parkkuen

This question will sound very stupid, but I would like some explanation... it's logical that you must put more gas to make the car go faster...... but my question is this..

Let's say RPM (Revolution Per Minute right?) at 25 MPH at 3rd gear is 2750 RPM... and RPM at 6th gear at 75 MPH is also at 2750 RPM... So technically RPM per engine is turning at the same rate right??? so am I wasting more gas if I would to drive 25 MPH at 2nd gear at 5000 RPM than 75 MPH at 6th gear at 2750 RPM?

I understand that there are wind resistance and more of other factors... but I am just curious ... curious mind wanted to know...

chill_dog

Hell, even I (a financial analyst) know this one...yes. Proof is easy, take your car out and hold a low gear...watch your gas gauge drop like a rock. Of course, just like you said there's wind resistance and what not, there's also whether you're racing someone, trying to merge, other reasons to get your RPM's up...but the bottom line is, the higher the RPM's, the more gas you're going to burn.

rbf351

it's kind of an obvious answer

the fast your engine turns (higher RPMS) the more gas is needed

parkkuen

that's what I figured... thank god for the 6th gear...

Legend2TL

It depends on engine load and RPM. In general the higher the RPM the higher the engine friction. But engine load is the primary factor here. Just because the engine is turning 2750 RPM in 3rd doesn't mean it is using the same rate of fuel as 6th gear. The 6th gear will be higher rate of consumption.

You are not necessarily wasting more fuel at 5K revs in 2nd doing 25MPH versus 2.75K revs in 6th at 75MPH. More than likely the 75MPH case you are using much more fuel since the aerodynamic load and friction load from the tires is higher.

Engine revs are park of the fuel consumption equation but vehicle load dominates (meaning how fast, how much weight, how much friction). For the same speed lower revs typically means lower fuel consumption. I.E. using 6th gear instead of 5th at 75MPH. The engine load is the same, but you save fuel due to less internal engine friction and intake pumping losses.

gregtomash

Interesting question! As far as I understand it, all engine power goes to overcome friction - nothing else (assuming steady speed - no acceleration). Wind resistances, friction in the cylinders, tire rolling resistance (btw, I think it doesn't depend on speed) etc - it's all just friction. Without friction you wouldn't need any power once you get going (which would be impossible without friction!). I would think that because it takes more RPM’s per mile at lower gear (i.e. more friction), you burn more gas. Of course engine efficiency at different RPM comes into play, but I don’t know much about it…

Legend2TL

On a flat surface this is true, all engine power goes to friction (not accounting for thermal losses since you said engine power which is the output of the engine).

Going up a hill this is not true, since now the engine must do potential energy = work (Force * Distance = Gravity * Mass * vertical distance in this case). Since when you go up hill and you want to maintain the same speed, you depress the accelerator down. That extra power is the work.

crazymjb

Wind resistance at higher speeds, takes more gas to keep the engine at that RPM.

avs007

If you connect an OBD scanner to your car, it will tell you what the calculated engine load is at any given time. I don't remember it being terribly high at cruising speed. I think it was something like 14% or something like that.

You are going to use more gas at 25mph at 5000rpm, then at 70mph at 2000rpm in 6th gear. The forces associated with wind resistence are exponential, so at like 90+mph, I think it would be tremendous, but not at 60-70mph.

I would think that's why City fuel economy is so much worse than highway, becuase you spend more time at higher rpms.

Ron A

This has been driving me crazy all day, but I have to ask if my reasoning is sound because I keep thinking it may be right and then again it may be wrong. And I don't think I read this theory in any of the previous posts.

I also think many people missed the original question, which was if you go a certain distance at the same (2,750) rpm, which is more economical?

If you drive 25mph in 3d gear at 2,750 rpm, in one hour you would only go 25 miles.

If you drive 75mph in 6th gear at 2750 rpm, in one hour you would go 75 miles.

Assuming that since you engine is turning 2,750 rpm in both cases, and for the sake of argument setting aside any friction or wind resistance losses, you would be using the same amount of fuel to go 25 miles in 3d as you would to go 75 miles in 6th.

Which do you think is more economical and yields higher mpg results? I would say the one that takes you a distance of 75 miles in an hour, rather than a distance of 25 miles in an hour.

