Everett writes in:
I know that each motor, transmission, and car combination has optimums areas of performance for mileage. All the Jetta drivers that swear by idling in 5th gear are a testament to that. With some of the newer V-8 motors, however, finding the sweet spot is a little more problematic. Take for example a GMC Yukon XL, paired with a 5.3L V-8. Because GM designed the motor to shut down four cylinders whenever it could to increase fuel economy, the motor does just that when coasting, during braking, and for short stretches where the motor can maintain the desired speed (provided the driver holds a steady pedal). Is it possible that because the motor would produce more horsepower under the V-4 mode at a higher rpm/road speed, that there is a sweet spot of efficiency at a HIGHER than expected speed? Having a power curve for the V-4 mode of the motor would be a good starting point to know if there is a point where the V-4 mode is revved higher and produces enough power to better maintain highway speeds, but saves gas because four cylinders are shut down. Any ideas if that exists for the actively managed motors out there and what speed that would be?
Everett, you ever heard of OCD?
It’s more simple than that; the most economical speed is a slower one.
If you had two GMC Yukon’s, one powered by a V8, and the other powered by a V-Twin (if that were possible), doing a steady 50mph would require the same POWER from both motors.
There would be differences in fuel consumption for both because of their design; mass, friction etc, but the power at the tail-shaft would be the same.
Going back to cylinder shutdown, fuel is consumed to produce the required POWER regardless of the configuration of firing. The very small fuel saving comes from not compressing the charge of air in the cylinder that won’t be fired.
If you were to look at the fuel consumption curves in various deactivation states, I’d be sure the deactivations are calibrated from those curves already, so there is no “sweeter” spot.
I think that slower is almost always going to be more efficient. Less drag and friction. As long as you’re going fast enough not to lug in top gear.
I’m skeptical of the ‘go faster to save fuel thing’. Maybe with carbs that was possible, but I think that modern motor controls have made that situation obsolete.
I get 24MPG in my 2006 Silverado Crew Cab….with the 5.3 and a 4-speed at 60 mph. GM V8s are just as efficient as other automakers 3.5 V6s.
And GM Trucks will get THE BEST fuel economy next year from Di an other tweaks. No automaker will match it.
Really? hypermiling a Yukon XL?
You want better mileage? Trade it in for an Odyssey. With the money saved, buy an Insight.
With the gas savings, rent a pickup for when you actually need to haul something > 5,000 lbs.
Speaking to the power curves, shouldn’t they be roughly equivalent regardless of how many cylinders are being used because it’s the same camshaft that’s timing everything. That is to say that while 4 cylinders will clearly make less power for a given rpm, the rpm at which the power and torque peak should be about the same irrespective of how many cylinders are firing. I’m no engineer, so please set me straight if I’m way off here.
That may be true, but the power necessary to maintain a given speed will occur at different rpm. It may well be that a higher revving 4 cyl with a more open throttle will be more fuel efficient than a V8 at a lower rpm with a more restricted throttle. When those situations occur, you want the engine operating as a 4 cyl. I would hope that GM already determined those points as part of developing the cylinder deactivation program. If they didn’t they’re bigger idiots than I thought.
A throttled gasoline engine is always most efficient running at wide open throttle at the torque peak engine speed, because the lack of pumping losses and the fact that the torque peak implies highest power output per revolution.
Note that I said efficient; that’s different than economical.
A small four running hard under those ideal conditions is optimal, but leaves too little reserve. That explains the Prius; it can tap its batteries for the reserve power.
So theoretically, your Yukon should run most efficiently at the torque peak, with a large throttle opening, in V4 mode. The problem is that the software doesn’t allow that; it undoubtedly shifts to V8 mode above a fairly low throttle setting.
Given the complexity of modern engines, with variable valve timing, etc., it’s probably difficult to ascertain an answer to your question theoretically. If you have an instant mpg readout, use the trial and error method.
I would actually expect the optimum speed for gas mileage to decrease in cylinder deactivation mode. When running as a V-8, the throttle is more restricted at a given speed, so opening it a little may give more speed for little increased fuel flow.
In 4 cylinder mode, in order to hold the same speed as in V-8 mode, the throttle is already more open, and the engine may need to run at a higher RPM in order to supply the required power, where in V-8 mode it could still be loafing along. As an example, witness the Corvette with its excess power and tall gearing: it gets highway mileage comparable to much lower-powered cars.
