Better MPG at 62MPH then at 55MPH
 

I have a 1999 Camry LE 4 Cyl 2.2 Automatic


I have an ultragauge hooked up and it has been calibrated

I have found that at just above 60MPH I get right round 40 to 42 mpg this is steady. On my way from Bakersfield to OC I got an average of 36 MPG on this trip. From what others have told me I should get better MPG at 55 but I don't isn't this strange?

Could it just be my car weight type of engine aerodynamics that cause right around 60 mph to be best speed for gas milage

Its highly doubtful. There are too many variables to consider one trip's worth of data significant. There is wind, traffic, temperature just to name a few that can change your mileage a ton.

The best way to test it would be A-B-A over the same stretch of road on the same day. Pick out a stretch of road (say 5-10 miles). Drive it both ways at 62 mph, then both ways at 55 mph, then both ways at 62 mph again. If desired, you can keep on repeating. The difference between the two 62 mph runs gives you an idea of the inherant variability in the testing, as well as whether things are changing over time.

If there actually is a real difference, I would suspect the auto tranny. You don't go through a shift between 55 mph & 62 mpg do you?

There's actually quite a few threads already on this topic. Increading FE with increasing speed is very rare:
http://ecomodder.com/forum/showthrea...u-15182-7.html

I've driven several generations of Camry, and they all went into top gear at 45 mph.

Metro had similar results in the camry he tested:

http://ecomodder.com/forum/showthrea...2-a-23818.html


http://ecomodder.com/forum/attachmen...1&d=1351467646

broski499: full points for monitoring your trip MPG, but it's a case of apples & oranges.

There must have been different factors on each leg of the trip affecting the results - traffic, wind, elevation change, temperature etc.

If you graphed bi-directional runs (to cancel wind/grade effects) and with no other traffic around, you'd find that MPG at 55 MPH is going to be a little better than at 60.

If this is your 'permanet' eco car, then you should do as Dave suggests.
Find a flat piece of land/freeway (should be easy in your neck of ca) and drive 10 miles up and 10 miles back using the gauge and cruise contrl.
Do 50, 55, 60,65 heck even 70.
Now you have a baseline for your car.

I always used the same strip of hiway (when I was in Sacrmento) to test my mods

As others have said, many variables to consider...another is the engines output curve. As with many production cars, dyno graphs show power delivery is not always smooth with often dips / flat spots at certain rpms not to mention steep increases. Could be quite simply that the engine is at a more efficient power output rpm at 62 MPH cruise speed than at 55 MPH cruise speed. Since the speed difference is very small, this could be another answer to the question of why it is better at 7 MPH faster. A dyno of the car against speed might indeed show this. Probably the case for the other Camry results difference not being linear with speed increase.
I was bored and did a quick workout in excel based on aero drag alone, starting at 50 MPH and 10 HP required to maintain that speed right upto 100 MPH and the 80 HP (80.00003 as could not get it any closer than that) required to maintain that speed (as say just based on aero drag alone requiring 8 times the power to double the speed). So on my little mess about table I saw 12.31144 HP required to cruise at 50 MPH and 16.47182 HP for 62 MPH. So what if at that given rpm the engine is only producing 11 HP at 55 MPH and then a bit of a spike to 18 HP at 62 MPH. Just a thought. While aero drag can be plotted as it increases with the square of speed so you know exactly what it will be...engines outputs can not. While looking for that optimum run for best MPG, speed might have to be adjusted based on the specific engines power curve.

Ok I am waiting to be told off now.

DirtyTorquer, you're on to the root of this.

Work output of the powertrain necessarily increases with speed because the aero drag is increasing with the square on speed. So for your mpg to get better at higher speed, something about the powertrain operation has to change to allow you to produce more work that you were before with less fuel than you were before. I.e. if the higher speed means you need X% more work output, the powertrain needs to have become >X% more efficient. That scenario is possible, but pretty rare.

Can it display LOAD of the engine ?
See if there's much difference between the load at 55 and 60.

Try the different speeds repeatedly - for a full tank - so outside influences start to even out a bit.

If the car stays in the same gear, the lower speed should get better FE.
If it drops a gear, you can expect FE to suffer.

And once upon a time I got 41 MPG out of a 1995 Dodge Neon at 80 MPH. But it was a warm summer day, I was moving with traffic in the fast lane, and a tail wind.

Wind has a huge effect. Most people cannot even feel a five MPH wind. But, at 60 MPH, that is the difference between 55 MPH and 65 MPH airspeed depending whether it's a tailwind or a headwind. The difference in aerodynamic drag is proportional to the square of the airspeed, or 40%.

Say that again?

Airspeed squared is the aero difference?