Be gentle if I'm wrong, because I am using pure logic rather than engineering expertise.

avs007

I think you are right. I say that based on the mpg indicator in my car. I have done that exact same scenario before. I drove 35mph at 2000rpm, and I drove 70 at 2000rpm. My instant MPG reading and average mpg reading (I reset them), was significantly higher at 70 than at 35.

Ron A

I concentrated so much on this part of the post

that I missed this part.

But the theory is still the same, that you would get better mpg going 75 miles in an hour in 6th at 2,750 rpm than you would get going 25 miles in an hour in 2nd gear at 5,000 rpm, which is even more true when you consider the engine is turning twice as fast in 2nd gear in this second illustration.

NoRespect

Is the question the amount of fuel being burned per unit time or per mile? They are different...

crazymjb

Per unit of time your best would be in neutral(especially with a manual). It is probably puts less stress on an auto to have it creep along instead of keeping your foot on the break at a stop as that causes resistance in the TC.

6speedv6

NO.

Wrong again. At any point over 4,800 rpms (V-Teck kicks in), the fuel system is wide open.

And because the engine is doing 800 or so rpms less in 6th gear.

Remember that once the car is moving at a certain speed, keeping it there is relatively easy. Once a body is in motion, it will keep moving untill a force is applied in the opposite direction. There is no way that under 70 mph the combination of wind drag, tire drag, and internal engine friction losses will be higher than the amount of resistance of a 3,500 pounds car has while standing still.

A simple experiment.
1. From a stop, on a totaly flat surface, accelerate your car up to 70mph but short-shifting it (let's say 3,500 rpm) and see how long it takes and how much distance have you covered.
2. Then, put the car in neutral, and let it roll to a stop.
3. Now repeat 1 and 2 but keep the car in gear (without pressing the gas pedal) and see how long it takes now.

You can also try this accelerating the car up to redline. What do you think will be the result ???

2_FastTLs

Are you trying to differentiate between static and kinematic coefficient drag constants? For a mass on wheels the static is nearly the same as kinematic. You also cannot compare a mass at constant velocity to a mass being accelerated. For constant V mass there are 2 force vectors, each being equal in magnitude but in opposite directions. Whatever force is being put down to the ground to keep the car at V their is a equal-opposite force (aka drag) pushing back. "drag" is a general term that describes numerous forces.

Your "experiment" is meaningless because it gives no qualitative data. The acceleration vectors are changing as a function of gear ratio and engine rpm.

cpurick

To answer this, you'd have to know whether the friction losses of driving around at 5000rpm in 2nd gear are more than the aerodynamic losses of driving at 75mph in 6th. I doubt anyone here can answer that without testing it on the MIL.

All I can say with certainty is that you'd be better off shifting into the highest possible gear at 25mph instead of holding it in 2nd.

youngTL

Also the original poster says:
so am I wasting more gas if I would to drive 25 MPH at 2nd gear at 5000 RPM than 75 MPH at 6th gear at 2750 RPM?

I'm in geophysics, but I hope I can provide some insight. Where's Road Rage and Scalbert when you need them eh?
The answer is yes, because even though the total drag on your car is less at 25mph, you're incurring much more pumping losses and other stuff at 5000RPM, all while covering 1/3 of the distance. That's highly inefficient. Now let's say you drive those speeds at equal RPM. 75mph is still more efficient because you are traveling triple the distance in an equal amount of time, therefore EVEN with wind and tire rolling resistance, there is no way to overcome the fact that you're only running your car for 1/3 the time, unless you travel at such a speed (ie: 90mph or so) where those exponential drag losses exceed the gains made by the reduced time.

Legend2TL

No what I wrote was quite correct. I was writing for constant speed (the orginal thread did not get into acceleration and deceleration losses).

The fuel system is also not "wide open" at any point over 4800 RPM. The ECU will look at the Manifold Atmospheric Pressure (MAP) sensor, intake air temperature sensor and engine RPM to determine what amount of air is entering the engine. Then the amount of fuel is calculated to keep the appropiate air/fuel ratio.

