Is there a calculation that a non swimmer can make when choosing bits of wood that are floating past so that he can grab one and stay afloat.
Lets say the man is 65 kgs in weight
What minimum size or weight of wood will keep him afloat for long periods and save his life.
As big as possible with as much out of the water as possible.
That’s the wrong question. The density of the wood is much more important. Remember the floats they have in swimming pools? A big bit of balsa wood will keep you afloat better than an equally sized (but much denser) piece of mahogany. At least in the short term. But if a piece of wood is floating then it’s buoyant and that’s the prime criterion.
To survive a long period you really need to get yourself out of the water otherwise desquamation and hypothermia set in and you are more likely to attract predators.
The density of the person is also important. Fat is less dense than muscle, so a person with a lot of body fat may not need any external buoyancy at all to keep their head above water.
Very true but the OP specified 65 kg which is very light for an adult man and therefore said man has very little fat.
It’s so highly variable. For a lot of people a toothpick will work. They just have to think of it like Dumbo’s feather: the toothpick will keep them afloat, they relax and indeed they float, no problem.
A minority will indeed need flotation aid. But in can range from a modest branch to a long skinny (so they can grip it) log.
Quartz, Thanks for teaching me a new word, “desquamation”.
The type of wood, not just its density but its propensity for absorbing water, will be the biggest determining factor. A well-aged, kiln-dried piece of soft wood will float pretty well until it becomes saturated. Then, not so much.
Without further information, it’s unanswerable.
If the objective is to get the 65 kg person out of the water, then if my basic physics (half remembered from decades ago) is correct, the following formula could be used.
V(wood) m[sup]3[/sup] X density(wood)kg/m[sup]3[/sup] + 65 kg = V(wood) m3 X density(water)kg/m[sup]3[/sup]
V(wood) m[sup]3[/sup] X density(wood)kg/m[sup]3[/sup] + 65 kg = V(wood) m[sup]3[/sup] X 1000 kg/m[sup]3[/sup]
V(wood)m[sup]3[/sup]= 65kg/(1000 kg/m[sup]3[/sup]– density(wood) kg/m[sup]3[/sup])
Balsa has a density of 112 kg/m[sup]3[/sup], pine has a density of 350 kg/m[sup]3[/sup] to 850 kg/m[sup]3[/sup]
For balsa, only 0.7 m[sup]3[/sup] is necessary. For the lightest pine, 0.1 m[sup]3[/sup], for the heaviest, then 0.43 m[sup]3[/sup].
For non-swimmers, a trial and error method is suggested rather than measuring the wood, estimating the density and then calculating the required volume.