Area needed to support 300 lbs

Hi, Can anyone do the math and let me know how much water need to be displace to support 300 pounds? Ideal what is the minimum size two hulls would need to be to support the weight?
thanks

 

There have been a couple of these 101 threads lately. Calculate the displacement of water needed for your load + the weight of what you build to support that weight + a safety factor / how high you want the deck to be above the surface of the water so one additional ounce doesn't sink her. 1 cubic foot of fresh water weighs 62.4 lbs. If this is a barge it's a simple calculation of L X W X D plus a couple calculations of the angled portions. If pontoon, pie x radius squared x length. Remember where you want the hull to float when calculating how much displacement (hull under the water surface) you can use to offset your weight. If more complex shape, slightly more complex to get the volume or mockup a model in freeship/delftship free to get the displacement.

See

http://www.boatdesign.net/forums/boat-design/need-help-equations-31480.html

http://www.boatdesign.net/forums/bo...acement-16-35-houseboat-31369.html#post343024

 

I wonder how many of these types of questions are from students looking for the answer to an assignment, particularly when they are the first post by the person.

 

Weatherbound; you are asking a question about volume required not area required. Volume is calculated in three dimensions while area uses only two dimensions. I suggest that you look up Archimedes principle on wikipedia or some other source. That will give you a good idea of how flotation works.

 

17.357 square feet.

 

Answer = (drum roll please) 300 lbs. Messabout gave some good advice. Archimedes told us that an object floating in a fluid will displace a volume of that fluid equal in weight to the object's weight. So, a 300 lb floating object displaces 300 lbs of whatever it's floating in. Use 62.4 lb/ft^3 for the density of fresh water, or 1 kg/liter and you can figure out the volume of the displaced water.
Amolitor, you sure about your calcs? Might want to double check 'em.

 

A sneaky way to calculate cocktail napkin displacement is as follows....Use the constant 0.03611 Call it whatever you wish. It is simply the result of dividing the weight of one cubic foot of fresh water by the cube of 12 or 1728
To find the volume of 300 pounds of water, simply divide 300/0.03611 = 8307.9 cubic inches. If we use Amolitars figure of 17.357 ft. sq., then we must have a third dimension. Convert his area to inches square.....17.357 x 144 = 2499.4 in^2. what then will be the draft, or third dimension? Easy...We needed 8307.9 cubic inches of displacement. Divide that number by the Amolitars area number....8307.8/2499.4 = 3.323 inches draft. Be careful to keep the same units in your calculations. That is to say, do not mix inches and feet. Do whatever conversion you need to stay within the same units.

Suppose that we wanted our boat to draw exactly 12 inches. Use the same method. Divide 8307.9 by 12 = 692.3 and that would be the area ....4.8 feet square if you prefer. if this was some kind of float rather than a boat you could take the square root of 692 and find that the float would measure 26.3 inches x 26.3 inches and it would sink 12 inches into the water. You could use any combination of three dimensions whose products are 8307.9

This elementary method is useful for small boats. For larger boats I would use a constant derived for cubic feet. With apologies to the rest of the world who cleverly use metric measure.
It works for metrics too but metrics are one helluva lot easier to deal with. One liter of water is one kilogram. Wow! that is too good a reality to ignore.

There Weatherbound, is that enough arithmetic to satisfy your need?

 

I believe the troll has come and gone.

 

This just seems to be someone in the depth of Winter with plans for Spring asking what to some may be a too simple question, not an actual troll. This is boatdesign.net, we don't have no steenking amateur trolls here.

 

17 sf of water will not support any weight.

 

I think maybe there's 18 tons of atmosphere settiing on it.?

 

17 sf of steel will not support any weight, either. Without volume there is no strength or weight to displace.

 

Probably some kid who needed help with his homework.....Hard to tell, but just in case it was a legitimate question we replied. Someone suggested that we have a separate section devoted to 101 type questions. Good idea except that most of you will not read that section and then the whole thing goes wanting.

 

Good answers all, thanks

I was looking for the 64 pounds/gallon cubed thing, I googled it... Anyway, I was wondering how small I could make an outrigger without my *** sinking into the water when I lean out trying to make my canoe into a proa and using a kite to pull me along.
Your right, I cant wait till spring.
Thanks all

 

Whats a troll?

No really, I don't know.