My 1000cc (fuel injected) sport bike is a similar story. Because the engine is so oversized and overpowered for regular cruising, at lower speeds most of this winds up as wasted energy. 75 MPH has the throttle just barely cracked from idle, even though it’s on the verge of lugging in top gear (4000 RPM on an engine that redlines at 10500). As a consequence, my optimum gas mileage seems to happen at at least 80-85 MPH (it’s hard to go much faster over long distances without things getting expensive), despite the fact that a motorcycle’s drag coefficient is absolutely dismal.
I recall that the owners manual on my Folks 300SL gullwing 1955 with a 325 rear end had a chart in it that showed the most economical speed in a no wind condition by charting MPG verses speed in each gear
My perhaps flawed memory recalls this chart as having its peak at between 45 and 50 mph in 4th and the MPG at that speed near 30
Strange that charts like this are no longer provided when the were there in some cars in the mid 50s when Ethel was .25 a gal
Cheers
Windy
since no one has brought this up by name…
http://www.scangauge.com/
MPG readouts should be standard (ideally required) on every car. Better yet you should be able to input your gas cost and have the readout show an instantaneous $ per mile.
Is there a point to hypermiling a big V-8?
Yeah, I think the only reliable answer will be empirical. Conventional wisdom is that the tradeoff between engine thermodynamic efficiency vs. aerodynamic drag meets at around 55 mph, but who’s to say a priori where the inflection point is with a variable displacment V8 Yukon XL? If you’re OCD and you really want to know, run a series of experiments on a straight, well known sequence between highway exits, and measure the fuel used at the pump. Don’t trust the trip MPG computer. 50, 60, 70, 80 mph.
“that the torque peak implies highest power output per revolution.”
Nope
“that the torque peak implies highest work output per revolution.” is a statement that a high school science student would probably find acceptable.
How about Kaiser’s “Power On Demand”? This commercial makes it sound like they had Variable Displacement in 1954. Kinda reminds me of the hype surrounding the Volt
If we’re going to get precise about hypermiling, in my experience, I get by far the best mileage (nearly 40 mpg) in my ’99 Accord 2.4 liter 5 speed by quickly accelerating (in 4th) to 70 and coasting back down to 60, and repeat ad infinitum. I could probably do even better at lower speeds because of less wind drag, but the rationale is that pumping losses are minimized, and air drag isn’t nearly as great as if I were keeping the throttle open to the same degree but going at a steady speed in fifth.
I would note that I get somewhere around 35 mph if I can drive at a steady 45-maybe50.
Everyone’s making it too complicated.
Baring a stupendous stuff-up from GM, the motive power required to maintain speed X requires fuel flow Y. You’ll get differences that result if you swap in a different motor (V8, I6, I4, V2) due to the way they develop power per unit of fuel.
Rather than swapping in a different motor, GM have the gimmicky deactivation. If you turn a V8 into a V4 delivering the same amount of power (to maintain the same speed) then GM have already done the work in their engine management unit.
So my advice is; drive at the minimum speed to enable V4 operation to save the most fuel. You’ll find that minimum is 0 mph.
@ Greg Locock
I’d like to hear from my friend Maxb49.
Paging Maxb49.
“How about Kaiser’s “Power On Demand”? This commercial makes it sound like they had Variable Displacement in 1954.”
I think that’s referring to the supercharger option Kaiser offered that year. They had no V8, so they stuck a blower on their six (by Continental, right?) in an attempt to compete.
Hypermiling a Yukon. Sorry, Yukon XL. Next up, we’ll discuss the best option for fitting a Class-5 hitch to a Mini Cooper…
@ Jdizzle:
Yes, It was supercharged. I was just reading up on that. The last few days I have been reading up on the independents. Credit Paul’s Curbside Classic Packard and my own “Curbside Classic Experience”. Yesterday, I walked out to my car in the parking lot and parked next to me was an white, Late 30’s Auburn 851/852 Boattail Speedster. In just an average, Boca Raton, FL shopping mall parking space. Gotta Love it. You know I was snapping pics with my iPhone ;-)
SherbornSean: The Odyssey uses cylinder deactivation as well.
Greg Locock :
“Nope…that the torque peak implies highest work output per revolution.” is a statement that a high school science student would probably find acceptable”
You’re saying two different things: “Nope” implies my answer is flat wrong. Please correct me, if so.
“a high school science student would find acceptable” implies something else. I learned during my in high school days (not actually in school) that the torque peak is when BMEP is the greatest, and thus is when an engine generates its greatest torque. It has so far withstood the test of time. But I’m never too old to re-learn.
Edit: Wiki seems to agree with me:
http://en.wikipedia.org/wiki/Mean_effective_pressure
BMEP is when an engine makes its greatest power irrespective of its rotational speed/maximum overall power, and it always corresponds to maximum torque.
