Bouyancy formula

[Bruce1947]

I am trying to determine how much an all aluminum boat will sink in fresh and salt water. Is there a formula that can be used? Are there any books I could buy or websites I could visit to help me to determine this for various boats?

[rasorinc]

generally fresh water weighs 62 pounds per cubic foot and salt water about 64 pounds per cubic foot. any thing that weighs less then those numbers will float. Weigh your boat and provide flotation material for 150 percent of the boats weight. Salt water floats slightly more weight than fresh. Aluminum only floats because of displacement.

[nukisen]

I do usually count with 1,028 /m3 for salt water and 1 for fresh water.
If possible put the hull into freeship and it will calculate it for u.

[nukisen]

And then you can calculate the weight of the ship

[Ad Hoc]

It is just the ratios of FW to SW or if you wish mathematically:

Displacement = water density x volume

Water density or SG
FW = 1.000
SW = 1.025

vol = LxBxTxCb.

L= length
B= Beam
T = Draft
Cb = block coefficient. (if you not sure just use say 0.6 for both calc's, doesn't matter if constant)

Units SI/metric, not imperial.

[Fanie]

Geezzz guys, didn't we decide on the other thread we are going to use metrics :D

Weigh your boat and provide floation material for 150 percent of the boats weight.
This is very conservative, I would think the ratio should be much bigger, I would think more to 300 or 400 percent at least, but that would also depend on the boat size and purpose.

[nukisen]

Me and AdHoc gave a little different dens of saltwater.
Both can be used and will not make a very big different.
The salt water dens. is cause by salinity. High salinity higher density.

AdHoc i Have seen that you always know ur things.
:)

Have a nice day!!

[Ad Hoc]

nukisen
"..AdHoc i Have seen that you always know ur things..."

Well, it is my daily job....I'd be worried of didn't know these "things" :P

Thanks for the compliment though ..:)

[Bruce1947]

It is just the ratios of FW to SW or if you wish mathematically:

Displacement = water density x volume

Water density or SG
FW = 1.000
SW = 1.025

vol = LxBxTxCb.

L= length
B= Beam
T = Draft
Cb = block coefficient. (if you not sure just use say 0.6 for both calc's, doesn't matter if constant)

Units SI/metric, not imperial.

Ad Hoc, thanks for the reply. I guess I am a little dense, but I have some more questions.

1. What is the block coefficient? Is it different for different boat materials, or maybe it doesn't matter?
2. If I have a cabin on the deck of the boat with doors, do I include it in the volume calculations?
3. How do you know what the draft is if you haven't a boat to check it?
I assume draft is defined as how much of the boat sits underwater?
4. What is "Units SI/metric?

New question on stern or bow heavy: How can I determine whether an all aluminum boat will sit evenly in the water? Do you know any websites or books that can help me learn about buoyancy and displacement?

Thanks, Bruce

[Ad Hoc]

Bruce1947

Not dense, no. You're just not familiar with the all this "boat" terminology and its definition.

Most of your questions can be answered by simply buying a book on naval architecture, a simple one. Since as i ahve just noted, most of the "words", such as block coefficient etc are all basic definitions with exact definition in relation to a boat. Answering one of these question swill lead to another Q, because it is all new terminology.

Therefore, it is best to get some basic book that explains the basics of naval architecture. Otherwise we could be here a long time!!:eek:

[yipster]

It is just the ratios of FW to SW or if you wish mathematically:

Displacement = water density x volume

Water density or SG
FW = 1.000
SW = 1.025

vol = LxBxTxCb.

L= length
B= Beam
T = Draft
Cb = block coefficient. (if you not sure just use say 0.6 for both calc's, doesn't matter if constant)

Units SI/metric, not imperial.
thats the "formula" you asked for Bruce but i too had to think a sec on FW / SW
if you want it real easy do a metric quick rough one: 1 klogram displaces 1 liter water

[Fanie]

Bruce, you don't have to learn a formula for it, it is rather common sense. If you push a 2L coke bottle under the water you are displacing 2L of water.

If you dunk a 1000kg hull on the water then you displace 1000kg or 1 cu meter of water, even if the hull is not 1m x 1m x 1m. It is the weight of the floating item displacing the 1000kg.

The size of the thing you have, when submerged will be the total volume of the water you can displace. Volume is length x width x height, you know that already.

Just keep in mind you want the boat to stick out on top of the water some, so don't calculate the full volume for UNDER the water :D

Even heavy boats can seem to sit on top of the water, and not sink into it.
Only keeled mono hulls does that and there, I've just started the next forum fight :D

[Scott Carter]

Bruce,
A very basic but IMPORTANT distinction you need to remember to make (this is in reference to your question #2 above) is that the volume we refer to has nothing to do with the cabin or other appendages of the boat (even a lot of the hull itself). The volume which is to do with displacement is the volume of the submerged section of the hull, referred to as the underbody. The rest of the vessel (cabin, etc) does however affect the overall weight of the boat, and so therefore how far it sinks into the water as a function of its displacement.
If the boat has a waterline evident you can get a pretty close aproximation of the boat's displacement just by envisioning the underbody as a series of triangles, calculating the area of these various size triangles along the length of the hull and multiplying by the length of the hull they represent to determine their volumes, the sum of which equals (approximately) the volume of the underbody of the hull. This volume, multiplied by the density of water (as given in the above responses) will give you the boat's displacement, and thus its weight. Conversely, if you know the weight of the boat, divide this number by the density of the water you'll put her in to, and this will yield the volume of water displaced by her partially sunken hull. It'll then be up to your math to determine at what depth of the hull (draft) this volume is reached.

