center of flotation calculation and implications?

[daiquiri]

Yeah, I'd say it sums up the NA's job drama pretty well.

[Leo Lazauskas]

Originally Posted by daiquiri

How about seing the NA as a cook who uses the above-mentioned tools for preparing a nice meal?

Nah, I like my set of titanium Sporks. And I like the analogy that NA's have to spoon with others occasionally to get things done

[Fanie]

Eh you drivel maniacs watch it Uncle Jeff will spank you...




What happens when the Cp exceeds 0.7 ? I was under the impression it was mentioned somewhere but I couldn't find it.

If the Cp goes higher and higher, where does it lean towards or what is the destination when the Cp keeps going up ?

[terhohalme]

The log has Cp = 1.0.

[ancient kayaker]

Originally Posted by Leo Lazauskas

You'll get over it.

Naval architects ... can't all be experts on ... rheology, tribology, vibrations, fluid dynamics, etc etc ...

Speaking seriously on behalf of NA's I worked as a Systems Engineer for many years. That was in the older sense of SE not the computer variety. In order to lead a team of specialists I had to be knowledgeable in a wide range of technologies so I could at least speak all of their languages and occasionally put 2 and 2 together and spot that they made 5 on occasion. I am sure a NA also spends his time multitasking like crazy in addition to managing his own speciality. Gentlemen, let us raise out hats to them just this once, and never, ever again less they become conceited ...

[TeddyDiver]

Just out of curiosity, and of course to get some additional insight.. What did the original Colin Archers had as their Coefficients and Factors? They based anyway to waveform theory so whats the difference in the outcome?

[Eric Sponberg]

Originally Posted by Fanie

What happens when the Cp exceeds 0.7 ? I was under the impression it was mentioned somewhere but I couldn't find it.

If the Cp goes higher and higher, where does it lean towards or what is the destination when the Cp keeps going up ?

Beyond about 0.7, vessels lose the benefit of lower resistance for increasing speed. Hullforms start to approach that of a perfectly rectangular barge which has a Cp = 1.0. Think what power would be required to move such a barge at a high speed?? It's pointless because the power required is astronomically high. But this is why barges do indeed have rectangular planforms and cubic bodies, although with pram bows to minimize wave drag at very low speeds. They have tremendous capacity but require large, powerful tugboats to either push or pull them.

Eric

[M-Sasha]

Very nice effort Mr Sponberg, thank you!

Sasha

[mcollins07]

Eric,
If we have two boats of close designs, i.e. same displacement, Lwl, and similar hull shape, there primary difference is the Cp values, then ….
Is it correct to say that less energy is required to get the boat with Cp = 0.7 to a speed/length ratio of 2.0 is less than energy required to get the boat with Cp= 0.61 to that same speed?

Where as, for two similar boats with Cp = 0.7 and Cp= 0.9 it will require nearly identical amounts of energy to get each boat to speed/length ratio of 3.0?


~ Michael

[mcollins07]

This graph might be helpful.

The data that Eric supplied us for Speed/length ratios correlated to optimum Cp values was plotted, Cp vs Speed/Length.

Just looking at the numbers, I think one would tend to see a linear correlation. A linear fit would lead one to think that a Cp value above 0.7 would be optimum for a Speed/Length ratio in the range of 2.0 to 3.0.

A non-linear fit to the data shows a much better fit (or correlation coefficient), thus the correlation is not linear. We should avoid our tuition to extrapolate linearly into the range of Speed/length ratios of 2.0 or above.

~ Michael

[ancient kayaker]

I agree a curve fits better than a straight line but it is unlikely to have a maximum or even be asymptotic. Both imply hull forms with Cp greater than about 0.7 either do not exist or do not have a solution for speed/length ratio.

I have trouble understanding this whole discussion of the impact of Cp on SLR. Surely length/beam ratio has far more impact? Or am I trying to make sense of a purely theoretical discussion?

[Fanie]

Hi Ancient,

I have the same queery. I have drawed yet another hull. It seems the hull is more blunt if the Cp is around 0.7 than when it is 0.6.

The draft to carry the same weight between the two is about 100mm, so in the case of the Cp at 0.7 the draft is 400mm for 2300kg's and the draft for the Cp at 0.6 is 500mm also for about 2300kg.

This means the hull with a Cp of 0.7 has less water friction, the wetted area is smaller. The sharper hull with Cp of 0.6 and more draft has a larger wetted area.



Now the question is -

If speed increase, does the wetted area drag increase as well ?

How much ??? this will be drag per sq area and speed then ??

[Leo Lazauskas]

Originally Posted by Eric Sponberg

Beyond about 0.7, vessels lose the benefit of lower resistance for increasing speed. Hullforms start to approach that of a perfectly rectangular barge which has a Cp = 1.0. Think what power would be required to move such a barge at a high speed?? It's pointless because the power required is astronomically high. But this is why barges do indeed have rectangular planforms and cubic bodies, although with pram bows to minimize wave drag at very low speeds. They have tremendous capacity but require large, powerful tugboats to either push or pull them.
Eric

I agree, but one aspect of your barge example might be potentially confusing to some people here.

A rectangular block and a log with circular cross-sections both have Cp=1.0.
Both have rectangular planforms, but different cross-section shapes.
However, the distribution of volume along the length is the same for both bodies.

But maybe I have now confused more people.
Leo.

[apex1]

Originally Posted by Leo Lazauskas

But maybe I have now confused more people.
Leo.

Nono, not too many I guess.

[Leo Lazauskas]

Originally Posted by Fanie

Now the question is -
If speed increase, does the wetted area drag increase as well ?

That is a deep issue and certainly not something that can be answered by resorting to arguments involving just the Cp. The Cp is too blunt an object to help out here.

At low speeds (technically low Froude numbers) a displacement hull will sink down into the water slightly, and this might increase its wetted area a little. (At some low Froude numbers it can also rise slightly above its at-rest level). If the water level close to the hull also drops, the wetted area of the hull could remain exactly the same as its value at rest.

Then there is the whole (related) issue of the wave profile on the hull and how that increases the wetted area on some parts of the hull, and decreases it at other locations.

Confusing? You betcha!
Leo.