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Buoyancy for Divers: Floating Baseball March 4, 2009

Posted by Chris Sullivan in Training.
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I was musing about this problem a few days ago and undertook to work out the solution. This is the problem of how much of an object floats above the water vs. below. I’ve written some posts on buoyancy before, including why objects are more buoyant in salt water than in fresh, the weight of concrete underwater, and the difference in weight required between salt water and fresh, but these are just the tip of the iceberg. Like icebergs, all objects that float, including divers, will be partially submerged, except those which are so light that they can take advantage of the surface tension of the water.

A baseball is a sphere, and spheres are good shapes to try this out on because they are symmetrical in all 3 dimensions. If the ball floats, it will be in the same position no matter what the orientation. I’m going to make things easier by assume that the ball is a perfect sphere and that it is non-wettable – i.e. it will not become even slightly waterlogged.

The first thing to discover is whether it will sink or not. I’m going to use inches and ounces as my measurement units, because that’s what the baseball regulation specify for the ball’s physical properties. A regulation baseball is between 5 and 5¼ ounces in weight, and between 2 7/8” and 3” in diameter. Because the buoyancy of the baseball depends on the displacement of the water it is immersed in, for convenience I’ll convert the density of water from pounds per cubic foot to ounces per cubic inch. So sea water converts from 64 pounds per cubic foot to 0.592488 ounces per cubic inch, and fresh water converts from 62.4 pounds per cubic foot to 0.578037 ounces per cubic inch. If you tried this calculation yourself you might have found something slightly different because I used more precise density values than 64 and 62.4 which I found by using Calchemy.

So all we need to do is find out the weight of water displaced by the baseball. However, the regulations specify a range of legal dimensions for the baseball, so we need to make sure that the prediction is valid for all baseballs. The volume of a sphere is given by the formula V = 4/3 x PI x r3, where r is the radius which is ½ of the diameter. For a diameter of 2 7/8” the volume is 12.4426 cubic inches, and for the 3” ball it’s 14.1372 cubic inches. These respectively displace 7.1923 and 8.1718 ounces of fresh water 7.3721 and 8.3761. With the lowest of these displacements at 7.1923 ounces of upward buoyant force, the maximum weight of a baseball at 5¼ ounces will easily be buoyed by either fresh or salt water. The ball will float.

How high will it ride in the water? When it is floating, the buoyant force is equal to the weight, meaning that the submerged part of the ball displaces the same weight of water as the entire ball weighs. For our example, let’s assume the ball is 3” in diameter, and the one we use in fresh water is the heaviest possible at 5¼ ounces, and so will be the most submerged example, while the one we used in salt water is the lightest possible at 5 ounces, and so will be the least submerged possible.

The shape of the submerged part of the ball will be a sphere with the top sliced off. A book that I bought in first year university (the University of Western Ontario) in 1971 called “the Mathematical Handbook of Formulas and Tables” came in handy as it has the formula for the volume of this very shape. It is V =⅓πh2(3r-h), where h (height) is the distance between the waterline and the bottom of the sphere. So all we need to do is set V to the volume of water that weighs as much as the ball, and solve the equation for h.

For the salt water example, were using a 5 ounce ball. How much sea water is that? It’s 8.43899 cubic inches, less than the 14.1372 cubic inches that our floating baseball takes up, as you’d expect. Now that we know V, and r is simply ½ the diameter or 1.5 inches, our equation becomes:

8.643899=⅓π x 2.25 x h2 x (4.5 – h).

This is a cubic equation and rather than solving the equation algebraically, I used “goal seek” in MS-Excel to crunch the result by successive approximation and found that h=1.6949 inches. Just over 1.3 inches will show above the waterline.

For the fresh water example, we’re using a 5.25 ounce ball displacing 9.08246 cubic inches. This results in an height of 1.7885, or about .09” lower in the water. That was a lot of work for a baseball, and it would have been much easier to just to toss one in the water. But if we’re dealing with larger objects, like buoys, ships, and so forth, we might find be able to calculate height in the water quite useful.

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