The statement about 70MPH versus standing still also is not correct. The amount of force to move a average car through air friction, tire friction, drivetrain losses typically being between 10-30HP at 70MPH would also easily move a car from a standing start.

FWIW Honda/Acura engines they do not use air flow meters (hot-wire or other types) for their fuel injection system. They monitor the engine's intake manifold pressure and RPM, and from measuring in the lab setup at the factory measure the airflow. This is then put into a look-up table with other factors such as intake air temperature, Engine RPM, MAP pressure.

This thread started as a simple stattic speed question relating to engine RPM. If you mix into it dynamic speed (acceleration/deceleration) then we could start a college physics leason.

In terms of city versus highway mileage, one very important factor comes to play here. In the city quite often the car is sitting still at traffic lights and heavy stop/go traffic. It is going no-where so you're gas mileage is 0. Factor in losses from acceleration up to speed, then deceleration (where the car's kinetic speed energy goes to thermal engine heating up the brakes) and it's easy to see why mileage is less.

Yet another FWIW, most hybrids do better in their EPA city mileage over highway mileage since they are in a dynamic situation. The gas engines most of the time do not run at stop lights and stop and go traffic unless the battery bank is low. Also they turn most of their kinetic energy back into electrical energy by "reversing" the electic motor into a generator when you lift off the accelerator.

avs007

I see you guys are still arguing despite the fact that I have actually tested this . When going about 35mph, my fuel economy guage hovered around 18mpg. When I was going 70 at the same rpm, it shot up to 30mpg. And this was at the same rpm. Doing 5000rpm at 25mph would probably be much worse. I say that, because my instant mpg reading, when my car was at 5000rpm in 2nd gear was a pitiful 8mpg.

Legend2TL

We're not arguing, we're discussing

I didn't state anything as absolute since there are too many variables at play here. But as you indicated for a TL your testing proved the engine mechanical friction and pumping losses were greater than the vehicle friction at 75MPH versus 25MPH.

One thing I was not thinking much about was actual efficiency (gas mileage) versus gas rate usage. As someone else already pointed out gas mileage takes into account distance traveled versus rate at a given time.

Actuary

Another thing I noticed. Even you are crusing at 2500RPM, if you give gas moderately to bring RPM up a little, you will be using a lot of gas.

I noticed mileage instantly goes bad when I was at 2500RPM and gave moderate gas to go to 3000RPM in 5th gear.

So lower the RPM and easier on the foot consumes less gas.

6speedv6

My point is that after the car is accelerated to a certain speed, it is relatively easy for the engine to keep the car moving. Then, it is just a matter of keeping the engine operating at the lowest possible rpm to save fuel.

grudax

wouldn't all this be assuming the same final gear ratio? wouldn't that have an effect as well?


going start bleeding out the ears if I keep thinking about this stuff

2_FastTLs

Wrong, and, maybe. There is no such thing as "relatively easy" in physics world. Everything is clearly defined. It takes X amount of energy to keep a mass moving at constant V under known environment. As a general blanket statement, it takes less energy at 20mph than it does at 75mph.

As for the maybe answer to your rpm statement. I would agree with you under one condition, that all parasitic drag functions related to engine and powertrain components do not have variables with exponents anywhere in the denominator of the function. It may be possible that some drag forces actually decrease as engine and powertrain rpm's increase. In other words, as example, it might be better to run 75mph in 5th gear as opposed to 6th gear.

KJSmitty

If you lived in a vacuum/world with no drag, yes :-)

Always remember - Inherent and parasitic drag (resistance in general) cause the force/energy to move an object faster and faster to increase exponentially. It may take 100HP to get/maintain the car at 100mph but 400HP to get/maintain the same car at 200mph...

Also, using a 6MT for example. If you were to maintain a constant 50 mph on a level stretch in 6th gear you may very well be using more gas than if doing the same in 5th. In order for the speed/lowest engine rpm example to achieve optimum mpg the engine would have to be designed for maximum efficiency at that rpm/load.
--IE, just because the rpm's are lower doesn't mean the mpg would be better. The engine could be under more load at that lower rpm thus burning more fuel. Using one of my older 6sp Vett's for example - It would maintain speeds in excess of 120mph in either 5th or 6th, but not only used less pedal but also would give you the better mpg computer info while in 5th. RPM's were definitely lower in 6th but at those speeds the engine had to work harder in that gear thus burned more fuel. Also, it would pull over 155 in 5th but if you downshifted and floored it in 6th,,, you would start slowing down...

