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bhec10's avatar

How to calculate buoyancy?

Asked by bhec10 (6457points) February 5th, 2013

I’ve been looking at the hundreds of formulas in this Wikipedia page but I have no idea which one is appropriate.

Basically I want to calculate how much air is needed to bring a person back to the surface after it is submerged.
Let’s say a person wearing only a wetsuit dives to 10 or 15m and then inflates an air bladder with a CO2 cartridge.

I’d like to know how much air is needed or what is the force needed to bring that person back to the surface? How do I calculate that?
I’d appreciate any thoughts or calculations that can help me with this.

Anyone willing to help me out?

Thanks in advance!

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14 Answers

jerv's avatar

It will take enough pressure to counter the water pressure at whatever depth you are at. If memory serves, that is about 1 Bar every 10m. However, that assumes an unresisting bladder; if you must stretch the suit material, you must account for it’s elasticity as well.

Volume depends on the diver’s weight.

Luiveton's avatar

The answer probably depends on the exact information given to you.
Can you tell us the exact information given to you?

bhec10's avatar

Thanks for trying to help guys!

There is no information given to me, this is just a project I’m working on. Assuming an 85kg person dives to 15m and wants to inflate a bladder with air, how much air does he need so that the bladder takes him back to the surface?

Yes, the material of the bladder would stretch.

josie's avatar

A cubic foot of water weighs 60 pounds. Bladder would have to inflate enough to displace a volume of water that weighs more than what you want to float. Whatever the source of the inflating gas is, you should be able to calculate the cubic feet of gas compressed into it by checking with the supplier or manufacturer.

Adirondackwannabe's avatar

Well, I can submerge a basketball, but not an inner tube. So the answer must be in between those two.

LostInParadise's avatar

Why is any additional buoyancy needed? A person naturally floats on water. I am not familiar with diving, but it seems to me that effort is needed to stay submerged. If the person is wearing a weight to be able to get to lower depths then you will need to make the density of person plus weight less than the density of water.

jerv's avatar

@josie 62.5 pounds for freshwater, 64 for saltwater.

@LostInParadise Not all of us. You assume everybody has enough fat to float, but some of us are denser than water. True, most divers need weights to achieve neutral buoyancy, but how much, if any, varies.

Luiveton's avatar

@bhec10 I think there are many steps to this project you’re working on. Is this a/n (thought) experiment? Start by drawing a diagram and deciding the assumptions you’re going to make (ie. is body in equilibrium etc) and the potential errors you will overlook.
There are no absolutes/certainties with such little information provided (ie so many potential calculations) , I think you need to obtain more information, and I definitely think more than a few equations are going to be used if you’re willing to make it somewhat accurate.
Anyways. Possible methods to use: (You may need to add in extra information)
By doing so, you’ll be considering forces on the cartridge, and I think mechanically, this is the best way to go on about this. Treat both bodies separately. Assume it is spherical.
Use Archimedes principle [when a body is partially or totally submerged in a fluid it experiences an upthrust equal to the weight of fluid displaced.] Also you need to know that Volume of solid body= volume fluid displaced.
Since viscous drag is always against direction of motion, your equation will depend on whether calculations are derived as diver sinks or swims towards the surface.
This is if diver is moving downwards:
W(solid)= U(fluid)+D (<—viscous drag)
you can keep simplifying (eg. W is a product of mass by gravity of about 9.81)
mg= mg + 6(pi)rnv (r is radius of spherical object in which case it is the cartridge, assume its perfectly spherical unless you want to immerse in more complex means of figuring this out. n is coefficient of viscosity, v is terminal velocity. pi is pi.)
Regarding stokes law, note:

“Stokes’ law makes the following assumptions for the behavior of a particle in a fluid:
Laminar Flow
Spherical particles
Homogeneous (uniform in composition) material
Smooth surfaces
Particles do not interfere with each other.”

if you want to further simplify: d is density V is volume
since m=dV and V for a sphere= 4/3pi*r^3

d*4/3*pi*r^3g(solid)= d*4/3*pi*r^3g (fluid) + 6pirnv
Substitute known values in and see what you need.
You can search things like density of water etc.
If you already have the values to begin with then just go with
W=U+ D

For the human you can resolve forces horizontally and vertically (Fsintheta)—>vert (Fcostheta)—> horiz

You can also try using Boyle’s law (gas laws)
P1V1= P2V2
p is pressure V is volume T is temperature in (K)
1 is for initial data and 2 is final conditions
but either way you’ll definitely need more information.