It’s really the best way to understand torque, which many folks find difficult to grasp. It’s at the speed at which each power stroke generates the maximum potential power, i.e. torque; at higher engine speeds, torque drops, but horsepower is greater, due to higher engine speed.
It is flat wrong. The work produced by a torque is T*2*pi per revolution, in sensible units. At max torque T is maximised, hence the work per rev is maximised.
Note that this tells you NOTHING about power, because power per revolution is a meaningless term.
I don’t know why you misquoted me, your misquote makes no sense.
I wrote
“that the torque peak implies highest work output per revolution.” is a statement that a high school science student would probably find acceptable.
P71_CrownVic,
Is that a tank measurement or are you reading something instantaneous from the dash? Our Aeorsaur often told us “stretchers” as Mark Twain would call them but the tank fills tell the truth.
You may find this article interesting:
Edmunds on the Silverado, Titan and Tundra
I am not smart enough to explain it, but on a dyno chart the H.P. line always crosses the torque line at exactly 5252 rpm.
Fuel economy vs. speed is a complicated subject and every vehicle has it’s own peculiarities.
Have a look at this chart which summarizes some of the published studies:
http://www.theoildrum.com/uploads/12/mpg_vs_speed.jpg
Modern vehicles generally have a wider nearly flat point on the curve, but for any vehicle there is some speed above which fuel economy declines. That optimum speed has ranged from 35 to 60 mph in various tests I’ve looked at. The only way to know what the Yukon XL w/variable displacement would do is to test it. However, as a really big vehicle pushing a whole lot of air out of the way, I would expect the Yukon’s fuel economy to start dropping off at a lower speed than the “average”. As speeds increase, aerodynamic drag becomes the dominant fuel efficiency determinant, and a Yukon XL has a whole lot of aerodynamic drag.
BTW, the fuel economy benefits of so called variable displacement designs are relatively minor. Measurable and real, but minor. The Honda Odyssey is one of the few vehicles available with either a conventional 3.5l V-6 or a variable displacement version of the same engine. In EPA test cycles it improves from 18 mpg to 20 mpg average with the fancier engine, or just over 10%. Not nothing, but not earth shattering either. A six speed automatic transmission with appropriate gearing might return a similar improvement over the Honda’s now aging five speed.
PeteMoran, in cylinder deactivation mode, exhaust gases are still being compressed in the dead cylinders because of the design. Somehow this causes smoother operation, and prevents oil from being sucked into the combustion chamber.
@ mfgreen40
That’s because;
Power (hp) = Torque (ft-lb) * RPM / 5252
Greg Locock, I don’t know why you misquoted me, your misquote makes no sense.
How could I misquote you, when I cut and pasted your writing? I suggests that you’re having a hard time distinguishing between fact and your chosen interpretation of it. Here it is again:
Nope
“that the torque peak implies highest work output per revolution.” is a statement that a high school science student would probably find acceptable.
All I did was put three periods between your “Nope” and the rest of the line. That’s the acceptable way of separating two lines of writing within one quote.
That somehow makes it a misquote? Nope.
Now what else are you having a problem with?
And I just remembered why I make it a point not to comment, unless its for a very good reason.
Good idea. Your use of power per revolution is wrong.
We can argue about the formatting thing another time.
For the poster questioning hyper miling a V8…I drove my previous ’03 Durango 2WD/5 speed 4.7V8 on a 150 mile trip and driving 50~55 MPH I averaged 27 MPG.
Doing the same trip in a Grand Caravan 3.8 V6 I can average 31 MPG.
Yes, it can be done if you are willing to drive 50 MPH.
Is there a point to hypermiling a big V-8?
Yes. In fact, moreso than hypermiling an economy car – much greater potential fuel savings.
“Yes, it can be done if you are willing to drive 50 MPH.”
I’ve looked at that. For our lowest-mpg vehicle, a minivan, the fuel savings are about equal to saving, at most, $5/hour (at 55 vs 67, 28 vs 24mpg) over the course of a trip (fuel cost saved divided by extra hours). If there’s more than one person in the vehicle (we typically had 6), the savinger per person per hour are on the order of $1.
Balanced against that, on a 2720 mile trip, the extra time certainly puts you on the road through an extra meal, most likely an extra two meals and possibly an additional night in a motel.
slateslate: The ScanGauge won’t work with “variable-displacement” engines as it cannot tell when the engine is in reduced cylinder operating mode.
John Horner: a 10% fuel efficiency increase is pretty major to the automakers and their CAFE ratings.
