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*From*: John Denker <jsd@av8n.com>*Date*: Mon, 29 Mar 2021 02:11:49 -0700

Hi Folks --

There are some amusing physics calculations that one can do that

shed some light on the Ever Given / Suez incident.

It must be emphasized that I am not claiming to say anything about

the cause of the accident. A proper investigation takes time. The

initial reported facts are often wrong.

Even so, there are some calculations we can do that involve basic

physics principles that apply no matter what, in this situation

and many others.

1) Crosswind is a topic that has come up. So let's calculate the

force. The formula is:

F = ½ ρₐ vₐ² Cd Aₐ

where

F = force

ρₐ = density of the air

vₐ = crosswind component of velocity (relative to ship)

Cd = coefficient of drag

Aₐ = broadside area exposed to air

In some sense this formula implicitly defines Cd, but it's better

than that, because Cd is more-or-less independent of the other

factors on the RHS.

The Cd for a big flat plate is on the order of 1 ... in contrast

to a streamlined shape, which would have a much smaller Cd.

Reported peak gust was 50 km/hour. That's only 27 knots. That's

windy, but not astonishingly so. I've dealt with lots worse.

Again, beware the initial reports are often wrong. I calculate

a force of 250 tons. That seems like a lot, but it's actually

small compared to the mass of the ship. That gives an acceleration

on the order of 0.001 g. That force acting for an entire minute,

unopposed, would not move the ship very far. So it doesn't seem

like human reaction time is a much of a factor.

2) To oppose the force of the wind, the ship would have to head into

the wind a little bit. The formula is

F = ½ ρ v² Cl A

where

ρ = density of water

v = velocity of water (relative to ship)

Cl = coefficient of sideways lift

which depends on angle of attack

A = area exposed to water

The speed through the water is less, but the density is 1000× larger,

so it really shouldn't take much angle of attack to compensate for

the crosswind turning force. I'm not sure about this, because the

hull might have a really lousy lift/drag polar, more like a Huck Finn

raft than a racing cat.

3) One annoying thing about steering a boat is that in the classical

arrangement, you don't have a way of pushing the bow in the direction

you want to turn. Instead you use the rudder to push the stern in the

direction you /don't/ want to go.

As long as the angles are small and the off-course distance is small,

this is only a minor annoyance. OTOH it also means there is a coffin

corner; that is, if the boat is too far outside the envelope, trying

to fix it just makes things worse. You have to proactively stay out

of such situations.

4) Interestingly enough, the Ever Given has bow thrusters. That means

it can maneuver to some extent, even when it is making little or no

headway.

There are two bow thrusters, each rated at 2,500 kW. The question

arises, how much force is that?

It is an interesting physics exercise to figure that out. You can

pretty quickly surmise that the force depends on the power (P),

the density of the water, and the diameter (D) of the thruster.

Using nothing but dimensional analysis we find:

F³ ∝ ρ D² P²

It's IMHO an amusing formula. The 3/2 power law is probably not what

you would have guessed.

You can figure out the constant of proportionality from the fact

that you can buy a tugboat with a 1500 kW powerplant and a 1.3m

diameter thruster rated for a 17 ton static thrust (aka bollard

pull).

5) The same result can be obtained with less wizardry, i.e. without

dimensional analysis, just by considering the following scenario:

Take a parcel of water initially at rest, then accelerate it and

shoot it out a nozzle of diameter D. Calculate the kinetic energy

and momentum. This is mostly just high-school physics, although it

takes some judgment about what terms to keep and what to ignore.

You will never get the constant of proportionality exactly right,

because there are too many nonidealities, but you'll come close.

A similar calculation can be done for the static thrust of an airplane

propeller, but there are even more nonidealities. Among other things,

airplanes (unlike tugboats) are not optimized for static thrust. With

occasional exceptions, they are optimized for high-speed cruising.

Anyway ... I calculate that the bow-thruster force is pretty small

compared to the force of the crosswind. AFAICT this class of boats

was not really designed to operate in windy conditions, at least

not in close quarters.

6) I am not suggesting this had anything to do with the recent

incident — I'm virtually certain it didn't — but it's something to

think about. In any situation where the boat has to stop in a canal

or otherwise in close quarters, it will not be able to keep itself

out of trouble in windy conditions. It can't turn around, it can't

drop anchor, and it can't rely on the bow thrusters to opposed the

force of the crosswind.

This is a sufficiently foreseeable problem that I'm surprised they

don't already have limits on the forecast crosswind allowed for

operating in the canal.

The Panama canal uses "mules" (powerful cog-train engines) to guide

ships through the locks. I betcha the Suez guys wish they had mules

and/or hefty tugboats to escort big ships through their canal.

===========

I really must emphasize how much I do not want to opine about the

recent incident. I vividly remember the TWA 800 disaster. There

were eyewitnesses who said, on camera, that they heard a boom and

looked up just in time to see a missile hit the plane. There were

also eyewitness reports of survivors in life rafts. Those particular

reports were easy to reject since they were physically impossible,

but there were others that were equally wrong but less obviously so.

There's no point in doing a clever analysis if it's based on bogus

facts.

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