Wednesday, July 18, 2007

Heavy Lifting

How much does that 10,000 pound keel weigh? Divers know that the answer begs another question: "submerged or dry?" Regardless of how much something weighs on dry land, once it is submerged, it is displacing water, and because water gets denser as it gets deeper, a displacement of water by any object, no matter how heavy, creates a force towards the surface.
Of course, metals, and all materials denser than water will still sink when submerged, but the interesting question for salvors is: how much force (or lift) will it take to get this material back to the surface? The answer may surprise you.

Say you have a wood sailboat on the bottom with a big hole in it. The boat's total dry weight is #20,000s (not displacement, but actual dry weight). Assume the boat has a #10,000 lead keel. So, you have to lift 10,000#s of lead and 10,000#s of Mahogany. How much lift do you need? We know the mahogany wants to come to the surface on its own, but how much lifting force can you expect the wood to add to the operation? Will the lead need #10,000s of lift?

If you know the specific gravity of a material, you can do some calculations to figure out how "heavy" that material is when submerged in FRESH water, which has a specific gravity of 1. A number of organizations publish a table, Factors for Converting Various Boat Materials from Dry to Submerged Weight, that makes figuring this out really easy. The one below is from ABYC, but may not be very readable. [click here for a readable version]

The raw calculations aren't all that complicated, but using this table and the "K Factor" is darn near foolproof. All you need to do is find the material on the table, and multiply the total dry weight of that material times the K Factor from the third column of the table to find the submerged weight. Be careful with the minus numbers of "buoyant" materials. Stuff that has a specific gravity less than water will have a negative K factor, meaning that your result will be a negative number, aka lift.

Apply this easy math to our sailboat problem. Honduras Mahogany has a K factor of minus (-0.78), so #10,000 of submerged mahogany will weigh 10,000 x (-0.78), which equals -7,800: minus 7,800, so the result is #7,800 of lift!

Now do the same for the lead keel: the K factor for lead is 0.91, so 10,000 x 0.91 = 9,100 pounds for the keel. Add the two submerged weights together: #9,100 of lead and a minus #7,800 of wood leaves only #1,300 of stuff to lift. If you add just #1,301 pound of lift to this boat, up she comes!

Obviously, the numbers aren't so easy in the real world. Boats aren't made of just two materials. They have cabinetry, wiring, pluming, machinery...but knowing how much lift you will need just to raise the hull material is really important; why waste time rigging #10,000 of lift when half that might do the job? My example sailboat doesn't exist, with half the weight of the boat in the keel, but the calculations are all the same.

A close look at the table reveals some very useful data. Fiberglass Laminate has a K factor! So does gasoline and Ferrocement. Suddenly, raising a #50,000 Ferrocement (K factor .58) sailboat with #9,000 of steel ballast (K factor .88) and 400 gallons of gasoline (K factor -0.37) in an aluminum tank might be possible with lift bags instead of a crane barge.

Feel free to print the table out and use it. The table is based on FRESH water. As salt water is a little denser than fresh, submerged objects in salt water will actually need just a tad less lifting force than they would in fresh. The difference is a pretty small number, and using the K factor in salt water will work just fine.