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Formula for fuel consumption
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Nerby
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PostPosted: Thu Mar 09, 2006 10:32 am    Post subject: Formula for fuel consumption Reply with quote

Hi All,

doea anyone know which is the formula used to compute fuel consumption at different speeds for each ship?

I assume it comes from the ship's maximum speed and fuel consumption XML parameters (<max_geschw> and <treibstoffverbrauch>).

Does it also involve other parameters?

Thanks,

Nerby
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Bearsie
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PostPosted: Fri Mar 10, 2006 5:35 am    Post subject: Reply with quote

I would assume 200 grams HP/h but since the game does the calculating for me I never pay much attention. Since horsepower drops with speed it should be an easy enough formula.
Never measured the actual value used in the game.
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rdklein
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PostPosted: Fri Mar 10, 2006 8:01 am    Post subject: Reply with quote

depends on the ship, there is a list of ships with the fuel consumtion (ships.xml)
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Christian Todt
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PostPosted: Fri Mar 10, 2006 9:07 am    Post subject: Reply with quote

hi,

does the fuel consumption only depend on the ship?
doesn´t play the speed a role also?

would be nice to have the formula to calculate the most profitable speed
for each freight.

CT
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PostPosted: Fri Mar 10, 2006 9:25 am    Post subject: Reply with quote

Christian Todt wrote:
doesn´t play the speed a role also?


rdklein wrote:
depends on the ship, there is a list of ships with the fuel consumtion (ships.xml)


Indeed it depends on the ship's speed and type. But I was curious about:

- apart speed, does it depend only on these two characteristics (from the ships.xml file): <max_geschw> and <treibstoffverbrauch>, and not load, age, maintenance, etc.?
- which is the exact formula?

I've come to an experimental conclusion (but I have started test only yesterday), that the formula is a cubic function of the speed, but the difficult part is determining the coefficient depending on the two (?) XML parameters.

Christian Todt wrote:
would be nice to have the formula to calculate the most profitable speed for each freight.


This is exactly why I am asking. Of course the problem is more complex, since you have to take into consideration fuel price, daily fixed costs, distance and canals' fees (and these depend on the load ...). And finally, if you want to determine which is te most profitable cargo, you have to take into consideration time to wait outside ports and to load and unload the ship.

Nerby
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PostPosted: Tue Mar 21, 2006 6:29 pm    Post subject: Reply with quote

Beware: this is a definitively technical post.


I think I have found the formula used in Ports of Call for fuel consumption. I got it studying the experimental values that appear when a ship leaves an harbour.

I have studied the following ship classes, but the result should hopefully be the same for all classes: Canberra, Jahre Viking, Manukai, Rembrandt, Seaguardian.

Given the following parameters (available in the ship0.xml file, name of parameter in angle brackets):

Code:
Vmax : Maximum velocity <max_geschw>
P : Engine power <kw>
F : Fuel ratio <treibstoffverbrauch> (in grams per engine KW per hour)


and the ship's speed V, then fuel consumption in tons per day is

Code:
(24 * P * F / 1000000) * (V / Vmax)^3 + 1


In addition to interpolating game values quite well, this is also a fully logical formula, so I assume it is the correct one.

I must admit that in a few of the above mentioned classes I got a better approximation by using for Vmax only the integer part of <max_geschw>, for the ship classes that have a maximum speed with some decimal digits, but this is another story.

Hope you find this info interesting.

Nerby
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PostPosted: Tue Mar 21, 2006 9:02 pm    Post subject: Reply with quote

Almost the same or the correct formula used in POC...

I haven’t had the time to check your formula or calculate my own, but shouldn’t there be a loaded cargo parameter in such a formula…


Cheers, Novex
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Nerby
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PostPosted: Wed Mar 22, 2006 8:25 am    Post subject: Reply with quote

Novex wrote:
Almost the same or the correct formula used in POC...


Well, actually I think this IS the formula used in PoC.

Novex wrote:
I haven’t had the time to check your formula or calculate my own, but shouldn’t there be a loaded cargo parameter in such a formula…


Indeed. There's one note though: I didn't notice before (so, thanks Novex for pointing out), but when a ship nears the full load (i.e. cannot reach its full speed), its fuel consumption increases. This is independent of the chosen game mode (Beginner, Advanced, Expert).

For a complete simulation I think that fuel consumption should depend not only on load, but also on the age of the engine (and possibly allow the player to lay up the ship and have special engine maintenance to improve its consumption). ... and what about fuel quality? And about going above maximum speed and deteriorating abnormally the engine?

