Calculating buoyancy of inflatable pontoons

Discussion in 'Boat Design' started by gridesign, Sep 6, 2011.

  1. gridesign
    Joined: Sep 2011
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    gridesign New Member

    Hi guys,

    Im in the process of designing a small 1 man inflatable pontoon craft for a uni project , and most likely beacuse I have no experience in design inflatables, I'm having a bit of trouble about finding it's buoyancy/ how much it can carry at the waterline just under halfway. Using the Hydrostatics function in Rhino, it tells me it is a 367 L volume and as seen here (with the waterline on the red axis), and with a mostly 370mm diameter size, (3d veiw here).

    It says it displaces only 107 L of water, and by assumption can carry 107 kg of weight to reach that line. I know that's very basic and may not be indicative of surface tension and other factors. But ideally I'd like to carry a 110kg man and 20kg of other weight factoring in the frame, oars, seat and day gear.

    I was told by a lecturer that a craft of this volume/length could easily hold 6 people and reducing the volume would be fantastic, because i'm trying to go for maximum portability

    So all in all, is there a pen and paper formula that i can use to figure out how much volume i need to carry 130kg of weight easily and still be above the half way?

    The Austrailian standard for inflatable boats say the max rated weight shouldn't exceed 0.75 of the buoyancy chamber volume, which comes out at 282kg.

    So i'm getting mixed results/opinions, and there seems like there is no sure fire way to get an answer, so any help from you guys would be very much appreciated.

    cheers
     
  2. CatBuilder

    CatBuilder Previous Member

    I'm no designer, but I think you need to go back to basics:

    http://en.wikipedia.org/wiki/Buoyancy

    You definitely don't need a computer for this, other than to calculate the volume of the inflatable pontoon if it's a complex shape. (If it's a cylinder, you should be able to calculate this yourself)

    Archimedes figured out the answer quite some time ago. Ask the Physics department for a little assistance if you are unable to follow the Wikipedia page.
     
  3. messabout
    Joined: Jan 2006
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    messabout Senior Member

    You want a pencil and paper method for figuring displacement or bouyancy. Here goes!

    Start by drawing a circle with a dot at the exact center. Place a horizontal line somewhere below the centerline and let the line extend across the circle. The horizontal line will represent the waterline that you have chosen.

    Now draw a line from the center of the circle to the point on the circumference of the circle where the waterline intersects it. Do that on both the right and left sides of the centerpoint. You will have a triangle sitting on top of the immersed part of your circle.

    Now we must calculate the angles of the triangle. Example: suppose that the circle is 40 cm diameter and you have drawn the waterline 5 cm below the center of the circle. The triangle will be five cm high and 20 cm wide. The tangent of that angle is 5/20 = 0.25 Use your calculator or a trig table to find the angle for that tangent......it is about 14 degrees. The figure below the center point (the dot) will sweep through an angle of 180-14-14 = 152 degrees. What proportion of the entire circle is that? calculate 152/360 = 0.422......Stay with me we are getting there.

    Find the area of the entire circle. 40cm x 40 cm x 0.7854 = 1256 square centimeters. (the number 0.7854 is merely pi divided by four) Go back to the proportion of the circle that contains the the triangle and the rounded part that will be under water. That was 0.422 part of the whole circle. Multiply 1256 time 0.422 = 535 sq. cm. That is good but we know that the little triangle will not be below the waterline, it will not provide any bouyancy, so we must subtract the area of the triangle. The triangle is 5 cm high and 40 cm wide. Find that area. (5 x 40)/2 = 100............ Subtract 100 from 535 to get an immersed area of 435 square centimeters.

    Now you will consider the length of all the tube that is immersed. Suppose that the actual wet part is 400 cm long. Multiply the area by the length. 435 x 400 = 174,000 cubic centimeters or 174 liters or 174 kilograms of flotation.

    Ignore that part of the horse shoe that is above the water and also ignore the pointy ends. You will have a reasonably accurate assessment of the static bouyancy of your boat.

