1
$\begingroup$

I am a scuba diver and, hence, need to take into account how buoyant the materials are I take with me into the water. Given that I have a LOT of diving equipment, it is practically impossible to test the buoyancy of all of my materials while diving.

Therefore, I am interested in figuring out the buoyancy (in grams) of each of these items (different lights, several sets of undergarments, et cetera) separately, in the comfort of my home.

What I am looking for is some kind of experimental setup that allows me to measure the buoyancy of each item (accurate to, say, 100 grams or so) at home, using items I have available at home (so without the luxury of access to a science lab).

I have contemplated filling a bucket with water and then inserting each item (possible made more heavy by adding weights on top, is the item is fully positively buoyant) and measuring the weight of the water that flows of the bucket, but I don't know how to do this practically (for example, how do I actually weigh that water correctly, and how do I remove the bucket that contains the item without spilling more water, et cetera). So, I am hoping there may be a more practical way of measuring buoyancy of each item, only relying on material that I have in my home.

Does anyone have some ideas how to make this practically feasible?

Steve

$\endgroup$
  • $\begingroup$ What do you think? $\endgroup$ – QuIcKmAtHs Jan 26 '18 at 10:12
  • $\begingroup$ open water or fresh water diving? because the buoyancy will of course vary $\endgroup$ – Alex Robinson Jan 26 '18 at 10:39
  • $\begingroup$ Fresh water at room temperature would be fine. I can then always figure out buoyancy in various levels of salinity afterwards. $\endgroup$ – Steve G. Jones Jan 26 '18 at 13:46
1
$\begingroup$

You should find the volume of your objects and then do the necessary arithmetic which takes account of the water into which you are diving to fine the buoyancy.

If the objects sinks that is relatively easy because all you need to do is fill a bath with water and weight the object (luggage scales) when in air and when immersed in water.
The difference in the two readings in $kg$ can then be converted into a volume by noting that a difference of $1000 \, \rm kg$ represents a volume of $1 \, m^3$.
You could use a more accurate value for the density of water by measuring the temperature of the water in the bath and looking the density up in the appropriate text.

If the object floats then you can add weight(s) to just sink the sink the object but if this is difficult then add a weight which sinks the object and then use the luggage scales to find find the volume of the object and the added weight and then repeat the process with the added weight alone to find its volume.

The upthrust (buoyancy) is equal to the weight of water displaced and I do not think you need to make any correction for using mass rather than weight in your calculations.
The salinity and the temperature of the water at your diving location should possibly be taken into account?

Update as a result of a comment.

For an object which sinks:
scale reading with object in air (kg) - scale reading with object in water (kg) = mass of displaced water (kg)

$\dfrac{\text{mass of displaced water (kg)}}{1000} = \text{volume of object }(\rm m^3)$

assuming density of water is $1000 \, \rm kg\, m^{-3}$

For an object which floats.

Find the volume of an object which floats:
Use the above procedure to find the volume of the weight used as the "sinker" alone and then the volume of the object and the "sinker" but make sure that the object and the sinker are totally immersed in the water.

There are a lot of experiments to find densities using this method eg here and you are using a known density to find a volume.

$\endgroup$
  • $\begingroup$ This sounds great. Let me see if I understand. If it sinks: weigh the object in air and weigh the bucket of water with the object submerged. Would I need to take into account the amount of water it has displaced? If it floats: do the same, but add weight until it sinks (and correct for the amount of weight added). Of course, the water in which it is submerged also has a weight. How do I figure that out? Remove the object and weigh the bucket without it? This will be hard with clothing, as that will absorb a lot of the water. The difference between the weights = buoyancy? $\endgroup$ – Steve G. Jones Jan 26 '18 at 14:02
  • $\begingroup$ @SteveG.Jones I have added a bit to hopefully clarify my answer. $\endgroup$ – Farcher Jan 26 '18 at 15:59
  • $\begingroup$ Thanks @Farcher, this is really clear and useful. Just one final question. Some of the items I want to check are pieces of clothing (undergarment). I wear these underneath my drysuit, so they remain dry while diving. However, if I submerge them in a bucket, they will absorb water, and I think that will change their buoyancy. I considered putting the clothing in a plastic bag, but that would add air). Is there a way I could keep them dry while weighing, or to correct for the water absorbed? $\endgroup$ – Steve G. Jones Jan 27 '18 at 13:02
  • $\begingroup$ If those items of clothing remain dry then the volume which is going to determine the upthrust is the outer volume of the suit with you insider it which you could estimate with an assistant marking the level of the bath with you totally immersed in the bath and and then the level of the bath with you out of it. My answer was more geared to those items, eg air cylinders, which excluded the suit with you inside it. There are calculators foe the weights that you may need. After getting some theoretical estimates I think that an empirical approach is possibly the one to use? $\endgroup$ – Farcher Jan 27 '18 at 14:06
  • $\begingroup$ Probably an uninformed question, but how does dry undergarment affect buoyancy? If you wear it underneath a drysuit, It seems to me that it only affects buoyancy through its volume (as the drysuit will have to be bigger when the undergarments are bugger) and its weight-in-air. Hence, knowing the buoyancy of the undergarments in a bucket of water would not be informative. Or am I mistaken? $\endgroup$ – Peter Verbeet Jan 28 '18 at 18:06
0
$\begingroup$

weigh that water correctly

Well, since Archimedes principle states that buoyancy is equal to the weight of the water displaced, to measure the weight of the water displaced, we can use the classic experiment of placing the bucket filled full to the brim in a larger, empty basin. Then we insert the object into the bucket, and the displaced water will flow into the basin. All we need to do then is to remove the bucket, measure the mass of the basin(with water), and subtract the mass of the basin alone.

how do I remove the bucket that contains the item without spilling more water, et cetera)

This is impossible to do so what you can try is to be as careful as possible. After all, human error cannot be eliminated, only minimised.

$\endgroup$
  • $\begingroup$ This would work for objects that float (or some weight would need to added to make it sink), but not if the object is negatively buoyant and ends up sitting on the bottom of the bucket. The amount of water it has displaced will then only partially capture the buoyancy of the object, right? $\endgroup$ – Steve G. Jones Jan 26 '18 at 13:52
  • $\begingroup$ @SteveG.Jones The buoyant force on the object is the weight of the displaced water. If the object is denser than water then this force is not enough to overcome gravity (so it won't float), but its apparent weight while submerged will be equal to its actual weight minus this buoyant force. It's possible that you're using a technical definition of "buoyancy" which is specific to divers, though. Is that what you're looking for? $\endgroup$ – J. Murray Jan 26 '18 at 16:07
0
$\begingroup$

Get a hanging spring scale--they're cheap. Then just set the scale up over a bucket and weigh each item when it is submerged in water. If it floats, add weight until it sinks, and then subtract the weight (in water) of those added weights from the weight shown on the scale.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.