# How to Release Ever Given – Using Buoyancy Force!

Now looking for some values. Let’s start with a check. Consider *Never Given* the box is completely flat (no sharp arcs) so I can use the equation in my box from above. What should be the draft? First, I need the area below. The vessel is 399.94 meters long and 59 meters wide with a surface area of 2.36 x 10^{4} m^{2}. Now all I have to do is connect the mass of the vessel and the density of the water. This gives the helmet a depth of 8.5 meters. Yes, it is less than the value indicated above. Why is it different? There are two possible reasons. First, I took a completely rectangular base for the vessel. Obviously, it’s not correct (but roughly rough). Second, the indicated value can be a maximum draft instead of the depth of the current helmet.

But what if I want to lower the draft by 1 meter? How much mass should I remove from the boat? We can set a depth value of 7.5 and then fix the mass. This results in a reduction in mass of 23 million kilograms. OK. I didn’t expect much of a mass difference. I’m actually amazed.

Well, where can you get this mass? *Never Given*? There are two easy options for removing water ballast or fuel. Diesel fuel has a lower density than water (about 850 kg / m)^{3}), so you should remove more fuel than water. But removing water with a mass of 23 million kilograms would have a volume of 23,000 m^{3}. If changed to the ruler, it would have a volume of 27,000 m^{3}.

It’s hard to imagine large volumes. Let’s go to the different units—volume in the Olympic pools. These pools are about 50 mx 25 mx 2 m long, reaching a volume of 2,500 m^{3}. So if you want to go up *Never Given* 1 meter, you should discharge enough water to fill about 10 Olympic pools. That’s crazy. Well, I think it’s not as crazy as a boat *Never Given*“Because the ship is so large, its length is wider than the Suez Canal.”

Wait! All of these deck vessels are on the deck. If you remove a bunch of them to reduce the draft? Excellent. Let’s see how much you should remove. Of course, there is a small problem: all of these containers have different things inside. Some have TVs, others have clothes. So they could all be different masses. This means that I can calculate the mass of the shipping container.

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

If I had to remove a total mass of 23 million kilograms, it would be the equivalent of 605 containers *Never Given* It can hold 20,000 containers. Oh boy, that’s not good. How do you get the container out of the boat in the middle of a canal? A heavy lift helicopter? That would work, but how long would it take? Suppose a helicopter can remove a container every 30 minutes. I mean, this seems reasonable, you have to fly and then tie a container and then release it. This would put a total download time of 12 days. Fly straight.

Okay, last note. Yes, they are approximate estimates (the back of the envelope calculations) so they can be off. However, you can get useful information. Although my calculations for removing containers are 2 factors off, it would take 6 days to unload these things. I would guess that the best solution for this sailing boat is to use a cliff / fuel removal combination. What they do, I hope they fix soon.

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