We have discussed this before, very recently. RPM is NOT the dominant factor. Fuel consumption depends on your power output.
Lower RPM does not always mean lower fuel consumption. Higher RPM does not always equal higher fuel consumption. You cannot reply on RPM to determine your fuel consumption.
berkut did you see my post in that thread?
It also seems obvious to me that given the same load, a lower RPM (higher gear) will give better gas mileage. This would also explain why cars now have 5 and 6 speed transmissions instead of 3 or 4.
Also in light of my posts in this thread you QED and engineer_comp_geek are all wrong when it comes to the fuel consumption when coasting (at least when discussing a modern EFI car) The fuel is shut off when coasting down. The longer the fuel is shut off, the better the fuel mileage. I teach this stuff for a living. I know what you guys think you know. :rolleyes:
Quite possibly. I’ve managed to swap a couple of engines in my life, and for the most part do all of my own repair work on all of my cars, but I am definately not a mechanic nor have I ever been trained by anyone who was. Most of my knowledge comes from “oh **** something broke, where’s the manual so I can figure out how to fix it…”
However, if you go back and read what I posted, I thought I made it clear that it would be the same either way for a carburated car, but you would probably use less fuel if you used the engine to brake instead of the brakes if the car was computer controlled. Are you disagreeing with this? If so, please explain. Are you disagreeing with my statement that I thought the computer would still inject some fuel into the engine, which I plainly said was guesswork on my part?
If I’m wrong, please enlighten me. This forum is all about fighting ignorance, not rolling eyes and bragging rights.
Well, I knew there was still air being pumped through. It’s just weird to me to think there’s no combustion going on when on my exhaust tone is almost as loud during deceleration in gear than it is during acceleration.
What I was objecting to was this:
As I have explained when in fuel cut off mode (engine warm, throttle shut, and above about 1800 RPM) the computer is not injecting ANY fuel. None NADA zip. So there is not a minimum amount of fuel being injected. (Unless you consider zero to be a minimum amount, I don’t). Sorry, E_C_G I wasn’t trying to beat you over the head, but it sometimes bugs me when a person who is a subject matter expert gives a correct and definitive answer on a subject and people keep arguing with it. I was trying to fight ignorance, but sometimes after 3 posts in the same thread saying the same thing, I want to break out ole :rolleyes: .
I am sorry if pissed you off, as that was not my intention, I was only trying to fight ignorance, and make a point. But I posted late, and I was tired. This is my only excuse. I hope you will accept my apology, and we can still be friends across the Internet.
As Rick’s experiment showed, higher gear equals better mileage.
When you chose a higher gear, two things happen at the same time 1) engine rpm is LOWERED 2) engine load is INCREASED. This shows that engine rpm is more important than load.
Internal combustion engines consume a lot of fuel even when they have minimal load.
No, the information I posted is 100% correct. I’m not talking about a car’s fuel economy. I’m talking about fuel consumption vs. RPM. I’m saying that you cannot determine an engine’s fuel consumption from RPM. Do NOT make any more from that statement than it is.
I have here a power chart from a 360 cubic inch engine. All measurements were made in a instrumented test cell, with the engine mounted on a dynamometer.
At 55% rated power (99 bhp):
RPM Fuel Consumption
2100 9.3 gal/hr
2200 9.3 gal/hr
2300 9.3 gal/hr
2400 9.3 gal/hr
At 65% rated power (117 bhp)
RPM Fuel Consumption
2100 10.3 gal/hr
2200 10.3 gal/hr
2300 10.3 gal/hr
2400 10.3 gal/hr
At 75% rated power (135 bhp)
RPM Fuel Consumption
2200 11.2 gal/hr
2300 11.2 gal/hr
2400 11.2 gal/hr
Note how the RPM changes while fuel consumption remains exactly the same for any given power setting. I can make that engine burn anywhere from 9.3 gal/hr to 11.2 gal/hr at 2200 RPM.
What I have stated in this thread and the other one is 100% correct, but I’ll say it again, hoping someone reads and understands it this time:
You cannot determine an engine’s fuel consumption from the RPM alone. A low RPM does not always mean a lower fuel consumption. A high RPM does not always mean a high fuel consumption.
Here is the key: Your fuel consumption varies with power, not RPM. However, you may produce that power more efficiently at a lower RPM. Fuel efficiency is NOT THE SAME AS FUEL CONSUMPTION.
Or, to put it another way (again):
If you climb a very steep hill (throttle wide open) at 3000 RPM, is your car going to burn the same amount of fuel going down the hill (throttle closed) at 3000 RPM? Of course it isn’t. So you cannot determine your fuel consumption from RPM.
