Buoyancy salt vs fresh calculation

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KTDB

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My camera set up is negatively buoyant. I have ultralight arms but jumbo floats attached aren’t enough, so I plan to get the floaty arms.

I need to weigh my set up in water to find its water weight, however I’ll be weighing in fresh when I normally dive in salt.

Is there a calculation to find out what the weight would be in salt?
 
Whatever weighting you choose in freshwater, increase the total weight by 2.5% to account for the 2.5% increased density of seawater.
 
1. Weigh your camera rig in air in KG
2. Weigh your camera rig in fresh water in KG
3. Subtract the two figures from each other (this will give you the relative volume of your rig)
The figure in step 3 also equals the amount of water displaced by the volume of your camera rig.
4. You will need to add a buoyant force equal to the difference between the volume and the in-water weight for your rig to be neutral in fresh water.
5. To make your rig neutral in seawater multiply the amount calculated in step 4 by 1.0256 and then subtract the amount calculated in step 4 from this figure...this is how much more weight you will need for your rig (that is neutral in freshwater) to be neutral in seawater.
6. To convert the final figure to pounds, divide the final figure in KG by 2.2


Example:
RIg weight in air = 10kg
Rig weight in water = 8kg
volume = 2 liter
Note 1 liter of fresh water = @ 1Kg
Amount of buoyant force needed to be added to rig for it to be neutral in fresh water = 6kg or 13.2lbs of buoyancy
6 Kg * 1.0256 = 6.15
6.15 - 6 = 0.15kg or .33lbs
You would need to add .15kg or .33lbs more weight to your rig for it to be neutral in seawater.



1. Find volume of object: weight of object in air - weight of object in water

2. Find buoyant force acting on object: volume of object x specific gravity of water
***fresh water has a specific gravity of 1
***sea water (on average) has a specific gravity of 1.0256

3. Find how much object will weigh in water: weight of object in air - buoyant force

example:
weight of camera rig in air: 10kg
weight of camera rig in fresh water: 8kg
volume of camera rig: 2L
(fresh water: 1L = 1Kg)
buoyant force for fresh water: 2kg x 1 =2kg
buoyant force for sea water: 2kg x 1.0256 = 2.05kg
Weight of camera in sea water: 10kg -2.05kg = 7.95kg
7.95kg x 2.2 = 17.5lbs

-Z

edited to more directly answer the OP's question
 
My way requires only an empirical determination of step 4 in freshwater, and skips 1, 2, and 3, since density of rig remains constant and only requires dry weighing of rig once afterwards. How does one easily weigh something in the water?
 
How does one easily weigh something in the water?

One can suspend the object in the water with a cord attached to a luggage or fish scale held above the water.

-Z
 
Thanks everyone, I’ll give this a go ‍♀️
 
I believe I edited the formulas/directions in my initial response to finally be correct.

This is all based on Archimedes Principle:
The buoyant force acting on an object is equal to the amount of water the object displaces.
The amount of water an object displaces is equal to the volume of the object.
Volume the object can be calculated by subtracting the mass of the object in water from the mass of the object in air.


To make an object float neutrally you would need to add a buoyant force equal to the the mass of the object in water - the buoyant force already acting on the object.

Because of the greater density of seawater, one multiplies the freshwater calculation results by a factor of 1.0256 (or add @ 2.5%) to convert from freshwater to seawater.

Disregard this post,I believe it is flawed.....I have edited my original response above to more directly address the OP's question.

-Z
 
I don't follow your example. You've got a 10 kg rig that weighs 8 kg in water. The buoyant force is 2kg so the volume is 2 l (water being 1 kg/litre). which is all good. Archimedes principle is that the buoyant force is equal to the weight of water displaced.

where does the 6kg come from? If you have displaced 2kg of water You want to multiply your original 2kg of water displaced by 1.0256. which is 2.05 kg of salt water so the weight of the rig in salt water is calculated to be 10-2.05 = 7.95kg.

In practice you probably won't get to exactly neutral and may find the rig is easier to handle if slightly negative unless you use something like the Kraken adjustable float arms, which allow you to fine tune when you dive. You definitely don't want it positive. In this case just using the fresh water weights for your calculation is probably close enough as the difference will be something like 50 gr or so and getting the floats to exactly match may prove difficult. It will depend on your rig of course the bigger the volume the more the weight changes between fresh and salt. The volume displaced is what gives you the buoyancy and you need to convert that volume to weight to get the buoyancy. So a bigger rig displaces more water and bigger difference between fresh and salt.

Then look up the buoyancy of of your arms. INON for example publishes UW weights of all their accessories as well as float arms: INON Underwater Weight and their float arms are here: INON Arm System [Arm]

I ended up with the rig weighting 100-200gr UW depending on ports used and have one setup I use with the 170mm dome port and another to use with the fisheye dome and macro port.
 
I don't follow your example. You've got a 10 kg rig that weighs 8 kg in water. The buoyant force is 2kg so the volume is 2 l (water being 1 kg/litre). which is all good. Archimedes principle is that the buoyant force is equal to the weight of water displaced.

where does the 6kg come from? If you have displaced 2kg of water You want to multiply your original 2kg of water displaced by 1.0256. which is 2.05 kg of salt water so the weight of the rig in salt water is calculated to be 10-2.05 = 7.95kg..

My thinking was that since there is 2kg of buoyant force acting on the 8kg camera rig, one would need to add an additional 6kg of buoyant force to get the rig neutral.....

.....After looking over the maths I believe that this is flawed. I have edited my original response to more directly answer the OP's question.

-Z
 
Also depends where you are diving, you can pretty much double your estimations if you are in the Red Sea.
 

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