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Syntactic Foam:

WHAT IS SYNTACTIC FOAM?

Syntactic foam is a class of material created using pre-formed hollow spheres (commonly made of glass, ceramic, polymer or even metal) bound together with a polymer. The “syntactic” portion refers to the “ordered structure” provided by the hollow spheres. The “foam” term relates to the cellular nature of the material. Thanks to its unique properties of high strength at low density, syntactic foam has become widely used for a variety of subsea and flotation applications.

HOW TO CALCULATE BUOYANCY

The scientific principle of buoyancy is calculated by measuring the weight of fluid displaced by the object in it. An object placed in a fluid displaces a certain amount of the fluid equal to the object’s volume. While the object’s weight pushes down on the fluid (liquid or gas), an upward force is exerted on the object, creating a resistance to gravity. The weight of the fluid is equal to the buoyancy force acting upon the object. As a typical example, one might want to know how much syntactic foam would be required make 100 lbs of steel neutrally buoyant in seawater.

First you would need to know how much the steel “weighs” in water. This is calculated by the simple equation of:Weight in Sea Water = (Density of Sea Water – Density of Steel) x Volume of Steel.For example, assume a Sea Water density of 64.0 lbs/ft3 and a Steel density of 495 lbs/ft3,This means the 100 lbs of steel has a volume of 0.202 ft3.Water weight then becomes: Weight in Sea Water = (64.0 – 495) x 0.202 = – 87.06 lbsThis means we would need to provide 87.06 lbs of lift to make the steel neutrally buoyant. If we used a 24 lb/ft3 syntactic foam we can plug into the same equation to solve for the volume: Weight in Sea Water = (Density of Seawater – Density of Syntactic) x Volume of Syntactic 87.06 = (64.0 – 24.0) x Volume of Syntactic; 2.177 ft3 = Volume of Syntactic

Any volume of syntactic greater than this and the structure would become buoyant and float. Any less and the combination would sink.

First you would need to know how much the steel “weighs” in water. This is calculated by the simple equation of:Weight in Sea Water = (Density of Sea Water – Density of Steel) x Volume of Steel.For example, assume a Sea Water density of 64.0 lbs/ft3 and a Steel density of 495 lbs/ft3,This means the 100 lbs of steel has a volume of 0.202 ft3.Water weight then becomes: Weight in Sea Water = (64.0 – 495) x 0.202 = – 87.06 lbsThis means we would need to provide 87.06 lbs of lift to make the steel neutrally buoyant. If we used a 24 lb/ft3 syntactic foam we can plug into the same equation to solve for the volume: Weight in Sea Water = (Density of Seawater – Density of Syntactic) x Volume of Syntactic 87.06 = (64.0 – 24.0) x Volume of Syntactic; 2.177 ft3 = Volume of Syntactic

Any volume of syntactic greater than this and the structure would become buoyant and float. Any less and the combination would sink.

WHAT DENSITIES ARE AVAILABLE FOR ESS SYNTACTIC FOAM?

Please

__click here__for Density Chart.