MAX Update No. 70: MAX's Theoretical Max

| 4/4/2011 11:44:34 AM

What a miserable month, as is typical of the end of a Western Oregon winter. Rain rain rain snow rain, definitely not road test weather. But thanks to the HBFCI I've been able to learn a thing or two in the relative comfort (it's not raining indoors and that's good enough for me) of my well ventilated shop.

Two updates ago (MAX Update No. 68: Idle Speculation) I reintroduced the Honey Bear Fuel Consumption Indicator, and did some engine efficiency comparisons between MAX and my store-bought minivan. I was curious what we'd save if we shut our engines down at idle, plus I was curious what MAX's minimum fuel consumption might be. Well, I learned that it takes a fair bit of fuel just to make an engine run, and to nobody's surprise, as engine size and horsepower increase, so does the fuel required, even if the engine is just sitting there idling. 

But we're interested in fuel economy here — the fuel it takes to get someplace — so fuel burn at idle isn't the main issue. A more important question is, how much fuel does it take to run the engine at cruise? 

So with the van idling in park, I gradually increased throttle until my ScanGauge showed I was burning 1 gallon an hour, and noted the rpm. I did it a few times and 2150 looked about right; I double checked by holding 2150 rpm and sure enough, the gallons per hour nibbled around between 0.98 and 1.02 so I'm calling it 1 gph.

The sobering thing about this is, this van goes 54 miles an hour at 2150 RPM, so even if the van had no aerodynamic drag and no rolling resistance and no transmission losses at all, the best mileage it could ever hope to get would be 54 miles per gallon. There is no car available — not even in theory — that is so light and streamlined and friction free that it could get 55 mpg with that engine and gearing.

Interestingly, it takes less fuel for each turn of that engine at 2150 than at idle. Idling at 760 rpm on 0.43 gallons per hour, it takes about 1.21 ounces of fuel to turn the engine over 1000 times (you can trust me or do the math: gph divided by 60 gives gallons per minute, divide that by rpm for gallons per revolution, times 128 converts gallons to fluid ounces, times 1000 gives ounces per thousand revs, so 0.43 / 60 / 760 x 128 x 1000 = 1.2070... and since my measurements aren't likely to be perfect, I've rounded to the nearest 1/100 of an ounce). At 2150 RPM and 1 gallon per hour, it's 0.99 ounce of fuel per thousand revolutions (1 / 60 / 2150 x 128 x 1000 = 0.9922...) so for a very simplistic estimate, that engine is about 20% more efficient at cruise than at idle. That's what we want, of course, we want best efficiency while driving, not while idling, and presumably the folks at Kubota also aim for best efficiency in the engine's normal operating range.

jeff dean
5/12/2011 10:51:15 PM

Jack, You have got a great opportunity to change to a taller gear in Max (at least for testing) because you started with such a short tire. I know you will have to make a few adjustments, but the knowledge will be invaluable, and now looks like the time to do it. If you keep going to a taller tire until you get the rpm down as low as possible, then improvements in aerodynamics will give it drivability along with fuel economy. With that knowledge you could then work on the proper gear ratio for the tires you have to run. Ideally Max should have a very tall differential gear with five or six speed transmission with a 1:1 high gear because overdrives are less efficient than the 1:1. I know that's a lot of change and experimenting for Max but for Max II or Max III or Max XVII in my shop...well I think it would be great. You do all the work, and we get the benefit. Actually I'd love to physically help if we could figure out how to get me to Oregon or Max to Ohio. Keep up the good work.

Jack McCornack
5/6/2011 11:34:26 AM

jeff_58, you may be exactly right. You're definitely right in concept: the most efficient RPM for a given power output moves up as power demand goes up, down as power demand goes down. As far as what RPM is most efficient for the (as yet unknown) power needed to cruise, I can't get any info from the manufacturer except the peak efficiency curve. But if best RPM for 0 horsepower is 900 or so (idle) and the best for 32 horsepower is 3000, then best RPM for the (estimated) eight horsepower I need for freeway cruising could well be 1600 RPM. I could be 2150 too, and it'll take much more testing to and I are both working from educated guesses. Anyway, if I could find a taller final drive gear set for MAX's differential, I'd sure try it.

jeff dean
5/2/2011 8:56:34 PM

Jack, If 2150 is the most efficient rpm,why not set the idle at 2150? Maybe because it's the most EFFICIENT rpm for producing maximum power,but you don't need maximum power for just idling. You need maximum power for acceleration, not idling and not cruising at 55 mph. With max's light weight and aerodynamics, you should only need about half the maximum available hp from the diesel. That amount of power should be available at a much lower rpm than the maximum efficient 2150. 1600 rpm is not the most "efficient" rpm in terms of fuel per 1000 rpm, but it would be more efficient in terms of miles per gallon. The difference of rolling resistance between 35 mph to 50 mph would not be that much but the air resistance would be significant. If you could simulate that extra drag at the slower speed (lower rpm) you could determine the best rpm for fuel consumption and still have drivability. I thought that's what you did with the first non-aerodynamic version. You said you could cruise around at 35 mph in high gear (with the same drag as new version at 55 mpg) at low rpm, very fuel efficient.

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