Big picture is,,,,
understanding the above, car manufacturers like Honda/Acura go through painstaking equations etc. to achieve the perfect gear ratio/engine performance, torque and power curves to maximize fuel economy in "drive" and/or "6th" gear under cruise conditions at average national speed limits etc. (That's a long sentance..) If driving at 65 in 4th gear at whatever rpm resulted in the best mileage, thats probably where they would have you....

As usual, just my humble

Cheers

avs007

The vette is a bad example though, because it has 2 overdrive gears. It's 6th gear is VERY tall. The engine doesn't have enough power to top out 6th in that car.

KJSmitty

Actually the example is valid. Whether or not it can "top out" in any gear is irrelevant to the original discussion. The fact remains, if you select a given gear at a given speed just because it offers the lowest rpm, this does not always give you the best mpg. The reason however is just as you stated, the engine doesn't have the power/thus may have to work much harder thus burn more gas to maintain that speed etc.

I mainly used my vett experience to point out that once you get a car going fast, keeping it their is not relatively/consistently easy on the engine the faster the speed you're trying to maintain. 6th gear/engine power will maintain the car great at 80mph, but the same power and 6th gear will not maintain the same car at 160. The forces/drag acting upon the car at that speed are too great.


Not trying to argue with ya just wanted to clarify

Cheers

parkkuen

Yes.. that's why I threw this question... because.. even though I am cruising at 2750 rpm in 6th gear.. sometimes it feels like it's working harder..

RonA proposed nice example as well.. and I think I finally understand where efficiency of high speed comes from.....

in a constant speed... 2750 rpm at 2nd gear and 2750 rpm at 6th gear would indeed pretty much use same amount of gas... but engine efficiency of traveling distance comes from difference of circumference(SP?) of gear...

I finally realized when I read your post and 10 speed mountain bike.... right? it makes more sense.. NO?

avs007

I completely agree with you. I meant it was a bad example becuase the TL's 6MT doesn't have 2 overdrive gears like the Vette, and the TL's top speed is achieved in 6th gear, which means you won't have the same scenario of drag forces overpowering the engine in 6th gear, like the vette does, so I don't think the engine would be "working harder" in the TL doing 2750rpm in 6th, vs 2nd.

Besides, I did post that I got 30+mpg on the indicator cruising at 70 vs 18mpg at the same RPM in a lower gear right? Because of the exponential forces I was talking about earlier, I really don't think drag and such would effect your economy much if you were going less than 90mph.

mp3car

No, 2750 in 6th gear is using more gas. Laws of physics require it.

Look at it this way. If I am carrying 100Lbs on my back and walk 100 feet on a flat surface, and i do 60 steps per minute, I am doing a lot more "work" than I am if I just walk that same 100ft without carrying 100Lbs and still at 60steps per minute.

If I then walk those same number of steps again without the 100Lbs, this time taking larger steps, but at the same steps per minute, then I will go more than 100ft, in the same amount of time and my "rpms" were the same. It still took more overall energy for me to take the larger steps even though I still took the same number of steps in the same amount of time.


Let's say you are going up a steep hill at 2500rpms in second, there is a force opposing your direction of travel, it is a component of gravity... Therefore, it takes more gas (energy) to go up the hill than it would to just drive flat at 2500rpms in the same gear. Now, suppose you speed up... you’re in a higher gear going 2500rpms, let's say you are now going a speed in which the drag force is equal to the gravity (and the drag) that was acting against you when you were on the hill... you would now be using about the same amount of energy/gas as you were before when you were on the hill, but now you are going faster.

