Who would think a single container ship could cause a global shipping problem. But that's exactly what happened when the *Ever Given* ran into a bank of the Suez Canal and blocked the passage of other ships. The Suez Canal is a great short cut for shipping—without it, cargo would have to be sailed all the way around the southern tip of Africa. A blocked canal can cause some serious problems.

So, how do you get a grounded ship to float again? You have a couple of options. You could use tug boats to pull it out, or you could use excavators and dig it out. Another method would be to decrease the ship's total mass so that it would require less water depth to remain buoyant. This is what we are going to do—estimate the amount of mass you would need to remove to get the thing to float again.

Let's start with some rough estimations. The Ever Given is a Golden-class container ship. So from that we can get its dimensions. It's 399.94 meters long with a width of 59 meters. For a ship, the draft is the distance from the water line down to the bottom part of the hull. I'm not sure about the current draft for *Ever Given*, but Wikipedia lists it at 14.5 meters. It also seems like these large container ships have a flat-bottom hull—at least in the middle part of the vessel. That will make our calculation easier. Oh, one more thing. Most places report a total ship mass of 224,000 tons.

What does the ship mass have to do with the draft? Ah ha! Here is the real physics. OK, let's look at why ships float. I will start with a block of water floating in water. Yes, water floats. Here is a diagram.

This block of water has a depth of *d* and a bottom area of *A*. Since the water block is stationary, the total force acting on this block must be zero (zero vector). We know there is a downward-pulling gravitational force that is equal to the gravitational field (*g*) multiplied by the mass (*m*), so there must be another force pushing up. Let's call this upward pushing force the buoyancy force. It's clearly an interaction from the water on the bottom and sides of our arbitrary block of water.

Right there you can see an expression for the buoyancy force. The magnitude of this force must be equal to the gravitational force on the floating water block. But what is the mass of this water block? Let's assume the water has a uniform density of ρ. In that case the mass is equal to the volume of the block multiplied by the density. I can use this for the gravitational force on the water and set it equal to the buoyancy force.

What if we replace the water block with a ship? If the part of the ship that's underwater has the same shape as the original water block, then it has to have the same buoyancy force. Since we really care about the draft of the ship, we can use this to solve for *d*.

Here are some important comments.

- Notice that the units are the same on both sides of the equation. On the left, the draft should be in units of meters. On the right, it's kg divided by kg/m
^{3}multiplied by area (m^{2}). Yup, this gives meters too. - I'm going to need the density of water. Let's use fresh water at 1,000 kg/m
^{3}. I will also need the ship mass in kg instead of tons—how about around 200 million kilograms? - What about gravity? Since both the buoyancy force and the weight depend on
*g*, it cancels. This means that if we get a giant container stuck in a canal on Mars, the calculations will be the same even though the gravitational field is lower. - This calculation assumes that the hull has a flat bottom. Just imagine it had a V-shaped hull. In that case the volume of water displaced would not be linearly proportional to the draft. You can consider different hull shapes as a homework assignment.

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.