I drive a 4×2 2002 Ram 1500 with the 4.7L V8 and a 5spd manual and quasi-hypermile it. My long-term average economy including hauling exceeds the original EPA-rated highway economy (which has since been revised significantly downward) at about 19.1 mpg. I took a 450-mile trip on mostly 2-lane roads with lights driving 55-60 mph and averaged 21mpg.
I think that the multi-displacement gimmics are the automaker’s way of allowing the vehicle to “hypermile” itself without requiring intelligent inputs from the driver. Steady-state cruising, especially at highway speeds, is sub-optimal for fuel efficiency because of the typically low throttle opening. All multi-displacement engines that I know of use throttle-by-wire so the ECU can open the throttle wider when it shuts down cylinders to maintain an even power output. Throttle openings approaching WOT, but closed enough to prevent the ECU from entering open-loop fuel enrichment, are the most efficient due to reduced pumping losses. Multi-displacement engines further will typically remove a valve per cylinder (at least the Hemi does) so there is no through-flow to further reduce pumping losses.
If you want to hypermile a vehicle that has such an engine, you will probably find yourself *never* entering a scenario where the engine shuts off cylinders. You will instead accelerate at 80% load using all 8 cylinders until you reach a desired speed (like 65 for the highway), and then shift to neutral and/or shut off the engine until you need to accelerate back to speed again. I don’t know, maybe it’ll go to 4cyl operation while idling in neutral, but I doubt it. The “accelerate-coast ad infinitum” as one poster put it, or pulse and glide, is more efficient than steady state cruising.
@ MBella
Thanks. I was a bit unsure how the system worked.
Wackypedia had a useful explanation for me.
“Is it possible that because the motor would produce more horsepower under the V-4 mode at a higher rpm/road speed, that there is a sweet spot of efficiency at a HIGHER than expected speed?”
If such a pronounced ‘sweet spot’ really existed, don’t you think the manufacturerers, especially those attacked for their poor fuel economy, would have been bragging about it in their manuals?
If you want better MPG, the recipe is simple;
–make sure your tires are fully inflated (I would actually raise them 1-2 psi above the recs, since most people check tire pressure about once a month and if it is perfect the day they check, it goes down from there, so on average..)
–slow down. My 7 gets 24-26 MPG at 50-60 MPH, of course I always go 80 and get 22. Big deal, saving a buck and losing HOURS..
–no jackrabbit starts and burned rubber,
–no wasteful braking, anticipate stops and slow down first!
and the like. I am sure most here know all about the above.
We thoroughly map out the engine’s efficiency in V4 and V8 mode so the efficiency at any speed and load is clearly understood. This efficiency map is part of the engine’s calibration, and V4 mode is enabled when it will result in an improvement in fuel economy. In general, engine friction is the biggest enemy of fuel economy on an engine basis, and friction increases significantly with engine speed. As a result, best fuel economy is obtained at lower engine speeds, which is one of the reasons you see V4 mode most prevalent at lower engine speeds. Good topic Everett. Thanks Jordan L.
Tom Read
GM Advanced Technology Communications, Powertrain
Multi-displacement engines further will typically remove a valve per cylinder (at least the Hemi does) so there is no through-flow to further reduce pumping losses.…
Not sure what you mean here. When a variable displacement engine reduces the number of powered cylinders, the de-powered cylinders have all valves closed, no? The energy used to compress the air inside the cylinder is simply transferred back to the piston on the downstroke, or that’s how I thought it worked.
P71_CrownVic
I can get 87 mpg in my 08 Silverado Ext Cab going down hill with a tail wind towing 6,000 lbs of car and trailer. I also have the 5.3 with the variable cylinder deactivation. Stating a mileage you can get in a static state does little. What is your true MPG in your 06 Crew Cab from daily driving? – it has to be more like 16-18 mpg.
@Paul Niedermeyer:
The maximum torque of an IC engine occurs when the bmep is highest, as you said. For each engine, at WOT, there is an rpm at which this occurs.
In the out-of-date US system of measurements, most people write torque as ft-lbs, although in more engineering circles it is lbf-ft. As Mr. Locock tersely puts it, the work or energy (they are the same here) is maximum when bmep is maximum, and is the torque times 2 times pi per revolution. For engineering purposes, this is expressed as ft-lbf (the exact reverse order of torque, so that people can converse with one another, and know whether it’s torque or work we’re speaking about in a certain context)
There is no power involved – power is the rate of doing work, so to compute that one needs to know the rpm at which the engine is operating.
