From: firstname.lastname@example.org (George Herbert)
Subject: Re: [?] Force needed to propel ships thru water
Date: 10 Feb 2000 00:56:32 -0800
Andrew C. Toppan <email@example.com> wrote:
>Jeff (JHD5@psuvm.psu.edu) was seen to write:
>> It's my understanding that it takes progressively less power to propel
>> progressively larger displacements of vessels through the water. I've
>Not displacement as such, but length. The longer the ship, the "easier" it is
>to move it - in short you need less force because the water flows better
>around the longer object. The details of the fluid dynamics will be left to
>the naval architects and the CFD folks....
Sigh. Everyone should know this one...
There are two parts to ship drag. Wavemaking drag, and Frictional Drag.
For both types, there's a general equation where drag is equal
to half of the wetted area times the density of seawater times
square of velocity times the coeficient of drag. However, Cd
is not a constant for either type of drag. Reynolds number and
turbulent flow theory describe frictional drag. Froude number
and wave theory describe wavemaking drag.
A ship has a local minimum in energy lost to waves as it moves
at a speed at which a deep (or shallow, if appropriate) water wave
wavelength is the length of the hull. After much hydrodynamics
involved in how water waves propogate and at what speed, you find
that a wave's length and speed match:
v(knots) = 1.3 * sqrt(waterline length(feet))
For a given hull length, there's a corresponding speed given by
that formula known as "hull speed". There's also a related Froude Number,
which is Fn = v(knots) / sqrt(waterline length(feet)) .
Exceeding hull speed takes a lot of power. Note that waterline
length is sometimes referred to as Length between Perpindiculars
or Lpp (the frame where the bow enters the water at nominal draft,
and the rudder hinge position at the stern). Lpp is usually but
not always very close to the actual length immersed in the water.
Using overall hull length will give incorrect results.
If you have a hard time visualizing why this is true, draw a ship
side profile and sketch in a wave starting at zero height at the
bow, rising up to 1/4 length of the vessel, zero height amidships,
trough at 3/4 of the length and zero height again at the stern.
If you speed up the wave gets longer, and you end up with part
of the trough hanging off the back end of the ship, pulling you
back into the trough (adding drag). There are other effects
involved, it's not just that simple, but that's part of it.
The frictional drag is not very affected by hull shape.
The curves used to describe it are just dependent on the
Reynolds number and surface roughness of the ship.
Hull shape (and presence or absence of a bow bulb,
and block coeficient, and...) greatly affects the
wave drag coeficient. Long, thin ships with a bow
bulb have very little wave drag compared to short,
As an example, take two ships. A carrier might have waterline
length of 950 feet say. 1.3 * sqrt(950) is about 40.1 knots.
That says that a carrier hull will be hydrodynamically fairly
efficient up to around 40 knots. That does *not* mean that
it can go that fast, but that's as fast as it can go with
a reasonably steady increase in drag. Past that, it gets
much harder to go any faster.
A Destroyer which is 550 feet long has a hull speed of 30.5 knots.
If that DD can actually move 35 knots, it's moving past hull speed
and requires significant additional power to do so.
Numerically, a CVN weighing about 95,000 tons 950 ft Lpp uses
250,000 SHP of power to go (mumble) something like 37 knots.
A destroyer weighing around 6,000 tons 550 ft Lpp uses around
100,000 SHP of power to go something like 35 knots.
-george william herbert