Depth of Immersion of a Cylinder

Hello All,

I am looking for the equation(s) to calculate the depth of immersion of a cylinder. I do not have an engineering back ground and I have searched the internet for the principles/equations to calculate it but to no avail.

From what I have found I believe to calculate the immersion depth of a cube it is the volume of water displaced by the cube divided by the area of the cube; but am not sure if that is even close or where you go when trying to calculate it for a cylinder?

Any help would be greatly appreciated.

Thanks,

Nick

 

You've come to the right place with your question, we have tons of answers. It would help though to better explain what you need. It's unclear, but I bet you want to know how much weight a pontoon will support. ?

 

Find the area of the circle times the length of the cylinder. A cylinder with a radius of 1 ft and a length of 1 ft has a volume of Pi x r(squared) or 3.14 x 1 x1 or 3.14 cubic ft which will displace 195.94 lbs of fresh water or 201 lbs of salt water when completely submerged. The depth of immersion will be dictated by the weight of what the cylinder wall is composed of assuming it is a closed cylinder. An open cylinder will probably sink.

 

Nick,

This is a very strange request indeed.

You can not calculate the depth of immersion you have to measure it with a tape measure.

However deep it is, is in fact the true depth of immersion.

You may measure to the top, bottom or centre of the cylinder.

Is this really what you want to know?

Please explain.

 

For some further clarity I’m not looking for the complete displacement of a cylinder or what it can support at 50% capacity; I can calculate that; I’m interested in calculating how deep a cylinder will sit/float in fresh water.

For example, if I had a cylinder that was 24” x 25’ with a wall thickness of 0.1875” and weighed 47.86#/ft and I rolled it into the lake how many vertical inches of the cylinder would be under water? And what equation would I use to calculate it.

I ask because we recently built a large steel dock for our cottage and are looking to float it into place this summer. As as result, a very passionate debate with beers/money on the line has started as to where the water line will be; I said with math it could be calculated and now have come within ½ inch to win the bet.

So any help on the calculation and I’ll share the beers

Nick

 

From the very very basic comprehension of fluid mechanics I’ve researched on the matter I think the metacentric heights or center of buoyancy calculations could be used to calculate the emersion height but do not have the foggiest on how to apply them to a cylinder.

 

This is very serious stuff now that beer is involved. I think you have to find out how to find the area of a segment of a circle. I have to go to the post office and will look when I get back.

 

Hahaa.. I suppose your cylinder is lying on it side so you need to integral function of the cylinder volume. Some serious highschool math
Do you bring my part of the beer to here?
http://en.wikipedia.org/wiki/Integral

 

Here you have formulas for the case of a circular cylinder of radius "R" and length "L".
Valid only when the draft "T" is less than the radius R of the cylinder.
If you know the weight, the first formula deduce the "Area". With several attempts by trial and error, you can deduced the draft.
When you know the depth, calculate the area and therefore weight, is simpler.
NOTE : 1,025 is the density of salt water. For fresh water use 1.

 

I bet the blue one would get him in beer/money range.

 

This here repeats TANSL's explanation.

http://www.mathsisfun.com/geometry/circle-sector-segment.html



I think you calculate the area of the circle and use that to get the area of the sector, (the whole piece of pie shape) then subtract the area of the blue triangle from the area of the sector and and what is left is the area of the segment. From there you figure out volume and displacements.

If you do it right, you get beer and money. Those two things plus solving the problem will probably impress women and you will be a babe magnet. I think it's on to fame and fortune for you.

 

Lucky I survived high school; but I am failing to connect how to apply cylinder volume integrals to immersion levels; if the Riemann integral is ideal for calculating it I'd greatly appreciate additional guidance as I’m clearly missing something.

Thanks,

Nick

 

Thanks guys for the quick and helpful replies! I'll plug the numbers in and see what I get... and no doubt impress some women along the way

 

Don't forget things like if you have two cylinders you have to divide the weight and displacement in two, if three cylinders in thirds, etc.

 

Again, I'm not sure what you are talking about.

The weight of fresh water is about 62.4 lbs a cubic foot (Salt water is about 64)
If you had a box 12" x 12" x 12" or 1 cubic foot, it would take 62.4 pounds to sink the box level with the water. That's called displacement. If the box itself weighed 10 lbs, you could put 52.4 lbs of sand in it and it would sink level with the water.

To sink the box half way, or 6", you need 31.2 lbs. Dividing the sand in half and putting that in the box wouldn't work. 26.2 lbs of sand and 10 lbs of box equal 36.2 lbs, so the box would sink more than half way. You need 31.2 lbs, so you subtract the weight of the box from that and get 21.2 lbs. You put 21.2 lbs of sand in a 10 lb box and it will weigh 31.2 lbs and will sink half way because it displaces 1/2 a cubic foot of water.

So in your case you have to figure out what the dock (the "box" above) weighs. That's the weight of everything, the cylinders, the wood, the nuts and bolts, etc. Then divide it by how many floats (cylinders) you have.

Next, calculate the cubic foot volume of "x" amount of segment of one of your floats and multiply that by 62.4 lbs, which will tell you the weight of water you would be displacing at that depth of cylinder immersion. You'll have to keep trying different size segments until the answer matches the weight of the dock.