Gary,

Some observations on your analysis of the Electra's buoyancy:

Your airframe displacement buoyancy calculation uses the total empty weight of the aircraft to find the volume of seawater displaced, assuming everything in the plane was aluminum.

...So, it's by no means clear that the plane would simply flaot away from the island, never to be seen again. While that's a possibility, it's also possible that whatever buoyancy the plane had was compromised during the chain of events leading to its departure from the reef, and the plane sank in the vicinity of the island.

Bob

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All good points, Bob.

Responding in the same order.

1. I assumed that the cabin and the wings were flooded and provided no buoyancy, the only buoyancy being provided by the empty 1151 gallon fuel tanks plus 2/3rds empty 75 gallon oil tanks (50 gallons of air in the oil tanks) plus the buoyancy of the aircraft structure itself. The empty tanks total 1201 gallons so produce a buoyancy, in seawater, of 10,208 pounds equal to the weight of 1201 gallons of seawater. So, without even considering the contribution of the structure, there would be excess buoyancy of 3,208 pounds and approximately one-half of the plane, whatever its attitude, would be floating above the waterline.

2. I used the simplifying assumption that the structure was all aluminum but, obviously, some parts were made of steel and other denser materials. Steel is more than twice as dense as aluminum but there was a lot more aluminum in the structure than steel so the average density would have been closer to aluminum than to steel. Had the plane been entirely of aluminum, then 7,000 pounds of aluminum immersed in seawater would have produced an additional 2,676 pounds of buoyancy added to that of the empty tanks for a total 5,884 pounds, almost the entire weight of the plane, so the plane would have floated with most of it above the waterline. But, as you point out, the portion out of the water would not contribute to the total buoyancy (I did subtract 400 pounds for this factor in my prior post) so the plane would actually float lower in the water but no lower than approximately half way since that much buoyancy was provided by the empty tanks alone. Even if the entire plane had been made of steel, then the structure would still have provided 918 pounds of buoyancy for a total of 4,126 pounds so the plane would have ridden lower in the water but most of it would still be above the waterline. To take this to extremes, even if the plane had been made entirely out of lead, 7,000 pounds of lead immersed in seawater still provides 636 pounds of buoyancy so the total buoyancy would have been 3,844 pounds causing the plane to float quite nicely.

3. I do assume that the tanks were intact but this seems quite reasonable since the landing on the reef was gentle enough to end up with the plane still standing on its landing gear so that the engine could be operated. According to FAR 23.303 and FAR 23.337 and the parallel 1937 regulations, aircraft structures were required to be much stronger than needed and the Electra's structure, and its components, had to be strong enough to support 3.8 times its weight plus a 50% margin or 5.7 times its designed gross weight of about 10,000 pounds so the structure could support 57,000 pounds before suffering permanent deformation. The tanks structures had to be designed to deal with the loads imposed with them full of fuel and with them empty the forces developed by the tanks in a sudden deceleration would be trivial and easily contained by the structure. And this applies to jolts from swells too.

4. Still using the simplification that the structure was entirely aluminum providing 2,676 pounds of buoyancy, to keep the 7,000 pound plane afloat would only require an additional 4,324 pounds of buoyancy provided by empty fuel tanks meaning that it would take only 509 gallons total of intact fuel tanks to keep it afloat. This means that it would take damaging many of the tanks to make the plane sink. There were 10 fuel tanks total so if even only the smallest 6 held air the plane would float and this does not take into consideration the oil tanks. If the oil tanks were intact then it would take only the 5 smallest fuel tanks to keep the plane afloat. If only the four largest fuselage tanks held air then the plane would float, again without taking into consideration the oil tanks. Taking into consideration the oil tanks, then even the 4 smallest fuselage tanks would keep the plane afloat.

5. Once the tanks were submerged there would be an upward strain on the tiedowns equal to the buoyancy being created by the air in the tanks, they act like inflatable life vests. The very minimum load that must be designed for is in an upward direction and is 1.52 times the weight of the tank itself plus the fuel contained in the fuel tank. To this must be added the 50% safety factor making the tank tiedowns capable of holding 2.28 times the weight of the full fuel tanks. Aviation gas has a density of 6 pounds per gallon and seawater has a density of 8.5 pounds per gallon, only 1.41 times that of gas. Thus the buoyancy exerted by empty fuel tanks immersed in seawater can never exceed the 1.52 minimum structural design limit without even taking into consideration the required 50% safety margin. An example will make this clear. The 148 gallon fuselage tanks hold 888 pounds of gas so the tiedowns must be designed to restrain 1.52 times this weight (plus the weight of the tanks themselves) so must be strong enough to hold down at least 1,350 pounds. Add the 50% safety factor and the minimum strength must have been 2,025 pounds. Since seawater has a density of 8.5 pounds per gallon, the maximum amount of buoyancy that could be produced if this empty tank was entirely submerged is 1,258 pounds, equal to the weight of 888 gallons of seawater, well within the design requirements of the structure.

6. So if the plane were swept off the reef it is unlikely that it sank rapidly so a search limited to the sea bottom in close proximity to the edge of the reef is likely to prove fruitless.

gl