Chronology Current Month Current Thread Current Date [Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

# [Phys-L] floating fiasco physics

• 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

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.