However, someone mentioned that this was originally a discussion of static engine rpms, i.e. not driving. In this case, if you are sitting there in neutral, and you rev it to 4K and hold it there, versus reving it to 6K, yes, 6K would be using more gas.

also... think about this... the injectors are pulse width modulated, pwm, to control the amount of fuel. If you are holding the gas so that in neutral you're at 6000rpms, the pulse width is very close to the pulse width than if you are holding it at 3000rpms, there are now just twice as many pulses per second or minute. The pulse width comes in to play during loads. The heavier the load, the wider the width. In some instances, the pulse width could be almost the entire time the intake valve is open, which is one reason why people get larger, higher flow injectors... Also, if you are engine breaking, or, you just rev the engine up and let it come back down on it's own, the pulse width is almost zero, again b/c there is zero load on the engine, not even the load that it takes to keep the engine turning. A fun thing to do is watch the pulse width on an O'scope, while under the hood, if you put your hand on the throttle (which i guess we can't really do on the TL) and "blip" it and then let go, you see the pulse width stretch out, then drop to almost nothing, then go back to normal. if you quickly bring up the throttle just a little bit, but hold it there, you see the pulse width stretch for a moment, but then go back down to 'normal' once the engine has reached it's new RPM, now just the frequency of the pulses is higher.

One more tidbit of thought that some people may have never thought about... think about driving a car that doesn't have a speed limitter with a 20mph tailwind (wind going the same direction). Theoretically, your top speed is now 20mph faster... and likewise driving at 20mph now has no drag, but this is of course assuming gears will let you go 20mph faster than your "normal" drag limitted top speed

2_FastTLs

Hmmm, interesting odd example. We are talking about constant velocity on a level surface. As the incline increases it becomes very obvious that the vehicles need to generate more torque in order to overcome the force of gravity pulling back on the vehicle at that angle. To generate more torque the motor needs more energy, hence more fuel.

You also cannot compare a car in 6th @ 2750rpm to a car in 5th @ 2750 because they would each be doing different velocities!

We can compare a car, as example, in 5th @ 3000 rpm to say the same car in 6th @ 1900 rpm so that their velocities were equal. I suspect that when all is said and done the fuel consumed to run each 1000 miles is almost identical. I also suspect that @ 3000 rpm a tad more fuel will be used simply because there will be more heat generated by internal friction as the speeds of the parts increase.

faxmonkey

Not to mention if you had an atomic clock in your car it would be slightly off from the atomic clock in your kitchen if they were both synchronized before you went on your trip.

Legend2TL

In Modern Physics in college we had to calculate how much in time the astronauts that went to the moon missions gained due to their rather large velocity. I can't recall the number but I thought it was in the nanoseconds.

mp3car

That is exactly my point though. Even though RPMs are the same, huge difference in fuel economy... well no... let me rephrase that, huge difference in fuel consumed during an equal time period, BIG difference from when someone says fuel economy.

Other than a more friction/loss in the drivetrain (wheel bearings, certain parts of the transmission, etc.), my point was to overcome the addidional air drag, which would be a tremendous amount more at whatever speed you are in at 6th @2500 compared to 2nd @ 2500 (i compared 2nd to a "higher gear" such as 5th/6th, i wasnt comparing 5th to 6th) , it would be using a lot more fuel in 6th than 2nd, even though their RPMS were the same, in the example in 6th, the manifold vacuum would 'probably' be less (yes, less), but i don't want to get into all that, other than to say high load, low vaccum, low load, high vacuum.... I am not talking about the amount of air, just the pressure in the manifold.

mp3car

I hope you're joking about the atomic clock in our cars.... right???
(we do of course have "atomic time" available to us though (those of us with Nav)... from the multiple atomic clocks on board each gps satelite, which i think are even sychronized regularly to an even more accurate atomic clock back on earth.)

KJSmitty

Hopefully no-one here has pressure in their manifold.... Unless it's their exhaust manifold... Even then we hope the pressure is minimal.

Legend2TL

In relative terms we all have pressure in the manifold.
All depends what it is compared to (atmosphere pressure at sea level, 0 torre,...)

gt1

Guys you still don't count on some significant factors, for example, operating an engine with closed throttle is not very efficient. The topic of fuel consumption has been discussed forever, and the correct answer is that the lowest consuption is achieved by a steady driving on the highest gear with a lowest sustainable speed. The speed factor has limitation- if you go too low, the engine life may be affected as it has to deal with high torque load with low oil pressure.
The bottom line- drive steady 45-50MPH in 6th gear.

faxmonkey

Yup.... It's all "Relative."