tomlinson619

was jsut wondering for the DD sake, if i was to switch off the GSR tran to R's, am i looking at a significant difference in gas?

VTEC_PRODUCTION

they are pretty much the same...both are close ratio...

tomlinson619

really? on freeway the R revs higher, and the one im getting also has a 4.7 FD.. so would i still be losing any significant gas?

VTEC_PRODUCTION

no if anything you are getting better mileage

alexisthemovie

gas mileage on honda engines is tuned by throttle position, not rpm.

VTEC_PRODUCTION

which goes back to what i say is that they are pretty much the same...

GoldCoin

iTrader: (1)

higher rpms will put more wear on the engine though

95 integra

iTrader: (2)

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by StayFreshPlan &raquo;</TD></TR><TR><TD CLASS="quote">higher rpms will put more wear on the engine though</TD></TR></TABLE>


Not necessarily.....if this was entirely true, the 4k a GSR sits on at 80mph would have them blown out after 80k. Driving around town at 3k puts much more wear on the engine then sitting at a steady 4500 on interstate, thus its more load dependant. With that in mind, higher rpm at the same speed means you have more gearing, which means you have a greater torque multiplier which means you're putting less load on the engine at a higher rpm....which doesnt necessarily mean more wear.

Dogginator

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by 95 integra &raquo;</TD></TR><TR><TD CLASS="quote">


Not necessarily.....if this was entirely true, the 4k a GSR sits on at 80mph would have them blown out after 80k. Driving around town at 3k puts much more wear on the engine then sitting at a steady 4500 on interstate, thus its more load dependant. With that in mind, higher rpm at the same speed means you have more gearing, which means you have a greater torque multiplier which means you're putting less load on the engine at a higher rpm....which doesnt necessarily mean more wear.</TD></TR></TABLE>

Do you have supporting data or should we all just have a group hug from the warm, fuzzy feeling? LOL

Interstate cruising is certainly easier easier per mile on components than stop and go traffic. A gasoline engine operates at peak efficiency at WOT, where the dynamic compression ratio is the highest. A lighter load at higher RPM will consume more fuel. How much more? I do not have the data. Check the ITR forum, as several people they have run the 4.785:1 FD and 4.9:1 FD.

95 integra

iTrader: (2)

Yep, its that warm fuzzy feeling inside.

Now, lets see what we get....first off, lets define dynamic compression ratio. There are 2 ideas on this. The first one says the DCR is your static compression w/ some variable for how late the intake valve closes (take degrees of the crankshaft on the compression stroke and subtract out how many degrees the intake cam is left open after BDC). Then there is the second idea on DCR which is that you take the first idea and then factor in how much air volume makes it into the cylinder (it might be 4%, it might be 400% if you're boosting...NA cars even have a ram air effect that can cause over 100% fill). This shows how many times over the air is really compressed is at TDC compared to air at atmospheric pressure. So at at partial throttle not much air actually makes it into the combustion chamber (ie it creates a vacuum....ie, at idle your motor pulls a ???). Notice, when you do a compression test (even though the engine is spinning slowly), if you dont hold the throttle WOT, what happens? You get crappy results......why? If you can't figure that out after reading the first bit there....stop now.

Now on to the next step (being that you followed the first), the highest point on the DCR occurs where? At maximum torque, which on your integra is where? Yes, thats right, around 6200 rpms on a stock GSR. So as dogginator said, peak efficiency (the point where your are using the least amt of fuel per hp) occurs at your peak torque point. Now then, at non-WOT ranges, your most efficient fuel/hp point falls (thus at 35% throttle, peak efficiency is at a lower rpm).

Now that we understand that the highest DCR point = max torque point = max/ fuel efficiency we can go on to the next step.

HP = Torque*rpm/5252 Pretty simple, torque does not have a time factor involved....hp does. To put it simply, hp explains how twice as much work is being done if your rpms are double (50ft lbs at 500rpms is twice as much power as 50ft/lbs at 250rpms).

Gearing.....oh god the gearing topic. Gearing takes rpms and turns them into torque. Yes, thats right, I swear. Take for example. 50ft/lbs @ 500rpms going into a 10:1 gear ratio will yield you 500ft/lbs at 50rpms (think about it....10:1 means the input shaft is turning over 10Xs the spin of the output shaft....and what happens to torque)? Now with that being said.....the power output is the same (go back up and reread the part about hp = torque*rpm/5252 and do the math....power is the same on both setups). Thus, gearing only manipulates the way power is percieved.

