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I have always been curious about this one.

Since the gas has some weight, the car will have to burn some extra gas to carry it's own fuel around.

How can I calculate how much that extra gas is?


  • car lifetime of 300,000km
  • 50lt tank, topped up on each refuel
  • steady gas price along the years
  • steady gas density despite climate conditions
  • linear relationship between car weight and fuel consumption

What would be the formula to calculate the extra fuel needed?

In my specific case, the car weighs 1200kg and burns 7.2lt/100km.

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You may be interested in the rocket equation. This exact consideration is critically important in rockets. You need fuel to carry payload up, then more fuel to carry that fuel, so on and so forth. The saving grace for rockets is that you burn fuel as you go, so you don't have to lift it all into space. This is also one reason why you see the fuel tanks get jettisoned as it gets higher up. Gets rid of useless weight and saves fuel. – zibadawa timmy Aug 7 '14 at 14:13
A similar problem is food depots when using dog sleighs to reach a given pole. You need a previous dog sleigh to put the last depot, so that trip also requires depots. If you need to go really far, the last depot becomes quite expensive to put. – Thorbjørn Ravn Andersen Aug 8 '14 at 9:46
up vote 20 down vote accepted

Assuming the fuel tank is half full on average, the density of fuel is $0.75 \space kg/l$ (ref.), so the mass of fuel is $25 \times 0.75 = 18.75 \space kg$

The car and fuel weighs $1218.75 \space kg$. The amount of energy used to transport the fuel alone is $18.75 \div 1218.75 = 1.5385\%$

The fuel used to carry fuel over the car's lifetime is $7.2 \times 3000 \times 0.015385 = 332.3 \space l$

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@Ryan - from the OP's question: 7.2lt/100km, car lifetime of 300,000km – Chris Degnen Aug 7 '14 at 15:45
Kramer's Collorary: The fuel-optimal driving strategy is to keep just enough fuel in the tank to make it to the next gas station. – chriswynnyk Aug 7 '14 at 19:51
@chriswynnyk: My corrolary: even with gas at the ridiculous price of \$4/gallon, the computation above works out to less than $400 over the life of the car. Amortized over 10+ years, it's literally not worth the hassle to follow Kramer's strategy. – Mason Wheeler Aug 7 '14 at 22:12
This assumes that the entire energy cost of running the car is proportional to the mass of the car. This is a pretty big assumption; I'd strongly suspect that there is a decent-sized mass-invariant cost to run bearing friction and accessories. – Doug McClean Aug 8 '14 at 1:29
@NathanCooper I just think you made the problem NP-complete by introducing price flunctuations ;) – Thorbjørn Ravn Andersen Aug 8 '14 at 14:19

Gas is about $3/4$ the density of water, so a $50$ liter tank (on average half full) has a mass of about $18$ kg. If your dry car has a mass of $1200$ kg (easy to calculate with) and (as you ask) we assume fuel consumption is proportional to mass, you are using about $1.5\%$ of your fuel to move fuel. In fact, much of the fuel goes to overcome friction and air resistance, so the real loss will be lower.

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+1 good point on friction and air resistance – Chris Degnen Aug 7 '14 at 14:00
@ChrisDegnen: once you are up to speed, the only way extra weight hurts is increasing the rolling resistance because the tires flex more (and this is a small effect). The air resistance and engine friction are the same. – Ross Millikan Aug 7 '14 at 14:16

With the information given, we can't do better than a good guestimate.

Fuel consumption comes mostly from three components: Rolling resistance (linear with speed), the energy used to run the motor (square of rpm), and wind resistance (fourth power of speed). Each of these has a constant which depends on the weight, on the construction of the engine, and the shape of the car. Note that the distance also grows linear with speed, so going faster doesn't increase the energy needed to overcome the rolling resistance over a mile, just over a second or minute.

The fuel consumption was given as 7.2 ltr / 100km. This is shared between the three components. 7.2 is a lifetime average. If you drive to save fuel (hyper-miling), you can't reduce the rolling resistance, but wind resistance and engine resistance by driving slow and in a high gear. The fuel used for the rolling resistance could be approximated by taking the fuel consumption when the driver tries their best to save fuel (minimising engine and wind resistance), and then subtracting some more because there is still engine and wind resistance left. With 7.2 ltr average, 5.5 ltr / 100 km is probably achievable, and I would estimate 4 ltr / 100 km for rolling resistance.

The car weighs 1200 kg. A full tank weighs 37.5 kg. On average, the fuel in the car will weigh about 20 kg (because nobody empties their fuel tank completely). Additional weight for driver and passengers may be 100kg on average (people mostly drive alone, but not always). So the total weight on average is 1320 kg, fuel weight on average is 20 kg, and with 300,000 km I'd estimate the fuel spent to carry the fuel is

(4.0 / 100) * 300,000 * (20 / 1320) = 12,000 / 66 = 182 liter.

So the guesswork involved was: How much of the fuel consumption is weight dependent? (Less than the total fuel consumption). What percentage of the total weight is fuel (need to take into account weight of car + passengers + stuff in the car).

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Good point. Seems to be the correctest answer given. The fuel consumption is not linear to the car's weight. It's getting interesting.. – Christoph Bühler Aug 8 '14 at 14:42

This is NOT an answer. I put my thought in here just because the description has exceeded the comment limit.

I have seen the two valid answers but the question was treated in an arithmetical way. I, somehow, think that the question should be treated in a calculus manner as follows.

We assume that (1) the car is travelling at constant speed at all times; (2) in every $d$ seconds, the car has traveled a fixed distance of s km.

Initially, when $t = 0$, the total weight of the car is $1200 + 37.5$ kg. Suppose that it takes $x$ liters of gas to move the car (weighing $1237.5$ kg) forward for that interval.

At the start of the next interval, i.e. $t = \Delta t$, the car now weighs $1237.5 – x$ kg. For the next fixed interval, because the car is lighter now, it requires less fuel, (say $x – \Delta x$) liters to push it forward for the same distance.

The scenario goes on ….

A calculus problem?

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Using the mass of half a tank averages over the range from full to empty. As the problem is stated, the additional fuel consumption is linear in the fuel carried. Doing what you suggest will only correct for the fact that the fuel is consumed (a tiny bit) faster when the tank is full than when it is empty. At the level of this problem, this is an unnecessary complication. You should start with $1200+37.5$ kg, as that is a full tank if you want to do what you are suggesting. – Ross Millikan Aug 7 '14 at 19:26
@RossMillikan What if the problem was stated in my way? – Mick Aug 8 '14 at 1:14
Your approach will be more accurate. If fuel is a large fraction of car mass and if consumption rises more than linearly with car mass it could be needed to get an accurate answer. You would get a differential equation for distance traveled as a function of fuel burn. Solving it would give the range on one tank, which you could compare to the range assuming the fuel weighs nothing and ascribe the difference to be the consumption of carrying the fuel. – Ross Millikan Aug 8 '14 at 1:39
Assuming a half full tank should be accurate enough to do the calculation for a car. For a rocket, such an approximation would not be accurate enough. With that approximation the conclusion would be that a rocket could never make it to orbit. Doing the proper differential equation would OTOH give you the correct solution, which is known as the rocket equation. – kasperd Aug 8 '14 at 16:16
@kasperd Right. The rocket equation is exactly what I have in mind. It is delighted to see that the same matter was treated in a calculus way. – Mick Aug 8 '14 at 17:15

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