Buoyant Fabrications

http://en.wikipedia.org/wiki/Seawater

http://en.wikipedia.org/wiki/Archimedes%27_principle#Archimedes.27_principle

The volume of the equipment itself, including all the non-wetted voids inside it, will determine the uplift force from that volume of displaced seawater. Applying Archimedes' principle, the density of seawater multiplied by the above volume, will determine the uplift force.

The uplift force subtracted from the weight of the equipment in air determines the additional buoyancy material required to achieve neutral buoyancy. Do remember to use the density difference between the buoyancy material and the seawater when calculating the amount of buoyancy material to be added to the equipment!

The rest is arithmetic.

_{[Psst: There is no need to drill holes in it (doing so would mean more buoyancy material compared to not drilling holes in it).]}

My sincere thanks for posting the replies above.

I get how to calculate the items' weight in water etc. it's the volume of trapped air that's really tieing mein knots as I cannot find anything that tells me what lift this provide.

I have no doubt that its going to be soo obvious as to make me look stoopid, but I can't get my head round it.

I could easily drill my holes and be done with it, but the final cost of the buoyancy is not going to be cheap vis a vis the cost of the tool itself.

is air 100% buoyant? does 1 litre of air = 1Kg of up thrust?

as I could reduce my bguoyancy bill by half if that was the case.........

again, thank you.

Woody

and Dvader, Buzz says 'fit like?'

<...is air 100% buoyant? does 1 litre of air = 1Kg of up thrust?...>

More or less, though it does depend upon the temperature and composition of the water and the vessel that the air is contained in. Seawater is denser than fresh water, so the buoyancy of 1 litre of air is slightly greater in seawater than in fresh. However, the air itself weighs something, and its volume varies with immersion depth if the air were contained inside something flexible like a balloon; its volume would vary with depth and therefore its contribution to buoyancy would vary with depth also.

This aluminium framework thingy isn't in that category, though, so the lift force will not vary much with depth.

Be aware that as the temperature rises, water becomes less dense, so the buoyancy of the object that is immersed in it will vary with temperature. If the volume of the object varies more with temperature than the density of the water then the object will rise with increasing temperature; if it varies less then it will sink.

Hi PWSlack,

too much too much........but good points.

being as this is going to be used in a sub-tropical area, then sea water temperature varience could be quite high.

eeek.

time to admit defeat then, as I'm no buoyancy guru as you probably noticed.

thanks again,

Woody

Nonononononononononono! Don't admit defeat!

• George & Robert Stephenson didn't.
• Ismabard Kingdom Brunel didn't.
• The Wright brothers didn't.
• The Apollo 13 team at Nasa didn't.

So Woody 100 mustn't!

Generally speaking, air is 800 times less dense than water i.e. 1,000 liters of air ≈ 1.25kg. This might help you determine the volume of air you need to achieve your required buoyancy.

I just completed designing a hollow aluminum float for a level sensor. The lift is just equal to the external volume of the 'box' x density of water surrounding it.

Hi JohnDG,

your second calcaulation is depicting the correct scenario, and so would I be correct in saying this then;

0.149m³ @ 402Kg +0.31m³ @ 0Kg = 0.459m³ ≈ 402Kg? as the 0.31m³ is air.

the air only adds volume but no mass............I just understood. penny dropped; happy days...........I think.

your saying that it doesn't matter that its air or aluminium, its purely the volume and it's displacement that matters?

Looks like you interpreted perfectly. It was a twofold problem cos I didn't know any better.

Cheers,

Woody

So about -57 kg in fresh water, and, about -68 kg in sea water (depending on the density of the sea water).

Good point. This should not have been off topic. I don't think that the struts will implode but: the aluminium is only 1 mm thick so over a distance of 4 inches it will certainly bend, probably to the point at which the two surfaces touch.

(EDIT: Sorry to use mixed units it's just easier for me to "visualise")

Think you'll find it's ¼" ally plate (to stick with Imperial units ).

You guys really are awesome.

your coming at this from angles I hadn't even thought of never mind considered. I'm watching and reading everything your saying and taking it on board.

Looks like I'm sorted though as I've addressed this problem and added the sheet profiles of buoyancy that I need and it looks really good. Just got to hope;

a. it works, b. my machine shops and fabricators all get it done on time, c. I have enough techs available to build and test it, d. I can ship it 9 time zones to the east in time for trials mid Dec. and e. there are no issues when it gets there.

if I had hair, I'd be pulling it out, as I'm committed to one scary tight time line on this one.

Hope you guys like this sort of thing, as this could be the start of many.

thank you,

Woody

My calcs (aforementioned) suggest you need to add about 60kg to get the thing to sink!

