Archimedes Principle
Learning Objectives
•
•
•
•
•
Describe Archimedes’ Principle.
Define density, buoyancy, and specific
gravity.
Correctly calculate the buoyancy of an
object in either fresh or salt water.
Correctly solve lifting problems.
Correctly calculate surface air volume
equivalents.
Main Points
•
•
•
•
•
•
Density
Buoyancy
Specific gravity
Archimedes’ Principle
Surface Equivalent air volume
Lifting problems
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.
Buoyancy
•
Force that allows an object to float.
Specific Gravity
Density of a substance vs. density of pure
water.
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
When placed in seawater, what is the state
of buoyancy for each of these objects?
Where will they end up?
Positive
_______________________________________________
64 lbs
1 cu. Ft.
32 lbs
1 cu ft
96 lbs
1 cu. ft
Neutral
________________________________________________Negative_
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
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?
Answer to Example 1
Displaced wt.=
.22 cu. ft. x 64 lbs. per cu. ft.
-Dry wt.
14.08 lbs.
100
lbs.
- 86
lbs
Buoyancy
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.
Answer to Example 2
Displaced weight =
3.1 cu. ft. x 62.4 lbs./ cu. ft.
-Dry weight
193.4 lbs.
289 lbs.
Buoyancy
- 95.6 lbs.
Lift capacity = 50 lbs - 2 lbs = 48 lbs of lift / bag.
Use how many bags?
2 bags.
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?
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
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
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
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?
Calculate the Buoyancy of the
Object
Volume =
2 ft3
Weight of the water displaced =
2 ft3 x 64 lbs/ft3 =
Dry weight =
128 lbs
900 lbs
Buoyancy of the Object
128 lbs – 900 lbs = -772 lbs
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
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?
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?
Answer to Example3
The volume of the diver and his equipment will not change
SW displacement =
224 lbs./64 lbs. per cu. ft. =
FW displacement =
3.5 cu. ft. x 62.4 lbs./cu. ft. =
Wt. system Adjustment =
3.5 cu.ft.
218.4 lbs.
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
Have we covered:
•
•
•
•
•
•
Density
Buoyancy
Specific gravity
Archimedes’ Principle
Surface Equivalent air volume
Lifting problems
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?
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.