Cheers,

Nerby
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PostPosted: Wed Jul 05, 2006 11:10 pm    Post subject: Simple Fuel Consumption Formulas Reply with quote

Well simple formulas are :

Fuel consumption per day (tonnes) = (Disp^(2/3) x V^3 )/ Fuel coefficent

or

Daily fuel con. (tonnes) = SFC (kg/kW hr) x Power (kW) x [ 24 / 1000 ]

Where SFC is specific fuel consumption.

But using the first formula if one knows the Fuel Coefficient for the vessel concerned then as one knows the displacement and speed required one can work out the consumption.
I'm not too sure on the coefficients but will be directly related to block coefficient of the vessel and the age.
But hope this helps
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PostPosted: Wed Nov 22, 2006 12:00 pm    Post subject: Reply with quote

Hallo,
ich habe ein paar Berechnungen angestellt und habe ein kleines Perl-Skript geschrieben, welches recht nahe an die Werte bei 'Speed control' heran kommt. Im Skript habe ich versucht die beschreibenden Kommentare in englisch zu halten, hoffentlich ist es auch verständlich geworden. Bei meinen Tests bin ich zu dem Schluss gekommen, dass das Alter des Schiffs nicht extra berücksichtigt werden muss.
Hier nun das Skript zum experimentieren:
Code:
#!/usr/bin/perl
use warnings;
use strict;
use Getopt::Std;
my ($kw, $load, $tratio, $vmax, $dwt);
our($opt_k,$opt_l,$opt_t,$opt_v,$opt_d);
getopts('k:l:t:v:d:');
#
# check mandatory parameters
# if no parameters are given then the values from 'seaguardian' are default
if(defined($opt_k) && defined($opt_t) && defined($opt_v) && defined($opt_d)) {
   #
   # fill variables with parameters
   # I dont check plausibility of parameters!
   $kw     = $opt_k;
   $tratio = $opt_t;
   $vmax   = $opt_v;
   $dwt    = $opt_d;
}
else {
   #
   # the dafault values are taken from 'seaguardian'
   $kw     = 11500; #<kw> from ship0.xml
   $tratio = 170;   #<treibstoffverbrauch> from ship0.xml
   $vmax   = 16;    #<max_geschw> from ship0.xml
   $dwt    = 50250; #<dwt> from ship0.xml
}
#
# check the optional parameter
if(defined($opt_l) && $opt_l>=0) {
   $load = $opt_l;
}
else {
   $load = 0;
}
#
# calculate Vmax with load bigger or equal than 95% of DWT
my $loadmult = 0.95 + (($dwt - $load) / $dwt);
if($loadmult < 1) {
   $vmax = $vmax * $loadmult;
}
#
# calculate Vmin
my $vmin  = round($vmax/4, 1);
#
# calculate scale from 'speed control'
# i think it is 266
# i calculate it to test different rounding
my $scale = int(800/3);
#
# calculate the steps between scale
my $step  = ($vmax - $vmin) / $scale;
#
# calculate all fuelconsumptions between Vmax and Vmin
# there are diverse roundingproblems
# the results only near 'Speed control'
my $v;
for (my $i=$scale; $i>=0; $i--) {
   $v = $vmin + ($step * $i);
   my $verbrauch = ((24 * $kw * $tratio) / 1000000) * ($v / $vmax)**3 + (1 - ($v / $vmax)**3);
   printf("%7.1f t/day %5.1f kn\n", $verbrauch, $v);
}
exit;
sub round {
   my ($zahl, $stellen) = @_;
   $stellen = 0 unless(defined($stellen));
   $stellen = 0 if($stellen < 0);
   $zahl = 0 unless(defined($zahl));
   return int(($zahl*10**$stellen)+ (.5 * ($zahl <=> 0)))/10**$stellen;
}


Viel Spaß
Gruß Andy

PS: vielen Dank an den freundlichen Übersetzer. Ich bin froh wenn ich einige Worte sinngemäß ins englische übersetzt bekomme, bei ganzen Sätzen dürfte es in unverständlichem Gestammel enden.
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PostPosted: Thu Nov 23, 2006 1:20 am    Post subject: Reply with quote

