# How much weight has weight in fuel efficiency?

Comparing the official fuel efficiency figures for several cars, I was surprised to discover that the Smart Fortwo has a weight between 750 to 800 kg and needs 4.3 L/100 km, but the VW Polo with a weight of 1650 kg reaches 3.8 L/100 km. That is, the Smart has half the weight, but still a worse fuel efficiency.

When driving up and downhill (but not on a flat surface) does one cancel the other? That is, driving uphill with more weight costs more fuel, but you need less fuel when going downhill.

Are the major fuel efficiency improvements rather on other places, like better tires, better injection, better streamlining and hence lower weight cannot outweigh poor performance in these fields?

• Those fuel consumption figures are based on government test cycles aka rarely achieved in reality. Feb 8, 2020 at 22:12
• @SolarMike: sure, but here is relevant how the two values compare. Feb 9, 2020 at 0:42

Weight does indeed have weight:

Note that fuel efficiency is given in miles per gallon, not in l/100 km, so higher values mean better efficency. Source of graph.

The autoher of the linked piece gives the following, IMO reasonable, reasons as for why weight matters:

• higher weight means more rolling resistance
• higher weight means more energy expenditure for acceleration
• larger cars with more air resistance are also heavier

My guess would be the the second factor is the most relevant, but then you'd expect city milage to be more affected by weight and that does not appear to be the case.

This answer is really based on one minute of googling, I'd keep the question open for a bit - maybe an automotive engineer chimes in with a more comprehensive answer.

• The question is not about weight having weight, but how much the weight of other features can outweigh a heavy weight. Jul 9, 2020 at 23:39

Driving downhill or breakings only generate energy in cars that have regenerative systems, battery, transducer, electrical motor.

Weight may not be as big a factor in fuel consumption as the other factors such as a streamlined body, efficient drivetrain, and better tire design.

Weight to engine power ratio counts mostly in allowing faster acceleration.

$$\Sigma f=0 \quad F_{engine}- F{friction} = ma \quad a= \frac{F_{engine}- F{friction}}{m}$$

It means same engine will produce more acceleration in the lighter cae.

• Aerodynamics is very important for highway mileage. Smart cars have bad aerodynamics. Feb 9, 2020 at 5:19