Do your own homework and save fuel

[DCockey]

The Breguet range formula which Leo posted assumes constant resistance to drag ratio, constant specific fuel consumption and propeller efficiency, and constant velocity. How reasonable those assumptions are for a given marine vessel would need to be determined for any application of the formula. Presumably the fuel weight would need to be a not too large fraction of empty weight.

In the aircraft world a different version of the formula is used for jet aircraft, which leads to higher speeds for maximum range than for propeller driven aircraft due to the difference is efficiency characteristics of jets vs propellers. I wonder if something similar would hold for vessels with waterjet propulsion?

[Leo Lazauskas]

Originally Posted by rxcomposite

Leo,

Glad this came up. I found the breguet formula in one of your papers. I tried to feed the formula into my excell spreadsheet a long time ago but am not sure if I have done it correctly. Excell does not have a log e function so I have to insert a base number of 2.718 for the log. I also have to multiply the constant by a neg 1 because I am getting a negative value for the answer.

Any hope I can use Breguet's formula for the English System? What constant do I use? I have to write both in English and metric.

Thanks,
Rx

The constant at the front of the equation I gave is
367097.8 = 60 X 60 X 1000 / 9.80665.

gravitational acceleration is 9.80665
60 seconds per minute
60 minutes per hour
1000 to get grams

In Excel you can use the LN function for logarithms to base e.

To convert from lbs/hp-hr to g/kW-hr multiply by 608.277

Leo.

[Leo Lazauskas]

Originally Posted by DCockey

The Breguet range formula which Leo posted assumes constant resistance to drag ratio, constant specific fuel consumption and propeller efficiency, and constant velocity. How reasonable those assumptions are for a given marine vessel would need to be determined for any application of the formula. Presumably the fuel weight would need to be a not too large fraction of empty weight.

In the aircraft world a different version of the formula is used for jet aircraft, which leads to higher speeds for maximum range than for propeller driven aircraft due to the difference is efficiency characteristics of jets vs propellers. I wonder if something similar would hold for vessels with waterjet propulsion?

Yes, it is a little crude, but it is not as pessimistic as other formulas that are routinely used by naval architects and that do not include the "log factor".
See, for example, the discussion by L.J. Doctors following the "High Speed Sealift Technology Workshop, 1997".
The discussion is in the report:
"Hull Form and Propulsor Technology for High Speed Sealift", ed. Chris B. McKesson, 13 Feb. 1998", John J. McMullen and Assoc.
I think it is still around somewhere on the net.

A better way is to calculate the drag separately for, say, the full fuel condition, the half-fuel state, and the "burned-out" state. Each calculation should be made for the appropriate displacement weight after reducing by the weight of fuel loss.

Whether OPC and sfc remain constant is, as you say, open to question.

Leo.

[daiquiri]

Just found the answer to my doubts in the previous page, regarding the variable ship displacement: http://www.mckesson.us/mckwiki/index...21&redirect=no

Citation:

"Finally, note that this effect (Variable Displacement) is only realistic if the owner uses it: if he doesn’t refuel, and doesn’t ballast. The military practice, for example, of never allowing the ship to get below ½ or ¾ “tank” will obviate the benefits of this calculation: In effect the owner is running his ship in a Constant Displacement mode, and it behoves the Naval Architect to perform the calculations accordingly."

[rxcomposite]

Thanks Leo. I used LN and the numbers came out the same. I did right except for the method. I feel more confident now.

However, If I use
g= 32 feet/sec
60 sec/min
60 min/hour
608.277 to get lbs

In the English system, I get 68,112 for the constant. The number 74,445 seems to work best. Anyway, I used;

Np= 0.6
sfc= lbs/hr
Rt - lbs
W = Long tons
Wf= Long tons
Range= Nm

I guess all this convertion added to the error or complexity of the equation.

[Adler]

Dear All,

I would like to attach on this thread some information.
Last two years we participate on a similar design/project that will be applied on FAC
(High Powered Semi - Displacement Boats).

[Adler]

Dear All,

On the attached file there are some data regarding the relation between Fuel Load and the corresponding measured Speed that were extracted due to Sea Trials on a semi-displacement vessel.
Should be interesting to graph the Derivative Curve of the presented Curve there.
Following this method you can define the range that the variance of fuel load could be have less influence to the fuel consumption rate.
Important when is examined the possible adding advantage of a Stern Flap or any other Hull's factor regarding to reduce the drag at a given speed.

[Leo Lazauskas]

Originally Posted by rxcomposite

Thanks Leo. I used LN and the numbers came out the same. I did right except for the method. I feel more confident now.

However, If I use
g= 32 feet/sec
60 sec/min
60 min/hour
608.277 to get lbs

In the English system, I get 68,112 for the constant. The number 74,445 seems to work best. Anyway, I used;

Np= 0.6
sfc= lbs/hr
Rt - lbs
W = Long tons
Wf= Long tons
Range= Nm

I guess all this convertion added to the error or complexity of the equation.

I was brought up with feet,tons, versts, pounds, shillings and pence etc but I am hopeless with them now. You should give up that Masonic nonsense and go metric

[rxcomposite]

I did went metric a long time ago but the USAnians in this forum have a hard time deciphering what I write so I have to do both.

Anyway, thanks a lot Leo. I know my program is correct.

[Frosty]

I went metric Oh donkeys years ago. I went the whole 9 yards.

But I often do both --like 14 inches x 140mm.

In for a penny in for a pound I say .

Its the beer that confuses me --is 3x 330 Ml bottles 1 pint. or half a Quart.

Maybe its neither maybe its just 3x 330 Ml bottles.

On the hole I prefer mm for measuring drills though. 3 mm or 13mm conjures an immediate size in my head rather than a ridiculous 11/64ths. mind you I still keep good 1/8 for a pop rivet or two.

[rxcomposite]

I grew up using the English system but have to convert to metric due to standardization. Funny thing is my subconcious still pray tricks on me.

I am working on a drawing on ACAD. Metric setting of course. This afternoon, I went down to measure the additional part, I started measuring in inches. Only when I started drawing the part in ACAD did I realize my mistake. My mind is playing tricks on me.

It is like language. You learn to speak it but then, you start learning a second language.