Buoyancy Calculation

Maybe this will help.

http://www.ehow.com/how_5149859_calculate-buoyant-force.html

36" long x 2" dia. cylinder.

V =(Pi)(h)(r)(r)

V= 3.142 x 36" x 1"x1" = 113

FB = (V)(v) x (SPH2O)

FB = 113" x .95 ( amount of cylinder submerged) x 62.4 = 6698 lbs. ???

Where am I messing up?

I know that piece of pvc can not hold up 6698 lbs.

Your volume computation is in cubic inches; divide by 1728 to get cubic feet. Then it looks like about 3.8 pounds (minus the PVC weight, but plus the weight of the volume of water displaced by the steel.)

Tornado, very good observation. Also note that the 2" sched 40 is 2" ID. V needs to also include the thickness of the pipe and end caps, plus the internal dome volume if the cap is curved rather than flat.

62.4 is the weight of water in lbs. per cubic foot

You've figured out the volume of the pipe in cubic inches. You need to divide by the number of cubic inches in a foot, which I can't recall atm--something like 1728 (12 x 12 x 12).

The mistake you make is one of the reasons I was taught to carry the engineering units along with the numbers in a calculation.

I have a feeling it's going to be 1 lb or less. PVC is pretty heavy and not buoyant without the end caps.

Apart from what Tornado wrote, 2" nominal pipe is 2.375" OD, so you need to base volume of water displaced on that.

You should be able to find the weight of pipe and fittings in the catalogue, but if not, SG of uPVC is about 1.4 - pretty dense as plastics go. If you're desperate for buoyancy you could use PE or PP instead, SG 0.9.

Cheers..........Codey

You didn't convert cubic inches to cubic feet. 1 cubic foot = 1728 cubic inches.

Me thinks you forgot the part about lb/Cf Ft and lb/cu in. The formula where you came up with 6600+ lbs shows you using 62.4 which lb/Cu Ft. you need to divide the value by 12 cubed. Bringing the value supported down to around 3 LBs. I didn't do the calculation, just wagged it. You can do the math.

Thank you.

AS I always said, I am NOT an engineer and I have limited math skills in such things. Its why I come to this forum.

If we were using foam cell , EVA or PE injected floats, I'd know what size and how many.....but this contraption calls for linear floatation with capped pvc.

Thank you again.

Yes, the simple formula is the mass of the materials must be less than or equal to the mass of displaced water. Happy homework ;)

yes there is, its just the ratio of the densities, water (either fresh or sea) and your floating stuff. if ratio ρ(pvc)/ρ(water)<1 it will float, otherwise if ρ(pvc)/ρ(water)>1 it will sink.

Simplest formula is ρ = mass/volume

I worried about your plans, just don't steal that rolled steel, ok? I am not interested to be a conspirator to any of such ideas, if it might be the case.

The drawing is just simplified. Actually, the piece is shaped like a rectangle frame made with steel.

There will be wood and chain added. I wanted to have a safety factor in place hence the idea that if the flotation tube buoyed the steel frame work, it would indeed support the finished piece.

Thank you all again.

One step could help here:

Convert the units to S.I. and see what the specific gravity of the pipe system appears to be as compared to water as sp gr of 1.0

English (american) units are hard enough to carry through the calculations.

PI*(OD^2-ID^2)/4 * LENGTH * S.G. (1.4)

(((6.0325^2)-(5.20446^2))/4)*pi * 60.325 *1.4 = 96.67 grams of pvc

Volume of outside of pipe = ((6.0325^2)/4)*pi = 288.82 cc * 1.0 s.g. of water = 288.82 grams displacement

288.82 grams - 96.67 grams of pvc = 192.15 of steel for neutral bouyancy or about .423# of steel

Agreed - far easier in metric.

And about the only 'good use' of cm - as cm gives 'cc', which is gram of water - 1000 of is liter. So x SG = kg of 'sink or float'.

Though I'm not sure what "or about .423# of steel" means.

"#" is shorthand for pound, I could have used lb instead.

Allen

Well I learn something every day. Thanks Allen

GA! That is ingeniously simple. I love simple.

not only elegant, but accurate in a pro rata form. (scales well)

Kudos!

JE,

I don't understand, please explain this to me. It seems that your procedure will provide the weight of water that the tube can hold. I don't understand why this value will necesarrily be equal to the weight of steel the assembly is capable of floating?

Imagine you had a balloon with the same internal volume of your 2" PVC tube.

Lets say it took about 3 pounds of weight to submerge the balloon.

that weight would directly correlate to the weight of water that would fill the balloon.

If the weight of the hardware is about 1 pound, you will have about 2 pounds of lift.

Except for the wall thickness and SG of that 'balloon'.

I.e the outside creates the displaced volume.

Given PVC is about 1.4 SG, the wall thickness is 'gaining you water fill', compared to a wall of 1 SG.

Filling with water is a use full approximation, rather than 'accurate'.

But it's probably good enough, for this - as presumably this is to 'float' (not be neutrally buoyant) - so the 'steel payload' will need to be 'significantly less' than the 'lift' anyway.