So, if I were going 10 mph faster, it would be 10*10...oh, I give up :o

Correct. In this case, the two airspeeds are 55 mph and 65 mph.

65^2 = 4225
55^2 = 3025
4225/3025 = 1.397, or about 40% higher.

Or to put it another way, 10 mph over 55 is an 18% difference. 1.18*1.18 = 1.39, or again almost 40%.

Reminded me of several of my crosscountry trips from CA to texas when I would be on the high plains on hwy 40.
Once I could literally feel the car being pushed along!!!!! :thumbup:
Another time the cross winds had all the cb anteneas leaning at 45degree angles!!! :eek:

Something that wasn't touched on here is the possibility of the engine running more efficiently at the higher speed (higher rpm) than the lower rpm.
It could be the engine "coming on the cam", getting the best scavenging from the exhaust, both the tuning of the intake/exhaust plumbing, thermal dynamics inside the combustion chamber...etc.

Makes sense to me to have a target rpm range where a typical engine runs the most efficient...especially if this engine is connected to a CVT transmission...which isn't the case here, but just saying...

wungun - It's possible, of course. Diesel Dave touched on that idea in post #8.

But it's pretty unlikely. Actually it would be the first modern, conventional 4-cyl car anyone's seen around here that behaves that way.

Guess i speak a foreign language

I have driven my car on the same road. At the same speed. At the same time ( reseting the trip mpg every 2 minutes ) and got different numbers every time because its almost impossible to have the same exact conditions even if they are only 2 minutes apart

Wow, I knew that air resistance grew exponentially, but I didn't realize that a difference in 10 mph could make such a difference.

I wonder: What if trees were planted along all the highways? Once the trees grow, wouldn't that dramatically reduce wind resistance?

Although that might also reduce the benefits of a tailwind.

Whoops! Yes, you talked about it too. :D

Exponentially means just that...
And easy way to look at it is, to double your speed requires 4times the power.
So if your increasing your speed from 10mph to 20mph, it'll still require 4 times the power...except at those speeds, you might only be using 3hp to maintain 10mph (or 12hp for 20mph.)
But if you're doing 50mph, and you want to do 100mph, where it might be 50hp to do 50, you'll need 200hp to do 100mph...and exponential jump!

Someone correct me if I'm wrong....

I would add that even when a road looks level, it may have a very slight grade, which does make a considerable difference. A/C use is also a factor, (even your defroster will cycle the A/C). slight wind, previously mentioned, will make a difference. all these things together can make a big impact, easily explaining your difference.

Stop using that word!

I have the same thing with my car, it gets better cruising MPG at a touch over 60 rather than at 50-55. It's not just a one-time thing, either, it's very consistent.

Doubling your speed requires 8 times the power. Drag force is 4X as much at double the speed.
Power is a function of force * velocity.

So your saying, in my example, if it requires 50hp to do 50 mph, you'll need 400hp to do 100??
I think you're wrong....
Are you thinking of kinetic energy an object has at speed perhaps?
Or am I!?
Lol

Yes, if all your drag was aero. Look at the ecomodder tools
The default is for a small economy car, but the results chart goes up to 200 mph. At 50 mph aero power is 5.79, at 100 mph aero power is 46.36. hp. At 200 mph it would require 370.86.
-mort

You are.

From Wikipedia:

"Power

The power required to overcome the aerodynamic drag is given by:
http://upload.wikimedia.org/math/e/3...82a89503b1.png

Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula. Exerting four times the force over a fixed distance produces four times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, four times the work done in half the time requires eight times the power."

This is well known among salt flat racers.

Doubling your speed requires xpower cubed....End of story.
Think about what is being said...
Exerting four times the force over a fixed distance produces four times as much work.
The distance is not fixed...
The distance is the variable in the work calculation...In this case, how many miles is covered in x amount of time. This is how we measure speed.
At twice the speed, you are covering TWICE as much distance over the same time frame, right?
And the last sentence...
Since power is the rate of doing work, four times the work done in half the time requires eight times the power.
Again, this does not add up go me.
Rate means time, correct? The rate at which you do work. Think about it.
Where does "half the time" come from?
Obviously from the "fixed distance" comment...
Again, the time frame in these calculations is based on the speed measurement, and since speed is a product of time and distance , you are not doing 4 times the work in half the time...you are doing 4 times the work in the SAME amount of time. And the timing in question is one hour.
Make sense?

It does to me!

If I'm wrong, and I very well could be, someone please explain it better to me!?

Horsepower= torque {lb_ft} * RPM/5252. Universal hot rodding formula.

The following assumes the same gear ratio:

Torque required is directly proportional to drag force, and therefore, proportional to the square of the speed.
RPM is directly proportional to speed.

