Now for some value. Let’s start with the check. Think *Ever given* A completely flat box (no bow to bow) so I can use the equation of my box from the top. What should the draft be? First, I need the area below. The ship is 399.94 meters long and 59 meters wide for an area of 2.36 x 10^{D} M^{D}. Now I need to plug in the mass and water density of my vessel. It gives a depth of 8.5 m hall. Yes, it is lower than the standard described above. Why is it different? There are two possible reasons. First, I guessed a completely rectangular base for the ship. Obviously, this is not accurate (but it is still a subtle guess). Second, the value described may be the maximum draft instead of the depth of the current hall.

But what if I want to reduce the draft by 1 meter? How much mass do I need to remove from the ship? We can only keep the depth value of 7.5 and then solve for the mass. This gives a massive reduction of 23 million kilograms. Okay mass I didn’t expect such a big difference. I’m actually shocked.

Well, where do you get this mass from? *Ever given*? The two easiest options are to remove the water droplets or fuel. Diesel fuel density is lower than water (about 850 kg / m)^{D}), So you need to remove more fuel than water. But if you remove water with a mass of 23 million kilograms, it will be 23,000 meters.^{D}. If you switch to fuel, it will amount to 27,000 meters^{D}.

Large sizes are somehow difficult to imagine. Let’s switch to different units-Volume in Olympic swimming pools. These pools are about 50 mx 25 mx 2 m for an area of 2500 m2^{D}. So if you want to raise *Ever given* By 1 meter, you need to offload enough water to fill about 10 Olympic swimming pools. It’s madness. Okay, I guess it’s not as big a craze as a ship *Ever given*It is a ship so large that it is wider than the Suez Canal

Wait! There are all those shipping containers on the deck. What if you just removed a bunch of them to reduce the draft? Great let’s see how many you need to remove. Of course there is a small problem. All these containers have different things inside them. Someone has a TV, someone can have clothes. So they can all be different masses. This means I can estimate the mass of the shipping container.

This container has a fairly standard size. The larger ones are 2.4 mx 12.2 mx 2.6 m for a total volume of 76.4 m^{D}. For the folks, let’s float these things quite well in the water (I’ve seen pictures in floating containers). If the average container floats above half the volume of water, they must have a density of half that of water. Yes, salt water has a slightly higher concentration than fresh water – but that’s just a guess so I’m going to say the container has a density of 500 kg / m^{D}. This means that each container will have a mass of 36,000 kg.

If I had to remove a total of 23 million kilograms of mass, it would be the equivalent of 605 pots *Ever given* Can hold 20,000 holders Oh boy, not good. How can you get a container of a ship in the middle of a canal? A heavy lift helicopter? It will work, but how long will it take? Suppose the helicopter can remove a container every 30 minutes. I mean, it seems like a reasonable time, since you have to fly to the top and then hook up a container and then open it. This will take a total of 12 days to unload. Flying straight.

Okay, one final note. Yes – these are approximate estimates (Behind the envelope count), So that they can be closed. However, you can still find useful information. Even if my guess of removing the containers is stopped by a factor of 2, it still takes 6 days for these things to load. I guess the best solution for this stranded ship is to use the combination of ballast / fuel removal as well as digging on shore. Whatever they do, I hope they fix it soon.

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