The other red herring that always comes up in these discussions is the magic number 5252. It is a constant that falls out of the calculations in the US fps system. Nothing to do with any particular engine characteristic. In the US system, the acceleration due to gravity is initially 32.17 ft/sec/sec, but 9.81 metres/sec/sec in SI units. Same thing, different measurement systems.
Paul Niedermeyer :
October 22nd, 2009 at 1:09 am
And I just remembered why I make it a point not to comment, unless it’s for a very good reason.
But, but…best and brightest!
I have noticed on my 2008 Impala AFM 3900 and nearly all AFM V8 trucks that I have rented the best mileage comes at a steady 62-65 MPH in AFM mode on a flat road. I have seen over 32 MPG on several pure highway trips with the Impala and 24 with 08 and 09 extended cab Chevy 5.3 pickups and that was with the older 4 speed automatic!
Didn’t Cadillac have a variable displacement engine in the 80s?
Its easier to get more power out of smaller engines than to get less out of larger ones. Power to weight. All V engines will have double variable timing gear and chains, extra cats, O2s…on and on. Dead cylinders will always have friction and other losses.
Smaller I engines with compressorS wont be carrying all the dead weight and complication around. And Detroit needs to make a GLOBALLY COMPETITIVE PRODUCT. And I want to see HON and BWA stock go up.
A V8 will NEVER have superior economy because of several packaging problems.
FIRST – in order to maintain acceptable NVH levels, manufacturers are required to very heavily weight the crankshaft, and because the V8 has unbalanced firing between the cylinder banks (producing the distinctive and much-loved V8 “burble”), they’re required to add balance shafts as well. All of this saps the efficiency and power delivery of the motor.
SECOND – V8s tend to be very large in displacement, which is never good for fuel economy – even if you can shut down half the cylinders of the average V8 motor (we’ll say they land around 5 liters for average) you’ve still got a relatively large 2.5L operating.
THIRD – V8s tend to be packaged as RWD vehicles. This requires power to run from the front all the way to the back, increasing drivetrain loss and thus decreasing efficiency of power delivery and overall fuel economy.
So there you have it. Want fuel efficiency – go with small displacement, inherently smooth motors driving the front wheels.
What’s that you say? By adding a balance shaft Honda smoothed out their 1.5L straight 4 motor in the mid-90s Honda Civic which weighed only 2000 lbs, and delivered 45 mpg without hybrid power, variable valve timing or any other trickery?
Speaking of which…GM discontinued the variable displacement version of the 3.9L V6 that was in the Impala. Does anyone know why? Was it just slow sales? I would have thought that it that feature made a lot of difference in mileage, all else being the same, it would be either an easy sell or the default option if you were getting the 3.9L.
@ cretinx :
Lets look at you points:
1: You say V8s are bad because they require extra weight to be smooth. True, that cross plane crank V8s require balance weights on the crank shaft, but they do not require balance shafts that you mention. But, you then you suggest an inline four? An inline four has even more vibration problems than a cross plane V8. So what are you even suggesting.
2: Yes large displacement engines tend to use more fuel at a given power level than a smaller displacement engine. But what does this have to do with V8s? Want about small displacement V8s?
3: What does the distance have to do with power loss. Lets say you have an engine hooked up to a dyno through a driveshaft. Run it with two different length drive shafts. Would the longer drive shaft require more fuel during steady state conditions? Why would that be the case? Where does the power go?
So yes, a small light car with a low power engine gets better fuel economy than a large SUV. Thanks for pointing that out. I am sure nobody figured that out yet.
chaparral66 :
It’s not the length of the driveshaft. It’s the turning of the power through 90 degrees that’s so inefficient. Around 30% of torque is lost in that function. That is why especially thick oil is used in the rear differential. It is expected to handle the occasional high temperature of a hill climb although some towing situations may require the additional use of an oil cooler radiator.
In a pickup the driveshaft may be on the same plane as the wheel axle halfshafts allowing the use of a helical bevel to drive the Final gear but in most vehicles it enters lower than that so that the transmission tunnel through which it runs can be lower – for the benefit of rear seat passengers.
This entails the use of a much more complex geartooth shape, the Hypoid, which exacts the roughly 30% torque penalty just mentioned.
Generally hypoids are to be avoided wherever possible which is why the more powerful race cars are mid engined allowing their engine crankshafts to be parallel to the wheel axle. When this is done Parallel Spur gears can be used. They have a worst case coupling loss of about 7% gear to gear.
I would expect road cars using this configuration to employ Parallel Helicals in order to cut down on gear whine and “cogging”.
Of course, I cannot leave without broadcasting my evangel that the very best way to use a car engine is not to couple it to the wheels at all.