The project needs to be planned out thoroughly first if you want it to make any sense.

PS to find the amount of air you’ll have to work out the Volume from one of the methods. You can always assume the diver’s body and the cartridge are one big spherical object, so that you combine the diver’s information in the equation (ie the spherical object is going to weigh >85 kg, add some data to your exp) and all is hypothetically solved.

In retrospect I see that there are too many uncertainties related to this project, because the V of air won’t be constant, it’ll be instantaneously changing so you’ll need certain specialized equipment that measures V of air over frequent intervals of time. The wiki page you linked us to even has a section concerning divers:
Underwater divers are a common example of the problem of unstable buoyancy due to compressibility. The diver typically wears an exposure suit which relies on gas filled spaces for insulation, and may also wear a buoyancy compensator, which is a variable volume buoyancy bag which is inflated to increase buoyancy and deflated to decrease buoyancy. The desired condition is usually neutral buoyancy when the diver is swimming in mid-water, and this condition is unstable, so the diver is constantly making fine adjustments by control of lung volume, and has to adjust the contents of the buoyancy compensator if the depth varies.”

bhec10's avatar

@LostInParadise Let’s say your caught under a big wave and are running out of breath, you are going to need help to get back to the surface. I am trying to figure out how much air I need to transfer into an expandable bladder so that that person is brought up to the surface.

@Luiveton Thanks a lot for your time on that reply! I am still a bit confused as to which formula I should use. Let’s get one thing clear. Do I need anything else other than the mass of the diver (let’s say 85kg) and the depth at which he is (let’s say 15m)? Other than volume and density of water, which are given, is there any other variable or parameter that I need to clarify?

Luiveton's avatar

@bhec10 Yep if you’re going to use the W=U+D one you’ll need mass & volume of cartridge, coefficient of viscosity of water (is it sea water?) and radius of cartridge and terminal v of cartridge. Stokes law will only work if we assume the cartridge is perfectly spherical. (Or if you’re going to combine the diver and the cartridge and assume they’re both spherical.)

For the air specifically I think it might be useful to measure pressure, volume, temperature over an interval of time in certain depths. (initial and final) As I previously mentioned the V of air definitely won’t be constant because of the inflation and deflation. Also going to be helpful if for each depth you have the corresponding measurements for water as relative values to compare to, this might give you more ideas for ways to obtain measurements.

Anyway it’s always better to gather as much info as possible so that 1) you clarify the situation to yourself and others (more importantly yourself) 2) you know how exactly you’re going to approach the question

Have fun finding the answer

bhec10's avatar

@Luiveton Yes, it will be needed in sea water. Well, the diver and the cartridge are obviously not perfectly spherical but I really just need a rough guideline of a number by which I should guide myself.

I don’t mean to sound rude or anything but I’m not sure how the temperature over an interval and all that is going to influence this, it’s really just the amount of air that is needed to lift an 85kg object that is more or less round from 15m deep till the surface.

I’ll give it a try with some of the formulas.

dabbler's avatar

It also depends on whether the diver was buoyant in the first place, or how he/she was or was not in the first place at 15m, wearing whatever it is they’re wearing and with however much air is in the lungs.

Sounds like you are recreating an emergency life vest. They are available that are thin and compact but inflate when a cord is pulled.

LuckyGuy's avatar

I have a BMI of 21 and dive with snuba equipment (tanks on the surface enabling long durations when maintained by a second person.
With a 10 pound weight belt I am neutrally buoyant when I’m at mid-breath. If I inhale, I will naturally rise slowly. If I exhale all the way I will sink slowly. If I were to drop the 10 pound weight I estimate I would rise to the surface at a rate of about 1 ft per second. So if I had a gas cartridge that offset 10 pounds of water it would pull me to the surface at about that rate.
Hopefully you will find those numbers helpful. Your mileage may vary.

jerv's avatar

Temperature affects density, which affects buoyancy.

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