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Auto/Motor Diy measurement of car efficiency

  1. Jun 18, 2009 #1
    It seems to me a typical gasoline car has wastes about 95% of the gasoline energy before any rotating energy gets to the wheels. Here's my reasoning.

    I once measured my cars efficiency (an old Renault 5).
    I drove 100 km/h (28m/s) on a flat and level freeway, with no wind, and set the gears in neutral. It took the car about 30 seconds to slow down to 90 km/h (25m/s). The car weighs about 900kg.
    So we have E0=0.5*m*v2 = 353kJ and E1=281kJ. The car lost 72kJ in 30 seconds or 2.4kW
    So it takes just 2.4kW to keep a small car cruising at 100kph on a freeway. The stated gasoline consumption of that car is about 1 liter/18 km at 90 kph so 1.3 ml/s of gasoline. Gasoline has ca 32MJ/l energy content, so 1.3ml/s is equivalent to 44kW.
    The system efficiency of a car cruising on a flat freeway is therefore about 5%; I define system efficiency as the power that gets applied to the wheels divided by the gasoline energy content used per second.

    Another way to calculate this:
    I lost 3 m/s in 30 seconds, so my deceleration is 0.1m/s^2. The car weighs 900kg, so F= 90N. If I had applied 90N, the car would have stayed at speed. 90N * 28m/s = 2.5kW. This gives the same figure.

    Am I making some big mistake in my reasoning?
  2. jcsd
  3. Jun 18, 2009 #2


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    Not quite the correct method, that's the average resistance integrated over speed.
    A better method is to find a gentle downhill slope if the car runs down the hill with no input power (ie. foot off the gas) at constant speed then you can work out the power you are using to overcome engine friction + aerodynamic friction.
    From the slope of the hill you know how many vertical metres you travel in 1 forward km and you know the speed in km/h you can work out the vertical m/s and so the power (from rate of change of potential energy).
    You can also do the same experiment not in gear (assuming you have a manual) and get just the tire and wind resistance.

    The answer should come out at about 20-30kW, = 30-40hp

    eg if you do 100km/h rolling down a 5% slope in a 2000kg car
    Then in 1km you will have dropped 50m in 1/100 of an hour = 36s
    Power = change in mgh / time = 2000 * 9.8 * 50 / 36 = 25kw = 33hp
    Last edited: Jun 18, 2009
  4. Jun 18, 2009 #3
    With only 10% change in speed it seems that the averaging should not create major non-linearity problems.
  5. Jun 18, 2009 #4


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    Sorry yes I hadn't noticed that.
    My guess is that the speed was wrong, the speedometer has a certain response time so it may be that you had slowed down much more but the speedo was lagging behind.

    My old 1.1L Peugeot 106 needed pretty much all of it's 50kw to stay at freeway speeds.
  6. Jun 18, 2009 #5
    I don't think the speedometer was wrong at all. What my test shows is that the 50kW of your Peugeot 106 goes almost completely into turning over the engine and the gearbox. By switching your gearbox to neutral, you take those losses out of the car friction, and you find out what road and air friction actually are. That's the whole point of my experiment
    I'm having a similar discussion at slashdot at the moment, and most people find the concept hard to get. It's just simple physics though, and an experiment that everyone can perform. I'd love to see this entered into highschool physics experiments.
  7. Jun 18, 2009 #6


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    Above about 50km/h most of the power goes into aerodynamic drag.
    Power from aerodynamic drag is 1/2 density * v^3 * area * Cd

    For a new VW golf (similair) front width=1700mm height=1500 mm and Cd is around 0.28

    then at 25m/s, Drag power = 1/2 * 1.2kg/m^3 * 25^3 * 1.7m * 1.5m * 0.28 = 6.7kw
    So you are only a factor of 3-4 out. Still can't spot where though.

    Yours is a valid method used by hyper-mileage people to calculate drag for their car.
    It's an interesting experiment to try with things like the windows open vs air conditioning and to see the effects of things like bike racks (huge) or even roof bars.
  8. Jun 18, 2009 #7


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    30 seconds to slow down 10km/h? Does the drag force make sense for that gradual of a slowdown? I don't know much about drag so I'm not confident in attempting to make the calculation.
  9. Jun 18, 2009 #8


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    I'm not sure you were on level ground, because 30 seconds is an awfully long time to slow down 10 km/h.

    And if you measured that correctly, your calculation actually overstates the efficiency, since you didn't differentiate between the proportion of that power used to defeat drag and the proportion used to defeat the friction in the drivetrain (which is what you are trying to calculate).

    Anyway, I'm not sure of the usefulness of the entire exercise. You're looking for the energy dissipated via wind resistance. Yeah, it isn't much compared to all other forms of energy dissipation. Heck, imagine how bad it is if you're going slow!
  10. Jun 18, 2009 #9


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    The way to test that is to do two measurements, going in opposite directions along the same section of road. If the times come out to be very different then try the experiment again, somewhere else.
  11. Jun 18, 2009 #10


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    Another aspect to look at in terms of efficiency of converting gasoline's energy to mechanical energy would be to get a complete thermal profile of the car. Your efficiency picture is incomplete without looking at how much goes to heat out the tail pipe and the rest of the surroundings.
  12. Jun 19, 2009 #11


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    Not only heat, but also sound and vibration, and probably a couple of other things that I'm not thinking of yet. Would light from the combustion within the cylinders count as a loss?
  13. Jun 19, 2009 #12
    I was actually trying to figure out how inefficient an ICE + gearbox are, compared to an all electric drive. So I figured, I need the wheels, axle and suspension, and the car body, and I wanted to know how much power it would take to keep that combination moving.

    As far as the road being flat, it was in 'Waterland', a Dutch polder, which is so flat that Kansas should be called mountainous :-).
  14. Jun 19, 2009 #13


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    I got some numbers from my car today (air conditioning on):
    .4 gph at idle
    .9 gph at 3000 rpm, in neutral. This corresponds to about 74 mph.
    2.0-2.2 gph @ 70 mph

    I drive a Mazda 6i, which is rated at 32 mpg highway. I get close to that cross-country, and these numbers are in line with that.

    Now straight from those numbers, you can say that at ~70 mph it takes 2.2-.9=1.3 gal/hr to defeat the wind and a little bit of the drivetrain drag. At 34 MJ/L, that's 46 kW.

    Perhaps I could compare the fuel flow at 50 mph and 70 mph and measure directly....

    On my way home, there is a stretch of road 1.1 miles long that is perfectly straight. It dips, then rises again. The ends are at elevations of 142 and 133 ft and in the middle it dips as low as 100 ft (this is largely just notes so I remember....)
  15. Jun 20, 2009 #14
    Funny you talking about a bit of road that is straight and only dips a little. Where I live in Dutch reclaimed land, most roads are straight as an arrow and completely level. Boring!

    I'd love to see others do the same experiment, and see the results. The trick is in setting the gears in neutral, otherwise your engine will be braking the car.
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