Rule of Thumb for Weight-Displacement Relationship

[lady pirate]

Hi I'm new in boat/ship design. Can anyone help me if there is a rule of thumb as how much of the total weight is the displacement? I am currently doing some manual calculations for my draft estimates, and I need something to countercheck it. Thank you very much.

[Ad Hoc]

Originally Posted by lady pirate

Can anyone help me if there is a rule of thumb as how much of the total weight is the displacement?.

I take it you are new to naval architecture?

There is no rule of thumb, there is Archimedes theorem, it ia proven fact.

The total weight equals the total displacement…doesn’t float otherwise!

[lady pirate]

Yes, I am totally new. I am aware of the Archimedes Principle. But how do I get compute (manually) the Draft at Bow & Stern if I have a lot of unknowns? All I have is the weight and shape of my ship. I can only use archemedes' if I am able to compute my submerged volume, but with my available data, I am lost :-(

[Ad Hoc]

Originally Posted by lady pirate

But how do I get compute (manually) the Draft at Bow & Stern if I have a lot of unknowns?(

You need a set of hydrostatics of the boat inquestion. You need the hydrostatics to show the LCF, LCB, MCT 1cm, KM(L) and KM(T) at a range of drafts. The correct KG (or close to it) should be used when computing or presenting the hydrostatic data. (It effects some of the values calculated).

Then you’ll need your weight, which has the correct LCG for that condition.

Then it is a matter of using the two sets of data to find the draft.

For a naval architect this is a simple task. For a novice like yourself..it may as well be brain surgery. No disrespect meant, but unless you know what you’re doing, you’ll make a huge mess.

So, if you can post the data here, or seek a friendly local naval architect to help you.

[gonzo]

Once you have the weight/displacement and the Center of Gravity, you can calculate the submerged volume. As AdHoc says, it is not so simple. You can do it manually or use on of several software programs.

[lady pirate]

Thanks Ad Hoc. Currently, I dont have those data as I am only doing manual calculations using Excel spreadsheet. What I have right now is the total weight of the ship and the center of gravity. My boss handed me an old spreadsheet stating that the displacement should be at least 10% of the total weight. I dont know where he got the rule, and I am not really convinced with it. I am really, really new in Marine Engineering/Ship Building, I used to be a Building Structural Engineer, so what I know is just basic physics for floating objects. I started reading about Bon Jean Curves last week, but people here told me to skip the integral and just stick with the basic shape-volume compuation.

[gonzo]

Displacement is 100% of the total weight; they are identical. If you are an engineer, the math should be fairly straightforward. The term displacement comes from the volume of water the hull displaces. There are several though: fresh water, salt water and winter and summer for each. The density of each is different. Those circles with lines on ships show the maximum load line for each condition.

[lady pirate]

I understand that displacement should be the weight of the submerged part. The problem is, how can I compute my volume if I dont have my drafts and water length since water length is dependent to my drafts? My boss's spreadsheet was a trial and error one, where he inputs all data for drafts and lengths then check if the displacement equals 10% of the total weight.

So far, I have came up with this equation:

Vol Submerged = [{B * L * (Ds+Db)/2} + {0.5Ds^2*B/tanØ} + {.5Ds^2*B/tanØ} + {L*((Ds+Db)/2)tanØ} + {2*(1/3*Ds*(Ds/tanØ)^2)}
+ {2*(1/3*Db*(Db/tanØ)^2}]

The only available data in that solution is B, L and Ø.

I am really swimming in dark waters right now

[Alik]

With this level of understanding, You better not touch it, blonde pirate

[lady pirate]

Alik, what I need is basic knowledge & principle. What I am doing is rough counter-checking. I am TOTALLY NEW here, I wont be begging for your advice if I am NOT interested and willing to learn. I hope you understand.

[Ad Hoc]

Originally Posted by lady pirate

The problem is, how can I compute my volume if I dont have my drafts and water length since water length is dependent to my drafts?

Volume is the most basic calculation that one is taught at school. Volume = length x breadth x height.

You have your sketch. The rest is easy once you understand how to caulcate volume; as you "hull" is the most basic of shapes. If you are unable to calculate the volume of the shape you have provided, then the task at hand for you, is way way beyond your ability. And thus, you should seek help for someone to do the calculations for you..

[lady pirate]

Ad Hoc, with all due respect I know how to get the volume that's why I was able to come up with the equation. The problem is, I dont have the drafts for bow & stern and the water line nor the freeboard.

[liki]

The question should be to decide the dimensions in a way that produces the required displacement for the submerged portion and fulfills other requirements given, right?

[lady pirate]

Something like that liki... they are following a 10% of total weight rule = displacement, then everything is a trial and error basis already. I've been reading since last week and I have not come across about the 10% thing. I just want to make sure that I will be producing an effective spreasheet, not a trial & error one that is why I asked if there is a rule of thumb or any standards regarding the weight-displacement relationship or a standard for draft-hull height.

[JRMacGregor]

Hi Lady Pirate

A statement like "displacement should be ten percent of the weight" does not make sense.

Maybe what your bosses are saying is that "the light weight of the barge is ten percent of the total volume of the barge"

That is a reasonable rule of thumb - if worded a little incorrectly. For example if your barge was 10m deep x 30m wide and 60m long it would have a total volume of 18,000m3. The self weight (light weight) could be in the region of 1,800 tonnes. That is the only way the 10% "rule" makes sense.

Then you need to get back to your original problem. How deep will it float ?

If your barge was really a cuboid and really did weigh 1800 tonnes when empty it would float in fresh water at a draught of 1m because 60 x 30 x 1 = 1800 cubic metres. And Archimedes tells us weight = displaced mass.

If you load your barge with stuff it will of course be heavier and will float deeper.