Sailboat weight

[sailortonyb Allied Mis]

The Listed Displacement of a boat

is equal to the boats weight at the designed waterline. This assumes that the designed waterline is correctly placed. Most manufacturers will also tell you if asked, how much the boat will sink for each 1,000 lbs addded. That is b-e-c-a-u-s-e, when the boat sinks a few inches, it takes up more space under water (displacing more water) and weighs more. Plain and simple, the weight of a boat's displacement is equal to the weight of the water it displaces. Tony B

 

[newly anonymous]

displacement is a theoretical calculation...

...weight is an empirical quantity.Try figuring this one out: would an aluminum barge displace more or less than a steel barge built to the same dimensions?

 

[sailortonyb Allied Mis]

If the hull thicknesses were

also the same, the steel would displace more. The steel is heavier and will sit lower in the water, therefore displacing more water.Displacement is not a theoretical calculation in itself. As for boats, yes, then it is theoretical because the calculations are way too complicated to be able to factor in everything at a reasonble cost because naval architects charge for their time.Weight is NOT emperical. It is measurable and if you want to get really anal about it, it can be done accurately at any altitude or barometric pressure if need be. Ever hear of the abbreviation STP, meaning Standard Temp and Pressure. Tony B

 

[Ross]

Newly, You need a good dictionary.

If an object floats on water it will diplace a quantity of water equal to its weight. The volume of the object doesn't enter into the equation. A 100 tom aluminum barge will displace 100 tons of water and a 100 ton steel barge will displace 100 tons of water. The aluminum barge MAY have a greater or lesser volume than the steel barge depending upon the scantlings of each vessel.

 

[Steve Braverman]

Which weighs more....

a pound of feathers, or a pound of bricks?

 

[Ross]

Steve if you put a pound of feathers and a pound of lead into plastic bags and d

river which one will displace the most water? edited because fat fingers hit 2 keys at one time.

 

[Scott]

I think Ed makes a good point ...

And his milk carton analogy is a good illustration. I always figured that displacement is the same as weight but now I'm not so sure. Here is where Ed was going with his analogy.If the designer of the milk carton drew a line around the milk carton about 3/4's of the way up, he could say that the displacement of the milk carton is 6 pounds (a gallon of water weighs about 8 pounds, right?). Of course the milk carton itself may weigh less than an ounce, but that isn't important to the designer. The designer wants the carton to rest in water (and maybe be sailed in water) at a particular waterline for stability and performance characteristics. If he wanted the carton to rest in water just barely at the surface, he might say the displacement is 8 pounds.So if a barge weighs 30,000 tons but the waterline is calculated with a load of 50,000 tons, then I bet the displacement that the barge designer says his barge is would be 80,000 tons.In a sailboat, I suppose that the weight of the boat and the hardware is maybe 85% to 90% of the displacement at waterline (I'm guessing). I bet that could mean that the stated displacement of a cruising sailboat could be the combined weight of the boat itself, plus the assumed weight of the crew and mucho staples, or extra batteries, or generators, or whatever a cruiser could decide is necessary.I'll bet that the stated displacement of a racing sailboat would be much closer to its actual weight because the designer assumes merely the weight of the crew and rigging and sails. The designer would assume that all the cruising gear would not be in the boat.So you might see differing stated displacements in similar boats that are intended for different purposes, and the performance of the boat at a particular waterline is the governing factor in the stated displacement.Oh, to get the weight of your boat ... put it on a trailer and put the whole thing on a scale, then weigh the trailer.

 

[Ross]

Scott , The barge designer would give a light displacement

figure and a maximum loaded displacement figure. The strength of the barge structure also has to be considered. It may have a design loaded displacement of 8,000 tons but still have four feet of free board( one half of its overall depth) if it were loaded to two feet of free board it might break in half because it wasn't designed to carry 12,000 tons. When the barge breaks in two and sinks it will displace the volume of the material used in construction and of the load.

 

[T37 Phil]

Itch

Can you hear the heads being scratched? I just had my Tartan 37 transported overland. The stated displacement of the boat is 15,500. The trucker told me it came across the scales at 22,000. It was never clear to me what gear is included in the stated design displacement and what was left to "load weight". Perhaps it even varies from manufacturer to manufacturer.

 

[Ross]

6500 pounds is a lot of extras. Take away the

spars and standing rigging and the sails and you probably don't have more than 1500 pounds. Batteries, engine, plumbing and electrical are a part of the boat. I wonder where the extra 5000 pounds came from.

