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Boys this is a scuba diving forum, the advanced section too
weighting- freshwater vs. seawater
- Thread starter sandanbob
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The difference in weight needed to go from fresh water to salt water can be calculated using the buoyancy equation:
B = VD - W
Where B = buoyancy force in lb (neutral buoyancy = 0 lb), V = volume in ft3, D = density of water in lb/ft3,
W = weight in lb.
Rearranging the equation to solve for weight gives:
W = VD - B
The average volume of a human is 2.26 ft3, the estimated volume of a 80 ft3 tank plus equipment is 0.459 ft3, the density of sea water is 64 lb/ft3, the density of fresh water is 62.4 lb/ft3. Assuming the target buoyancy is -5 lbs (slightly heavy) lets calculate the weight needed in salt water (Ws) then the weight needed in fresh water (Wf).
Ws = (2.26 + 0.459) x 64 + 5 = 179.0 lb
Wf = (2.26 + 0.459) x 62.4 + 5 = 174,7 lb
The difference in weight is: 179.0 - 174.7 = 4.3 lb. You need to add about 5 lb going from fresh to salt water.
B = VD - W
Where B = buoyancy force in lb (neutral buoyancy = 0 lb), V = volume in ft3, D = density of water in lb/ft3,
W = weight in lb.
Rearranging the equation to solve for weight gives:
W = VD - B
The average volume of a human is 2.26 ft3, the estimated volume of a 80 ft3 tank plus equipment is 0.459 ft3, the density of sea water is 64 lb/ft3, the density of fresh water is 62.4 lb/ft3. Assuming the target buoyancy is -5 lbs (slightly heavy) lets calculate the weight needed in salt water (Ws) then the weight needed in fresh water (Wf).
Ws = (2.26 + 0.459) x 64 + 5 = 179.0 lb
Wf = (2.26 + 0.459) x 62.4 + 5 = 174,7 lb
The difference in weight is: 179.0 - 174.7 = 4.3 lb. You need to add about 5 lb going from fresh to salt water.
The principle involved in buoyancy is called Archimedes Principle and is captured in this equation:
B = VD - W
According to the equation the buoyant force B is the difference between the weight of the water displaced by the object (VD) that causes you to float, and the dry weight (W) of the object that causes you to sink. If B is zero the object is neutrally buoyant and will neither sink nor float. If B is negative (-) the object will sink and if B is positive (+) the object will float.
From the above equation volume plays a huge roll in our buoyancy. Any increase, however small, is multiplied by a factor of 64 for sea water. What is the weight needed if our volume increases by 10%?
New volume = 2.26 + (2.26 x 10%) = 2.49 ft3
Using the same value above for our equipment (0.459) our new calculated weight is:
W = VD - B = (2.49 + 0.459) x 64 + 5 = 193.74 lbs.
This is an increase of 193.74 - 179.02 = 14.72 ~ 15 lbs. We need an additional 15 lb of weight to offset the buoyancy gained from a 10% increase in our volume.
The weight of the diver without the buoyancy contributed by the equipment is:
W = VD - B = 2.26 x 64 - 0 = 145 lbs.
The water in the Dead sea has a density of 77.42 lb/ft3. Applying this to the weight found above the net buoyant force becomes:
B = VD - W = 2.26 x 77.42 - 145 = 30 lbs. This means there is a buoyant force of 30 lbs in excess of your body weight that keeps you afloat. No wonder people can take a newspaper and sit down in the water without getting the newspaper wet!
B = VD - W
According to the equation the buoyant force B is the difference between the weight of the water displaced by the object (VD) that causes you to float, and the dry weight (W) of the object that causes you to sink. If B is zero the object is neutrally buoyant and will neither sink nor float. If B is negative (-) the object will sink and if B is positive (+) the object will float.
From the above equation volume plays a huge roll in our buoyancy. Any increase, however small, is multiplied by a factor of 64 for sea water. What is the weight needed if our volume increases by 10%?
New volume = 2.26 + (2.26 x 10%) = 2.49 ft3
Using the same value above for our equipment (0.459) our new calculated weight is:
W = VD - B = (2.49 + 0.459) x 64 + 5 = 193.74 lbs.
This is an increase of 193.74 - 179.02 = 14.72 ~ 15 lbs. We need an additional 15 lb of weight to offset the buoyancy gained from a 10% increase in our volume.
The weight of the diver without the buoyancy contributed by the equipment is:
W = VD - B = 2.26 x 64 - 0 = 145 lbs.
The water in the Dead sea has a density of 77.42 lb/ft3. Applying this to the weight found above the net buoyant force becomes:
B = VD - W = 2.26 x 77.42 - 145 = 30 lbs. This means there is a buoyant force of 30 lbs in excess of your body weight that keeps you afloat. No wonder people can take a newspaper and sit down in the water without getting the newspaper wet!
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