Displacement to (efficiency per ton) ratio at a given speed - SailNet Community
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Displacement to (efficiency per ton) ratio at a given speed

Maybe this isn't the best place to ask, but I was trying to figure out how increasing the size of a boat decreases the energy required to push it along, per ton.

If the displacement is increased by a factor of 8, that means that the surface area in the water is expected to increase by 4. So 8 times as many tons, and only 4 times as much energy to move it along. The energy used per ton is half (as long as the speed is the same.)

Doubling the speed should create 8 times as much drag. I'm not sure how having the speed approach the hull speed of the vessel affects that number. 8 times as much drag at double the speed means 4 times the energy used per distance traveled. So if the size is increased by a factor of 64, a vessel should be able to double her speed and only use double the energy per ton to maintain that speed. For power boats, arriving in half the time would mean that the miles per gallon per ton would be half, so increasing the size by a factor of 64 means that the miles per gallon per ton is the same, at twice the speed.

If a 1.5 ton 21' can go 6 knots with a 6 HP engine, then a 96 ton 84' should go 12 knots with 384 HP.

The prop gets less efficient at higher speeds, and of course for sail boats the sails would perform differently with that 5 knot difference in wind speed, but I don't know how to consider those things.

No wonder boats are so big, and little ones go so slow.

What is a more accurate way to do this calculation? My search didn't reveal any previous topic on this.

It seems that drag is actually less than 4 times as much with 8 times the displacement. The proper calculation is probably way more complicated.
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Old 12-31-2012
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Re: Displacement to (efficiency per ton) ratio at a given speed

I seem to have lost my first answer. What you're missing is that only at very low speeds is the skin friction the driver in hull resistance. At the speeds most boats go at it's the wavemaking resistance that is the driver. A boat rides in a wave that starts at the bow and ends at the stern. The boat's power is used to push that wave. The longer the boat is the larger that wavelength and the faster the wave travels, so the faster the boat goes at a given power. If you put on enough power the boat rides over that wave and planes, and the resistance equations all change. For non-planing boats the speed of that wave is called "hull speed" which equals "SL" X (square root of the waterline length) where SL is the speed length ratio. That ranges from 1.1 to 1.4 and is most commonly expressed as a function of pounds per horsepower. Most you see 1.3 used which is about 550 lb/hp. This is largely empirical. A good discussion of this including semi-displacement and plaining hulls is chapter 2 of David Gerr's Propeller Handbook>

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Old 12-31-2012
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Re: Displacement to (efficiency per ton) ratio at a given speed

Most of the time you will find Your sailboat is a displacement hull. A few sailboats are not Displacement hulls. The bow wave will not be passed up. The stern wave is a distance from the bow wave. The larger LWL the more speed can be made before the two waves start to interfere with each other. The sailboat is not going to plan or (Surf) most of the time... This is your key search to understanding why a longer LWL means a faster boat for a displacement hull water must be displaced. SHEARCH WIKIPEDIA Hull SPEED Regards. Lou

Last edited by Lou452; 12-31-2012 at 09:50 PM.
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Re: Displacement to (efficiency per ton) ratio at a given speed

Thanks for the replies so far. I don't want anyone to overlook that the hull surface area increases at the displacement ^ 2/3 power as displacement increases. If you drop a large rock and a small one the large one will fall faster through water or air even when both are the same density.

I know that going on plane or approaching the hull speed will change the linearity of the hull drag. In this case I am just thinking of displacement hulls which are being used under or well under hull speed when the mathematical characteristics of drag are most linear. I am thinking more about boats in excess of 12 tons or 42' which have long water lines to push the hull speed out to where it doesn't need to be worried about.

Last edited by steel; 12-31-2012 at 10:25 PM.
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Old 01-01-2013
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Re: Displacement to (efficiency per ton) ratio at a given speed

Quote:
Originally Posted by steel View Post
Maybe this isn't the best place to ask, but I was trying to figure out how increasing the size of a boat decreases the energy required to push it along, per ton.

If the displacement is increased by a factor of 8, that means that the surface area in the water is expected to increase by 4. So 8 times as many tons, and only 4 times as much energy to move it along. The energy used per ton is half (as long as the speed is the same.)

Doubling the speed should create 8 times as much drag. I'm not sure how having the speed approach the hull speed of the vessel affects that number. 8 times as much drag at double the speed means 4 times the energy used per distance traveled. So if the size is increased by a factor of 64, a vessel should be able to double her speed and only use double the energy per ton to maintain that speed. For power boats, arriving in half the time would mean that the miles per gallon per ton would be half, so increasing the size by a factor of 64 means that the miles per gallon per ton is the same, at twice the speed.

If a 1.5 ton 21' can go 6 knots with a 6 HP engine, then a 96 ton 84' should go 12 knots with 384 HP.

The prop gets less efficient at higher speeds, and of course for sail boats the sails would perform differently with that 5 knot difference in wind speed, but I don't know how to consider those things.

No wonder boats are so big, and little ones go so slow.

What is a more accurate way to do this calculation? My search didn't reveal any previous topic on this.