You’ve said so yourself, with the throttle closed:
Berkut Splain me this. Your chart shows no change in fuel consumption over a 200 RPM varrience. The engine has to consume more air to turn @ 2400 RPM then it does at 2200. RPM, right?
Now assuming that the mixture stays at 14.7:1 (Oxygen sensor feedback system) for both of the given RPMs putting more air into the engine will require more fuel to to maintain a 14.7:1 ratio. For every additional 14.7 lbs of air you put through the engine, you will need 1 additional lb of gas. So if you increase the amount of air going into the engine, you will have to increase the fuel flow. This is common sense.
Either your fuel measuring device is not exact enough, or it needs calibration.
No, it does not have to consume more air at 2400 RPM than at 2200 RPM. Assuming that would mean neglecting throttle position and the resulting manifold pressure. To maintain the same power at a higher RPM requires less manifold (and cylinder) pressure. In this case, taking the 55% power situation, the manifold pressure at 2200 RPM is 21.7 inches of mercury. The manifold pressure at 2400 RPM is 20.7 inches of mercury (these figures are measured). The net result is the same flow of the fuel/air mixture, the same power being developed, and the same fuel consumption, even though the RPM varies. The chart is accurate.
Now, please consider the following situation. I’ve said something like it before, but I’m going to put it in different terms. This is very important to the point I’m trying to make, so please humor me here.
You have an engine, running at 3000 RPM, with no load on it, no real power being developed. It doesn’t take much throttle to maintain this 3000 RPM, right? I mean, you can sit in your car in neutral and just a little throttle will bring it up to 3000 RPM. Let’s say it’s burning 2 gallons per hour of gasoline.
Now. You apply a load to the engine. Assuming that you don’t open the throttle, the RPM will drop under the load. But we want a constant RPM, so we open the throttle more to bring the engine back up to 3000 RPM. More throttle, more airflow, more fuel being consumed. SAME RPM. It natural to expect more fuel to be consumed, because this power we are now generating to oppose the load isn’t free. It has to come from somewhere. The fuel consumption goes up, the RPM is still 3000.
Consider those two situations. Both running at the same RPM, putting out different amounts of power and with different amounts of fuel being burned. You can see here that fuel burn will not be constant at any given RPM.
Your fuel consumption depends on how much power you are making. How much power you are making depends on two things:
So RPM alone is not the determining factor in fuel consumption. RPM and BMEP must be used together. You cannot make a blanket statement than an engine will use X amount of fuel at Y RPM. You can, however, make a blanket statement that an engine will use X amount of fuel at Y amount of power. Hence the term we use to rate the fuel efficiency of an engine, Brake Specific Fuel Consumption. BSFC, as you probably know, is how many pounds of fuel you burn per hour for a given power output. Power output, not RPM.
Now, where does the advantage of “gearing to reduce the RPM” come in? The BSFC of most engines is lower at a lower RPM. The engine is more efficient at a lower RPM. That does not mean it has a constant fuel consumption at a lower RPM. Power will still vary, but you will be producing that power more efficiently. Note that I don’t disagree with you on this point. You are correct about it being better to use a lower RPM. I haven’t said or meant otherwise.
But, a constant fuel consumption at a given RPM is just not true. I appreciate your last response being kinder than the first. 
Some questions in no specific order:
If so, how do you explain this?
The engine is producing the same amount of power in all three cases, but consumption is varying (or does the power output change?)
2.
Mileage is better when a higher gear is selected, not because of the drop in RPM, but because of the bigger throtle angle. Bigger throtle angle equates to smaller manifold pressure. Is that statement even remotely correct?
Brake Mean Effective Pressure (BMEP) - Some older cars had an analog meter marked “Economy”. I think it was connected to the intake manifold. Did it measure the BMEP?
Berkut it took me about 20 minutes of thinking, but I understand why the numbers you posted are the way they are.
Work with me here. If I am driving down the road at a constant road speed, my engine must produce a certain amount of torque to maintain that speed. As long as:
[ul]
[li]I Maintain the same road speed[/li][li]The grade of the road remains unchanged[/li][li]The headwind / tailwind remains unchanged[/li][li]Friction / rolling resistance remains unchanged[/li][/ul] The amount of torque produced by the engine must remain the same. If the amount of torque goes down, the car will slow down, if the amount of torque goes up, the car will speed up.
Now let’s assume that it takes 100 FtLbs of torque to drive my car at 70MPH on a particular stretch of road. It does not matter what gear I am in (assuming I can get to 70 in that gear of course) my engine must produce 100 FtLbs of torque to travel 70. If my engine produces less I will slow down. More I will speed up. So to maintain 70 the engine must produce 100 FtLbs.
