Best RPM for most efficient fuel consumption.

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.

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.

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

I'd suggest measuring your fuel consumption more carefully.

-John O

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.

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.

No one said that it did.

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.

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.

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

Oh, criminy!

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

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

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.

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.

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.

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.

indeed they are.

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

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

On Thu, 08 Sep 2005 23:13:36 GMT, Bruce Hoult wrote in news: snipped-for-privacy@news.clear.net.nz:

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.

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.

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

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

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

-John O