2. Index
Density
Buoyancy
Specific gravity
Archimedes’ Principle
Surface Equivalent air volume
Lifting problems

3. Density
 Definition
 Mass per Unit Volume
 Density of air at sea level
 .08 lbs. per cu. ft.
 Hydrostatic Density
 Salt Water
 64 lbs. per cu. ft.
 Fresh Water
 62.4 lbs. per cu. ft.

4. Buoyancy
 Force that allows an object to float.

5. Specific Gravity
Density of a substance vs. density of pure
water.

6. Archimedes’ Principle
 An object partially or wholly immersed in a fluid is
buoyed up by a force equal to the weight of the
fluid displaced by the object.
 Buoyancy of an object =
 Weight of the water displaced by the object - Weight of the object

7. buoyancy for each of these objects?
Where will they end up?
Positive
_______________________________________________
Neutral
________________________________________________Negative_
32 lbs
1 cu ft
64 lbs
1 cu. Ft.
96 lbs
1 cu. ft

8. States of Buoyancy
Positive buoyancy
 Specific Gravity of the object is less than that of the fluid
Neutral
 Specific gravity of the object is equal to the specific gravity of
the fluid
Negative
 Specific gravity of the object is greater than that of the fluid

9. Example 1
 What is the buoyancy of an anchor with a dry
weight of 100 lbs., and a volume of .22 cu. ft.,
when it is dropped in the ocean?

10. Answer to Example 1
Displaced wt.=
.22 cu. ft. x 64 lbs. per cu. ft. 14.08 lbs.
-Dry wt.
100 lbs.
Buoyancy
- 86 lbs

11. Example 2
How many 50 lb. lift bags will it take to lift
an object with a volume of 3.1 cu. ft. and a
dry weight of 289 lbs.?
Each lift bag weighs 2 lbs. and the object is
in fresh water.

12. Answer to Example 2
Displaced weight =
3.1 cu. ft. x 62.4 lbs./ cu. ft. 193.4 lbs.
-Dry weight
289 lbs.
Buoyancy
- 95.6 lbs.
Lift capacity = 50 lbs - 2 lbs = 48 lbs of lift / bag.
Use how many bags?
2 bags.

13. Surface Equivalent Air Volume
How much air must you bring down from the surface
if the object in example 2 is located at a depth of 120
ffw?

14. Surface Equivalent Air Volume cont.
Buoyancy of the object -95.6 lbs
How much lifting force must be generated to lift the
object to the surface?
 95.6 lbs

15. Surface Equivalent Air Volume cont.
How much freshwater must be displaced to generate
the required lifting force?
How is this calculated?
 Force required/density of fresh water
Density of fresh water
 62.4 lbs. per cu. ft.
95.6 lbs/62.4 lbs. per cu. ft. =
 1.53 cu. Ft. of water must be displaced

16. Surface Equivalent Air Volume cont.
How much air must we bring down from the
surface to displace 1.53 ft3 of fresh water at
a depth of 120 ffw.?
Calculate Pata at a depth of 120 ffw.?
 {Depth + 34}/34 = atm
 {120+34}/ 34 = 4.5 atm
Multiply Pata x Vol h20 to be displaced
 1.53 x 4.5 = 6.93 cu. ft. at the surface

17. Lifting problem
You have been enlisted to salvage an outboard motor
lost at sea. You locate the outboard, which displaces
2 ft3
of water and weighs 900 lbs in air, at a depth of
66 ft. How much air will you need to add to a lift bag
to bring the outboard to the surface? How much air
will be in the lift bag once at the surface?

18. Calculate the Buoyancy of the Object
Volume =
2 ft3
Weight of the water displaced =
2 ft3 x 64 lbs/ft3 = 128 lbs
Dry weight =
900 lbs
Buoyancy of the Object
128 lbs – 900 lbs = -772 lbs

19. Calculate the Volume of Water to be
Displaced
How much lifting force is necessary?
772 lbs
How much water must be displaced
772 lbs / 64 lbs/ft3 = 12.06 ft3

20. Calculate How Much Air You Need to
Bring Down from the surface
Calculate Pata
(66 / 33) + 1 = 4 ata
Multiply P ata x volume H20 to be displaced
4 ata x 12.06 ft3 = 48.24 ft3
How much air will be in the bag at the
surface?

21. Example 3
When properly weighted for diving in the
ocean, a diver and his gear weigh 224 lbs.
How must the diver adjust the amount of
weight in his weight system to be properly
weighted in fresh water?

22. Answer to Example3
The volume of the diver and his equipment will not change
SW displacement =
224 lbs./64 lbs. per cu. ft. = 3.5 cu.ft.
FW displacement =
3.5 cu. ft. x 62.4 lbs./cu. ft. = 218.4 lbs.
Wt. system Adjustment =
224 lbs.- 218.4 lbs.
Answer:
Remove 5.6 lbs
Shortcut
Adjust up or down by 2.5% of total diver weight.
This is the difference in density between ocean water and
fresh water

23. Have we covered:
Density
Buoyancy
Specific gravity
Archimedes’ Principle
Surface Equivalent air volume
Lifting problems

24. Can You
 Describe Archimedes’ Principle?
 Define density, buoyancy, and specific gravity?
 Correctly calculate the buoyancy of an object in
either fresh or salt water?
 Correctly solve a lifting problem?
 Correctly calculate Surface Air Volume
Equivalents?

25. Last Thoughts
Understanding and applying Archimedes’
Principle enables you to weight yourself properly
and to achieve and maintain the appropriate state
of buoyancy.
Combining Archimedes’ Principal with Boyle’s Law
enables you to correctly calculate the volume of gas
and number of lift bags you will need to bring from
the surface to ensure you can lift and object off the
bottom.

26. Physics project
BY
ADITYA SANGWAN
05
9D