Hmmmm.
I never understood why the line fuel consumption shows up to 170%...
100 is good enough
basically I disagree with the game set up in a few area's
fuel consumption of a diesel engine is based on output and output alone !
it doesnt care what brand paint is used if any...
since the engine is "fixed" i.e. max output is a known quantity
at full output there will be full consumption, no more, no less.
What will change is speed, based on draft, trim, weight, resistance, blah blah.... the only reason fuel consumption drops at lower speeds
is not because we lose speed actually, but because we reduce engine output, now there is one way to increase consumption dramatically at lower speed even and that is to put the wrong propeller on the ship.
Either lugging the engine down or revvying it to max with no advance...
Roughly, specific fuel consumption should be 150 to 200 grams per HP/Hr
Then I tweak the game until it is about right.
Well, that's my 2 cents Very Happy
All of which still doesn't explain why in the game my coasters (any engine under 1000kw) use dramatically more fuel when going slow, and "save" fuel when going all out lol
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PostPosted: Thu Nov 23, 2006 11:27 am    Post subject: Reply with quote

the most fuel you can save,when the vessel is in the port.but maybe the money from the crew will be finish soon(in former times in good ports) Rolling Eyes brgds mooringman
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PostPosted: Fri Nov 24, 2006 3:01 am    Post subject: Reply with quote

Christian Todt wrote:
hi,

does the fuel consumption only depend on the ship?
doesn´t play the speed a role also?

would be nice to have the formula to calculate the most profitable speed
for each freight.

CT
Most profitable speed depends on speed, fuelconsumption, maintenance costs, shipcosts.
I did some calculations with excel on different ships. It resulted with Maintenance OFF, the most profitable speed is about 75% of the maximum, no matter the kind of ship. If you go faster, the fuelconsumption is getting too high. If you go slower, the daily shipcosts make you pay more.
With maintenance set to Excelent, The maintenance costs per day are of such a big influence, you want to go as fast as posible.

I used this formula for Total Freight costs (TF):

TF = (days of transport * FuelConsumption * fuelprice) + (days of transport * (Shipcosts per day + Maintenance costs per day)).

Shipcosts per day is mentioned in the ship0.xml file.
Maintenance costs you can check in your Office. This parameter is related to the <preis_max> of the ship.

Greetings,
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PostPosted: Fri Nov 24, 2006 3:42 am    Post subject: Reply with quote

@ Filips Hollandicus

Now that makes sense !

My only caveat would be that the added days for the trip at a given fuel savings, do not cost more in potential freight income then I save on fuel...
example: if my daily average freight earnings is 20 000 dollars and I add 4 days to a 20 day trip and save 500 a day in fuel but drop my daily freight income to 18 000 dollars, I am actually losing 1500 a day...
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PostPosted: Fri Nov 24, 2006 12:01 pm    Post subject: Reply with quote

filips hollandicus wrote:
I used this formula for Total Freight costs (TF):

TF = (days of transport * FuelConsumption * fuelprice) + (days of transport * (Shipcosts per day + Maintenance costs per day)).


Hi All,

my original interest in the fuel formula, was to select the speed that minimizes travel costs. I will now go on mathematics. The itneresting results for everybody are in double angle brackets <<>>.

First of all, I will shorten the formula using the following acronyms:

dot - days of transport
fc - fuel consumption per day
fp - fuel price
tcd - total costs per day = Shipcosts per day + Maintenance costs per day

The first two values directly depend on speed (s):

dot(s) = d / s

fc(s) = k * s^3

where d is the travel distance, and k is a coefficient that includes various values (see formula above) (k = (Disp^(2/3) / Fuel coefficent) )

The total travel cost formula becomes:

TF = dot(s) * ( fc(s) * fp + tcd)

Now, going to calculus, the optimal speed s is the one for which:

dot'(s) * (fc(s) * fp + tcd) + dot(s) * fp * fc'(s) = 0

Substituting the functions depending on speed and differentiating:

- (d / s^2) * (k * fp * s^3 + tcd) + (d / s) * fb * 3 * k * s^2 = 0

Distance can be factored out, and disappears from the formula.

<< The optimal speed that optimizes travel costs does not depend on how far the ship travels. >>

The formula becomes:

fp * (1 / s) * 3 * k * s^2 = (1 / s^2) * (k * fp * s^3 + tcd)

Since speed is greater than 0, I can multiply by s^2:

3 * fp * k * s^3 = k * fp * s^3 + tcd

then

2 * k * fp * s^3 = tcd

s^3 = tcd / (2 * k * fp)

<< The speed that minimizes travel cost is the cubic root of (TotalDayCosts * FuelCoefficient / 2 / FuelPrice / Displacement^0.666) >>

Notes:
- This formula does not take into account day spent for loading and unloading, and days spent outside harbours.
- Also, on long journeys, there might be a very slight increase of maintenance costs, which we know depend on ship condition, which degrades with time. Probably this increase is neglegible for our purposes.
- This is not really the optimal speed, because costs are computed daily, so there is no need for our ship to arrive at 1 PM, when it could arrive at 11 PM on the same day, i.e. taking the same number of days but travelling a bit slower, consuming less fuel.

Cheers,

Nerby
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