Now that we understand, hp, torque and gearing, lets move on. Lets say I want to keep my car at 80mph on a flat road. This means its going to take X amt of power to overcome the forces that be (gravity, air resistance, etc) and keep the car moving at 80 (lets assume 200lbs of force at a constant rate). We will do so through a LS transmission and a Type R transmission. On the LS transmission our motor is turning out 3600rpms (or something like that) and on the type R we're pumping out like 4400 (4350 as I recall). Now, being that it takes the same amt of force (at a constant rate) to keep the car at 80, how much torque is being applied from the engine on each transmission? Thats right, the ITR is applying less torque at a higher rpm and letting the gears do their job which is to turn rpms into torque. The LS is applying more torque at less rpms......same amt of power from both....just in differnt forms (ah the power of a simple tool....yes, like a wedge or pulley you talked about in 5th grade....simple and complex machines (a complex tool being 2 simple machines put together)).

Now....as of this point.....I give up on linking all these subjects together because its my dinner time.

At a constant speed higher rpms = less torque being required = more or less wear and tear???

Blahblah718293

wow, i need to go back to school.

ahhem: my nubby .02:

if the engine is spinning at higher rpm, that creates more friction = more wear?

GoldCoin

iTrader: (1)

wow. after reading all that i felt like i was being in a lecture at school.

Sam92Teg

So one engine due to taller gearing uses more torque at lower rpms to acheive the same speed.

While the other engine uses less torque at higher rpms to acheive that same speed.

If more torque is required on internals, then wear increases per revolution, but you have less revolutions.

Conversely, if less torque is required on internals, then wear decreases per revolution, but the amount of revolutions are increased.

Sounds like a wash to me. Just thinking out loud. BTW, we have Honda's, they last forever anyway. My 92 RS has 303K miles and it still performs like new.

mrdeadman

Nice, 95 integra.

95 integra

iTrader: (2)

I'm impressed that people actually read all that crap. Very good thinking Sam92Teg.....indeed from what you said it is a wash. Friction and the force need to overcome it (kinetic energy) have a formula that looks like this. KE = (1/2) (mass X velocity)^2, Thus, if the mass (in this case the force being applied downward on the crank....ie the torque being made) decreases but the velocity (crank rpm/piston speed) increases they will cancel each other out.....being that they increase and decrease proportionally to each other.

Oh but wait, we have oversimplified this. We have just considered force and velocity. What about mechanical loses, wind resistance (on the crank, rods and pistons), engine efficiency and combustion quality? So if the engine is turning more, you have more mechanical loses (wind resistance on the crank, pistons, rods, etc etc). Now we have a higher mechanical loss, so whats going to counter-act it? Engine efficiency is. We previously stated that the engines efficiency is greatest at its peak torque (at WOT) which basically shows how well the engine is turning the energy from the fuel into a mechanical force. This is a much longer formula which you have to figure how much force is being applied downward on the crankshaft vs exerted sideways and upwards on the cylinder walls and head.

Now we need to figure in what type of cylinder pressure (not compression type pressure, but when the fuel actually ignites and creates tons o' force) is being applied. Obviously, less torque requires less cylinder pressure; more torque requires more. We'll basically leave it at that.

So the question is; which one creates more wear. More rpms create more fiction, but higher cylinder pressure and greater load on the pistons/rods/crank also creates more friction (at a rate that basically cancels each other out). So we must move onward to the last step Im going to take. Which is all about the metal. Metal has properties. The most important for this very example is shear strength. Shear strength is the metals capacity to resist shearing. To simplify this we'll just state that the shear strength of metal is expontional (being that under double the load, it doesnt shear twice as fast, it shears 3 times as fast) (its a non-linear fuction as well, so the curve gets stepper the more force is places on the metal). Thus, if I place a 500 lbs of force on the lubricated rod bearing, nothing happens and I can do it over and over again. However if I place 1000lbs of force on the rod bearing I can spin it over 3Xs before its half its original thickness.