The formula for Horsepower is HP= RPM * Torque /5252
So if we fill in some numbers here
4th gear 70 MPH 2500 RPM
3rd gear 70 MPH 3000 RPM
HP generated by my engine at a constant 70 MPH in 4th =2500 * 100 /5252 = 47.60HP
HP generated by my engine at a constant 70 MPH in 3rd =3000 * 100 /5252 = 57.12 HP
Generating that extra almost 10 horsepower requires extra air, which requires extra fuel. This agrees with the data I collected in the other thread (quoted above)
So why does your data disagree?
Well let’s go back and look at your numbers
Torque = 5252 * HP /RPM
Solving for the lowest and highest values gives us
T = 5252 * 99 / 2100 = 247.59 FtLbs
T = 5252 * 99 / 2400 = 216.64 FtLbs
This engine was being run at a constant throttle opening, not at a constant load. This results in the engine RPM increasing when load is removed from the engine, and engine speed decreasing with load is applied. You cannot maintain a constant road speed if you reduce the torque of the engine. To do so would require a special dispensation from the laws of physics. Or a change in one of the 3 other items I listed above (Grade, wind resistance, friction). This would be the same as taking a non computerized car (No idle speed controls) and comparing the fuel consumption at idle out of gear and in gear. Out of gear idle = 750 RPM, in gear = 600RPM. What difference in fuel consumption would be noticed? Answer none. The same amount of air is flowing into the engine in both cases. The difference is that the transmission is a larger load then just the engine, so the idle speed goes down. Do I need to mention that cars do not drive at a constant throttle opening?
The OP in the Other thread asked the following question
I took that to mean that the same road speed was to be maintained. Maintain the same speed in a lower gear will result in a higher RPM, and more HP being generated. More HP = more fuel burned. If you were working with a set throttle opening, then speed would go down, which would also result in a lowered wind resistance, and might actually see an improvement in gas mileage. I have never meet anyone who drives to achieve a set HP output from their engine regardless of gear/driving conditions. Everyone I know drives to achieve a set speed. Under those conditions the higher the gear the better the fuel mileage, and the higher the RPM, the lower the fuel mileage.
While I concede that your numbers are correct for an engine on a dyno, they bear little relationship to numbers that will be generated by a doper driving his or her own car.
I’m with ya buddy.
No, that is not correct. If you are driving down the road at a constant road speed, your engine must produce a certain amount of power, not torque. Torque and power are not the same. Unfortunately, most of the remainder of your post is based on this, so your examples will not be accurate. I’ll prove it a little later in this post.
This is all based on the incorrect assumption that torque has a direct relationship with speed. It just doesn’t.
If you wanted to change that to a correct statement, you would have to add one more condition, which I’ve placed in bold:
Agreed.
Not correct. You require constant power, not torque. If you reduce the torque, you can increase the RPM to maintain the speed. Note that this is reflected in your calculations for the torque values on my engine. When the torque dropped to 216.64 ft/lbs, increasing the RPM to 2400 gave us the same power output. Since we have the same power output, we have the same speed, even with a reduced torque. To show that I’m not talking out of my ass on this, here is another chart of actual measured values for you (I calculated the torque from RPM and HP. Oh, and it isn’t the same engine as before, but the relationships are what matter anyway):
RPM Manifold pressure HP Torque Speed Fuel Consumption
2200 24.8 in/Hg 110 262.6 110 10.2 gph
2500 22.2 in/Hg 110 231.1 110 10.2 gph
Note the torque changing, but the speed staying the same. Your speed does not depend on torque. It depends on torque and RPM, or power.
NOW we are getting closer to agreement.
Exactly, which proves the point I’ve been trying to make all along. You decreased the RPM, but the fuel consumption did not decrease. When you raised the RPM, the fuel consumption did not increase. That is exactly my point! You simply cannot say that fuel consumption varies with RPM. Here the RPM changed, fuel flow and power output stayed the same.
Yep.
No need to mention it. So lets say in your idle example that when your engine drops to 600 RPM, you decide to bring it back up to 750 RPM without removing the load. So what do you do? You open the throttle until it hits 750 RPM again. I could then apply even more load to drop the RPM back down again. And then I could open the throttle even more to bring it back up. I could keep increasing this load and then opening the throttle more to compensate for it. I can keep doing this until the engine starts to lug, which anyone who has attempted to drive a car up a hill in too high a gear can attest to.
Now look at what you have in your idle example. More power being generated, more airflow, more fuel being consumed, and the exact same RPM as what you started with before you applied the load.
Fuel consumption varies with power, not RPM.
Berkut when you say that engines are more efficient at low RPMs it’s because most car engines’ torque peak is in the low rpms, correct? So is any engine most efficient at its torque peak?