center of flotation calculation and implications?

Using metric world and your message (though fine and thorough) woud be a half length only...

A metric equivalence is DLR = V / (L/30.48)^3, where V is volume in m3 and L is waterline length in m

 

just for clarity:
when the 'metric equivalence' would be DLR = V / (L/30.48)^3 it is strechting the word equivalence by far... when you divide meter by 30.48 you end up doing the whole calculation in foot...
so what you do in the end is dividing Volume by cubic-foot...
1 foot = 0.3048 meter = 30.48 cm

and this might be the reason why in the metric world LDR - as mentioned by eric - is more commonly used...

 

if we compare that with the formula given by terhohalme

we see that L in m is converted to foot/100... the exactly same as in erics formula...
what still puzzeled me where the m³ for the volume... how can you get similar numbers if you have pounds/long tons on the one hand and m³ on the other both divided by the same stuff - (foot/100)^3?

now the volume or displacement in erics formula is given in 'long tons'...
we know that 1 m³ equals 1 metric ton of freshwater and some where in the area of 1.02-1.03 (depending on salinty and temperature) ton of seawater...
1 kg now equals 2.20462262 pounds...
multiply the pounds figure with 1020 and you get 2248.71 pounds...
thats quite close to a long ton - don't you think?...
so we see that a long ton is roughly the empirial equivalence in pounds for 1 m³ of seawater which is the displacement in m³ times 1.02...

 

Close. Calculating another way:
1 M^3 = 35.32 cubic feet

35.32 cu.ft. * 64 lbs/cu.ft. sea water = 2,260.5 lbs

which is 9% larger than 1 long ton. That is generally considered too great an error. And there is no need for this type of thinking because...

You should always use consistent units. Don't mix imperial with metric or vice versa. If you want to use metric units, use the LDR formula as in Principles of Yacht Design, and if you want to use imperial units, use DLR.

This is a cardinal rule of all scientific and technical endeavor--Keep Units Consistent!

Eric

 

i completely agree with you...
i just tried to point out that with the 'metric' formula for DLR you get roughly the same figures although mixing up m³ with (ft/100)³...
they are of course too far off to use data taking from a modell having given a DLR in imperial to accurately determine everthing you so brilliantly explained in your post...

and actually - DLR is a ratio and therefore it should not matter how you calculate it if you are going to compare different hulls - PROVIDED that you calculate it for every hull the same way - isn't it?

and i know that seawater has an average density of 1.025 g/cm³ which then gives us 2259,7 lbs per m³ of seawater....
i was just a bit curious of why you get similar figures although messing up with units...

 

Eric: there was no error. You calculated that 1 cu meter (of sea water) weighs more than a tonne. Perfectly true!

 

By error, I meant that to say that a cubic meter of water "is roughly the empirial equivalence in pounds for 1 m³ of seawater" is a little too rough, in my opinion, the weight being off by 9% from a long ton; it is too big an error of estimation.

Eric

 

One long tonne is 1016 kg, one cubic meter of seawater weights 1025 kg, so the error is only 0.9 %.

DLR is only for classification: wheter the lady is fat, normal or thin, nothing more really. Making comparisons doesn't need absolute numbers.

If you move your boat from seawater to fresh water, you should use longer waterline, because the boat floats deeper. So what happens to that DLR number ??


Eric,
Why I sent this was to make it easier for readers familiar with metric system to find similar numbers easy, nothing more. I have found your writings very informative, so sorry to confuse things.

yours Terho

 

It will change by something of the order of 0.3 - 0.6%

 

I stand corrected--I errored on the error by misplacing a decimal point. 0.9% it is. Still, I believe you should not mix units.

As far as moving from sea water to fresh water, pick a reference and stick with it. The change will be minimal and not really enough to affect your overall design. When comparing DLRs among a population of boats, you are looking for gross differences, by values of 10 or more (180, 190, 200, 210, etc.). Most boats are designed for sailing in seawater, and so comparing them on sea water would be the most consistent. However, if you are judging your design against a population of fresh water designs, and you know that their DLRs were calculated on fresh water, then you can do the same reliably.

Eric

 

In other areas of scientific and technical endeavor dimensionless coefficients are prefered where appropriate. Then it doesn't matter what units are used for the calculations. For probably historical reasons naval architects seem to prefer coefficients with "interesting" units such as "knots / square root of feet" and "tons / feet cubed".

A dimensionless form of DLR would have the displacement expressed as the volume of water displaced, using the cubic form of the length units used for LWL, i.e. Volume dispaced / (LWL ^ 3) Then it makes no difference if the length and volume are in feet/cubic feet, meters/cubic meters, inches/cubic inches, millimeters/cubic millimeters, etc. The only drawback of using the dimensionless form is the numerical value will be different than the common dimensional form with units of "tons / feet cubed". If you don't like where the decimal point winds up in the non-dimensional form multiply it by a scaling factor such as 100 or 1000. It's still non-dimensional. Just make sure it's clear what scaling factor was used.

Froude number is dimensionless while the usual speed-length ratio has units of "knots / square root of feet". But the latter is what is commonly used which becomes confusing when using wave theory which Froude number fits in a very natural manner.

 

Just for interest sake -

I fiddled with a hull design here, I actually scaled a 10m hull to a 9m size and was quite amazed at the result. The smaller hull was faster than the bigger hull with the same power input. When comparing the differences it seems the smaller hull's CP was 6.9 where the bigger hull was 6.2. Both are 2000kg displacement and the same beam.

This just means one thing - the 10m hull is not optimised !

 

Fanie,

What is the basis for this conclusion about speed and power input? Did you build a model, or is this some simulation, or else?

 

I use Freeship's software 2.82 I think it is. It gives a graph telling you what the expected power is for different speeds.

This means with a 28Hp = 20kW motor this hull can expect around 16.8kn or 31km/hr...

 

Thanks Fanie