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Dave and Anke
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Sunday, December 22, 2013

Box Barge Displacement: Archimedes 101

Plimsoll Lines from the Wider World of Displacement
 Box Barge Displacement: Archimedes 101

Story goes that, as Archimedes eased himself into his bath, the water he displaced ran over the edges and onto the floor. 

Nothing but a mess, for most of us, but Archimedes made a sudden connection; he knew, of course, that the volume of water he displaced was equal to his own, immersed volume. But what was not so obvious; The weight of the water he displaced must equal the weight of his entire self... specs, toupe, false teeth and all!

He got so excited by his discovery that he ran down the streets of Syracuse, buck-nekkid, shouting Eureka!!! Must have been somethin'... we're still talking about it 2000+ years later!

The actual Archimedes principle goes on to state that the floaty force acting on a floating object is equal to the weight of the displaced fluid.

So why do we care?

Well, we live aboard floaty objects. We want to know how much they can carry; how deep their hulls reach into the water; how much freeboard will stick out; how much it will settle if we move our anvil collection aboard; how much sail it takes to drive them; how they will float on their sides or upside-down. These and many other questions are informed by calculating displacement.

It involves a wee bit of math (Eeeeek!), but don't be frightened!!  Curvy Dogs require elaborate calculus, but we - box bargers - need only simple 'rithmetic. [Niener!]

The gist is that we want to convert our underwater (immersed) shape into a simple slab, whose volume is easy to work out. Being square sectioned (slab-sided), and rectilinear in plan view, box barges are already half-way there! One cheap trick is all we need.

Here's a walk through: 

Cheap Trick is close enough for Jazz... 
What we're doing, here, is combining the two wedgy slices at the ends into a single slablet, and adding that to the middle slab over the deadflat. We don't really need to flip one, as seen in step three... it's enough to understand that this is, in effect what we're about.

NOTE: Due to plywood standard dimensions, TriloBoat math is a snap in the Imperial System (feet, inches, eighths and pounds). The following can be done in Metric, but sadly, generates funny numbers and mistakes. So we console ourselves with a pint and work in the yoke of vanished Empire.

A, B and C are all linear distances between points along the waterline. Beam, Draft and Total are dimensional distances (at right angles to one another). When we multiply these together, we end up with Volume in cubic feet (ft3).

Once we have Volume, we multiply it by the weight of water in pounds per cubic foot (lbs/ft3). 

Fresh water weighs about 62.4lbs/ft3. Sea(salt)water is heavier at about 64.3lbs/ft3. A designer would choose one or another based on where the boat is expected to be used. It's not a huge difference, but does add up. The upshot is that the boat will float a little higher in saltwater. [Figures like these can be found in the Pocket Ref or searched for on-line.]

Displacement = Volume x Weight.of.Water / ft3

That's the general picture. Let's try an example from my point of view as designer:

Let's say I draw out a T32x8 on graph paper, for use in salt water. I decide that the draft will be 1ft and draw that in. Next I count squares along the water line, and find that:

A = 5.5ft, B = 16ft and C = 5ft

Total = (A+B)/2 + C
         = (5.5ft+5ft)/2 + 16ft
         = (10.5ft)/2 +16ft
         = 5.25ft  + 16ft
         = 21.25ft

Tell ya the truth, if the ends are this similar (which they usually are in TriloBoats), I'll cheat again and just call the wedges equal, and therefore A and C are too. Since they're the same, we don't have to average them; C simply completes A.So the above simplifies even further:

A = 5ft and B = 16ft
Total = A + B
         = 5ft + 16ft
         = 21ft

The difference between them only rounds a skosh downwards; ultimately, a bit of extra displacement is a pleasant surprise.


Volume = Beam x Draft x Total
              = 8ft x 1ft x 21ft
              = 168ft3

And (I reach for my calculator):

Displacement = Volume Weight.of.Water / 1ft3
                         = 172 ft3 x 64.3lbs / 1ft3
                         =  10802.4 lbs                         ... notice that the ft3s cancel out.


This boat displaces about 10800lbs. That is, by Archimedes Principle, how much she'll weigh, fully loaded, including the boat itself, gear, supplies, crew and the dog, when loaded to her Design Water Line (DWL) - the waterline as intended by the designer.

But lo! In the course of time, we tend to accumulate stuff, and each gimcrack and doohickey settles us a little deeper in the water.

For extra credit, let's calculate a displacement related number, Pounds Per Inch of immersion (PPI... somebody, somewhere dropped 'I' for Immersion!). This number is the amount of weight required to lower the vessel one inch deeper in the water. 

Here's the formula for our 32ft box barge:

PPI = DWL x Beam x (1in x 1ft/12in) x Weight.of.Water/ft3

That odd bit in parenthesis is a ratio, which merely converts inch units to foot units (one inch being 1/12th of a linear foot, or about 0.833ft). In our example:

PPI = 26ft x 8ft x 1in x (1ft/12in) x 64.3lbs/1ft3
       = 26ft x 0.833ft x 64.3lbs/1ft3
       = 1392lbs

That's a lot of seashells!

Note: Box barges with high ends load gracefully with weight secured low in the hull. But there are limits, and exceeding them can be deadly. Be aware of them! If in doubt, the Coast Guard is happy to conduct free stability tests for your design or boat.


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