It seems that drag is actually less than 4 times as much with 8 times the displacement. The proper calculation is probably way more complicated.
I suggest you get a copy of Skenes or another book on the fundamentals of yacht design. Part of your answer is in the law of mechanical similitude.

I, myself, personally intend to continue being outspoken and opinionated, intolerant of all fanatics, fools and ignoramuses, deeply suspicious of all those who have "found the answer" and on my bad days, downright rude.
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Old 01-01-2013
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Re: Displacement to (efficiency per ton) ratio at a given speed

[QUOTE=steel;969481]Thanks for the replies so far. I don't want anyone to overlook that the hull surface area increases at the displacement ^ 2/3 power as displacement increases. If you drop a large rock and a small one the large one will fall faster through water or air even when both are the same density. QUOTE]

Didn't Galileo prove otherwise ? or am I missing something ?

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Re: Displacement to (efficiency per ton) ratio at a given speed

If Skenes is a little too deep I'd suggest David Gerr's "the nature of boats" - it's a pretty thorough book that covers this and much more at a level that is easily understood.
As JonB said above it's about mechanical similitude - but also relativity.
For example, although wetted area and surface friction are critical to speed, so in fact is hull shape (not just a fine bow as opposed to a barge, but also length to beam and the overall under body shape).
ergo - You can't calculate the HP required to achieve hull speed without taking into account the prismatic coefficient (shape) of the boat.

Just going by displacement and HP (forgive me if I use sail power vs HP, this is SailNet not MotorNet) -

I've got a 20,000 pound Irwin 38 CC - Displacement to the 2/3 = roughly 48, and Sail Area (771 sq feet) to displacement (SA/D) = 16.7
Per D. Gerr, you multiply the two (48 * 16.7) and get 801 - so I'm slightly under - canvassed for 'good' performance.

For the die hard crowd:
We can also calculate the HP as a factor of sail power, there are numerous charts out there to do so, I'll just use the 'standard' of .02 HP per square foot at force four wind (13-15)
My Irwin 38 CC has 771 square feet of sail (100 percent fore triangle). So in a force four breeze I can expect the sails to generate 15.4 HP.
Thankfully force five winds produce twice the power

You can do the math backwards - 48 *16.7 = 801 sq feet of sail for good performance - 801 * .020 = 16hp

Here's the real life - at 2200 RPM she makes 6.3kts in 1 foot chop - and according to the RPM - HP charts about 20 HP - meaning about 16HP delivered to the prop (altenator loss and transmission loss take the rest).
Funny how math just seems to work out isn't it..

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Old 01-01-2013
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Re: Displacement to (efficiency per ton) ratio at a given speed

Steel's maths seems wrong.
"If the displacement is increased by a factor of 8, that means that the surface area in the water is expected to increase by 4. So 8 times as many tons, and only 4 times as much energy to move it along. The energy used per ton is half (as long as the speed is the same.)"
If the surface area increases by 50% of the weight more or less, the drag component increases proportionately but not the energy to achieve a certain momentum which is proportionate to mass. Also the drag is partly friction but partly resistance due to the water displaced by the 2 dimensional forward plane. That will not increase just because of length.
"Doubling the speed should create 8 times as much drag."
Um no. I thought this was V squared not cubed. Sorry I didn't bother with the rest.
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Old 01-01-2013
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Re: Displacement to (efficiency per ton) ratio at a given speed

Quote:
Originally Posted by chris_gee View Post
Steel's maths seems wrong.
"If the displacement is increased by a factor of 8, that means that the surface area in the water is expected to increase by 4. So 8 times as many tons, and only 4 times as much energy to move it along. The energy used per ton is half (as long as the speed is the same.)"
If the surface area increases by 50% of the weight more or less, the drag component increases proportionately but not the energy to achieve a certain momentum which is proportionate to mass. Also the drag is partly friction but partly resistance due to the water displaced by the 2 dimensional forward plane. That will not increase just because of length.
"Doubling the speed should create 8 times as much drag."
Um no. I thought this was V squared not cubed. Sorry I didn't bother with the rest.
Resistance DOES vary as the cube of speed - resistance and wavemaking aren't the same thing.

Here are the pertinent ratios from Skene for displacement hulls;

Speed varies as the Sq. root of LWL
beam, draft & length vary as the LWL
S/A and wetted surface vary as the Sq. of LWL
Disp. varies as the cube of LWL
Heeling moment of wind pressure varies as cube of LWL
Resistance varies as cube of LWL
Stability varies as 4th power of LWL
Moment of Inertia varies as fifth power of LWL

I, myself, personally intend to continue being outspoken and opinionated, intolerant of all fanatics, fools and ignoramuses, deeply suspicious of all those who have "found the answer" and on my bad days, downright rude.
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Old 01-01-2013
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Re: Displacement to (efficiency per ton) ratio at a given speed

I wonder, also, if underwater shape scales differently for large changes in displacement. For the same displacement, fin keels will have less wetted surface than full keels. However, is this a bigger factor with a 24' boat vs. a 42' boat?

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