[pamarine]

Rhinomarine works great for calculating static and dynamic waterlines. But, you would need a 3-d model of the vessel with weight and cg information.

Basically what Ad Hoc and Scott have said is the answer. The displacement of a vessel is fairly easy to compute. It is equal to the vessel weight.

Finding the static waterline (which I am assuming is what you are after here) is determined by the hull shape and size (volume) that causes the displacement of an amount of water equal to the weight of the vessel. Ad Hoc's formula will tell you this.

The next step is probably the most difficult to do without a computer program (not really difficult, just labor intensive) and that is to find what portion of the hull gives you the volume of water calculated above.

To find the actualy WL you will need to know the Center of Gravity of the vessel. As the CG moves fore and aft, the hull will pivot around the lateral axis (bow will rise and fall) and as the CG moves athwartship the hull will pivot longitudinally (list port or starboard). Ideally you want the CG on the hulls centerline and I typically will try to place it in the middle third of the hull's legth but abaft the geometric center (for a planing hull). This will give you a slight bow high attitude at rest. once you know the vessel's attitude you can then figure out where the WL needs to be to give you the desired displacement.

Now I am self taught so excuse the lack of polish in my answer, but I hope it helps

[trevormlb]

So if I have a boat with a 500lb weight capacity in fresh water what is the weight capacity in salt?

[apex1]

So if I have a boat with a 500lb weight capacity in fresh water what is the weight capacity in salt?

Was´nt the above quite clear enough???
then:


http://www.consoft.de/Default.aspx?art=multicalc

[Fanie]

1.025 x 500

[conceptia]

the density of fresh water is 62.3 lb/cubic feet, while that of sea water is 64 lb/ cubic feet. so 500 X (64/62.3) will give you the weight capacity in sea water.

[conceptia]

1.025 x 500

Fanie, dont get confused over with the units. 1.025 is in SI and 500 is in US units.

[Ike]

not much more 500/60.4 = 8.278
8.278 X 62.4 = 516.556 (517 lb)

However small boats are generally rated for worst case which would be fresh water and that gives the lower figure.

Most of the answers above have gotten way to complex.

Scott and pamarine came closest.

For boats under 20 feet (6.96 meters) finding bouyancy and weight capacity is pretty simple. Actually the easiest way is if you have a lot of old weights laying around. For 500 lb (227 KG) you'll need 2500 to 300 pounds. Put them in the boat distributed evenly until water starts to come in at the lowest point. This takes a little juggling of weights.

The total weight is the displacement weight. You can compute the volume by dividing the weight by the wieght of a cubic foot of water or a cubic meter if that's your flavor.

But the maximum weight capacity is 1/5 of that for outboards and 1/7 of that for inboards.

displacement weight/5 or displacement weight/7

You can do this in reverse by calculating the volume of the boat below what is called the static float plane, that is, the waterline where water would start coming in. As was said if you have the boats lines in a computer program, many of them will do this for you, you just specify the waterline and it tells you the volume.

Maximum weight capacity minus engine weight gives person weight (for outboards)

For Inboards the engine weight is part of the boat weight so Maximum weight capacity can equal persons capacity but most builders use a smaller persons weight to leave a safety margin.

Persons weight + 32 /141 = persons. (yeah that's a screwy formula but it is the one that the USCG, ABYC, Canada and ISO all use to compute number of persons)

[Paul Kotzebue]

1.025 is in SI

That is not correct. 1.025 is the specific gravity of sea water and has no units. It is applicable to both SI and US systems, so Fanie is right.

[conceptia]

do we get weight of an vessel by multiplying its volume with specific gravity?

[Paul Kotzebue]

do we get weight of an vessel by multiplying its volume with specific gravity?

No. Weight = volume * density. And density of sea water = specific gravity of sea water * density of fresh water, regardless of units. So, weight of sea water displaced = specifc gravity of sea water * weight of fresh water displaced, regardless of units.

[Wynand N]

do we get weight of an vessel by multiplying its volume with specific gravity?

Very elementary issue as Paul explained and one of the very basics one learns in boat design studies regardless of institution as a boat designer or by going a bit better, Naval Architect.

However, what mystifies my mind; you clearly state at your name "Naval Architect"... :confused:

[hoytedow]

No. Weight = volume * density. And density of sea water = specific gravity of sea water * density of fresh water, regardless of units. So, weight of sea water displaced = specifc gravity of sea water * weight of fresh water displaced, regardless of units.The weight of sea water displaced would equal the weight of fresh water displaced. the VOLUME of the sea water displaced would be less, so the boat would float higher.

[Paul Kotzebue]

The weight of sea water displaced would equal the weight of fresh water displaced. the VOLUME of the sea water displaced would be less, so the boat would float higher.

True for constant weight.

I stand by my statement, which is correct for constant volume. I was considering the question asked in post #15:

"So if I have a boat with a 500lb weight capacity in fresh water what is the weight capacity in salt?"

And the response in post #17:

"1.025 x 500"

[hoytedow]

Fanie is a reliable source of information.

[PhotoBoatGuy]

Trevormlb:

If your boat will hold 500 lb in fresh water and you want to maintain the same waterline in saltwater, it will hold 500 * 1.028 = 514 lb.

Cliff W Estes
BaseLine Technology

[murdomack]

Trevormlb:

If your boat will hold 500 lb in fresh water and you want to maintain the same waterline in saltwater, it will hold 500 * 1.028 = 514 lb.

Cliff W Estes
BaseLine Technology

At last, a clear answer. I must admit I was a bit lost till I read your post.

[Submarine Tom]

And it only took 28 posts to get there.

LOL!

[PhotoBoatGuy]

"Every once in a while, I get a little wood on the ball..."

Glad I could help.