
Bruce Hoult

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.

Reply to
Bruce Hoult

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.

Reply to
Grolsch

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.
Reply to
Bruce Hoult

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.

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Reply to
Bruce Hoult

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?

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

Reply to
JohnO

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

Reply to
JohnO

An interesting study was done here:

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

Reply to
David Patnaude

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?

Reply to
Cam Penner

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

Reply to
Larry Van Wormer

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

Reply to
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.

Reply to
Mike Lloyd

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.)

Reply to
Larry Van Wormer

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.

Reply to
Clifford Heath

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

Reply to
CompUser

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

Reply to
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.

Reply to
Gilles Gour

On Sat, 10 Sep 2005 18:28:42 GMT, Gilles Gour wrote in news:wBFUe.110266$ snipped-for-privacy@weber.videotron.net:

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.

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.

Reply to
Dave Morrison

On Sat, 10 Sep 2005 20:24:30 GMT, Dave Morrison wrote in news:Xns96CDA6B57E288nOsiRrOmevAd@216.196.97.142:

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.

Reply to
Dave Morrison

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