    I hope that this will help.
     
  4. Submarine Tom

    Submarine Tom Previous Member

    367 litre volume = 367 kg buoyancy in fresh water.

    Times 0.4 = 147 kg for maximum loading while maintaining 60% reserve buoyancy or positive buoyancy.

    So, the total load including boat, motor, oars, lines, safety equipment, people, gear, dogs, food, everything must not exceed 147 kg (320 pounds).

    The pen and paper way to figure it out is to calculate the area of the cross-section of the tube, then multiply that by the length of the tube equalling the volume. If you do it in cubic feet, you should know that one displaces ~62.5 pounds of fresh water.

    -Tom
     
  5. portacruise
    Joined: Jun 2009
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    portacruise Senior Member

    This has been done commercially with weights below 10#, depending on how streamlined you want it. Here's an example with hard frame, but there are some others which are all fabric construction with 2 distinct toons for less water resistance: http://www.supercat.us/site/412251/page/118491 You can google kickboats for more information.

    I have accumulated several of these over the years which I use in my fishing hobby. Some other things to consider are the contraction and loss of volume when they are put into colder water and the sag if using longer pontoons. Drop stitch construction allows pressures above 6.5 psi which makes for a much more rigid construction than even the zodiac tubes rated at about 3.5psi.

    Porta

     
  6. portacruise
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    portacruise Senior Member

  7. DCockey
    Joined: Oct 2009
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    DCockey Senior Member

    Surface tension effects are completely negligable for any "boat" larger than several centimeters.


    A 2.1 m long boat able to hold 6 people. Very, very interesting. :rolleyes:
     
    Last edited: Sep 7, 2011
  8. gridesign
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    gridesign New Member

    thanks for all the replies, i think i'll have to give up on the idea of making it any smaller, will there be a significant difference in load bearing allowance at the waterline if i have a shallower angle on the 'horseshoe' shaped part, to actually have it touching the water?

    I have seen dave scadden's craft before, but i'll have a closer look to what he's done

    thanks messabout for doing those calculations, i'll give it a go myself.

    Just a quick edit; when i was trying the calculations I've got to the point to find the area of the entire circle. You've got 40x40x0.7854=1256cm^2, but shouldn't it be pi x r^2 = area? or am I missing something here :)

    cheers guys.
     
  9. portacruise
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    portacruise Senior Member




    YES, quite an increase in load if the "horseshoe" part is in the water. But the drag for propulsion of the boat will greatly increase IF that part touches the water. But that may not be a factor of concern, since you have only mentioned load, and weight and size of the boat.

    P.
     
  10. Submarine Tom

    Submarine Tom Previous Member

    You know, these formulas and calculations should be very basic for a university level student. Is this for a fun water game or part of the curriculum?

    -Tom
     
  11. Poida
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    Poida Senior Member

    Sub Tom

    He's from New Zealand!!
     
  12. gridesign
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    gridesign New Member

    Well it is for a project, but not part of the curriculum if you know what I mean. As I am an Industrial Design student, marine/transport design isn't taught as a specialised subject, so a lot of my information I have gathered is through self taught, research, and trial-and-error, and for as much as I know I can understand the calculations, but it is the application and checking if I was on the right track, was the part I was having trouble with, which is why I came here. And don't get me wrong, it's all been massively helpful.
     
  13. messabout
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    messabout Senior Member

    Gridesign; Good observation. Your formula, r^2 x pi, for the area of a circle, will yield the same result as D^2 x .7854. Try it. The only reason for using the alternate method is that really basic calculators do not have a pi key. Also I do not know how to find the pi symbol on my keyboard.
     
  14. Submarine Tom

    Submarine Tom Previous Member

    LOL... Wait a minute, that's not funny.

    Poor Kiwi...

    -Tom
     

  15. Poida
    Joined: Apr 2006
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    Poida Senior Member

    You're right Tom, that wasn't funny. Gridesign I am sorry.
    But I would have thought you would have learned a bit about Archemedes Principle dipping sheep.
     
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