Sail Area / Power

Discussion in 'Boat Design' started by dustman, Sep 2, 2020.

  1. Will Gilmore
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    Will Gilmore Senior Member

    You don't get that at Sailing Anarchy?

    -Will (Dragonfly)
     
  2. ryanonthebeach
    Joined: Mar 2007
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    ryanonthebeach Junior Member

    a whole other level of flak over there, but mostly not condescending
    somehow that's better

     
  3. dustman
    Joined: Jun 2019
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    dustman Junior Member

    Did you read the original post?

    "There are a great many factors that go into how much power your sails will produce in various conditions, arrangements, trims, aspect ratios, what angle you are sailing relative to the true wind direction, etc, etc. And how fast your boat will go with a given amount of power has a number of variables as well.
     
  4. dustman
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    dustman Junior Member

    I would be grateful if you chose to expand on my post in a constructive way.
     
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  5. fastsailing
    Joined: Sep 2017
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    fastsailing Junior Member

    A sail of it's own will (or would) never produce any power at all. It would drift downwind at the speed of truewind, thus the apparent wind would be zero, therefore aerodynamic forces are also zero, as is the power.
    Sail will only deliver propulsive power when used in combination of some useful part immersed in the water, typically including but not limited to keel or daggerboard or centerboard, and a rudder. In that case delivered propulsive power is the product of boat speed relative to whatever you want, and the component of total aerodynamic force taken in direction of boat velocity. Of course this assumes special relativity is not needed, because boatspeeds are typically well below speed of light in vacuum. under that assumption the previous sentence can be used as a definition for propulsive power.

    Because all of the above, delivered propulsive power is not very useful merit for sails, aerodynamic force for a unit sail area is as long as apparent wind speed and direction, as well as boatspeed is known.
     
  6. Blueknarr
    Joined: Aug 2017
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    Blueknarr Senior Member

    I think asking how much HP per square foot of sail is equivalent to asking how much HP per gallon of fuel.

    Depends on the engine it is pumped thru
     
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  7. Dolfiman
    Joined: Aug 2017
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    Dolfiman Senior Member

    (I deleted my previous quote which included a stupid conversion error, here below is the good version)
    As it was already commented, there is a lot of parameters to take into account and the answer depends both of the boat and its sailing condition. Yet VPP tools has been developed to deal with this complexity at the best and can give you the main trends. I did the exercise with a VPP I am developing and a numerical sailboat, the "Blue Water 39" inspired by the Corbin 39 with its mk2 sailplan. "Net power per m2" is here defined as : Thrust(=Drag) x Boat speed per m2 of sails area .
    Two sailing conditions were studied : beam reaching, upwind
    When beam reaching : the net power per m2 appears independent at first order of the sails area, reaching ~ 0,25 to 0,28 kW/m2 with wind 20 knots
    When upwind : more complex, the net power per m2 actually appears dependent of the Flat adjustement when this parameter is involved, i.e. the efficiency (in terms of power) decreases with the sails camber despite the fact that the speed continue to increase a bit (which is the main goal), net power reaching ~ 0,08 to 0,13 kW/m2 with wind 20 knots.
    Data and more details in the document attached.
     

    Attached Files:

  8. Will Gilmore
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    Will Gilmore Senior Member

    Not that I'm anyone to pay attention to for these type of questions, but I don't find the asking of the question to be unreasonable.

    The fact is, more sail area translates to more power. It isn't straight forward and it can't be easily isolated, but if you increase the sail area for a given design, you get more power out of the sail.

    Sometimes that increase in power means more heel, and more heel can mean less speed and sometimes it means other issues that have to do with the practical handling of the boat. However, the old J class AC boats flew huge sail plans for exactly that reason. When you have a dynamic relationship, such as sail area to velocity, like that, the question of how much does each unit squared increase the power became a very reasonable question to explore. Even if the ultimate answer is, it's to complex to give a good answer, it is still a question worth wrestling with.

    When designing a sailboat, the location of the center of effort is balanced with the location of the center of lateral resistance. Most boat designers use to just use rule-of-thumb and broad category estimation as to the location of CE and CLR and the relationship between those two varied according to a number of factors, both controlled and uncontrolled. Today, with modern computer-based design, I imagine it is more precise, but if you can't know the force to expect from a given sail area, how would you even begin to balance a new design?

    -Will (Dragonfly)
     
  9. dustman
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    dustman Junior Member

    The power a given sail area/trim will produce at a given apparent wind angle and speed is independent of other characteristics of the boat. Of course the dynamic nature of the sailing environment and how the characteristics the boat interact with it matter, but it doesn't change that simple fact. A larger sail will produce more force, and smaller sail will produce less force, all else being equal. At given trim and 10 degrees apparent wind angle and 45 degrees of heel and 20 knots wind speed a 200ft2 sail will produce about twice as much force as a 100ft2 sail of the same relative dimensions(yes I know, there will be minor deviations due the larger diameter mast, height above the water, etc).
     
  10. Blueknarr
    Joined: Aug 2017
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    Blueknarr Senior Member

    The problem is HP is a measurement of work.
    Square footage is a measurement of area.
    Wind speed is a measurement of energy.

    To convert from energy to work one had to know the mechanical efficiency of the engine involved.

    So, what percentage of the wind's available energy will the sail and rigging transfer to the hull in the desired direction?

    Way to many variables.

    Back to my comparison of how my HP per gallon of fuel.

    Depends on how efficiently it is converted.

    More gallons or more sq ft is more available power. But that is all that can be determined.
     
  11. dustman
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    dustman Junior Member

    How is it possible for the force to be independent of area? This is incomprehensible to me.
     
  12. dustman
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    dustman Junior Member

    Yes, you choose sail area/layout based on the parameters of your hull and the desired outcome. Resistance, stability.
     
  13. messabout
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    messabout Senior Member

    Blueknarr, you are correct to say that fuel efficiency depends on the particular engine that is using the fuel. An old Lunenberg, for example, is not admirably efficient. A modern racing type engine and more than a few modern automobile engines have a fuel consumption ratio of about 0.45 pounds of fuel per horsepower hour. Less efficient ones are in the 0.50 to 0.55 region. Note that the fuel consumption is measured in pounds of fuel, not volume. Engine people call this measure; BSFC (brake specific fuel consumption)

    We might stretch that generality to apply to sails. The efficiency of the sail is dependent on how well it is cut, the condition of the sail, and how well it is utilized. I have some dog sails in my locker and some identical ones that deliver an easily measurable better result, most observably on weather legs. Flat cut sails work only a little bit better than rusty beer signs when on a windward leg. But then we can dispute that last claim by varying the wind speed. I do not want a full cut sail when in a force four.

    Until we learn more, I am inclined to go with Dolfiman's numbers for sail force. As has been mentioned, sail power can not be directly translated into boat speed. Maybe we need to tinker with a power to resistance ratio or some such thing. We already have a SA/displacement ratio. That cannot be translated to performance except for a narrow range of hull types. Are we having fun yet?
     
  14. Dolfiman
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    Dolfiman Senior Member

    I interpreted and answered to your question as being about output power for the propulsion of the boat but I think you have in mind the input power of the sails system, like it was the installed power of a motor made of sails with the wind as fuel and independent of the boat behaviour.
    I have no idea to do that properly. By analogy with the rationale of a wind turbine, can be something like : ½ Rhoair S Vwind^3 , avoiding the use of Vapp the apparent wind ?
     

  15. gonzo
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    gonzo Senior Member

    It is missing the information necessary to make sense of it.
     
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