bouyancy

1. Calculate enclosed volume (= pi x r² x l, where r is radius of pipe, l is length).
2. Calculate weight of that volume of seawater (= volume x density).

To a first approximation, the result is the buoyancy.

If it's on the surface (I assume you mean floating), the buoyancy equals the weight.

What I am trying to figure is how much steel weight will these sizes of a closed system , support at the surface without sinking.

3x36 would float "?" of steel weight.

Thank you.

Take the enclosed volume. Multiply that by the weight density of the water you're gonna put this in. Subtract the PVC weight. That's your answer.

D'oh. Take the TOTAL volume, not just enclosed.

You're right, but I don't think a little bit of PVC's going to make much difference either way (in this context).

Yep, it's close enough to water that it's not a big deal. I just thought I should pick on my answer as much as I pick on others. 'Course, I'm closer to the dumb, so I see it easier.

Hello TVP45 - it's not the weight compared with water that matters, it's the weight compared with air. SG of PVC is about 1.1, and with a few assumptions I make the weight of each float about 4.5 lb. That needs to be subtracted to get the net (pun intended!) buoyancy.

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

Sounds kinda heavy - what wall thickness did you take?

8 mm (5/16")

Ah! I was thinking more of 2.5 - 3mm. Wonder if netmaker's still around? I'd better pm him to try to get the facts straight.

Eureka! I think it's water, but I'm open to convincing. Why do you think air?

As you said in #4 and 9, it's the total volume that determines water displaced, hence the upthrust. But if you just use that, it's like assuming the pipe is of zero wall thickness (or zero wall material density). Need to subtract weight of the pipe (in air) to get net upthrust.

If the pipe were solid, total volume = material volume. If you just subtract the apparent weight in water from the gross you get net upthrust 90% of gross (for SG 1.1) but actual net upthrust is negative (or in simple terms, it sinks!)

There's probably a neater explanation, but that's all I can think of at the minute.

Cheers......Codey

Build a pipe capped only at one end. Go to the ocean. Begin to add weight until the tube sinks vertically below the surface to the desired level.

That's your number.

I have a 35 lb. steel weight that has to be bouyed just under the surface in sea water.

I can not use more than 36" in length.

How many 3" diameter x 36" lengths do I need to keep the weight supported?

Nothing exact , just a close approx. of how many sections.

OK, for one length,

V = pi x r² x l

= 3.14 x 1.5 x 1.5 x 36 cubic inches

≈ 255 in³

Density of seawater ≈ 64 lb/ft^{3 [1]} ≈ 0.037 lb/in³

So each pipe can support about 255 x 0.037 = just under 9½ lb.

So a 35 lb steel weight would sink with 3 lengths, and float with 4.

[1] I looked up this figure, but didn't check it.

That's what I needed. 4 lengths is the number.

Thank you.

nm