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Assume the water bottle is the most sturdily built standard water bottle one could acquire in an american supermarket.

This is the first question I've asked, so I'm sorry if I'm not in the right place or asking it the right way. If I may I'd like to throw in these follow up questions:

Is there a good way to approximate how good an object would be at serving as a long term floatation device, like a pontoon or bouy?

Is there a way to calculate what the optimum size or shape of a 'water bottle like structure/material' (pocket of air encased in plastic) would be for displacing weight lifted out of the water by the buoyancy?

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    $\begingroup$ You're asking three different questions here. Which do you want us to answer? $\endgroup$ – Wasabi May 23 '18 at 21:40
  • $\begingroup$ is it deformable? i.e. will it compress then re-gain its original shape once the pressure is reduced? $\endgroup$ – Solar Mike May 23 '18 at 21:40
  • $\begingroup$ @SolarMike Yes should be fairly deformable. Thanks for asking! $\endgroup$ – Jake May 23 '18 at 22:36
  • $\begingroup$ @Wasabi ideally all three, sorry. but dealers choice for sure, Thanks. Any information I can get I will be very happy to receive. $\endgroup$ – Jake May 23 '18 at 22:38
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How deep would a water bottle have to be submerged in the sea to burst?

It won't burst. It will be crushed.

Pressure increases by 1 bar approximately every 10 m in salt water.

Is there a good way to approximate how good an object would be at serving as a long term floatation device, like a pontoon or bouy?

Yes. 1 L of displaced water gives 1 kg (10 N) lift.

Is there a way to calculate what the optimum size or shape of a 'water bottle like structure/material' (pocket of air encased in plastic) would be for displacing weight lifted out of the water by the buoyancy?

Yes. Use finite element analysis on a model of the shape.

The strongest structure for a given weight of material and will probably be a sphere or egg-shape. Cylindrical shapes with vertical axis may be easier to manufacture.

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  • $\begingroup$ That 10 N of lift pr liter is constant regardless of depth? I had thought maybe the further it was pushed down the more lift it would create in the opposite direction? Or do you mean it is displacing more liters the further it descends? Well, bottles are initially manufactured by a process similar to blowing a balloon inside a mold. where the balloon is hot/wet plastic, so a sphere/egg seems really plausible for mass construction to me. $\endgroup$ – Jake May 23 '18 at 22:24
  • $\begingroup$ The 10 N of lift per litre is constant regardless of depth (if the volume remains constant and the bottle doesn't get crushed) because the mass of the water displaced is constant and does not vary with pressure. The lift is described by Archimedes Principle. $\endgroup$ – Transistor May 23 '18 at 22:28
  • $\begingroup$ This page: education.seattlepi.com/… Says a 2 liter bottle can withstand about 150lb per square inch. I'm betting thats internal rather than external pressure, I'm not sure if that makes a difference. But if a bar is added for every 10 meters of depth, then google's conversion calculator makes me think it would max out near 100 meters. $\endgroup$ – Jake May 23 '18 at 22:31
  • $\begingroup$ Yes, I've made bottle rockets with 2 L PET bottles. The bottles are designed to be pressurised with CO$_2$ and the shape is very strong in tension. "I'm not sure if that makes a difference." It certainly does. Go metric, man. Nearly everything is simpler. $\endgroup$ – Transistor May 23 '18 at 22:42
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As Transistor already wrote, according to Archimedes' principle, the lift is identical to the weight of the superseded material, reduced by the weight of the superseding material. Since water is nearly incompressible, you can use 1 kg per liter (or 1000 kg/m³) lift for most cases.

The main failure mode of a submerged standard supermarket plastic bottle is plate buckling. The corresponding German wikipedia article gives a pressure limit of $$ p_{\mathrm{buckling}} = 0.65 \cdot E \cdot \frac{r}{l} \cdot \left(\frac{t}{r}\right)^{\frac 52} $$ as maximum external pressure $p_{\mathrm{buckling}}$ for buckling of a cylinder, with $E$ being Young's modulus, $r$ the mean radius and $l$ the length of the cylinder and $t$ the wall thickness. Please note that in this equation there are no safety factors included, e.g. against a sudden local change of external pressure due to waves or contact with other solids (ground, ships, whatever) and so on.

For more complex shapes, as Transistor already wrote, finite elements are a suitable way to estimate the buckling resistance.

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  • $\begingroup$ The wall thickness of plastic bottles can be so thin that they buckle merely from a person's grip during drinking. That doesn't mean that they're damaged. $\endgroup$ – Chemomechanics May 25 '18 at 0:22
  • $\begingroup$ @Chemomechanics, I didn't define failure - sorry. Failure for me means that the volume of the plastic bottle decreases significantly, so that the buoyancy also decreases significantly. $\endgroup$ – Robin May 26 '18 at 4:52

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