I'm with you on this. I don't think it works, GA or no GA . It assumes upthrust = weight of water inside the pipe, but it should be the total weight of water displaced, including the pipe/fittings. Then of course subtract the weight of the pipe/fittings (in air) to get the nett upthrust. It would work for a pipe of negligible wall thickness.

But apart from that, can only do this after the float has been made. Maybe more useful to have a calculated figure, then if the buoyancy isn't enough for the OP he can redesign.

Cheers.......Codey

it works fine..

"about"

obv there is some wiggle room..

simple question. simple answer

I thought about the hardware displacement.. and decided it was rather inconsequential ... for me

___________________________________________

...how about this?.. stick the whole works in the water.

push it down into the water using an inverted scale..

read the scale... How's that?

It does work fine. Just to repeat, weigh all of the hardware minus the metal weight at the bottom, which we don't know the value of. Fill PVC tube with water, weigh again. Subtract dry weight from wet weight. This is the answer. To be on the safe side, subtract a couple of ounces from that answer. Bingo.

Sorry JE, I'm not going to give you more GA's for explaining this.

It works fine if you don't mind the answer being a fair way off. Assuming wall thickness 1/4" (probably about right for Sch. 40) the pipe wall volume adds over 60% to the water volume, so the buoyancy figure is well down.

If there's a tank of water available, may as well tie the piece of steel to the pipe, and can see whether it floats or not.

Cheers.........Codey

Before I go on, I've got some PVC, caps, and a scale. I'll give it a try and see what happens. Taking the wife out to lunch right now. I'll do it. Might be tomorrow.

2 inch PVC is almost exactly 1/8" wall thickness.

The area of that 2" OD PVC is R squared (1) times pi (3.14) or 3.14

The ID is 1.75, and the area is .875 squared (.7656) times pi or 2.40 or about 24% smaller. The specific gravity of PVC is 1.4, meaning an equal volume of pvc has 40% more mass than an equivalent volume of water. 1.40% of 24% is 33.6%. But incorrect, because while you do have 24% less air to provide buoyancy (increased displacement without increased mass), the mass of the pipe did not change. The wall thickness mass was accounted for in both measurements, was it not? Weighing the water inside the tube was a reflection of the pipe wall thickness impact on reduced volume. I believe the correction is to multiply 24% by the increased density (40%) Your answer will over estimate the buoyancy by about 9.5%., due to the increased density of PVC over water.

Another view Your measurement of water mass in the tube was reduced by wall thickness. When you empty the tube to weigh the water, you are underestimating both the displacement, (because the PVC wall thickness did not pour out, which is the ID, not OD) and the weight of that displaced water (water now in a bucket plus that which would have been occupying space taken by PVC)

By the way, I'm guessing you can google this a thousand places, but you phrased it unusually.

I bet it is less.

Schedule 40 2-inch pipe is 2.375" o.d., 2.067" i.d., and 0.154" wall. It weighs 0.68 lb/ft. The end cap dimensions depend on the style, mfr, and schedule.

The water displaced depends on the outside dimensions.

OK guys, apologies . I was ignoring the net weight of the pipe material when submerged. If the plastic is same SG as water, JE in C's method works OK. If the SG is different, additional buoyancy = plastic volume*(1 - SG) (so reduction in buoyancy if SG > 1).

Just using water weight would still be a bit out if it were steel pipe, but OP said PVC, and buoyancy overestimated, but not by much.

As Tornado said, thickness of 2" Sch. 40 pipe is 0.154". Until checking on the web I didn't realise (or had forgotten) that Schedule determines thickness for any material. Till then I thought maybe it was related to pressure rating, so thicker for plastic than steel.

Cheers.........Codey

I don't know why that's off topic. Seems like an inverted scale would work also..............just enough pressure to push the pvc below, or almost below, the surface of the water.

Oh well. I've done enough talking now, that if you're wrong, I'm going down in flames with you.

As said, I think filling it is near enough - but obviously, if you have one to fill, sinking it would be nearer - as later suggested.

The only 'oddity' is; I use calculation to put me in the 'ballpark' - before I go spend money and time in empirical adventures.

So my inclination is to read the title as Buoyancy Calculation, no so much as 'prototype then measure'.

OP,

Are you looking forthe weight of the steel when submerged, or do you want to know it's mass?

I came up with another very non technical method.

Build your contraption, fill a one gallon jug with dry sand, cap it and tie it to the bottom, take it to the water and float it, pour off sand until it works the way you want.......weigh it.

You may have to use a larger container, but I doubt it. This will give you the exact weight that you need, and you will know it will work in the real world application.

That's off topic? Oh well, that's the method I would use. It may be off topic, but it would work. I'm assuming it wasn't the OP that marked me off. He was the one I was talking to.

Oops! There would be air trapped in the jug along with the sand, which would throw off that number. Use a burlap sack and allow the sand to get wet. Adjust as necessary. Sorry.

wait another minute! The weight of the burlap would change upon removal from the water.. ...too much volumetric aqueous capillary retention..

While I enjoy a good sarcastic comment, telling someone you are being sarcastic by indicating it with a symbol is not being sarcastic. It is being bombastic, or passive aggressive, although that term is usually meaningless. Anyway, don't denote it at all, and take your lumps, or take a bow for an excellent turn of phrase! You don't get both.