Suppose a car requires 35 lb_ft torque to maintain 50 MPH, and is running 1500 RPM.

35*1500/5252=9.996HP

Same car requires (100/50)^2 or 4times the torque (140 lb_ft) to overcome the drag @ 100MPH. It's also running 3000 RPM.

140*3000/5252=79.969HP (8 times what it takes to run 50 MPH).

Maybe someone else can explain it in simpler terms.

Hi wungun,
There are 3 engineering properties to consider. Force is pushing on something. If you weigh 150 lbs you exert a force against the earth of that many pounds.
If you want to move an object, like a car, you can push on it with a force, say 90 lbs. If you push and jog along at 5 mph you are using 1.2 hp, that's power. Horsepower is just a conventional way of saying lb*mph (with a conversion factor).
Now to go anywhere you need energy. Burning gasoline releases energy, in units of horsepower hours. A gallon has about 45 hp hr of energy. Energy and work have the same units.

Imagine a car that only has wind resistance. (OK - that's an airplane.)
At 60 mph the wind resistance force on my imaginary car is about 90 lbs.
If my imaginary car had an engine that was 33% efficient at 15 hp, and it takes 15 hp to go 60 mph Then I could go for an hour on 1 gallon of gas and that would move me 60 miles. That's 60 mpg.

Now if I went 120 mph the force needed to overcome wind resistance is four times more than at 60 mph. The power rquired to overcome wind resistance is 8 times higher. The energy in a gallon of gas is not changed.

So in my imaginary car I also have an engine that can produce 120 hp at 33% efficiency. With that engine I can go 120 mph. With 1 gallon of gas delivering 15 hp hr through the engine, I can go for 7.5 minutes. That's 15 miles. Thats 15 mpg.

So at twice the speed I can travel for 1/8 as much time, and go 1/4th the distance on the same amount of fuel. I've done the same amount of work, but the engine needed to be 8 times more powerful. If I wanted to go faster, say 240 mph it would take 960 hp. I'd have just less than a minute of fuel, and I'd move 3.75 miles. (3.75 mpg)
-mort

That's right, 8x the power for 2x the speed. That's why on the track, while a huge amount of energy is lost to braking, probably a bigger amount of energy is lost to pushing the car through the air at high speeds, which is why the brakes aren't being cooked with anywhere near the full output of the engine. It's why you need only 300-400hp in a "typical" sports car to hit 180mph, but to hit 250mph you need over 1000hp.

Ok......

So the solution is to do nothing?

sure there are variables but the idea is to get a BASE idea of what the car does.

1. warm it up before first drive.
2. do it after noon (temp stay consistant for a couple of hours)
3. Up and down negates wind
4. don't do it on a windy day.
5. do it on a day over 70 degrees.

Wow! Talk about a response. Yes this is my Eco car, mainly because its just the car I have right now but it holds it own at 40mpg cruising

On that trip I talked about there is a lot of up and down and I coasted in neutral down some big hills which really boosted my mpg

I will do some A B A testing and let you guys know what I find and see if I can find the engine load gauge

Thanks for all the help guys!!

ok all I am going to test the speed verses mpg quandry this weekend and will post the results. my last trip of 200+ miles one way yielded unexpected results. at 1950-2000 rpm mileage was mid 20s . I increased the speed to make my grandsons soccer game to 68-70mph and I had a mileage jump a big jump 6-8mpg at 2400rpm. I suspect the turbo sweet spot. I am going to try it again and will keep all informed

You probably already know this, but measurements taken during a trip aren't the best way to do this.

You ideally want to: repeatedly drive the same segment of road one run after the other, in the same conditions, with a fully warmed up vehicle, changing only your speed on each run.

That's how to get the most credible figures/results. Minimize the number of variables that could be affecting your fuel consumption to just one: speed.

Drag is a second-order function of velocity.

...but,

Power is a third-order function of velocity.

well I hope that I have found the point where all mechanicals are at their most efficient and I am using the fuel not wasting it at a lower speed. I am going to top the tank and drive by the scangauge for max mileage then when I exit the freeway I will top it off again and see the result, I already know that stop and go in L.A. is a fuel suck. Its worth a check .

OK here it is we just got back from huntington beach .down to H.B. 304.4 miles 16.621 gal mixed driving 18.314 mpg running 67-68 mph. return trip home to S.L.O. ca 224.5 miles 10.149 gal all freeway 60 mph 22.12 mpg. see fuel log. bottom line is speed kills/stop and go is bad for mpg

We don't notice air as having resistance really because we are accustomed to it, but it can be treated like any fluid. The faster you go the more you compress the air and the more it wants to act like a solid.