 

[sailortonyb Allied Mis]

Tartan

Part of the weight difference could be in tankage and also part of the weight could be in a waterlogged hull from 30 years in the water. Also remember that the displacement weight is for the calculated waterline, not the whole entire vessel underwater...I think. If you add 5000 lbs to the boat by putting sandbags on the deck, the boat may go down another 12 inches into the water. At this point, it will be displacing more water (greater volume under water) but it is no longer at the designed waterline. If this were the norm, which its not, the manufacturer would change the designed waterline to the new mark and publish the new displacement figure at the new waterline.Tony B

 

[newly anonymous]

think about it for a moment

In the world of physics, displacement is the weight of the volume of water displaced by the immersed object. If the displacement were less than the actual weight of the vessel, the boat would sink, because the boat is buoyed up by a force equal to the weight of the water displaced. In the real world of physics, displacement would always exceed actual weight in any boat that floats. However, if you go to the stats page on this website and look up your boat's "displacement," what you'll see is a theoretical computation--based upon assumptions about a theoretical waterline--that was computed by the naval architect who designed your boat prior to when the first hull was molded.In the world of naval architecture, "displacement" has never, never, ever equaled the weight of the boat. It's merely a calculated element of the boat's design. And, Ross, the dictionary in my office is 9" thick. I send my students to it frequently when they make arguments like the one you just made.

 

[Ross]

Newly, You're talking through your

hat.

 

[Ross]

Newly , if you take one pound of lead and roll it into a thin sheet and

bend and fold it into a tray shape and place it in water it will float ane will displace one pound of water. but if you sink that tray it will displace its volume which is about 5.5 pounds of water.

 

[Ross]

Ross reply 33 is Ross reply 32 somehow the site dropped my sign-in.

I am not being sly here. I posted an error in 32 The displacement of one pound of lead submerged in water is 1.4 ounces.

 

[Ross]

I found a 2 pound block of lead and submerged it

in a measuring cup of water. It displaced just under 3 ounces of water.

 

[Stillraining]

We actually

Have three subtopics here...1) Do or do not Designers calculate the actual weight as displacement or not? I believe they do...Displacement numbers however will be given in empty or loaded capacity depending on designer..You must remember that this is a best guess by the designer...if 1500lbl more welding rod or 1/4 " more fiberglass was laid up then "Designed" the displacement figure will be WRONG! 2) Is "Measured" displacement the actual weight of the boat? Yes! ....in other words once the boat is floated it will equal exactly how much water it is displacing to float it...no matter what the designer says..or where the water line has been painted. 3) Is volume of that displacement equal to its weigh? As stated by others "Only when afloat" it will be less when sunk UNLESS it has positive bouncy WHEN sunk and requires outside forces to sink it as in Ross's wood in a bucket analogy...in that case the wood requiring you to physically force it under will displace MORE then its weight...but not more then its volume, which will be equal.Edit to add...My designed displacement is 25,000 its actual displacement is 29,600..no provisions...the designer either did not calculate for such things as aux generators, air conditioning, hot water tanks. anchor road and chain, cabinetry lay out etc ..etc...etc...or he is a lousy guesser..."or" the boat got built much heaver then he designed plane and simple...I choose to believe the later is most often the case.

 

[newly anonymous]

back to Archimedes

If the weight of the displaced water is greater than the weight of the object, the object will float. Boats float, unlike the two-pound lead block that only displaced three ounces of liquid. According to the Archimedes Principle, the lead block in Ross' example weighs three ounces less when submerged; however, it still sinks because its mass exceeds the weight of the volume displaced. How this is different in a solid object from a hollow object, such as a boat hull, is that the total density equals the mass of the hull plus the mass of the enclosed air divided by the volume. Note that this total density is different than the hull's weight.A boat is designed around a waterline. The "displacement" is the theoretical calculation as to the weight of the volume of water that will be displaced for that vessel to sit on its waterline.

 

[Stillraining]

The air

that the boat makes room for IS what gives you the positive displacement...It dosent matter if your hull has 1" gunnel's and a surface area of a foot ball field or 30' gunnel's and is shaped like a canoe...if it weighs 30,000 lbs it will require a displacement of 30,000 pounds of water to flote it regardless of surface area in which to impart its force against the water...but the foot ball field will sure require less draft...and anything other then a glass calm day will sink it..:Designing a boat ship whatever that does not factor in enough positive buoyancy ( which gives you free board ) due to and by that air volume is designing a potentially dangerous boat...Think of it this way a Racing hull weighing 30,000 and a blue water cruiser weighing 30,000 will have a very different internal air volume for the same weight displacement....and one will sit in the water, sail and handel wave and sea state very differently then the other...Why? because one is more buoyant due to the larger volume of air ratio to weight then the other..or Positive buoyancy..We call this a light or heave built boat...our nomenclature has nothing to do with actual physical weight...but the character of its displacement....If you will...a cruiser designed with less positive flotation then a racer. (ie. less internal air space per weight)...it will be slower..sit deeper, be steadier..stronger and cut through thoes waves instead of pounding on top of them because of that less positive flotation at the same 30,000 lbs.Just dont design to little positive floation into your designs or you end up with a anchor

 

[Ross]

Newly you are using ambiguous language that is meaningless

The placement of the water line determines the reserve buoyancy the the hull possesses. Submarines are equipped with ballast tanks that allow them to achieve neutral buoyancy. For a submarine to remain on the surface while not moving it must have positive buoyancy. With neutral buoyancy a bird landing on its periscope would sink it.