Ofcourse then you have to take into consideration the oil/lubricate.....which Im going to touch o' so briefly on. There is a thin layer of oil on all your metal parts, if the metal never touch anything but the oil, then it wears much slower then if its hitting metal on metal. At a greater rpm and lower load, there is more oil and less force trying to squeeze it out bewteen the metal. Thus, at higher rpms (and a lesser load) you have better lubrication and less friction.

Bottom line: Shear force shows us that metal wears faster under higher loads and less velocity then lower loads and higher velocity. Thus, its possible for your engine to wear less at 4k then at 2k (omg, the V8 peeps just dont understand.....maybe this tis the reason their cars that sit at 2k on interstate blow up after 100k and integras last 250k+).

I also hit on trying to explain how/why your car can get better fuel mileage at a higher rpm if thats where the maximum efficiency of the engine lies (thus the reason a type R tranny got me about .5mpg better then the gsr tranny).

Dogginator

The torque peak in a Honda engine usually occurs at a higher RPM than the peak fuel economy. Both occur at WOT, but the peak torque is affected by the intake "supercharging" resonance effect. So the intake effect exceeds the additional friction to achieve the peak torque.

Higher torque at lower RPM vs. lower torque at higher RPM essentially exchanges where the wear is going to occur. Higher torque could lead to more main and rod bearing wear. Higher RPM could lead to more piston ring, valve train, and rod bearing wear.

Regarding the FD vs. fuel economy, check out this link in the ITR forum.

https://honda-tech.com/zerothread?id=1567957

As you can see from that thread, many factors are involved, and 92TypeR's results are counterintuitive. Tuning and the ITR intake manifold are two reasons why the ITR could operate more efficiently at higher RPM and partial throttle than at lower RPM and greater throttle. I'm skeptical of the results.

Maybe for my next 3 tanks, I'll not shift my GSR into fifth gear at all and see how that affects the fuel mileage. I'll report MPG not miles per tank, which is what 92TypeR reported. I have vast MPG data for my car and can make a statistically significant apples to apples comparison.

Here is a plot from Honda of Japan for the JDM Si B18C engine. The lowest line is the fuel efficiency, which shows a peak near the 3500 RPM range. Keep in mind that this is for the engine, not engine plus drivetrain.



Here are more plots from their factbook.

http://www.honda.co.jp/factboo....html



Modified by Dogginator at 12:26 PM 4/29/2007

95 integra

iTrader: (2)

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Dogginator &raquo;</TD></TR><TR><TD CLASS="quote">The torque peak in a Honda engine usually occurs at a higher RPM than the peak fuel economy. Both occur at WOT, but the peak torque is affected by the intake "supercharging" resonance effect. So the intake effect exceeds the additional friction to achieve the peak torque.

Higher torque at lower RPM vs. lower torque at higher RPM essentially exchanges where the wear is going to occur. Higher torque could lead to more main and rod bearing wear. Higher RPM could lead to more piston ring, valve train, and rod bearing wear.

Regarding the FD vs. fuel economy, check out this link in the ITR forum.

https://honda-tech.com/zerothread?id=1567957

As you can see from that thread, many factors are involved, and 92TypeR's results are counterintuitive. Tuning and the ITR intake manifold are two reasons why the ITR could operate more efficiently at higher RPM and partial throttle than at lower RPM and greater throttle. I'm skeptical of the results.

Maybe for my next 3 tanks, I'll not shift my GSR into fifth gear at all and see how that affects the fuel mileage. I'll report MPG not miles per tank, which is what 92TypeR reported. I have vast MPG data for my car and can make a statistically significant apples to apples comparison.

Here is a plot from Honda of Japan for the JDM Si B18C engine. The lowest line is the fuel efficiency, which shows a peak near the 3500 RPM range. Keep in mind that this is for the engine, not engine plus drivetrain.



Here are more plots from their factbook.

http://www.honda.co.jp/factboo....html

Modified by Dogginator at 12:26 PM 4/29/2007</TD></TR></TABLE>

Very nice link, minus the little curves on the front and rear of the fuel curve, it still appears that peak economy is fairly close to the peak torque (looks like 5700 vs 6200 or so). It is also true that turning higher rpms will cause more wear on the valvetrain, however maybe not so much on the rings (higher rpms = less load, less load = lower combustion chamber pressures, which should cause less wear on the rings)....going back to the shear strength property of metal.