Go through it yourself - check & re-check all the figures. Take the total volume of your rig (making sure it's all enclosed - no air leaks into those voids!), and convert this to the weight of the same volume of seawater in the region where it's going to be deployed. Try searching "density of seawater" & suchlike, & narrow it down to the target region. This calculator is good if you can estimate the local water temperature & salinity.

If the weight of the displaced seawater is greater than the weight of the rig (which I think it is) - the rig will float, and you'll need to add ballast to get it under the surface.

Good luck.

My best suggestion - hedging your bets & all that - is to make up a ballast tank. Take a length of your 100x100 box section, weld a plate on one end, & fix it so that you can close off the other end with a plate fixed with toggle-bolts or whatever & sealed with an NBR gasket.

Calculate the length (& hence volume) such that when it is empty, it will give about 80kg of buoyancy, and when full of sand/gravel (density: dry 1600 kgm^-3, wet ~2080kgm^-3) will give about -80kg (sinkancy!).

I mention dry & wet ballast, because you will have to find out what's available locally - can't go spreading nasty bio-contaminants - and work with that figure. Probably better to use wet, because the easiest thing to do would be to shovel damp sand off the local beach, but if you can only get dry material, there's plenty of water around).

When you come to deploying the rig, strap the ballast tank to it. Try first with the tank empty - maximum buoyancy. It will float!!! Now add a bit of wet ballast, & try again. Iterate.

Thanks, I "off topic" myself as i felt i was not answering the original question.Mixed units are cool, i dance with both of them.

Take care

I think getting your mind adjusted to boyancy calculations is a good thing, however, bear in mind that the density of fresh and seawater changes with temperature. Temperature changes with time and depth. The bottom line is that if you want your device to be neutrally boyant at 150 meters, you are going to need to be prepared to provide boyancy compensation anyway.

Note that the density of seawater is also going to change with pressure, which means depth as well. It is possible to design a system that gives you neutral buoyancy at any depth you choose- and at depth, temperature is not going to vary as much as on the surface...

"What will this weigh in water.... ?"

Working on the mass and volume you've provided and assuming the density ρ of seawater is about 1030 kg/m³.

When fully submerged, the assembly will be subjected to a bouyant (upward thrust?) force F_b, acting through it's centre of mass and equal in magnitude (oppsite in direction) to the weight W, of the volume V, of seawater that the assembly displaces - which of course would be equal to the volume you gave (Archimedes' Principle?).

The weight W₁, of the assembly in air is the product of its mass m, and the acceleration due to gravity g,

i.e. W₁ = mg = 402 kg x 9.807 m/s² ≈ 3942 N

The bouyant force F_b = W = ρVg = 1030 kg/m³ x 0.149 m³ x 9.807 m/s² ≈ 1505 N

The (apparent) weight, W₂ of the assembly when fully submerged is then found knowing that,

W₁ = W₂ + F_b → W₂ = W₁ - F_b ≈ 2437 N.

Is that what you were after?

"The buoyant force F_b = W = ρVg = 1030 kg/m³ x 0.149 m³ x 9.807 m/s² ≈ 1505 N ..."

You can't ignore the enclosed volume!!!!! (which the OP thoughtfully specifies as 0.31m³ ).

Yes indeed!

Thanks JohnDG, I missed that even though it was right underneath the frame volume - I think I'll need to replace my glasses!

so, F_b = W = ρVg = 1030 kg/m³ x (0.149 + 0.31) m³ x 9.807 m/s² ≈ 4636 N?

→ W₂ = W₁ - F_b ≈ -694 N

→ >71 kg of ballast material to keep it under?

Yes indeed!

Thanks JohnDG, I missed that even though it was right underneath the frame volume - I think I'll need to replace my glasses!

so, F_b = W = ρVg = 1030 kg/m³ x (0.149 + 0.31) m³ x 9.807 m/s² ≈ 4636 N?

→ W₂ = W₁ - F_b ≈ -694 N

→ >71 kg of ballast material to keep it under?

no need for ballast on this one guys as the tooling contained with this chassis weighs another 580Kg or there abouts.

If I can figure a way to do so I'll try and post a pic of what we've been 'speaking' about.........and here we have one

Woody

The structure as originally described is buoyant, and would need to be pushed to the sea floor, which would be difficult. The extra 580 kg of equipment (depending on how much volume it occupies) may compensate enough to make the whole assembly sink. DaveB's post 3 states the principles correctly. Add up the total weight (in air) and volume of this assembly. Compare the total weight to that of seawater of the same volume. This will tell whether the entire assembly will float/sink.

As a peripheral matter, the picture shows no diagonal bracing in the structure. I hope it doesn't rack when bumping on the bottom. I looked briefly at the 100mm x 6.3 wall tube, and I think it's okay under the external pressure, but this should be further checked.