# How to unleash the never given – using the force of buoyancy!

Now for some values. Let’s start with a check. Suppose the *Never given* is a completely flat box (no sharp arc) so I can use my box equation from above. What should the project be? First of all, I need the bottom area. The vessel has a length of 399.94 meters and a width of 59 meters for an area of 2.36 x 10^{4} m^{2}. All I have to do is connect the mass of my vessel and the density of the water. This gives a hull depth of 8.5 meters. Yes, it is less than the value stated above. Why is it different? There are two possible reasons. First of all, I made the assumption of a completely rectangular base for the ship. Clearly that’s not correct (but it’s still a nice approximation). Second, the value shown could be the maximum draft instead of the actual depth of the hull.

But what if I want to reduce the draft by one meter? What mass would I need to remove from the vessel? We can just put a depth value of 7.5 and then solve for the mass. This gives a mass decrease of 23 million kilograms. OK – I did not expect such a mass difference. I am actually stunned.

Well where could you get so much mass from the *Never given*? Two simple options are to remove the water ballast or the fuel. Diesel fuel has a lower density than water (about 850 kg / m^{3}), you will therefore have to remove more fuel than water. But if you remove water with a mass of 23 million kilograms, it would have a volume of 23,000 m^{3}. If you switch to fuel it would be a volume of 27,000 m^{3}.

It is quite difficult to imagine such large volumes. Let’s move on to different units –volume in Olympic swimming pools. These pools measure approximately 50 mx 25 mx 2 m for a volume of 2500 m^{3}. So if you want to increase the *Never given* per 1 meter, you would need to discharge enough water to fill about 10 Olympic size swimming pools. It’s crazy. Well I guess it’s not that crazy for a ship as big as the *Never given*– a ship so large that its length is in fact wider than the Suez Canal.

Wait! There are all these shipping containers on the bridge. What if you just deleted a few to reduce the draft? Awesome. Let’s see how many you will need to remove. Of course, there is a small problem: all of these containers contain different elements. Some have a TV, others may have clothes. So they could all be different masses. It just means that I can estimate the mass of the shipping container.

These containers have a fairly standard size. The large ones measure 2.4 mx 12.2 mx 2.6 m for a total volume of 76.1 m^{3}. For the masses, let’s say these things float pretty well in water (I’ve seen photos of floating containers). If the average container floats with half the volume above water, it should have a density two times that of water. Yes, salt water has a slightly higher density than fresh water, but this is only an estimate, so I will say the container has a density of 500 kg / m^{3}. This means that each container would have a mass of 38,000 kg.

If I were to remove a total mass of 23 million kilograms, that would equate to 605 containers – the *Never given* can hold 20,000 containers. Oh boy, this is not good. How do you get a container off a ship in the middle of a canal? A heavy transport helicopter? It would work, but how long would it take? Let’s say the helicopter can remove a container every 30 minutes. I mean, that seems like a reasonable amount of time because you have to hover over and then plug in a container and then unhook it. This would take a total unloading time of 12 days. Fly straight.

OK, one last note. Yes, these are rough estimates (back envelope calculations), so they could be turned off. However, you can still get some useful information. Even though my estimates for container removal are divided by 2, it would still take 6 days for these things to be unloaded. I imagine the best solution for this stranded vessel is to use a combination of ballast / fuel removal and dig the shore. Whatever they do, I hope they fix it soon.

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