Proof the K factor works

Remember this blog back in July where I posted a chart and the formula for calculating submerged weight? The chart shows a simple mulitplier, called the "K Factor". Well, Ham Gale and I had a perfect opportunity to prove out the theory back in October up in Annapolis. A Soling sank in about 30 feet of water. According to this spec sheet

the Soling displaces 2277 lbs. Because Soling's are usually dry sailed (stored on a trailer), they are equipped with four short cables that remain attached to the keel bolts, and the entire boat is designed to be lifted by these cables (how handy for us salvage divers, huh?). So, I was able to access the lifting cables down there in 30 feet of cloudy Chesapeake Bay, and attach a single 2000lb lift bag. Both Ham and I were pretty confident that the 2000# bag would bring the 2277# boat to the surface.

Guess what? Here it is, after the initial lift, hanging by one single 2000# lift bag, just barely submerged. Notice that the lift bag is well out of the water, which means that at this point it is generating significantly less than 2000# of lift. Solings do not have any internal floatation, other than two small compartments (one fore, one aft under the decks) that are supposed to be kept closed. The reason this boat sank is that the skipper had failed to close the forward one. They are not really water tight compartments; but it would take a while for water to seep in there in the event of a capsize. When we hauled the boat at the crane, we found both compartments totally flooded.

Lets do the math. Unfortunately, I don't find a seperate listing of how much ballast they carry, but lets say its 1/3 of the total, which means that the keel weighs about 750# The K Factor for lead is .91, so 750# of lead submerged will only need 682# of lift. Lets say the mast, boom & rigging weighs 200#. In the picture, you'll notice that virtually all the mast and rig is out of the water, so we'll leave its weight at 200#, because we're going to lift it completely out of the water.

Now, the remaining fiberglass hull would weigh 1327# dry. Multiply the K Factor for fiberglass of .33, and the hull only needs about 440# of lift.

Lets add it together: 440 for the hull, 682 for the lead keel, and 200 for the rig = 1322# of lift calculated to bring this mass to the surface. Look once again at the picture; doesn't it look like about one third of the bag is out of the water? That leaves two thirds of a 2000# bag creating lift. Lets see, two thirds of 2000 is 1320!

IT WORKS! (thanks to Capt Cory Deere of TowBoat/US Annapolis for the nice picture. And thanks Ham for the nice day's work!)