I'd also like to point out that in 92TypeRs post (even though it makes it even more true in his particular case), injector duty cycle is how long the injector stays open compared to how long it can possibly stay open on one revolution. The problem with this is that its not linear. When the injector starts to open and is closing, the pintle or disc is allowing less fuel to pass by then when its fully open. Thus, if you are comparing 12% to 24% you dont get twice as much fuel because more time is spent opening/closing on the 12% then the 24%. Take for example, 6% of the time is spent opening/closing the pintle, thus at 12% duty cycle the injector is only fully open 6% of the time while on 24% its open fully for 18% of the time. It is the same as far as milliseconds the injector stays open and thus the reason if you check out hondatas site, they say that their software corrects for this and they have some random *** # that corrects for this and that way in their software a value twice as large is equal to twice as much fuel. Like I said, not that this disproves anything, however it should be taken into consideration.

Dogginator

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by 95 integra &raquo;</TD></TR><TR><TD CLASS="quote">

Very nice link, minus the little curves on the front and rear of the fuel curve, it still appears that peak economy is fairly close to the peak torque (looks like 5700 vs 6200 or so). </TD></TR></TABLE>

You are reading the plot backwards: the lower g/(PS*h) number is better for economy. That occurs in the ~3200-3800 RPM range.

rjgerolaga

all this to save a buck or two. you switch for performance point of view. if your gonna worry about gas. dont bother and just use 87 octane. another useless thread.

95 integra

iTrader: (2)

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Dogginator &raquo;</TD></TR><TR><TD CLASS="quote">

You are reading the plot backwards: the lower g/(PS*h) number is better for economy. That occurs in the ~3200-3800 RPM range. </TD></TR></TABLE>

Damn, I sure am.....oh well, there goes that. You just need to get to the point right around 6k and cruise there......bwhaha.

AW

Real world experience. When I put the ITR tranny into my old GSR, mileage was not affected much. When I installed teh 4.785 final drive it went down alot because it was reving at a much higher rpm for the same speed. Now that I have a GSR again I can't wait to put in my old ITR 4.785 tranny!

motoman

I think everyone here has over analyzed this situation. The simple fact of the matter is that the more throttle a gasoline engine is operated under, the more fuel it is going to consume, per amount of power generated at the crank. If you need 20 hp to maintain whatever speed it is you want to maintain, the engine is going to consume more fuel producing 20 hp at 4000 rpm and 50 percent throttle, than it is going to at 2000 rpm and 0 percent throttle.

When you throttle an engine, it takes energy to create the pressure differential between the the atmosphere and the intake manifold. This energy is essentially a parasitic drag on the engine. Because gasoline has a very narrow a/f mixture range of flammability (relative to other fuels, such as diesel), the power output of a gasoline engine cannot be controlled by simply varying the amount of gasoline burned during each cycle; the amount of air has to be limited as well. By rarefying (throttling) the air going into each cylinder, we have now limited the amount of oxygen in the cylinder. We can now inject less gasoline, and still have it burn correctly.

The lack of throttling is one major reason why a diesel engine is more efficient than a gasoline engine (along with a massive expansion ratio, compared to a gasoline engine). This is why we are seeing all these drive by wire (dbw) systems on gasoline cars now. The whole goal of these dbw systems is to limit the amount of throttle the engine operates under. The dbw system controls the throttle, along with the cam timing, to create an illusion of a linear throttle/power response.

Dogginator

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by motoman &raquo;</TD></TR><TR><TD CLASS="quote">I think everyone here has over analyzed this situation. The simple fact of the matter is that the more throttle a gasoline engine is operated under, the more fuel it is going to consume, per amount of power generated at the crank. If you need 20 hp to maintain whatever speed it is you want to maintain, the engine is going to consume more fuel producing 20 hp at 4000 rpm and 50 percent throttle, than it is going to at 2000 rpm and 0 percent throttle.

