How much HP to push a barge??

Discussion in 'Boat Design' started by JamesG, Jan 24, 2010.

  1. JamesG
    Joined: Aug 2009
    Posts: 82
    Likes: 4, Points: 8, Legacy Rep: 55
    Location: Virginia

    JamesG Junior Member

    How do you calculate the horse power needed for a boat to push a barge?

    The barge will be totally square, all right angles, so basically a box.

    I need a formula or graph that lets me input the weight of the barge and the dimentions as variables.

    Thanks a lot!
    -james

    PS- Is torque a major factor or is it mainly HP?
     
  2. Landlubber
    Joined: Jun 2007
    Posts: 2,640
    Likes: 125, Points: 0, Legacy Rep: 1802
    Location: Brisbane

    Landlubber Senior Member

    2 hp will move it, but how fast do you want to move it?

    ....it would help greatly if you would consider telling us just how big it is and what it displaces.....
     
  3. JamesG
    Joined: Aug 2009
    Posts: 82
    Likes: 4, Points: 8, Legacy Rep: 55
    Location: Virginia

    JamesG Junior Member

    I don't know how big it will be, but it will probably be somewhere in this range:

    It might be as small as 10'X30' with a weight of 13,500lbs and a draft of about 14".
    Or as large as 50'X100' with a weight of 200,000lbs and a draft of about 14".

    These are all really rough numbers.

    -james
     
  4. Landlubber
    Joined: Jun 2007
    Posts: 2,640
    Likes: 125, Points: 0, Legacy Rep: 1802
    Location: Brisbane

    Landlubber Senior Member

    These are all really rough numbers.


    ...yep....

    surely if you are for real you have some idea of just what you are designing mate?
     
  5. Submarine Tom

    Submarine Tom Previous Member

    James,

    A box, a square, or a cube?

    10 X 30 and 50 X 100 are not squares.

    I think you may be in real trouble here?

    But, if you insist on a "formula", here's one:

    1 HP per 10 square feet.

    Good luck!

    -Tom

    P.S. It's torque.
     
  6. PAR
    Joined: Nov 2003
    Posts: 19,126
    Likes: 498, Points: 93, Legacy Rep: 3967
    Location: Eustis, FL

    PAR Yacht Designer/Builder

    Everyone's a naval architect. Some concept of what you're doing would be a reasonable idea JamesG, don't you think? Are you familiar with Archimedes?

    How hard could it be, right? Lets say you've "designed" up a nice, basic 50'x100' barge. You've got it lightly loaded and she's sunk down and now draws about 12" of water. This little barge is displacing about 311,000 pounds or about 141 ton. It's PPI is about 26,000 pounds or about 12 ton, for each inch you further sink her down. Well my boy, this is a fair bit of engineering to handle all this weight, on a lightly loaded barge, so I'll ask again, are you up to the engineering necessary to keep this puppy from folding in half, when you load it up?
     
  7. JamesG
    Joined: Aug 2009
    Posts: 82
    Likes: 4, Points: 8, Legacy Rep: 55
    Location: Virginia

    JamesG Junior Member

    All i'm asking for is a formula for the amount of horsepower needed to push a barge at a certain speed. If there is no formula then maybe there's a graph because this is probably a hydrodynamic/fluid dynamic equation.

    The inputs would probably be:
    Weight of the barge
    Displacement of the barge OR dimensions of the barge
    Drag coefficient (it is a square, blunt, hull in this case (imagine a shoebox))
    And Speed

    The output would be :
    Horse Power of an engine.

    My apologies for not being clear enough last time. Let me know if you need more info.
     
  8. JamesG
    Joined: Aug 2009
    Posts: 82
    Likes: 4, Points: 8, Legacy Rep: 55
    Location: Virginia

    JamesG Junior Member

    Who says I wanna keep it from folding in half?
     
  9. haru
    Joined: Jan 2010
    Posts: 50
    Likes: 1, Points: 0, Legacy Rep: 21
    Location: Terra

    haru Junior Member

    We have many skeptics here so until you come clear help comes slow.

    This is physics 101 you could just have asked in a physics forum.
    Anyway, here's the formula simplified:
    P = 0.00067 * v^3 * cos(alpha) * rv * A
    P: power in HP
    v: relative velocity
    rv: river flow
    alpha: angle to river flow; 0 = same direction
    A: area

    You need to know torque in terms of pressure from the pushing boat and the transported loads. Pushing gently will do.
     
  10. gonzo
    Joined: Aug 2002
    Posts: 16,790
    Likes: 1,714, Points: 123, Legacy Rep: 2031
    Location: Milwaukee, WI

    gonzo Senior Member

    There is also wind resistance. In a lightly loaded barge it is often more than the wave and skin friction resistance. If you fold it in half, it depends if it is lengthwise or sidewise. You need to ask the right questions to get useful answers. That means studying the basics first.
     
  11. jonr
    Joined: Sep 2008
    Posts: 721
    Likes: 11, Points: 0, Legacy Rep: 57
    Location: Great Lakes

    jonr Senior Member

    HP will be the major factor. One can get whatever torque desired with gearing.
     
  12. Submarine Tom

    Submarine Tom Previous Member

    James,

    In post #5 above I gave you a formula.

    Why are you asking for one again?

    -Tom
     
  13. JamesG
    Joined: Aug 2009
    Posts: 82
    Likes: 4, Points: 8, Legacy Rep: 55
    Location: Virginia

    JamesG Junior Member

    Tom,

    Thank you for the formula. I am actually am using it along with that other one.


    -james
     
  14. JamesG
    Joined: Aug 2009
    Posts: 82
    Likes: 4, Points: 8, Legacy Rep: 55
    Location: Virginia

    JamesG Junior Member

    Thanks Huru!

    Just curious. Are you sure rv stands for "river flow" or does it stand for "relative velocity"? I just wanna make sure i use it correctly.

    And is "relative velocity" the barge's speed through the water, not the ground speed? Kind of like with airplanes, "airspeed" vs "ground speed"?

    thanks,
    james
     

  15. haru
    Joined: Jan 2010
    Posts: 50
    Likes: 1, Points: 0, Legacy Rep: 21
    Location: Terra

    haru Junior Member

    I made rv up. :lol: since I converted it to your particular case.
    Cos(alpha) * rv = velocity vector of the river
    v is relative to the water so it's like airspeed

    As gonzo said, no wind is accounted. Also side way movement isn't really accounted either so it will sidetrack, but it should get to the other side.
    Having more HP certainly won't hurt at least it won't strain the engine.
     
Loading...
Forum posts represent the experience, opinion, and view of individual users. Boat Design Net does not necessarily endorse nor share the view of each individual post.
When making potentially dangerous or financial decisions, always employ and consult appropriate professionals. Your circumstances or experience may be different.