Low Cost Scuba Weights

Theory - Buoyancy

Will steel’s lower density affect how useful it is as a weight? Consider the following densities:

Ref: http://www.simetric.co.uk/si_metals.htm Ref: http://hypertextbook.com/facts/2002/EdwardLaValley.shtml

Calculate buoyant force of an object:

m(b) = m(object) x (1- (p(fluid)/ p(object)))

m(object)= true mass of the object p(object)= average density of the object p(fluid)= average density of the surrounding fluid

For an m(object) of unity, you can calculate the factor (1 - (p(fluid) / p(object))), based on p(fluid) of seawater:

This for 1 kg of lead, your buoyant force is -0.909 kg. Thus to achieve this same -0.909 kg buoyancy, you would need:

The amount of weight required for steel is almost the same as for lead - maybe 0.5 kg more. Also notice how much more ‘useful’ a steel tank’s weight is for you…

Theory - Size

A second question is how this affects the required size of the weight.

1kg of lead takes up 1/11300 = 88.5 cm3. Assume our cylinder is 4cm in diameter (r=2), using A = pi * r * r, you find A = 12.56 cm2. Thus to achieve 88.5 cm3, it would have to be 7 cm long.

1kg of steel takes up 1/8000 = 125 cm3. The same 4cm cylinder would need to be 10 cm long, a little bulkier. Remember the density of steel is 0.708 times lower, so even if you need the same weight as with lead, it will be a lot bulkier.

Theory - Size of Random Pack

With random packing of spheres your factor is 0.64 at best (ref: http://en.wikipedia.org/wiki/Random_close_pack). Our spheres won’t be perfect, so let’s say the factor is 0.6. Thus the density of packed steel shot becomes around 4800 kg/m3.

For a 2KG weight with steel, that requires 2/4800 = 0.000417 m3, or 417 mL of shot in theory. Seems reasonable…

Theory - Building the Weights

Alright, from the above I needed some way to make the weights. As I’m using steel, it should be waterproof. You could alternatively use stainless steel, or coat the shot, but I decided to go with sealing it. The easiest thing: bike inner-tube. My bike’s inner-tube is around 50 mm in diameter, thus has a cross-section area of (2 * pi * 25E-3 * 25E-3) = 0.003925 m2. So 2KG of weight, which was found to require 0.000417 m3, would require around (0.000417 / 0.003925) = 0.106m = 10.6 cm of filled tube.

Thus if I required around 16 kg of weight, I found fill a tube ((16 / 2) * 10.6) = 85 cm long. Seems pretty reasonable to me!