Calculating volume

I know how to determine the volume of a 6" pipe x 12'. I wish to test out using qualty inner tubes between my side frames as flotation. They will get no sunlight. If I determine the circumference of a 13" tire tube inflated to 5" in diameter then I can use the same math as a rigid pipe to get square inches then cubic feet. Would I then just multiply the cubic feet times 62 pounds to get to totally flotation? Higer inflation just increases the the diameter only??
Thanks much, Stan

 

Stan,

The circumference of the tube must be at the centroid of the section you are using to calculate volume. Also, the tube must be completely immersed for its entire volume considered as flotation. The pressure in the tube must be at least equal to the hydrostatic pressure it will be subjected to.

You should consider the trim and stability of the boat in the flooded condition. This is standard procedure for professional naval architects, but is almost always overlooked by amateur designers. It is possible to have positive flotation and still have a vessel that will capsize (transversely or longitudinally) when flooded.

 

Paul, thank you. Would the hydrostatic pressure you refer to be Gravity or around 17 pounds pressure in the tube. I'm not concerned about floating upside down--just floating. Thanks again, Stan

 

Hydrostatic pressure is the pressure exterted on a submerged (or partially submerged) object by the surrounding water. It is equal to the weight density of the water times the depth of the object (in consistent units).

If an object is submerged 3 feet in fresh water, the hydrostatic pressure on that object is:

62 lb/ft^3 * 3 ft = 186 lb/ft^2 or 1.3 psi.

 

volume of your pipe

13ft x 6 inch diameter you say?

First thing you need to do is come to Australia and learn the metric system and how to drink real beer. After that you can return with a real education.

13ft = 4m (a fraction under, but very close to 4m)
6 inche = .15m (a tiny fraction over but very close)

volume = 4m x pi x .15 x .15 divided by 4
volume = 3.141 x .15 x .15 = .0707 cubic meters
which = 70.7L (equivalent to 70.7kg of water)

you prefer pounds, then 70.7 x 2.2 = 155 pounds

if it was 5 inches, not 6, then multiply 155 x (25/36)
this is (5x5) divided by (6x6)


metric is easy!!!!!!!!

 

my bet is once you've been upside down for a while you might feel differently

 

Peter you are rubbing it in. Would that we yanks had the good sense to use the SI method of calculation. Sure enough it would be easier. How convenient that a liter of water is a Kg. In our case we say that "a pint is a pound the world around". A cute homily but not quite accurate. Oh well! Perhaps we will come to our senses one day.

One more thing. I am not so sure that Oz beer is better than ours, although we do put ours in containers that are calibrated in silly units such as ounces.

 

Not quite, Peter.

The problem is, that due to international standards, a NPS 6" (DN 150mm) pipe is actually 6.625" (168.3mm) OD. So it's not just an imperial vs metric thing. Also 1kg = 1L only with 100% fresh water and only at 4+/- degrees C, significantly below the temperature of most bodies of water, so you should use a density of ~.999 for most fresh water and 1.0259 for salt water for real calculations. Finaly 1 lb is defined as 0.4535924 kg so the conversion would be 2.2046

We need to avoid that whole Mars Climate Orbiter thing again...

 

Now that I'm thoroughly confused can someone tell me what a 13" diameter round tire inner tube will float when inflated to 6" round(diameter)in fresh water?

 

hmmm
lets see if I can rip that out of my ***

ok
13 outside diameter of a 6 inch thick tube floating in fresh water

in short
yes it will float if inflated with air

if inflated with concrete then
no
it will not float

its volume would be the same as for a tube of the same length as the torus when measured at its center for diameter

so the radius of a circle of 13" diameter is 26
subtract 6 to abtain the center of torus diameter and you get 20
20 x pi = ~63
an 63" long tube 6" diameter
3.1416 x 3^2 x 63 =1781 cu/inchs
/231 cu/inchs pr gallon = 7.7 gallons at 8.35 lbs = 64.39 lbs

so what does the material you made the torus out of weight and I can tell you how many you will need to make that Kayak float

 

Thanks Boston

 

Boston, you can save a couple of steps in the bouyancy calcs by dismissing gallons. Find the total volume in cubic inches and multiply by 0.03611 and you get pounds of bouyancy. Or if you want to know the required volume for a particular weight...divide the weight by the same number. 100 pounds needs 100/0.03611= 2769 cubic inches of volume....etc This is suitably accuate only for fresh water and if more precision is required for salt water you change the number slightly, as follows 64/1728= 0.03703 All we are doing is finding pounds per cubic inch and using the number to play with.