PLEASE HELP! Buoyancy Calculations

I am going on a river trip with some friends in June and I am wanting to construct a PVC cooler float. Ideally, it will be a rectangular frame with webbing straps as a bottom to set the cooler on to hold all the beer and what not.

I found a lightweight cooler that I want to use that weighs 5lbs. It holds 60 cans with room for ice.
Cooler- 5lbs
60 cans of beer @ 0.807 lbs = 48.42 lbs
10 lb bag of ice.
so total weight minus the PVC pipe would be 63.42 lbs

The coolers dimensions are H13"xW23"xD14.5"

so the PVC frame will be 24"X15.5" to give 1" of space on each side. total linear length 77"= 6.42'. That's the interior dimension of the frame. not a center line or exterior dimension.

I am trying to use 4" PVC pipe but I can't find any definitive calculations on what it can hold.

Ideally, I wouldn't want the entire pipe to be submerged for stability but anywhere from 1/4- 1/2 of the pipe being submerged would be good.

Another thought I had was if the 4" pipe wasn't enough, I can make 2 frames out of 3" PVC pipe with the same dimensions and have them connected by "TEE" joints so they can be attached on top of each other.

I found one calculation but when I ran the numbers it didn't seem to be accurate, so I want some clarification if it is accurate or not.
the calculation is this.

Also if I did use the 2 3" frames connected would this formula still work with the pipe frames being stacked together or is it a whole other formula.

Multiply the outside pipe diameter in feet by itself using a calculator and call it "D." Assuming the diameter to be 1.5 feet, D = 1.5 X 1.5 = 2.25.

Multiply "D" with 62.4 (the weight of water in pounds-per-cubic-foot) and call it "D1." For the example being considered, D1 = D X 62.4 = 2.25 X 62.4 = 140.4.

Multiply "D1" with 0.78 (the percentage of the pipe which is submerged in the water) and call it "W." For the example being considered, W = D1 X 0.78 = 140.4 X 0.78 = 109.51.

Multiply "W" with the length of the pipe to get the required buoyant force in pounds. Assuming the length of the pipe to be 5 feet, the force will be F = W X 5 = 109.51 X 5 = 547.55 pounds.

So using this formula with the dimensions of the pipe.
.3X.3=.09
D=.09X 62.4=5.616
D1=5.616 X 0.60=3.3696 (.60 is 60% pipe submersion)
3.3696 X 6.42' =21.6LBS


ANY HELP YALL CAN PROVIDE WOULD BE GREATLY APPRECIATED!

 

You need to do some research.
Volume of a cylinder is not hard to find.
Your cooler and contents displace one cubic foot of water at 63.5 pounds/cubic foot and has a volume of about 2 cubic feet.
But as you remove beer, it's going to get significantly lighter and tip over.
Capped, sealed 3" pipe will be plenty I think.

Try Google 'cause it's not likely anyone here is going to design this thing for you.

 

Your cooler displaces 2.509 cubic ft. of water fully submerged so this is 156.56 lbs of water since according to your calculations it will weigh 63.42 lbs when full you only need the PVC pipe to work as outriggers to keep it from tipping over so you don't need to do any calculations just make a frame from the PVC pipe that will go around the cooler and as long as the water isn't too rough it will be fine. Remember as the cooler gets emptier the higher it will float but the weight will be in the bottom. The further out the PVC pipe is from the cooler the better it will work.

 

As DaveT alludes to, most actual coolers float. The Wylie Wabbit was interesting because in order for the sailboat to float swamped, the cooler had to be in place and strapped down to support the hull.

<edit: spell Wylie correctly>