Not for License Exam Fuel Consumption

It’s really important to note that to do this right you need to know the specific fuel consumption (usually found in the OEM manual or commissioning data) and the percent load of the engine at those speeds.

Fuel consumption may or may not be fairly linear compared to load, but higher speed may require significantly more load and therefore fuel.

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Yes - proportional to the cube of speed. Hence the equation KC applied.

Yes, but the writers of the USCG license exams are retards and probably still think in terms of boilers and steam plants and lighting additional burners and such.

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Normally the approximate rate of daily fuel consumption is assumed to be proportional to the cube of the speed. However, researchers have found that this figure should be between 2.7 and 3.3 for container ships. Similarly, they showed that it is equal to 3.5 for the feeder container ships and equal to 4 for medium sized container ships and even 4.5 for very large sizes. So there is a fuel dependency on the displacement of the ship. If you think about it this is a logical conclusion.

For those interested in this theory there is a research article “A new method for calculating fuel consumption and displacement of a ship in maritime transport” that gives more information on this subject.


For a rough estimate. And with consideration to what @Dutchie mentioned below.

If I’m looking for an accurate estimate to provide the office for a transit I’m still going to use specific consumption curve data gathered from actual running of the specific engines, as I’ve yet to see one match up to the formula.

Most definitely. Which is why engine load and sp-consumption is a good component for this calculation. A ship only in ballast but empty of cargo would presumably require less engine load for a given speed (and thus less fuel per kt) than one fully laden.

In principle you could calculate your own exponent of the base number with on board gathered data. For instance calculate the daily average speed and distance. These days with a Doppler log or especially with a GPS on board this can be done with a high accuracy. I suppose that also the fuel consumption can be accurately measured.

A problem could be that you will have to steam at least one day, preferably a couple of days at a slower speed to get sufficient data to carry out the calculation of the exponent. See also this website.

There probably also is a much smaller draft dependency as the underwater hull form can change with the draft. Same for the trim.

That’s the issue though, when would that ever be a given that C2=C1? If that’s the way the exams are written, fine.

But it is pretty unrealistic since we know from the first question that C2 doesn’t equal C1, that’s the whole point of learning the first equation.

As a follow-up, if you ever really did want to use the first question as a basis for the second, that is figuring out what speed will get you that distance with the same fuel and taking into account the consumption at different speeds, here is the interpolation:

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There is more to it than meets the eye.

An important factor is the weather that is encountered during the voyage. This makes a correct calculation a bit erratic.


See also ‘Math for Mariners’ and especially the chapter ‘Endurance’ starting at page 92.


You’re trying to bring the real world to a test question? I bet you believe in essay questions too. :wink:


Oops, it was a test question…:slightly_frowning_face:

Sir, where did u get this chart from

I just put the formulas in an excel spreadsheet so it’s easy to see each step. Then just copied the formulas down to start showing fuel burn for each given speed. Once finding that 15kt looked close I put in the 15.x decimal speeds to find the closest one.

Ok i guess the question i should have asked was were did u find the amount of bbls being burned. Is it in the merchant marine officers handbook ?

Ahh, sure. With the understanding that this is purely academicif the Exams ask you use the equation KC listed where Con1=Cons2 then use that.

But for the realistic version using the first formula con2/con1= (v2/v1)^3 :

  1. You already had the information that 17.5kt burns 274bbl/d. Using that and the first formula you can solve for the consumption at 17kt. This comes to 251bbl/d
  2. You said you’re going 775nm at 17kt. So 17kt = 17nm/hr. 775nm divided by 17nm/hr = 45.6 hr. 45.6 hr divided by 24hr/day = 1.89 days
  3. 251bbl/day multiplied by 1.89 days = 477 bbl

All models are wrong, some models are useful.

or expanded:

… all models are approximations. Essentially, all models are wrong, but some are useful. However, the approximate nature of the model must always be borne in mind…
– George Box

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I plotted the function S2 =Sqrt [(S1^2 *D1)/D2] which is a simple linear one as the square root of a square results in 1.

I agree wholeheartedly, a model is never perfect that should always be born in mind. However, it can be helpful and give a direction and guidance.

con1/con2= (S1^3) / (S2^3)

for con1 = con2
con1/con2 =1 (for con ≠ 0)

S1^3 = S2^3
S1 = S2

Is 775/977 the slope of the line?

Looks like the slope is square root of 775/977

The plot is S1 = S2 (Sqrt(D1/D2)). Correct?