Rule of Thumb for Weight-Displacement Relationship

here is what I got....

Dw * L * B = rectangular section of hull, if (Ds+Db)/2 will equal Dw


PLUS.....

B' * Dw * (L+((Xb+Xs)/2)) = the angled sections of hull running down both sides added together assuming they are more or less trianglar in cross section....two B'/2 added together to make one whole B'. The Xb+Xs)/2 added to Length treats front and back sections as two right angle triangles.

PLUS.....

Ds * (Xs/2) * (B+B') = the front angled section. I figured tacking on B' to B would be about right to estimate the corner volume.

PLUS.....


Db * (Xb/2) * (B+B') =rear angled section, just using Xb and Db instead of Xs and Bs.




So just add those all together.......


(Dw * L * B) +

(B' * Dw * (L+((Xb+Xs)/2)) +

(Ds * (Xs/2) * (B+B')) +

(Db * (Xb/2) * (B+B')) +
----------------------------------

should be fairly close

 

Squidly that looks like the weight Vs displacement of a Colins Class Submarine under full ballast.

 

Two conditions need to be satisfied when a vessel is floating at rest.

1) The displaced mass of the water equals the entire weight of the vessel including cargo.

2) The center of buoyancy, which is at the location of the center gravity of the displaced water, is aligned vertically with the center of gravity of the vessel including cargo, etc.

Generally interation is required to find the draft, trim and heel which satisfy these conditions.

 

Firstly, if you work for someone who insists that displacement is 10% of the weight, you work for someone with absolutely no clue of physics. Unless your boss learns fifth grade sience, it will be impossible to satsify his stupid request.
Also, I don't understand what data you have. Either you have the weight, and thus displacement, or you have an existing ship in which case you can measure and then calculate.

 

Draft and freeboard can be measured with a measuring tape. Alternatively if nobody wants to get wet keelhaul a dive computer..

 

Perhaps there is a communications issue here, with the "boss" using the term displacement for other than the volume/weight of the water displaced. The suggestion that the boss may have meant that the empty weight of the vessel should be no more than 10% of the total weight when fully laden may be reasonable for a barge without propulsion or accomodations.

I've seen some ads for not inexpensive boats which had different values listed for weight and displacment. As far as I could figure out, one was probably the weight/displacement of the basic boat, while the other was either the maximum weight/displacment the boat shuld have, or possibly a typical displacment. Quite confusing.

 

Even more confusing when manufacturers give displacement (total weight for calculation of performance) in a light condition, empty tanks, one person on board. In smaller vessels this gets much heavier under actual use with people, food, water and fuel and performance degrades proportionally. I agree that the xxxxxxxxxxx, excuse me, boss who is asking you to do something out of your job description is making it difficult due to his own lack of precision in the terminology. You sound like a calm and smart person so should be able to work this out.
Excel is probably not the ideal program.
Weight of immersed volume (displaced liquid = displacement) equals total weight of vessel. If the weight of vessel exceeds weight of displaced volume of water the vessel sinks deeper until all is equal again.
The built construction weight or light displacement of the vessel, unloaded, is fixed, so this is one specification and depends on design and materials.
Let's say it's a steel barge 10 meters long and 2.5 meters deep, 4 meters wide that draws 0.5 meters totally empty.
10x4x0.5=20 metric tons light displacement. (1 cubic meter water = 1000 kg) This is what the barge weighs alone.
Now load it totally down with gravel so it draws 2.5 meters and the water is just even with the deck. 10x4x2.5=100 tons loaded displacement and therefore maximum 80 ton dead weight capacity.
Now add one small rock, making the weight 100.00001 tons.
You have exceeded the available displacement holding you up, there is no reserve buoyancy above the waterline to compensate, and the barge sinks, slowly, but sinks nevertheless out of sight and down to the bottom. This would be a 20 percent displacement/weight difference example maybe if this is what your boss is demanding of you.
To really get a handle on this stuff find a copy of the book FROM MY OLD BOAT SHOP by Weston Farmer. He shows very simple methods proven over many years for calculating displacement and draft.
Boats are simple and straightforward things and sometimes we all get carried away with the math.
What is the hull shape you are dealing with? You haven't been very specific.

 

Your Xb and B' corners get computed separately as prisms without the radius, then added to the remaining rectangular volume.
Knowing the hull weight gives the required total volume to float the light vessel, making it possible to calculate the draft from that. Break the big problem into small ones by breaking the volume into easily calculated units then adding them together.
If you strike a trial waterline and calculate the volume below that and compare it to the known weight of the light ship, this should quickly lead you to the answer from the difference in results. Strike a new waterline where you think it should be and recalculate. The error will be much smaller and the next waterline should be spot on, giving theoretical draft.
Known hull weight = X tons = X cubic meters of displaced water required for light ship condition or light displacement.
Known hull weight X plus known loading Y = X+Y cubic meters of water required for loaded displacement.
These should show you the light and loaded drafts.
Don't try to make a real vessel fit some 10% rule that is nebulous at best.

 

Your screen name suggests that you are a female person. Question; Is it possible that your boss or bosses are male chauvinist pigs (MCP) who are giving you a hard time or attempting to intimidate you with that weight/displacement gimmick?

 

Thank you all for your replies. The project is currently under construction. We have added some add'l strengthening elements (our barge is a concrete one) thus making it heavier but keeping the same hull shape. I got the displacement but still with the trial & error way and equating it with 10% of the Total Lightship Weight. I am still researching for ways as how to make my spreadsheet better for our upcoming projects regardless of what shape our barge/ship hull will be. Thank you all very much

 

A way to experiment and learn a lot, get a feeling of things, etc, is to play with the free software Freeship -and be patient to learn how it works.
Regards
APP