Volume of a Half-Diamond For Hull Volume Displacement

Discussion in 'Boat Design' started by Free Pirate, Mar 9, 2005.

  1. Free Pirate
    Joined: Mar 2005
    Posts: 24
    Likes: 0, Points: 0, Legacy Rep: 10
    Location: Here and There

    Free Pirate Junior Member

    I'm trying to figure out where to put my 1290 lbs of displacement below the waterline. The boat is canoe-shaped (no transom) and V-bottomed. It's about 19.7 cubic feet of mass displacement in salt water (am I correct?).

    I don't know the beam to length ratio or the draft, is it a bad idea to find it out by moving around the volume that needs to be displaced until I find something that seems to fit, and designing the freeboard up from that? Or am I on the right track? Or just a weird one?

    What formula is there to calculate the volume of a v-bottomed canoe-shaped hull? (It's basically a half-diamond, a diamond being two pyramids resting on each other's bases). I can clarify what I mean with drawings if it's confusing.

    I'm avoiding finding it out through Simpson's because I'm trying to get a rough, basic shape without having to measure and redraw, and also because I don't know it well enough to see if I'm doing it wrong.

    Should I just use Simpson's or is there a faster way for a volume of a half-diamond?

    Sorry for being new and unknowing. ;) Your answers are helping me out a lot. Tell me if I'm asking too many questions.
     
  2. Free Pirate
    Joined: Mar 2005
    Posts: 24
    Likes: 0, Points: 0, Legacy Rep: 10
    Location: Here and There

    Free Pirate Junior Member

    What I need to know is how to find the volume of an irregular tetrahedron.

    If you put two identical irregular tetrahedrons together so that they are shaped kind of like a canoe, it will make the rough shape of the wetted surface. I need to be able to find the volume of one of them so I can find out the possible proportions between the draft, LWL, and BWL.

    Is there a better way? It seems to me like 2 identical irregular tetrahedrons should be easier to find the volume of than a complex hull shape.

    Please help!
     
  3. Raggi_Thor
    Joined: Jan 2004
    Posts: 2,457
    Likes: 64, Points: 0, Legacy Rep: 711
    Location: Trondheim, NORWAY

    Raggi_Thor Nav.arch/Designer/Builder

    I suppose you have some books?
    Try to figure out a waterline length, say 16 feet for a simple calculation.
    What speed do you intend to cruise at?
    If you plan to paddle at 4 knots or slower, go for a Cp ~ 0.5
    If you want to go faster than 2x (lwl)^0.5 (faster than 8 knots) go for Cp ~ 0.7
    Then you can decide an area for your midship section, or the larges section, normally a few % aft of "midship".
    Area x Cp x LWL = Displacement
    So if LWL = 16 , Cp = 0.5 and Displacement = 20 your section area is 2.5 square feet. For a simple V shape that could be 5 feet beam (at wl) x 1 feet draft, for example.
    I suppose all this is found in any of the popular books.
     
Loading...
Similar Threads
  1. cpo1
    Replies:
    9
    Views:
    4,540
  2. thenavalarch
    Replies:
    1
    Views:
    1,470
  3. mustafaumu sarac
    Replies:
    2
    Views:
    2,512
  4. Adarsh Edakkote
    Replies:
    7
    Views:
    6,958
  5. Squidly-Diddly
    Replies:
    24
    Views:
    4,518
  6. Saqa
    Replies:
    42
    Views:
    5,055
  7. appusree888
    Replies:
    27
    Views:
    5,208
  8. Paddlelite
    Replies:
    4
    Views:
    2,178
  9. vignesh
    Replies:
    13
    Views:
    2,459
  10. Kyle Motiv8Labs
    Replies:
    23
    Views:
    500
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.