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Subaru Forester MPG-Real World Numbers

174 messages, Last post on May 13, 2009 at 4:45 PM
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Averaging about 20 miles per day or about 7,300 miles per year. Mostly short trips, but I make my commute a bit longer to allow the engine to warm fully. Steady diet of 92 octane from Super America. Reformulated fuels and winter oxygenated blends (10% ethanol year-round). Shifts almost always between 3-4k rpm and not afraid to let the rpm's go higher if/when necessary. best: 26 mpg (all highway at 75 mph) worst: 18 mpg (winter) average: 21 mpg Average mpg up until this past winter was 1 mpg better than my Outback. Partial thanks to winter fuel, no doubt. -Brian |
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(my first Subie and was prior to my '04 F-XT) Averaged 26 miles per day, or about 9,500 miles per year. 87 octane from Super America. Reformulated fuels and winter blends both with 10% ethanol. best: 28 mpg (highway trip) worst: 17 mpg (again, winter) average: 22 mpg -Brian |
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Never did better than 21mpg, whether I go 65 or continuous 85mph. - 80 miles per day 95% highway. - - Use 87 octane, with that lousy ethanol. - - Tried all brands of gas - no difference. Tried Octane booster to a tank of gas, new special, copper plugs - drove 65-70mph - 20 mpg. Winter milage is even worse. My '92 Legacy LS Wagon got 25 and it's >= 400 lbs heavier. ( It's still on the road BTW ) How are you achieving those type of numbers? Dom
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Replying to: cinemod1 (Jun 28, 2006 6:31 pm) -Brian |
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I've had my 01 Forester (S+ Automatic) for 102K now, and am getting 24 MPG. I did notice that mileage improve markedly after ~7500 miles or so. |
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Replying to: cinemod1 (Jun 28, 2006 6:31 pm) Not by driving 85mph. Speed kills gas mileage. The sweet spot seems to be about 45-55 MPH for most cars. The FXS isn't going to win any prizes for minimum air resistance either. It's too bad the EPA doesn't do an experiment with the MPG test by doing the standard test for highway speed, and then repeating the test at a higher speed, say 80-85 mpg. It would be interesting indeed to see the MPG difference.
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Replying to: growler5 (Jul 11, 2006 6:49 am) The air resistance is proportional to the speed cubed so it rises strongly with increasing speed. But this doesn't mean that the proportionality factor is 1. Link.
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Replying to: jim314 (Jul 28, 2006 6:46 pm) That's actually the power required to balance the force of air drag - which varies as the square of the speed. tidester, host
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Replying to: tidester (Jul 28, 2006 7:56 pm) That third power bothered me, but I didn't think it out. In the past the usual reference has been to the retarding "force" of air resistance, which for simple physics problems I thought was proportional to speed, so power (speed times retarding force)should be proportional to the second power of speed. But in air, and with turbulence, etc. the retarding force must be proportional to the square of the speed in order to give a third power dependence of the power. As I recall the old statement was that the "air resistance" at 70 mph was twice that at 50 mph, which fits with the force being proportional to the square of the speed since (70/50)^2 = 1.4^2 = 2.0. The power (metric unit of power is the Watt, where 1000 W = 1 kW = 1.34 HP) is the energy consumed per unit time, but what we usually want to know is energy consumed per unit distance travelled. In Europe this is the litres of fuel consumed per 100 km travelled, but in the US we use the reciprocal, that is, distance travelled per unit of fuel consumed (mi/galUS or simply "mpg"). FYI the conversion between the European and US measures of fuel economy is y mpgUS is related to the equivalent z L/100km by yz = 235. That is, you divide whichever one you have into 235 to get the other one. But the point is that to get the extra vol of fuel consumed per unit of distance caused by driving faster we should divide the vol of fuel consumed per unit time by the speed (distance/unit time). So it would seem that the extra fuel per unit distance required to go faster would be proportional to the square of the speed, or maybe proportional to the square of the increment in speed, or maybe proportional to the difference in the squares of the speeds. You'd need to do some algebra to get it exactly straight. But as I said the proportionality factor is presumably much less than one. The formula given in the link could be used to calculate the factor for a given vehicle. But for any given vehicle, the best way to determine this would be driving different steady speeds (40 mph, 50 mph, . . .80 mph) and recording the instantaneous mpg values (or L/100 km values) shown on the fuel mileage computer display. I think these are pretty accurate, aren't they? Then you would just subtract to get the benefit from driving slower and the penalty from driving faster than the posted speed. To get the actual fuel and money saved you'd express the fuel use as gal/mi not mi/gal as we usually do in the US. |
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I don't have a car with a mpg readout, but if anyone wants to do this here is some madeup sample data to show how it would be done, assuming level ground and the tranny in top gear: Speed(mph)..mi/galUS.....galUS/100mi....Fuel-Std 70mph* 45..........35...........2.86...........-0.71 gal/100mi 50..........35...........2.86...........-0.71 55..........34...........2.94...........-0.63 60..........32...........3.13...........-0.44 65..........30...........3.33...........-0.24 70..........28...........3.57............0.00 75..........26...........3.85............0.28 80..........23...........4.35............0.78 85..........19...........5.26............1.69 So if you want to know how much fuel you'd save on a 500 mi trip at 65 mph as opposed to 80 mph, Fuel saved = (0.78-(-0.24)) x 500/100 = 1.02 x 5 = 5.1 gal |
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