Did plot triumph over physics in the climax of the film “*Titanic*?” Could Jack have shared the raft of flotsam with Rose, or would both actors have sunk beneath the waves?

Let’s examine the math calculations for floating this improvised raft, made from a door that broke away from the *Titanic*.

## Setting the Stage and the Goal

The primary question is whether they sink or float under ideal conditions, but we’ll also explore exactly how big the raft would need to be, in order to support 2 people.

We set the target of “floating” to mean that the total volume of the wood could be just at the water line, but that the people remain dry on top.

The requirement will be that the weight of the sea water that would be displaced by the entire raft must equal or exceed the weight of the raft, plus any survivors.

## Weight, Mass, Density, and Buoyancy via the Archimedes Principle

Normally, it is easy to measure the weight of an object, whether using a spring-loaded weigh scale or a balance.

The density of an object is equal to its weight divided by its volume. Again, it is easy to calculate the volume of a regular solid such as a cube, which equals the length times width times height.

The Greek philosopher Archimedes realized that the volume of water displaced by any fully-submerged object is the same as the volume of the object. Therefore, an irregularly-shaped crown displaces the same volume of water as a cube of the metal from which it was cast.

The weight of the displaced water equals the volume of water times its density.

In general, the buoyancy provided by any fluid equals the mass that was displaced, multiplied by the acceleration due to gravity. This translates mass into “dynes.” However, the downward force opposed by the buoyancy is *also* the mass of the object *multiplied by gravity*.

This article will, therefore, omit multiplying weights by “9.8M/s^2,” since the acceleration due to gravity will be irrelevant to the question of sinking or floating.

This leaves a slight error due to the buoyancy of the object in air. This is about 0.08 pounds/foot^3 or (0.08lb/2.2lb/Kg) per (1foot/3.28M)^3 = .0363636… Kg/0.02834M^3 = 1.283Kg/M^3. Compared to water’s density of 1000Kg/M^3, this error just over 0.1%. That is extremely minor, compared to the assumptions about the type of wood or size of the door we’ll be making.

The *Archimedes Principle* is that a submerged object’s effective weight is reduced by the weight of the displaced fluid. If an object is less dense than the fluid it displaces, it will float, partially submerged. The volume that is submerged will displace enough fluid to match the weight of the whole object.

Decoding Science. One article at a time.