# How to Free the Ever Given—Using Buoyancy Force!

Now for some values. Let’s start with a check. Suppose the *Ever Given* is a completely flat box (no pointy bow) so that I can use my box equation from above. What should the draft be? First, I need the area of the bottom. The ship has a length of 399.94 meters and a width of 59 meters for an area of 2.36 x 10^{4} m^{2}. Now I just need to plug in my ship mass and density of water. This gives a hull depth of 8.5 meters. Yes, this is less than the value stated above. Why is it different? There are two possible reasons. First, I made the assumption of a completely rectangular base for the ship. Clearly, that’s not correct (but it’s still a fine approximation). Second, the stated value might be the maximum draft instead of the current hull depth.

But what if I want to decrease the draft by 1 meter? How much mass would I need to remove from the ship? 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 that large of a mass difference. I’m actually stunned.

Well, where could you get this much mass from the *Ever Given*? Two easy options are to remove water ballast or fuel. Diesel fuel has a lower density than water (about 850 kg/m^{3}), so you would 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’s kind of difficult to imagine volumes that large. Let’s switch to different units—volume in Olympic swimming pools. These pools are about 50 m x 25 m x 2 m for a volume of 2,500 m^{3}. So if you want to raise the *Ever Given* by 1 meter, you would need to offload enough water to fill about 10 Olympic swimming pools. That’s crazy. Well, I guess it’s not so crazy for a ship as big as the *Ever Given*—a ship so big that its length is actually wider than the Suez Canal.

Wait! There are all of those shipping containers on the deck. What if you just remove a bunch of those to decrease the draft? Great. Let’s see how many you would have to remove. Of course there is one small problem—all these containers have different things inside of them. Some have TVs, some might have clothes. So they could all be different masses. That just means I get to estimate the shipping container mass.

These containers have a fairly standard size. The big ones are 2.4 m x 12.2 m x 2.6 m for a total volume of 76.1 m^{3}. For the mass, let’s say these things float in water fairly well (I’ve seen pictures of floating containers). If the average container floats with half the volume above the water, they would have to have a density half that of water. Yes, salt water has a slightly higher density than fresh—but it’s just an estimate so I’m going to say the container has a density of 500 kg/m^{3}. That means each container would have a mass of 38,000 kg.

If I need to remove a total mass of 23 million kilograms, that would be equivalent to 605 containers—the *Ever Given* can hold 20,000 containers. Oh boy, that’s not good. How do you get a container off a ship in the middle of a canal? A heavy lift helicopter? That would work, but how long would it take? Let’s say the helicopter can remove one container every 30 minutes. I mean, this seems like a reasonable time, since you have to fly over and then hook up a container then unhook it. That would put a total unload time of 12 days. Flying straight.

OK, a final note. Yes—these are rough estimates (back of the envelope calculations), so they could be off. However, you can still get useful information. Even if my estimates for the removal of containers is off by a factor of 2, it would still take 6 days to get those things unloaded. I would guess that the best solution for this stuck ship is to use a combination of ballast/fuel removal along with digging out the shore. Whatever they do, I hope they fix it soon.

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