Is the Hull Speed Debate Misleading: Online vs. Textbook Perspectives?

[Jezza]

There is a well known rule of thumb amongst sailors that a boat has a theoretical maximum speed called the hull speed, given by v_h \approx 1.34 \sqrt{L_{wl}}, where LWL is the waterline length of the hull. This so-called rule is pretty much complete rubbish; a boat can overcome this speed with enough power, and modern hull designs basically subvert the problem. In any case, it's not a hard speed limit. Having said that, it's a useful indication of how fast a boat can be expected to go, because increasing amounts of power are required to increase one's speed beyond hull speed.

The online literature surrounding the matter is largely qualitative, and where numerical, any constants are empirically obtained numbers. Today I discovered why upon consulting a fluid dynamics textbook (T. E. Faber, Fluid Dynamics for Physicists); the analytical treatment is beyond my understanding just yet, and is at any rate beyond the scope of a 15 minute talk I have to give about boats in a few weeks.

Having said this, I did pick up a useful way of looking at the topic of hull speed from the book. The issue is that this seems to be completely at odds with any information I can find online.

Everywhere online says the hull speed is reached when the second crest of the bow wave coincides with the stern. Beyond this speed, the stern begins to dip down as it falls down the 'other side' of the crest, and the boat is left trying to climb up its own bow wave. It follows that hull speed is reached when the length of the boat reaches the wavelength of the waves. I have a few issues with this explanation. The main one is this notion of the boat trying to 'climb up its bow wave.' The boat is still moving horizontally; the only thing happening to the boat is that it's rotating; and it will never succeed in climbing its own bow wave because the boat is creating it!

The textbook, on the other hand, offers an explanation which seems more concrete but, as I say, is at odds with everything I've read online. It begins by explaining that the bow must coincide with the crest of its wave, and that there is also a separate stern wave. It explains that the stern always coincides with the trough of its wave. This seems familiar from experience, but also seems to make qualitative sense by analogy with the bow; the bow pushes the water in front of it, whereas the stern pulls water along with it. The book then asserts, reasonably enough, that these waves carry energy away from the boat. The two waves inevitably interfere with each other. The energy transported away, and hence the drag, is therefore maximised when the bow and stern waves interfere constructively. By this argument, hull speed is reached when the wavelength of the waves is twice the length of the boat, so that the trough can line up with the stern. The book never explicitly refers to this speed as the 'hull speed' however.

My question really amounts to who's right? The talk I'm giving is part of my degree and the audience includes professors, so I'd really rather not have to turn up with the first explanation, and so if the internet is right (and I think it frankly must be), can anyone provide a more concrete explanation?

Many thanks for any help :)

 

[anorlunda]

Boats vary a lot in size and design. I think that you expect too much of you want a clean definition that applies to small sailboats as well as giant container ships.

As a sailor who has been cruising full time on my Westsail 32 for 11 years, let me give you my version.

The faster I go, the more the bow wave moves stern. Initially, it has little effect because there is not so much buoyancy up in the bow. But eventually, it starts lifting the bow and the stern squats. Going still faster makes the whole hull start lifting out of the water. So it depends on water dynamics and also the shape of the hull. That's why it takes so much more energy to exceed hull speed. Not impossible, but difficult.

My hull speed is 7.2, but I once averaged 8.6 for more than 24 hours on a beam reach.

My boat is double ended, so my stern wave has negligible effect compared to the stern wave of a square transom boat with loads of buoyancy in the stern.

Planing boats reach the point where the crest of the bow wave comes under the COG of the vessel.

If you want an accurate engineering definition, plot speed thought the water versus net force. The whole curve is your answer. Resist the temptation to label features of the curve with words. Discard all verbal descriptions and words originating in days of yore.

EDIT: The curve you plot will be hull specific.

 

[sophiecentaur]

This so-called rule is pretty much complete rubbish;

Not if you apply it with sense. Its describes the speed at which any further increase in speed requires a significant increase in input power and it applies only to a displacement hull and, probably to a limited range of hull profiles. You call it a rule of thumb - which is exactly what it is. It gives you somewhere to start if you don't want to pay an expert or to take a course in advanced hydrodynamics and marine hull design. You can choose to use it or not. It is what it is.

The boat is still moving horizontally; the only thing happening to the boat is that it's rotating;

It's only "rotating" during a change in speed. Once it is established, the engine is effectively having to lift the tilted hull up as fast as it is falling down the 'slope' it has created. A uniform section 5 tonne hull 10m long has a design speed of around 4m/s. When this is exceeded, it is 'motoring uphill'. Very crudely, if it going fast enough to be tilted upwards at an angle of, say 5° (which may be only a bit higher than at 4m/s), it is being constantly 'lifted' at a rate of say 2m every second would correspond to 50 kW of Power. That Power is in addition to what's needed to deal with all the other 'resistive' forces. When you think that many yachts have engines that don't produce anything like 50kW, you are really dealing with an unassailable Wall.
Sorry for using SI units but they are much easier for back of a fag packet calculations.

 

[Jezza]

Thank you for the responses, they're very helpful.

Sophiecentaur, your response has prompted me to construct a very simplistic model involving a uniform rod.

In terms of presenting this to my professors, I'm much happier to go in with this! I'm afraid I don't quite understand how you reached a ball-park 50kW though, could you please elaborate?

 

[sophiecentaur]

I don't quite understand how you reached a ball-park 50kW though, could you please elaborate?

I was basing it on rate of Energy transfer. The boat is, effectively being lifted up the 5° slope as it moves forwards (it would be the same for being pulled up an inclined plane)
I was assuming that the vertical movement would be 10 (the boat length) sin 5 per second but, of course the movement is for just 4m of forward motion so the height lifted would be 0.35m
Work per second would be
mgh = 5X10³ X10X0.35 J/s
=17.5kW
My earlier figure is based on the wrong rate of lift.
As I see it, you are assuming that the boat needs to be constantly rotated - but it doesn't change angle. It is lifted up an equal amount along its whole length. That, I think, accounts for your difference of a factor a 1/2 compared with my answer. I would always steer clear of a 'forces' approach to this sort of problem, in favour of a straight 'energy' approach. I think my (revised) estimate is more likely to be correct.
Either way, it is a significant amount of extra Power from a yacht engine. Of course, I have no idea as to whether the 5° slope is realistic or even what proper geometry should be used to determine bow and stern levels in the water. I do know that, in a canal boat, going up a shallow and narrow canal, the impression was very much of going uphill. I soon throttled back to save the canal bank from wash.
If the slope is just proportional to speed then I think it's correct to say that the power would go up as the square of the speed - or worse - based only on the above argument

 

[Jezza]

Ah ok I understand now :)
Thank you for your help; I think that's enough for me to be getting on with.