# How do you mathematically convert a ship's displacement of water to its weight on land?

This assumes it’s floating on fresh water (density = 1). The only thing i can understand is Archimedes’ rule that a sinking ship’s weight is reduced by its volume x density of the water. But if the weight is less than this amount, it will float. Corrections are very welcome.

It’s very simple. A floating object displaces a certain amount of liquid (in which it is floating). The weight of this displaced liquid is equal to the weight of the floating object.

That’s everyone’s hunch. But lemme just clear my head:

1. The density of the medium is irrelevant since a different density simply changes the amount displaced.

2. Now this is what confuses me. Given two ships of the same weight on land, same load. But one’s hull is smaller in terms of volume so it will ride lower in a given medium, since it’s more dense overall, and therefore will displace more medium. How do you fix that now?

You just explained it.
The hull is smaller, so it rides lower.
Same displacement.

To add some additional detail: As the density of the liquid in which the object is floating increases, the smaller the volume of liquid that is displaced.

Seawater is more dense than fresh water, so a floating object in seawater will displace a smaller volume of liquid than that same object floating in fresh water. The weight of both liquids that is displaced will be the same (because it will always be equal to the weight of the floating object).

An object floating in liquid mercury (which is a very dense liquid) will displace a correspondingly smaller volume of liquid. The weight of this displaced liquid will still be equal to the weight of the floating object.

:smack::smack::smack:

It’s more than a hunch. Indeed, this has all been well-established for quite some time.

Changing the density of the liquid mean that the volume will be different, but the weight of the displaced liquid will be unchanged.

You were doing fine until the last [incorrect] statement. Both ships will displace the same amount of water. The more dense ship with the smaller hull will ride lower in the water in order for its smaller hull to displace that same amount of water.

In short, what beowulff wrote while I was typing this…

Still more detail: what about a fully submerged object?

A fully submerged object will displace a volume of liquid equal to that of the object. If the weight of the object is greater than that of the displaced liquid, then it will sink. Its apparent weight will be reduced by the weight of the displaced liquid. (The weight of the displaced liquid is also referred to as the force of buoyancy.)

If the weight of the object is equal to that of the displaced liquid, it will be neutrally buoyant, and will neither rise nor sink.

If the weight of the object is less than that of the displaced liquid, it will rise until it reaches the surface of the liquid, and will continue to rise until it only displaces a volume of liquid whose weight is equal to that of the object. The object is now floating, and you are back to the original case of a floating object.

Positive or negative buoyancy, within limits, can be cancelled out (or augmented) by hydrodynamic effects achieved through forward motion. This includes the diving planes on a submarine, but it also applies to hydrofoil boats.

Good point. All of the previous discussion was, of course, in the absence of hydrodynamic forces.

Incidentally, this has a practical consideration onboard a military nuclear submarine. Submarines are constantly changing weight (e.g. by shooting torpedoes or other weapons, emptying sanitary holding tanks, and shooting trash). The latter two practices are only done in deep water that far from shore, and the trash is encased and compacted in a perforated metal container.

The sub is also constantly bringing on seawater for the distiller/evaporator to make fresh water for shipboard use.

The submarine adjusts its net weight to remain neutrally buoyant (or close to this) by filling or emptying the variable ballast tanks with seawater. This is not always an exact science, and the crew uses various “thumb rules” and their own observations as to how buoyant the sub appears to be.

Anyway, when a nuclear submarine is traveling submerged at high speed for an extended period of time (such as during a high speed transit across an ocean), it can be easy for this to all get out of whack, as the hydrodynamic forces become a much larger factor during the run. The personnel operating the sub can find that the sub is very heavy (or light) as it slows down, which requires quick corrective action.

Wouldn’t it be easy to know that this situation is developing? ISTM that the vessel’s net buoyancy could be estimated at any given speed by knowing the actual angles of the dive planes required to maintain a given depth, and comparing those actual angles to the angles that would be required if the vessel were in fact neutrally buoyant. A large discrepancy means a large excursion from neutral buoyancy; this knowledge would allow buoyancy correction to be applied before slowing to a speed at which the planes provide inadequate control authority.

What am I missing?

Anvil in Mercury.

Does not the shape of the hull come into play? Would not two ships of the same actual weight displace a different volume of water if one was flat bottomed (like a barge) an the other a traditional “V” shaped hull?

I was reading a WW2 submariner’s autobiography, and he stated that as a submarine dove? dived? descended further into the ocean, the sub’s hull would be more and more compressed by the pressure of the water into a smaller volume, meaning that it now displaces less water, meaning… it’s buoyancy changes?

Nope, volumes will be the same, just shaped differently.

Definitely - there’s no way to make a perfectly rigid and incompressible submarine.

It’s also true that as you descend, the density of the water increases - though not by much: Googling suggests a depth of 4km yields a density increase of about 1.8%.

Not in a steady state while standing still, the basic principal of hydrostatics. When the vessels are moving, it’s possible that hydrodynamic forces can result in a difference between displacement and weight. That’s negligible generally for larger slower surface vessels of any hull shape, but can be non-negligible for some small relatively high (for their size) speed vessels. In the extreme, a hydrofoil is displacing almost no water when going fast enough to get up on its foils, but when standing still it displaces its weight in water like every other floating object.

This is an effect that is quite noticeable for divers (SCUBA). As a diver descends, the air in the buoyancy compensator (or dry suit) is under greater pressure and decreases in volume. Since it displaces less water (which retains basically the same density at all depths), it becomes less buoyant and air has to be added to compensate for it. The opposite occurs when a diver goes toward the surface. Air has to be released periodically or the BC will expand dramatically and accelerate the movement upwards, which is usually not desirable.

Ok, thanks.

The book, as I stated, was in reference to WW2 submarines, which typically displaced 1000 to 2000 tons. Their smaller size may play a factor here.

I only encountered this fact (the buoyancy change due to cruising depth) to be stated in only one of many biography/autobiographies that I read from that period.

I also recall many of these books discussing how the crewman/officer responsible for keeping the sub’s trim may have to pump water from tank to tank as the center of gravity in the boat shifts (as crewman and cargo are shifted).

It just occurred to me that if the trim officer had to compensate for the few hundred pound weights of two or three crewman moving aft, then the change in buoyancy due to depth may have to also addressed by this same man.

:eek: Do they have the help of gauges for that, or do they do it “by feel”?