When you throttle an engine, it takes energy to create the pressure differential between the the atmosphere and the intake manifold. This energy is essentially a parasitic drag on the engine.</TD></TR></TABLE>
I agree. You are using 0% throttle as WOT, where most would call that 100%.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by motoman &raquo;</TD></TR><TR><TD CLASS="quote">Because gasoline has a very narrow a/f mixture range of flammability (relative to other fuels, such as diesel), the power output of a gasoline engine cannot be controlled by simply varying the amount of gasoline burned during each cycle; the amount of air has to be limited as well. By rarefying (throttling) the air going into each cylinder, we have now limited the amount of oxygen in the cylinder. We can now inject less gasoline, and still have it burn correctly.</TD></TR></TABLE>

You throttle a gasoline engine to avoid detonation with less fuel. Some of the direct injection engine run a leaner mixture with faster kinetics by injecting the fuel after TDC to control detonation.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by motoman &raquo;</TD></TR><TR><TD CLASS="quote">The lack of throttling is one major reason why a diesel engine is more efficient than a gasoline engine (along with a massive expansion ratio, compared to a gasoline engine). This is why we are seeing all these drive by wire (dbw) systems on gasoline cars now. The whole goal of these dbw systems is to limit the amount of throttle the engine operates under. The dbw system controls the throttle, along with the cam timing, to create an illusion of a linear throttle/power response.</TD></TR></TABLE>

The massive expansion ratio equates to high compression ratio, which is thermodynamically more efficient. For a diesel without a throttle plate, the static and dynamic compression ratios are nearly identical if there is no forced induction.

Drive by wire systems on gasoline engines are designed to produce port turbulence to properly mix the gasoline with the air. WOT at low RPM is not always efficient due to the lack of turbulence.

motoman

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Dogginator &raquo;</TD></TR><TR><TD CLASS="quote">

You throttle a gasoline engine to avoid detonation with less fuel. Some of the direct injection engine run a leaner mixture with faster kinetics by injecting the fuel after TDC to control detonation.

The massive expansion ratio equates to high compression ratio, which is thermodynamically more efficient. For a diesel without a throttle plate, the static and dynamic compression ratios are nearly identical if there is no forced induction.

Drive by wire systems on gasoline engines are designed to produce port turbulence to properly mix the gasoline with the air. WOT at low RPM is not always efficient due to the lack of turbulence.</TD></TR></TABLE>

A leaner mixture does not always mean detonation. In fact, detonation has nothing to do with the fact that it is lean. Detonation occurs with lean mixtures because a lean mixture burns slower than a stoich mixture. Because it burns slower, the mixture has more time to heat up. This eventually leads to a spontaneous combustion of the mixture (i.e., detonation).

Even if throttling was used to control the mixture ratio, and hence detonation, a gasoline spark ignition engine could only operate in a a/f range of about 10:1 to 18:1, which is not enough range to control power output. Of course with direct injection, and other stratified mixture techniques, you could get an average a/f leaner than 18:1 (e.g., civix vx, ~22:1 a/f), but you will still need to throttle the engine.

It is not the compression ratio that dictates the thermodynamic efficiency, but the expansion ratio that dictates the efficiency. The compression ratio is simply a consequence of the expansion ratio. The more you allow the combustion gases to push on the piston, (i.e., greater expansion ratio), the more mechanical work you have extracted from the expanding gases. The compression ratio and the expansion ratio are not necessarily the same in a non-forced induction engine. Engines using the atkinson cycle have a greater expansion ratio than compression ratio. The effective compression ratio can be controlled by valve timing events. So, for example, if during the intake stroke, I wait until the piston has traveled half way back up before I close the intake valve, I have effectively cut the compression ratio in half, compared to the static compression ratio. But now, during the expansion stroke, the valves remain closed for the entire downstroke. Because the cylinder was only half filled, but now it can expand to twice the volume, the heat of combustion has now done more work than it would have done, had the compression ratio been the same as the expansion ratio.

Because limiting cylinder filling also has a detrimental effect on power output, we could actually use this to our advantage, as a way to limit power output, sort of like throttling, but without the mechanical loses. So, with less compression ratio, we could run less throttle, and still get the desired lack of power. But what happens when we need power? If we were not throttling, but then revert back to otto cycle, there would be a sudden increase in power. This is where dbw comes in. When we are switching between modes, we also apply or reduce throttle to make the transitions smoother.

Dogginator

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by motoman &raquo;</TD></TR><TR><TD CLASS="quote">I think everyone here has over analyzed this situation. The simple fact of the matter is that the more throttle a gasoline engine is operated under, the more fuel it is going to consume, per amount of power generated at the crank. </TD></TR></TABLE>

Enough said. The other details are going to confuse people. Integra engines run the Otto cycle. Restricting the intake with the throttle will reduce efficiency.